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 ALGEBRA
 PROBLEMS
   Proven to IMProve Your AlgebrA SKIllS
3 Gain algebra confidence with targeted practice
3 Acquire new problem-solving skills
3 Master the most commonly tested algebra problems

   Mark A. McKibben, PhD



     L EARNING E XPRESS        ®
1001 ALGEBRA
   PROBLEMS
        OTHER TITLES OF INTEREST FROM
                LEARNINGEXPRESS
                       1001 Math Problems
                      501 Algebra Questions
                     501 Math Word Problems
                   Algebra Success in 20 Minutes
     Algebra in 15 Minutes a Day (Junior Skill Builders Series)
                 Express Review Guides: Algebra I
                 Express Review Guides: Algebra II
                         Math to the Max




ii
1001
ALGEBRA
PROBLEMS



  Mark A. McKibben, PhD


                                       ®




                          NEW   YORK
Copyright © 2011 LearningExpress, LLC.

All rights reserved under International and Pan-American Copyright Conventions. Published in the United
States by LearningExpress, LLC, New York.

Library of Congress Cataloging-in-Publication Data:
McKibben, Mark A.
1001 algebra problems / [Mark McKibben].
      p.cm.
  ISBN: 978-1-57685-764-9
  1. Algebra—Problems, exercises, etc. I. LearningExpress (Organization) II. Title. III.
Title: One thousand and one algebra problems. IV. Title: One thousand and one algebra problems.
  QA157.A16 2011
  512.0078—dc22
                                                                       2010030184

Printed in the United States of America

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           ABOUT THE AUTHOR


Dr. Mark McKibben is currently a tenured associate professor of mathematics and computer science at Goucher
College in Baltimore, Maryland. He earned his Ph.D. in mathematics in 1999 from Ohio University, where his
area of study was nonlinear analysis and differential equations. His dedication to undergraduate mathematics edu-
cation has prompted him to write textbooks and more than 20 supplements for courses on algebra, statistics,
trigonometry, precalculus, and calculus. He is an active research mathematician who has published more than
25 original research articles, as well as a recent book entitled Discovering Evolution Equations with Applica-
tions Volume 1: Deterministic Equations, published by CRC Press/Chapman-Hall.




                                                                                                           v
                                        CONTENTS




INTRODUCTION                                                       ix

SECTION 1      Pre-Algebra Fundamentals                             1

SECTION 2      Linear Equations and Inequalities                   17

SECTION 3      Polynomial Expressions                              65

SECTION 4      Rational Expressions                                77

SECTION 5      Radical Expressions and Quadratic Equations         89

SECTION 6      Elementary Functions                           101

SECTION 7      Matrix Algebra                                 123

SECTION 8      Common Algebra Errors                          143

ANSWERS & EXPLANATIONS                                        153
GLOSSARY                                                      277




                                                             vii
1001 ALGEBRA
   PROBLEMS
                                  INTRODUCTION



M                 any of the questions you ask in everyday life, such as “How many MP3 downloads can I buy with
                  a certain amount of money?” or “What percentage reduction in price would lower the cost of a
                  particular shirt to $20?” are solved using algebra. Although you might not have realized it, you’ve
been doing algebra for quite some time, believe it or not!
      The set of rules and techniques that has come to be known as algebra revolves around finding values of some
unknown quantity that, when used, make a given mathematical statement true. Such a value might be the length
of the side of a fence, the number of minutes a jogger needs to run in order to catch the nearest opponent, or the
original cost of an item. Mastery of the rules and techniques embodied in the problem sets in this book will arm
you with the tools necessary to attach applied problems accurately and with ease.



                                      How to Use This Book
This book has been designed to provide you with a collection of problems to assist you in reviewing the basic
techniques of algebra. It has been written with several audiences in mind. Anyone who has taken an algebra course
and needs to refresh skills that have become a bit rusty—this book is for you. Instructors teaching an algebra course
might find this repository of problems to be a useful supplement to their own problem sets. Teachers and tutors
might use the problems in this book in help sessions. Or, if you are a student taking algebra for the first time, this
book will provide you with some extra practice. Whatever your background or reason for picking up this book,
we hope that you will find it to be a useful resource in your journey through algebra!




                                                                                                               xi
                1                                PRE-ALGEBRA
S E C T I O N




                                                 FUNDAMENTALS




T           he basic arithmetic properties of whole numbers, integers, exponential expressions, fractions, and dec-
            imals are fundamental building blocks of algebra. In fact, the properties used to simplify algebraic expres-
            sions later in the text coincide with the rudimentary properties exhibited by these number systems. As
such, it is sensible to first gain familiarity with them and to then determine how to adapt them to a setting in which
variables are involved. These properties are reviewed in the first five problem sets in this section. Translating ver-
bal statements into mathematical ones and learning to deal with elementary algebraic expressions involving vari-
ables are the focus of the remaining four problem sets in this section.




                                                                                                                 1
                                   –PRE-ALGEBRA FUNDAMENTALS–



        Set 1      (Answers begin on page 153)           6. Which of the following whole numbers is divis-
                                                            ible by both 7 and 8?
The arithmetic properties of the set of whole numbers       a. 42
are reviewed in this set.                                   b. 78
                                                            c. 112
  1. (15 + 32)(56 – 39) =                                   d. 128
     a.   142
     b.   799                                            7. What is the estimated product when both 162
     c.   4,465                                             and 849 are rounded to the nearest hundred
     d.   30                                                and then multiplied?
                                                            a. 160,000
  2. What is the value of 65,715    4 rounded to the        b. 180,000
     nearest thousand?                                      c. 16,000
     a. 20,000                                              d. 80,000
     b. 16,000
     c. 16,428                                           8. Which of the following choices is equivalent to
     d. 16,429                                              5    5    5?
                                                            a.   3 5
  3. Estimate the value of 7,404    74.                     b.   10 5
     a.   1                                                 c.   15
     b.   10                                                d.   125
     c.   100
     d.   1,000                                          9. Which of the following choices is equivalent
                                                            to 35 ?
  4. 12(84 – 5) – (3    54) =                               a. 8
     a.   786                                               b. 15
     b.   796                                               c. 243
     c.   841                                               d. 125
     d.   54,000
                                                        10. The whole number p is greater than 0, a multiple
  5. Which of the following expressions is equal to         of 6, and a factor of 180. How many possibilities
     60,802?                                                are there for the value of p?
     a. 600 + 80 + 2                                        a. 7
     b. 6,000 + 800 + 2                                     b. 8
     c. 60,000 + 80 + 2                                     c. 9
     d. 60,000 + 800 + 2                                    d. 10
                                                            e. 11




    2
                                 –PRE-ALGEBRA FUNDAMENTALS–


11. Which of the following is the prime factoriza-             Set 2          (Answers begin on page 154)
    tion of 90?
    a. 9 10                                              The arithmetic properties of the integers are reviewed
    b. 90 1                                              in this set.
    c. 2 3 3 5
    d. 2 5 9                                              17. –25         4–9 =
    e. 3 3 10                                                 a.   –30
                                                              b.   –20
12. Which of the following is the set of positive fac-        c.   –5
    tors of 12 that are NOT multiples of 2?                   d.   5
    a. { }                                                    e.   13
    b. {1}
    c. {1,3}                                              18. –4      –2     –6     3=
    d. {1,2,3}                                                a.   –144
    e. {2,4,6,12}                                             b.   144
                                                              c.   –9
13. Which of the following operations will result in          d.   9
    an odd number?
    a. 36 + 48                                            19. 5 – (–17 + 7)2        3=
    b. 20 8                                                   a.   –135
    c. 37 + 47                                                b.   315
    d. 7 12                                                   c.   –295
    e. 13 + 12                                                d.   –45
                                                              e.   75
14. Which of the following equals 24 ?
    a.   10                                               20. (49        7) – (48   (–4)) =
    b.   15                                                   a.   19
    c.   32                                                   b.   5
    d.   16                                                   c.   –5
                                                              d.   –19
15. Which of the following expressions is equal to 5?
    a.   (1 + 2)2                                         21. In the equation y = 6p – 23, if p is a positive
    b.   9 – 22                                               whole number, which of the following is the
    c.   11 – 10 5                                            least value of p for which y is positive?
    d.   45 3 3                                               a. 1
                                                              b. 2
16. Which of the following is a prime number?                 c. 3
    a.   6                                                    d. 4
    b.   9                                                    e. 5
    c.   11
    d.   27



                                                                                                            3
                                 –PRE-ALGEBRA FUNDAMENTALS–



22. –(5 3) + (12      (–4)) =                   28. If g   0 and h 0, which of the following
    a.   –12                                        quantities is always positive?
    b.   –18                                        a. gh
    c.   12                                         b. g + h
    d.   18                                         c. g – h
                                                    d. |h| – |g|
23. –2 (–2)2 – 22 =                                 e. h2
    a.   4
    b.   –4                                     29. If g   0 and h 0, which of the following
    c.   –12                                        quantities cannot be negative?
    d.   12                                         a. gh
                                                    b. g + h
24. (32 + 6)   (–24       8) =                      c. –g –h
    a.   –5                                         d. 2g + 3h
    b.   5
    c.   4                                      30. If g   0 and h 0, which of the following
    d.   –4                                         quantities is the largest?
                                                    a. –g + h
25. (–2[1 –2(4 – 7)]2 =                             b. g – h
    a.   –36                                        c. g + h
    b.   36                                         d. –g –h
    c.   28
    d.   196                                    31. If g   0 and h 0, which of the following
                                                    quantities is the smallest?
26. 3(5 – 3)2 –3(52 – 32)=                          a. –g + h
    a.   9                                          b. g – h
    b.   –36                                        c. g + h
    c.   15                                         d. –g –h
    d.   0
                                                32. If g   –2, which of the following quantities is
                      2                             the largest?
27. –(–2 –(–11 – (–3 – 5) – 2)) =
    a.   3                                          a. g
    b.   –3                                         b. –g
    c.   4                                          c. –g2
    d.   –4                                         d. (–g)2




   4
                                       –PRE-ALGEBRA FUNDAMENTALS–


                                                            17        5
           Set 3        (Answers begin on page 156)     36. 20 – 6 =
                                                                 1
                                                            a.   5
The arithmetic properties of the set of fractions are            12
reviewed in this set.                                       b.   14
                                                                  1
                                                            c.   60
       5    1
 33.   9 – 4=                                               d. none of these
          11
       a. 36                                                18             9
          4                                             37. 5             20 =
       b. 5                                                       63
          3                                                 a.   100
       c. 4                                                      300
           5                                                b.    45
       d. 18
                                                            c. 8
       2        1   1    3                                  d. 10
 34. 15 + 5 + 6 + 10 =
             7
       a.   36                                          38. Which of the following fractions is the largest?
            4                                                    5
       b.   5                                               a.   8
              1                                                  2
       c.   750                                             b.   3
                                                                  8
       d. none of these                                     c.   11
                                                                  4
                                                            d.   10
 35. What fraction of the following figure is shaded?
                                                                                                           1
                                                        39. Which of the following fractions is between 4
                                                                  2
                                                            and 3 ?
                                                                 5
                                                            a.   8
                                                                 5
                                                            b.   6
                                                                  8
                                                            c.   11
                                                                  7
                                                            d.   10

                                                                                 3
                                                        40. Irma has read 5 of the novel assigned for her
            1                                               English class. The novel is 360 pages long. How
       a.   2                                               many pages has she read?
            1
       b.   4                                               a. 216
            2                                               b. 72
       c.   3
            3                                               c. 300
       d.   8
                                                            d. 98




                                                                                                       5
                                   –PRE-ALGEBRA FUNDAMENTALS–


    5        4
41. 8        7=                                                     –5
                                                                     3     –2
          5                                                                                7   10
    a.                                                45. 1 +            –10               5    3   =
         14
                                                                          7
         20
    b.    8                                                    3
         25                                               a.   2
    c.   32                                                    1
          9                                               b.   2
    d.   16                                                        1
                                                          c. – 49
                              21                               99
42. What is the reciprocal of 42 ?                        d.   49
         1
    a.   2
             21
    b. – 42                                           46. Judy’s math class, there are m men in a class of
          1                                               n students. Which expression gives the ratio of
    c.   –2
                                                          men to women in the class?
    d. 2                                                       m
                                                          a. n
                                                               n
43. Which of the following pairs of fractions is in       b. m
    the ration 4:5?                                                m
       1 1                                                c. m – n
    a. 4 , 5
                                                                   n
         1 1                                              d. n – m
    b. 5 , 4
         1 4
                                                                   m
    c. 5 , 5                                              e. n – m
         4 5
    d. 5 , 4
              4                                       47. Which of the following fractions is closest in
    e. 1, 5                                                                1
                                                          value to 2 ?
                                                               2
                                                          a.   3
44. Danny addressed 14 out of 42 envelopes.                     3
    What fraction of the envelopes still needs            b.   10
                                                               5
    to be addressed?                                      c.   6
         23                                                    3
    a.   42                                               d.   5
         13
    b.   21                                                                        2
         2                                            48. 7 5 – 3 1                    =
    c.   3                                                     6               2
         4                                                     17
    d.   7                                                a.   24
                                                               17
                                                          b.    6
                                                               61
                                                          c.   12
                                                               5
                                                          d.   4




   6
                                       –PRE-ALGEBRA FUNDAMENTALS–



        Set 4           (Answers begin on page 157)   54. –5(–1 –5–2) =
                                                          a. –45
The basic exponent rules in the context of signed              26
                                                          b.    5
arithemetic are reviewed in this set.                          24
                                                          c.    5
                                                                 24
 49. –53 =                                                d.   –5
       a.   –15
                                                                    3 –2       2 2
       b.   15                                        55. – – 2           –    3     =
       c.   125
       d.   –125                                          a. – 5
                                                               3
                                                                   8
                                                          b. – 9
 50. (–11)2 =                                             c. 0
       a.   121                                                    18
                                                          d. – 4
       b.   –121
                                                                               1 2
       c.   –22                                                 1 –3          –3
                                                      56. –    –2 –                  =
       d.   22                                                                9–2

                                                                   1
                                                          a. – 54
 51. What is the value of the expression 5(4˚)?
       a.   0                                             b. –1
                                                               7
       b.   1                                             c.   8
                                                                   9
       c.   5                                             d. – 8
       d.   20
                                                                    0
                                                      57. – 2           (–32 + 2–3)–1=
                                                               5
            2 –3
 52. (2 ) =
             1                                            a. 0
       a.   64
               1                                          b. –17
       b.   – 32                                                   71
                                                          c. – 9
       c. –12                                                   8
                                                          d.   71
       d. 2–5
                                                                                     –2
                2                                     58. 4–2 1 – 2(–1)–3                 =
       (1 – 3)
 53.      –8        =
       a. 1                                               a. 32
                                                                    1
       b. –1                                              b. – 144
                1
       c. – 2                                             c. 12–2
            1                                                       1 2
       d.   2                                             d.       28




                                                                                              7
                                         –PRE-ALGEBRA FUNDAMENTALS–



                   (–13 + (–1)3)–2                        64. If p is a fraction strictly between 1 and 2, which
59. –2–2 +               –22         =
           1                                                  of the following has the smallest value?
    a. – 10                                                   a. p
    b. 3                                                      b. p2
    c. –5                                                     c. p–2
           5
    d. – 16                                                   d. p–1

60. Which of the following quantities has the great-
    est value?                                                 Set 5        (Answers begin on page 158)
           1 –1
    a. – 4                                               Arithmetic involving decimals and percentages is the
               3                                         focus of this set.
    b. –           1
           8(– 4 )
                                                          65. On an exam, Bart is asked to choose two ways
               1
    c. 4 – 4 + 3                                              to determine n% of 40. He is given these four
               1 0
                                                              choices:
    d. – – 4
                                                                     I. n 100 40
                                                                    II. (n 0.01) 40
61. If p is a fraction strictly between 0 and 1, which
                                                                   III. (n 100) 40
    of the following has the largest value?
                                                                   IV. (n 0.01) 40
    a. p
    b. p2                                                     Which two ways are correct?
    c. p3                                                     a.   I and II
    d. p–1                                                    b.   I and IV
                                                              c.   II and III
62. If p is a fraction strictly between 0 and 1, which        d.   II and IV
    of the following has the smallest value?                  e.   III and IV
    a. p
    b. p2                                                 66. What is the result of increasing 48 by 55%?
    c. p3                                                     a.   26.4
    d. p–1                                                    b.   30.9
                                                              c.   69.6
63. If p is a fraction strictly between –1 and 0,             d.   74.4
    which of the following has the largest value?
    a. p
    b. p2
    c. p3
    d. p–1




   8
                                   –PRE-ALGEBRA FUNDAMENTALS–


67. Which of the following expressions show how        71. Which of the following inequalities is true?
    to determine the sale price of a $250 car stereo       a.   0.52 0.0052
    that is being offered at a 25% discount?               b.   0.52 0.052
           I. 0.25 $250                                    c.   0.00052 0.052
          II. 0.75 $250                                    d.   0.052 0.0052
         III. (1 + 0.25) $250
         IV. (1 –0.25) $250                            72. Which of the following is 400% of 30?
                                                           a.   1.2
    a.   I and III
                                                           b.   12
    b.   I and IV
                                                           c.   120
    c.   II and III
                                                           d.   1,200
    d.   II and IV
    e.   III and IV                                                                5         9
                                                       73. If 0.34   x 0.40 and 16 x 20 , which of
                                                           the following is a possible value for x?
68. Which of the following is the value of the point            1
    labeled as A on the following number line?             a.   3
                                                                2
                                                           b.   5
                                                                3
                                                           c.   8
                                                                3
                                                           d.   7
                           A                                    4
          –10         –5       0    5        10            e.   9

    a.   2.5                                           74. 22.5% is equivalent to which of the following
    b.   –2.5                                              decimals?
    c.   –1.5                                              a. 2.25
    d.   –3.5                                              b. 0.225
                                                           c. 0.025
69. Rounding 117.3285 to the nearest hundredth             d. 0.0225
    results in which of the following decimals?
    a. 100                                                                                         2
                                                       75. Which of the following is not less than 5 ?
    b. 117.3                                                    1
                                                           a. 3
    c. 117.33                                              b. 0.04
    d. 117.329                                                  3
                                                           c.   8
                                                                3
70. What percentage of 300 results in 400?                 d.   7

    a.   200%                                              e. 0.0404
             1
    b.   133 3 %
    c.   500%
    d.   1,200%




                                                                                                         9
                                  –PRE-ALGEBRA FUNDAMENTALS–


76. Which of the following decimals is between            Set 6      (Answers begin on page 159)
    –0.01 and 1.01?
    a. –0.015                                       This set contains problems that focus on evaluating
    b. –0.005                                       algebraic expressions at numerical values.
    c. 1.5
    d. 1.15                                          81. What is the value of the expression –2x2 + 3x –
                                                         7 when x = –3?
77. Which of the following decimals is equivalent        a. –34
         5    2                                          b. –27
    to 8 – 5 ?
    a. –0.25                                             c. –16
    b. 0.225                                             d. –10
    c. 0.25                                              e. 2
    d. 0.275                                                                                    7a
                                                     82. What is the value of the expression a2 + a
78. (3.09         1012) 3 =                              when a = –2?
    a.   1.03       104                                  a. –14
    b.   3.09       104                                  b. –7
                                                                7
    c.   1.03       1012                                 c. – 4
    d.   1.03       3.3312                               d. 7
                                                            4
                                                         e. 7
79. 0.00000321 is equivalent to which of the
    following?                                       83. What is the value of the expression 2ax – z
    a. 3.21 10–6                                         when a = 3, x =6, and z=–8?
    b. 3.21 10–5                                         a. 28
    c. 3.21 106                                          b. 44
    d. 3.21 105                                          c. 288
                                                         d. 20
                              8     1
80. What percentage of 9 results in 3 ?
         1
    a.                                               84. If y = –x3 + 3x –3, what is the value of y
         3%
    b.   29.6%                                           when x = –3 ?
    c.   37.5%                                           a. –35
            1
    d.   40 3 %                                          b. –21
                                                         c. 15
                                                         d. 18
                                                         e. 33




  10
                                    –PRE-ALGEBRA FUNDAMENTALS–



85. What is the value of the expression bx + z       y                                               7        3 a
                             1
                                                         90. What is the value of the expression 5a2 + 10a
    when b = –5, x =6, y =   2   and z = –8?                 when a = –2?
    a. 46                                                           1
                                                             a. – 10
    b. –46                                                          2
    c. –76                                                   b. – 25
    d. 76                                                    c. – 2
                                                                  5
                                                             d. 25
                                           m2
86. What is the value of the expression 3 – 4m +
    10 when m = 6?                                                                                 6x2   4x
                                                         91. What is the value of the expression 2y2 + 3y
    a. –12                                                   when x = 2 and y =3?
    b. –2                                                         4
                                                             a.   9
    c. 6
                                                                  4
    d. 12                                                    b.   3
    e. 22                                                         20
                                                             c.    9
                                                                  21
                                                             d.    9
87. What is the value of the expression
                                                                  13
    4(x –y)(2xy)(3yx) when x = 2 and y = –2?                 e.    3
    a. 6
    b. 8                                                 92. What is the value of the expression
                                                                       a
    c. 12                                                    ab + b + a2 – b 2 when a = 1 and b = –1?
    d. 24                                                    a. –4
    e. 384                                                   b. –3
                                                             c. –2
                                                12
88. What is the value of the expression 7x + x – z           d. –1
    when x = 6 and z = –8?                                   e. 0
    a. 52
                                                                                                          x
    b. 36                                                93. What is the value of the expression (xy) y
    c. 58                                                    if x = 2 and y = –x?
    d. 46                                                    a. –4
                                                                    1
                                                             b.   256
89. What is the value of the expression (3xy + x)
                                                                   1
     x
          when x = 2 and y = 5?                              c.   16
     y
                                                             d. 4
    a.   16
    b.   12.8                                                e. 16
    c.   32.4
    d.   80




                                                                                                         11
                                         –PRE-ALGEBRA FUNDAMENTALS–


 94. What is the value of the expression                   99. Simplify the expression 6(e–2)–2.
          x                                         1
     y    2    – 3 – 4a when a = 3, x = 6, and y = 2 ?         a.   6e –4
                                                               b.   6e 4
     a.   6
                                                               c.   36e–4
     b.   –6
                                                               d.   36e4
     c.   12
     d.   –12
                                                          100. What is the simplified result of the operation
                                                               (–45a4 b9 c5)   (9ab3c3)?
 95. What is the value of the expression z2 – 4a2 y
                                     1                         a. –5a3b6c2
     when a = 3, z =–8, and y = 2 ?
                                                               b. –5a4b3c3
     a. 4
                                                               c. –5a4b36c2
     b. –28
                                                               d. –5a5b12c 8
     c. –82
     d. 46
                                                          101. Simplify the expression 4(3x3)2.
                                                               a.   12x5
 96. What is the value of the expression 3x2b(5a – 3b)
                                                               b.   144x6
     when a = 3, b = –5, and x = 6?
                                                               c.   12x6
     a. –16,200
                                                               d.   36x6
     b. –1,800
     c. 0                                                                                  3
                                                                                                      4
                                                          102. Simplify the expression (ab) .
     d. 1,800                                                                               b
                                                               a.   a7
                                                               b.   a12
                                                               c.   a7b6
         Set 7       (Answers begin on page 160)
                                                               d.   a12b8
The problems in this set focus on simplifying algebraic        e.   a12 b11
expressions using the exponent rules.                                                      x 2       y –2
                                                                                           y         x
                                                          103 Simplify the expression                       .
                                                                                                xy
                                 (3x2)3
 97. Simplify the expression                                        1
                                  x2x4                         a.   xy
     a. 9                                                           x3
     b. 27                                                     b.   y5
          9                                                         x3
     c.                                                        c.   y3
          x

     d.
          27                                                   d. x3y3
           x
                                                               e. x5y5
 98. Simplify the expression (4w9)3.
                                                                                           2a         a–1
     a.   4w12                                            104. Simplify the expression     b         (2b)–1     .
     b.   4w27                                                 a.   1
     c.   12w27                                                b.   2
     d.   64w27                                                c.   2a
                                                               d.   4
                                                                    a2
                                                               e.   b2




    12
                                        –PRE-ALGEBRA FUNDAMENTALS–



105. Simplify the expression 3x2y(2x3y2).                 111. If 3x2 is multiplied by the quantity 2x3y raised to
     a.   6x6y2                                                the fourth power, to what would this expression
     b.   6x5y2                                                simplify?
     c.   6x5y3                                                a. 48x14y4
     d.   6x6y3                                                b. 1,296x16y4
                                                               c. 6x9y4
                                    2       –2       –1
106. Simplify the expression a          b        1   .         d. 6x14y4
                                b       a        a
     a. a
     b.
          1                                               112. Express the product of –9p3r and the quantity
          a
          a3                                                   5p – 6r in simplified form.
     c.   b4                                                   a. –4p4r – 15p3r2
          a4
     d.   b4                                                   b. –45p4r + 54p3r2
          a5                                                   c. –45p4r–6r
     e.   b4
                                                               d. –45p3r + 54 p3r2

107. Simplify the expression (3xy5)2 – 11x2 y2 (4y 4)2.
     a.   –82x2y10
     b.   6x2y7 – 88x 2y8
                                                                 Set 8        (Answers begin on page 161)
     c.   –167x2y10                                       The problems in this set focus on simplifying arith-
     d.   9x2y7 – 176x 2y8                                metic combinations of algebraic expressions by using
                                                          exponent rules and combining like terms.
                               2(3x2y)2 (xy)3
108. Simplify the expression      3(xy)2      .
     a. 6x5y3                                             113. Simplify the expression 5ab4 – ab4.
     b. 4x5y3                                                  a.   –5ab4
               7   5                                           b.   –5a2b8
     c. 4x 2 y2                                                c.   4ab4
               7 5
     d. 6x 2 y2                                                d.   The expression cannot be simplified further.

                               (4b)2x –2
109. Simplify the expression (2ab2x)2 .                   114. Simplify the expression 5c2 + 3c – 2c2 + 4 – 7c.
            4                                                  a.   3c 2 – 4c + 4
     a.   a2b2x4
                                                               b.   –3c 4 – 4c 2 + 4
           4
     b.   a2b2                                                 c.   –10c2 – 21c + 4
            4                                                  d.   The expression cannot be simplified further.
     c.   a2b2x4
           2
     d.   a2b2                                            115. Simplify the expression –5(x–(–3y)) + 4(2y + x).
                                                               a.   x + 7y
110. The product of 6x2 and 4xy2 is divided by 3x3y.           b.   x – 7y
     What is the simplified expresson?                         c.   –x – 7y
     a. 8y                                                     d.   –x + 7y
          4y
     b. x
     c. 4y
          8y
     d.   x


                                                                                                            13
                                    –PRE-ALGEBRA FUNDAMENTALS–


                                                                                                     (–3x–1)–2        8
116. Simplify the expression                             123. Simplify the expression                   x–2      + 9 (x2)2
     3x2 + 4ax – 8a2 + 7x2 – 2ax +7a2.                             7
                                                              a. 9 x4
     a. 21x2 – 8a2x – 56a2
     b. 10x2 + 2ax – a2                                       b. x4
                                                                   62 4
     c. 10x4 + 2a2x2 – a4                                     c.    9x
     d. The expression cannot be simplified further.          d. The expression cannot be simplified further.

                                                                                                                              b   –2
117. Simplify the expression 9m3n + 8mn3 + 2m3n3.        124. Simplify the expression –(–a–2bc–3)–2 + 5 a2c3 .
     a.   19m7n7                                                   a4c6
                                                              a.   24b2
     b.   19(m3n + nm3 + m3n3)
                                                                   4a4c6
     c.   17(mn)3 + 2m3n3                                     b.    b2
     d.   The expression cannot be simplified further.             6a4c6
                                                              c.    b2
                                                                   a4c6
118. Simplify the expression –7g6 + 9h + 2h – 8g6.            d.   4b2
     a.   –4g6
            h
           6
     b. –2g – 4h                                         125. Simplify the expression
                                                                                       2w(z + 1)
     c. –5g6 + h                                              3(z + 1)2w3 –                          .
                                                                                     ((z + 1)w2)–1
     d. –15g6 + 11h
                                                              a. 3(z + 1)2w3 – w3
                                                                                        1
                                                              b. 3(z + 1)2w3 –
119. Simplify the expression (2x 2)(4y 2) + 6x 2y 2.                                    w

     a.   12x 2y 2                                            c. (z + 1)2w3
     b.   14x 2y 2                                            d. The expression cannot be simplified further.
     c.   2x 2 + 4y 2 + 6x 2y 2
     d.   8x 2 + y 2 + 6x 2y 2                           126. Simplify the expression
     e.   8x 4y 4 + 6x 2y 2                                                                  2y(4x + 1)2         –2
                                                               –2(4x + 1)5 y–5 – ((4x + 1)y–2)–3                   .

120. Simplify the expression (5a2 3ab) + 2a3b.                         y10
                                                              a.   16(4x +1)10
     a.   15a2b + 2a3b
                                                              b. 8(4x + 1)–10 y10
     b.   17a6 b2
     c.   17a3b                                               c. 8(4x + 1) –3 y–7
     d.   The expression cannot be simplified further.        d. none of the above

                                      3x–1
121. Simplify the expression 2x–3 – x4 – (x3)–1.         127. Simplify the expression                                  –1
                                                                                                         1 2y6
     a.   –2x –3                                              4z((xy–2)–3 + (x–3y6))–1 –                 z x3             .
     b. x–3 – 3x –5                                                8zy6        zx3
                                                              a.    x3     –   2y6
     c. –x–3 – x –2
                                                                   3zx3
     d. The expression cannot be simplified further.          b.   2y6
                                                                   2zx3
                                                              c.   3y6
122. Simplify the expression (ab2)3 + 2b2 – (4a)3b6.
                                                                   8zy6        zx3
     a.   2b2 – 63a3b6                                        d.    x3     +   2y6
     b.   2b2 – 11a3b6
     c.   a3b5 + 2b2 – 12a3b6
     d.   The expression cannot be simplified further.

    14
                                    –PRE-ALGEBRA FUNDAMENTALS–


                                            2      5x4
128. Simplify the expression (0.2x–2)–1 + 5 x2 – (2x)2 .   132. Which of the following expressions represents
          83 2
                                                                nine less than three times the sum of a number
     a.   20 x                                                  and 5?
          2 2
     b.   9x                                                    a. 3(x + 5) – 9
          –21 2                                                 b. (3x + 5) –9
     c.    4 x
     d. none of the above                                       c. 9 – 3(x + 5)
                                                                d. 9 – (3x + 5)

                                                           133. A hotel charges $0.35 for the first minute of a
       Set 9      (Answers begin on page 162)
                                                                phone call and $0.15 for each additional
This problem set focuses on interpreting verbal math-           minute of the call. Which of the following
ematical statements as symbolic algebraic expressions.          equations represents the cost y of a phone call
                                                                lasting x minutes?
129. Two less than four times the square of a number            a. y = 0.15(x – 1) + 0.35
     can be represented as which of the following?              b. x = 0.15(y – 1) + 0.35
     a. 2 – 4x2                                                 c. y = 0.15x + 0.35
     b. 4x2 – 2                                                 d. x = 0.15y + 0.35
     c. (4x)2 – 2
     d. both b and c                                       134. Which of the following expressions represents
                                                                half the difference between a number and five?
130. If the volume V in a water tank is increased by            a. x – 5
     25%, which of the following expressions repre-                1
                                                                b. 2 (x – 5)
     sents the new volume of water?                                1
             1                                                  c. 2 x – 5
     a. V + 4 V
                                                                           1
     b. 1.25V                                                   d. 5 – 2 x
     c. V + 0.25V
     d. All three choices are correct.                     135. Which of the following expressions describes
                                                                the sum of three numbers multiplied by the
131. Jonathon is paying a math tutor a $30 one-time             sum of their reciprocals?
                                                                                        1       1   1
     fee plus $40 per hour for time spent tutoring.             a. (a + b + c) + ( a )( b )( c )
     Which of the following equations indicates how                    1       1            1
                                                                b. a( a ) + b( b ) + c( c )
     to compute x, the total amount Jonathon will                                   1     1      1
     be charged for h hours?                                    c. (a + b + c) ( a +      b + c)
                                                                                     1 1 1
     a. x = $30h + $40                                          d. (a)(b)(c) +     ( a )( b )( c )
     b. x = $30 + $40h
     c. x = ($30 + $40)h                                   136. Which of the statements below represents the
     d. x = $30h – $40                                          equation 3x + 15 = 32?
     e. ($30 – $40)h                                            a. 15 less than 3 times a number is 32.
                                                                b. 32 times 2 is equal to 15 more than a number.
                                                                c. 15 more than 3 times a number is 32.
                                                                d. 3 more than 15 times a number is 32.



                                                                                                          15
                                    –PRE-ALGEBRA FUNDAMENTALS–


137. Suppose that a desk costs D dollars, a chair           141. If q is decreased by p percent, then the resulting
     costs E dollars, and a file cabinet costs F dollars.        quantity is represented by which of the follow-
     If an office needs to purchase x desks, y chairs,           ing expressions?
     and z file cabinets, which of the following                 a. q – p
     expressions can be used to calculate the total              b. q – 1p
                                                                         00
     cost T?                                                           pq
                                                                 c. – 100
     a. xF + yE + zD                                                       pq
                                                                 d. q –   100
     b. xE + yD + zF
     c. xD + yE + zF                                             e. pq   – 1p0q0
     d. xF + yD + zD
                                                            142. Two brothers decide to divide the entire cost of

138. The value of d is increased by 50%, and then
                                                                 taking their father out to dinner evenly between
     the resulting quantity is decreased by 50%.                 the two of them. If the three meals cost a, b,
     How does the resulting quantity compare to d?               and c dollars, and a 15% tip will be added in for
     a. It is 25% smaller than d.                                the waiter, which of the following equations
     b. It is 25% larger than d.                                 represents how much each brother will spend?
     c. It is 50% smaller than d.                                a. 0.15(a + b + c)           2
     d. It is 50% larger than d.                                 b.   1.15(a + b +c)
                                                                            2
     e. It is the same as d.                                          (a + b + c) + 0.15(a + b + c)
                                                                 c.                 2

139. There are m months in a year, w weeks in a                  d. both b and c
     month, and d days in a week. Which of the
     following expressions represents the number            143. If the enrollment E at a shaolin kung fu school
     of days in a year?                                          is increased by 75%, which of the following
     a. mwd                                                      expressions represents the new enrollment?
     b. m + w + d                                                a. 0.75E
     c. mw                                                       b. E – 0.75E
           d
               w                                                 c. 3 E
                                                                     4
     d. d +    d
                                                                 d. E + 3 E
                                                                         4

140. If 40% of j is equal to 50% of k, then j is
                                                            144. Mary gets a 15% discount on all orders that
     a.   10% larger than k.
                                                                 she places at the copy store. If her orders cost W
     b.   15% larger than k.
                                                                 dollars, X dollars, Y dollars, and Z dollars
     c.   20% larger than k.
                                                                 before the discount is applied, which of the fol-
     d.   25% larger than k.
                                                                 lowing expressions represents how much it will
     e.   80% larger than k.
                                                                 cost her after the discount is deducted from her
                                                                 total?
                                                                 a. 0.85(W + X + Y +Z)
                                                                 b. 0.15(W + X + Y +Z)
                                                                 c. (W + X + Y + Z) + 0.15(W + X + Y + Z)
                                                                 d. (W + X + Y + Z) – 15(W + X + Y + Z)



    16
                2                                  LINEAR EQUATIONS
S E C T I O N




                                                   AND INEQUALITIES




E           quations and inequalities and systems thereof, made up of expressions in which the unknown quantity is
            a variable that is raised only to the first power throughout, are said to be linear. Elementary arithmetic prop-
            erties (e.g., the associative and distributive properties of addition and multiplication), properties of inequal-
ities, and the order of operations are used to solve them.
        A graph of a line can be obtained using its slope and a point on the line; the same is true for linear inequal-
ities, with the additional step of shading the region on the appropriate side of the line that depicts the set of ordered
pairs satisfying the inequality. Systems are handled similarly, although there are more possibilities regarding the
final graphical representation of the solution. These topics are explored in the following 13 problem sets.




                                                                                                                     17
                            –LINEAR EQUATIONS AND INEQUALITIES–



     Set 10        (Answers begin on page 163)       150. What value of p satisfies the equation
                                                          2.5p + 6 = 18.5?
This set is devoted to problems focused on solving        a. 5
elementary linear equations.                              b. 10
                                                          c. 15
145. What value of z satisfies the equation               d. 20
     z – 7 = –9?
     a. –2                                                                                       3x
                                                     151. What value of x satisfies the equation 10 = 15 ?
                                                                                                      25
     b. –1                                                a.   2
     c. 2                                                 b.   2.5
     d. 16                                                c.   3
                                                          d.   3.5
                                              k
146. What value of k satisfies the equation 8 = 8?
     a. 1
        8                                            152. What value of y satisfies the equation
     b. 8                                                 2.3(4 – 3.1x) = 1 – 6.13x ?
     c. 8                                                 a. 8.5
     d. 16                                                b. 451
     e. 64                                                c. 8.1
                                                          d. –8.5
147. What value of k satisfies the equation
     –7k – 11 =10?                                   153. If 11c – 7 = 8, what is the value of 33c – 21?
     a. –3                                                a.    15
                                                                11
     b. –1                                                      8
                                                          b.    3
     c. 2
                                                          c. 16
     d. 21
                                                          d. 24
                                                          e. 45
148. What value of a satisfies the equation
     9a + 5 = –22?
                                                     154. What value of x satisfies the equation
     a. –27                                               x
                                                          2    + 1 x = 4?
                                                                 6
     b. –9                                                       1
     c. –3                                                a.    24
                                                                1
     d. –2                                                b.    6
           17                                             c. 3
     e.   –9
                                                          d. 6
149. What value of p satisfies the equation
     p                                               155. What value of b satisfies the equation
     6 + 13 = p – 2?                                            5
                                                          b–    2 =   –2?
                                                                       3
     a. 6
     b. 12                                                a.   – 10
                                                                 16
     c. 15                                                b. –3
     d. 18                                                      5
                                                          c.    3
                                                                11
                                                          d.     6



    18
                            –LINEAR EQUATIONS AND INEQUALITIES–


156. What value of c satisfies the equation        Set 11             (Answers begin on page 164)
     3c
     4  – 9 = 3?
     a. 4                                     This problem set is focused on linear equations for
     b. 12                                    which obtaining the solution requires multiple steps.
     c. 16
     d. 20                                    161. What value of v satisfies the equation
                                                   –2(3v + 5) = 14?
157. What value of a satisfies the equation        a. –4
     – 2 a = –54 ?                                 b. –2
       3
     a. –81                                        c. 1
     b. 81                                         d. 3
     c. –36
     d. 36                                    162. What value of x satisfies the equation
                                                   5
                                                   2 (x –    2) + 3x = 3(x + 2) – 10?
                                                       1
158. What value of x satisfies the equation        a. 5
     1.3 + 5x – 0.1 = –1.2 – 3x?                       2
                                                   b. 5
     a. 0.3                                              1
                                                   c. – 5
     b. 3                                                2
     c. 3.3                                        d. – 5
     d. –0.3
                                              163. Twice a number increased by 11 is equal to 32
159. What value of v satisfies the equation        less than three times the number. Find the
     4(4v + 3) = 6v – 28?                          number.
     a. 3.3                                        a. –21
                                                        21
     b. –3.3                                       b. 5
     c. –0.25                                      c. 43
     d. –4                                              43
                                                   d.    5

160. What value of k satisfies the equation
                                              164. What value of a satisfies the equation
     13k + 3(3 – k) = –3(4 +3k) – 2k?               4a + 4        2 – 3a
                                                      7       =     4 ?
     a. 1
                                                        30
     b. –1                                         a. – 37
     c. 0                                             12
                                                   b. 5
     d. –2                                         c. 4
                                                   d. 6
                                                   e. 16




                                                                                                19
                            –LINEAR EQUATIONS AND INEQUALITIES–


165. The sum of two consecutive even integers is        171. What value of x satisfies the equation
                                                             2x + 8         5x – 6
     126. What are the integers?                               5        =     6      ?
     a. 62, 64                                                  14
                                                             a. 3
     b. 62, 63
                                                             b. 6
     c. 64, 66
                                                                  76
     d. 2, 63                                                c.   20
                                                                  –14
                                                             d.    3
166. What value of x satisfies the equation
     0.8(x + 20) – 4.5 = 0.7(5 + x) – 0.9x ?            172. When ten is subtracted from the opposite of a
     a. 8                                                    number, the resulting difference is 5. What is
     b. –8                                                   the number?
     c. 80                                                   a. 15
     d. –80                                                  b. –15
                                                             c. 12
167. If 4x + 5 = 15, then 10x + 5 =                          d. –52
     a.   2.5
     b.   15                                            173. What value of x satisfies the equation
     c.   22.5                                                        8       8
                                                             9x +     3   =   3x     + 9?
     d.   25                                                      3
     e.   30                                                 a.   8
                                                             b. 1
                                                                  8
168. Ten times 40% of a number is equal to four less         c.   3
     than six times the number. Find the number.             d. 9
     a. 12
     b. 8                                               174. Convert 50˚ Fahrenheit into degrees Celsius
     c. 4                                                    using the formula F = 5 C + 32.
                                                                                            9

     d. 2                                                    a. 45˚C
     7
                                                             b. 2˚C
169. 8 of nine times a number is equal to ten times          c. 10˚C
     the number minus 17. Find the number.                   d. 122˚C
     a. 18.6
     b. 80                                              175. Determine a number such that a 22.5%
     c. 1.86                                                 decrease in its value is the number 93.
     d. 8                                                    a. 27
                                               7b – 4
                                                             b. 114
170. Solve the following equation for b: a =     4           c. 115
          a
     a.   7                                                  d. 120
          4a
     b.   7
          a+1
     c.     7
          4a + 4
     d.      7
          7a – 4
     e.     7

    20
                                    –LINEAR EQUATIONS AND INEQUALITIES–


176. Negative four is multiplied by the quantity         180. The sum of four consecutive, odd whole num-
     x + 8. If 6x is then added to this, the result is        bers is 48. What is the value of the smallest
     2x + 32. What is the value of x?                         number?
     a. There can be no such number x.                        a. 9
     b. 1                                                     b. 11
     c. 0                                                     c. 13
     d. 16                                                    d. 15
                                                              e. 17

     Set 12                (Answers begin on page 166)   181. If PV =nRT, which of the following represents
                                                              an equivalent equation solved for T?
In this problem set, we consider more advanced linear                       PV
                                                              a. T =        nR
equations and word problems that can be solved using          b. PVnR = T
linear equations.                                                  PVR
                                                              c.    n       =T
                                                                             1
                                                              d. T =        PV       nR
177. What value of x satisfies the equation
      1
      2x   –4        x+8
          3      =    5    ?                             182. Solve the following equation for A:
     a. –8                                                             C+A
                                                              B=       D–A
     b. 8
                                                                            BD – C
     c. –88                                                   a. A =         1+B
     d. 88                                                                  D–C
                                                              b. A =        1+B
                                                                            B–C
178. What value of x satisfies the equation
                                                              c. A =        C+B
                                                                            B+D
     5x – 2[x – 3(7 – x)] = 3 – 2(x – 8) ?                    d. A =        C+B
     a. 23
     b. –23                                              183. If 30% of r is equal to 75% of s, what is 50% of
           23
     c.     5
                                                              s if r = 30?
           –23                                                a. 4.5
     d.      5
                                                              b. 6
179. Assuming that a      b, solve the following equa-        c. 9
     tion for x: ax + b = cx + d                              d. 12
           d+b                                                e. 15
     a.    a+c
           d–b
     b.    c–a
                                                         184. If fg + 2f – g = 2 –(f + g), what is the value of
           b–d                                                g in terms of f ?
     c.    a–c                                                a. –1
           d–b                                                     1
     d.    a–c                                                b.   f
                                                                   4
                                                              c.   f
                                                              d. 2 – 2f
                                                                   2 – 3f
                                                              e.      f


                                                                                                           21
                             –LINEAR EQUATIONS AND INEQUALITIES–


185. The length of a room is three more than twice      190. Solve the following equation for x:
                                                             5x – 2
     the width of the room. The perimeter of the              2–x     =y
     room is 66 feet. What is the length of the room?        a. x =
                                                                      2 – 2y
                                                                       5+y
     a. 10 feet                                                       2 + 2y
     b. 23 feet                                              b. x =    5–y
     c. 24 feet                                                       2 + 2y
                                                             c. x =    5+y
     d. 25 feet                                                          2 + 2y
                                                             d. x =   – 5+y
186. Solve the following equation for y:
     4 – 2x            1–y                              191. A grain elevator operator wants to mix two
       3       –1=      2
               1 – 4x
                                                             batches of corn with a resultant mix of 54
     a. y =      3                                           pounds per bushel. If he uses 20 bushels of
               1 + 4x
     b. y =       3                                          56 pounds per bushel corn, which of the
               4x – 1                                        following expressions gives the amount of
     c. y =      3
                  1 + 4x
                                                             50 pounds per bushel corn needed?
     d. y =    – 3                                           a. 56x + 50x = 2x 54
                                                             b. 20 56 + 50x = (x + 20) 54
187. The average of five consecutive odd integers is         c. 20 56 + 50x = 2x + 54
     –21. What is the least of these integers?               d. 56x + 50x = (x + 20) 54
     a. –17
     b. –19                                             192. What value of x satisfies the equation
     c. –21                                                  –5[x – (3 – 4x – 5) – 5x] – 22 = 4[2 –(x–3)]?
     d. –23                                                  a. 11.5
     e. –25                                                  b. 10.5
                               a
                                                             c. 9.5
188. If –6b + 2a – 25 = 5 and b + 6 = 4, what is the         d. 8.5
                       2
     value of      b    ?
                   a
          1
     a.   4
     b. 1                                                    Set 13           (Answers begin on page 169)
     c. 4
     d. –4                                              Solving basic linear inequalities is the focus of this
                                                        problem set.
189. If three more than one-fourth of a number is
     three less than the number, what is the value of   193. What is the solution set for 3x + 2      11?
     the number?                                             a.   {x : x   3}
          3                                                  b.   {x : x   –3}
     a.   4
                                                             c.   {x : x   3}
     b.   4
                                                             d.   {x : x   –3}
     c.   6
     d.   8
     e.   12




    22
                                –LINEAR EQUATIONS AND INEQUALITIES–



194. What is the solution set for 5x    23?              200. What is the solution set for x + 5     3x + 9?
                                                                            7
     a.   {x : x    115}                                      a.   {x : x   2}
     b.   {x : x    23}                                       b.   {x : x   –2}
     c.   {x : x    4.6}                                      c.   {x : x   –2}
     d.   {x : x    23}                                       d.   {x : x   2}

195. What is the solution set for 1 – 2x      –5?        201. What is the solution set for –6(x + 1)     60?
     a.   {x : x    3}                                        a.   {x : x   –9}
     b.   {x : x    3}                                        b.   {x : x   –9}
     c.   {x : x    –3}                                       c.   {x : x   –11}
     d.   {x : x    –3}                                       d.   {x : x   –11}

196. What inequality is represented by the following     202. Which of the following statements accurately
     graph?                                                   describes the inequality 2x – 4 7(x – 2)?
                                                              a. The sum of seven and the quantity two less
 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10
                                                                 than a number is greater than four less than
                                                                 two times the number.
     a.   x    –4
                                                              b. The product of seven and the quantity two
     b.   x    –4
                                                                 less than a number is greater than four less
     c.   x    –4
                                                                 than two times the number.
     d.   x    –4
                                                              c. The product of seven and the quantity two
                                                                 less than a number is less than four less than
197. What is the solution set for the inequality
                                                                 two times the number.
     4x + 4 24?
                                                              d. The product of seven and the quantity two
     a. {x : x 5}
                                                                 less than a number is greater than four less
     b. {x : x 5}
                                                                 than two more than the number.
     c. {x : x 7}
     d. {x : x  7}                                                                           –x
                                                         203. What is the solution set for 0.3     20?
                                                              a.   {x : x   –6}
198. What is the solution set for –8x + 11      83?
                                                              b.   {x : x   –6}
     a.   {x : x    –9}
                                                              c.   {x : x   –60}
     b.   {x : x    –9}
                                                              d.   {x : x   –60}
     c.   {x : x    9}
     d.   {x : x    9}
                                                         204. What is the solution set for
                                                              –8(x + 3)      2(–2x + 10)?
199. What is the solution set for
                                                              a. {x : x     –10}
     –4(x –1)       2(x + 1)?
                     1
                                                              b. {x : x     –10}
     a. {x : x      –3}                                       c. {x : x     –11}
                     1
     b. {x : x      –3}                                       d. {x : x     –11}
                    1
     c. {x : x      3}
                    1
     d. {x : x      3}
     e. {x : x      3}

                                                                                                          23
                            –LINEAR EQUATIONS AND INEQUALITIES–


205. What is the solution set for                     Set 14           (Answers begin on page 170)
     3(x – 16) – 2 9(x – 2) – 7x?
     a. {x : x –32}                              This problem set focuses on solving linear equations
     b. {x : x 32}                               and inequalities that involve the absolute value of
     c. {x : x 32}                               certain linear expressions.
     d. {x : x –32}
                                                 209. What values of x satisfy the equation
206. What is the solution set for                     |–x| – 8 = 0?
     –5[9 + (x – 4)]   2(13 –x)?                      a. 8 only
     a. {x : x 17}                                    b. –8 only
     b. {x : x –17}                                   c. both –8 and 8
     c. {x : x –17}                                   d. There are no solutions to this equation.
     d. {x : x 17}
                                                 210. How many different values of x satisfy the
207. What is the solution set for the compound        equation 2|x| + 4 = 0?
     inequality –4 3x – 1 11?                         a. 0
     a. {x : –1 x 4}                                  b. 1
     b. {x : 1 x –4}                                  c. 2
     c. {x : –4 x 1}                                  d. more than 2
     d. The solution set is the empty set.
                                                 211. How many different values of x satisfy the
208. What is the solution set for the compound        equation –3|x| + 2 = 5|x| – 14?
     inequality 10 3(4 – 2x) – 2      70?             a. 0
     a. {x : 0 x   10}                                b. 1
     b. {x : –10 x 0}                                 c. 2
     c. {x : 0 x 10}                                  d. more than 2
     d. {x : –10 x 0}
                                                 212. What values of x satisfy the equation
                                                              2    1
                                                      |3x – 3 | – 9 = 0?
                                                            5      7
                                                      a.   27 and 27
                                                            5        5
                                                      b.   27 and – 27
                                                            7        7
                                                      c.   27 and – 27
                                                              5        7
                                                      d.   – 27 and – 27




    24
                               –LINEAR EQUATIONS AND INEQUALITIES–


213. What values of x satisfy the equation          219. What is the solution set for –|–x –1|     0?
     |3x + 5| = 8?                                       a.   (– , 1) (1, )
          13                                             b.   (– , –1) (–1, )
     a.    3 and 1
            13                                           c.   The only solution is x = –1.
     b.   – 3 and 1
                                                         d.   The solution set is the empty set.
            13
     c.   – 3 and –1
          13                                        220. What is the solution set for |8x + 3|     3?
     d.    3 and –1
                                                         a.   [0, )
                                                                     3
214. How many different values of x satisfy the          b.   (– , – 4 ]
                                                                              3
     equation –6(4 – |2x + 3|) = –24?                    c.   [0, ), (– , – 4 ]
     a. 0                                                d.   none of these choices
     b. 1
     c. 2                                           221. What is the solution set for |2x – 3|     5?
     d. more than 2                                      a.   (– , 4)
                                                         b.   (4, )
215. How many different values of x satisfy the          c.   (–4, 1)
     equation 1 – (1 – (2 –|1–3x|)) = 5?                 d.   (–1, 4)
     a. 0
     b. 1                                           222. What is the solution set for
     c. 2                                                2 – (1 –(2 –|1 – 2x|))       –6?
     d. more than 2                                      a. (–4, 5)
                                                         b. (– , –4) (5, )
216. How many different values of x satisfy the          c. (–5, 4)
     equation |2x + 1| = |4x – 5|?                       d. (– , –5) (4, )
     a. 0
     b. 1                                           223. What is the solution set for
     c. 2                                                –7|1 – 4x| + 20      –2|1 – 4x| – 15?
                                                                      3
     d. 3                                                a. (–     , –2]
                                                         b. [2, )
217. What is the solution set for |x|     3?                          3
                                                         c. (– , – 2 ]     [2, )
     a.   (3, )                                                  3
     b.   (–3, )                                         d.   [– 2 , 2]
     c.   (–3, 3)
     d.   (– , –3)     (3, )                        224. What is the solution set for
                                                         |1 – (–22 + x) – 2x | |3x – 5|?
                                                              5
218. What is the solution set for |–2x|        0?        a. ( 3 , )
                                                                  5
     a.   (– , 0) (0, )                                  b. (– , 3 )
     b.   the set of all real numbers                    c. The solution set is the empty set.
     c.   (– , 0)                                        d. the set of all real numbers
     d.   The solution set is the empty set.




                                                                                                        25
                                –LINEAR EQUATIONS AND INEQUALITIES–



     Set 15          (Answers begin on page 173)         226. What coordinates are identified by point J
                                                              shown in the following Cartesian plane?
The basics of the Cartesian coordinate system are
explored in this problem set.                                                          y


225. What are the signs of the coordinates of points                         J
     in the shaded quadrant?

                            y
                                                                                                         x




                                                     x


                                                              a.   (–4,–3)
                                                              b.   (–4,3)
                                                              c.   (–3,–4)
                                                              d.   (3,–4)
                                                              e.   (–3,4)
     a.   x value is negative, y value is positive
     b.   x value is positive, y value is negative       227. Consider the following graph and assume that
     c.   x value is negative, y value is negative            ABCD is a square. What are the coordinates of
     d.   x value is positive, y value is positive            point B?
     e.   none of these choices
                                                                                       y




                                                                                 B          C
                                                                                             (6,4)


                                                                                                     x



                                                                                 A          D
                                                                             (–1,–3)




                                                              a.   (–1,–4)
                                                              b.   (–1,4)
                                                              c.   (–1,6)
                                                              d.   (–3,1)
                                                              e.   (–3,4)

    26
                                   –LINEAR EQUATIONS AND INEQUALITIES–


228. Consider the following graph and assume that       231. For all real numbers x    –2, points whose
     ABCD is a square. What are the coordinates of           coordinates are given by (|–x – 2|, –|–x – 1|)
     point D?                                                must lie in which quadrant?
                                                             a. Quadrant I
                               y
                                                             b. Quadrant II
                                                             c. Quadrant III
                                                             d. Quadrant IV
                         B               C
                                          (6,4)
                                                        232. If x is a positive real number and y is any real
                                                             number, which of the following is an accurate
                                                  x
                                                             characterization of the point (x, y)?
                                                             a. The point (x, y) can be in Quadrant I, in
                         A               D
                     (–1,–3)                                    Quadrant II, or on the x-axis.
                                                             b. The point (x, y) can be in Quadrant I, in
                                                                Quadrant IV, or on the x-axis.
                                                             c. The point (x, y) can be in Quadrant I, in
                                                                Quadrant II, on the x-axis, or on the y-axis.
     a.   (6,–4)                                             d. The point (x, y) can be in Quadrant I, in
     b.   (–6,4)                                                Quadrant IV, on the x-axis, or on the y-axis.
     c.   (–6,–4)
     d.   (–4,6)                                        233. If x is any real number and y is a nonnegative
     e.   (6,–3)                                             real number, which of the following is an
                                                             accurate characterization of the point (x, y)?
229. The point (2,–5) lies in which quadrant?                a. The point (x, y) can be in Quadrant I, in
     a.   Quadrant I                                            Quadrant II, or on the x-axis.
     b.   Quadrant II                                        b. The point (x, y) can be in Quadrant I, in
     c.   Quadrant III                                          Quadrant IV, or on the x-axis.
     d.   Quadrant IV                                        c. The point (x, y) can be in Quadrant I, in
                                                                Quadrant II, on the x-axis, or on the y-axis.
230. For all nonzero real numbers x and y, points            d. The point (x, y) can be in Quadrant I, in
     whose coordinates are given by (x2,(–y)2) lie in           Quadrant IV, on the x-axis, or on the y-axis.
     which quadrant?
     a. Quadrant I                                      234. Assume a     0. Which of the following points
     b. Quadrant II                                          lies in Quadrant IV?
     c. Quadrant III                                         a. (–a,a)
     d. Quadrant IV                                          b. (a,–a)
                                                             c. (a,a)
                                                             d. (–a,–a)




                                                                                                         27
                            –LINEAR EQUATIONS AND INEQUALITIES–



235. Assume a     0. Which of the following points      240. If y is a nonpositive real number, which of the
     lies in Quadrant III?                                   following is an accurate characterization of
     a. (–a2,a2)                                             points of the form (1,–y)?
     b. (a,–a2)                                              a. For some values of y, the point (1,–y) will lie
     c. (a2,a2)                                                 in Quadrant IV.
     d. ((–a)2,–a2)                                          b. There is no value of y for which the point
                                                                (1,–y) is on the x axis.
236. Assume a     0. Which of the following points           c. Both a and b are true.
     lies in Quadrant II?                                    d. Neither a or b is true.
     a. (–a,a)
     b. (a,–a)
     c. (a,a)                                                Set 16          (Answers begin on page 174)
     d. (–a,–a)
                                                        The problems in this set deal with determining the
                                                        equations of lines using information provided about
237. For all negative integers x and y, points whose
                                                        the line.
     coordinates are given by (–x3, xy2) lie in which
     quadrant?
                                                        241. What is the slope of the line whose equation is
     a. Quadrant I
                                                             3y – x = 9?
     b. Quadrant II                                               1
     c. Quadrant III                                         a. 3
     d. Quadrant IV                                          b. –3
                                                             c. 3
                                                             d. 9
238. For all negative integers x and y, points whose
                                –x2   1
     coordinates are given by (–y)3 , xy lie in which   242. What is the slope of the line whose equation is
     quadrant?                                               y = –3?
                                                             a. –3
     a.   Quadrant I
                                                             b. 0
     b.   Quadrant II
                                                             c. 3
     c.   Quadrant III
                                                             d. There is no slope.
     d.   Quadrant IV

                                                        243. What is the y-intercept of the line whose equa-
239. If x is any real number, which of the following
                                                             tion is 8y = 16x – 4?
     is an accurate characterization of points of the
                                                                       1
     form (–x,–2)?                                           a.   (0,– 2 )
     a. For some values of x, the point (–x,–2) will         b.   (0,2)
         lie in Quadrant III.                                c.   (0,8)
     b. The point (–x,–2) is never on the x-axis             d.   (0,16)
     c. Both a and b are true.
     d. Neither a or b is true.




    28
                              –LINEAR EQUATIONS AND INEQUALITIES–


244. Which of the following lines contains the point        247. Transform the equation 3x + y = 5 into slope-
     (3,1)?                                                      intercept form.
     a. y = 2x + 1                                               a. y = 3x + 5
     b. y = 2x + 2                                               b. y = –3x + 5
                                                                        1
              2
     c. y = 3 x – 2                                              c. x = 3 y + 5
                                                                            1
              2                                                  d. x = – 3 y + 5
     d. y =   3x   –1
     e. none of the above
                                                            248. What is the equation of the line that passes
                                                                 through the points (2, 3) and (–2, 5)?
245. A line is known to have a slope of –3 and a y-inter-
                                                                 a. y = x + 1
     cept of (0, 2). Which of the following equations                     1
                                                                 b. y = – 2 x + 4
     describes this line?                                                   1
     a. y = 2x – 3                                               c. y = – 2 x
                                                                            3
     b. y = –3x + 2                                              d. y = – 2 x
     c. y = –2x + 3                                                         3
                                                                 e. y = – 2 x + 2
     d. y = 3x – 2
                                                                                                2    3
                                                            249. Transform the equation y = – 15 x – 5 into
246. Which of the following equations was used to                standard form.
     construct this input/output table?                          a. –2x + 15y = –9
          x        y                                             b. 2x + 15y = 9
          1        7                                             c. 2x + 15y = –9
          2        10                                            d. 2x – 15y = –9
          3        13
          4        16                                       250. What is the slope of the line whose equation is
          5        19                                            –3y = 12x – 3?
                                                                 a. –4
     a.   y = 3x + 4                                             b. –3
     b.   y = 4x – 1                                             c. 1
     c.   y = 5x – 2                                             d. 4
     d.   y = 7x                                                 e. 12

                                                            251. Which of the following lines has a negative
                                                                 slope?
                                                                 a. 6 = y – x
                                                                 b. y = 4x – 5
                                                                 c. –5x + y = 1
                                                                 d. 6y + x = 7




                                                                                                              29
                            –LINEAR EQUATIONS AND INEQUALITIES–


252. Determine the missing value of z that com-         254. Which of the statements is true?
     pletes the following table, assuming that all of        a. A vertical line need not have a y-intercept.
     the points are collinear.                               b. A horizontal line need not have a y-intercept.
                                                             c. A line with positive slope need not cross the
          x         y                                           x-axis.
          –4        15                                       d. A line with negative slope need not cross the
          –2        11                                          x-axis.
          2         z
          5         –3                                  255. A line has a y-intercept of (0, –6) and an
          7         –7                                       x-intercept of (9, 0). Which of the following
     a.   –11                                                points must also lie on this line?
     b.   0                                                  a. (–6,–10)
     c.   3                                                  b. (1, 3)
     d.   8                                                  c. (0, 9)
                                                             d. (3, –8)
                                                             e. (6, 13)
253. A line is known to pass through the points
     (0, –1) and (2, 3). What is the equation of
                                                        256. Determine the value of y if the points (–3, –1),
     this line?
                1                                            (0, y), and (3, –9) are assumed to be collinear.
     a. y = 2 x – 1
                                                             a. 1
                1
     b. y = 2 x + 1                                          b. –1
     c. y = 2x – 1                                           c. –3
                                                             d. –5
     d. y = 2x + 1




    30
                                            –LINEAR EQUATIONS AND INEQUALITIES–



                                                 Set 17      (Answers begin on page 176)

The problems in this set deal with graphing straight lines.


257. Which of the following is the graph of y = –3?

a.                                                                     c.
                                        y                                                                      y
                                  10                                                                     10
                                    9                                                                     9
                                    8                                                                     8
                                    7                                                                     7
                                    6                                                                     6
                                    5                                                                     5
                                    4                                                                     4
                                    3                                                                     3
                                    2                                                                     2
                                    1                                                                     1
                                                                   x                                                                      x
     –10 –9 –8 –7 –6 –5 –4 –3 –2 –1         1 2 3 4 5 6 7 8 9 10            –10 –9 –8 –7 –6 –5 –4 –3 –2 –1         1 2 3 4 5 6 7 8 9 10
                                   –1                                                                     –1
                                   –2                                                                     –2
                                   –3                                                                     –3
                                   –4                                                                    –4
                                   –5                                                                    –5
                                   –6                                                                    –6
                                   –7                                                                    –7
                                   –8                                                                    –8
                                   –9                                                                    –9
                                  –10                                                                   –10




b.                                                                     d.
                                        y                                                                      y

                                  10                                                                     10
                                   9                                                                      9
                                   8                                                                      8
                                   7                                                                      7
                                   6                                                                      6
                                   5                                                                      5
                                   4                                                                      4
                                   3                                                                      3
                                   2                                                                      2
                                   1                                                                      1
                                                                   x                                                                      x
     –10 –9 –8 –7 –6 –5 –4 –3 –2 –1         1 2 3 4 5 6 7 8 9 10            –10 –9 –8 –7 –6 –5 –4 –3 –2 –1         1 2 3 4 5 6 7 8 9 10
                                   –1                                                                     –1
                                   –2                                                                     –2
                                   –3                                                                     –3
                                  –4                                                                     –4
                                  –5                                                                     –5
                                  –6                                                                     –6
                                  –7                                                                     –7
                                  –8                                                                     –8
                                  –9                                                                     –9
                                 –10                                                                    –10




                                                                                                                                    31
                              –LINEAR EQUATIONS AND INEQUALITIES–


258. What is the slope of the line segment shown in   260. What is the slope of the line segment in the fol-
     the following graph?                                  lowing graph?




                                                                                                (10,2)
                 (–3,0)
                                              x
                                                                                                         x
                                                                                   (0,0)




                              (0,–5)                                     (–2,–6)



                          y                                                        y

          5
     a.   3
                                                           a. –2
     b.   –3
            5                                              b. 2
            3                                              c. – 2
                                                                3
     c.   –5
                                                           d. 2
                                                              3
          3
     d.   5


259. Which of the following is an accurate charac-
     terization of the slope of the y-axis?
     a. It has a slope of zero.
     b. Its slope is undefined.
     c. It has a positive slope.
     d. It has a negative slope.




    32
                                        –LINEAR EQUATIONS AND INEQUALITIES–


261. Graph y = 2x + 3.

a.                                                               c.
                               10                                                               10

                                8                                                                8

                                6                                                                6

                                4                                                                4

                                2                                                                2

                                                             x                                                             x
     –10   –8   –6   –4   –2            2   4   6   8   10            –10   –8   –6   –4   –2         2   4   6   8   10
                                –2                                                               –2

                                –4                                                               –4

                                –6                                                               –6

                                –8                                                               –8

                               –10                                                              –10

                                 y                                                                y


b.                                                               d.
                               10                                                               10

                                 8                                                               8

                                 6                                                               6

                                 4                                                               4

                                 2                                                               2

                                                             x                                                             x
     –10   –8   –6   –4   –2            2   4   6   8   10            –10   –8   –6   –4   –2         2   4   6   8   10
                                –2                                                               –2

                                –4                                                               –4

                                –6                                                               –6

                                –8                                                               –8

                               –10                                                              –10

                                    y                                                             y




                                                                                                                           33
                                         –LINEAR EQUATIONS AND INEQUALITIES–


262. Graph y = –2x + 9.

a.                                                                c.
                               10
                                                                                                 10

                                8
                                                                                                  8

                                6
                                                                                                  6

                                4
                                                                                                  4

                                2
                                                                                                  2

                                                              x
     –10   –8   –6   –4   –2             2   4   6   8   10                                                                    x
                                                                       –10   –8   –6   –4   –2            2   4   6   8   10
                                –2
                                                                                                  –2

                                –4
                                                                                                  –4

                                –6
                                                                                                  –6

                                –8
                                                                                                  –8

                               –10
                                                                                                 –10
                                 y
                                                                                                   y


b.                                                                d.
                               10                                                                10

                                 8                                                                 8

                                 6                                                                 6

                                 4                                                                 4

                                 2                                                                 2

                                                              x                                                                x
     –10   –8   –6   –4   –2             2   4   6   8   10            –10   –8   –6   –4   –2            2   4   6   8   10
                                –2                                                                –2

                                –4                                                                –4

                                –6                                                                –6

                                –8                                                                –8

                               –10                                                               –10

                                     y                                                                y




     34
                                                –LINEAR EQUATIONS AND INEQUALITIES–


263. Which of the following is the graph of y = – 5 x – 5?
                                                  2

a.                                                                         c.
                                            y                                                                      y
                                    10                                                                       10
                                        9                                                                     9
                                        8                                                                     8
                                        7                                                                     7
                                        6                                                                     6
                                        5                                                                     5
                                        4                                                                     4
                                        3                                                                     3
                                        2                                                                     2
                                        1                                                                     1
                                                                       x                                                                       x
       –10 –9 –8 –7 –6 –5 –4 –3 –2 –1           1 2 3 4 5 6 7 8 9 10            –10 –9 –8 –7 –6 –5 –4 –3 –2 –1         1 2 3 4 5 6 7 8 9 10
                                     –1                                                                       –1
                                     –2                                                                       –2
                                     –3                                                                       –3
                                     –4                                                                      –4
                                     –5                                                                      –5
                                     –6                                                                      –6
                                    –7                                                                       –7
                                    –8                                                                       –8
                                    –9                                                                       –9
                                   –10                                                                      –10




b.                                                                         d.
                                         y                                                                         y
                                   10                                                                        10
                                    9                                                                         9
                                    8                                                                         8
                                    7                                                                         7
                                    6                                                                         6
                                    5                                                                         5
                                    4                                                                         4
                                    3                                                                         3
                                    2                                                                         2
                                    1                                                                         1
                                                                       x                                                                       x
      –10 –9 –8 –7 –6 –5 –4 –3 –2 –1            1 2 3 4 5 6 7 8 9 10            –10 –9 –8 –7 –6 –5 –4 –3 –2 –1         1 2 3 4 5 6 7 8 9 10
                                    –1                                                                        –1
                                    –2                                                                        –2
                                   –3                                                                        –3
                                   –4                                                                        –4
                                   –5                                                                        –5
                                   –6                                                                        –6
                                   –7                                                                        –7
                                   –8                                                                        –8
                                   –9                                                                        –9
                                  –10                                                                       –10




                                                                                                                                              35
                                      –LINEAR EQUATIONS AND INEQUALITIES–


264. What is the equation of the line shown in the              265. What is the equation of the line in the follow-
     following graph?                                                ing graph?
                                                                                                         y
                                10
                                                                                                   10
                                 8                                                                  9
                                                                                                    8
                                 6                                                                  7
                                                                                                    6
                                 4                                                                  5
                                                                                                    4
                                 2                                                                  3
                                                                                                    2
                                                            x                                       1
     –10    –8   –6   –4   –2          2   4   6   8   10                                                                           x
                                                                      –10 –9 –8 –7 –6 –5 –4 –3 –2 –1         1 2 3 4 5 6 7 8 9 10
                                 –2                                                                 –1
                                                                                                    –2
                                 –4                                                                 –3
                                                                                                   –4
                                 –6                                                                –5
                                                                                                   –6
                                 –8                                                                –7
                                                                                                   –8
                                –10                                                                –9
                                                                                                  –10
                                  y

     a.    y=x+7                                                     a.   y = –2x + 5
                                                                                3
     b.    y=x–7                                                     b.   y = 2x + 5
                                                                              3
     c.    y = –x – 7                                                c.   y = –3x + 5
                                                                                2
     d.    y = –x + 7                                                d.   y = 3x + 5
                                                                              2




    36
                                              –LINEAR EQUATIONS AND INEQUALITIES–


266. Graph 2 y – 1 x = 0.
           3     2

a.                                                                                      c.
                                    10                                                                                 10


                                     8                                                                                   8


                                     6                                                                                   6


                                     4                                                                                   4

                                     2                                                                                   2

                                                                                    x                                                                x
     –10    –8    –6    –4    –2              2       4       6       8       10             –10   –8   –6   –4   –2            2   4   6   8   10

                                     –2                                                                                 –2


                                     –4                                                                                 –4

                                     –6                                                                                 –6


                                     –8                                                                                 –8


                                    –10                                                                                –10

                                      y                                                                                     y


b.                                                                                      d.
                                     10                                                                                10

                                         8                                                                              8

                                          6                                                                             6

                                         4                                                                              4

                                          2                                                                             2

                                                                                    x                                                                x
      –10    –8    –6    –4    –2                 2       4       6       8    10            –10   –8   –6   –4   –2            2   4   6   8   10

                                      –2                                                                                –2

                                      –4                                                                                –4

                                      –6                                                                                –6

                                      –8                                                                                –8

                                     –10                                                                               –10

                                          y                                                                              y




                                                                                                                                                     37
                              –LINEAR EQUATIONS AND INEQUALITIES–


267. Which of the following lines has a positive slope?


a.                                                        c.
                          y                                         y




                                                 x                          x




b.                                                        d.
                          y                                    y




                                                 x                      x




     38
                             –LINEAR EQUATIONS AND INEQUALITIES–


268. Which of the following lines has an undefined slope?


a.                                                      c.
                         y                                         y




                                               x                        x




b.                                                      d.
                         y                                         y




                                               x                        x




                                                                       39
                           –LINEAR EQUATIONS AND INEQUALITIES–


269. The equation 0.1x – 0.7y = 1.4 is shown in which of the following graphs?


a.
                                                      10


                                                          8


                                                          6


                                                          4

                                                          2

                                                                                      x
                          –10   –8   –6    –4   –2             2   4   6    8    10

                                                       –2


                                                       –4


                                                       –6


                                                       –8


                                                     –10

                                                           y

b.
                                                     10


                                                      8


                                                      6


                                                      4

                                                      2

                                                                                      x
                         –10    –8   –6   –4    –2             2   4   6   8     10

                                                      –2


                                                      –4


                                                      –6


                                                      –8


                                                     –10

                                                       y

     40
       –LINEAR EQUATIONS AND INEQUALITIES–


c.

                                 10


                                  8


                                  6


                                  4

                                  2

                                                                 x
     –20   –16   –12   –8   –4            4   8   12   16   20

                                  –2


                                  –4


                                  –6


                                  –8


                                 –10

                                   y

d.
                                 10


                                   8


                                   6


                                   4

                                   2

                                                                 x
     –20   –16   –12   –8   –4            4   8   12   16   20

                                  –2


                                  –4


                                  –6


                                  –8


                                 –10

                                      y




                                                                     41
                            –LINEAR EQUATIONS AND INEQUALITIES–


270. Which of the following describes a possible               Set 18           (Answers begin on page 177)
     scenario?
     a. The graph of a line with positive slope can       This set focuses on more advanced properties of linear
        cross into both Quadrants II and IV.              equations, as well as more advanced word problems
     b. The graph of a line with negative slope can-      modeled using linear equations.
        not cross into both Quadrants I and II.
     c. The graph of y = c, where c ≠ 0, can cross        273. To which of the following lines is y = 2 x –5
                                                                                                      3
        into only two of the four quadrants.                   perpendicular?
     d. The graph of a vertical line cannot cross into         a. y = 2 x + 5
                                                                      3
        both Quadrants II and III.                             b. y = 5 – 2 x
                                                                           3
                                                               c. y = – 2 x – 5
                                                                        3
271. Which of the following describes a possible               d. y = 3 x –5
                                                                      2
     scenario?                                                 e. y = – 3 x + 5
                                                                        2
     a. The graph of x = c, where c ≠ 0, cannot cross
        the x-axis.                                       274. The graphs of which of the following pairs
     b. The graph of y =c, where c ≠ 0, must have a            of linear equations would be parallel to
        y-intercept.                                           each other?
     c. A line with an undefined slope can cross into          a. y = 2x + 4, y = x + 4
        both Quadrant I and Quadrant II.                       b. y = 3x + 3, y = – 1 – 3
                                                                                    3
     d. A line whose graph rises from left to right            c. y = 4x + 1, y = –4x + 1
        has a negative slope.                                  d. y = 5x + 5, y = 1 x + 5
                                                                                  5
                                                               e. y = 6x + 6, y = 6x – 6
272. Which of the following describes a possible
     scenario?                                            275. The line y = –2x + 8 is
     a. A line whose equation is of the form y = –x + c        a.   parallel to the line y = 1 x + 8
                                                                                              2
        can cross into three of the four quadrants.            b.   parallel to the line 1 y = –x + 3
                                                                                         2
     b. A line with positive slope need not cross the          c.   perpendicular to the line 2y = – 1 x + 8
                                                                                                      2
        x-axis.                                                d.   perpendicular to the line 1 y = –2x – 8
                                                                                                 2
     c. A line with negative slope need not cross the          e.   perpendicular to the line y = 2x – 8
        y-axis.
     d. A horizontal line has an undefined slope.         276. Which of the following is the equation of the
                                                               line perpendicular to y = 3 x – 2 and passing
                                                                                         4
                                                               through the point (–6, 4)?
                                                               a. y = 3 x +4
                                                                      4
                                                               b. y = 3 x – 4
                                                                      4
                                                               c. y = 4 x – 4
                                                                      3
                                                               d. y = – 4 x – 4
                                                                        3




    42
                               –LINEAR EQUATIONS AND INEQUALITIES–


277. Which of the following is the equation of the      282. A 60-foot piece of rope is cut into three pieces.
     line parallel to y = 3x + 8 and passing through         The second piece must be 1 foot shorter in
     the point (4,4)?                                        length than twice the first piece, and the third
     a. y = 3x + 4                                           piece must be 10 feet longer than three times
     b. y = 3x – 8                                           the length of the second piece. How long
     c. y = 1 x + 8
            3                                                should the longest piece be?
     d. y = – 1 x + 8
              3                                              a. 37 feet
                                                             b. 40 feet
278. Which of the following is the equation of the           c. 43 feet
     line that has y-intercept (0,12) and is parallel        d. 46 feet
     to the line passing through the points (4,2)
     and (–5,6)?                                        283. At Zides Sport Shop, a canister of Ace tennis
     a. y = 4 x + 12
             9                                               balls costs $3.50 and a canister of Longline
     b. y =   – 4 x + 12
                9
                                                             tennis balls costs $2.75. The high school tennis
     c. y =   –9x + 2                                        coach bought canisters of both brands of balls,
                4
              9                                              spending exactly $40.25 before the sales tax. If
     d. y =   4 x + 12
                                                             he bought one more canister of Longline balls
                                                             than he did Ace balls, how many canisters of
279. Which of the following is the equation of the
                                                             each did he purchase?
     line perpendicular to y = – 13 x + 5 and passing
                                 18
                                                             a. 6 canisters of Ace balls and 7 canisters of
     through the origin?
                                                                Longline balls
     a. y = – 18 x
              13
              13
                                                             b. 6 canisters of Ace balls and 7 canisters of
     b. y =   18 x                                              Longline balls
              18
     c. y =   13 x                                           c. 5 canisters of Ace balls and 6 canisters of
     d. y =   – 13 x
                18
                                                                Longline balls
                                                             d. 7 canisters of Ace balls and 8 canisters of
280. Which of the following lines must be perpen-               Longline balls
     dicular to a line with an undefined slope?
     a. x = –2
     b. y = –2
     c. both a and b
     d. neither a nor b

281. Which of the following lines must be parallel to
     a line with zero slope?
     a. x = –2
     b. y = –2
     c. both a and b
     d. neither a nor b




                                                                                                        43
                            –LINEAR EQUATIONS AND INEQUALITIES–


284. One essential step to ensure the success of a       286. Kari invested some money at 10% interest and
     microgravity bean seed germination project is            $1,500 more than that amount at 11% interest.
     that 10 gallons of a 70% concentrated nutrient           Her total yearly interest was $795. How much
     solution be administered to the bean seeds. If           did she invest at each rate?
     the payload specialist has some 90% nitrogen             a. $2,000 at 10% interest and $3,500 at
     and some 30% nitrogen, how many gallons                     11% interest
     (accurate to 2 decimal places) of each should            b. $2,500 at 10% interest and $4,000 at
     she mix in order to obtain the desired solution?            11% interest
     a. 2.50 gallons of the 30% nitrogen solution with        c. $4,000 at 10% interest and $5,500 at
        7.50 gallons of the 90% nitrogen solution                11% interest
     b. 7.50 gallons of the 30% nitrogen solution with        d. $3,000 at 10% interest and $4,500 at
        2.50 gallons of the 90% nitrogen solution                11% interest
     c. 6.67 gallons of the 30% nitrogen solution with
        3.33 gallons of the 90% nitrogen solution        287. A piggy bank is full of just nickels and dimes. If
     d. 3.33 gallons of the 30% nitrogen solution with        the bank contains 65 coins with a total value of
        6.67 gallons of the 90% nitrogen solution             5 dollars, how many nickels and how many
                                                              dimes are in the bank?
285. How long would it take a girl bicycling at 17            a. 32 nickels and 33 dimes
     mph to overtake her instructor bicycling at 7            b. 30 nickels and 35 dimes
     mph along the same path, assuming that her               c. 28 nickels and 37 dimes
     instructor had a 3-hour head start?                      d. 25 nickels and 40 dimes
     a. 2 hours 6 minutes
     b. 2 hours 15 minutes                               288. Lori is twice as old as her sister, Lisa. In 5 years,
     c. 3 hours                                               Lisa will be the same age as her sister was 10
     d. 3 hours 12 minutes                                    years ago. What are their current ages?
                                                              a. Lisa is 12 years old and Lori is 24 years old.
                                                              b. Lisa is 15 years old and Lori is 30 years old.
                                                              c. Lisa is 20 years old and Lori is 40 years old.
                                                              d. Lisa is 23 years old and Lori is 46 years old.




    44
                                           –LINEAR EQUATIONS AND INEQUALITIES–



     Set 19                   (Answers begin on page 180)           290. Which inequality is illustrated by the following
                                                                         graph?
The problems in this set consist of graphing linear
                                                                                                       10
inequalities in the Cartesian plane.
                                                                                                        8


289. Which inequality is illustrated by the following                                                   6


     graph?                                                                                             4

                                                                                                        2
                                     10

                                                                                                                                  x
                                      8                                  –10       –8   –6   –4   –2         2   4   6   8   10

                                                                                                        –2
                                      6

                                                                                                        –4
                                      4

                                                                                                        –6
                                      2

                                                                                                        –8
                                                                x
     –10       –8   –6   –4     –2         2   4   6   8   10
                                                                                                       –10
                                      –2
                                                                                                         y
                                      –4
                                                                         a.    y        2x + 7
                                      –6
                                                                         b.    y        2x + 7
                                      –8
                                                                         c.    y        –2x + 7
                                     –10
                                                                         d.    y        –2x + 7
                                       y

     a.    y        –2                                              291. Which inequality is illustrated by the following
     b.    y        –2                                                   graph?
     c.    y        –2
                                                                                                       10
     d.    y        –2
                                                                                                        8


                                                                                                        6


                                                                                                        4

                                                                                                        2

                                                                                                                                  x
                                                                         –10       –8   –6   –4   –2         2   4   6   8   10

                                                                                                        –2


                                                                                                        –4

                                                                                                        –6

                                                                                                        –8


                                                                                                       –10

                                                                                                         y

                                                                         a.    y        4x – 3
                                                                         b.    y        –4x – 3
                                                                         c.    y        –4x – 3
                                                                         d.    y        4x – 3




                                                                                                                                  45
                                                     –LINEAR EQUATIONS AND INEQUALITIES–


292. Which inequality is illustrated by the following                                    294. Which inequality is illustrated by the following
        graph?                                                                                graph?
                                  y                                                                                        y

                             10                                                                                          10

                              8                                                                                           8

                              6
                                                                                                                          6
                              4
                                                                                                                          4
                              2
                                                                                                                          2
                                                                                     x
   –9         –6        –3             3         6       9   12       15        18
                                                                                                                                                       x
                              –2                                                              –10   –8    –6   –4   –2            2   4   6   8   10

                                                                                                                          –2
                              –4


                              –6                                                                                          –4

                              –8                                                                                          –6

                             –10
                                                                                                                          –8


        a.    y         8                                                                                                –10

        b.    y         8
                                                                                              a.    y–x        0
        c.    x         8
                                                                                              b.    x–y        0
        d.    x         8
                                                                                              c.    y–x        0
                                                                                              d.    x–y        0
293. Which inequality is illustrated by the following
        graph?
                                             y                                           295. Which inequality is illustrated by the following
                                           10                                                 graph?
                                                                                                                              y
                                            8
                                                                                                                         10
                                            6
                                                                                                                          8
                                            4
                                                                                                                           6
                                            2
                                                                                                                           4
                                                                                x
        –10        –8   –6   –4       –2             2   4   6    8        10
                                                                                                                           2
                                            –2
                                                                                                                                                       x
                                                                                              –10   –8    –6   –4   –2            2   4   6   8   10
                                            –4
                                                                                                                          –2
                                            –6
                                                                                                                          –4
                                            –8
                                                                                                                          –6
                                           –10
                                                                                                                          –8

        a.    x+y            2                                                                                           –10

        b.    x–y            2                                                                      1
                                                                                              a.    3x + 2y –1
        c.    x–y            –2
                                                                                              b.    x +2y –3
        d.    x+y            –2
                                                                                              c.    x + 6y –1
                                                                                                    1
                                                                                              d.    3x + 2  –1




    46
                                             –LINEAR EQUATIONS AND INEQUALITIES–


296. Which inequality is illustrated by the following                 298. Which inequality is illustrated by the following
     graph?                                                                graph?
                                         y                                                                  y

                                    10                                                                    10


                                     8                                                                     8


                                     6                                                                     6


                                     4                                                                     4


                                     2                                                                     2

                                                                  x                                                                     x
     –10    –8   –6   –4       –2            2   4   6   8   10            –10       –8   –6    –4   –2            2   4   6   8   10

                                     –2                                                                    –2


                                     –4                                                                    –4


                                     –6                                                                    –6


                                     –8                                                                    –8


                                    –10                                                                   –10



     a.    2y + 6x         8                                               a.    y        3x + 1
     b.    2y – 6x         8                                               b.    y        3x + 1
     c.    2y + 6x         8                                               c.    y        3x + 1
     d.    2y – 6x         8                                               d.    y        3x + 1

297. Which inequality is illustrated by the following                 299. Which of the inequalities is illustrated by the
     graph?                                                                following graph?
                                      y                                                                        y

                                    10                                                                    10


                                     8                                                                      8


                                     6                                                                      6


                                     4                                                                      4


                                     2                                                                      2


                                                                  x                                                                      x
                 –6   –4                     2   4   6   8                 –10       –8    –6   –4   –2            2   4   6   8   10
     –10    –8                 –2                            10

                                     –2                                                                    –2


                                     –4                                                                    –4


                                     –6                                                                    –6


                                     –8                                                                    –8


                                    –10                                                                   –10



     a.    3x + y +2 0                                                     a.    3x + y         7x + y – 8
     b.    3x – y + 2 0                                                    b.    3x – y         7x + y + 8
     c.    3x – y –2 0                                                     c.    3x + y         7x + y – 8
     d.    3x + y – 2 0                                                    d.    3x – y         7x + y – 8




                                                                                                                                        47
                        –LINEAR EQUATIONS AND INEQUALITIES–


300. For which of the following inequalities is the point (3,–2) a solution?
       a.   2y –x 1
       b.   x+y 5
       c.   3y –3x
       d.   9x – 1 y

301. Which of the following graphs illustrates the inequality y       4?

a.                                                        c.
                            y                                                           y

                       10                                                        10




                                                   x                                                  x
     –10                                      10                –10                          10




                       –10                                                      –10




b.                                                        d.
                            y                                                       y

                       10                                                      10




                                                   x                                              x
     –10                                      10               –10                          10




                       –10                                                     –10




48
                                –LINEAR EQUATIONS AND INEQUALITIES–



302. Which of the following graphs illustrates the inequality x     4?

a.                                                       c.
                            y                                                  y

                       10                                                 10




                                                 x                                      x
     –10                                    10                –10                  10




                       –10                                               –10




b.                                                       d.
                            y                                                 y

                       10                                                10




                                                 x                                      x
     –10                                    10                –10                  10




                       –10                                               –10




                                                                                        49
                               –LINEAR EQUATIONS AND INEQUALITIES–


303. Which of the inequalities is illustrated by the following graph?

                                                               y




                                                         18



                                                         15



                                                         12



                                                          9



                                                          6



                                                           3



                                                                                                                 x
–20   –16    –14   –12   –10    –8   –6   –4   –3   –2             2   4   6   8   10   12   14   16   18   20


                                                          –3



                                                          –6



                                                          –9



                                                         –12



                                                         –15



                                                         –18




      a.   28y –2x – 14(y + 10)
      b.   –28y  2x – 14(y + 10)
      c.   28y 2x – 14(y + 10)
      d.   –28y 2x – 14(y – 10)




50
                             –LINEAR EQUATIONS AND INEQUALITIES–


304. Which of the following describes a possible         307. Given that both of the following equations must
     scenario?                                                be satisfied simultaneously, use the elimination
     a. Points of the form (x, 2x), where x 0, are in         method to determine the value of x.
        the solution set of the linear inequality y x.
                                                                           3(x +4) – 2y = 5
     b. The solution set of a linear inequality can
                                                                           2y – 4x =8
        intersect all four quadrants.
     c. The solution set of a linear inequality               a.   –2
        y – 2x < –1 includes points on the line               b.   –1
        y = 2x –1.                                            c.   1
     d. Points of the form (8, y) satisfy the linear          d.   13
        inequality x 8.                                       d.   15

                                                         308. Given that both of the following equations must
     Set 20         (Answers begin on page 103)               be satisfied simultaneously, use the elimination
                                                              method to determine the value of x.
Systems of 2 2 linear equations are solved using the
elimination method in this problem set.                                    2x + y =6
                                                                           y
                                                                           2 + 4x = 12
305. Given that both of the following equations must
                                                              a.   –2
     be satisfied simultaneously, use the elimination
                                                              b.   0
     method to determine the value of b.
                                                              c.   1
                    5a +3b = –2                               d.   3
                    5a – 3b = –38                             e.   6
     a.   –6
     b.   –4                                             309. Given that both of the following equations must
     c.   6                                                   be satisfied simultaneously, use the elimination
     d.   12                                                  method to determine the value of x .y
     e.   13
                                                                           4x + 6 = –3y
                                                                           –2x + 3 = y + 9
306. Use the elimination method to determine the
     solution of the following system of linear equa-         a.   –6
     tions:                                                   b.   –1
                                                              c.   0
                    –x + 3y = 11
                                                              d.   1
                    x – 5y = –3
                                                              e.   6
     a.   x = 17, y = 4
     b.   x =1, y = 4
     c.   x = 1, y = –4
     d.   x = –23, y = –4




                                                                                                        51
                            –LINEAR EQUATIONS AND INEQUALITIES–


310. Given that both of the following equations must    313. Given that both of the following equations must
     be satisfied simultaneously, use the elimination        be satisfied simultaneously, use the elimination
     method to determine the value of b.                     method to determine the value of y.
                         b                                                1
                  –7a + 4 = 25                                            2 x + 6y = 7
                  b + a = 13                                              –4x – 15y = 10

     a.   –3                                                 a.   –10
     b.   4                                                  b.   –12
     c.   12                                                 c.   2
     d.   13                                                 d.   5
     e.   16                                                 e.   6

311. Given that both of the following equations must    314. Given that both of the following equations must
     be satisfied simultaneously, use the elimination        be satisfied simultaneously, use the elimination
     method to determine the value of n.                     method to determine the value of a + b.

                  2(m + n) + m =9                                         4a + 6b = 24
                  3m –3n = 24                                             6a –12b = –6

     a.   –5                                                 a.   2
     b.   –3                                                 b.   3
     c.   3                                                  c.   4
     d.   5                                                  d.   5
     e.   8                                                  e.   6

312. Given that both of the following equations must    315. Given that both of the following equations must
     be satisfied simultaneously, use the elimination        be satisfied simultaneously, use the elimination
     method to determine the value of a.                     method to determine the value of a + b.
                                                                          1
                  7(2a + 3b) =56                                          2 (a + 3) – b = –6
                  b + 2a = –4                                             3a – 2b = –5

     a.   –5                                                 a.   5
     b.   –4                                                 b.   15
     c.   –2                                                 c.   20
     d.   4                                                  d.   25
     e.   6                                                  e.   45




    52
                            –LINEAR EQUATIONS AND INEQUALITIES–


316. Given that both of the following equations must    319. Given that both of the following equations must
     be satisfied simultaneously, use the elimination        be satisfied simultaneously, use the elimination
                                         c
     method to determine the value of d .                    method to determine the value of (p + q)2.
                  c–d
                   5   –2=0                                                   8q + 15p = 26
                  c – 6d = 0                                                  –5p + 2q = 24

     a.   2                                                  a.   4
     b.   6                                                  b.   5
     c.   8                                                  c.   25
     d.   12                                                 d.   49
     e.   14                                                 e.   81

317. Given that both of the following equations must    320. Use the elimination method to determine
     be satisfied simultaneously, use the elimination        the solution of the following system of linear
     method to determine the value of xy.                    equations:

                  –5x + 2y = –51                                              4x – 3y = 10
                  –x –y = –6                                                  5x + 2y = 1

     a.   –27                                                a.   x = 4, y = –3
     b.   –18                                                b.   x = 1, y = –2
     c.   –12                                                c.   x = –1, y = – 13
     d.   –6                                                 d.   x = 2, y = – 2
                                                                               3
     e.   –3

318. Given that both of the following equations must         Set 21           (Answers begin on page 185)
     be satisfied simultaneously, use the elimination
                                                        Systems of 2 2 linear equations are solved using the
     method to determine the value of (y – x)2.
                                                        substitution method and graphical techniques in this
                  9(x – 1) = 2 –4y                      problem set.
                  2y + 7x = 3
                                                        321. Given that both of the following equations must
     a.   1
                                                             be satisfied simultaneously, use the substitution
     b.   4
                                                             method to solve the following system:
     c.   16
     d.   25                                                                  x = –5y
     e.   36                                                                  2x –2y = 16

                                                             a.   x = 10, y = –2
                                                             b.   x = –2, y = 10
                                                             c.   x = 20, y = –4
                                                             d.   x = –5, y = 1




                                                                                                        53
                        –LINEAR EQUATIONS AND INEQUALITIES–


322. Given that both of the following equations       325. Given that both of the following equations
     must be satisfied simultaneously, use the sub-        must be satisfied simultaneously, use the
     stitution method to determine the value of x.         substitution method to determine the value
                                                           of a.
                  2x + y = 6
                  y
                  2 +4x = 12                                            7(2a + 3b) = 56
                                                                        b + 2a = –4
     a.   –2
     b.   0                                                a.   –5
     c.   1                                                b.   –4
     d.   3                                                c.   –2
                                                           d.   4
323. Given that both of the following equations            e.   6
     must be satisfied simultaneously, use the sub-
     stitution method to determine the value of       326. Given that both of the following equations
     √a .
        b                                                  must be satisfied simultaneously, use the
                  a                                        substitution method to determine the value
                  2 =b+1                                      c
                                                           of d .
                  3(a – b) = –21
                                                                        c–d
          4                                                              5   –2=0
     a.   9                                                             c – 6d = 0
          2
     b.   3
          3                                                a.   2
     c.   4
          4
                                                           b.   6
     d.   3                                                c.   8
          3
     e.   2                                                d.   12
                                                           e.   14
324. Given that both of the following equations
     must be satisfied simultaneously, use the sub-   327. Given that both of the following equations
     stitution method to determine the value of b.         must be satisfied simultaneously, use the
                         b
                  –7a + 4 = 25                             substitution method to determine the value
                  b + a = 13                               of xy.

     a.   –3                                                            –5x + 2y = –51
     b.   4                                                             –x – y = –6
     c.   12                                               a.   –27
     d.   13                                               b.   –18
     e.   16                                               c.   –12
                                                           d.   –6
                                                           e.   –3




54
                               –LINEAR EQUATIONS AND INEQUALITIES–


328. Given that both of the following equations must     331. How many solutions are there to the system of
     be satisfied simultaneously, use the substitution        equations shown in the following graph?
     method to determine the value of ab.
                                                                                       y

                  10b – 9a = 6
                  b–a=1

     a.   –12
     b.   –7
     c.   1
     d.   7                                                                                                   x

     e.   12

329. Given that both of the following equations must
     be satisfied simultaneously, use the substitution
     method to determine the value of x – y.
                  x+y
                   3   =8
                  2x – y = 9                                  a.   0
                                                              b.   1
     a.   –24                                                 c.   2
     b.   –2                                                  d.   infinitely many
     c.   0
     d.   1
                                                         332. Determine the number of solutions of the
     e.   2
                                                              following system of equations:

330. Which of the following linear systems contains                          y = 3x + 2
     two parallel lines?                                                     y – 3x = –2
     a. x = 5, y = 5
                                                              a.   1
     b. y = –x, y = x – 1
                                                              b.   0
     c. x – y = 7, 2 – y = –x
                                                              c.   infinitely many
     d. y = 3x + 4, 2x + 4 = y
                                                              d.   none of the above




                                                                                                         55
                               –LINEAR EQUATIONS AND INEQUALITIES–


333. Given that both of the following equations           335. Determine the number of solutions of the linear
     must be satisfied simultaneously, use the substi-         system that has the following graphical depiction:
     tution method to determine the value of 2yx .                                           y

                   3x – y = 2                                                           10


                   2y – 3x = 8

     a. 4
        3
     b. 4
        5
     c. 5
     d. 8                                                                                                      x
                                                                    –10                                   10
     e. 12

334. Determine the number of solutions of the linear
     system that has the following graphical depiction:
                               y                                                    –10

                          10

                                                               a.   1
                                                               b.   0
                                                               c.   infinitely many
                                                               d.   none of the above

                                                    x
         –10                                   10         336. Determine the number of solutions of the
                                                               following system of equations:

                                                                             y = 3x + 2
                                                                             – 3y + 9x = –6

                         –10                                   a.   1
                                                               b.   0
     a.   1                                                    c.   infinitely many
     b.   0                                                    d.   none of the above
     c.   infinitely many
     d.   none of the above




    56
                               –LINEAR EQUATIONS AND INEQUALITIES–



                                      Set 22    (Answers begin on page 188)

The problems in this set consist of graphing systems of linear inequalities.

337. The graphs of the lines y = 4 and y = x   2 form the boundaries of the shaded region. The solution set of
     which of the following systems of linear inequalities is given by the shaded region?
                                      y


                                     14

                                     12

                                     10


                                     8


                                      6


                                      4

                                      2

                                                                     xx
      –10     –8     –6   –4   –2         2    4     6     8    10

                                     –2


                                     –4


                                     –6


                                     –8


                                    –10


                                    –12

                                    –14


     a.   y   4, y    x+2
     b.   y   4, y    x +2
     c.   y   4, y    x+2
     d.   y   4, y    x+2




                                                                                                       57
                                      –LINEAR EQUATIONS AND INEQUALITIES–


338. The graphs of the lines y = 5 and x = 2 form                 339. The graphs of the lines y = –x + 4 and y = x + 2
     the boundaries of the shaded region. The solu-                    form the boundaries of the shaded region. The
     tion set of which of the following systems of                     solution set of which of the following systems of
     linear inequalities is given by the shaded                        linear inequalities is given by the shaded region?
     region?
                                                                                                           y
                                     y                                                                10

                                10
                                                                                                       8

                                 8
                                                                                                       6

                                 6
                                                                                                       4

                                 4
                                                                                                       2

                                 2                                                                                                  x
                                                                        –10     –8     –6   –4   –2            2   4   6   8   10
                                                              x                                        –2
      –10   –8   –6   –4   –2            2   4   6   8   10

                                 –2
                                                                                                       –4

                                 –4
                                                                                                       –6

                                 –6
                                                                                                       –8

                                 –8
                                                                                                      –10

                                –10
                                                                       a.   y        x +2, y     –x + 4
     a.   y 5, x 2                                                     b.   y        x +2, y     –x + 4
     b.   y ≤ 5, x ≤ 2                                                 c.   y        x +2, y     –x + 4
     c.   y 5, x 2                                                     d.   y        x +2, y     –x + 4
     d.   y 5, x 2




    58
                                           –LINEAR EQUATIONS AND INEQUALITIES–


340. The graphs of the lines y = 1 x and y = –4x form
                                 4                                     341. The graphs of the lines 2y – 3x = –6 and y =
                                                                                5
     the boundaries of the shaded region. The solu-                         5 – 2 x form the boundaries of the shaded
     tion set of which of the following systems of                          region. The solution set of which of the follow-
     linear inequalities is given by the shaded                             ing systems of linear inequalities is given by the
     region?                                                                shaded region?

                                          y                                                                 y

                                     10                                                                10


                                      8                                                                 8


                                      6                                                                 6


                                      4                                                                 4


                                      2                                                                 2


                                                                   x                                                                 x
     –10    –8     –6     –4    –2            2   4   6   8   10            –10    –8   –6   –4   –2            2   4   6   8   10

                                      –2                                                                –2


                                      –4                                                                –4


                                      –6                                                                –6


                                      –8                                                                –8


                                     –10                                                               –10




     a. y        1                                                          a.   2y – 3x     –6, y      5 – 5x
                 4 x, y        –4x                                                                          2
                 1
                                                                            b.   2y – 3x     –6, y      5 – 5x
                                                                                                            2
     b. y        4 x, y        –4x
                                                                            c.   2y – 3x     –6, y      5 – 5x
                                                                                                            2
                 1
     c. y        4 x, y        –4x                                          d.   2y – 3x     –6, y       5 – 5x
                                                                                                             2
                 1
     d. y        4 x, y        –4x




                                                                                                                                59
                                   –LINEAR EQUATIONS AND INEQUALITIES–


342. Which of the following graphs depicts the
      solution set for the following system of linear
      inequalities?

                   y    2
                   y    2x + 1
                                                         c.
a.
                               y                                           y
                         10                                          10




                                                     x                              x
     –10                                        10            –10              10




                         –10                                         –10




b.                                                       d.
                               y                                           y

                         10                                          10




                                                     x                              x
     –10                                        10            –10              10




                         –10                                         –10




     60
                                      –LINEAR EQUATIONS AND INEQUALITIES–


343. The graphs of the lines 5y = 8(x + 5) and                   344. The graphs of the lines y = 3x and y = –5 form
     12(5 – x) = 5y form the boundaries of the                        the boundaries of the shaded region. The solu-
     shaded region. The solution set of which of the                  tion set of which of the following systems of lin-
     following systems of linear inequalities is given                ear inequalities is given by the shaded region?
     by the shaded region?
                                                                                                                y
                                  y                                                                        10


                                 14                                                                         8


                                 12                                                                         6


                                 10                                                                         4


                                 8                                                                          2


                                  6                                                                                                       x
                                                                       –10     –8     –6    –4        –2            2   4   6   8    10

                                  4                                                                         –2


                                  2                                                                         –4


                                                             x
                                                             x                                              –6
     –10    –8   –6   –4   –2          2   4    6   8   10

                                 –2                                                                         –8


                                 –4                                                                        –10


                                 –6
                                                                      a.   y        3x, y        –5
                                 –8                                   b.   y        3x, y        –5
                                –10                                   c.   y        3x, y        –5
                                –12
                                                                      d.   y        3x, y        –5

                                –14


     a.    5y    8(x + 5), 12(5 – x)       5y
     b.    5y    8(x + 5), 12(5 – x)       5y
     c.    5y    8(x + 5), 12(5 – x)       5y
     d.    5y    8(x + 5), 12(5 – x)       5y




                                                                                                                                    61
                                                    –LINEAR EQUATIONS AND INEQUALITIES–


345. The graphs of the lines 9(y – 4) = 4x and –9y =                                   346. The graphs of the lines y – x = 6 and 11y =
         2(x + 9) form the boundaries of the shaded                                         –2(x + 11) form the boundaries of the shaded
         region. The solution set of which of the follow-                                   region. The solution set of which of the following
         ing systems of linear inequalities is given by the                                 systems of linear inequalities is given by the
         shaded region?                                                                     shaded region?
                                                y                                                                                             y

                                          10                                                                                            10


                                           8                                                                                             8


                                           6                                                                                             6


                                           4                                                                                             4


                                           2                                                                                             2


                                                                                   x                                                                                        x
                                                                                           –16   –14   –12   –10    –8   –6   –4   –2             2   4   6   8   10   12
   –14   –12    –10   –8   –6   –4   –2             2   4   6   8   10   12   14
                                                                                                                                        –2
                                          –2

                                                                                                                                        –4
                                          –4

                                                                                                                                        –6
                                          –6

                                                                                                                                        –8
                                          –8
                                                                                                                                        –10
                                          –10




         a.    9(y – 4)         4x, –9y             2(x + 9)                                a.   y–x               6, 11y          –2(x + 11)
         b.    9(y – 4)         4x, –9y             2(x + 9)                                b.   y–x               6, 11y          –2(x + 11)
         c.    9(y – 4)         4x, –9y             2(x + 9)                                c.   y–x               6, 11y          –2(x + 11)
         d.    9(y – 4)         4x, –9y             2(x + 9)                                d.   y–x               6, 11y          –2(x + 11)




    62
                                                  –LINEAR EQUATIONS AND INEQUALITIES–


347. The graphs of the lines 5x– 2(y + 10) = 0 and                                       348. The graphs of the lines 7(y – 5) = –5x and
         2x + y = –3 form the boundaries of the shaded                                        –3 = 1 (2x – 3y) form the boundaries of the
                                                                                                    4
         region. The solution set of which of the follow-                                     shaded region below. The solution set of which
         ing systems of linear inequalities is given by the                                   of the following systems of linear inequalities is
         shaded region?                                                                       given by the shaded region?
                                                      y                                                                       y
                                                 10
                                                                                                                         10
                                                  9

                                                                                                                          8
                                                  6
                                                                                                                          6
                                                  3
                                                                                                                          4
                                                                                     x
   –16    –14   –12   –10   –8   –6   –4   –2             2   4   6    8   10   12
                                                                                                                          2
                                                 –3
                                                                                                                                                               x
                                                                                              –10    –8   –6   –4   –2              2      4      6   8   10
                                                 –6
                                                                                                                          –2

                                                 –9
                                                                                                                          –4

                                                –12
                                                                                                                          –6

                                                –15                                                                       –8


                                                –18                                                                      –10


                                                                                                                                  1
         a.   5x – 2(y + 10)               0, 2x + y              –3                          a.    7(y–5)     –5x, –3            4 (2x   – 3y)
                                                                                                                                  1
         b.   5x – 2(y + 10)               0, 2x + y              –3                          b.    7(y–5)     –5x, –3            4 (2x   – 3y)
                                                                                                                                  1
         c.   5x – 2(y + 10)               0, 2x + y              –3                          c.    7(y–5)     –5x, –3            4 (2x   – 3y)
                                                                                                                                  1
         d.   5x – 2(y + 10)               0, 2x + y              –3                          d.    7(y–5)     –5x, –3            4 (2x   – 3y)




                                                                                                                                                          63
                             –LINEAR EQUATIONS AND INEQUALITIES–


349. For which of the following systems of linear      351. For which of the following system of linear
     inequalities is the solution set the entire            inequalities does the solution set consist pre-
     Cartesian plane?                                       cisely of the points in Quadrant III, not includ-
     a. y x + 3, y x – 1                                    ing either axis?
     b. 2y – 6x ≤ 4, y 2 + 3x                               a. x 0, y 0
     c. y x, y ≥ x                                          b. x 0, y 0
     d. none of the above                                   c. x 0, y 0
                                                            d. x 0, y 0
350. For which of the following systems of linear
     inequalities is the solution set the empty set?   352. For which of the following systems of linear
     a. y x – 3, y x – 1                                    inequalities does the solution set consist of the
     b. y x + 3, y x – 1                                    points on a single line?
     c. y x, y 2x                                           a. 2y – 6x 4, y 2 – 3x
     d. y 3x + 4, y 3x +6                                   b. 2y – 6x 4, y 2 + 3x
                                                            c. 2y – 6x 4, y 2 + 3x
                                                            d. 2y – 6x 4, y 2 + 3x




    64
                3                                  POLYNOMIAL
S E C T I O N




                                                   EXPRESSIONS




A               lgebraic expressions consisting of sums of constant multiples of nonnegative powers of the variable are
                called polynomials. Simplyfying polynomials and understanding their graphical properties rely heavily
                on the use of factoring. These topics are the focus of the following seven problem sets.




                                                                                                                65
                                             –POLYNOMIAL EXPRESSIONS –



      Set 23                 (Answers begin on page 190)   358. Compute (2 – 3x3) – [(3x3 + 1) – (1 – 2x3)].
                                                                a.   2 + 8x3
The problems in this set focus on the basic definition
                                                                b.   –2 + 8x3
of and addition/subtraction of polynomials.
                                                                c.   –2 – 8x3
                                                                d.   2 – 8x3
353. Compute (x2 – 3x + 2) = (x3 – 2x2 + 11).
      a.   x3 + x2 + 3x + 13
                                                           359. What is the degree of the polynomial
      b.   x3 – x2 + 3x + 13
                                                                –5x8 + 9x4 – 7x3 –x2 ?
      c.   x3 + x2 – 3x + 13
                                                                a. –5
      d.   x3 – x2 – 3x + 13
                                                                b. 8
                                                                c. 9
354. Compute (3x2 – 5x + 4) – (– 2 x + 5).
                                 3                              d. 2
                    13
      a. 3x2 –       3x–1
                    7
      b. 3x2 –      3x – 1                                 360. What is the degree of the polynomial
      c. 3x2 –      13
                     3x–9                                       – 3 x + 5x4 –2x2 + 12?
                                                                  2
      d. 3x2 –      7                                           a. 5
                    3x – 9
                                                                b. – 32
355. Compute ( 1 x2 – 1 x – 2 ) – ( 2 x2 – 170 x + 1 ).         c. 4
               3      5     3       3              2
                                                                d. 1
      a. – 1 x2 –
           3
                    9     1
                   10 x – 6
      b.   – 1 x2 – 1 x – 7
             3      2     6                                361. What is the degree of the constant polynomial 4?
      c.     1 2     9
           – 3 x – 10 x + 1 6                                   a.   0
      d.   1 2     9      1                                     b.   1
           3 x – 10 x + 6
                                                                c.   A constant polynomial does not have a degree.
356. Compute (9a2b + 2ab – 5a2) – (–2ab – 3a2 + 4a2b).
                                                                d.   none of the above
      a.   5a2b + 8a2
      b.   13a2b + 4ab – 8a2                               362. Which of the following is not a polynomial?
      c.   5a2b + 4ab – 2a2                                     a.   2
      d.   13a2b – 4ab –8a2                                     b.   2 – 3x – x2
                                                                c.   x – 3x–2
357. Compute ( 1 x2 + 2 x + 1) + (2x – 2 x2 + 4) –
                                                                d.   1 – [1 – x2 – (2 – x)]
               6      3                3
      ( 7 + 3x + 1 x2).
        2        2
                                                           363. Which of the following is not a polynomial?
      a. –x2 – 1 x +
               3
                             3
                             2
                                                                a.   –2–2x – 3–1
      b.   – 4 x2
             3      +   5
                        3x   +   3
                                 2                              b.   (2x0)–3 + 5–2x2 – 3–1x
      c.   –x2 +5x + 3
                3    2                                          c.   (–2x)–1 – 2
      d.   –x2 + 5x + 3                                         d.   All of the choices are polynomials.
                 3    2




    66
                                    –POLYNOMIAL EXPRESSIONS –


364. Which of the following statements is always             Set 24         (Answers begin on page 192)
     true?
     a. The difference of two polynomials is a          The problems in this set focus on the multiplication of
        polynomial.                                     polynomials.
     b. The sum of three polynomials is a polynomial.
     c. A trinomial minus a binomial is a polyno-       369. Compute 3x3        7x2.
     mial.                                                   a.   21x5
     d. All of the above statements are true.                b.   21x6
                                                             c.   10x5
365. Which of the following statements is not true?          d.   10x6
     a. The quotient of two polynomials is a
        polynomial.                                     370. 2x(5x2 + 3y) is equivalent to which of the
     b. The product of a constant and a polynomial           following expressions?
        is a polynomial.                                     a. 5x3 + 6xy
     c. The degree of the polynomial, in simplified          b. 10x2 + 6xy
        form, is the highest power to which the vari-        c. 10x3 + 6xy
        able is raised in the expression.                    d. 10x3 + 6y
     d. The degree of a constant polynomial is zero.
                                                        371. Which of the following expressions is equiva-
366. Write the expression –(–2x0)–3 + 4–2x2 – 3–1x –2        lent to x3 + 6x?
     in simplified form.                                     a. x (x2 + 6)
     a. 1 x2 – 1 x – 185                                     b. x (x + 6)
        8      3
           1 2 1         15                                  c. x (x2 + 6x)
     b.   16 x – 3 x – 8
                                                             d. x 2(x + 6)
     c.   – 1 x2 – 1 x – 8
            8      3
     d.   – 1 x2 + 1 x – 185
            8      3                                    372. Compute 2x2(3x + 4xy – 2xy3).
                                                             a.   6x3 +8x2y – 4x3y3
367. Compute –(2 – (1 – 2x2 –(2x2 – 1))) –                   b.   6x3 +8x3y – 4x3y3
     (3x2 – (1– 2x2)).                                       c.   6x3 +8x3y – 4x2y3
     a. –9x2 – 1                                             d.   6x2 +8x2y – 4x3y3
     b. 9x2 – 1
     c. 9x2 + 1                                         373. Compute 7x5(x8 + 2x4 – 7x – 9).
     d. –9x2 + 1                                             a.   7x 13 + 9x9 – 14x6 – 16x5
                                                             b.   7x 40 + 14x20 – 49x5 – 63
368. Compute –22(2–3 – 2–2x2) + 33(3–2 – 3–3x3).             c.   7x 13 + 2x4 – 7x – 9
                     5
     a. x3 + x2 +    2                                       d.   7x 13 + 14x9 – 49x6 – 63x5
                         5
     b. –x3 + x2 +       2
                         5
     c. –x3 – x2 +       2
                     5
     d. x3 – x2 –    2




                                                                                                          67
                                      –POLYNOMIAL EXPRESSIONS –



374. Compute 4x2z(3xz3 – 4z2 + 7x5).               379. What is the product of (2x + 6)(3x – 9)?
     a.   12x3z4
               –    8x2z3
                       +    28x7z                       a.   5x2 – 54
     b. 12x2z3 – 16x2z2 + 28x10z                        b.   6x2 – 54
     c. 12x3z4 – 16x2z3 + 28x7z                         c.   6x2 + 18x – 15
     d. 12x3z4 – 4z2 + 7x5                              d.   6x2 – 18x – 15
                                                        e.   6x2 + 36x – 54
375. What is the product of (x – 3)(x + 7)?
     a.   x2 – 21                                  380.Compute –3x(x + 6)(x – 9).
     b.   x2 – 3x –21                                   a.   –3x3 + 6x – 54
     c.   x2 + 4x – 21                                  b.   -x3 + 3x2 + 24x
     d.   x2 + 7x – 21                                  c.   -3x3 – 3x2 – 54
     e.   x2 – 21x – 21                                 d.   –3x2 + 6x – 72
                                                        e.   –3x3 + 9x2 + 162x
376. What is the product of (x – 6)(x – 6)?
     a.   x2 + 36                                  381. Compute (x – 4)(3x2 + 7x –2).
     b.   x2 – 36                                       a.   3x3 + 5x2 – 30x – 8
     c.   x2 – 12x – 36                                 b.   3x3 + 5x2 – 30x + 8
     d.   x2 – 12x + 36                                 c.   3x3 – 5x2 – 30x + 8
     e.   x2 – 36x + 36                                 d.   3x3 – 5x2 – 30x – 8

377. What is the product of (x –1)(x + 1)?         382. Compute (x – 6)(x – 3)(x – 1).
     a.   x2  –1                                        a.   x3 – 18
     b.   x 2 +1                                        b.   x3 – 9x – 18
     c.   x2 – x – 1                                    c.   x3 – 8x2 + 27x – 18
     d.   x2 – x +1                                     d.   x3 – 10x2 – 9x – 18
     e.   x2 – 2x –1                                    e.   x3 – 10x2 + 27x – 18

378. What is the value of (x + c)2?                383. Which of the following equations is equivalent
     a.   x2 +c 2                                       to (5x + 1)(2y +2) = 10xy + 12?
     b.    2
          x + cx + c2                                   a. 10x + 2y + 2 = 10
     c.   x2 + c2x2 + c2                                b. 10x + y = 10
     d.   x2 + cx2 + c2x + c2                           c. 5x + y = 5
     e.   x2 + 2cx + c2                                 d. 5x – y = 5

                                                   384. Compute (2x3 – 2x2 + 1)(6x3 + 7x2 – 5x – 9).
                                                        a.   12x6 + 2x5 – 24x4 – 2x3 + 25x2 – 5x – 9
                                                        b.   12x6 – 2x5 – 24x4 + 2x3 + 25x2 – 5x – 9
                                                        c.   12x6 – 2x5 – 24x4 – 2x3 – 25x2 + 5x – 9
                                                        d.   12x6 + 2x5 – 24x4 + 2x3 – 25x2 + 5x – 9




    68
                                     –POLYNOMIAL EXPRESSIONS –



     Set 25         (Answers begin on page 193)          391. Factor out the GCF: 125x3 – 405x2
                                                              a.   –5x 2(25x – 81)
The method of factoring out the greatest common fac-
                                                              b.   5x(25x2 – 81)
tor (GCF) from a polynomial is the focus of this prob-
                                                              c.   5x 2 (25x – 81)
lem set.
                                                              d.   This polynomial cannot be factored further.

385. Factor out the GCF: 15x – 10
                                                         392. Factor out the GCF: 73x3 – 72x2 + 7x – 49
     a.   –5(3x – 2)
                                                              a.   –7(49x3 + 7x2 + x – 7)
     b.   5(3x + 2)
                                                              b.   7(14x3 – 7x2 – x + 7)
     c.   –5(3x + 2)
                                                              c.   7(49x3 – 7x2 + x – 7)
     d.   This polynomial cannot be factored further.
                                                              d.   This polynomial cannot be factored further.

386. Factor out the GCF: 9x5 + 24x2 – 6x
                                                         393. Factor out the GCF: 5x(2x + 3) – 7(2x + 3)
     a.   3(3x 5 + 8x2 – 2x)
                                                              a.   (2x + 3)(7 – 5x)
     b.   3x(3x 4 + 8x – 2)
                                                              b.   (2x + 3)(5x – 7)
     c.   x(9x 4 + 24x – 6)
                                                              c.   (2x + 3)(5x + 7)
     d.   This polynomial cannot be factored further.
                                                              d.   This polynomial cannot be factored further.

387. Factor out the GCF: 36x4 – 90x3 – 18x
                                                         394. Factor out the GCF: 5x(6x – 5) + 7(5 – 6x)
     a.   9x(4x3 – 10x2 – 2)
                                                              a.   (5x – 7)(5 – 6x)
     b.   18(2x4 – 5x3 – x)
                                                              b.   (5x + 7)(6x– 5)
     c.   18x(2x3 – 5x2 – 1)
                                                              c.   (5x – 7)(6x– 5)
     d.   This polynomial cannot be factored further.
                                                              d.   This polynomial cannot be factored further.

388. Factor out the GCF: x3 – x
                                                         395. Factor out the GCF:
     a.   x(x2 –1)
                                                              6(4x + 1) – 3y(1 + 4x) + 7z(4x + 1)
     b.   –x(x2 + 1)
                                                              a. (6 – 3y + 7z)(4x + 1)
     c.   –x(x2 – 1)
                                                              b. (6 – 3y – 7z)(4x + 1)
     d.   This polynomial cannot be factored further.
                                                              c. (–6 + 3y – 7z)(1 + 4x )
                                                              d. This polynomial cannot be factored further.
389. Factor out the GCF: 5x2 + 49
     a.   5(x2 + 49)
                                                         396. Factor out the GCF: 5x ( 2 x + 7) – ( 2 x + 7)
                                                                                       3            3
     b.   5(x2 + 44)
                                                              a. 5x( 2 x + 7)
                                                                     3
     c.   5x(x + 49)
                                                              b. (5x – 1)( 2 x + 7)
                                                                           3
     d.   This polynomial cannot be factored further.
                                                              c. (5x + 1)( 2 x + 7)
                                                                           3

390. Factor out the GCF: 36 – 81x2                            d. This polynomial cannot be factored further.
     a.   9(4 – 9x2)
     b.   9(4 – x2)
     c.   9(x2 – 4)
     d.   This polynomial cannot be factored further.



                                                                                                          69
                                      –POLYNOMIAL EXPRESSIONS –



397. Factor out the GCF: 3x(x + 5)2 – 8y(x + 5)3 + 7z(x + 5)2
     a.   (x + 5)(3x – 8yx – 40y + 7z)
     b.   (x + 5)2(–3x + 8yx + 40y – 7z)
     c.   (x + 5)2(3x – 8yx – 40y + 7z)
     d.   This polynomial cannot be factored further.

398. Factor out the GCF: 8x4y2(x – 9)2 – 16x3y5(x – 9)3 + 12 x5y3(9 – x)
     a.   4x3y2(x – 9)[2x2 – 18x – 4y3x2 + 72y3x – 324y3 – 3x2y]
     b.   4x3y2(x – 9)[2x2 – 18x + 4y3x2 – 72y3x +324y3 – 3x2y]
     c.   4x3y2(x – 9)[2x2 + 18x – 4y3x2 + 72y3x – 324y3 + 3x2y]
     d.   This polynomial cannot be factored further.

399. Factor out the GCF: 8x4y2z(2w – 1)3 – 16x2y4z3(2w – 1)3 + 12x4y4z(2w – 1)4
     a.   4xyz(2w – 1)2[2x3 –4y3z3 + 6x3y3w – 3x3y3]
     b.   4x2y2z(2w – 1)2[2x2 –4y2z2 + 6x2y2w – 3x2y2]
     c.   4x2y2z(2w – 1)3[2x2 –4y2z2 + 6x2y2w – 3x2y2]
     d.   This polynomial cannot be factored further.

400. Factor out the GCF: –22a3bc2(d –2)3(1 – e)2 + 55a2b2c2(d – 2)2(1 – e) – 44a2bc4(d – 2)(1 – e)
     a.   11a2bc2(d –2)(1– e)[2a(d – 2)2(1 – e) + 5b(d – 2) + 4c2]
     b.   11a2bc2(d –2)(1– e)[–2a(d – 2)2(1 – e) + 5b(d – 2) – 4c2]
     c.   11a2bc2(d –2)(1– e)[–2a(d – 2)2(1 – e) + 5b(d – 2) + 4c2]
     d.   This polynomial cannot be factored further.




    70
                                     –POLYNOMIAL EXPRESSIONS –



     Set 26         (Answers begin on page 194)          407. Factor completely: 32x5 – 162x
                                                              a.   2x(4x2 + 9)
The problems in this set focus on factoring polynomi-
                                                              b.   2x(2x – 3)2(2x +3)2
als that can be viewed as the difference of squares or
                                                              c.   2x(2x – 3)(2x + 3)(4x2 + 9)
as perfect trinomials squared.
                                                              d.   This polynomial cannot be factored further.

401. Factor completely x2 – 36
                                                         408. Factor completely: 28x(5 – x) – 7x3(5 – x)
     a.   (x – 6)2
                                                              a.   7x(x – 2)(x + 2)(5 – x)
     b.   (x – 6)(x + 6)
                                                              b.   7x(2 – x)(2 + x)(5 – x)
     c.   (x + 6)2
                                                              c.   7x(5 – x)(x2 + 4)
     d.   This polynomial cannot be factored further.
                                                              d.   This polynomial cannot be factored further.

402. Factor completely: 144 – y2
                                                         409. Factor completely: x2(3x – 5) + 9(5 – 3x)
     a.   (12 – y)(12 + y)
                                                              a.   (x – 3)(x + 3)(3x – 5)
     b.   (11 – y)(11 + y)
                                                              b.   (x2 + 9 )(3x – 5)
     c.   (y – 12)(y + 12)
                                                              c.   (x2 + 9 ) (5 – 3x)
     d.   This polynomial cannot be factored further.
                                                              d.   This polynomial cannot be factored further.

403. Factor completely: 4x2 + 1
                                                         410. Factor completely: x(x2 + 7x) – 9x3(x2 + 7x)
     a.   (2x + 1)2
                                                              a.   x2(1 – 3x)(1 + 3x)(x + 7)
     b.   (2x + 1)(2x – 1)
                                                              b.   x2(x + 7)(1 + 9x2)
     c.   (2x + 1)2
                                                              c.   x2(3x –1)(3x + 1)(x + 7)
     d.   This polynomial cannot be factored further.
                                                              d.   This polynomial cannot be factored further.

404. Factor completely: 9x2 – 25
                                                         411. Factor completely: 1 + 2x + x2
     a.   (3x – 5)
                                                              a.   (x – 1)2
     b.   (3x –5) (3x + 5)
                                                              b.   (x + 1)2
     c.   (5x – 3)(5x + 3)
                                                              c.   (x + 1)(x +2)
     d.   This polynomial cannot be factored further.
                                                              d.   This polynomial cannot be factored further.

405. Factor completely: 121x4 – 49z2
                                                         412. Factor completely: 4x2 – 12x + 9
     a.   (11x2 – 7z)(11x2 + 7z)
                                                              a.   (2x – 3)(2x + 3)
     b.   (12x2 – 7z)(12x2 + 7z)
                                                              b.   (2x + 3)2
     c.   (7z – 11x2 )(7z + 11x2)
                                                              c.   (2x – 3)2
     d.   This polynomial cannot be factored further.
                                                              d.   This polynomial cannot be factored further.

406. Factor completely: 6x2 – 24
     a.   (6x – 2)(x + 2)
     b.   6(x –2)2
     c.   6(x – 2)(x + 2)
     d.   This polynomial cannot be factored further.



                                                                                                          71
                                           –POLYNOMIAL EXPRESSIONS –



413. Factor completely: 75x4 + 30x3 + 3x2               419. Factor completely: 6x2 + 11x – 2
     a.   3x 2(5x + 1)(5x – 1)                               a.   (2x + 2)(3x – 1)
     b.   3x 2(5x – 1)2                                      b.   (3x + 2)(2x – 1)
     c.   3x 2(5x + 1)2                                      c.   (x + 2)(6x – 1)
     d.   This polynomial cannot be factored further.        d.   This polynomial cannot be factored further.

414. Factor completely: 9x2(3 + 10x) –24x(10x + 3)      420. Factor completely: 12x2 – 37x – 10
     + 16(3 + 10x)                                           a.   (4x – 10)(3x + 1)
     a. (3 + 10x)(3x – 4)2                                   b.   (3x – 10)(4x + 1)
     b. (3 + 10x)(3x – 4)(3x + 4)                            c.   (3x – 2)(4x + 5)
     c. (3 + 10x)(3x + 4)2                                   d.   This polynomial cannot be factored further.
     d. This polynomial cannot be factored further.
                                                        421. Factor completely: 7x2 – 12x + 5
415. Factor completely: 1 – 6x2 + 9x4                        a.   (7x –1)(x – 5)
     a.   (1 + 3x2)2                                         b.   (7x +1)(x + 5)
     b.   (1 – 3x2)2                                         c.   (7x –5)(x – 1)
     c.   (1 – 3x2)(1 + 3x2)                                 d.   This polynomial cannot be factored further.
     d.   This polynomial cannot be factored further.
                                                        422. Factor completely: 9 – 7x – 2x2
416. Factor completely:   8x2   –   24x4   + 18x             a.   (9 + 2x)(1 – x)
     a.   2x(2x3
               +   3)2                                       b.   (3 + 2x)(3 – x)
     b. 2x(2x3 – 3)2                                         c.   (3 + x)(3 – 2x)
     c. 2x(2x3 – 3)(2x3 + 3)                                 d.   This polynomial cannot be factored further.
     d. This polynomial cannot be factored further.
                                                        423. Factor completely: 2x3 + 6x2 + 4x
Set 27       (Answers begin on page 194)                     a.   2(x + 2)(x2 + 1)
Factoring polynomials using the trinomial method is          b.   2(x2 + 2)(x + 1)
the focus of this problem set.                               c.   2x(x + 2)(x + 1)
                                                             d.   This polynomial cannot be factored further.
417. Factor completely: x2 + 2x – 8
     a.   (x + 4)(x – 2)                                424. Factor completely: –4x5 + 24x4 – 20x3
     b.   (x – 4)(x + 2)                                     a.   4x3(5 – x)(1 – x)
     c.   (x + 1)(x – 8)                                     b.   4x3(x – 5)(x – 1)
     d.   This polynomial cannot be factored further.        c.   –4x3(x –5)(x – 1)
                                                             d.   This polynomial cannot be factored further.
418. Factor completely: x2 – 9x + 20
     a.   (x – 4)(x – 5)                                425. Factor completely: –27x4 + 27x3 – 6x2
     b.   (x + 2)(x – 10)                                    a.   –3x2(3x + 1)(3x + 2)
     c.   –(x + 4)(x + 5)                                    b.   –3x2(3x – 1)(3x – 2)
     d.   This polynomial cannot be factored further.        c.   –3x2(3x – 1)(3x – 2)
                                                             d.   This polynomial cannot be factored further.



    72
                                       –POLYNOMIAL EXPRESSIONS –



426. Factor completely: x2(x + 1) – 5x(x + 1) +               432. Factor completely:
     6(x + 1)                                                      6x2(1 –x4) + 13x(1 – x4) + 6(1 – x4)
     a.   (x + 1)(x + 3)(x – 2)                                    a. (1 – x)2(1 + x)2(2x + 3)(3x + 2)
     b.   ( x + 1)(x – 3)(x – 2)                                   b. (1 – x)2(1 + x)2(2x + 3)(3x + 2)
     c.   (x – 1)(x – 3)(x + 2)                                    c. (1 – x)(1 + x)(1 + x2)(3x + 2)(2x + 3)
     d.   This polynomial cannot be factored further.              d. This polynomial cannot be factored further.


427. Factor completely:                                       Set 28       (Answers begin on page 196)
     2x2(x2 – 4) – x(x2 – 4) + (4 – x2)                       This problem set focuses on finding roots of poly-
                                                              nomials using factoring techniques and the Zero Fac-
     a.   (x – 2)(x + 2)(2x + 1)(x – 1)
                                                              tor Property.
     b.   (x – 2)(x + 2)(2x – 1)(x + 1)
     c.   (x – 2)(x + 2)(2x – 1)(x – 1)
                                                              433. Which of the following is a complete list of
     d.   This polynomial cannot be factored further.
                                                                   zeroes for the polynomial 9x2 – 36?
                                                                   a. 6
428. Factor completely: 27(x – 3) + 6x (x – 3) – x2 (x – 3)
                                                                   b. –6
     a.   –(x – 3)(x + 3)(x + 9)                                   c. –6 and 6
     b.   –(x – 3)(x + 3)(x –9)                                    d. 4 and 9
     c.   (x – 3)(x + 3)(x –9)
     d.   This polynomial cannot be factored further.         434. Which of the following is a complete list of
                                                                   zeroes for the polynomial 9x2 – 25?
429. Factor completely:                                            a. – 5 and 5
                                                                        3     3
     (x2 + 4x + 3)x2 + (x2 + 4x + 3) 3x + 2(x2 + 4x + 3)           b. –3 and 3
     a.   (x + 1)2(x – 2)(x – 3)                                   c. – 3 and 3
                                                                        5     5
     b.   (x – 1)2(x + 2)(x + 3)                                   d. –3 and 5
     c.   (x + 1)2(x + 2)(x + 3)
     d.   This polynomial cannot be factored further.         435. Which of the following is a complete list of
                                                                   zeroes for the polynomial 5x2 + 49?
430. Factor completely: 18(x2 + 6x + 8) – 2x2(x2 +6x + 8)          a. 0
     a.   2(x + 2)(x + 4)(3 – x)(3 + x)                            b. –1 and 0
     b.   2(x + 2)(x – 4)(x – 3)(x + 3)                            c. –2 and 0
     c.   2(x – 2)(x + 4)(3 – x)(3 + x)                            d. There are no zeroes for this polynomial.
     d.   This polynomial cannot be factored further.
                                                              436. Which of the following is a complete list of
431. Factor completely:                                            zeroes for the polynomial 6x2 – 24?
     2x2(16 + x4) + 3x(16 + x4) + (16 + x4)                        a. –2 and 4
     a.   (16 + x4)(2x + 1)(x + 1)                                 b. –2 and 2
     b.   (4 + x2)(4 – x2)(2x + 1)(x + 1)                          c. 2 and –4
     c.   (4 + x2)(2 – x2)(2 + x )(2x + 1)(x + 1)                  d. There are no zeroes for this polynomial.
     d.   This polynomial cannot be factored further.




                                                                                                             73
                                       –POLYNOMIAL EXPRESSIONS –


437. Which of the following is a complete list of            442. Which of the following is a complete list of
     zeros for the polynomial 5x(2x + 3) – 7(2x + 3)?             zeros for the polynomial 12x2 – 37x – 10?
     a. 2 and – 5
        3        7                                                a. 130 and –4
          2         7                                                   3        1
     b.   3   and   5                                             b.   10 and – 4
     c. – 2 and
          3
                        7
                        5                                         c.   10
                                                                        3 and – 4
                                                                                 1

     d. – 3 and
          2
                        7
                        5                                         d.   – 130 and –4

438. Which of the following is a complete list of            443. Which of the following is a complete list of
     zeros for the polynomial 5x( 2 x + 7) – ( 2 x + 7)?
                                  3            3                  zeros for the polynomial 9 – 7x – 2x2?
     a. – 1 and – 221
          5                                                       a. – 2 and –1
                                                                       9
          1         21
     b.   5   and    2                                            b. – 2 and 1
                                                                       9
     c. – 1 and – 221
          5                                                       c. – 9 and –1
                                                                       2
          1
     d.   5   and – 221                                           d. – 9 and 1
                                                                       2


439. Which of the following is a complete list of            444. Which of the following is a complete list of
     zeros for the polynomial 28x(5 – x) –      7x3
                                                  (5 – x)?        zeros for the polynomial 2x3 + 6x2 + 4x?
     a. 0, –2, 2, and 5                                           a. 0, 1, and 2
     b. –2, 2, and 5                                              b. –2, –1, and 0
     c. 0, –2, 2, and –5                                          c. –1 and 2
     d. –2, 2, and –5                                             d. 1 and 2

440. Which of the following is a complete list of            445. Which of the following is a complete list of
     zeros for the polynomial   75x4        3
                                       + 30x + 3x ?   2           zeros for the polynomial –4x5 + 24x4 – 20x3?
     a. 1 and – 1
        5        5                                                a. 0, 2, and 4
                    1                                             b. 0, –2 and 4
     b. 0 and       5
     c. 0 and – 1                                                 c. 0, 1, and 5
                5
              1                                                   d. 1, 2, and 5
     d. 0,    5   and – 1
                        5

                                                             446. Which of the following is a complete list of
441. Which of the following is a complete list of
                                                                  zeros for the polynomial 2x2(x2 – 4) – x(x2 – 4) +
     zeros for the polynomial x2 – 9x + 20?
                                                                  (4 – x2)?
     a. 4 and 5
                                                                  a. 1, 2, –2 and – 1
                                                                                    2
     b. –4 and 5
                                                                  b. –2, –1, and 2
     c. 4 and –5
                                                                  c. –2 and 2
     d. –4 and –5
                                                                  d. –2, 1, and 2




    74
                                          –POLYNOMIAL EXPRESSIONS –


447. Which of the following is a complete list of       452. Which of the following is the solution set for
     zeros for the polynomial       2x2       4
                                       (16 + x ) +           6x2 – 24 0?
     3x(16 + x4) + (16 + x4)?                                a. (–∞,–2]∪[2,∞)
     a. –1, – 1 , 2, and –2
              2                                              b. (–∞,–2)∪(2,∞)
     b.   –1 and 12                                          c. (2,∞)
     c.   1 and – 1
                  2                                          d. [–2,2]
     d. –1 and       –1
                      2
                                                        453. Which of the following is the solution set for
448. Which of the following is a complete list of            5x(2x + 3) – 7 (2x + 3)        0?
     zeros for the polynomial 18(x2 + 6x + 8) –              a. (–∞,– 3 )∪( 7 ,∞)
                                                                      2     5
     2x2(x2 + 6x + 8)?                                       b. (–∞,– 3 ]∪[ 7 ,∞)
                                                                      2     5
     a. 2, 3, and 4                                          c. (– 3 , 7 )
                                                                   2 5
     b. –4, –2, –3, and 3
                                                             d. [– 3 , 7 ]
                                                                   2 5
     c. –2, –3, and 4
     d. –2, –3, and 4
                                                        454. Which of the following is the solution set for
                                                             5x( 2 x + 7) – ( 2 x + 7)
                                                                 3            3            0?
     Set 29               (Answers begin on page 197)        a. (– 221 , 1 )
                                                                         5
                                                             b. [ 1 ,∞)
                                                                  5
This problem set focuses on solving polynomial
inequalities.                                                c. [– 221 , 1 ]
                                                                         5
                                                             d. (–∞,– 221 )∪( 1 ,∞)
                                                                              5
449. Which of the following is the solution set for
     x2 – 36 0?                                         455. Which of the following is the solution set for
     a. (6,∞)                                                28x(5 – x) – 7x3(5 – x) 0?
     b. (–∞,–6)∪(6,∞)                                        a. (–∞,–2]∪[0,2]∪[5,∞)
     c. (–∞,–6]∪[6,∞)                                        b. [–2,0]∪[2,5]
     d. the set of all real numbers                          c. (–∞,–2]∪[–2,0]∪[5,∞)
                                                             d. the set of all real numbers
450. Which of the following is the solution set for
     9x2 – 25 0?
                                                        456. Which of the following is the solution set for
     a. (– 5 , 5 )
           3 3
                                                             75x4 + 30x3 + 3x2        0?
     b. [– 5 , 5 ]
           3 3                                               a. (–∞,0)
     c. (–∞,– 5 )∪( 5 ,∞)
              3     3                                        b. (– 1 ,0)
                                                                   5
     d. the empty set                                        c. {– 1 ,0}
                                                                   5
                                                             d. the empty set
451. Which of the following is the solution set for
     5x2 + 49 0?
     a. (–∞,– 459 )
     b. (–∞,0)
     c. the empty set
     d. the set of all real numbers

                                                                                                       75
                                     –POLYNOMIAL EXPRESSIONS –


457. Which of the following is the solution set for    461. Which of the following is the solution set for
     x2 –9x + 20 0?                                         –4x5 + 24x4 – 20x3     0?
     a. (–∞,4)∪(5,∞)                                        a. (–∞,0]∪[1,5]
     b. (–∞,4]∪[5,∞)                                        b. (–∞,1]
     c. (–∞,5]                                              c. [0,1]∪[5,∞)
     d. (4,5)                                               d. the set of all real numbers

458. Which of the following is the solution set for    462. Which of the following is the solution set for
     12x2 – 37x – 10        0?                              2x2(x2 – 4) – x(x2 –4) + (4 – x2)   0?
     a. (–∞,– 1 )∪( 130 ,∞)
              4
                                                            a. (–2,– 1 )∪(1,2)
                                                                     2

     b. (–∞,– 1 ]∪[ 130 ,∞)                                 b. (– 1 ,1)∪(1,2)
                                                                  2
              4

     c. [– 1 , 130 ]                                        c. (–∞,– 1 )∪(2,∞)
                                                                     2
           4
                                                            d. (–∞,–2)∪(1,2)
     d. (– 1 , 130 )
           4

                                                       463. Which of the following is the solution set for
459. Which of the following is the solution set for
                                                            2x2(16 + x4) + 3x(16 + x4) + (16 + x4)     0?
     9 – 7x –2x2       0?
                                                            a. (–∞,–1]∪[– 1 ,∞)
     a. [– 9 ,1]
           2
                                                                           2
                                                            b. (–∞,–1)∪(– 1 ,∞)
     b. (–∞,– 9 )∪(1,∞)
              2
                                                                          2
                                                            c. [–1,– 1 ]
     c. the set of all real numbers except – 9 and 1
                                             2
                                                                     2
                                                            d. (–1,– 1 )
     d. (– 9 ,1)
           2
                                                                     2


                                                       464. Which of the following is the solution set for
460. Which of the following is the solution set for
                                                            18(x2 + 6x + 8) – 2x2(x2 + 6x + 8) 0?
     2x3 + 6x2 +4x 0?
                                                            a. (–∞,–4)∪(–2,∞)
     a. [–2,0]
                                                            b. (–4,–3)∪(–2,3)
     b. [–2,–1]∪[0,∞)
                                                            c. [–4,–3]∪[–2,3]
     c. (–∞,–2]∪[–1,∞)
                                                            d. the set of all real numbers except –4, –3, –2,
     d. (–∞,–1]∪[0,∞)
                                                               and 3




    76
                4                                    RATIONAL
S E C T I O N




                                                     EXPRESSIONS




Q               uotients of polynomials are called rational expressions.The arithmetic of rational expressions closely resem-
                bles that of fractions. Simplifying and understanding the graphical properties of both polynomials and
                rational expressions relies heavily on the use of factoring. This is the focus of the following six problem sets.




                                                                                                                         77
                                                         –RATIONAL EXPRESSIONS–



      Set 30                (Answers begin on page 202)                                      2x +     2
                                                                          470. Simplify: 4x3 – 16x24x 48x
                                                                                                    –
                                                                                     x+2
This problem set focuses on basic properties and sim-                           a.   x–6
plification of rational expressions.                                                       x
                                                                                b.   (x + 2)(x + 6)
                                                                                        1
                                                                                c.   2x – 12
                        2z2 – z – 15
465. Simplify: z2 + 2z – 15                                                            x+2
                                                                                d.   4x(x – 6)
           2z – 5
      a.    z–5                                                                      2x(x + 2)
                                                                                e.     x–6
           2z + 5
      b.    z–5
           2z – 5                                                         471. Which of the following makes the fraction
      c.   z+5
           2z + 5                                                                 x2 + 11x + 30
      d.    z+5                                                                 4x3 + 44x2 + 120x
                                                                                                      undefined?

                        25(–x)4
                                                                                a.   –6
466. Simplify: x(5x2)2                                                          b.   –4
      a. – 1
           x                                                                    c.   –3
            5
      b.   2x                                                                   d.   –2
      c.   – 25x                                                                e.   –1
           1
      d.   x
                                                                          472. The domain of the expression x3 2x4x is
                                                                                                               –
                3
467. Simplify: z8z – 16z                                                        a.   (–∞,–2)∪(2,∞)
                   – 32
           z(z + 4)                                                             b.   (–∞,2)∪(2,∞)
      a.       8
           z(z – 4)
                                                                                c.   (–∞,-2)∪(2,0)∪(0,2)∪(2,∞)
      b.      8                                                                 d.   (–,∞-2)∪(¬2,2)∪(2,∞)
           –z(z + 4)
      c.       8
           –z(z – 4)                                                                              2
      d.       8
                                                                                            x 16
                                                                          473. Simplify: x3 + x– – 20x
                                                                                               2

                                                                                      4
                                                                                a.   x+5
468. Simplify:
                                                                                     x+4
      y2 – 64          (y – 8)(y + 8)       (y – 8)(y + 8)                      b.    x
       8–y         =        8–y         =      –(y – 8)
                                                             = –(y + 8)
                                                                                     x+4
                                                                                c.   x+5
      a.   –y + 8
                                                                                      x+4
      b.   –(y + 8)                                                             d.   x2 + 5x
      c.   –(y – 8)                                                                     16
                                                                                e. – x3 – 20x
      d.   y+8

                         x2 + 8x                                          474. Which of the following could be equal to 4xx ?
469. Simplify: x3 – 64x
      a.    1                                                                   a. – 1
                                                                                     4
           x–8
                                                                                     0
      b.    x                                                                   b.   4
           x–8
      c.   x+8                                                                  c. 0.20
           x–8
                                                                                      4
      d. x – 8                                                                  d.   12
                                                                                      5
      e. x + 8                                                                  e.   20




    78
                                               –RATIONAL EXPRESSIONS–


475. Which of the following values make the expression
       x – 16
                                                           480. (2x – 5)(x +4)( ––(2x – 5)(x + 1) =
                                                                            9(2x 5)
       x2 – 16
                  undefined?
                                                                          1
      a.   –16                                                   a.   3(2x – 5)
      b.   –4                                                    b.       1
                                                                      9(2x – 5)
      c.   –1
                                                                      1
      d.   1                                                     c.   3
      e.   16                                                         1
                                                                 d.   9

476. Which of the following lists of values makes the
                       x2 + 7x + 12
                                                                 Set 31               (Answers begin on page 203)
      expression       x3 + 3x2 – 4x
                                       undefined?
      a.   –4, 1                                           This problem set focuses on adding and subtracting
      b.   –4, 0 1                                         rational expressions.
      c.   –4, –1, 0
      d.   –1, 0 ,4
                                                                                                    –      x+
                                                           481. Compute and simplify: 4x – 45 + 2xx– 99 – 3x – 91
                                                                                       x–9
      e.   0, 1, 4                                                    3x – 55
                                                                 a.    x–9
                                                                      3x – 53
                                                                 b.    x–9
                5x2
477. Simplify: 10x2(x – 1) – 3x(x – 1) –2(x ––1)
                    (x – 1) + 9x(x – 1) + 2(x 1)                      3x – 55
                                                                 c.   3x – 27
            x+1
      a.   2x + 1                                                     3x – 53
                                                                 d.   3x – 27
            x–1
      b.   2x + 1
            x–1                                                                                 5a        2a
      c.   2x – 1                                          482. Compute and simplify: ab3 + ab3
           x+1
      d.   2x – 1                                                a.   7
                                                                      b3

                      6x3 – 12x                                  b. 7b3
478. Simplify:          24x2                                           7
           x2 + 2                                                c.   ab3
      a.     4x
                                                                      7a
           x2 – 2                                                d.   b3
      b.     4x
              2
      c. – x 4– 2
              x
                                                                                                     3 – 2x    2–x
                                                           483. Compute and simplify: (x + 2)(x – 1) – (x –1 )(x +2)
              2
      d. – x 4+ 2
              x                                                  a.         –1
                                                                      (x – 1)(x + 2)
                                                                            1
                       4ab2 – b2
479. Simplify: 8a2 + 2a – 1
                                                                 b.   (x – 1)(x +2)

             b2                                                         1
      a.                                                         c. – x – 2
           2a + 1
                2
      b.   – 2ab+ 1                                              d. – x 1
                                                                        +2
             b2
      c.   2a – 1
                2
      d.   – 2ab– 1




                                                                                                                79
                                                     –RATIONAL EXPRESSIONS–


484. Compute and simplify: s43 + r22
                            r     s                                     489. Compute and simplify:
          4(s + r2)                                                            3y + 2            7y – 3           5
     a.                                                                       (y – 1)2
                                                                                         –   (y – 1)(y +1)
                                                                                                             +   y+1
            s2r3
          2s + r2
     b.    s2r3                                                                         +4
                                                                             a. – (y –y1)(y +1)
          4s + r2
     c.    s2r3
          2(2s + r2)
                                                                                       y+
                                                                             b. – (y – 1)(y4+ 1)2
     d.      s2r3
                                                                                   y+4
                                                                             c.   (y – 1)2
485. Compute and            simplify: x(x2- 2)   –       5 – 2x
                                                     (x – 2)(x – 1)                   y+4
                2
                                                                             d.   (y – 1)(y +1)
           2x – 3x +1
     a.   x(x – 1)(x – 2)
                                                                                                                            –1
     b.    2x2 + 3x – 1                                                                                        2z –
                                                                        490. Compute and simplify: 64zz+ 132 + 4z + 6
                                                                                                       +            3
          x(x – 1)(x – 2)
                                                                                  1
           2x2 – 3x – 1                                                      a.   2
     c.   x(x – 1)(x – 2)
                                                                             b. 2
           2x2 + 3x + 1
     d.   x(x – 1)(x – 2)                                                    c. –2
                                                                             d. – 1
                               4
486. Compute and simplify: t(t + 2) – 2
                                      t
                                                                                  2

           –2t
     a.   t+2
                                                                                                     4         5
                                                                        491. Compute and simplify: x – 3 + x + x
                                                                                                           3–
           –2
     b.   t+2
                                                                                      1
                                                                             a. – x + 3
                                                                                  x–
            2
     c.   t+2                                                                b. – x + 1
                                                                                  x+3
            2t                                                                    x+1
     d.   t+2                                                                c.   x–3
                                                                                  x–1
                                                                             d.   x–3
                                         1             3   2x
487. Compute and simplify: x(x + 1) – (x + 1)(x + 2) + x
                                                                                                                     3 x
           x2 + 10x – 8                                                 492. Compute and simplify: x2 – 10x + 24 – x – 6 + 1
     a.   x(x +1)(x +2)                                                           x–6
                                                                             a.   x+4
            2
           x + 10x + 8
     b.   x(x +1)(x +2)                                                      b.   x+6
                                                                                  x+4
           x2 – 10x + 8                                                           x–6
     c.   x(x +1)(x +2)                                                      c.   x–4
                                                                                  x+6
     d.    x2 – 10x – 8                                                      d.   x–4
          x(x +1)(x +2)

                                                                                                                 –x2 + 5x
                                          x
488. Compute and simplify: 2x + 1 – 2x – 1 + 4x2 – 1
                                                       1          2x2   493. Compute and simplify: (x – 5)2 + x + 1
                                                                                                              x+5
           (4x – 1)(x – 1)
                                                                                       x )
                                                                             a. – (x x(5)– 9+5)
                                                                                     – (x
     a.   (2x – 1)(2x + 1)
                                                                                     x( x + 9 )
           (4x + 1)(x – 1)                                                   b.   (x – 5)(x +5)
     b.   (2x – 1)(2x + 1)
          (4x + 1)(x + 1)
                                                                                       x    )
                                                                             c. – (x x(5)+ 9+5)
                                                                                     – (x
     c.   (2x – 1)(2x + 1)                                                           x(x – 9)
                                                                             d.   (x – 5)(x +5)
           (4x – 1)(x + 1)
     d.   (2x – 1)(2x + 1)




    80
                                          –RATIONAL EXPRESSIONS–



                                  2x2       1
494. Compute and simplify: x4 – 1 – x2 – 1 + x2 + 1
                                                         1   499. Compute and simplify: 9x––42
                                                                                        8 x
                                                                                                            10 – 5x
                                                                                                            2 – 9x
                                                                                      2
     a. – x2 2 1                                                  a. – 5(2 – x))
                                                                       (9x – 2
             +
                                                                        5
            2                                                     b.   –4
     b.    2
          x +1                                                                            2
                                                                  c. – 29(2 ––x) )2
                                                                        0(9x 2
                   2
     c.   – (x – 1)(x + 1)                                              9(2 – x)2
                                                                  d.   20(9x – 2)2
                 2
     d. – (x – 1)(x + 1)
                                                                                        12x2                –24xy3
                                                             500. Compute and simplify: –18xyy               56y3
                                 x–1     3x – 4                        2x2
495. Compute and simplify: x – 2 – x2 – 2x                        a.   7y
     a. – x – 2
            x                                                     b.   2x
                                                                       7y
          x+2
     b.     x                                                     c.   2x2
                                                                       7y2
          x–2
     c.     x                                                     d.   2x
                                                                       7y2
     d.   – x –x+ 2
                                                                                              x2 – x – 12
                                                             501. Compute and simplify: 3x2 – x –2 ÷ (3x2 – 10x – 8)
                                       3x – 3
496. Compute and simplify: 1 + x – 1 – x2 + 3x
                                 x                                          x+3
                                                                  a.   (3x + 2)2(x – 1)
          2x + 1
     a.    x +3                                                        (x – 4)2(x + 3)
                                                                  b.
     b.       x+
          – 2(x +31)                                                       (x – 1)
                                                                            x+3
     c.   2(x + 1)                                                c.   (3x + 2)(x – 1)2
            x +3
            2x + 1                                                         (x + 3)2
     d.   – x +3                                                  d.   (3x + 2)2(x – 1)

     Set 32            (Answers begin on page 204)                                      x –3   2
                                                                                                 –
                                                             502. Compute and simplify: 2x3 ÷ x 4x3x
                                                                                  2
This problem set focuses on multiplying and dividing
                                                                         – 3)
                                                                  a. – (x8x3
rational expressions.                                             b.   (x – 3)2
                                                                         8x3
                                 4x3y2   y3z4                          2
497. Compute and simplify:                                        c.   x3
                                  z3     2x3
          2y 6z2                                                       (x – 3)
     a.    x2                                                     d.     x3
          2y 6z
     b.    x3                                                                            2           2
                                                                                                          8x
                                                             503. Compute and simplify: xx2– 694 ÷ 6x x+ 46
                                                                                            –       2 –
          2y 5z
     c.    x2                                                     a.     x–8
                                                                       3(x + 3)
               6
          2y z                                                          x+8
     d.    x2
                                                                  b.   3(x + 3)
                                                                         x–8
                                                                  c.   3x(x + 3)
                                  8a4    5a2 + 13a – 6
498. Compute and simplify: 9 – a2         24a – 60a2                     x–8
                                                                  d.   3x(x + 3)
            2a3
     a.   3(3 – a)
            2a3                                                                           x – 6)2
     b.   3(3 + a)                                           504. Compute and simplify: 2(x + 5               –(5 +x)
                                                                                                              4(x –6)
             2a3                                                       –(x – 6)
     c.   –3(3 + a)                                               a.       2
             2a3                                                       (x – 6)
     d.   –3(3 – a)                                               b.      2
                                                                       –x – 6
                                                                  c.      2
                                                                       x + 6)
                                                                  d.      2


                                                                                                                        81
                                                    –RATIONAL EXPRESSIONS–


                                                                                                                  2
                            x2y3
505. Compute and simplify: 914x                    21y
                                                  15xy2
                                                            10x
                                                            12y3
                                                                                                                     x
                                                                            510. Compute and simplify: (x – 3) ÷ x +3x – 18
          9x                                                                           x
     a.   4y
                                                                                 a.   x+6
          3x                                                                           x
     b.   4y
                                                                                 b.   x–3
          3x                                                                          (x –3)2(x + 6)
     c.   16y                                                                    c.         x
           9x                                                                         x–3
     d.   16y3                                                                   d.   x+6

                                         4x2 + 4x + 1       2x2 + 3x + 1                                2
                                                                                                                           10xy2               3x2 + 3x
506. Compute and simplify:                 4x2 – 4x
                                                        ÷     2x2 – 2x      511. Compute and simplify: x 4– x
                                                                                                          y                2x – 2      ÷        15x2y2
          (2x + 1)(x + 2)                                                               5x3y3
     a.   2(x – 1)(x + 1)                                                        a.   2(x + 1)
          (2x + 1)(x – 2)                                                              25x3y3
     b.   2(x – 1)(x + 1)                                                        b.   4(x + 1)
     c.   (2x + 1)(x – 2)                                                               5x2y2
             2(x2 + 1)                                                           c.   4(x – 1)
          (2x – 1)(x – 2)                                                              25x2y3
     d.      2(x2 + 1)                                                           d.   4(x – 1)

                                         x2 – 1    2x + 2      x2 + x – 2
507. Compute and simplify: x2 +x                   1 – x2        x2 – x
                                                                            512. Compute and simplify:
          2(x + 2)                                                                  x+2          2x2 + 7x +3    6x2 + 5x + 1        x2 – 4
     a.      x2                                                                  x2 + 5x +6      4x2 + 4x + 1     3x2 + x          x2 + 2x

     b.    –2(x + 2)                                                             a. x + 2
                                                                                       x
              x2                                                                     –x + 2
          2x + 2
                                                                                 b. x
     c.     x2                                                                         +
                                                                                 c. x x2 2
          –2x + 2
     d.     x2
                                                                                       –
                                                                                 d. x x2 2

508. Compute and simplify:

     (4x2 – 8x – 5) ÷         –(x – 3)
                               x+1
                                           2x2 – 3x – 5
                                               x –3
                                                                                 Set 33             (Answers begin on page 205)
            2x + 1                                                          This problem set focuses on simplifying complex
     a.    – (2x – 5)
     b. 2x + 1                                                              fractions and performing multiple operations involv-
                                                                            ing rational expressions.
     c. –(2x + 1)
                                                                                                                      3
          2x + 1
     d.   2x – 5
                                                                                                                      4        1       5        1 2
                                                                            513. Compute and simplify: 1 –             9       4   +   2   –    4
                                                                                                                      16
                                                                                      275
                                                                                 a.    36
509. Compute and simplify:
                                                                                      275
                                                                                 b.    45
            a2 – b2         2a2 – 7ab + 3b2          ab – 3b2
        2a2 – 3ab + b2          a2 +ab        ÷   a2 + 2ab + b2                       245
                                                                                 c.    48
                                                                                      245
          (a + b)2                                                               d.    36
     a.      ab
          a2 + b2
     b.     ab
          2(a + b)
     c.      ab
          a – 3b
     d.     ab




   82
                                                  –RATIONAL EXPRESSIONS–



                                        2 3                   519. Compute and simplify:
                                        3+4
514. Compute and simplify:                                                5          2
                                        3 1
                                        4–2                            (x – 1)3 – (x –1)2
          16
     a.    3
                                                                          2          5
                                                                       (x –1)3 – (x –1)4
          17
     b.    3                                                      a. –(x + 1)
          17
     c.    6                                                      b. (x – 1)
          15
     d.    4                                                      c. –(x – 1)
                                                                          1
                                                                  d.   (x – 1)
515. Compute and simplify:
      3x2 + 6x           2+x   3x – 1                                                            x
       x–5          +    5–x   25 –x2                         520. Compute and simplify:
                                                                                            1–   5
                                                                                                 x
                                                                                            1+   5
     a. (x + 2)(x + 5)                                            a.     5
                                                                       x+5
     b. –(x + 2)(x + 5)                                           b.    1
                                                                       x+5
     c. –(x – 2)(x + 5)                                           c. 1
     d. (x + 2)(x – 5)                                            d. 0

                              1       1
516. Compute and simplify: (x + h)2 – x2              h       521. Compute and simplify:
           2x +h
     a.   x2(x +h)                                                       a–2 a+2
                     2
                                                                         a+2 – a–2
          –(2x +h)
     b.    x2(x +h)                                                      a–2 a+2
                                                                         a+2 + a–2
            2x +h
     c.   x2(x +h)2                                                         4a2
                                                                  a.   (a +2)(a – 2)
          –(2x +h)
     d.   x2(x +h)2                                               b.        –4a
                                                                       (a +2)(a – 2)
                                       1
                                    a+ b                          c.    4a
                                                                       a2 +4
517. Compute and simplify:             1
                                    b+ a                                –4a
          a
                                                                  d.   a2 +4
     a.   ab   +1
     b. ab                                                    522. Compute and simplify:
          ab + 1
     c.   ba – 1                                                       4
          a                                                          4 – x 2 –1
     d.   b
                                         3 1                         1        1
                                         x–2                        x+2 + x–2
518. Compute and simplify:              5    1
                                                                  a.   –x
                                        4x – 2x                        2
          2(6 – x)
     a.       3                                                   b.   x
                                                                       2
          6–x
     b.     3
                                                                         x
                                                                  c. – x – 4
     c.       –
          –6 3 x                                                       x+4
                                                                  d.   x–4
          6+x
     d.     6




                                                                                                     83
                                           –RATIONAL EXPRESSIONS–



523. Compute and simplify: (a –1 + b –1)–1                   Set 34           (Answers begin on page 207)
           ab
     a.   b+a                                          This problem set focuses on solving rational equations.
          b+a
     b.    ab
     c. ab                                             529. Solve: 3 = 2 + x
                                                                   x
     d. a + b                                                a.   –3 and 1
                                                             b.   –3 only
                           x –1 – –1
524. Compute and simplify: x –1 + y –1
                                  y                          c.   1 only
     a. 0                                                    d.   There are no solutions.
          y–x
     b.   y +x
     c.   y+x                                          530. Solve: 2 – 3 = 1
                                                                   3   x   2
          y–x
                                                                   7
     d. 1                                                    a.   18
                                                                  18
                                                             b.    7
525. Compute and simplify:                                   c. –18
         x2 + 4x –5    2x + 3         2                      d. 18
         2x2 + x – 3    x +1    –   x +2

     a.     x 2 – 5x + 8                                                     1
                                                       531. Solve: t 2t7 + t – 1 = 2
                                                                     –
          (x + 1)(x + 2)
           x2 + 5x + 8                                       a. –2
     b.   (x + 1)(x + 2)
                                                             b. 2
            x 2 – 5x – 8
     c.   (x + 1)(x + 2)                                          7
                                                             c.   5
               2
     d.   – (xx+–15xx+ 82)
                  )( +                                       d.   5
                                                                  7

                               5
526. Compute and simplify: x + 3 – x
                           x–
                                                1
                                              x –3
                                                                              12
                                                       532. Solve: x + 8 + x2 + 2x = 2
                                                                   x+2               x
     a.   (x + 5)(x – 1)                                     a.   –4
     b.   (x – 5)(x + 1)                                     b.   4
     c.   –(x – 5)(x + 1)                                    c.   –4 and –2
     d.   –(x + 5)(x + 1)                                    d.   4 and –2

                                  1          x+3                       x      2
527. Compute and simplify: [3 + x + 3 ]      x–2
                                                                                   3
                                                       533. Solve: x – 3 + x = x – 3
           x+3
     a.   3x + 10                                            a.   3
     b.   3x + 10                                            b.   2 and 3
           x–2
          3x + 10
                                                             c.   –2
     c.    x+3                                               d.   –2 and 3
          x+3
     d.   x–2

                                   3    1
528. Compute and simplify: 1 – 2 – 2x – 6x
                               x
          x+2
     a.     x
          x–2
     b.     x
          3x + 2
     c.     3x
          3x – 2
     d.    3x


    84
                                                                –RATIONAL EXPRESSIONS–


                    3                         6                                              x            3                   3
534. Solve: x + 2 + 1 = ( 2 – x)(2 + x)                                     540. Solve: x + 1 – x + 4 = x2 + 5x + 4
      a.   –4 and 1                                                               a.   –3 and –2
      b.   1 and 4                                                                b.   2 and 3
      c.   –1 and 4                                                               c.   –3 and 2
      d.   –4 and –1                                                              d.   –2 and 3

                      10                      3                                          2            4
535. Solve: (2x – 1)2 = 4 + 2x – 1                                          541. 1 + x – 3 = x2 – 4x – 3
             9
      a. –   8                                                                    a.   –1 and 1
           –1      9                                                              b.   3
      b.    2 and 8
           1     9                                                                c.   –1
      c.   2 and 8
           1       9                                                              d.   –1 and 3
      d.   2 and – 8
                                                                                             3       x–3
                           1                      1        1                542. Solve: x + 2 = x – 2
536. Solve for q:          f   = (k – 1)          pq   +   q
                                                                                  a.   2 and 4
                   f (k – 1)
      a. q =           p                                                          b.   4
                   f (k – 1)(1 + p)                                               c.   0
      b. q =               p                                                      d.   0 and 4
                   f (k + 1)(1 + p)
      c. q =               p
                                                                                             t+1              4
                   f (k + 1)                                                543. Solve: t – 1 = t2 – 1
      d. q =           p
                                                                                  a.   –1 and –3
                   x–1             4                                              b.   1
537. Solve: x – 5 = x – 5
                                                                                  c.   –3
      a.   3
                                                                                  d.   –1
      b.   –5
      c.   5                                                                                                      v1 +v2
      d.   There are no solutions.                                          544. Solve for v1: v =                     v1v2
                                                                                                                  1+    c2

                                          3                2                                 c2(v2 – v)
                         22
538. Solve: 2p2 – 9p – 5 – 2p + 1 = p – 5                                         a. v1 =
                                                                                              vv2 – c2
      a.   –5 and 1                                                                          c2(v – v2)
                                                                                  b. v1 =
                                                                                              vv2 – c2
      b.   –5
                                                                                             c2(v – v2)
      c.   5                                                                      c. v1 =
                                                                                              c2 – vv2
      d.   There are no solutions.                                                               2
                                                                                  d. v1 = – c (v2 + v2)
                                                                                               2
                                                                                                 c – vv2
                    x+1                 1                         1
539. Solve:        x3 – 9x
                               –   2x2 + x – 21        =   2x2 + 13x + 21
           7
      a.   9
               7
      b. – 9
               9
      c. – 7
      d. There are no solutions.




                                                                                                                                  85
                                          –RATIONAL EXPRESSIONS–



      Set 35            (Answers begin on page 211)          549. Determine the solution set for the inequality
                                                                  2z 2 – z – 15
                                                                  z 2 + 2z – 15        0.
This problem set focuses on solving rational inequalities.
                                                                  a. (–∞,–5]∪[– 5 ,∞)
                                                                                2
545. Determine the solution set for the inequality                b. [–5,– 5 ]
                                                                           2
      (x – 1)(x + 2)
         (x + 3)2           0.                                    c. (–∞,–5)∪[– 5 ,)
                                                                                2

     a.   (–2,1)                                                  d. [– 5 ,∞)
                                                                        2
     b.   [–2,1]
     c.   [–3,–2]                                            550. Determine the solution set for the inequality
     d.   [–3,–2]∪[1,∞)                                           25(–x)4
                                                                  x (5x2) 2       0.

546. Determine the solution set for the inequality                a.   the empty set
        x2 + 9                                                    b.   the set of all real numbers
      x2 – 2x – 3
                       0.
                                                                  c.   (0,∞)
     a.   (–1,∞)
                                                                  d.   (–∞,0)
     b.   (–1,3)
     c.   (–∞,–1)∪(3,∞)
                                                             551. Determine the solution set for the inequality
     d.   (–∞,3)                                                  z3 – 16z
                                                                  8z – 32         0.
547. Determine the solution set for the inequality                a. (0,4)
            2
         –x –1
      6x4 – x3 – 2x2
                     0.                                           b. (–∞,–4)
            1        2                                            c. (4,∞)
     a. (– 2 ,0)∪(0, 3 )
            1 2                                                   d. (–4,0)
     b. [– 2 , 3 ]
                  1  2
     c. (–∞,– 2 ]∪[ 3 ,∞)                                    552. Determine the solution set for the inequality
            1 2                                                   y2 – 64
     d. (– 2 , 3 )                                                 8–y            0.
                                                                  a.   [–8,8]
548. Determine the solution set for the inequality                b.   [–8,8)∪(8,∞)
      2     1                                                     c.   (–8,8)
       2 –
      x    x–1                                                    d.   (–8,8)∪(8,∞)
                ≥ 0.
        1     4
           – 2
      x+3 x
                                                             553. Determine the solution set for the inequality
     a. [–3,–2)∪[1,6)                                             x2 + 8x
                                                                  x 3 – 64x   > 0.
     b. [–3,–2]∪[1,6)
                                                                  a.   (8,∞)
     c. [–3,–2]∪[1,6]                                             b.   [8,∞)
     d. (–3,–2)∪(1,6)                                             c.   (8,–∞)
                                                                  d.   (–∞,8]




    86
                                                   –RATIONAL EXPRESSIONS–


554 Determine the solution set for the inequality              558. Determine the solution set for the inequality
     5x2(x – 1) – 3x(x – 1) – 2(x – 1)                               x+5                   1
    10x2(x – 1) + 9x(x – 1) + 2(x – 1) 0.                                    –x                          0.
                                                                     x–3                  x–3

     a. (– 1 ,– 2 )∪(– 2 ,1]
           2 5         5                                            a.   (–1,5)
     b. [– 1 ,– 2 )∪(– 2 ,1]
           2 5         5                                            b.   (–1, 3)∪(3,5)
     c. (–∞,– 1 ]∪(– 2 ,1]
              2      5
                                                                    c.   (–∞,3)∪(3,∞)
     d. (–∞,– 1 ]∪[– 2 ,∞)                                          d.   (–∞,3)∪(3,5)
              2      5

                                                               559. Determine the solution set for the inequality
555. Determine the solution set for the inequality
                                                                      x            1           2x 2
          3
     6x – 24x
                        0.                                          2x + 1   –   2x – 1   +   4x 2 – 1        0.
       24 x 2
     a.       [–2,0]∪[2,∞)                                          a. [– 1 ,– 1 ]∪( 1 ,1]
                                                                          2 4        2
     b.       (–∞,0]∪[2,∞)                                          b. [– 1 ,– 1 ]
                                                                          2 4
     c.       [–2,0)∪[2,∞)                                          c. [– 1 ,– 1 ]∪[ 1 ,1]
                                                                          2 4        2
     d.       [0,2]
                                                                    d. (– 1 ,– 1 ]∪( 1 ,1]
                                                                          2 4        2

556. Determine the solution set for the inequality
                                                               560. Determine the solution set for the inequality
     (2x – 5)(x + 4) – (2x – 5)(x + 1)                               3y + 2              7y – 3                5
                                                   0.               (y – 1)2
                                                                                 –                       +          0.
                 9(2x – 5)                                                           (y – 1)(y + 1)           y+1

     a.       ( 5 ,∞)
                2
                                                                    a.   (–∞,4]
     b.       (–∞, 5 )
                    2
                                                                    b.   (–∞,–4]
     c.       the empty set                                         c.   (–∞,4)
     d.       the set of all real numbers                           d.   (–∞,–4)


557. Determine the solution set for the inequality
         3 – 2x                   2–x
     (x + 2)(x –1)      –    (x – 1)(x + 2)   0.
     a.       (–∞,–2]
     b.       (–∞,1]
     c.       (–∞,–2)
     d.       [–2,1]




                                                                                                                         87
                5
                                              RADICAL EXPRESSIONS
S E C T I O N




                                              AND
                                              QUADRATIC EQUATIONS




A            n algebraic expression involving a term raised to a fractional exponent is a radical expression. The
             arithmetic of such expressions is really a direct application of the familiar exponent rules. Some-
             times, raising a negative real number to a fractional exponent results in a complex number of the
form a + bi, where a and b are real numbers and i = –1; the arithmetic of complex numbers resembles the
algebra of binomials.
     Various methods can be used to solve quadratic equations, and the solutions often involve radical terms.
These topics are reviewed in the seven problem sets in this section.




                                                                                                          89
                        –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–


                                                                                 4
      Set 36           (Answers begin on page 217)             566. Simplify:        312
                                                                    a.   27
The definition of fractional powers and the simplifica-
                                                                    b.   9
tion of expressions involving radicals are the focus of this
                                                                    c.   81
problem set.
                                                                    d.   243

561. –5 is a third root of what real number?                                     5

           3
                                                               567. Simplify:        515
      a. –5                                                         a.   5
      b. 25                                                         b.   15
      c. –125                                                       c.   125
      d. –625                                                       d.   625

562. Which of the following are second roots (ie.,             568. Find a number b that satisfies the following:
      square roots) of 49?                                           4
                                                                         (2b)4 = 8
      a. 7 only
                                                                    a.   2
      b. –7 only
                                                                    b.   3
      c. 7 and –7
                                                                    c.   4
      d. none of the above
                                                                    d.   There is no such value of b.

563 Which of the following is the principal fourth                                   1
                                                               569. Simplify: 64 6
      root of 625?                                                         6

      a. 5                                                          a. 2 2
      b. –5                                                         b. 2
      c. 25                                                         c. 664
      d. –25                                                        d. none of the above

                   5                                                                 5
564. Simplify:         –32                                     570. Simplify: 49 2
               5
      a.   2 2                                                      a. 245
                                                                        2
              5
      b.   –2 2                                                     b. 343
      c.   2                                                        c. 35
      d.   –2                                                       d. 16,807

                                                                                     –3
565. Which of the following is a value of b that satis-        571. Simplify: 81      4
                             3
      fies the equation b = 4?                                      a.    1
                                                                         27
      a. 64
           3                                                        b.   – 243
                                                                            4
      b. 4
                                                                    c. 27
      c. 16
                                                                    d. 9
      d. none of the above




    90
                        –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–



572. Simplify: 32 5
                    3
                                                    Set 37                    (Answers begin on page 217)
          96
     a.    5                                   The simplification of more complicated radical expres-
          1                                    sions is the focus of this problem set.
     b.   8
     c. 8                                                                3             3
     d. 64                                     577. Simplify:                9             –3
                                                    a. 3
                        –2
573. Simplify:
                   8     3                          b. –3
                   27                                   3
                                                    c. –12
     a. – 11365                                         3
                                                    d. –81
          4
     b.   9
          9                                                                                          x5
     c.   4                                    578. Simplify, assuming x                        0:   x7
     d.   3                                         a. x
          2
                                                         1
                                                    b.   x
                         –1                               1
574. Simplify:(–64)       3                         c.   x2
     a.   –1                                        d. x2
           4
     b. –4
     c. –16                                    579. Simplify: a3                  a3
     d. – 116                                       a.   a4 a
                                                    b.   a5
575. Simplify:(4x –4)
                          –1
                           2                        c.   a5 a
                                                    d.   a6
     a. x2
                                                    e.   a9
     b. 2x2
         2
     c. x
        2                                                                4    g
                                               580. Simplify:
     d. 2x –2                                                                4g
                                                    a. 2
576. Simplify: 4        x144                        b. 4
     a.   x36                                       c. g
                                                         2           g
     b.   4x72                                      d.       g
     c.   4x36                                      e. 2             g
     d.   2x72
                                                                         3
                                                                             27y3
                                               581. Simplify:
                                                                             27y2
                                                                 3
                                                    a.       3
                                                    b.           3
                                                         y 3
                                                    c.    3
                                                    d. y
                                                    e. y 3




                                                                                                          91
                                   –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–


                                                                                      4
                             a2b        ab2                                               243
582. Simplify:                                                587. Simplify:              4
                                   ab                                                         3
                                                                        4
               ab                                                  a. 9
     a.       ab                                                         4
                                                                   b. 3 3
     b. ab                                                             4
                                                                   c. 3
     c. ab
                                                                   d. 3
     d. ab ab
     e. a2b2                                                  588. Simplify:              x2 + 4x + 4
                                                                   a. x + 2
583. Simplify:               (4g2)3 (g4)                           b. x +2 x + 1
     a.   8g3                                                      c. x + 2 x + 2
     b.   8g4                                                      d. x + 2
     c.   8g5
     d.   8g10                                                589. Simplify:
                                                                                      4
                                                                                          32x8
     e.   8g12                                                               4
                                                                   a.   x2 8
                                                                           4
                           9pr                                     b.   x2 4
584. Simplify:                 3
                                                                   c.   2x 4
                                                                            4
                          (pr) 2
                                                                              4
     a.           3pr                                              d.   2x2 2
          3
     b.   pr                                                                          4

     c. 3 pr                                                  590. Simplify:              x21
                                                                             4
     d. 3pr                                                        a.   x4       x
                                                                             4
     e. 3p2r2                                                      b.   x5       x
                                                                             4
                                                                   c.   x4       x3
                                                                             4

585. If n = 20, what is the value of
                                              n+5   n
                                                        5 ?        d.   x3       x3
                                               n    2
     a. 5                                                                             3

     b.   5           5                                       591. Simplify:              54x2
               2                                                             3
                                                                   a.   2x 3x2
     c. 10                                                                 3

     d. 5 5                                                        b.   3x 2x2
                                                                             3
                                                                   c.   3x2 2x
     e. 25                                                                   3
                                                                   d.   2x2 3x
                            125
586. Simplify:               9
          5
                                                              592. Simplify:              x3 + 40x2 + 400x
     a.   3                                                        a.   (x + 20) x
     b.           5
              3                                                    b.   x x + 2x 10 + 20 x
          5           5                                            c.   x x + 20
     c.        3                                                   d.   This radical expression cannot be simplified
                  5                                                     further.
     d.       9




    92
                            –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–



     Set 38             (Answers begin on page 218)       599. Simplify: (5 –        3)(7 +          3)
                                                               a.   32 – 2      3
This problem set focuses on the arithmetic of radical
                                                               b.   32 + 2      3
expressions, including those involving complex numbers.
                                                               c.   16 – 4      3
                                                               d.   16 + 4      3
593. Simplify:         –25
     a.   5
                                                          600. Simplify: (4 +        6)(6 –          15)
     b.   5i
                                                               a.   24 + 4      15 + 6       6–3      10
     c.   –5
                                                               b.   24 – 4      15 + 6       6 –3    10
     d.   –5i
                                                               c.   24 + 4      15 – 6       6 –3    10
                                                               d.   24 + 4      15 – 6       6+3      10
594. Simplify:         –32
     a.   4i 2                                                                 –10 + –25
                                                          601. Simplify:            5
     b.   –4i 2
                                                               a.   –2 + i
     c.   –3i 2
                                                               b.   2+i
     d.   3i 2
                                                               c.   2–i
                                                               d.   –2 – i
595. Simplify: –            48 + 2 27 –       75
     a.   –3 3
                                                          602. Simplify: (4 + 2i)(4 – 2i)
     b.   –3 5
                                                               a.   12
     c.   5 3
                                                               b.   16
     d.   –5 5
                                                               c.   20
                                                               d.   20i
596. Simplify: 3            3+4 5–8       3
     a.   – 8                                             603. Simplify: (4 + 2i)2
     b.   –4 3 + 4 5                                           a.   12 – 16i
     c.   4 3–3 5                                              b.   16 + 16i
     d.   –5 3 + 4 5                                           c.   16 – 16i
                                                               d.   12 + 16i
597. Simplify: xy           8xy2 + 3y 2 18x3
     a.   11y 2x                                          604. Simplify:        21       3
                                                                                             +   7
                                                                                         7       3
     b.   11xy 2xy
                                                               a.   10    3
     c.   11x2y2 2
                                                               b.   10
     d.   11xy2 2x
                                                               c.   10    7
                       18       32
                                                               d.   10    21
598. Simplify:              +
                       25        9
          29       2
     a.        5                                          605. Simplify: (2 +        3x)2
          23 2                                                 a.   4 3x + 7x
     b.     15
                                                               b.   4 3x + 7 + x
          29 2
     c.     15                                                 c.   4 + 4 3x + x 3
          32 2                                                 d.   4 + 4 3x + 3x
     d.     15

                                                                                                           93
                         –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–


                                                                      2                     4
606. Simplify: (         3+    7)(2 3 – 5   7)              611. If a 3 =6, then a 3 =
     a.   –29 – 3        21
                                                                 a. 3
     b.   –29 + 3        21
                                                                 b. 6
     c.   29 – 3        21
                                                                 c. 3 6
     d.   –29 – 3        21
                                                                 d. 6 6
                      1                                          e. 36
607. Simplify:
                    3–5 2
          3 –5 2                                                                   4
     a.      41                                             612. If        p           = q–2 = – 1 , which of the following is
                                                                                                 3
             5
     b. – 3 –41     2
                                                                 a possible value of p?
          3+5 2                                                  a. – 1
                                                                      3
     c.     41                                                        1
                                                                 b.   9
     d.         5
          – 3 +41 2                                                   1
                                                                 c.   3
                                                                 d. 3
                                                       2x
608. Simplify by rationalizing the denominator:                  e. 9
                                                   2 –3 x

     a. – 2 4 2x9x
              +
                +6x
                                                                               3
                                                            613. Solve:            5x – 8 = 3
              2x + 2x
     b.       2 – 3x
                                                                 a.   49
          2    2x – 6x
                                                                 b.   –7
     c.       4 – 9x                                             c.   7i
     d.   2    2x + 6x                                           d.   –7
              4 – 9x
                                                                               3
                                                            614. Solve:            7 – 3x = –2
     Set 39              (Answers begin on page 219)             a.   –5
                                                                 b.   5
This problem set focuses on solving equations involving          c.   5i
radicals.                                                        d.   –5i

609. Solve:       7 + 3x = 4                                615. Solve: (x –3)2 = –28
     a.   3                                                      a.   3 2i 7
     b.   –3                                                     b.   –3 2i 7
     c.   3i                                                     c.   2 3i 7
     d.   There is no solution.                                  d.   –2 3i 7

610. Solve:       4x + 33 = 2x – 1                          616. Solve:     10 – 3x = x – 2
     a.   4                                                      a.   –2
     b.   –2                                                     b.   –2 and –3
     c.   –2 and 4                                               c.   3
     d.   There is no solution.                                  d.   There is no solution.




    94
                     –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–



617. Solve:     3x + 4 + x = 8              624. Solve: x2 + 81= 0
     a.   4 and 15                               a. 9
     b.   15                                     b. 9i
     c.   –4 and 15                              c. –9,–9i
     d.   4                                      d. 9,9i

618. Solve: (x – 1)2 + 16 = 0
     a.   1 2i                                   Set 40             (Answers begin on page 221)
     b.   1 4i
                                            Solving quadratic equations using the quadratic formula
     c.   –1 4i
                                            is the topic of this problem set.
     d.   –1 2i

                                            625. Solve using the quadratic formula: x2 – 7 = 0
619. Solve: x3 = –27
                                                 a.        7i
     a.   3i
                                                 b.       7
     b.   –3
                                                 c.       i 7
     c.   –3i
                                                 d.           7
     d.   3

                                            626. Solve using the quadratic formula: 2x2 – 1 =0
620. Solve: x2 = 225
     a. 15i                                                     2
                                                 a.         2
     b. –15i                                               i 2
                                                 b.
     c.   15                                                2

     d. 5 5                                      c.         2
                                                 d.       i 2
621. Solve: x3 = – 125
     a.   –5                                627. Solve using the quadratic formula: 4x2 + 3x = 0
     b.   5                                      a. 0,– 3
                                                        4
     c.   5i                                     b. – 3
                                                      4
     d.   –5i
                                                 c. – 4
                                                      3
                                                      3
622. Solve: (x + 4)2 = 81                        d.   4
     a.   –13
     b.   5                                 628. Solve using the quadratic formula: 5x 2 + 20x = 0
     c.   –13, 5                                 a.   –4
     d.   There is no solution.                  b.   0,4
                                                 c.   4,–4
623. Solve: x2 + 1 = 0                           d.   0,–4
     a. 1
     b. –1,–i
     c. 1,i
     d. i


                                                                                              95
                                 –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–


629. Solve using the quadratic formula:                 634. Solve using the quadratic formula:
     x2   + 4x + 4 = 0                                       x2 + 2 2x + 3 = 0
     a.   2                                                  a. 2          i
     b.   2i                                                 b. 2          1
     c.   2                                                  c. 1 i         2
     d.   –2i                                                d. i          2

630. Solve using the quadratic formula:                 635. Solve using the quadratic formula: x2 = –2x
     x2 + 5x – 6 = 0                                         a.   2,0
     a.   –2, –3                                             b.   –2,0
     b.   1,–6                                               c.   2i,–2i
     c.   –1,6                                               d.   0
     d.   3,2
                                                        636. Solve using the quadratic formula: (3x –8)2 = 45
631. Solve using the quadratic formula:                      a.   –3 8          5
                                                                     3
     3x2   + 5x + 2 = 0                                           –3        5
                                                             b.
     a. –1, 2
            3
                                                                       8
                                                                  –8 3          5
     b. –1, – 2
              3
                                                             c.      3
     c. 1, – 2
             3                                               d.   –8        5
                                                                       3
     d. 1, 2
           3
                                                        637. Solve using the quadratic formula:
                                               2
632. Solve using the quadratic formula: 5x – 24 = 0          0.20x2 – 2.20x + 2 + 0
                2           30                               a.   0.01, 0.1
     a.                 5
                                                             b.   10, 100
                2i 30
     b.           5                                          c.   0.1, 1
     c.         2        6                                   d.   1, 10
     d.        2i 6
                                                        638. Solve using the quadratic formula:
633. Solve using the quadratic formula:                      x2 – 3x – 3 = 0
     2x2 = –5x – 4                                                3 3 7
           –5           i 7                                  a.     2
     a.             4
                                                                  3 7 3
           5     i 7                                         b.     2
     b.         4
                                                                  –3        21
           –5     i 7                                        c.        2
     c.         –4
                                                                  3        21
           –7           i 5                                  d.        2
     d.             4




    96
                               –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–


639. Solve using the quadratic formula:                      644. Solve using radical methods: (3x – 8)2 = 45
     1 2
     6x    – 5 x +1 = 0
             3                                                         –8 3i 5
                                                                  a.      3
     a.   –5   19
                                                                       –8 3     5
     b.   5   19                                                  b.      3
     c.   5 i 19                                                  c.   8 3 5
                                                                         3
     d.   –5 i 19
                                                                       –8 3i 5
                                                                  d.      3
640. Solve using the quadratic formula:
     (x – 3)(2x + 1) = x(x – 4)                              645. Solve using radical methods: (–2x + 1)2 – 50 =0
                                                                       –1 5     2
     a.   –1              13                                      a.      2
                  2
                                                                       1 5i 2
     b.   1           13                                          b.     2
                  2
                                                                       1 5 2
     c.   1 i 13                                                  c.     2
             2
                                                                       –1 5i 2
     d.   –1 i 13                                                 d.      2
             2

                                                             646. Solve using radical methods: –(1 – 4x)2 – 121 = 0
     Set 41                    (Answers begin on page 222)             –1 11i
                                                                  a.     4
Solving quadratic equations using radical and graphical                1 11i
methods is the focus of this problem set.                         b.     4
                                                                       1 i 11
                                                                  c.      4
                                                2
641. Solve using radical methods: 4x = 3                               –1 i 11
                                                                  d.      4
     a.               3
                      2

     b.               2                                      647. Find the real solutions of the following equa-
                      3
                                                                  tion, if they exist, using graphical methods:
     c.       i       3
                      2                                           5x2 – 24 = 0
     d.       i       2                                           a. ≈ 2.191
                      3
                                                                  b. 4.8
642. Solve using radical methods: –3x2 = –9                       c. ≈ 2.191
     a.        3i                                                 d. The solutions are imaginary.
     b.       3
     c.          3                                           648. Find the real solutions of the following equa-
     d.       i 3                                                 tion, if they exist: 2x2 = –5x – 4
                                                                  a. 0.5, 1.5
643. Solve using radical methods: (4x + 5)2 = –49                 b. –1.5, 0
          5 7i
                                                                  c. –0.5, 0.5
     a.    4                                                      d. The solutions are imaginary.
          –5 7i
     b.     4
          –7 5i
     c.     4
          7 5i
     d.    4
                                                                                                             97
                    –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–


649. Find the real solutions of the following equa-   655. Find the real solutions of the following equa-
     tion, if they exist:   4x2   = 20x – 24               tion, if they exist: 1 x2 – 5 x + 1 = 0
                                                                                6      3
     a. 2, 3                                               a. ≈ 0.51, 10.51
     b. 16, 36                                             b. ≈ 0.641, 9.359
     c. –2, –3                                             c. 1, 4.2
     d. 16, –36                                            d. The solutions are imaginary.

650. Find the real solutions of the following equa-   656. Find the real solutions of the following equa-
     tion, if they exist: 12x –  =0 15x2                   tion, if they exist: (2x + 1)2 – 2(2x + 1) – 3 = 0
     a. 0, –1.25                                           a. –11, –1
     b. –1.25, –1.25                                       b. 1, 11
     c. 0, 1.25                                            c. –1, 11
     d. The solutions are imaginary.                       d. The solutions are imaginary.

651. Find the real solutions of the following equa-        Set 42                    (Answers begin on page 227)
     tion, if they exist: (3x     –8)2
                                  45
                                                      Solving equations that can be put in quadratic form via
     a. ≈ –3.875, 3.875
                                                      substitution is the focus of this problem set.
     b. –3, 5
     c. ≈ 3.875, 4.903
                                                      657. Solve: b4 – 7b2 + 12 = 0
     d. The solution are imaginary.
                                                           a.            2,           3
652. Find the real solutions of the following equa-
                                                           b.   2,   3
     tion, if they exist: 0.20x2 – 2.20x + 2 = 0           c.     2, 3
     a. –10, –1                                            d.   2, 3
     b. 1, 10
     c. –1, 10                                        658. Solve: (3b2 – 1)(1 – 2b2) = 0
     d. The solutions are imaginary.                       a.            2
                                                                                 ,        3
                                                                     2                3
                                                                         2                3
653. Find the real solutions of the following equa-        b.        3       ,        2
     tion, if they exist: x2 – 3x – 3 = 0                  c.            3
                                                                             ,            2
                                                                     2                3
     a. ≈ –0.791, 3.791
                                                                         2                3
     b. 1, 3                                               d.        3       ,        3
     c. –1, –3
     d. The solutions are imaginary.                  659. Solve: 4b4 + 20b2 + 25 = 0

                                                           a.    i       10
654. Find the real solutions of the following equa-                       3

                                                           b.        10
     tion, if they exist: x2 = –2x                                    3
     a. 0, 2                                                         10
                                                           c.
     b. –2, 0                                                         2

     c. –2, 2                                              d.   i        10
                                                                          2
     d. The solutions are imaginary.



    98
                                      –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–



660. Solve: 16b4 – 1 = 0                                     665. Solve: 3 + x– 4 + x
                                                                                                  1          –1
                                                                                                              2   =0
     a.        i     1,               1                                          –16                       –16
                     2                2                           a.                            ,
                                                                       (1 +              13)4         (1 –        13)4
                     1                1                                             16                       16
     b.        i     2    ,           2                           b.                            ,
                                                                       (1 + 13)4                      (1 – 13)4
                     1                                                     –16i                           –16i
     c.          i            ,               1                   c.                            ,
                     2                        2                        (1 + 13)4                      (1 –      13)4
                                                                           –16i                              16i
     d.     i        1,           1                               d.                            ,
                     2            2                                    (1 +              13)4         (1 –        13)4

                                                  1
661. Solve: x + 21 = 10x 2
                                                             666. Solve: (x3 + 5)2 – 5(x3 + 5) + 6                            0
     a.   –49,–9                                                        3             3
     b.   –49,9                                                   a.            2,        –3
                                                                        3                 3
     c.   9,49                                                    b.            –2,           3
                                                                        3             3
     d.   –9,49                                                   c.            2,        3
                                                                        3                 3
                                                                  d.            –2,           –3
662. Solve: 16 –56                            x + 49x = 0
          16                                                 667. Solve: 4x6 + 1 = 5x3
     a.   49
          49                                                      a. – 3 1 , –1
     b.   16                                                                    4
     c.   – 16
            49                                                    b. –      3   1, 1
                                                                                4
     d.   – 49
            16
                                                                  c.   3    1, 1
                                                                            4
663. Solve: x –                   x=6                             d.   3    1 , –1
                                                                            4
     a.   9
                                                                                                      2
     b.   –9                                                 668. Solve: x2 + x                           + 12 = 8 x2 + x
     c.   16
                                                                  a.   1, 2, 3
     d.   –16
                                                                  b.   –3, –2, 1, 2
                          1               1                       c.   –3, –2, –1, 2
664. Solve: 2x 6 – x 3 = 1
                                                                  d.   –3, 1, 2, 3
     a.   –1
     b.   i                                                                                                  2
                                                             669. Solve: 2 1 +                         w          = 13 1 +   w –6
     c.   1
                                                                  a.   25
     d.   –i
                                                                  b.   16
                                                                  c.   36
                                                                  d.   –16




                                                                                                                                    99
                                 –RADICAL EXPRESSIONS AND QUADRATIC EQUATIONS–


                         3                 3
670. Solve: (r – r )2 – (r – r ) – 6 = 0                                  2     1
                                                        672. Solve: 2a 3 – 11a 3 + 12   0
                    3            21                                        27
      a. –3, 1,              2                               a. –64, –      8
                    3            21                          b. –64, 287
      b. –1, 3,              2
                                                             c. 64, 287
                   3         i 21
      c. –3, 1,              2                               d. 64, –     27
                                                                           8
      d. –3, 1, –3           2
                                 21


                                      4
671. Solve: 6          x – 13             x+6   0
      a.   – 16 , – 81
             81     16
           16 81
      b.   81 , 16
      c.   – 16 , 81
             81 16
      d.   – 81 , 16
             16 81




   100
                6
S E C T I O N




                                                 ELEMENTARY
                                                 FUNCTIONS




T         he first functions to which you are typically exposed are those described by sets of ordered pairs that
          can be visualized in the Cartesian plane. Such functions are generally described using either algebraic
          expressions or graphs, and are denoted using letters, such as f or g. When we want to emphasize the input-
output defining relationship of a function, an expression of the form y = f(x) is often used. The arithmetic of real-
valued functions is performed using the arithmetic of real numbers and algebraic expressions.
      The domain of a function can be thought of as the set of all possible x-values for which there corresponds
an output y. From the graphical viewpoint, an x-value belongs to the domain of f if an ordered pair with that
x-value belongs to the graph of f. When an algebraic expression is used to describe a function y = f(x), it is con-
venient to view the domain as the set of all values of x that can be substituted into the expression and yield a mean-
ingful output. The range of a function is the set of all possible y-values attained at some member of the domain.
Basic functions and their properties are reviewed in the ten problem sets that make up this section.




                                                                                                              101
                                      –ELEMENTARY FUNCTIONS–



     Set 43        (Answers begin on page 231)            674. In the following graph of f(x), for how many
                                                               values of x does f(x) = 0?
The problems in this set focus on the notions of domain
                                                                                    y
and range and the basic arithmetic of elementary
functions.

673. In the following graph of f(x), for how many
     values of x does f(x) = 3?
                           y
                                                                                                        x




                                                  x
                                                               a.   2
                                                               b.   3
                                                               c.   4
                                                               d.   5
                                                               e.   8

     a.   0                                               675. In the following graph of f(x), for how many
     b.   1                                                    values of x does f(x) = 10?
     c.   2
                                                                                    y
     d.   3
     e.   4




                                                                                                        x




                                                               a.   0
                                                               b.   2
                                                               c.   4
                                                               d.   5
                                                               e.   8



   102
                                                     –ELEMENTARY FUNCTIONS–


                                                                                                                                    1
676. What is the range of the function f(x) = x2 – 4?             677. Which of the following is true of f(x) = – 2 x 2 ?
     a. the set of all real numbers excluding 0                        a. The range of the function is the set of all real
     b. the set of all real numbers excluding 2 and –2                    numbers less than or equal to 0.
     c. the set of all real numbers greater than or                    b. The range of the function is the set of all real
        equal to 0                                                        numbers less than 0.
     d. the set of all real numbers greater than or                    c. The range of the function is the set of all real
        equal to 4                                                        numbers greater than or equal to 0.
     e. the set of all real numbers greater than or                    d. The domain of the function is the set of all
        equal to –4                                                       real numbers greater than or equal to 0.
                                                                       e. The domain of the function is the set of all
                                                                          real numbers less than or equal to 2.

For questions 678–680, refer to the functions f and g, both defined on [–5,5], whose graphs are shown here.

                             y                                                                     y

                         5                    y =f(x)                                          5                   y =g(x)
                         4
                                                                            (–5,4)
       (–5,3)                                                                                  4
                         3
                                                                                               3
                         2               (2,2)                                                 2
                          1
                                                                                                1

           –5 –4 –3 –2 –1        1   2    3      4   5
                                                              x                                                               x
                         –1                                                      –5 –4 –3 –2 –1        1   2   3   4   5
                                                                                               –1
                         –2      (2,–1)
                                                                                               –2
                        –3                           (5,–2)                                   –3
                        –4
                                                                                              –4                           (5,–4)
                        –5
                                                                                              –5




678. The range of f is which of the following?                    680. 2 f(0) + [f(2) g(4)]2 =
     a.   [–2,2]∪{3}                                                   a.   18
     b.   (–2,–1]∪[0,3]                                                b.   10
     c.   (–2,–1]∪[0,2)∪{3}                                            c.   8
     d.   [–2,2)∪{3}                                                   d.   16

679. The range of g is which of the following?
     a.   [–4,4]
     b.   [–4,2]∪(2,4]
     c.   [–4,1]∪(1,4]
     d.   none of the above




                                                                                                                                    103
                                               –ELEMENTARY FUNCTIONS–


For questions 681–684, refer to the graphs of the fol-     684. Which of the following is the solution set for
lowing fourth-degree polynomial function y = p(x).               the inequality –1 p(x)         0?
                                                                 a. (1,3)
                                                                 b. [1,3]∪{–3}
                       y                                         c. (1,2)∪(2,3)
                                                                 d. [1,2)∪(2,3]∪{–3}
                   4
                   3                                                                                   2x + 1
                                            y =p(x)        685. The range of the function f(x) = 1 – x is which
                   2
                                                                 of the following?
                   1
                                                                 a. (–∞,1)∪(1,∞)
                                                  x
     –4 –3 –2 –1           1   2    3   4                        b. (–∞,–2)∪( –2,∞)
                –1
                               (2,–1)
                                                                 c. (–∞,– 1 )∪(– 1 ,∞)
                                                                           2     2
                  –2
                                                                 d.
                  –3

                  –4                                       For questions 686–688, use the following functions:
                                                                                                                   1
                                                           f(x) = –(2x –(–1 –x2))    g(x) = 3(1 + x)    h(x) = 1 + x2
681. The zeros of p(x) are x =
                                                                                                                9f(x)
     a.   –3, 0, 2                                         686. Which of the following is equivalent to g(x) ?
     b.   –3, 1, 3                                               a.   –g(x)
     c.   –3, 0, 1, 2, 3                                         b.   g(x)
     d.   none of the above                                      c.   –3g(x)
                                                                 d.   3g(x)
682. Which of the following is the range of p(x)?
     a.                                                    687. What is the domain of the function 2g(x)h(x)?
     b. [–1,4]                                                   a.   (–∞,–1)∪(–1,1)∪(1,∞)
     c. [–1,∞)                                                   b.   (–∞,–1)∪(1,∞)
     d. [–4,4]                                                   c.   (–∞,–1)∪(–1,∞)
                                                                 d.   the set of all real numbers
683. Which of the following is the domain of p(x)?
     a.                                                    688. Which of the following is equivalent to
     b. [–1,4]                                                   3f(x) – 2xg(x) – h(1x) ?
     c. [–1,∞)                                                   a. –10x2 + 6x + 2
     d. [–4,4]                                                   b. –2(5x2 + 6x + 2)
                                                                 c. 10x2 + 12x + 4
                                                                 d. 2(5x +2)(x + 1)




   104
                                        –ELEMENTARY FUNCTIONS–



                                   Set 44       (Answers begin on page 233)

This problem set focuses on compositions of functions, the simplification involved therein, and the general prin-
ciples of the graph of a function.

For questions 689–693, use the following diagrams:

                         y                                                          y
            A                                                          D

                                    x                                                          x




                   x2 + y 2 = 4                                                     1
                                                                                 y= x

                         y
                                                                                    y
           B
                                                                       E
                                    x
                                                                                                x




                    y= |x | –3
                                                                             y=(x – 3)2 + 1

                        y
                                                          689. Which of the coordinate planes shows the
           C                                                    graph of an equation that is not a function?
                                                                a. A
                                                                b. B
                                   x
                                                                c. C
                                                                d. D
                                                                e. A and D



                   y= √x



                                                                                                         105
                                     –ELEMENTARY FUNCTIONS–


690. Which of the coordinate planes shows the            695. Simplify f(x + h) – f(x) when f(x) = –(x – 1)2 + 3.
     graph of a function that has a range that con-           a.   h
     tains negative values?                                   b.   f(h)
     a. A                                                     c.   h(h – 2x + 2)
     b. B                                                     d.   –2hx + h2 – 2h
     c. D
     d. B and D                                          696. Compute (g ˚ h)(4) when g(x) = 2x2 – x – 1 and
     e. A, B, and D                                           h(x) = x – 2    x.
                                                              a. 0
691. Which of the coordinate planes shows the graph of        b. 1
     a function that has a domain of all real numbers?        c. –1
     a. B                                                     d. 4
     b. D
     c. E                                                697. Simplify (f ˚ f ˚ f)(2x) when f(x) = –x2.
     d. B and D                                               a.   16x
     e. B and E                                               b.   –16x
                                                              c.   64x8
692. Which of the coordinate planes shows the                 d.   –256x8
     graph of a function that has the same range as
     its domain?                                         698. If f(x) = 3x + 2 and g(x) = 2x – 3, what is the
     a. B and C                                               value of g(f(–2))?
     b. C and D                                               a. –19
     c. B and D                                               b. –11
     d. B and E                                               c. –7
     e. D and E                                               d. –4
                                                              e. –3
693. Of the equations graphed on the coordinate
     planes, which function has the smallest range?      699. If f(x) = 2x + 1 and g(x) = x – 2, what is the
     a. A                                                     value of f(g(f(3)))?
     b. B                                                     a. 1
     c. C                                                     b. 3
     d. D                                                     c. 5
     e. E                                                     d. 7
                                                              e. 11
694. Simplify f(2y – 1) when f(x) = x2 + 3x –2.
     a.   4y2 + 2y – 4                                   700. If f(x) = 6x + 4 and g(x) = x2 – 1, which of the
     b.   4y2 + 6y – 2                                        following is equivalent to g(f(x))?
     c.   4y2 + 6y – 3                                        a. 6x2 – 2
     d.   2y2 + 6y – 4                                        b. 36x2 + 16
                                                              c. 36x2 + 48x + 15
                                                              d. 36x2 + 48x + 16
                                                              e. 6x3 + 4x2 – 6x – 4

   106
                                                          –ELEMENTARY FUNCTIONS–


For questions 701–702, refer to the functions f and g, both defined on [–5,5], whose graphs are shown here.

                         y                                                                  y

                    5                     y =f(x)                                      5                    y =g(x)
                                                                      (–5,4)
                    4                                                                  4
 (–5,3)
                    3                                                                  3
                    2                (2,2)                                             2
                     1                                                                  1

    –5 –4 –3 –2 –1           1   2    3      4   5
                                                          x                                                            x
                                                                         –5 –4 –3 –2 –1         1   2   3   4   5
                  –1                                                                   –1
                    –2       (2,–1)                                                    –2
                    –3                           (5,–2)                                –3
                    –4                                                                 –4                           (5,–4)
                    –5                                                                 –5




701. (f ˚ g)(0) =                                                     704. If f(x) =  –3x and g(x) =                 2x2 + 18, then the
      a. 1
         2
                                                                           domain of g ˚ f is
      b. –1                                                                a. [0,∞)
      c. 2                                                                 b. (–∞,0]
      d. undefined                                                         c.
                                                                           d. none of the above
      (
702. f (f ( f (f(5)))) =     )
      a.   0                                                                   Set 45       (Answers begin on page 236)
      b.   –1
                                                                      This problem set explores some basic features of com-
      c.   3
                                                                      mon elementary functions.
      d.   undefined

                                                                      705. Determine the domain of the function
703. If f(x) =      x2 – 4x, then f(x + 2) =
                                                                           f(x) = –x.
      a.     x2 – 4x + 2                                                   a. (–∞,0]
      b.     x2 – 4x –4                                                    b. [0,∞)
      c. x2 –4                                                             c. (–∞,0)
      d. |x – 2|                                                           d. (0,∞)




                                                                                                                                 107
                                        –ELEMENTARY FUNCTIONS–


706. Determine the domain of the function                711. Which of the following is true of
                    1
     g(x) =     3          .                                  f(x) = 4x – 1?
                    –1–x
                                                              a. The domain of the function is all real num-
     a.   the set of all real numbers
                                                                 bers greater than 1 and the range is all real
                                                                                   4
     b.   (–1,∞)
                                                                 numbers greater than 0.
     c.   (–∞,–1)
                                                              b. The domain of the function is all real num-
     d.   (–∞,–1)∪(–1,∞)
                                                                 bers greater than or equal to 1 and the range
                                                                                                4
                                                                 is all real numbers greater than 0.
707. Which of the following is true of the function
                                                              c. The domain of the function is all real num-
     f(x) = 2?
                                                                 bers greater than or equal to 1 and the range
                                                                                                4
     a. It is not a function.
                                                                 is all real numbers greater than or equal to 0.
     b. It has a range of 2.
                                                              d. The domain of the function is all real num-
     c. It has no domain.
                                                                 bers greater than 0 and the range is all real
     d. It has a slope of 2.
                                                                 numbers greater than or equal to 1 .4
     e. It has no y-intercept.
                                                              e. The domain of the function is all real num-
                                                                 bers greater than or equal to 0 and the range
708. Which of the following is true about the function
                                                                 is all real numbers greater than or equal to 14
     f(x) = |x|?
                                                                 .
     a. It has one y-intercept and two x-intercepts.
     b. It has one y-intercept and one x-intercept.
                                                         712. Consider the graphs of f(x) = x2 and g(x) = x4.
     c. There exists precisely one x-value for which
                                                              Which of the following statements is true?
        f(x) = 1.
                                                              a. f(x) g(x), for all real numbers x.
     d. f(x) 0, for all real numbers x.
                                                              b. The graphs of y = f(x) and y = g(x) do not
                                                                 intersect.
709. Which of the following is true about the func-
                                                              c. The range of both f and g is [0,∞).
     tion f(x) = x3?
                                                              d. The graphs of both f and g are increasing on
     a. f(x) 0, for all real numbers x.
                                                                 their entire domains.
     b. The graph of y = f(x) crosses the line y = a
        precisely once, for any real number a.
                                                         713. Which of the following is the domain of the
     c. The graph of y = f(x) is decreasing on its                                  1
        domain.                                                function f(x) =          2   ?
                                                                                 (2 – x)5
     d. The range is [0,∞).                                   a.   (–∞,2)
                                                              b.   (2,∞)
710. Which of the following is true about the func-           c.   (–∞,2)∪(2,∞)
     tion f(x) = 1 ?
                 x                                            d.   none of the above
     a. f(x) 0, for all real numbers x.
     b. The graph has one x-intercept.                   714. How many x-intercepts does the function
     c. The graph of y = f(x) is decreasing on the            f(x) = 1 – |2x–1| have?
        interval (0,∞).                                       a. 0
     d. The range is (0,∞).                                   b. 1
                                                              c. 2
                                                              d. more than 2


   108
                                           –ELEMENTARY FUNCTIONS–


715.How many points of intersection are there of       720. What can you conclude about the graph
     the graphs of f(x) =    x2               4
                                  and g(x) = x ?            of y = f(x) if you know that the equation
     a. 0                                                   f(x) = 3 does not have a solution?
     b. 1                                                   a. 3 is not in the domain of f.
     c. 2                                                   b. 3 is not in the range of f.
     d. more than 2                                         c. The graph of the function cannot have
                                                               y-values larger than 3.
716. How many points of intersection are there of           d. The graph of the function cannot be defined
     the graphs of f(x) = 2x and g(x) = 4x3?                   for x-values larger than 3.
     a. 0
     b. 1                                                   Set 46           (Answers begin on page 238)
     c. 2
                                                       This problem set focuses on properties of more
     d. more than 2
                                                       sophisticated functions, including monotonicity,
                                                       asymptotes, and the existence of inverse functions.
717. How many points of intersection are there of
     the graphs of f(x) = 3 x 2 and g(x) =
                          4
                                               5 2
                                              16 x ?                                      2x
                                                       721. The domain of f(x) = x3 – 4x is
     a. 0
                                                            a.   (–∞,–2)∪(2,∞)
     b. 1
                                                            b.   (–∞,2)∪(2,∞)
     c. 2
                                                            c.   (–∞,–2)∪(–2,0)∪(0,2)∪(2,∞)
     d. more than 2
                                                            d.   (–∞,–2)∪(–2,2)∪(2,∞)

718. What is the y-intercept of the function
             –2 –|2 – 3x|                              722. Which of the following are the vertical and hor-
     f(x) = 4 – 2x2 |–x| ?
                                                            izontal asymptotes for the function
     a.   (0,0)                                                      (x – 3)(x2 – 16)
                                                            f(x) =   (x2 + 9)(x – 4)
                                                                                      ?
     b.   (0, –1)
                                                            a.   x = –3, x = 4
     c.   (–1, 0)
                                                            b.   x = –3, x = 4, y = 1
     d.   There is no y-intercept.
                                                            c.   x = 4, y = 1
                                                            d.   y=1
719. What can you conclude about the graph of
     y = f(x) if you know that the equation f(x) = 0
     does not have a solution?
     a. The graph has no x-intercept.
     b. The graph has no y-intercept.
     c. The graph has neither an x-intercept nor a
        y-intercept.
     d. There is not enough information to con-
        clude anything about the graph of f.




                                                                                                      109
                                                   –ELEMENTARY FUNCTIONS–


723. On what intervals is the graph of the following fourth degree polynomial function y = p(x) increasing?




                                                                         y



                                                                    4
                                                                    3
                                                                                               y =p(x)
                                                                    2
                                                                     1
                                                                                                     x
                                                       –4 –3 –2 –1           1   2    3    4
                                                                  –1
                                                                    –2           (2,–1)

                                                                    –3

                                                                    –4




     a.   (–3,0)∪(2,∞)
     b.   (–3,0)∪(3,∞)
     c.   (–∞,–3)∪(0,1)
     d.   (–∞,–3)∪(0,2)

724. Consider the functions f and g, both defined on [–5, 5], whose graphs are shown here.

                               y                                                                                 y

                           5                    y =f(x)                                                      5                   y =g(x)
                                                                                          (–5,4)
                           4                                                                                 4
          (–5,3)
                           3                                                                                 3
                           2               (2,2)                                                             2
                            1                                                                                 1

             –5 –4 –3 –2 –1        1        3      4
                                                                x                              –5 –4 –3 –2 –1        1   2   3   4   5
                                                                                                                                            x
                                       2                5
                           –1                                                                                –1
                           –2      (2,–1)                                                                    –2
                          –3                           (5,–2)                                               –3

                          –4                                                                                –4                           (5,–4)
                          –5                                                                                –5




     Which of the following statements is true?
     a. f has an inverse on the interval (0, 5).
     b. f has an inverse on the interval (–5, 2).
     c. g does not have an inverse on the interval
        (–5, –1).
     d. All of the above statements are false.



   110
                                                   –ELEMENTARY FUNCTIONS–


725. Which of the following is the inverse function             729. Which of the following are characteristics of
                   x–1                                                                      (2 – x)2(x + 3)
     for f(x) =   5x + 2 , x ≠ –2?
                                 5                                   the graph of f(x) =       x(x – 2)2 ?
                    2y–1          1
     a. f –1(y) =   5y +1 , y ≠ – 5                                         I. The graph has a hole at x = 2.
                    2y + 1             1                                   II. y = 1 is a horizontal asymptote and x = 0
     b. f –1(y) =   5y – 1 , y     ≠   5
                                                                               is a vertical asymptote.
     c. f –1(y) = –52yy––11 , y ≠      1
                                       5                                  III. There is one x-intercept and one y-intercept.
                    2y + 1                 1
     d. f –1(y) =   5y +1 , y      ≠ –5                              a.   I and III only
                                                                     b.   I and II only
726. Assume that the function f has an inverse f -1,                 c.   I only
     and the point (1, 4) is on the graph of y = f(x).               d.   none of these choices
     Which of the following statements is true?
     a. If the range of f-1 is [1,∞), then f(0) is not          730. Which of the following functions is decreasing
        defined.                                                     on (–∞,0)?
     b. f –1(4) = 1                                                  a. f(x) = x3
     c. The point (4, 1) must lie on the graph of                    b. f(x) = 2x + 5
        y = f –1(x).                                                 c. f(x) = 1
                                                                               x
     d. All of the above statements are true.                        d. f(x) = 3

727. Which of the following is the inverse of                   731. Which of the following functions is increasing
     f(x) = x3 + 2?                                                  on (0,∞)?
     a. f –1(y) =
                     3
                          y–2                                        a. f(x) = x3
           –1         3                                              b. f(x) = 2x + 5
     b. f (y) =           y–2
                     3                                               c. f(x) = |x|
     c. f –1(y) =         2–y
                     3         3
                                                                     d. all of the above
           –1
     d. f (y) =           2–       y
                                                                732. Which of the following statements is false?
728. Which of the following are characteristics of                   a. The domain of any polynomial function is
                                           x2 +1
     the graph of f(x) = 2 –               x–1?                         the set of all real numbers.
            I. The function is equivalent to the linear func-        b. There exists a rational function whose
               tion g(x) = 2 – (x + 1) with a hole at x = 1.            domain is the set of all real numbers.
           II. There is one vertical asymptote, no hori-             c. A rational function must have both a vertical
               zontal asymptote, and an oblique asymptote.              and a horizontal asymptote.
          III. There is one x-intercept and one y-intercept.         d. All of the statements are true.

     a.   I only
     b.   II only
     c.   II and III only
     d.   I and III only




                                                                                                                    111
                                     –ELEMENTARY FUNCTIONS–


733. Which of the following statements is false?             Set 47        (Answers begin on page xx)
     a. There exists a polynomial whose graph is
        increasing everywhere.                         This problem set focuses on translations and refelctions
     b. A polynomial must have at least one turning    of known graphs.
        point.
     c. There exists a polynomial whose graph          737. Which of the following sequence of shifts
        remains below the x-axis on its entire              would you perform in order to obtain the
        domain.                                             graph of f(x) = (x +2)3 – 3 from the graph
     d. All of the statements are true.                     of g(x) = x3?
                                                            a. Shift the graph of g up 3 units and then left
734. Which of the following statements is true?                two units.
     a. Linear functions with positive slopes are           b. Shift the graph of g down 3 units and then
        increasing.                                            right two units.
     b. There exists a rational function whose graph        c. Shift the graph of g up 3 units and then right
        intersects both Quadrants I and II.                    two units.
     c. All quadratic functions are decreasing on           d. Shift the graph of g down 3 units and then
        one side of the vertex and increasing on the           left two units.
        other side of the vertex.
     d. All of the statements are true.                738. Which of the following parabolas has its turning
                                                            point in the second quadrant of the coordinate
735. Determine the x-values of the points of inter-         plane?
     section of the graphs of f(x) =–4x and g(x) =          a. y = (x + 1)2 – 2
     2 x.                                                   b. y = (x–1)2 – 2
     a. 0, 4                                                c. y = –(x + 1)2 – 2
     b. 0, 1                                                d. y = –(x + 1)2 + 1
           4
     c. 2                                                   e. y = (x–2)2 + 1
     d. 1  2
                                                       739. Compared to the graph of y = x2, the graph of
736. Determine the x-values of the points of inter-         y = (x–2)2 – 2 is
     section of the graphs of f(x) = x and g(x) =           a. shifted 2 units right and 2 units down
     3 x.                                                   b. shifted 2 units left and 2 units down
     a. 0                                                   c. shifted 2 units right and 2 units up
     b. 0,9                                                 d. shifted 2 units left and 2 units up
     c. 0,3                                                 e. shifted 4 units left and 2 units down
     d. The graphs do not intersect.




   112
                                      –ELEMENTARY FUNCTIONS–


740. Which of the following sequence of shifts would      743. Which of the following sequence of shifts would
     you perform in order to obtain the graph of               you perform in order to obtain the graph of
     f(x) = (x – 4)3 + 1 from the graph of g(x) = x3?          f(x) = x – 5 – 3 from the graph of g(x) = x ?
     a. Shift the graph of g up 1 unit and then left 4         a. Shift the graph of g up 3 units and then left 5
        units.                                                    units.
     b. Shift the graph of g down 1 unit and then              b. Shift the graph of g down 3 units and then
        right 4 units.                                            left 5 units.
     c. Shift the graph of g up 1 unit and then right          c. Shift the graph of g down 5 units and then
        4 units.                                                  left 3 units.
     d. Shift the graph of g down 4 units and then             d. Shift the graph of g down 3 units and then
        right 1 unit.                                             right 5 units.

741. Which of the following sequence of shifts            744. Which of the following sequence of shifts
     would you perform in order to obtain the                  would you perform in order to obtain the
     graph of f(x) = (x – 2)2– 4 from the graph                graph of f(x) = 2 x+3 from the graph of g(x)
     of g(x) = x 2?                                            = 2 x?
     a. Shift the graph of g up 4 units and then left 2        a. Shift the graph of g up 3 units.
        units.                                                 b. Shift the graph of g down 3 units.
     b. Shift the graph of g down 4 units and then             c. Shift the graph of g right 3 units.
        left 2 units.                                          d. Shift the graph of g left 3 units.
     c. Shift the graph of g up 4 units and then right
        2 units.                                          745. Which of the following sequence of shifts
     d. Shift the graph of g down 4 units and then             would you perform in order to obtain the
        right 2 units.                                         graph of f(x) = |x + 6| + 4 from the graph of
                                                               g(x) = |x|?
742. Which of the following sequence of shifts                 a. Shift the graph of g up 4 units and then left
     would you perform in order to obtain the                     6 units.
     graph of f(x) = (x – 2)3 –1 from the graph                b. Shift the graph of g down 4 units and then
     of g(x) = x3?                                                right 6 units.
     a. Shift the graph of g up 1 unit and then left 2         c. Shift the graph of g up 6 units and then right
         units.                                                   4 units.
     b. Shift the graph of g down 1 unit and then              d. Shift the graph of g down 6 units and then
        right 2 units.                                            left 4 units.
     c. Shift the graph of g up 1 unit and then right
        2 units.
     d. Shift the graph of g down 2 units and then
        left 1 unit.




                                                                                                         113
                                       –ELEMENTARY FUNCTIONS–


746. Which of the following sequence of shifts would       749.Which of the following functions’ graphs can be
     you perform in order to obtain the graph of                obtained by shifting the graph of g(x) =    x
     f(x) = –|x – 1| + 5 from the graph of g(x) = |x|?          right 5 units and then up 2 units?
     a. Shift the graph of g left 1 unit, then reflect          a.   f(x)= x – 5 + 2
        over the x-axis, and then up 5 units.                   b.   f(x) = x + 5 + 2
     b. Shift the graph of g right 1 unit, then reflect         c.   f(x) = x – 2 + 5
        over the x-axis, and then up 5 units.                   d.   f(x) = x + 2 – 5
     c. Shift the graph of g left 1 unit, then reflect
        over the x-axis, and then down 5 units.
                                                           750. Which of the following functions’ graphs can
     d. Shift the graph of g right 1 unit, then reflect
                                                                be obtained by shifting the graph of g(x) = |x |
     over the x-axis, and then down 5 units.
                                                                left 3 units, then reflecting it over the x-axis,
                                                                and then shifting it down 2 units?
747. Which of the following sequence of shifts                  a. f(x) = –|x–3| + 2
     would you perform in order to obtain the                   b. f(x) = –|x + 2| –3
     graph of f(x) = –(x + 3)3 + 5 from the graph of            c. f(x) = –|x + 3| –2
     g(x) = x3?                                                 d. f(x) = –|x–2| + 3
     a. Shift the graph of g left 3 units then reflect
        over the x-axis, and then up 5 units.
                                                           751. Which of the following functions’ graphs can
     b. Shift the graph of g right 3 units, then reflect
                                                                be obtained by reflecting the graph of g(x) = 1 x
        over the x-axis, and then up 5 units.
                                                                over the x-axis, and then shifting it up 2 units?
     c. Shift the graph of g left 3 units, then reflect                           1
                                                                a. f(x) = 2 +     x
        over the x-axis, and then down 5 units.
                                                                                 1
     d. Shift the graph of g right 3 units, then reflect        b. f(x)   =2– x
                                                                                1
        over the x-axis, and then down 5 units.                 c. f(x)   = –x+2
                                                                               1
                                                                d. f(x)   =– x – 2
748. Which of the following functions’ graphs can
     be obtained by shifting the graph of g(x) = x4        752. Which of the following functions’ graphs can
     right 5 units and then reflecting it over the              be obtained by shifting the graph of g(x) =     1
                                                                                                                x2
     x-axis?                                                    right 2 units and then reflecting it over the
     a. f(x) = –x4 + 5                                           x-axis?
     b. f(x) = –x4–5                                                           1
                                                                a. f(x) = – (x +2)2
     c. f(x) = –(x + 5)4
                                                                            1
     d. f(x) = –(x–5)4                                          b. f(x) = – x2 +2
                                                                            1
                                                                c. f(x) = – x2 – 2
                                                                               1
                                                                d. f(x) = – (x – 2)2




   114
                                         –ELEMENTARY FUNCTIONS–



      Set 48            (Answers begin on page 243)      757. Simplify: (ex + e–x)2
                                                               a.   e2x + e–2x
This problem set focuses on the basic computations
                                                               b.   e2x + e–2x + 2
and graphs involving exponentials, as well as applica-                 2     2
                                                               c.   e x + e–x
tion of the exponent rules.                                            2     2
                                                               d.   e x + e–x + 2

753. If ex = 2 and ey = 3, then e3x – 2y =                                  (53x – 1)3 5x – 1
          9                                              758. Simplify:            52x
     a.   8
          8                                                    a.   58x – 1
     b.   9
                                                               b.   1253x – 2
     c. 1
                                                               c.   254x – 1
     d. –1
                                                               d.   6252x – 1

754. Simplify: 2x 2 x+1
              2                                          759. Simplify: e x(e x – 1) – e–x (ex – 1)
     a. 2x +x
                                                               a.   e2x – e x – 1 + e–x
     b. 22x + 1
          2x2                                                  b.   e2x + e x – 1 + e–x
     c.   2– x                                                 c.   e2x + 2e x – 1
          2–x2
     d.    2x                                                  d.   e2x + 1 – 2e–x

755. Simplify: (4x – 1)2 16                                                 e x(e x – e –x) + e–x (ex + e –x)
                                                         760. Simplify:
                                                                                            e–2x
     a.   42x + 2
                                                               a.   e 2x + 1
     b.   4x
                                                               b.   e 4x + 1
     c.   42x
                                                               c.   e –4x + 1
     d.   4–2x
                                                               d.   e –2x + 1
                              1
                     54x      2
756. Simplify:
                    52x – 6                              761. Which of the following statements is true?
              x–3
     a.   5                                                    a. If b 1, the graph of y = bx gets very close to
     b.   53x                                                     the x-axis as the x-values move to the right.
     c.   5x + 3                                               b. If b 1, the graph of y = bx gets very close to
     d.   5–3x                                                    the x-axis as the x-values move to the left.
                                                               c. If b 1, the graph of y = bx grows without
                                                                  bound as the x-values move to the left.
                                                               d. If b 1, the y-values associated with the
                                                                  graph of y = bx grow very rapidly as the x-
                                                                  values move to the right.




                                                                                                                115
                                             –ELEMENTARY FUNCTIONS–



                                                                                                           2x
                                                                                                       2
762. Which of the following statements is true?           766. What is the solution set for – 3                 ≤0?
     a. If 0 b 1, the graph of y = get very  bx                a.   (–∞,0)
        close to the x-axis as the x-values move to            b.   (–∞,0]
        the left, and the y-values grow very rapidly as        c.   the empty set
        the x-values move to the left.                         d.   the set of all real numbers
     b. If 0 b 1, the graph of y = bx get very
        close to the x-axis as the x-values move to       767. Which of the following is a true characterization
                                                                                               x
        the right, and the y-values grow very rapidly          of the graph of f(x) = –    3
                                                                                               ?
                                                                                           4
        as the x-values move to the right.
     c. If 0 b 1f, the graph of y = bx get very                a. The graph has one x-intercept and one
        close to the x-axis as the x-values move to               y-intercept.
        the right, and the y-values grow very rapidly          b. There exists an x-value for which f(x) = 1.
        as the x-values move to the right.                     c. The graph is increasing as the x-values move
     d. If 0 b 1, the graph of y = bx get very                    from left to right.
        close to the x-axis as the x-values move to            d. The graph is decreasing as the x-values move
        the right, and the y-values grow very rapidly             from left to right.
        as the x-values move to the left.
                                                          768. Which of the following is a true characterization
                                                                                               3x
                                                                                          15
763. Which of the following statements is true?                of the graph of f(x) = –    7       ?
     a. If 0 b 1, then the equation =–1 has a bx
                                                               a. The graph has one x-intercept and one
        solution.
                                                                  y-intercept.
     b. If b 1, then the equation bx = 1 has two
                                                               b. There exists an x-value for which f(x) = 1.
        solutions.
                                                               c. The graph is increasing as the x-values move
     c. If b 0, then the equation bx = 0 has no
                                                                  from left to right.
        solution.
                                                               d. The graph is decreasing as the x-values move
     d. If b 0, then only negative x-values can be
                                                                  from left to right.
        solutions to the equation bx = 0.

                                                               Set 49         (Answers begin on page 245)
764. Which of the following statements is true?
     a. 2x        3x, for all x    0.                     This problem set focuses on more advanced features of
              x           x
     b.   1           1
                          , for all x   0.                exponential functions, including solving equations
          2           x
                                                          involving exponential expressions.
          1 –x
     c.   2          0, for any real number x.
                                                          769. The range of the function f(x) = 1– 2ex is which
     d. All of the above statements are true.
                                                               of the following?
                                                               a. (–∞,1]
765. What is the solution set for the inequality
                                                               b. (–∞,1)
     1–3x ≤ 0?
                                                               c. (1,∞)
     a. [1,∞)
                                                               d. [1,∞)
     b. [0,∞)
     c. the empty set
     d. the set of all real numbers

   116
                                          –ELEMENTARY FUNCTIONS–




770. Determine the values of x that satisfy the equa-      776. Solve: 125x = 25
                2                                                     3
      tion 27x      –1
                         = 43x.                                  a.   2
      a. x =    –3
                     14
                           37
                                                                 b. – 3
                                                                      2

      b. x =    3         37                                     c. – 2
                                                                      3
                   14
                                                                      2
      c. x      1
             =– 7 and x = 1                                      d.   3

      d. x   = 1 and x =–1
               7                                           777. Solve: (e x) x – 3 = e10
                                                                 a.   –5 and –2
771. Determine the values of x, if any, that satisfy the
                                    1                            b.   –2 and 5
      equation 5 x + 1 =           25 .                          c.   –5 and 2
      a. 3
                                                                 d.   2 and 5
      b. –3
      c. log5 3
                                                           778. Solve: 163x – 1 = 42x + 3
      d. no solution
                                                                 a. – 5
                                                                      4

772. Which of the following are characteristics of               b. – 4
                                                                      5
                                                                      4
      the graph of f(x) = –e2 – x –3?                            c.   5
                                                                      5
      a. The graph of f lies below the x-axis.                   d.   4
      b. y = –3 is the horizontal asymptote for the
                                                                                        2x
         graph of f.                                                               1
                                                           779. Solve: 4x + 1 = 2
      c. The domain is .                                         a. –2–2
      d. all of the above
                                                                 b. 1
                                                                    4
                                                                 c. – 1
                                                                      2
773. Solve: 2x – 5 = 8
      a.   –3                                                    d. –2
      b.   3
      c.   2                                               780. Solve: x 3x + 5 3x = 0
      d.   8                                                     a.   –5
                                                                 b.   5
774. Solve: 32x = 9 3x – 1
                                                                 c.   0
      a.   1                                                     d.   0 and –5
      b.   0                                                                       2x
      c.   –1                                              781. Solve: 10x + 1          = 100
      d.   none of the above                                     a.   –1       5
                                                                           2
                                                                      1
                               1                                 b.   3
775. Solve: 42x – 3 = 4x
                                                                 c. – 1
                                                                      3
      a.   –1
                                                                           1
      b.   0                                                     d.        3
      c.   1
      d.   none of the above


                                                                                                117
                                         –ELEMENTARY FUNCTIONS–



782. Solve: 2      x   2=8                            788. log7       7=
     a.   16                                               a. 0
     b.   8                                                b. 1
     c.   4                                                c. 1
                                                              2
     d.   2                                                d. –1

783. Solve: 2x2 . ex – 7x ex + 6 ex = 0               789. log5 1 =
          2
     a.   3   and 2                                        a. 0
     b.   3
          2   and –2                                       b. 1
                                                              5
     c.   3
              and 2                                        c. 1
          2
     d.   –2   and 2                                       d. – 1
                                                                5
           3

                                                      790. log16 64 =
784. Solve: e2x +5ex– 6 = 0
                                                                2
     a.   2 and 3                                          a.   3
     b.   0                                                b. – 2
                                                                3
     c.   –3 and –2                                        c. 4
     d.   –3, –2, and 0                                         3
                                                           d.   2


                                                      791. If log6 x = 2, then x =
      Set 50            (Answers begin on page 247)
                                                           a.   6
This problem set focuses on basic computations             b.   12
involving logarithms.                                      c.   36
                                                           d.   –36
785. log3 27 =
     a.   –2                                          792. If 5      a = x , then loga x =
     b.   2                                                a. loga 5 –    1
                                                                          2 loga a
     c.   –3
                                                           b. loga 5    + 1 loga a
                                                                          2
     d.   3
                                                           c. loga 5 + 2 loga a
786. log3 9 =
               1                                           d. loga 5 – 2 loga a
     a.   2
     b.   –2                                          793. log3(34 93) =
     c.   3                                                a.   8
     d.   –3                                               b.   10
                                                           c.   6
787. log 1 8 =                                             d.   12
          2

     a.   16
     b.   3
     c.   –3
     d.   4



   118
                                     –ELEMENTARY FUNCTIONS–



794. If 53x – 1 = 7, then x =                                                                 e2y
            1                                        800. If ln x = 3 and ln y = 2, then ln         =
     a.           7)
            3 (1–log5                                                                           x
     b. –3(1 + log5 7)                                    a. 2
                                                             5

     c. – 1 (1–log5 7)                                    b. – 2
                                                               5
          3
     d. 1 (1 + log5 7)                                    c. 5
                                                             2
        3
                                                          d. – 5
                                                               2
                                             x
795. If loga x = 2 and loga y = –3, then loga y3 =
     a.     –23
     b.     11
                                                          Set 51          (Answers begin on page 249)
     c.     –32                                      This problem set focuses on basic features of logarith-
     d.     8                                        mic functions, and simplifying logarithmic expressions
                                                     using the logarithm rules.
                2
796. 3log 3 =
     a.     2                                        801. Which of the following is equivalent to
     b.     1                                             3 ln (xy2) – 4 ln(x2y) + ln(xy)?
     c.     0                                                       y3
                                                          a. ln [ x4 ]
     d.     –1
                                                          b. ln [y3x4]
                                                                     4
797. loga(ax) =                                           c. ln [ x3 ]
                                                                  y
     a.     ax                                            d. ln [y3] + ln [x4]
     b.     0
     c.     x                                        802. Simplify: log8 2 + log8 4
     d.     xa                                            a.   1
                                                          b.   –1
                            1
798. If 3 ln x = ln 8, then x =                           c.   2 log8 2
            1                                             d.   3
     a.     2
     b. 2
                                                     803. Simplify: 4 log9 3
     c. – 1
          2
                                                          a.   8
     d. –2
                                                          b.   –8
         – 1 ln 3                                         c.   2
799. e     2
                    =
                        3                                 d.   –2
     a. –           3
                    3
     b.         3                                    804. Which of the following is equivalent to
                3                                         ln 18x3 – ln 6x?
     c.             3                                     a. ln 3x2
                    3                                     b. 2 ln 3x
     d. –               3
                                                          c. ln (3x)2
                                                          d. ln (108x4)




                                                                                                        119
                                              –ELEMENTARY FUNCTIONS–


                             2         2
805. Simplify: log7 49 – log7 7                           811. Which of the following is equivalent to
     a. 2                                                      ln[(2          x + 1)(x2 + 3)4]?
     b. –1                                                          1
                                                               a.   2   ln 2(x + 1) – 4 ln (x2 + 3)
     c. 429                                                                 1                     2
                                                               b. ln 2 –    2 ln (x + 1) – 4 ln (x + 3)
     d. – 429
                                                               c. ln 2    + 1 ln (x + 1) + 4 ln (x2 + 3)
                                                                            2
                                                                    1
                                 2                             d.   2   ln 2(x + 1) + 4 ln (x2 + 3)
806. Simplify: 3 log4 3 + log4 27
     a. – 2
          3
          2
                                                          812. Which of the following is equivalent to
     b.   3                                                              x2    2x – 1
                                                               log3                     ?
          –3
                                                                                   3
     c.     2                                                             (2x + 1) 2
          3                                                                          1
     d.                                                        a. 2 log3 (x) +       2 log3 (2x – 1)(2x + 1)
          2
                                                                                     3
                                                               b. 2 log3 (x)       – 2 log3 (2x – 1)(2x + 1)
807. Which of the following is equivalent to                   c. 2 log3 (x)       + 1 log3 (2x – 1) – 3 log3
                                                                                     2                 2        (2x + 1)
     log (2x3)?                                                d. 2 log3 (x)                2x –
                                                                                   – 3 log3 2x + 1
                                                                                     2           1
     a. log 2 – 3 log x
     b. –log 2 + 3 log x
                                                          813. Which of the following is a true characteriza-
     c. log 2 + 3 log x
                                                               tion of the graph of f(x) = ln x?
     d. –log 2 – 3 log x
                                                               a. As the x-values decrease toward zero, the
                                                                  y-values plunge downward very sharply.
808. Which of the following is equivalent to                   b. As the x-values decrease toward zero, the
                8yz4
     log3        x2
                       ?                                          y-values shoot upward very sharply.
     a.   3 + log2 y – 4 log2 z – 2 log2 x                     c. As the x-values move to the right, the
     b.   3 – log2 y – 4 log2 z + 2 log2 x                        y-values decrease very slowly.
     c.   3 + log2 y + 4 log2 z + 2 log2 x                     d. As the x-values move to the left, the y-values
     d.   3 + log2 y + 4 log2 z – 2 log2 x                        increase very slowly.

                                                          814. What is the domain of k(x) = log3(–x)?
809. Which of the following is equivalent to
     3
       log2 4 –        2
                           log2 8 + log2 2?                    a.   (–∞,0)
     2                 3
     a. log2 2                                                 b.    (0,∞)
     b. –1                                                     c.   [0,∞)
     c. –2                                                     d.   the set of all real numbers
     d. 2
                                                          815. What is the domain of b(x) = log5(x2 + 1)?

810. Which of the following is equivalent to
                                                               a.   (–∞,0)
     3 logb (x + 3)–1– 2 logb x + logb (x + 3)3 ?              b.   (0,∞)
     a. 2 logbx                                                c.   [0,∞)
     b. –logbx                                                 d.   the set of all real numbers
     c. logbx(x + 3)
                  1
     d. logb      x2




   120
                                       –ELEMENTARY FUNCTIONS–


816. What is the x-intercept of f(x) = log2 x?            821. Which of the following choices for f and g are
     a.   (2, 0)                                               inverses?
     b.   (0, 1)                                               a. f(x) = e–x and g(x) = ln     x,x>0
     c.   (1, 0)                                               b. f(x) = e2x and g(x) = ln     x,x>0
     d.   This function does not have an x-intercept.          c. f(x) = e–x and g(x) = ln     –x , x > 0
                                                               d. f(x) = e2x and g(x) = ln     2x , x > 0

     Set 52           (Answers begin on page 250)         822. Solve: ln(x – 2) – ln(3 – x) = 1
                                                                    3e + 2
This problem set focuses on more advanced features of          a.    e+1
                                                                    3(e + 2)
logarithmic functions and solving equations and                b.     e+1
inequalities involving logarithms.                             c. 2
                                                               d. 3
817. The range of the function f(x) = ln(2x – 1) is
     which of the following?                              823. Determine the solution set for the inequality
     a.
                  1
          (–∞,– 2 )                                            5      4e2 – 3x + 1    9.
            1                                                         2 – ln 2 e–2
     b.   [ 2 ,∞)                                              a.        3 , 3
            1
     c.   ( 2 ,∞)                                                     2 – ln 2 2
                                                               b.        3 ,3
     d.
                                                                      2 –2 + ln 2
                                                               c.     3,   3
818. Which of the following, if any, are x-intercepts
     of the functionf(x) ln (x2–4x + 4)?                       d.
                                                                      e–2 –2 + ln 2
                                                                       3 ,    3
     a. (1, 0)
     b. (3, 0)                                            824. Determine the solution set for the inequality
     c. both a and b                                           ln(1–x2) 0.
     d. neither a nor b                                        a. (–1,0) (0,1)
                                                               b. (–∞,–1) ((1,∞)
819. The domain of the function f(x) = ln (x2 – 4x + 4)        c. (–1,1)
     is which of the following?                                d. [–1,1]
     a. (2,∞)
     b. (–∞,2)                                            825. Solve: log x + log(x + 3) = 1
     c. (–∞, 2) (2,∞)                                          a.   –2 and 5
     d.                                                        b.   2 and 5
                                                               c.   –5 and 2
820. Which of the following is a characteristic of the         d.   2
     graph of f(x) = ln(x + 1) + 1?
     a. The y-intercept is (e, 1).
     b. x = –1 is a vertical asymptote of f.
     c. There is no x-intercept.
     d. y = 1 is a horizontal asymptote.




                                                                                                            121
                                               –ELEMENTARY FUNCTIONS–


826. Solve: log2 (2x – 1) + log2 (x + 2) = 2               830. Solve for: 3 ln 4y + ln A = ln B:
                                                                          1       1
     a.   0                                                     a. y = 4 e3 (ln B–ln A)
     b.   1                                                               1       1
                                                                b. y = 4 e3 (ln B + ln A)
     c.   0 and 1                                                             1       1    B

     d.   none of the above                                     c. y =– 4 e3 (ln A)
                                                                              1       1
                                                                d. y =– 4 e3 (ln AB)
827. Solve: log(x – 2) = 2 + log(x +3)
          302                                              831. Solve for x: l + ln (x y) = In z
     a. 99
     b. 3                                                       a.   x = e ln z + ln y–1
     c. 2                                                       b.   x = e ln z–ln y + 1
     d. There is no solution to this equation.                  c.   x = e ln z–ln y–1
                                                                d.   x = e –(ln z + ln y–1)
828. Assuming that b             1, solve: b3 logbx = 1
     a.   b                                                832. Solve for t: P = Poe–kt
     b.   b2                                                                  1
                                                                a. t = – k ln ( P )
                                                                                           P
                                                                                            0
     c.   0
                                                                                           P
     d.   1                                                     b. t = –k ln              (P )
                                                                                            0
                                                                          1                P
                                                                c. t = k ln (– P )
                             –a(b + c)                                                         0
829. Solve for x: y = e
                    1                                                     1                P0
     a. x = – a (ab + ln y)                                     d. t   = –k       ln (     P       )
     b. x = –b + ln y
                1
     c. x = a (ab–ln y)
                –ab + ln y
     d. x =         a




   122
                7
S E C T I O N




                                              MATRIX
                                              ALGEBRA




S          ystems of linear equations can also be solved using Cramer’s rule, which involves the use of matrices.
           Matrix operations, including matrix arithmetic, computing determinants and inverse, and applying back
           substitution and Cramer’s rule to solve systems of linear equations, are reviewed in the seven problem
sets in this section.




                                                                                                         123
                                               –MATRIX ALGEBRA–



     Set 53              (Answers begin on page 252)                                        –1 –1 0
                                                          836. Compute, if possible: –3 >           H
                                                                                             0 –3 1
Basic features of matrices and the arithmetic of matri-
                                                                      –3 –3 0
ces are explored in this problem set.                          a. >            H
                                                                       0 –9 –3
833. What are the dimensions of the matrix
                                                                      –3 3 0
     [1 2      – 1 0 ]?                                        b. >          H
                                                                       0 9 3
     a. 4      4
     b. 1      4                                                      3 3 0
                                                               c. >          H
     c. 4      1                                                      0 9 –3
     d. 1      1
                                                               d. This computation is not well-defined.
834. What are the dimensions of the matrix
                                                                                                      R V
                                                                                                      S1W
     R      V                                                                                         S0W
     S 0 –2 W                                             837. Compute, if possible: 9 1 –1 –2 2 C + S W
     S0 1 W                                                                                           S1W
     S 0 –2 W ?                                                                                       S –5 W
     S      W                                                  a. 9 2 –1 –1 –3 C                      T X
     S0 0 W
     T      X                                                     R V
     a. 2 4                                                       S2W
     b. 4 2                                                       S –1 W
                                                               b. S –1 W
     c. 2 2                                                       S W
                                                                  S –3 W
     d. 4 4                                                       T X
                                                                  R                 V
                                                                  S 1    –1 –2     1W
835. Which of the following matrices has dimen-                   S 0    0 0       0W
     sions 3        2?                                            S 1    0 0       0W
                                                               c. S
                                                                  S –5 0 0
                                                                                    W
        R       V                                                                  0W
        S2     1W                                                 T                 X
     a. S 3    5W                                              d. This computation is not well-defined.
        S2     0W
        T       X
                                                          838. Compute, if possible: 2 > –3 –1 H –3 > 2   –1
                                                                                                             H
          –1    –1 0                                                                     0    1      2    2
     b. >            H
           0    –3 1                                           a. > –4 –6 H
                                                                     1 –12
            3 3
     c. >       H
            3 3                                                b. > 12 –1 H
                                                                    6 4
     d. none of the above
                                                                      –12 1
                                                               c. >         H
                                                                      –6 –4

                                                               d. This computation is not well-defined.




   124
                                            –MATRIX ALGEBRA–


                            R               V        842. How many ordered pairs x, y make the follow-
                            S –1 0       1 W
                          2                               ing equality true:
839. Compute, if possible: S 1 0         1 W
                          5S                W
                            S 0 1        –1 W               R     V      R       V
                            T               X               S0 2 W       S 0 –6 W
        R              V                                  3 S 1 1 W = –1 S –3 –3 W?
        S 2   0 –2W                                         S     W      S       W
        S 5          5W                                     S1 x W       S –3 6y W
        S– 2 0 – 2 W                                        T     X      T       X
     a.
        S 5          5W                                   a. 0
        S 0 –2 2 W                                        b. 1
        S       5 5 W
        T              X                                  c. 2
         R        V                                       d. infinitely many
         S2 0 2W
         S5      5W
     b.  S2 0 2W                                     843. Determine an ordered pair (x, y) that makes the
         S5      5W
         S0 2 2W                                          following equality true:
         S   5 5W
         T        X                                                                                        2x 10 6
                                                                                                       H=>         H
                                                               3 x – 2 0 –2          4x + 2   –5     1
        R            V                                  –4 >                  H –2 >
                                                                  –2   2 y –1          2    4 – 3 y –1     4 y 6
        S– 2 0 2 W
        S 5       5 W                                                  8
     c. S 2 0 2 W                                         a.(–2,– 3 )
        S 5       5 W                                                  8
        S 0 2 –2W                                         b. (2,– 3 )
        S     5     5W                                             8
        T            X                                    c. (2, 3 )
                                                                       8
     d. This computation is not well-defined.             d. (–2, 3 )

840. Determine the values of x, if any exist, that   844. Determine an ordered triple (x, y, z) if one
     make the following equality true:                    exists, that makes the following equality true:

         x –2   1 6 –4
     >        H= >     H
                                                              x 2y     2x 3y     3x 4y
         0 2    2 0 4                                     >        H–>       H=>       H
                                                              3z 4     4z 4      –2z 0
     a.   3                                               a.    (1, 1, 1)
     b.   –3                                              b.    (0, 1, 1)
     c.   6                                               c.    (0, 0, 0)
     d.   There is no such x-value.                       d.    (0, 1, 0)

841. Determine the values of x, if any exist, that   845. Which of the following statements is true?
     make the following equality true:                    a. The sum of two 4 2 matrices must be a
                                                             4 2 matrix.
         –1 x 2     –1 4
     >          H=>       H                               b. The sum of a 4 2 matrix and a 2 4
         3x –1       6 –1
                                                             matrix is well-defined.
     a.   –1 and 1                                        c. A constant multiple of a 3 1 matrix need
     b.   –2 and 2                                           not be a 3 1 matrix.
     c.   –2                                              d. All of the above statements are true.
     d.   2



                                                                                                           125
                                           –MATRIX ALGEBRA–


846. Which of the following statements is true?

               –1 2
          0>         H= 0
     a.
                2 –1

             2    –1      2 –1
     b. >      H+> H = >       H
            –1     3     –1 3

     c. 3 9 –1 0 0 C –2 9 0 1 0 C + 9 0 0 –1 C = – 9 3 2 1 C

     d. All of the above statements are false.

847. Which of the following statements is true?

            –1 –1          0 0
     a. >         H+ 1 = >     H
            –1 –1          0 0
                                                                       R              V R          V
                                                                       SX   1   0   0 W S 15 1 0 0 W
                                                                       S0   X   1   0 W S 0 15 1 0 W
     b. There is an X-value that makes the following equation true: –3 S               =
                                                                       S0   0   X   1 W S 0 0 15 1 W
                                                                                      W S          W
                                                                       S0   0   0   X W S 0 0 0 15 W
                                                                       T              X T          X
          1 1                       0 0
     c. >     H–9 1 1 C–9 1 1 C = >     H
          1 1                       0 0

     d. All of the above statements are false.

848. How many ordered triples (x, y, z) (where x, y, and z are real numbers) make the following equation true:

     R                 V R     2     V
     S 1 x – 2 –1 –1 W S 1 –x –1 –1 W
     S –3 –1 2 y 1 W S –3 –1 y 2 1 W
     S –2               =
     S     1    1 4z 2 W S –2 1 1 8z W ?
                       W S           W
     S0   –3   –4  0 W S 0 –3 –4 0 W
     T                 X T           X
     a. 8
     b. 4
     c. 2
     d. 0




   126
                                            –MATRIX ALGEBRA–



                                    Set 54 (Answers begin on page 254)
The multiplication of matrices and matrix computations involving multiple operations are the focus of this
problem set.

For questions 849–864, use the following matrices:
  R–1 2 V
  S       W                                              0
A=S0 2W                                               F=> H
  S –1 –1 W                                              0
  T       X
                                                          R–2 –1 0 1 V
     1 –2 –1
B =>         H
                                                          S            W
                                                      G = S –1 –2 –1 0 W
     3 5 0
                                                          S 1 –1 –2 –1 W
                                                          T            X
       0 1
C =>        H                                           R        V
       1 –4                                             S 3 1 –1 W
                                                        S 1 –2 1 W
    R         V                                       H=S        W
    S 3 2 1W                                            S 0 0 –2 W
D = S 0 1 2W                                            S –2 1 0 W
    S         W                                         T        X
    S –1 –1 0 W
    T         X                                           R2 V
                                                          S W
                                                      I = S2 W
E = 9 –4 –2 0 C                                           S1 W
                                                          T X
849. Express as a single matrix, if possible: CF       850. Express as a single matrix, if possible: (2G)(–3E)
                                                               R        V
     a. 9 0 0 C                                                S –1 1 0 W
                                                            a. S 0 1 0 W
                                                               S        W
         0                                                     S 1 0 1W
     b. > H                                                    T        X
         0
                                                                   1 0
                                                            b. >       H
            0 0                                                    0 1
     c. >       H
            0 0
                                                                 9 –1 –1 1 C
                                                            c.
     d. not possible
                                                            d. not possible




                                                                                                      127
                                            –MATRIX ALGEBRA–


851. Express as a single matrix, if possible: AB          854. Express as a single matrix, if possible: FF

                                                                    0 0
                                                               a. > 0 0 H
     a. 5 12
        >      H
          6 10

                                                                   0
          12 1
     b. > –3 1 H                                               b. > H
                                                                   0
        R          V
        S 5 12 1 W                                             c. 9 0 0 C
        S 6 10 0 W
     c. S          W
        S –4 –3 1 W
        T          X                                           d. not possible
     d. not possible
                                                          855. Express as a single matrix, if possible: IE + D

852. Express as a single matrix, if possible: 4BA
                                                                    R         V
                                                                    S 5 2 –1 W
                                                               a.   S 8 3 –2 W
             64 –4
     a. >          H                                                S
                                                                    S5 3 0 W
                                                                              W
            –12 0                                                   T         X
                                                                    R           V
             0  4                                                   S –5 –2 1 W
     b. >          H                                           b.   S –8 –3 2 W
            12 –64                                                  S           W
                                                                    S –3 –3 0 W
                                                                    T           X
             0 –4
     c. >          H                                                R
                                                                    S 1 –2 5 W
                                                                                  V
            –12 64
                                                               c.   S –8 –3 2 W
                                                                    S             W
     d. not possible                                                S 1 –3 –5 W
                                                                    T             X
                                                               d.   not possible
853. Express as a single matrix, if possible: (–2D)(3D)
          R              V                                856. Express as a single matrix, if possible: (BG)H
          S –48 –42 –42 W
     a.   S 12    6 –12 W
          S              W                                     a. > –3 –7 –3 H
          S 18 18 18 W                                             –52 18 8
          T              X
          R            V                                          R         V
          S –6 –7 –7 W                                            S –3 –52 W
          S 2 1 –2 W
     b.   S            W                                       b. S –7 18 W
          S3 3 3W                                                 S         W
          T            X                                          S –3 8 W
                                                                  T         X
          R              V                                        R       V
          S 48 42 42 W                                            S 3 52 W
          S –12 –6 12 W
                                                               c. S 7 –18 W
     c.
          S              W                                        S       W
          S –18 –18 –18 W                                         S 3 –8 W
          T              X                                        T       X
     d.   not possible                                         d. not possible




   128
                                           –MATRIX ALGEBRA–


857. Express as a single matrix, if possible: (EG)(HI)   861. Express as a single matrix, if possible:
     a.   22                                                  (2C)(2C)(2C)F
     b.   33                                                      0
     c.   66                                                  a. > H
                                                                  0
     d.   not possible
                                                              b. 9 0 0 C
858. Express as a single matrix, if possible: (ED)(AC)

     a. 9 36 –164 C
                                                                     0 0
                                                              c. >       H
                                                                     0 0

     b. 9 –36 164 C                                           d. not possible

             36
     c. >        H                                       862. Express as a single matrix, if possible:
            –164
                                                              (EAF)(CF)
     d. not possible                                                 0 0
                                                              a. >       H
                                                                     0 0
859. Express as a single matrix, if possible: E(G + A)
                                                              b. 9 0 0 C
          1 0
     a. >     H
          1 0
                                                                  0
                                                              c. > H
                                                                  0
            –2
     b. >      H
             1                                                d. not possible
     c. –1
                                                         863. Express as a single matrix, if possible:
     d. not possible
                                                              3D – 2AB + GH
                                                                 R            V
860. Express as a single matrix, if possible: 4B – 3FE           S 10 17 –2 W
                                                              a. S 17 14 –7 W
            2 –1 2                                               S            W
     a. >          H                                             S –9 –5 0 W
            1 1 –2                                               T            X
                                                                 R              V
        R2 1 V                                                   S –10 –17 2 W
        S      W                                              b. S –17 –14 7 W
     b. S –1 1 W                                                 S              W
        S 2 –2 W                                                 S 9     5 0W
                                                                 T              X
        T      X                                                 R                V
            0 0                                                  S –10 –17 –2 W
     c. >       H                                             c. S –17 –14 –7 W
            0 0                                                  S                W
                                                                 S –9 –5 0 W
                                                                 T                X
     d. not possible                                          d. not possible




                                                                                                         129
                                             –MATRIX ALGEBRA–



                                                                                        –1 2
864. Express as a single matrix, if possible:         869. Compute the determinant: >         H
                                                                                         2 –4
     (2F)(–2E) + 2B                                        a.   –10
             2 1                                           b.   –6
     a. >        H                                         c.   6
            –1 0
                                                           d.   0
            –2 –5
     b. >         H                                                                     6 3
             4 2                                      870. Compute the determinant: >       H
                                                                                        2 1
                                                           a.   9
          3 2
     c. >     H                                            b.   0
          1 5                                              c.   16
                                                           d.   –16
     d. not possible
                                                                                        –3 4
                                                      871. Compute the determinant: >        H
                                                                                         4 2
      Set 55          (Answers begin on page 257)          a.   –20
                                                           b.   20
This problem set is focused on computing determi-
                                                           c.   –22
nants of square matrices.
                                                           d.   22
                                    –3 7
865. Compute the determinant: >          H                                              1 –4
                                     1 5              872. Compute the determinant: >        H
     a.   –38                                                                           0 25
                                                           a.   25
     b.   –26
                                                           b.   –25
     c.   22
                                                           c.   –4
     d.   –22
                                                           d.   4
                                    a 0
866. Compute the determinant: >         H                                               3 –1
                                    0 b               873. Compute the determinant: >        H
     a.   0                                                                             1 –2
                                                           a.   1
     b.   a
                                                           b.   –1
     c.   ab
                                                           c.   –5
     d.   b
                                                           d.   5
                                    1 2
867. Compute the determinant: >         H                                                –2 0
                                    2 3               874. Compute the determinant: >         H
     a.   –1                                                                            –12 3
                                                           a.   24
     b.   1
                                                           b.   36
     c.   –3
                                                           c.   –6
     d.   3
                                                           d.   –24
                                    2 3
868. Compute the determinant: >         H                                                0 1
                                    1 1               875. Compute the determinant: >         H
     a.   –5                                                                            –2 –1
                                                           a.   –2
     b.   5
                                                           b.   2
     c.   1
                                                           c.   1
     d.   –1
                                                           d.   –1


   130
                                            –MATRIX ALGEBRA–



                                  –1 0                    Set 56       (Answers begin on page 257)
876. Compute the determinant: >         H
                                   2 –1
     a.   2                                          This problem set is focused on writing systems in
     b.   1                                          matrix form.
     c.   –2
                                                     881. Write this system in matrix form: *
                                                                                                –3x + 7y = 2
     d.   0
                                                                                                  x + 5y = 8
                                                                 3 –7    x   2
                                                          a. >        H > H=> H
                                  3 2
877. Compute the determinant: >       H                          1 5     y   8
                                  3 2
     a.   0
     b.   5                                                      –3 7     x   2
                                                          b. >         H > H=> H
     c.   12                                                     –1 –5    y   8
     d.   –12
                                                                 –3 1    x   2
                                3 –2                      c. >        H > H=> H
878. Compute the determinant: >      H                            7 5    y   8
                                9 –6
     a. 0
                                                                 –3 7    x   2
     b. 15                                                d. >        H > H=> H
                                                                  1 5    y   8
     c. 48

                                                     882. Write this system in matrix form: * y = b
     d. –15                                                                                     x=a

                                  –1 –1
879. Compute the determinant: >         H                        1 0    x   a
                                  –1 0                    a. >       H > H=> H
     a.   1                                                      0 1    y   b
     b.   0
                                                                 1 1    x   a
     c.   –1                                              b. >       H > H=> H
                                                                 0 0    y   b
     d.   2
                                                                 0 0    x   a
                                                          c. >       H > H=> H
                                  0 2
880. Compute the determinant: >       H
                                                                 1 1    y   b
                                  4 0
     a.   2
     b.   –2                                                     1 0    x   b
                                                          d. >       H > H=> H
     c.   8                                                      0 1    y   a
     d.   –8
                                                                                                x + 2y = 4
                                                     883. Write this system in matrix form: *
                                                                                                2x + 3y = 2
                                                                 3 2    x   4
                                                          a. >       H > H=> H
                                                                 2 1    y   2

                                                                 1 2    x   4
                                                          b. >       H > H=> H
                                                                 2 3    y   2

                                                                 1 2    x   2
                                                          c. >       H > H=> H
                                                                 2 3    y   4

                                                                 3 2    x   2
                                                          d. >       H > H=> H
                                                                 2 1    y   4


                                                                                                      131
                                             –MATRIX ALGEBRA–



                                           2x + 3y = 1
884. Write this system in matrix form: *
                                                                                                    –3x = 1 – 4y
                                                         887. Write this system in matrix form: *
                                           x + y = –2                                               2y + 3 = –4x
                                                                       2 4     x    1
                                                              a. >          H > H=> H
            2 1    x    1
     a. >       H > H=> H
            3 1    y   –2                                              4 –3    y   –3

            2 3    x   –2                                              –3 4    x   –3
     b. >       H > H=> H                                     b. >          H > H=> H
            1 1    y    1                                               4 2    y    1
                                                                        3 4    x    1
          2 3    x    1                                            >        H > H=> H
     c. >     H > H=> H                                       c.       4 2     y   –3
          1 1    y   –2
                                                                   >          H > H=> H
            2 1    x   –2
     d. >       H > H=> H
                                                                       4 –3      y   1
                                                              d.
            3 1    y    1
                                                                                                Z
                                         –x + 2y = 3                                            ] –2 = x – 4y
885. Write this system in matrix form: *                                                        [      1
                                         2x – 4y = –6    888. Write this system in matrix form: ] 5y =
                                                                                                       5
          –4 2       x      3                                                                   \
     a. >         H > H=> H                                        1 –4      x     –2
           2 –1      y     –6                                 a. >       H > H=> 1 H
                                                                   0 5       y      5

            –1 2     x   –6
     b. >         H > H=> H                                             1 0    x   –2
             2 –4    y    3                                   b. >          H > H=> 1 H
                                                                       –4 5    y    5


            –1 2     x    3
     c. >         H > H=> H                                             1 0    x    1
             2 –4    y   –6                                   c. >          H > H=> 5 H
                                                                       –4 5    y   –2
            –4 2     x   –6
     d. >         H > H=> H                                            1 –4    x    1
             2 –1    y    3                                   d. >          H > H=> 5 H
                                                                       0 5     y   –2
                                           6x + 3y = 8
886. Write this system in matrix form: *                 889. Write this system in matrix form:
                                           2x + y = 3
                                                                  x – 2 = 4x – y + 3
                                                              *
            6 3    x   3
     a. >       H > H=> H
            2 1    y   8                                          6 – 2 y = –3 – x

            6 2    x   3
     b. >       H > H=> H                                     a. >
                                                                       –2 1     x    5
                                                                             H > H=> H
            3 1    y   8                                                1 –3    y   –9

                                                                       –3 1     x    5
                                                              b. >           H > H=> H
            6 2    x   8
     c. >       H > H=> H
            3 1    y   3                                                1 –2    y   –9

                                                                       –3 1     x   –9
                                                              c. >           H > H=> H
            6 3    x   8
     d. >       H > H=> H
            2 1    y   3                                                1 –2    y    5

                                                                       –2 1     x   –9
                                                              d. >           H > H=> H
                                                                        1 –3    y    5



   132
                                         –MATRIX ALGEBRA–


890. Write this system in matrix form:            892. Which of the following systems can be written
         –1 = –3 – 2x                                  in the matrix form
     *
         2 – 3y = 6 (1 – 2x)
                                                           –1 0    x     −2
                                                       >        H > H = > H?
            –3 0    x   –2                                 2 –1    y     1
     a. >        H > H=> H
            12 2    y    4
                                                              –x = –2
                                                       a. *
             2 0     x   –2                                   –2x + y = 1
     b. >         H > H=> H
            12 –3    y    4
                                                              x = –2
                                                       b. *
            2 12    x   –2                                    2x – y = 1
     c. >        H > H=> H
            0 –3    y    4
                                                              –x = –2
                                                       c. *
                                                              2x – y = 1
            2 12    x    4
     d. >        H > H=> H
            0 –3    y   –2
                                                              –x = –2
                                                       d. *
891. Write this system in matrix form:                        2x + y = 1

         2x − 3y + 5 = 1 + 2x − 4y                893. Which of the following systems can be written
     *
         3 − x − 2y = –y + x + 3                       in the matrix form
                                                           3 2    x     −2
           0 1     x   −4                              >       H > H = > H?
     a. >       H > H=> H                                  3 2    y     1
          –2 –1    y   0
                                                              2 x – 3 y = –2
          –1 1    x   –4                               a. *
     b. >      H > H=> H                                      2x – 3y = 1
          –2 0    y    0
                                                              2x + 3y = –2
           0 1      x    0                             b. *
     c. >        H > H=> H                                    2x + 3y = 1
          –2 − 1    y   –4
                                                              3x – 2y = –2
          0 –2    x   –4                               c. *
     d. >      H > H=> H                                      3x – 2y = 1
          1 –1    y    0
                                                              3x + 2y = –2
                                                       d. *
                                                              3x + 2y = 1




                                                                                              133
                                         –MATRIX ALGEBRA–




894. Which of the following systems can be written   896. Which of the following systems can be written
     in the matrix form                                   in the matrix form
         3 –2    x    4                                       0 2    x     14
     >        H > H=> H?                                  >       H > H=>     H?
         9 –6    y   12                                       4 0    y    –20

            –3y + 2x = 4                                         2x = 14
     a. *                                                 a. *
            –9y + 6x = 12                                        4y =− 20

                                                                 2y = 14
     b. * 3x – 2y = 4                                     b. *
          9x – 6y = 12                                           4x = –20

                                                                 2y = 14
                                                          c. *
            3x – 2y = 4
     c. *
            9x – 6x = 12                                         4y = –20

                                                                 2y = 14
                                                          d. *
            3x – 2x = 4
     d. *
            9y – 6x = 12                                         4x = –20

895. Which of the following systems can be written
     in the matrix form ?                                  Set 57             (Answers begin on page 261)
         – 1 –1    x     –1
     >          H > H = > H?                         This problem set is focused on computing inverse
         –1 0      y      1                          matrices.

            x – y = –1
     a. *
                                                                                                 –3 7
                                                     897. Compute the inverse, if it exists: >        H
            x=1                                                                                   1 5
                                                                   5        7
                                                          a. >
                                                                         – 22
                                                                            3 H
                                                                  22
            –x – y = –1
     b. *
                                                                     1
                                                                 –  22   – 22
            –x = 1
                                                                    5     7
                                                          b. >
                                                                 – 22
                                                                              H
                                                                         22
            –x + y = –1
     c. *
                                                                   1      3
                                                                  22     22
            x=1
                                                                  5 –7
          –y + x = –1                                     c. >         H
     d. *                                                        –1 –3
          –y = 1
                                                          d. The inverse does not exist.




   134
                                              –MATRIX ALGEBRA–



                                                                                                   –1 2
898.      Assume that a and b are not zero. Compute    901. Compute the inverse, if it exists: >        H
                                                                                                   2 –4
       the inverse, if it exists:
                                                                   1 –2
       >
           a 0
               H                                            a. >        H
           0 b                                                     –2 4

                                                                   1 2
                                                            b. >       H
       a. > 0       1 H
                1
            a       0
                                                                   2 4
                    b


                                                                   –4 2
                                                            c. >        H
       b. > a    H
           –1 0
                                                                   2 –1
            0 –1
               b

                                                            d. The inverse does not exist.
            –a 0
       c. >       H
                                                                                                   6 3
                                                       902. Compute the inverse, if it exists: >       H
             0 –b
                                                                                                   2 1
       d. The inverse does not exist.
                                                                 1 3
                                          1 2               a. > 2 6 H
899. Compute the inverse, if it exists: >     H
                                          2 3
                                                                    6 –3
           –3 2                                             b. >         H
       a. > 2 –1 H                                                 –2 1

                                                                    1 –3
             3 –2                                           c. >         H
       b. >       H                                                –2 6
            –2 1
                                                            d. The inverse does not exist.
             3 –2
       c. >       H
            –2 1                                                                                   –3 4
                                                       903. Compute the inverse, if it exists: >        H
                                                                                                   4 2
       d. The inverse does not exist.
                                                                      1     2

                                                            a. >
                                                                   – 11
                                                                                H
                                                                           11
                                            2 3
900. Compute the inverse, if it exists: >       H
                                                                     2      3
                                                                    11     22
                                            1 1
                                                                   2 4
           –1 3                                             b. >        H
       a. > 1 –2 H                                                 4 –3

                                                                   2 –4
             1 –3                                           c. >         H
       b. >       H                                                –4 –3
            –1 2
            >           H
                                                            d. The inverse does not exist.
                1 2
       c.


       d. The inverse does not exist.




                                                                                                            135
                                              –MATRIX ALGEBRA–



                                            1 –4                                                    0 1
904. Compute the inverse, if it exists: >        H     907. Compute the inverse, if it exists: >         H
                                            0 25                                                   –2 –1
                         4
                                                               –1 –1
     a. > 0                                                   > 2
                                                            a. 1 0 H
          1
                             H
                        25                                         2
                         1
                        25



                                                                   –1 1
                                                            b. >        H
             1 0
     b. >         H
            –4 25                                                  –2 0

                                                                –1 –1
                                                            c. > 2 0 H
            25 4
     c. >        H
             0 1

     d. The inverse does not exist.                         d. The inverse does not exist.

                                                                                                   –1 0
                                                       908. Compute the inverse, if it exists: >         H
                                            3 –1
905. Compute the inverse, if it exists: >        H
                                            1 –2                                                    2 –1

                                                                 1 0
                                                            a. > 2 1 H
            –2 –1
     a. >         H
             1 3

                                                                   1 2
                                                            b. >       H
            –2 1
     b. >        H
            –1 3                                                   0 1

                2
                             –1                                    –1 0
     c. >                       H                           c. >         H
                5             5

            –       1
                    5        –3
                              5                                    –2 –1

     d. The inverse does not exist.                         d. The inverse does not exist.

                                                                                                   3 2
                                                       909. Compute the inverse, if it exists: >       H
                                            –2 0
906. Compute the inverse, if it exists: >         H
                                            –12 3                                                  3 2
           3  0                                                  2 2
     a. > –12 –2 H                                          a. > 3 3 H

            1
                                                                    2 –2
     b. >
              0
                 H                                          b. >         H
            2

            2 –1
               3                                                   –3 3

            3 0                                                     2 –3
     c. >         H                                         c. >         H
            12 –2                                                  –2 3

     d. The inverse does not exist.                         d. The inverse does not exist.




   136
                                                 –MATRIX ALGEBRA–



                                            3 –2               Set 58            (Answers begin on page 264)
910. Compute the inverse, if it exists: >        H
                                            9 –6
                                                          This problem set is focused on solving matrix equations
          –6 –2                                           of the form Ax = b.
     a. > 9 3 H
                                                          913. Solve this system by first converting to an
          –6 9
     b. >      H
                                                               equivalent matrix equation:
          –2 3
                                                                   –3x + 7y = 2
                                                               *
                                                                     x + 5y = 8
            –6 2
     c. >        H                                                       23        13
            –9 3                                               a. x =    11 , y = 11

                                                               b. x =    – 23 , y = – 13
                                                                           11         11
     d. The inverse does not exist.
                                                               c. There is no solution.
                                                               d. There are infinitely many solutions.
911. Compute the inverse, if it exists: > –1 –1 H
                                            –1    0
          0 –1                                            914. Solve this system by first converting to an
     a. > –1 1 H                                               equivalent matrix equation:
                                                                   x=a
                                                               *
          1         –1                                             y=b
     b. >              H
          –1        0
                                                               a.   There is no solution.
            1 1                                                b.   There are infinitely many solutions.
     c. >       H                                              c.   x = a, y = b
            1 0
                                                               d.   x = b, y = a
     d. The inverse does not exist.
                                                          915. Solve this system by first converting to an
                                            0 2
912. Compute the inverse, if it exists: >       H              equivalent matrix equation:
                                            4 0
                                                                    x + 2y = 4
           0 –2                                                *
     a. > –4 0 H                                                   2x + 3y = 2
                                                               a.   x = 8, y = –6
                1
                                                               b.   x = –8, y = 6
     b. >
            0
                    H
                4
            1
            2   0                                              c.   There is no solution.
                                                               d.   There are infinitely many solutions.
            0 – 41
     c. >          H
            –1 0
             2



     d. The inverse does not exist.




                                                                                                           137
                                            –MATRIX ALGEBRA–




916. Solve this system by first converting to an     920. Solve this system by first converting to an
     equivalent matrix equation:                          equivalent matrix equation:
         2x + 3y = 1                                      Z
     *                                                    ] –2 = x – 4y
         x + y = –2                                       [ 5y = 1
                                                          ]       5
     a.   x = 1, y = –5                                   \
     b.   x = –7, y = 5                                   a. x = – 46 , y = 215
                                                                   25
                                                                   46
     c.   There is no solution.                           b. x =   25 ,   y = – 215
     d.   There are infinitely many solutions.            c. There is no solution.
                                                          d. There are infinitely many solutions.
917. Solve this system by first converting to an
     equivalent matrix equation:
                                                     921. Solve this system by first converting to an
       –x + 2y = 3
     *                                                    equivalent matrix equation:
       2x – 4y = –6
                                                              x – 2 = 4x – y + 3
     a.   x = –3, y = 2                                   *
                                                              6 – 2y = –3 – x
     b.   x = 2, y = –3                                            19          22
     c.   There is no solution.                           a. x =    7,y=–7

     d.   There are infinitely many solutions.            b. x =   – 179 , y = 272
                                                          c. There is no solution.
918. Solve this system by first converting to an          d. There are infinitely many solutions.
     equivalent matrix equation:
         6x + 3y = 8                                 922. Solve this system by first converting to an
     *
         2x + y = 3                                       equivalent matrix equation:
                                                              –1 = –3 – 2 x
                                                          *
     a.   x = –4, y = –6
     b.   x = 4, y = –6                                       2 – 3y = 6 (1 – 2x)
     c.   There are infinitely many solutions.            a.   There is no solution.
     d.   There is no solution.                           b.   There are infinitely many solutions.
                                                          c.   x = –1, y = – 136
919. Solve this system by first converting to an          d.   x = 1, y = 136
     equivalent matrix equation:
         –3x = 1 – 4y
     *
                                                     923. Solve this system by first converting to an
         2 y + 3 = –4 x                                   equivalent matrix equation:
              – 171 , y = – 252                               2x – 3y + 5 = 1 + 2x – 4y
                                                          *
     a. x =
     b. x =    7         5                                    3 – x – 2y = –y + x + 3
              11 , y = 22
     c. There is no solution.                             a.   There is no solution.
     d. There are infinitely many solutions.              b.   There are infinitely many solutions.
                                                          c.   x = 2, y = –4
                                                          d.   x = –2, y = 4



   138
                                            –MATRIX ALGEBRA–



                            –1    0    x    –2            Set 59            (Answers begin on page 270)
924. Solve this system: > 2      –1
                                    H > H=>
                                       y    1
                                               H
                                                     This problem set is focused on solving matrix equations
     a.   x = –2, y = –3                             of the form Ax = b using Cramer’s rule.
     b.   x = 2, y = 3
     c.   There is no solution.                      929. Solve this system using Cramer’s rule:
     d.   There are infinitely many solutions.
                                                              –3x + 7y = 2
                                                          *
                            3 2    x    –2                      x + 5y = 8
925. Solve this system: >       H > H=>    H
                            3 2    y    1
                                                          a. There is no solution.
     a.   x = –2, y = –1                                  b. There are infinitely many solutions.
     b.   x = 2, y = –1                                   c. x = 23 , y = 13
                                                                 11       11
     c.   There are infinitely many solutions.
                                                          d. x = – 23 , y = – 13
                                                                   11         11
     d.   There is no solution.
                            3 –2     x     4
926. Solve this system: > 9 –9 H > y H = > 12 H
                                                     930. Assume that a and b are nonzero. Solve this sys-
                                                          tem using Cramer’s rule:
     a.   x = 9, y = –2                                       x=a
     b.   x = –9, y = 2                                   *
                                                              y=b
     c.   There is no solution.
     d.   There are infinitely many solutions.            a.   x = b, y = a
                                                          b.   x = a, y = b
                            –1 –1    x   –1
927. Solve this system: >         H > H=> H               c.   There is no solution.
                            –1 0     y    1               d.   There are infinitely many solutions.
     a.   There is no solution.
     b.   There are infinitely many solutions.       931. Solve this system using Cramer’s rule:
     c.   x = –1, y = 2                                       x + 2y = 4
     d.   x = 1, y = –2                                   *
                                                              2x + 3y = 2
                            0 2    x    14
928. Solve this system: >       H > H=>     H             a.   x = 8, y = –6
                            4 0    y    –20               b.   x = –8, y = 6
     a.   There is no solution.                           c.   There is no solution.
     b.   There are infinitely many solutions.            d.   There are infinitely many solutions.
     c.   x = –5, y = 7
     d.   x = 5, y = –7                              932. Solve this system using Cramer’s rule:
                                                              2x + 3y = 1
                                                          *
                                                              x+y= —   2
                                                          a.   x = –7, y = 5
                                                          b.   x = 7, y = –5
                                                          c.   There is no solution.
                                                          d.   There are infinitely many solutions.




                                                                                                      139
                                            –MATRIX ALGEBRA–


933. Solve this system using Cramer’s rule:          937. Solve this system using Cramer’s rule:
         –x + 2y = 3                                          x – 2 = 4x – y + 3
     *                                                    *
         2x – 4y = –6                                         6 – 2y = –3 –x
     a.   x = 2, y = 5                                    a. There is no solution.
     b.   x = –2, y = –5                                  b. There are infinitely many solutions.
     c.   There is no solution.
                                                          c. x = – 1 , y =
                                                                   5
                                                                             22
                                                                              5
     d.   There are infinitely many solutions.
                                                          d. x = 1 , y =
                                                                 5         – 252
934. Solve this system using Cramer’s rule:
                                                     938. Solve this system using Cramer’s rule:
       6x + 3y = 8
     *                                                        –1 = –3 – 2 x
       2x + y = 3                                         *
                                                              2 – 3y = 6 (1 – 2x)
     a.   x = –12, y = 4                                                  16
     b.   x = 12, y = –4                                  a. x = 1, y =    3

     c.   There are infinitely many solutions.            b. x = –1, y   = 136
     d.   There is no solution.                           c. There is no solution.
                                                          d. There are infinitely many solutions.
935. Solve this system using Cramer’s rule:
         –3x = 1 – 4y
     *                                               939. Solve this system using Cramer’s rule:
         2 y + 3 = –4 x
                                                              2x – 3y + 5 = 1 + 2x – 4y
                                                          *
     a. x = – 171 , y = – 252                                 3 – x – 2y = –y + x + 3
               7           5
     b. x =   11 , y   =   22                             a.   x = 2, y = –4
     c. There are infinitely many solutions.              b.   x = –2, y = 4
     d. There is no solution.                             c.   There is no solution.
                                                          d.   There are infinitely many solutions.
936. Solve this system using Cramer’s rule:
     Z                                               940. Solve this system using Cramer’s rule:
     ] –2 = x – 4y
                                                              –1 0     x   –2
     [ 5y = 1                                             >         H > H=> H
     ]      5                                                  2 –1    y    1
     \
     a. There is no solution.                             a.   x = –2, y = –3
     b. There are infinitely many solutions.              b.   x = 2, y = 3
     c. x = – 46 , y =        1                           c.   There is no solution.
              25             25
              46                                          d.   There are infinitely many solutions.
     d. x =   25 , y   =   – 215




   140
                                            –MATRIX ALGEBRA–


941. Solve this system using Cramer’s rule:          943. Solve this system using Cramer’s rule:
                                                              –1 –1    x   –1
                                                          >         H > H=> H
         3 2    x   –2
     >       H > H=> H
         3 2    y    1                                        –1 0     y    1

     a.   x = –5, y = –3                                  a.   x = 1, y = –2
     b.   x = 5, y = 3                                    b.   x = –1, y = 2
     c.   There are infinitely many solutions.            c.   There is no solution.
     d.   There is no solution.                           d.   There are infinitely many solutions.

942. Solve this system using Cramer’s rule:          944. Solve this system using Cramer’s rule:
         3 –2    x    4                                       0 2    x     14
     >        H > H=> H                                   >       H > H=>     H
         9 –6    y   12                                       4 0    y    –20

     a.   x = –1, y = 4                                   a.   x = 5, y = –7
     b.   x = 1, y = –4                                   b.   x = –5, y = 7
     c.   There is no solution.                           c.   There is no solution.
     d.   There are infinitely many solutions.            d.   There are infinitely many solutions.




                                                                                                      141
                                                 COMMON

                8
S E C T I O N




                                                 ALGEBRA
                                                 ERRORS




I   t is common and expected for those who are learning algebra for the first time or reviewing the subject after
    having been away from it for a while to make errors. Several of the most typical errors made are explored in
    the four sets in this section.
      For all of the questions in this section, identify the choice that best describes the error, if any, made in each
scenario.




                                                                                                               143
                                     –COMMON ALGEBRA ERRORS –



       Set 61         (Answers begin on page 274)          949.
                                                                  a+2
                                                                          = 2, for any nonzero value of a.
                                                                   a
Some common arithmetic and pre-algebra errors are                 a. This is incorrect because you cannot cancel
explored in this problem set.                                        members of a sum; you can cancel only fac-
                                                                     tors that are common to the numerator and
945. (–3)–2 = 6                                                      denominator.
       a. The answer should be –9 because (–3)–2 =                b. The correct result should be 3 because a = 1
                                                                                                            a
          –3 3.                                                      and a + 2 = a + 2 = 1 + 2 = 3.
                                                                            a     a
       b. The answer should be 1 because (–3)–2 =
                               9                                  c. There is no error.
              1
          (–3) (–3) =1.
                     9
                                                                  3   a     3+a
       c. There is no error.                               950. 4 + 2 = 6 , for any real number a.
                                                                  a. You must first get a common denominator
946.   (– 2 )0   =1                                                  before you add two fractions. The correct
          3                                                                                a
                                                                     computation is: 3 + 2 = 3 + 24a = 3 + 2a .
       a. The answer should be –1 because (– 2 )0 =
                                             3
                                                                                      4      4           8
                                                                  b. You must first get a common denominator
          –( 2 )0 = –1.
             3                                                       before you add two fractions. The correct
       b. The answer should be zero because you                                            a
                                                                     computation is: 3 + 2 = 3 + 24a = 3 + 2a .
                                                                                      4      4           4
          should multiply the base and exponent.                  c. There is no error.
       c. There is no error.
                                                           951. 4 is 200% of 8.
947. 0.00013 = 1.3       104                                      a. There is no such quantity as “200%.” You
       a. The statement should be 0.00013 = 1.3 104                  cannot exceed 100%.
          because the decimal point must move four                b. The placement of the quantities is incorrect. A
          places to the left in order to yield 0.00013.              correct statement would be “200% of 4 is 8.”
       b. The statement should be0.00013 = 1.3 10–3               c. There is no error.
          because there are three zeros before the deci-
          mal point.
                                                           952. 0.50% of 10 is 0.05.
       c. There is no error.
                                                                  a. In order to compute this percentage, you
                                                                     should multiply 0.50 times 10 to get 5, not
948. –42 = –16
                                                                     0.05.
       a. The answer should be –8 because 42 = 8, and             b. In order to compute this percentage, you
          this is then multiplied by –1.                             should multiply 50.0 times 10 to get 500, not
       b. The answer should be 16 because –42 =                      0.05.
          (–4)(–4) = 16.                                          c. There is no error.
       c. There is no error.




   144
                                                   –COMMON ALGEBRA ERRORS –


                                                                        3                 3        3
953.      3+ 6= 3+6= 9=3                                         957.       27x3 =            27       x3 = 3x, for any real number
       a. You must first simplify 6 as 6 = 2 . 3                        x.
          = 2 3, and then combine with 3 to con-                        a. The first equality is wrong because the radi-
          clude that 3 + 6 = 3 3.                                          cal of a product is not the product of the
       b. The sum 3 + 6 cannot be simplified fur-                          radicals.
          ther because the radicands are different.                     b. The second equality holds only if x is not
       c. There is no error.                                               negative because you can only take the cube
                                                                           root of a non-negative real number.
                3                                   5
954.      6 + –38 = (2+3) 6 – 38 = –32 = –2                             c. There is no error.
       a. The calculation is correct until the last line;               3
                                                                                3    8        6
          the fifth root of a negative number is not             958.   4
                                                                        8
                                                                            =   4    5    =   5
                                                                        5
          defined.
                                                                        a. The first equality is wrong because you must
       b. The first equality is incorrect: the radicals
                                                                           multiply the numerator by the reciprocal of
          cannot be combined since their indices are
                                                                           the denominator.
          different.
                                                                        b. The second equality is wrong because the
       c. There is no error.
                                                                           fraction on the far right should be 24 , which
                                                                                                               20
          1               1         2+    3        2+ 3                    cannot be simplified further.
955.             =                            =   (2 + 3)2
                                                             =
        2+ 3            2+ 3        2+    3                             c. There is no error.
         2+     3            2+ 3        2+ 3
       22 + (   3)2     =     4+3   =      7                            x12          12
                                                                                          = x–4, for
                                                                 959. x–3 = x –3                       any non-negative real
       a. The third equality is incorrect because the                   number x.
          binomial was not squared correctly. The cor-                  a. The correct answer should be x36 because
          rect denominator should be 22+2 3 +                                 x12
                                                                              x–3
                                                                                    = x12x3 = x12 3 = x36.
                    2
          ( 3)          = 7 + 2 3.                                      b. The correct answer should be x15 because
                                                                              x12
       b. The first equality is wrong because multiply-                       x–3
                                                                                    = x12x3 = x12 + 3 = x15.
                        2+    3                                         c. There is no error.
          ing by        2+    3
                                  changes the value of the
          expression. The rest of the equalities are
                                                                 960. (e4x)2 = e4x+2
          correct.
                                                                        a. The correct answer should be e8x because
       c. There is no error.                                                    2
                                                                           (e4x) = e4x 2 = e8x.
                                                                                                            2
           2                                                            b. The correct answer should be e16x because
956. (x5) = x7, for any real number x.                                          2       2     2 2     2
                                                                           (e4x) = e(4x) = e4 x = e16x .
       a. The exponents should be multiplied, not
                                                                        c. There is no error.
          added, so that the correct answer should be x10.
       b. The correct answer should be x25 because
              2    2
          (x5) = x5 .
       c. There is no error.




                                                                                                                             145
                                       –COMMON ALGEBRA ERRORS –


                                                                                         4         x+3
      Set 62        (Answers begin on page 274)               964. The solution of – x + 7 = x + 7 is x = –7.
                                                                   a. The equation obtained after multiplying
Some common errors in solving equations and                           both sides by x + 7 was not solved correctly.
inequalities, as well as simplifying algebraic expressions,           The correct solution should be x = 1.
are explored in this problem set.                                  b. x = –7 cannot be the solution because it
                                                                      makes the terms in the original equation
961. The solution set for the inequality –6x     24 is                undefined—you cannot divide by zero. As
      (–4,∞).                                                         such, this equation has no solution.
      a. The inequality sign must be switched when                 c. There is no error.
         multiplying both sides by a negative real
         number. The correct solution set should be           965. The solutions of the equation log5x + log5(5x3)
         (∞,–4).                                                   = 1 are x = –1 and x = 1.
      b. You should multiply both sides by –6, not
                                                                   a. Both solutions should be divided by       5;
         divide by –6. The correct solution set should                                                          1
                                                                      that is, the solutions should be x =              .
         be (–144,∞).                                                                                               5
      c. There is no error.                                        b. While x = 1 satisfies the original equation, x
                                                                      = –1 cannot because negative inputs into a
962. The solution set for the equation |x – 1| = 2                    logarithm are not allowed.
      is {–1}.                                                     c. There is no error.
      a. There are two solutions of this equation,
          namely x = –1 and x = 3.                            966. The complex solutions of the equation x2 + 5 = 0
      b. The solution of an absolute value equation
                                                                   obtained using the quadratic formula are given
          cannot be negative. The only solution is x = 3.                   0   02 – 4(1)(5)
                                                                   by x =                      =    2i 5.
      c. There is no error.                                                       1
                                                                   a. The denominator in the quadratic formula is
963. log31 = 0                                                        2a, which in this case is 2, not 1. As such, the
      a. 1 is an invalid input for a logarithm. As such,              complex solutions should be x = i 5.
         the quantity log3 is undefined.                           b. There are no complex solutions to this equa-
      b. The input and output are backward. The real                  tion because the graph of y = x2 + 5 does not
         statement should read log30 = 1.                             cross the x–axis.
      c. There is no error.                                        c. There is no error.

                                                              967. x2 – 4x – 21 = (x + 7)(x – 3)
                                                                   a. This is incorrect because multiplying the
                                                                      binomials on the right side of the equality
                                                                      yields x2 – 21, which is not the left side
                                                                      listed above.
                                                                   b. The signs used to define the binomials on
                                                                      the right side should be switched. The cor-
                                                                      rect factorization is (x – 7)(x + 3).
                                                                   c. There is no error.



   146
                                     –COMMON ALGEBRA ERRORS –


968. Since taking the square root of both sides of the     972. x2 + 25 = (x – 5)(x + 5)
     inequality x2 4 yields the statement x          2.         a. The correct factorization of the left side is
     Since both statements must be satisfied simul-                x2 + 25 = x2 + 52 = (x + 5)2.
     taneously, the solution set is (–∞, –2].                   b. The left side is not a difference of squares. It
     a. You must move all terms to one side of the                 cannot be factored further.
        inequality, factor (if possible), determine the         c. There is no error.
        values that make the factored expression
        equal to zero, and construct a sign chart to             2x–1 – y–1       2–1              1
                                                           973. x–1 + 4y–1 = 1 + 4 = 5
        solve such an inequality. The correct solution
                                                                a. Cancelling the terms x–1 and y–1 leaves 0
        set should be [–2,2].
                                                                   each time, not 1. So, the correct statement
     b. When taking the square root of both sides                               –1
                                                                                   – y–1
                                                                   should be 2x +4y–1 = 2 = 1 .
                                                                              x–1         4   2
        of an equation, you use only the principal
        root. As such, the correct statement should             b. You cannot cancel terms of a sum in the
        be x 2, so that the solution set is (–∞,2].                numerator and denominator. You can
     c. There is no error.                                         only cancel factors common to both. The
                                                                   complex fraction must first be simplified
969. (x – y)2 = x2 – y2
                                                                   before any cancelation can occur. The
     a. The left side must be expanded by FOILing.                 correct statement is:
                                                                                                           2y       x
        The correct statement should be (x – y)2 =                  2x–1 – y–1
                                                                                       2
                                                                                       x   –   1
                                                                                               y           xy   –   xy
                                                                                                                             2y – x
                                                                                                                               xy         2y – x
        x2 – 2xy + y2.                                              x –1 + 4y–1
                                                                                  =    1
                                                                                           +   4       =    y
                                                                                                                +   4x   =   y + 4x   =     xy
                                                                                       x       y           xy       xy         xy

     b. The –1 must be squared. The correct state-                       xy           2y – x
        ment should be (x – y)2 = x2 + y2.                             y + 4x     =   y + 4x

     c. There is no error.
                                                                c. There is no error.
970. The solution of the equation     x = –2 is x = 4,
                                                           974. ln(ex + e2y) = ln(ex) + ln(e2y) = x + 2y
     as seen by squaring both sides of the equation.
     a. The correct solution is x = –4 because when             a. The first equality is incorrect because the
        you square both sides of the equation, you                 natural logarithm of a sum is the product of
        do not square the –1.                                      the natural logarithms. So, the statement
     b. This equation has no real solutions because                should be ln(ex + e2y) = ln(ex) ln(e2y) = 2xy.
        the output of an even–indexed radical must              b. The first equality is incorrect because the
        be nonnegative.                                            natural logarithm of a sum is not the sum of
     c. There is no error.                                         the natural logarithms. In fact, the expression
                                                                   on the extreme left side of the string of
                                                                   equalities cannot be simplified further.
971. The solution set of the inequality |x + 2|     5
                                                                c. There is no error.
     is (–∞, –7)∪(3,∞).
     a. The interval (–∞, –7) should be deleted
         because an absolute value inequality cannot
         have negative solutions.
     b. You must include the values that make the
         left side equal to 5. As such, the solution set
         should be (–∞, –7]∪[3,∞)
     c. There is no error.
                                                                                                                                            147
                                     –COMMON ALGEBRA ERRORS –


975. log5(5x2) = 2log5(5x) = 2[log5(5) + log5(x)] =         978. The line x = a has a slope of zero, for any real
     2[1 + log5(x)]                                              number a.
     a. The first equality is incorrect because 2log5(5x)        a. The line is vertical, so its slope is undefined.
        = log5(5x2) = log5(25x2). The other equalities           b. The statement is true except when a = 0.
        are correct.                                                The y–axis cannot be described by such
     b. The very last equality is incorrect because                 an equation.
        log5 5 = 0. The other equalities are correct.            c. There is no error.
     c. There is no error.
                                                            979. The point (–2, 1) lies in Quadrant IV.
          2
976. ln(4x – 1) = ln[(2x – 1)(2x + 1)] + ln(2x – 1) +            a. The point is actually in Quadrant II.
     ln(2x + 1)                                                  b. The point is actually in Quadrant III.
     a. The “natural logarithm of a difference rule”             c. There is no error.
        was not applied correctly. The correct state-
        ment should be ln(4x2 – 1) = ln(4x2) – ln(1)        980. The inverse of the function f(x) = x2, where x is
        = ln(4x2) – 0 =ln(4x2). The last expression in           any real number, is the function f –1(x) = x.
        this string of equalities cannot be simplified           a. f cannot have an inverse because it doesn’t
        because the exponent 2 does not apply to the                pass the vertical line test.
        entire input of the logarithm.                           b. The domain of f must be restricted to [0,∞)
     b. Using the fact that the natural logarithm of a              in order for f to have an inverse. In such case,
        difference is the quotient of the natural loga-             the given function f –1(x) = x is indeed its
        rithms, we see that the expression ln(4x2 – 1)              inverse.
              4x2       4x2
        = lnl(n1 ) = ln(0 ) , so the expression is not           c. There is no error.
        well–defined.
                                                                                                  1
     c. There is no error.                                  981. The lines y = 3x + 2 and y = – 3 x + 2 are
                                                                 perpendicular.
                                                                 a. The lines are parallel since their slopes are
     Set 63        (Answers begin on page 275.)                     negative reciprocals of each other.
This problem set highlights common errors made in                b. The lines cannot be perpendicular since the
graphing, computing with, and interpreting functions.               product of their slopes is not 1.
                                                                 c. There is no error.
977. The vertical asymptote for the graph of
              x
     f(x) = x2 + 2 is y = 0.                                982. The slope of a line passing through the points
                +4
     a. The expression should be factored and sim-               (a, b) and (c, d) is m = b – d , provided that a ≠ c.
                                                                                          a–c
                                   1
        plified to obtain f(x) = x – 2 . Then, we can            a. The slope is actually equal to the quantity m
                                                                       a–c
        conclude that the vertical asymptote for f is               = b – d , provided that b ≠ d.
        x = 2.                                                   b. The slope is actually equal to the quantity m
     b. The line y = 0 is the horizontal asymptote for f.           = b – a , provided that c ≠ d.
                                                                       d–c

     c. There is no error.                                       c. There is no error.




   148
                                     –COMMON ALGEBRA ERRORS –


                                         8–x
983. The graph of the function f(x) = x2 – 64 has an       986. If f(x) = 5 and g(x) =   x, it follows that
     open hole at x = 8.                                        (f ˚ g)(–2) = 5.
     a. The graph actually has a vertical asymptote             a. The composition was computed in the
        at x = 8 because this value makes the denom-                wrong order. The correct output should
        inator equal to zero.                                       be 5.
     b. The graph actually has a horizontal asymp-              b. –2 is not in the domain of g, so that the
        tote at x = 8 because this value makes the                  composition is not defined at –2.
        denominator equal to zero.                              c. There is no error.
     c. There is no error.
                                                           987. The x–intercept of f(x) = x3 + 1 is (0, 1).
984. If f(2) = 5, then the point (5, 2) must be on the          a. The point (0,1) is the y–intercept of f, not
     graph of y = f(x).                                            the x–intercept.
     a. The coordinates of the point that is known              b. There are no x–intercepts for this function
        to lie on the graph of y = f(x) are reversed; it           because x3 + 1 is always positive.
        should be (2,5).                                        c. There is no error.
     b. The given information is insufficient to
        make any conclusion about a point being on         988. The graph of g(x) = 2–x is increasing as x moves
        the graph of y = f(x). All that can be said is          from left to right through the domain.
        that 2 is in the range of f.                            a. The graph of g is actually decreasing as x
     c. There is no error.                                         moves from left to right through the
                                                                   domain.
985. The range of the function f(x) = (x – 1)2                  b. There are intervals on which the graph of g is
     is [0,∞).                                                     increasing and others on which it is
     a. The graph of f is the graph of g(x) = x2                   decreasing.
         shifted vertically up one unit. Since the range        c. There is no error.
         of g is [0,∞), it follows that the range of f
         must be [1,∞).                                    989. The graph of y = f(x + 3) is obtained by shifting
     b. The graph of f is the graph of g(x) = x2                the graph of y = f(x) to the right 3 units.
         shifted vertically down one unit. Since the            a. The graph of y = f(x + 3) is actually obtained
         range of g is [0,∞), it follows that the range            by shifting the graph of y = f(x) to the left 3
         of f must be [–1,∞).                                      units.
     c. There is no error.                                      b. The graph of y = f(x + 3) is actually obtained
                                                                   by shifting the graph of y = f(x)vertically up
                                                                   3 units.
                                                                c. There is no error.




                                                                                                              149
                                        –COMMON ALGEBRA ERRORS –


990. The graph of y = f(x) – 2 is obtained by shifting         992. The graph of y = 5 does not represent a function
      the graph of y = f(x)down 2 units.                            because it does not pass the horizontal line test.
      a. The graph of y = f(x) – 2 is obtained by shift-            a. The graph of y = 5 passes the vertical line
         ing the graph of y = f(x) to the left 2 units.                test, so it represents a function. It is, how-
      b. The graph of y = f(x) – 2 is obtained by shift-               ever, not invertible.
         ing the graph of y = f(x) to the right 2 units.            b. The fact that y = 5 does not pass the hori-
      c. There is no error.                                            zontal line test does not imply it is not a func-
                                                                       tion. However, since the range of a function
991. If, f(x) = x4, then f(x – h) = f(x) – f(h) = x4 – h4.             must consist of more than a single value, we
      a. You cannot distribute a function across parts                 conclude that it y = 5 cannot represent a
         of a single input. As such, the correct state-                function.
         ment should be f(x – h) = (x – h)4.                        c. There is no error.
      b. The second equality is incorrect because
         you must also square the –1. As such, the
         correct statement should be f(x – h) =
         f(x) – f(h) = x4 + h4 .
      c. There is no error.


                                       Set 64         (Answers begin on page 276)

This problem set highlights common errors made when dealing with linear systems of equations and matrix algebra.

                     2x + 3y = 6
993. The system *                 has infinitely many solutions.
                     –2x – 3y = 2
      a. Since adding the two equations results in the false statement 0 = 8, there can be no solution of this
         system.
      b. The slopes of the two lines comprising the system are negatives of each other. As such, the lines are
         perpendicular, so the system has a unique solution.
      c. There is no error.

                     2x – 5y = –1
994. The system *                  has no solutions.
                     4x – 10y = –2
      a. Since multiplying the first equation by –2 and then adding the two equations results in the true state-
         ment 0 = 0, there are infinitely many solutions of this system.
      b. The two lines comprising the system intersect, so the system has a unique solution.
      c. There is no error.




   150
                                     –COMMON ALGEBRA ERRORS –



                                2 1 x       –2
995. The matrix equation >          H> H = > H has infinitely many solutions.
                               –4 –2 y       4
                                                             2   1
     a. Since the determinant of the coefficient matrix >           H is zero, the system has no solution.
                                                            –4   –2
                                                                     –1
                                                   x       2 1     –2
     b. The system has a unique solution given by > H = >       H > H.
                                                   y      –4 –2     4
     c. There is no error.

         0     0 1     0 1
996. > H + >       H=>     H
         1     0 0     1 0
     a. The two matrices on the left side of the equality do not have the same dimension. As such, their sum is
        undefined.
                                                                         0 0 1
     b. The matrices were added incorrectly. The right side should be >          H.
                                                                         1 0 0
     c. There is no error.

                               –2 1
997. 9 –1 2 C . 9 2 –1 C = >         H
                                4 –2
     a. The product should be a real number, namely 9 –1 2 C . 9 2 1 C = (–1) (2) + (2) (–1) = –4 .
     b. The inner dimension of the two matrices on the left side are not the same. As such, they cannot be
        multiplied.
     c. There is no error.

             4 2
998. det >        H= (2)(1) – (4)(–1) = 6
             1 –1
     a. The wrong pairs of entries are being multiplied to form the determinant. The correct statement
                        4 2
        should be det >      H= (4)(2) – (1)(–1) = 9.
                        1 –1
                                                                                             4 2
     b. The difference is computed in the wrong order. The correct statement should be det >       H = (4)(–1)
                                                                                             1 –1
        – (2)(1) = –6.
     c. There is no error.
              –1
         1 1        0 –1     0 1
999. >       H = –>      H=>      H
         1 0        –1 1     1 –1
     a. The constant multiple on the right side of the first equality should be 1, not –1. Therefore, the inverse
                     0 –1
          should be >      H.
                    –1 1
     b. The inverse does not exist because several of the entries are the same real number.
     c. There is no error.




                                                                                                             151
                                   –COMMON ALGEBRA ERRORS –



            1 –2          2 –1
1000.   >        H+ 1 = >      H
            2 3           3 4
    a. You cannot add a 2 2 matrix and a real number because their dimensions are different. Therefore,
       the sum is not well–defined.

    b. The 1 should be added only to the diagonal entries, so that the correct statement should be
         1 –2          2 –2
       >      H+ 1 = >      H.
         2 3           2 4
    c. There is no error.

          –1                         1 1 1 1
1001. >      H . 9 –1 –1 –1 –1 C = >         H
          –1                         1 1 1 1
    a. The product is not well–defined because the matrices must have the same dimensions in order to be
       multiplied.
    b. The correct product should be 9 1 1 1 1 C .
    c. There is no error.




  152
                                              ANSWERS &
                                              EXPLANATIONS




     Section 1—Pre-Algebra                                 Then, multiply left to right as such products
          Fundamentals                                     arise. Finally, compute sums and differences
                                                           from left to right as they arise, as follows:
Set 1 (Page 2)
                                                           12(84 – 5) – (3 54) = 12(79) – (162) =
  1. b. Multiply the contents of each set of paren-
                                                           948 – 162 = 786
     theses first. Then, multiply the resulting prod-
                                                        5. d. Computing the sum 60,000 + 800 + 2 yields
     ucts: (15 + 32)(56 – 39) = (47)(17) = 799
                                                           60,802.
  2. b. Dividing 65,715 by 4 results in 16,428 with a
                                                        6. c. Since 112   7= 16 and 112 8 = 14, we
     remainder of 3. Since the hundreds place is not
                                                           conclude that 112 is divisible by both 7 and 8.
     5 or greater, rounding the quotient to the
                                                        7. a. Rounding 162 to the nearest hundred yields
     nearest thousand yields 16,000.
                                                           200 (since the tens place is greater than 5), and
  3. c. Approximate 7,404 by 7,400. The quotient
                                                           rounding 849 to the nearest hundred yields
     7,400 74 = 100 is a good approximation of
                                                           800 (since the tens place is less than 5). Multi-
     the quotient 7,400 74.
                                                           plying 200 times 800 yields a product of
  4. a. Using the order of operations, compute the
                                                           160,000, which is an estimation of the product
     quantities within each set of parentheses first.      of 162 and 849.
                                                                                                    153
                                         ANSWERS & EXPLANATIONS–


 8. d. Multiplying 5 times 5 yields 25. Then, mul-            Set 2 (Page 3)
     tiplying this product by 5 results in 125. Thus,          17. c. Begin by simplifying the absolute value
     5 5 5= 125.                                                    quantity. Then, divide left to right:
 9. c. By the definition of an exponent, we have                    – 25 |4 – 9| = – 25 |–5| = –25           5 = –5.
     35 = 3     3   3    3    3 = 243.                         18. a. Since there are an odd number of negative
10. b. First, the following are the multiples of 6                  signs, the product will be negative. Computing
     between 0 and 180:                                             this product yields –4 –2 –6 3 = –144
     6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78,        19. c. Applying the order of operations, first sim-
     84, 90, 96, 102, 108, 114, 120, 126, 132, 138,                 plify the quantity enclosed in parentheses,
     144, 150, 156, 162, 168, 174, 180                              then square it, then multiply left to right, and
     Of these, the following are also factors of 180:               finally compute the resulting difference:
     6, 12, 18, 30, 36, 60, 90, 180                                 5– (–17 + 7)2 3 = 5 – (–10)2 3 = 5 – 100
     There are eight possibilities for the whole                       3 = 5 – 300 = –295
     number p.                                                 20. a. Applying the order of operations, first com-
11. c. The only choice that is a product of prime                   pute the quantities enclosed in parentheses.
     numbers equaling 90 is 2        3   3    5.                    Then, compute the resulting difference:
12. c. The factors of 12 are 1, 2, 3, 4, 6, 12. Of                  (49 7) – (48 (–4)) = (7) – (–12) =
     these, only 1 and 3 are not multiples of 2.                    7 + 12 = 19
     Thus, the set of positive factors of 12 that are          21. d. Note that substituting the values 1, 2, and 3
     NOT multiples of 2 is {1,3}.                                   in for p in the equation y = 6p – 23 yields –17,
13. e. The sum of 13 and 12 is 25, which is an odd                  –11, and –5, respectively. However, substitut-
     number. Each of the other operations produces                  ing 4 in for p results in the positive number 1.
     an even number: 20 8 = 160, 37 + 47 = 84,                      So, of the choices listed, the least value of p for
     7 12 = 84, 36 + 48 = 84                                        which y is positive is 4.

14. c. By the definition of an exponent, 24 = 2           2    22. b. Applying the order of operations, first com-

        2     2 = 16.                                               pute the quantities enclosed in parentheses.
                                                                    Then, compute the difference from left to right:
15. b. Applying the order of operations, we first
     perform exponentiation, then subtract from                     –(5 3) +(12       ( – 4)) = –(15) + (–3) =
     left to right to obtain 9 – 22 = 9 – 4 = 5.                    –15 – 3 = –18

16. c. The only choice that is divisible by only 1             23. c. Applying the order of operations, compute

     and itself is 11. Each of the other choices has                both exponentiated quantities first. Then,
     factors other than 1 and itself.                               multiply from left to right as products arise.
                                                                    Finally, compute the resulting difference:
                                                                    –2(–2)2 – 22 = –2(4) – 4 = –8 – 4 = –12




  154
                                      ANSWERS & EXPLANATIONS–


24. a. Applying the order of operations, first per-       28. c. Since h < 0, it follows that –h > 0. Since we
     form the exponentiation. Then, compute the                are also given that g > 0, we see that g – h = g +
     quantities enclosed with parentheses. Finally,            (–h) is the sum of two positive numbers and
     compute the resulting quotient:                           hence, is itself positive.
     (32 + 6) ( – 24      8) = (9 + 6)    ( –3) =         29. c. Observe that –g – h = –(g + h). Since g < 0
     (15) ( –3) = –5                                           and h < 0, it follows that g + h < 0, so –(g + h)
25. d. This one is somewhat more complicated                   is positive.
     since we have an expression consisting of            30. d. First, note that since g < 0 and h < 0, it fol-
     terms within parentheses which are, in turn,              lows that g + h must be negative, so –g – h =
     enclosed within parentheses, and the whole                –(g + h) is positive. As such, we know that
     thing is raised to a power. Proceed as follows:           –(g + h) is larger than g + h. Next, each of the
                                                               sums –g + h and –g – h consists of one positive
     (–2[1 – 2(4 – 7)])2 = (–2[1 – 2( – 3)])2
                                                               integer and one negative integer. Thus, while it
     = (–2[1 – (–6)])2
                                                               is possible for one of them to be positive, its
     = (–2[7])2
                                                               values cannot exceed that of –g – h since this
     = (–14)2
                                                               sum consists of two positive integers. As such,
     = 196
                                                               we conclude that – g – h is the largest of the
26. b. Applying the order of operations, first com-            four expressions provided.
     pute quantities enclosed within parentheses and
                                                          31. c. First, note that since g < 0 and h < 0, it fol-
     exponentiated terms on the same level from left
                                                               lows that g + h must be negative and so, –g – h
     to right. Repeat this until all such quantities
                                                               = –(g + h) is positive. As such, we know that
     are simplified. Then, multiply from left to right.
                                                               –(g + h) is larger than g + h. Next, each of the
     Finally, compute the resulting difference:
                                                               sums –(g + h) and g – h consists of one posi-
     3(5 – 3)2 – 3(52 – 32) = 3(2)2 – 3(25 – 9) =
                                                               tive integer and one negative integer. Thus,
     3(4) – 3(16) = 12 – 48 = –36
                                                               while it is possible for one of them to be nega-
27. a. Here we have an expression consisting of                tive, its values cannot be smaller than g + h
     terms within parentheses which are, in turn,              since this sum consists of two negative inte-
     enclosed within parentheses. Proceed as                   gers. As such, we conclude that g + h is the
     follows:                                                  smallest of the four expressions provided.
     –(–2 – ( –11 – (– 32 – 5) – 2)) =                    32. d. First, note that since we are given that g < –2,
     –(–2 – (–11 – (–9 – 5) – 2))                              it follows that both g and – g2 are negative, while
     = –(–2 –( –11 – (–14) – 2))                               both –g and (–g)2 are positive. Moreover, –g is
     = –(–2 –(–11 + 14 – 2))                                   an integer larger than 2 (which follows by mul-
     = –(–2 –(1))                                              tiplying both sides of the given inequality by
     = –(– 3)                                                  –1). Squaring an integer larger than 2 produces
     =3                                                        an even larger integer. As such, we conclude
                                                               that (–g)2 is the largest of the four expressions
                                                               provided.



                                                                                                          155
                                                               ANSWERS & EXPLANATIONS–



Set 3 (Page 5)                                                                             Comparison 2:         2 ? 8
                                                                                                                                Cross multiplying yields
                                                                                                                 3 > 11
 33. a. Express both fractions using the least com-                                        the false statement 22 > 24. This implies that
      mon denominator, then add: 5 – 9
                                                                   1
                                                                   4   =                    8
                                                                                                is larger.
      5 4                                                                                  11
      9 4 – 1 9 = 20 – 396 = 203– 9 = 11
            4 9   36            6      36                                                                         8 ? 4
                                                                                           Comparison 3:         11 > 10         Cross multiplying
 34. b. First, rewrite all fractions using the least                                       yields the true statement 80 > 44. This implies
      common denominator, which is 30. Then, add:                                                  8
                                                                                           that   11   is larger.
       2                                              1 6
      15    + + + 130 = 125 22
               1
               5
                        1
                        6                        +    5 6    +   1 5
                                                                 6 5   +      3 3
                                                                             10 3                                                  8
                                                                                           Thus, we conclude that                 11    is the largest of the
            4    6   5     9
      =    30 + 30 + 30 + 30                                                               choices.
           4+6+5+9                                                                                                  1       5      2
      =        30                                                                     39. a. The fact that 4 < 8 < 3 is evident from the
           24
      =    30                                                                              following two comparisons:
           4
      =    5                                                                                                      ?
                                                                                                                 1<5
                                                                                           Comparison 1:         4 8        Cross multiplying yields
 35. d. The square is divided into 8 congruent
                                                                                           the true statement 8 < 20, so the original
      parts, 3 of which are shaded. Thus, 3 of the
                                          8
                                                                                           inequality is true.
      figure is shaded.
                                                                                                                 5 ? 2
                                                                                           Comparison 2:         8 > 3          Cross multiplying yields
 36. c. First, rewrite both fractions using the least
                                                                                           the true statement 15 < 16, so the original
      common denominator, which is 60. Then,
      subtract:                                                                            inequality is true.
                                                                                                           3            3
                                                                                      40. a. Since 5 (360) = 5 (5 72) = 216, we conclude
      17       5        17 3         5 10        51       50       51 – 50        1
      20   –   6   =    20 3     –   6 10    =   60   –   60   =     60      =   60        that Irma has read 216 pages.
 37. c. Rewrite this as a multiplication problem,                                     41. a. Cancel factors that are common to the
      cancel factors that are common to the numer-                                         numerator and denominator, then multiply:
      ator and denominator, and then multiply:
                                                                                           5      4         5       4        5
                                                                                           8      7    =   4 2      7   =   14
      18            9       18   20          9 2       4.5
       5           20   =    5    9    =      5         9      =8
                                                                                                                            21         42
 38. c. A reasonable strategy is to begin with one of                                 42. d. The reciprocal of 42 is 21 , which is equiva-
      the fractions, say 5 , and compare it to the next
                          8
                                                                                           lent to 2.
      one in the list. Discard whichever is smaller                                   43. b. Two real numbers are in a ratio of 4:5 if
      and compare the remaining one with the next                                          the second number is 5 times the value of the
                                                                                                                 4
      in the list. Repeat this until you reach the end                                     first number. Observe that 1 is 5 times the
                                                                                                                      4    4
      of the list. Doing so results in the following                                       value of 1 .
                                                                                                    5
      three comparisons:                                                              44. c. The remaining 28 (of 42) envelopes need to
                                     5 ? 2                                                 be addressed. Thus, the fraction of envelopes
      Comparison 1:         Cross multiplying yields
                                     8 > 3
                                                                                           that needs to be addressed is 28 = 2 14 = 2 .
                                                                                                                         42   3 14    3
                          the false statement 15 > 16.
                        2
      This implies that 3 is larger.




   156
                                                                ANSWERS & EXPLANATIONS–


45. b. Apply the order of operations:                                                       Set 4 (Page 7)
                   5
                                                                                             49. d. The exponent applies only to 5, not to the
              –    3    ( – 2)                                       10
                                       7       10                                7   5 2
     1+                               (5        3)=1+
                                                                      3
                                                                                (5    3 )
                                                                                                  –1 multiplied in front. So, –53 = –(5                                         5      5)
                                                                    – 170
                   – 170                                                                          = –125.
                                                                                             50. a. By definition, (–11)2 = (–11)                                         (–11) = 121.
                   10
     =1+            3        (– 170 )          (7
                                                5
                                                          5 2
                                                           3    )
                                                                                             51. c. Using the fact that any nonzero base raised
     =1+           10
                           (– 170 )         ( 134 )                                               to the zero power is 1, we have 5(40) = 5(1) = 5.
                    3
               7           14
                                                                                             52. a. Applying the exponent rules yields:
     =1–       3           3                                                                                                                       1             1
              7          3
                                                                                                  (22)–3 = 2(2x – 3) = 2–6 =                       26
                                                                                                                                                            =   64
     =1      –3         14
                                                                                             53. c. Applying the order of operations and the
     =1      –1
              2                                                                                   definition of an exponent yields:
     =1
      2                                                                                           (1 – 3)2        (–2)2                (–2)        (–2)           4       –1
                                                                                                     –8      =     –8              =          –8            =    –8   =    2
46. e. Since there are m men in a class of n students,
                                                                                             54. b. Applying the order of operations and the
     there must be n – m women in the class. So, the
                                                                                                  definition of an exponent yields:
     ratio of men to women in the class is n mm .
                                             –                                                    –5(–1 – 5– 2) = –5(–1 – 215 ) = –5(– 25 –
                                                                                                                                       25
                                                                                                                                                                                 1
                                                                                                                                                                                25 )   =
                                                                            1
47. d. Compute the difference between        and each                       2                     –5(– 26 ) = 5( 5 5 = 256
                                                                                                       25
                                                                                                                 26

     of the four choices. Then, compare the absolute                                         55. c. First, apply the definition of a negative
     values of these differences; the choice that pro-                                            exponent to simplify the first term within the
     duces the smallest difference is the one closest                                             brackets. Next, rewrite the resulting first term
     to 1 . The differences are as follows:
        2                                                                                         using the fact that “a product raised to a power
     2   1   4   3    1                                                                           is the product of the powers.” Then, simplify:
     3 – 2 = 6 – 6 = 6
      3   1    3    5      2
     10 – 2 = 10 – 10 = – 10                     = –1                                                        –2                2                        2             2
                                                    5                                             – –3
                                                                                                     2            –     2
                                                                                                                        3              = – –2
                                                                                                                                            3               –    2
                                                                                                                                                                 3        =
     5   1   5   3    2   1
     6 – 2 = 6 – 6 = 6 = 3                                                                                   2         2               2                    2             2
                                                                                                                  2                2                    2             2
     3   1    6    5    1                                                                         – –1                     –               =–                   –              =0
     5 – 2 = 10 – 10 = 10
                                                                                                                  3                3                    3             3


     The smallest absolute value of these four dif-                                          56. b. First, apply the definition of a negative
     ferences is 110 . Of the four choices, the one                                               exponent to simplify the two terms to which it
     closest to 1 is 3 .
                2      5                                                                          applies. Then, apply the order of operations:
48. c. Applying the order of operations, first sim-                                                                   (– 1 )2                               (– 1 )2
                                                                                                  –(– 1 )–3 –
                                                                                                                         3
                                                                                                      2                9–2
                                                                                                                                   = –(–2)3 –                  3

     plify the exponentiated term. Then, multiply                                                                                                           ( 1 )2
                                                                                                                                                              9

     left to right. Finally, compute the resulting dif-                                                                                       –1 –1
                                                                                                                                               3  3

     ference by first rewriting both fractions using                                              = –[(–2)(–2)(–2)] –                          1
                                                                                                                                               9
                                                                                                                                                        1
                                                                                                                                                        9
     the least common denominator, which is 12:                                                                        1
                                                                                                                       9
                             2
                                                                                                  = –[–8] –
         5               1              5             1         35          3                                          1
     7   6    –3         2       =7     6    –3       4    =    6      –    4   =                                     81

                                                                                                  =8– 1
                                                                                                      9
                                                                                                                  81
                                                                                                                   1
     35 2          3 3
      6 2     –    4 3       = 70 –
                               12
                                        9
                                        12   =   61
                                                 12
                                                                                                  =8–9
                                                                                                  =–1


                                                                                                                                                                                157
                                                                ANSWERS & EXPLANATIONS–


57. d. Apply the order of operations and exponent                                          1. So, of the four expressions provided, the one
     rules:                                                                                with the largest value is p–1.
                                                                     1 –1             62. c. The reciprocal of a fraction p strictly
     –( 2 )0 (–32 + 2– 3)–1 = –1 ( – 9 +
        5                                                            23 )
                                                                                           between 0 and 1 is necessarily larger than 1.
                                               = –1         ( – 9 + 1 )–1
                                                                        8
                                                               72       1 –1
                                                                                           So, p– 1 1. Raising a fraction p strictly
                                               = –1         (– 8 + 8 )
                                                                                           between 0 and 1 to a positive integer power
                                               = –1         (– 781 )–1                     results in a fraction with a smaller value. (Try
                                               = –1         (– 781 )                       this out with p = 1 .) We know that 0 p3
                                                                                                               2
                                                   8
                                               =   71                                      p2 p 1. Therefore, of the four expressions
58. c. Apply the order of operations and exponent                                          provided, the one with the smallest value is p3.
     rules:                                                                           63. b. Note that the expressions p, p3, and p–1 are
                                                                  3 –2
                                         1                   1
     4– 2(1 – 2(–1)– 3)– 2 =             42     1–2         –1           =                 all negative since it is assumed that p is a frac-
                                                                                           tion between –1 and 0. Since squaring a nega-
      1                                1                     1 1
     16 (1     – 2( – 1))– 2 =         16 (3)
                                             –2
                                                    =       16 32    =                     tive fraction results in a positive 1, we conclude
      1
           =     1
                      = 12–2                                                               that p2 is positive and is, therefore, the largest
     144        122
59. d. Apply the order of operations and exponent                                          of the four choices.
     rules:                                                                           64. c. Raising a fraction strictly between 1 and 2 to

                    (–13 +(–1)3) – 2        1                   (– 1 + (–1))–2
                                                                                           a positive integer power results in a larger frac-
     –2 – 2 +             –2 2          = – 22 +                                 =
                                                                      –4                   tion. Thus, we know that 1 p p2. Moreover,
                                         1                       1
                (–2)–2                 (–2)2                                               the reciprocals of fractions larger than 1 are
     –1                       –1                    –1           4
      4    +      –4     =     4   +    –4      =       4   +   –4   =
                                                                                           necessarily less than 1. In particular, p–1 1,
     –1 –
      4
                1
               16   = – 156                                                                which shows that p–1 is smaller than both p
60. c. Simplify each expression:                                                           and p2. Finally, multiplying both sides of the
     (– 1 )–1 = (–4)1 = –4                                                                 inequality , p–1 1 by p–1 shows that p–2 = p–1
        4
                                                                                           p–1 p–1. Therefore, we conclude that the
     –   3     = – –32 = 3
                         2
           1
       8(– 4 )                                                                             smallest of the four expressions is p–2.
     4(– 1 ) + 3 = –1 + 3 =
         4                               2
     –(– 1 )0 = –(1) = –1                                                            Set 5 (Page 8)
         4
                                                                                      65. a. The quantity n% means “n parts out of
     Hence, the expression with the largest value is
                                                                                           100.” It can be written as 1n or equivalently as
                                                                                                                       00
     4(– 1 ) + 3.
         4                                                                                 n 0.01. Applying this to 40 yields the equiva-
61. d. The reciprocal of a fraction p strictly                                             lent expressions I and II.
     between 0 and 1 is necessarily larger than 1.
                                                                                      66. d. The result of increasing 48 by 55% is given
     So, p–1 1. Also, raising a fraction p strictly
                                                                                           by 48 + 0.55(48) = 74.4.
     between 0 and 1to a positive integer power
                                                                                      67. d. The price resulting from discounting $250
     results in a fraction with a smaller value. (To
                                                                                           by 25% is given by $250 – 0.25($250). This
     see this, try it out with p = 1 .) As such, both p2
                                   2
           3 are less than p and are not larger than
                                                                                           quantity is equivalent to both 0.75 $250 and
     and p
                                                                                           (1 – 0.25) $250.



  158
                                        ANSWERS & EXPLANATIONS–


68. b. The point A is exactly halfway between –2               Next, we compare these two fractions. To this
     and –3 on the number line; therefore, its value           end, note that – 11 – 21 is equivalent to 11
                                                                                  00     00                   00
                                                                    1
     is –2.5.                                                      200 . Cross multiplying in the latter inequal-
69. c. Since the digit in the thousandths place is 8,          ity yields the true statement 200 > 100, so the
     we round the digit in the hundredths place up             inequality is true. Since –0.005 is clearly less
     by 1, resulting in 117.33.                                than 1.01, we conclude that –0.005 is between
70. b. We must determine the value of n for which              –0.01 and 1.01.
      n                                                                              5    2   25   16   9
            300 = 400. The value of n that satisfies       77. b. Observe that 8 – 5 = 40 – 40 = 40 = 0.225.
     100
     this equation is 133 1 . So, we conclude that
                          3                                78. c. Observe that (3.09 1012) 3 = 3
                                                                                                   3.09
                                                                                                          1012
     133 1 % of 300 results in 400.
         3                                                     = 1.03 1012. Alternatively, you could first
71. d. Starting with 0.052, moving the decimal                 rewrite 3.09 1012 as 3,090,000,000,000 and
     place to the left one unit to obtain 0.0052 is            divide by 3 to obtain 1,030,000,000,000, which
     equivalent to dividing 0.052 by 10. Therefore,            is equivalent to 1.03 1012.
     0.0052 is smaller than 0.052.                         79. b. Move the decimal place to the right until
72. c. The phrase “400% of 30” is equivalent to the            just after the first nonzero digit; each place
     mathematical expression 400100     30. Simplifying        moved contributes an additional –1 power of
     this expression yields 120.                               10. Doing so in 0.0000321 requires that we
                         3
73. c. Note that x = 8 = 0.375, which satisfies the            move the decimal place 5 units to the right, so
     condition 0.34 < x < 0.40. It also satisfies the          that 0.0000321 is equivalent to 3.21 10–5.
     condition 156 < x < 290 , which is seen by per-       80. c. We must determine the value of n for which
                                                                 n     8
     forming the following two comparisons using                100    9   = 1 . Solve for n, as follows:
                                                                             3
     cross multiplication:                                       n     8       1
                                                                100    9   =   3
                             ?                                   n     9       1   3
     Comparison 1: 156 3 Cross multiplying yields
                           8                                    100=   8       3 = 8
                                                                               3   300
     the true statement 40 < 48, so the original               n = 100         8 = 8     = 37.5
     inequality is true.                                                                                8
                         ?                                     Thus, we conclude that 37.5% of          9   is 1 .
                                                                                                               3
     Comparison 2: 3 290 Cross multiplying yields
                      8
     the true statement 60 < 72, so the original          Set 6 (Page 10)
     inequality is true.
                                                           81. a. Apply the order of operations as follows:
                                 22.5
74. b. 22.5% is equivalent to 100 , which is equal
     to 0.225.                                                 –2(–3)2 + 3(–3) – 7 = –2(9) –9 – 7 = –18 – 9 –7
                   2             3                3            = –34
75. d. Note that 5 = 0.40 and 7 ≈ 0.42857. So, 7 is
     not less than 2 .                                     82. b. Apply the order of operations as follows:
                   5
                                                                   7(–2)           –14
76. b. To see that – 0.01 < – 0.005, first convert
                                                                (–2)2 + (–2)
                                                                               =   4–2   = – 14 = –7
                                                                                              2
     both to their equivalent fractional form:

     –0.01 = – 11 – 0.005 = – 10500 = – 21
                00                       00




                                                                                                                     159
                                                                     ANSWERS & EXPLANATIONS–


83. b. Apply the order of operations as follows:                                                 94. b. Apply the order of operations as follows:
                                                                                                       1 6                     1                  1
    2(3)(6) – ( – 8) = 36 + 8 = 44                                                                     2 [( 2   – 3) – 4(3)] = 2 [(3 – 3) – 12] = 2 [–12] = – 6
84. c. Apply the order of operations as follows:                                                 95. d. Apply the order of operations as follows:

    y = –(– 3)3 + 3(–3) – 3 = –(–27) –9 – 3 =                                                          (–8)2 – 4(3)2( 1 ) = 64 – 4(9)( 1 )= 64 – 18 = 46
                                                                                                                      2                2
    27 – 9 – 3 = 15                                                                              96. a. Apply the order of operations as follows:
85. b. Apply the order of operations as follows:
                                                                                                       3(6)2(–5)(5(3) – 3(–5)) = 3(36)(–5)(15                                                               15)
                                                                             1
    (–5)(6) + (–8)      = –30 – 8       (1)
                                         2                                   2   =                     = 3(36)(–5)(30) = –16,200
    –30 –8 2 = –30 – 16 = –46
86. b. Apply the order of operations as follows:                                                Set 7 (Page 12)
                                                                                                                  3
                                                                                                            3x2                  33x6
     62
      – 4(6) + 10 =                    36
                                             – 4(6) + 10 = 12 – 24 + 10                          97. b. x2x4 = x6 = 33 = 27
     3                                  3
    = –2                                                                                         98. d. (4w9)3 = 43w27 = 64w 27

87. a. Apply the order of operations as follows:                                                 99. b. Note that the power –2 does not apply to the 6
                                                                                                       since it is not enclosed in the parentheses to which
                                                                1            1
    4(2–2                     –2
             )(2(2) )(3( – 2) )=4                2
                                                                4    2       4       (3 4) =           the exponent applies. Therefore, 6(e– 2)– 2 =6e4.
     4 2 3 4
                          =6                                                                                                                                               –45a4b9c5
       4 4                                                                                      100. a. (–45a4b9c5)   (9ab3c3) = 9ab3c3 = –5a3b6c2
88. a. Apply the order of operations as follows:
                                                                                                101. d. 4(3x3)2 = 4(32x6) = 36x6
                     12                                                                                                      4                    4
    7(6) +            6   – (–8) = 42 + 2 + 8 = 52                                              102. d.
                                                                                                             (ab)3
                                                                                                                             =         a3b3
                                                                                                                                                       = (a3b2)4 = a12b8
                                                                                                               b                        b
89. b. Apply the order of operations as follows:
                                                                                                                   y
                                                                                                           ( x )2( x )–2
                                                                                                             y
                                                                                                                                      x2    y–2          x2           x2            x4
                                   2                   2            64                                                                y2    x–2          y2           y2            y4          x4             x3
    (3(2)(5) + 2)                  5    =(32)          5       =    5    = 12.8                 103. b.         xy               =         xy      =             xy        =    xy        =             =
                                                                                                                                                                                              y4(xy)           y5

90. d. Apply the order of operations as follows:                                                            2a         a–1                  2a          2b
                                                                                                104. d.     b         (2b)–1
                                                                                                                                       =    b           a    = 4baab = 4
                                       –2                                    –2
          7                 3                     7               3
        5(–2)2
                     +    10(–2)
                                        =        5.4       +    10(–2)            =
                                                                                                105. c. 3x 2y(2x 3y 2) = 6x5y3
                                            –2                 –2                                           a        b           1                a2         b–2            a            a2    a2      a        a5
         7
             –    3 –2
                              =     4
                                             =         1
                                                                   = 52 = 25                    106. e. ( b )2( a )–2( a )–1 = 2                                            1   =                      1   =
        20       20                20                  5                                                                       b                             a–2                         b2    b2               b4
                                                                                                                         2                                   2
                                                                                                107. c. (3xy5) – 11x2y2(4y4) = 32x2y10 –
91. c. Apply the order of operations as follows:
                                                                                                       11x2y2 42y8 = 9x2y10 – 176x2y10 = 167x2y10
     6(2)2            4(2)         6 4               4(2)           24       8
     2(3)2       +    3(3)     =   2 9       +       3(3)      =    18   +   9       =                     2(3x2y)2(xy)3                    2(32x4y2)(x3y3)                          18x7y5
                                                                                                108. a.
                                                                                                              3(xy)2
                                                                                                                                       =        3(x2y2)
                                                                                                                                                                                =     3x2y2
                                                                                                                                                                                               =6x5y3
     24          16           40        20
     18      +   18       =   18   =     9                                                                 (4b)2x –2                 42b2x –2               16b2                           16b2              4
                                                                                                109. c.                      =                     =                                =               =
92. c. Apply the order of operations as follows:                                                           (2ab2x)2                  22a2b 4x2           22a 2b 4x 2x 2                   4a2b4x4          a2b2x4

                                                                                                110. a. “The product of 6x2 and 4xy2 is divided by
    (1)(–1) + –11 + (1)2 – (–1)2 = –1 – 1 + 1 – 1 = –2                                                                                                                                         (6x2)(4xy2)
                                                                                                       3x3y” can be expressed symbolically as                                                     3x3y
                                                                                                                                                                                                                    ,
93. b. Note that if x = 2, then y = –2. Now, apply
                                                                                                       which is simplified as follows:
    the order of operations as follows:
                                                                                                        (6x2)(4xy2)                  24x3y2
    (((2)(–2)–2)2 = ((–4)– 2)2 = (–4)– 2                                         2
                                                                                     = (–4)–4              3x3y
                                                                                                                             =        3x3y
                                                                                                                                                  = 8y
            1           1
    =     (–4)4      = 256


  160
                                                                  ANSWERS & EXPLANATIONS–



111. a. The expression described by the phrase “3x2                                                                            2w(z + 1)
                                                                                        125. c. 3(z + 1)2w3 – ((z + 1)w2)–1
                                                                      3
      is multiplied by the quantity 2x y raised to the
      fourth power” can be expressed symbolically                                             = 3(z + 1)2w3 – 2w(z + 1) ((z + 1)w2)
      as (3x2)(2x3y)4, which is simplified as follows:                                        = 3(z + 1)2w3 – 2(z + 1)2w3
      (3x2)(2x3y)4 = (3x2)(24x12y4) =48x14y4                                                  = (z + 1)2w3
112. b. “The product of –9p3r and the quantity                                          126. a.
                                                                                                                                 2y(4x + 1)2         –2
      5p – 6r” can be expressed symbolically as                                                –2(4x +1)5y –5 – ((4x + 1)y–2)–3
      (–9p3r)(5p – 6r), which is simplified using                                                      2(4x +1)5             2y(4x + 1)2 –2
      the distributive property as follows:                                                   = –         y5
                                                                                                                        –   ((4x + 1)–3y 6
      (–9p3r)(5p – 6r) = –45p4r + 54p3r2.                                                              2(4x +1)5            2(4x + 1)5 – 2
                                                                                              = (–        y5
                                                                                                                        –       y5
                                                                                                                                      )
Set 8 (Page 13)                                                                                        4(4x +1)5 –2
                                                                                              = –         y5
113. c. 5ab4 – ab4 = 4ab4
                                                                                                           y5           2
114. a. 5c2 + 3c – 2c2 + 4 – 7c = (5c2 – 2c2) +                                               = – 4(4x + 1)5
      (3c – 7c) + 4 = 3c2 – 4c + 4
                                                                                                    (–1)2y10
115. c. – 5(x – ( – 3y)) + 4(2y + x) = – 5(x + 3y) +                                          =    2
                                                                                                  4 (4x + 1)10
      4(2y + x) = – 5x – 15y + 8y + 4x = – x – 7y                                                     y10
                                                                            2       2         =   16(4x + 1)10
116. b. Gather like terms, as follows: 3x + 4ax – 8a
      + 7x2 – 2ax + 7a2 = (3x2 + 7x2) + 4ax – 2ax) +                                    127. b.
                                                                                                                                                          –1
      (–8a2 + 7a2) = 10x2 + 2ax – a2                                                                                                         1 2y6
                                                                                              4z((xy–2)–3 + (x–3y6))–1 –                     z x3
117. d. The base expressions of the three terms used                                                                                                  –1
                                                                                                                                               2y6
      to form the sum 9m3n + 8mn3 + 2m3n3 are dif-                                            = 4z((x–3y6) + (x – 3y6))–1 –                    zx3
      ferent. So, they cannot be combined.                                                                                              –1
                                                                                                                                 2y6
                     6                              6            6              6             = 4z(2(x–3y6))–1 –                 zx3
118. d. – 7g + 9h + 2h – 8g =(–7g – 8g ) +
      (9h + 2h) = –15g6 + 11h                                                                                                     zx3
                                                                                              = 4z(2–1x3y – 6))–1 –               2y6
119. b. (2x2)(4y2) + 6x2y2 = 8x2y2 + 6x2y2 = 14x2y2
120. c. (5a2 3ab) + 2a3b = 15a3b + 2a3b = 17a3b                                                   4zx3          zx3
                                                                                              =   2y6
                                                                                                          –     2y6
                             3x–1                         2       3             1
121. b. 2x – 3 – x4 – (x3)–1 = x3 – x5 – x3 =
                                                                                                  3zx3
                                                                                              =   2y6
       1            3
       x3
            –       x5   = x–3 –3x–5
                                                                                        128. a.
122. a. (ab2)3 + 2b2 – (4a)3b6 = a3b6 + 2b2 – 43a3b6                                                                            5x4
                                                                                              (0.2x–2)–1 + 2 x2 –
                                                                                                           5                   (2x)2
                2              3 6
      = 2b – 63a b
                                                                                                                –1
                                                                                                    2                             5x4
123. b.
            (–3x–1)–2
                              + 8 (x2)2 =
                                9
                                                      x2
                                                                +         8x4
                                                                           9    =             =    10    x–2          + 2 x2 –
                                                                                                                        5         4x2
               x–2                                 (–3x–1)2
          x2                  8x4         x4       8x4
       (–3)2x–2
                         +     9    =     9    +    9     = x4                                = 5x2 + 2 x2 – 5 x2
                                                                                                      5      4
                                                b
                                                                                              = ( 12000 +       8       25 2
                                    –2                   –2
124. b. –(–a–2bc–3) + 5                                       = –(a4b–2c6) +                                   20   –   20 )x
                                               a2c3
                                                                                                  83 2
             b–2                    4 6
                                               5a4c6          4a4c6                           =   20 x
      5      –4 –6
            a c
                             = – abc +
                                   2
                                                b2
                                                         =     b2



                                                                                                                                                               161
                                       ANSWERS & EXPLANATIONS–



Set 9 (Page 15)                                             135. c. The sum of three numbers is represented by
129. b. According to the order of operations, we                  (a + b + c). The reciprocals of these numbers
      perform exponentiation first, then multiplica-
                                                                      1 1                                           1
                                                                  are a , b and 1 . The sum of the reciprocals is ( a
                                                                                c
      tion, and then subtraction. The square of a
                                                                    1
                                                                  + b + 1 ), and so, the product of these two
                                                                          c
      number x is x2. Four times this quantity is 4x2,
                                                                                                 1   1
                                                                  sums is given by (a + b + c)( a + b + 1 ).
                                                                                                         c
      and finally, two less is 4x2 – 2.                     136. c. The expression 3x is described by the phrase
130. d. First, note that 25% of V is equal to 0.25V,              “3 times a number.” So, the sum of 3x + 15 is
      which is equivalent to 1 V. Since the original              described by the phrase “15 more than 3 times
                              4
      volume of the tank is being increased by this               a number.” Finally, since the word “is” is inter-
      quantity, we add it to the original volume V.               preted as “equals,” we conclude that the given
      This results in the equivalent expressions V +              equation is described by choice c.
      0.25V and V + 1 V. Adding the coefficient of V
                       4
                                                            137. c. The cost for x desks, each of which costs D
      in the first expression yields another equiva-              dollars, is xD. Similarly, the costs for the chairs
      lent expression 1.25V. So, they are all correct             and file cabinets are yE and zF, respectively.
      choices.                                                    Thus, the total cost T is equal to xD + yE + zF.
131. b. The amount of money paid for the total              138. a. First, increasing d by 50% is described by
      number hours of tutoring is $40h. Adding the                the expression d + 0.50d,which is equal to
      one-time fee of $30 to this amount results in               1.50d. Now, a decrease of this quantity by 50%
      x = $30 + $40h.                                             is described by the expression
132. a. According to the order of operations, we                  1.50d – 0.50(1.50d) = 1.50d – 0.75d = 0.75d
      interpret parenthetical quantities first, then
      multiply, and then subtract. The quantity the               This value is 75% of the original value d. Hence,
      sum of a number and 5 is (x + 5). Then, three               it is 25% smaller than d.
      times this quantity is 3(x + 5), and finally nine     139. a. First, since there are w weeks in one month,
      less results in 3(x + 5) – 9.                               the number of weeks in m months must be
133. a. The total cost for a phone call lasting x min-            mw. Since we are told that there are m months
      utes is the cost for the first minute plus the cost         in one year, the quantity mw represents the
      for the remaining x – 1 minutes. The first minute           number of weeks in one year. Next, since there
      costs $0.35 and the cost for the remaining x – 1            are d days in one week, and the number of
      minutes is $0.15(x – 1). The sum of these results           weeks in a year is mw, we conclude that there
      in the total cost y = 0.15(x – 1) + 0.35.                   are mwd days in one year.
134. b. According to the order of operations, we first      140. d. The phrase “40% of j” is expressed symboli-
      interpret parenthetical quantities and then mul-            cally as 0.40j and the phrase “50% of k” is
      tiply. The difference between a number and five             expressed symbolically as 0.50k. Equating
      is represented by the expression (x – 5). Then,             these quantities yields the statement 0.40j =
                                 1
      half of this difference is 2 (x – 5).                       0.50k. Dividing both sides by 0.40 then results
                                                                  in the equivalent equality j = 0..50 k = 1.25k.
                                                                                                   0 40
                                                                  This says that the value of j is 125% of k. So,
                                                                  we conclude that j is 25% larger than k.


   162
                                       ANSWERS & EXPLANATIONS–


141. b. The phrase “p percent” can be represented           146. e.
                        p                                        k
      symbolically as         such, since we are
                       100 . As                                  8    =8
      decreasing q by this quantity, the resulting               k
                                                                       8=8 8
                                                                 8
      quantity is represented by q – 1p .
                                       00
                                                                 k = 64
142. d. The cost of the three meals is (a + b + c)
                                                            147. a.
      and a 15% tip is represented by. 0.15(a + b + c).
                                                                 –7k – 11 = 10
      This latter value is added to the cost of the three
                                                                 –7k – 11 + 11 = 10 + 11
      meals to obtain the total cost of the dinner,
                                                                 –7k = 21
      namely (a + b + c) + 0.15(a + b + c) = 1.15(a +
      b + c). Now, splitting this cost evenly between            –7k – 1 = 21 – 1
                                                                       7        7
      the two brothers amounts to dividing this                  k = –3
      quantity by 2; this is represented by choices b       148. c.
      and c.                                                     9a + 5 = –22
143. A 75% increase in enrollment E is represented               9a + 5 – 5 = –22 – 5
      symbolically as 0.75E, which is equivalent to              9a = –27
      3
      4 E. Adding this to the original enrollment E              9a     1
                                                                               = –27    1
                                                                        9               9
      results in the sum E + 3 E, which is the new
                              4
      enrollment.                                                a = –3
                                                            149. d.
144. a. The total cost of her orders, before the dis-
                                                                 p
      count is applied, is represented by the sum W              6    + 13 = p – 2
                                                                 p
      + X + Y + Z. A 15% discount on this amount                 6    + 13 – 13 = p – 2 – 13
      is represented symbolically as 0.15(W + X +                p
                                                                 6    = p – 15
      Y+ Z). So, her total cost is (W + X + Y+ Z).               p
                                                                 6    – p = p – p – 15
      (W + X + Y+ Z) – 0.15(W + X + Y+ Z), which
      is equivalent to 0.85 (W + X + Y+ Z).                      – 56p = –15
                                                                 –5 p – 6 = –15 – 6
                                                                    6   5         5

                                                                 p = 18
 Section 2—Linear Equations                                 150. a.
       and Inequalities
                                                                 2.5p + 6 = 18.5
Set 10 (Page 18)                                                 2.5p + 6 – 6 = 18.5 – 6
145. a.                                                          2.5p = 12.5
                                                                        12.5
      z – 7 = –9                                                 p=      2.5   =5
      z – 7 + 7 = –9 + 7                                    151. a.
      z = –2                                                     3x
                                                                      =   15
                                                                 10       25
                                                                 3x 10            15 10
                                                                 10 ( 3 )     =   25 ( 3 )
                                                                        150
                                                                 x=      75    =2




                                                                                               163
                                              ANSWERS & EXPLANATIONS–


152. b.                                                     158. d.
      2.3(4 – 3.1x) = 1 – 6.13x                                  1.3 + 5x – 0.1 = –1.2 – 3x
      9.2 – 7.13x = 1 – 6.13x                                    1.2 + 5x = –1.2 – 3x
      9.2 – 7.13x + 7.13x = 1– 6.13x + 7.13x                     1.2 + 5x – 1.2 = –1.2 – 3x – 1.2
      9.2 = 1 + x                                                5x = –2.4 – 3x
      9.2 – 1 = 1 – 1 + x                                        5x + 3x = –2.4 – 3x + 3x
      8.2 = x                                                    8x = –2.4
                    41                                           x = –0.3
      Since 8.2 =   5 , the      correct choice is b.
                                                            159. d.
153. d. An application of the distributive property
                                                                 4(4v + 3) = 6v – 28
      shows that 33c – 21 = 3(11c – 7). So, since
                                                                 16v + 12 = 6v – 28
      11c – 7 = 8, we conclude that 33c – 21 = 3(8) = 24.
                                                                 16v + 12 – 12 = 6v – 28 – 12
154. d.
      x   1
                                                                 16v = 6v – 40
      2 + 6x = 4                                                 16v – 6v = 6v – 6v – 40
      6x    2x
      12 + 12 = 4                                                10v = –40
      8x                                                         v = –4
      12 = 4
      8x 12       12                                        160. b.
      12 8 = 4 8
                                                                 13k + 3(3 – k) = –3(4 + 3k) – 2k
      x=6
                                                                 13k + 9 – 3k = – 12 – 9k – 2k
155. d.
            5     2
                                                                 10k + 9 = –12 – 11k
      b–    2 = –3                                               10k + 9 – 9 = – 12 – 11k – 9
      b   –5 + 5 = –2
           2    2    3   +   5
                             2                                   10k = – 21 – 11k
              4   15
      b   = –6 + 6                                               10k + 11k = – 21 – 11k + 11k
      b   = 161                                                  21k = – 21
                                                                        1            1
156. c.                                                          21k   21   = –21   21
      3c
      4 –9=3                                                     k = –1
      3c
      4 –9 + 9 = 3 +     9
      3c                                                    Set 11 (Page 19)
      4 = 12
      3c 4       4                                          161. a.
      4 3 = 12 3
                                                                 –2(3v + 5) = 14
      c = 16                                                     –6v – 10 = 14
157. b.                                                          –6v = 14 + 10 = 24
      2a
                                                                     24
      3 = –54                                                    v = –6 = –4
      –2a –3 =
       3   2        –54 – 3
                          2

      a = 81




   164
                                         ANSWERS & EXPLANATIONS–


162. b.                                                  166. b.
      5
      2 (x – 2) + 3x = 3(x + 2) –   10                         0.8(x + 20) – 4.5 = 0.7(5 + x) – 0.9x
      5                                                        8(x +20) – 45 = 7(5 + x) – 9x
      2 x – 5 + 3x = 3x + 6 – 10
      11                                                       8x + 160 – 45 = 35 + 7x – 9x
       2 x – 5 = 3x – 4
                                                               8x + 115 = 35 – 2x
      11
       2 x – 3x – 5 = –4                                       10x = –80
      5
      2 x = –4 + 5 = 1                                         x=–8
      x= 2 5
                                                         167. e. First, we solve the equation 4x + 5 = 15 for x:

163. c. Let x be the unknown number. The sentence              4x + 5 = 15
      “Twice a number increased by 11 is equal to 32           4x = 10
      less than three times the number” can be                 x = 140 = 2.5
      expressed symbolically as 2x + 11 = 3x – 32.
      We solve this equation for x, as follows:                Now, substitute x = 2.5 into the expression
                                                               10x + 5 to obtain 10(2.5) + 5 = 25 + 5 = 30.
      2x + 11 = 3x – 32
                                                         168. d. Let x be the unknown number. 40% of this
      2x = 3x – 32 – 11
                                                               number is represented symbolically as 0.40x.
      2x – 3x = –43
                                                               Therefore, the sentence “Ten times 40% of a
      –x = –43
                                                               number is equal to 4 less than six times the
      x = 43
                                                               number” can be expressed as the equation
164. d.                                                        10(0.40x) = 6x – 4. We solve this equation
      4a + 4
        7      = – 2 –43a                                      for x:
               4a + 4
      28 ( 7 ) = 28 (– 2 –43a )
                                                               10(0.40x) = 6x –4
      4(4a + 4) = –7(2 – 3a)
                                                               4x = 6x – 4
      16a + 16 = –14 + 21a
                                                               4x + 4 = 6x
      16 = –14 + 21a – 16a
                                                               4 = 2x
      16 + 14 = 5a
                                                               x=2
      30 = 5a
                                                         169. b. Let x be the unknown number. The sentence
      a=6                                                        7
                                                               “ 8 of nine times a number is equal to ten times
165. a. Let x be the smaller of the two unknown
                                                               the number minus 17” can be expressed as the
      integers. The next consecutive even integer is
                                                               equation 7 (9x) = 10(x – 17). Solve this equa-
                                                                          8
      then x + 2. The sentence “The sum of two con-
                                                               tion for x:
      secutive even integers is 126” can be expressed
                                                               7
      symbolically as x + (x + 2) = 126. We solve this         8 (9x) = 10(x   – 17)
      equation for x:                                          8 7 (9x) = 8
                                                                   8            10(x – 17)
                                                               63x = 80(x – 17)
      x + (x + 2) = 126
      2x + 2 = 126                                             63x = 80x – 1360
      2x = 124                                                 –17x = –1360
      x = 62                                                   x = 80

      Thus, the two integers are 62 and 64.
                                                                                                        165
                                      ANSWERS & EXPLANATIONS–


170. d.                                                 175. d. Let x be the unknown number. A 22.5%
           7b – 4
      a=     4                                                decrease in its value can be expressed symboli-
                    7b – 4                                    cally as x – 0.225x = 0.775x. We are given that
      4a = 4          4
                                                              this quantity equals 93, which can be expressed
      4a = 7b – 4                                             as 0.775x = 93. We solve this equation for x:
      4a + 4 = 7b
           4a + 4                                             0.775x = 93
      b=     7                                                        93
                                                              x=    0.775     = 120
171. b.
      2x + 8   5x – 6                                   176. c. The scenario described in this problem can
        5 = 6
                                                              be expressed as the equation 4(x + 8) + 6x =
             +
      30 2x 5 8 = 30         5x – 6
                               6                              2x + 32. We solve this equation for x:
      6(2x + 8) = 5(5x – 6)
      12x + 48 = 25x –30                                      4(x + 8) + 6x = 2x + 32
      12x = 25x – 78                                          4x + 32 + 6x = 2x + 32
      –13x = –78                                              10x + 32 = 2x + 32
      x=6                                                     8x = 0
172. b. Let x be the unknown number. The sentence
                                                              x=0
      “When ten is subtracted from the opposite of a
                                                        Set 12 (Page 21)
      number, the resulting difference is 5” can be
                                                        177. c.
      expressed symbolically as the equation –x –10 =
                                                              1
      5. We solve this equation for x as follows:             2x    –4        x+8
                                                                   3     =     5
                                                                         1
      –x – 10 = 5                                                        2x   –4           x+8
                                                              15             3      = 15    5
      –x = 15
      x = –15                                                 5 ( 1 x –4) = 3(x + 8)
                                                                  2
173. b.
                                                              5
                                                              2x   – 20 = 3x + 24
             8
      9x +   3   = 8x + 9
                   3
      3 (9x + 8 ) = 3 ( 8 x + 9)
               3        3                                     2 ( 5 x – 20) = 2(3x + 24)
                                                                  2
      27x + 8 = 8x + 27
                                                              5x – 40 = 6x + 48
      27x – 8x + 8 = 27
      19x = 27 – 8 = 19                                       5x – 88 = 6x
      x=1                                                     –88 = x
                                                  9
174. c. Substitute F = 50˚ into the formula F =   5C
                                                        178. a.
      + 32 and then solve the resulting equation for          5x – 2[x – 3(7 – x)] = 3 – 2(x – 8)
      C, as follows:                                          5x – 2x + 6(7 – x) = 3 – 2(x – 8)
      50 = 9 C + 32
            5
                                                              5x – 2x + 42 – 6x) = 3 – 2x – 16
      5 50 = 5 ( 9 C + 32)
                 5
                                                              –3x + 42 = 19 – 2x
      250 = 9C + 160                                          23 = x

      90 = C
      C = 10

   166
                                     ANSWERS & EXPLANATIONS–


179. d.                                                183. b. 30% of r is represented symbolically as 0.30r,
      ax + b = cx + d                                        and 75% of s is represented symbolically as
      ax – cx =d – b                                         0.75s. The fact that these two quantities are
      (a – c)x = (d – b)                                     equal is represented by the equation 0.30r =
       (a – c)x   (d – b)                                    0.75s. We are interested in 50% of s when r =
        (a – c) = (a – c)
            d–b
                                                             30. So, we substitute r = 30 into this equation,
      x = a–c                                                solve for s, and then multiply the result by 0.50:
180. a. Let x be the smallest of the four whole num-
                                                             0.30(30) = 0.75s
      bers. The next three consecutive odd whole
                                                             9 = 0.75s
      numbers are then x +2, x + 4, and x + 6. The
                                                             s = 0.9 = 12
                                                                   75
      sentence “The sum of four consecutive, odd
      whole numbers is 48” can be expressed as the           So, 50% of s is equal to 0.50(12) = 6.
      equation x + (x + 2) + (x + 4) + (x + 6) = 48.   184. e. We must solve the given equation for g:
      We solve this equation for x as follows:
                                                             fg + 2f – g = 2 – (f + g)
      x + (x +2) + (x + 4) + (x + 6) = 48                    fg + 2f – g = 2 – ( f + g)
      4x + 12 = 48                                           fg + 2f – g = 2 – f – g
      4x = 36                                                fg = 2 – f – g– 2f + g
      x=9                                                    fg = 2 – 3 f
      Thus, the smallest of the four whole numbers           g = 2 –f 3f
      is 9.                                            185. b. Let x be the width of the room. Then,

181. a. In order to solve for T, we must simply              the length of the room is equal to 2x + 3.
      divide both sides of the equation by nR. This          The perimeter of the room is given by 2x +
      results in the equation T = PV\nR.                     2(2x + 3). Since this quantity is known to be
                                                             66, we must solve the equation 2x + 2(2x + 3)
182. a.
            C+A                                              = 66 as follows:
      B=    D–A
      B = (D – A) = C + A                                    2x + 2(2x + 3) = 66
      BD – BA = C + A                                        2x + 4x + 6 = 66
      BD – C = A + BA                                        6x + 6 + 66
      BD – C = A (1 + B)                                     6x = 60
            BD – C                                           x = 10
      A=     1+B

                                                             Thus, the length of the room is 2(10) + 3 = 23
                                                             feet.




                                                                                                       167
                                                   ANSWERS & EXPLANATIONS–


186. b.                                                                         Plugging this in for b in the expression a = –2b
      4 – 2x      1–y
        3 –1= 2                                                                 yields a = –2(–3) = 6. Finally, we substitute
      6 4 –32x – 6 1         =6      1–y
                                      2                                         these numerical values for a and b into       b
                                                                                                                                  2
                                                                                                                                      to
                                                                                                                              a
                                                                                               2           2
      2(4 – 2x) – 6 = 3(1 – y)                                                  obtain    –3
                                                                                                   = –1        = 1.
                                                                                           6          2          4
      8 – 4x – 6 = 3 – 3y
                                                                          189. d. Let x be the unknown number. The sentence
      2 – 4x = 3 – 3y                                                           “Three more than one-fourth of a number is
      3y = 1 + 4x                                                               three less than the number” can be expressed
      y=    1 + 4x                                                              as the equation 1 x + 3 = x – 3. We must solve
                                                                                                 4
              3
                                                                                this equation for x as follows:
187. e. Let x be the smallest of five consecutive odd
                                                                                1
      integers. The next four consecutive odd integers                          4x +3=x–3
      are given by x +2, x + 4, x + 6, and x + 8. The                           4 ( 1 x + 3) = 4 (x – 3)
                                                                                     4
      average of these five integers is equal to their                          x + 12 = 4x – 12
      sum divided by 5, which is expressed symboli-                             x + 24 = 4x
      cally by   x + (x + 2) + (x + 4) + (x + 6) + (x + 8)
                                                           . Since              24 = 3x
                                     5
      this quantity is given as –21, we must solve the                          x=8

      equation       x + (x + 2) + (x + 4) + (x + 6) + (x + 8)
                                                                 = –21,   190. c.
                                         5                                                5x – 2
      as follows:                                                               (2 – x) 2 – x = (2 – x) y
                                                                                5x – 2 = y(2 – x)
       x + (x + 2) + (x + 4) + (x + 6) + (x + 8)
                           5
                                                   = –21                        5x – 2 = 2y – xy
       5x + 20
                 = –21                                                          5x + xy = 2 + 2y
          5
      x + 4 = –21                                                               x(5 + y) = 2 + 2y
                                                                                     2 + 2y
      x = –25                                                                   x=    5+y
                                                                          191. b. Solve this problem by determining the
      Thus, the least of the five integers is –25.
                               a
                                                                                weight of each portion. The sum of the
188. a. First, we solve b + 6 = 4 for a:
      a
                                                                                weights of the initial corn is equal to the
      b   +6=4
                                                                                weight of the final mixture. Therefore,
      a
      b = –2
      a = –2b
                                                                                                    56 pounds
                                                                                (20 bushels)          bushel     +
      Next, substitute this expression for a into the
      equation –6b + 2a – 25 = 5 and solve for b:                               (x bushels)        50 pounds
                                                                                                                =
                                                                                                     bushel
      –6b + 2a – 25 = 5
                                                                                                          54 pounds
      –6b + 2(–2b) – 25 = 5                                                     [(20 + x) bushels]          bushel
      –6b –4b – 25 = 5
      –10b – 25 = 5                                                             Suppressing units yields the equation 20          56
      –10b = 30                                                                 + 50x = (x + 20) 54.
      b = –3



   168
                                     ANSWERS & EXPLANATIONS–


192. d.                                                      (Note: Remember to reverse the inequality
      –5[x – (3 – 4x – 5) – 5x]–22  = 4[2 –(x –3)]           sign when dividing both sides of an inequality
      –5[x – 3 + 4x – 5) – 5x] – 4 = 4[2 – x + 3]            by a negative number.)
      –5[2] – 4 = 4[5 – x]                             199. c.
      –10 – 4 = 20 – 4x                                      –4(x – 1) 2(x + 1)
      –14 = 20 – 4x                                          –4x + 4 2x + 2
      –34 = –4x                                              4 6x + 2
      x = ––344 = 127 = 8.5                                  2 6x
                                                             1   2
                                                             3 = 6    x
Set 13 (Page 22)
                                                             The answer can be written equivalently as
193. c.
                                                             x 1.3
      3x + 2     11
      3x 9                                             200. c.

      x 3                                                    x + 5 3x + 9
                                                             5 2x + 9
194. c.
                                                             –4 2x
      5x    23
           23                                                –2 x
      x     5= 4.6
195. a.                                                      The answer can be written equivalently as
      1 – 2 x –5                                             x –2.
      –2x –6                                           201. d.
      x 3                                                    –6(x + 1) 60
                                                             –6x – 6 60
      (Note: Remember to reverse the inequality
                                                             –6x 66
      sign when dividing both sides of an inequality
                                                             x –11
      by a negative number.)
196. d. All values to the right of and including –4          (Note: Remember to reverse the inequality
      are shaded. Thus, the inequality that depicts          sign when dividing both sides of an inequality
      this situation is x –4.                                by a negative number.)
197. b.                                                202. b. The right side of the inequality 2x – 4   7
      4x + 4 24                                              (x – 2) can be described as “the product of
      4x 20                                                  seven and the quantity two less than a number,”
      x 5                                                    and the left side can be described as “four less
198. a.                                                      than two times the number.” Reading from
      –8x + 11        83                                     right to left, the quantity on the right side is
      8x 72                                                  greater than the one on the left. Hence, the
      x –9                                                   correct choice is b.




                                                                                                     169
                                      ANSWERS & EXPLANATIONS–


203. a.                                                        step is that when the coefficient of x is nega-
      –x
      0.3     20                                               tive, both inequality signs are switched. We
     x      (–0.3)(20) = –6                                    proceed as follows:

     (Note: Remember to reverse the inequality                 –4     3x – 1 11
     sign when dividing both sides of an inequality            –3     3x 12
     by a negative number.)                                    –1     x≤4
204. d.                                                  208. b. Using the same steps as in question 207,
     –8(x + 3) 2(–2x + 10)                                     proceed as follows:
     –8x – 24 –4x + 20
                                                               10 3(4 – 2x) –2 70
     –24 4x + 20
                                                               10 ( 12 – 6x – 2 70
     –44 4x
                                                               10 10 – 6x 70
     –11 x
                                                               0 – 6x 60
205. b.
                                                               0 x –10
     3(x – 16) – 2 9(x – 2) – 7x
     3x – 48 – 2 9x – 18 – 7x                                  The last compound inequality above can be
     3x – 50 2x – 18                                           written equivalently as –10 x 0.
     x – 50 –18
     x 32                                                Set 14 (Page 24)
206. b.                                                  209. c. Using the fact that |a| = b if and only if a =b,
     –5[9 + (x – 4)] 2(13 – x)                                 we see that solving the equation |–x| – 8 = 0, or
     –5[5 + x] 2(13 – x)                                       equivalently |–x| = 8, is equivalent to solving –x
     –25 – 5x 26 – 2x                                          = 8. We solve these two equations separately:
     –51 – 5x –2x                                              –x = 8               –x = –8
     –51 3x                                                    x = –8               x=8
     –17 x
                                                               So, both –8 and 8 are solutions of this equation.
     The answer can be written equivalently as
                                                         210. a. We rewrite the given equation as an equiva-
     x – 17.
                                                               lent one solved for |x|, as follows:
207. a. When solving a compound inequality for
     which the only expression involving the vari-             2|x| + 4 = 0
     able is located between the two inequality                2|x| = –4
     signs and is linear, the goal is to simplify the           |x| = –2
     inequality by adding/subtracting the constant             The left side must be nonnegative for any
     term in the middle portion of the inequality              value of x (since it is the absolute value of an
     to/from all three parts of the inequality, and            expression), and the right side is negative, so
     then to divide all three parts of the inequality          there can be no solution to this equation.
     by the coefficient of x. The caveat in the latter




   170
                                           ANSWERS & EXPLANATIONS–


211. c. We rewrite the given equation as an equiva-              214. b. First, we rewrite the equation in an equiva-
      lent one solved for |x|:                                         lent form:

      –3|x| + 2 = 5|x| – 14                                            –6(4 – |2x + 3|) = –24
      –3|x| + 16 = 5|x|                                                –24 + 6|2x + 3| = – 24
      16 = 8|x|                                                        6|2x + 3| = 0
      2 = |x|                                                          |2x + 3| = 0

      Using the fact that |a| = b if and only if a =                   Now, using the fact that |a| = b if and only if
        b, it follows that the two solutions of the                    a = b, we see that solving the equation |2x + 3|
      equation 2 = |x| are x = 2. Thus, there are                      = 0 is equivalent to solving 2x + 3 = 0. The
      two distinct values of x that satisfy the given                  solution of this equation is x = – 3 . So, we
                                                                                                          2
      equation.                                                        conclude that there is only one value of x
212. a. Using the fact that |a| = b if and only if a =      b,         that satisfies this equation.
                                                   2    1
      we see that solving the equation |3x –       3| – 9 = 0,   215. a. First, we rewrite the equation in an equiva-
                            2     1
      or equivalently |3x – 3 | = 9 , is equivalent to solving         lent form:
      3x – 2 = 1 . We solve these two equations
           3      9
                                                                       1 – (1 –(2 –|1 – 3x|)) = 5
      separately:
                                                                       1 – (1 – 2 + |1 – 3x|) = 5
            2    1                 2   1
      3x –  3 =– 9            3x – 3 = 9                               1 – (–1 + |1 – 3x|) = 5
      3x = 2 – 1 = 5
            3  9   9          3x = 2 + 1
                                   3   9   =   7
                                               9
                                                                       1 + 1 – |1 – 3x| = 5
           5                      7                                    2 – |1 – 3x| = 5
      x = 27                  x = 27
                                                                       –|1 – 3x| = 3
      So, both 257 and    7
                         27   are solutions to this                    1 –3x| = – 3
      equation.
                                                                       Since the left side is non-negative (being the
213. b. Using the fact that |a| = b if and only if a =      b,
                                                                       absolute value of a quantity) and the right side
      we see that solving the equation |3x + 5 | = 8 is
                                                                       is negative, there can be no value of x that sat-
      equivalent to solving 3x + 5 = 8. We solve these
                                                                       isfies this equation.
      two equations separately, as follows:
                                                                 216. c. Note that |a| =|b| if and only if a =    b.
      3x + 5 = –8             3x + 5 = 8                               Using this fact, we see that solving the equa-
      3x = –13                3x = 3                                   tion |2x + 1| = |4x – 5| is equivalent to solving
      x = – 133               x=1                                      2x + 1 = (4x – 5). We solve these two equa-
                                                                       tions separately:
      Thus, the solutions to the equation are x = – 133
      and x = 1.                                                       2x + 1 = (4x – 5)      2x + 1 = –(4x – 5)
                                                                       6 = 2x                 2x + 1 = –4x + 5
                                                                       1
                                                                       3 =x                   6x = 4
                                                                                              x= 23

                                                                       Thus, there are two solutions to the original
                                                                       equation.


                                                                                                                   171
                                       ANSWERS & EXPLANATIONS–



217. d. Note that |a|      c if and only if (a c or              –5        2x –3 5
      a –c). Using this fact, we see that the values             –2        2x 8
      of x that satisfy the inequality |x| 3 are pre-            –1        x 4
      cisely those values of x that satisfy either x 3
                                                                 Thus, the solution set is (–1,4).
      or x –3. So, the solution set is (–∞,–3)∪(3,∞).
                                                           222. a. First, we rewrite the given inequality in an
218. a. First, note that |–2x| = |–1| |2x| = |2x|.
                                                                 equivalent form:
      Also, |a| c if and only if (a c or a –c).
      The values of x that satisfy the inequality                2 – (1 – (2 – |1 – 2x|)) –6
      |2x| 0 are those that satisfy either 2x 0 or               2 – (1 – 2 + |1 – 2x|) –6
      2x 0. Dividing both of these inequalities by               2 – (–1 + |1 – 2x|) –6
      2 yields x 0 or x 0. So, the solution set is               2 + 1 – |1 – 2x| –6
      (–∞,0)∪(0,∞).                                              3 – |1 –2x| –6
219. c. First, note that the inequality –|–x – 1|      0         –|1 – 2x| –9
      is equivalent to |–x – 1| 0. Moreover, since               |1 – 2x| 9
      |–x – 1| = |–(x + 1)| = |–1| |x + 1| = |x + 1|,
                                                                 Now, note that |a| c if and only if if –c a
      this inequality is also equivalent to |x + 1| 0.
                                                                    c. Using this fact, we see that the values of x
      The left side must be nonnegative since it is
                                                                 that satisfy the inequality |1 –2x| 9 are pre-
      the absolute value of a quantity. The only way
                                                                 cisely those values of x that satisfy –9 1 – 2x
      that it can be less than or equal to zero is if it
                                                                    9. We solve this compound inequality:
      actually equals zero. This happens only when x
      + 1 = 0, which occurs when x = –1.                         –9        1 – 2x   9
220. c. Note that |a|     c if and only if (a c or a             –10        –2x     8
      –c). Using this fact, we see that the values of x
      that satisfy the inequality |8x + 3| 3 are pre-            5     x      –4
      cisely those values of x that satisfy either 8x +          So, the solution set is (–4,5).
      3 3 or 8x + 3 –3. We solve these two
                                                           223. c. First, we rewrite the given inequality in an
      inequalities separately:
                                                                 equivalent form:
      8x + 3    3        8x + 3 –3
                                                                 –7|1 – 4x| + 20        –2|1 – 4x| – 15
      8x 0               8x –6
                                                                 –7|1 – 4x| + 35        –2|1 – 4x|
      x 0                x –6 = –3
                               8 4                               35 5|1–4x|
      Thus, the solution set is [0,∞)∪(–∞,– 3 ]
                                            4
                                                                 7 |1 – 4x|
221. d. Note that |a|     c if and only if –c a c.               The last inequality is equivalent to |1 – 4x| 7.
      Using this fact, we see that the values of x that          Now, |a| c if and only if (a c or a –c).
      satisfy the inequality |2x –3| 5 are precisely             Using this fact, we see that the values of x that
      those values of x that satisfy –5 2x – 3 5.                satisfy the inequality |1 – 4x| 7 are precisely
      We solve this compound inequality as follows:              those values of x that satisfy either 1 – 4x 7




   172
                                           ANSWERS & EXPLANATIONS–



      or 1 – 4x –7. We solve these two inequalities               the y-coordinate of A, which is –3. So, the
      separately:                                                 coordinates of D are (6,–3).
                                                            229. d. Points in Quadrant IV have positive x-
      1 – 4x 7            1 – 4x –7
                                                                  coordinates and negative y-coordinates.
      –4x 6               –4x –8
                                                                  Therefore, (2,–5) lies in Quadrant IV.
      x –6 = –3
            4  2          x 2
                                                            230. a. For all nonzero real numbers, both x2 and
                                      3
      So, the solution set is (–   , –2]   [2, ).                 (–y)2 are positive, so points of the form (x2,(–y)2)
224. d. First, we rewrite the given inequality in an              must lie in Quadrant I.
      equivalent form:                                      231. d. First, note that |–x –2| = 0 only when x = –2

      |1 – (–22 + x) – 2x | | 3x – 5                              and |–x – 1| = 0 when x = –1. For all other val-
      1 – (–4 + x) – 2x | | 3x – 5                                ues of x, these expressions are positive. For all
      |1 + 4 – x – 2x | | 3x – 5                                  real numbers x –2, we conclude that |–x – 2|
      |5 – 3x| |3x – 5|                                              0 and –|–x –1| 0. Therefore, points whose
                                                                  coordinates are given by (|–x –2|, –|–x–1|)
      Now, note that the left side of the last inequal-           must lie in Quadrant IV.
      ity is equivalent to                                  232. b. The fact that x is a positive real number
      |5 – 3x| = |–1(3x – 5)| = |–1| |3x – 5)| = |3x – 5|         requires that the point (x,y) lie to the right of
                                                                  the y-axis, so it cannot lie in Quadrants II or
      Thus, the original inequality is actually equiva-           III, or be on the y-axis. It can, however, lie in
      lent to |3x – 5| |3x –5|. Since the left and                Quadrants I or IV, or be on the x-axis. The fact
      right sides of the inequality are identical, every          that y can be any real number does not further
      real number x satisfies the inequality. So, the             restrict the location of (x,y). Hence, the correct
      solution set is the set of all real numbers.                choice is b.
                                                            233. c. Because y is a non-negative real number, the
Set 15 (Page 26)
                                                                  point (x,y) must lie on or above the x-axis, so
225. c. The coordinates of points in the third
                                                                  it cannot lie in Quadrants III or IV. The point
      quadrant are both negative.
                                                                  can also be on the y-axis if it is the origin. The
226. e. The x-coordinate of J is –3 and the                       fact that x can be any real number does not
      y-coordinate is 4. So, J is identified as the               further restrict the location of (x,y), so the
      point (–3, 4).                                              correct choice is c.
227. b. Since ABCD is a square, the x-coordinate            234. a. We need to choose the selection that has a
      of B will be the same as the x-coordinate of A,             positive x-coordinate and negative y-coordinate.
      namely –1, and the y-coordinate of B will be                Since a 0, it follows that –a 0. Thus, the
      the same as the y-coordinate of C, namely 4.                choice that lies in Quadrant IV is (–a,a).
      So, the coordinates of B are (–1,4).
                                                            235. b. The correct selection will have a negative x-
228. e. Since ABCD is a square, the x-coordinate of               coordinate and negative y-coordinate. Note
      D is the same as the x-coordinate of C, which               that for any nonzero real number, that –a2 0
      is 6, and the y-coordinate of D is the same as              and (–a)2 0. The choice that lies in Quadrant
                                                                  III is (a,–a2).

                                                                                                              173
                                         ANSWERS & EXPLANATIONS–


236. a. Look for the selection that has a negative x-               and the y-coordinate is not negative, the point
      coordinate and positive y-coordinate. Since                   must be in Quadrant I or on the
      a 0, it follows that –a 0. The choice that                    x-axis. It can be on the x-axis if y = 0. Neither
      lies in Quadrant II is (–a, a).                               a nor b is true.
237. d. Note that if x is a negative integer, then –x3 =
      –(x)(x)(x) must be positive (because it is a            Set 16 (Page 28)
      product of an even number of negative inte-             241. a. Convert the given equation 3y – x = 9 into
      gers). Likewise, since x and y are both negative              slope-intercept form by solving for y, as follows:
      integers, xy2 is negative (because it is a product
                                                                    3y – x = 9
      of an odd number of negative integers). Hence,
                                                                    3y = x + 9
      the x-coordinate of (–x3, xy2) is positive and its
                                                                    y= 1 x+3
                                                                        3
      y-coordinate is negative. So, the point lies in
      Quadrant IV.                                                  The slope of this line is the coefficient of x,
                                                                    namely 1 .
                                                                            3
238. b. Using the fact that x and y are both assumed
                                                              242. b. The line whose equation is y = –3 is hori-
      to be negative integers, we must determine the
                                               –x  2    1           zontal. Any two distinct points on the line
      signs of the coordinates of the point ( (–y)3 , xy ).
                                                                    share the same y-value, but have different
      To this end, note that –x2 is negative, (–y)3 is
                                                                    x-values. So, computing the slope as “change
      positive (since –y is a positive integer and the              in y over change in x” results in 0, no matter
      cubes of positive integers are positive integers),            which two points are used.
      and xy is positive (since it is a product of an         243. a. Convert the given equation 8y = 16x – 4
      even number of negative integers). The x-                     into slope-intercept form by solving for y:
      coordinate of the given point is therefore neg-
                                                                    8y = 16x – 4
      ative (since the numerator is negative and
                                                                    y = 2x – 1
                                                                             2
      denominator is positive, thereby creating a
      quotient involving an odd number of negative                  So, the y-intercept is (0,– 1 ).
                                                                                                2

      integers) and the y-coordinate is positive. So,         244. d. Substituting x = 3 and y = 1 into the equa-
                                                                    tion y = 2 x – 1 yields the true statement 1 =
                                                                              3
      the point lies in Quadrant II.                                2
                                                                    3 (3) – 1, which implies that the point (3,1) is
239. c. Since the y-coordinate of the point (–x, –2)
                                                                    on this line.
      is –2, it follows that for any real number x, the
                                                              245. b. The slope-intercept form of a line with
      point must lie somewhere strictly below the x-
                                                                    slope m and y-intercept (0,b) is y = mx + b.
      axis. If x 0, the x-coordinate of the point is
                                                                    So, the equation of the line with slope –3 and
      negative, so that it lies in Quadrant III, while it
                                                                    y-intercept of (0,2) is y = 3x + 2.
      lies in Quadrant IV if x 0 and on the y-axis
      if x = 0. So, the correct choice is c.                  246. a. First, choose two of the five points listed and
                                                                    compute the slope. We will use the first two
240. d. The phrase “y is nonpositive” can be expressed                                                          0
                                                                    listed, (1,7) and (2,10). The slope is m = 12 ––17
      symbolically as y 0. As such, –y 0. Since
                                                                    = 3. Next, use one of the points, such as (1,7),
      the x-coordinate of the point (1, –y) is positive
                                                                    and the slope m = 3 to write the equation of


   174
                                            ANSWERS & EXPLANATIONS–


      the line using the point-slope formula y – y1 =                        where (x1, y1) is the point on the line. Applying
      m(x –x1), where (x1,y1) is the point on the line.                      this yields the equation y – 15 = –2(x – (–4)),
      Applying this yields the equation y – 7 = 3(x – 1),                    which simplifies to y – 15 = –2x – 8, or equiva-
      which simplifies to y – 7 = 3x – 3, or equivalently                    lently y = –2x + 7. Now, to determine the miss-
      y = 3x + 4.                                                            ing value z, we simply substitute x = 2 into this
247. b. Transforming the equation 3x + y = 5 into                            equation; the resulting value of y is equal to
      slope-intercept form simply requires that we                           the missing value of z. The substitution yields
      solve for y to obtain the equation y = –3x + 5.                        y = –2(2) + 7 = 3.
248. b. First, the slope of the line containing the                    253. c. First, the slope of the line containing the
                                             5–3          2
      points (2,3) and (–2,5) is m =        (–2)–2   =   –4   =              points (0,–1) and (2,3) is m =       3 – (–1)
                                                                                                                             =   4
                                                                                                                                     = 2.
                                                                                                                    2–0          2
      – 1 . Next, use
        2               one of the points, such as (2,3),                    Next, use one of the points, such as (2,3), and
      and the slope m = – 1 to write the equation of
                          2                                                  the slope m = 2 to write the equation of the line
      the line using the point-slope formula y – y1 =                        using the point-slope formula y – y1 = m(x – x1),
      m(x –x1), where, (x1 –y1) is the point on the                          where (x1, y1) is the point on the line. This
      line. This yields the equation y – 3 = – 1 (x – 2),
                                               2                             yields the equation y – 3 = 2(x – 2), which
      which simplifies to y – 3 = – 1 x + 1, or equiva-
                                    2                                        simplifies to y – 3 = 2x – 4, or equivalently
      lently y = – 1 x + 4.
                   2                                                         y = 2x – 1.
249. c. We transform the equation y =       x– – 125          3
                                                              5        254. a. Consider the line whose equation is x = 2. All
      into standard form Ax + By = C, as follows:                            points on this line are of the form (2,y), where y
      y=   – 125
              x    –3
                    5
                                                                             can be any real number. However, in order for
      –15y = 2x + 9                                                          this line to have a y-intercept, at least one of the
      0 = 2x + 15y + 9                                                       points on it must have an x-coordinate of 0,
      2x + 15y = –9                                                          which is not the case. A vertical line need not
250. a. We must solve the equation –3y =12x – 3                              have a y-intercept.
      for y, which can be done by simply dividing                      255. a. First, we must determine the equation of
      both sides by –3. This yields y = –4x + 1. The                         the line. The slope of the line is given by m
                                                                                            6
      slope of this line is –4.                                              = 0 – (–6) =
                                                                                 9–0        9   = 2 . Since the y-intercept of the
                                                                                                  3
251. d. Solving the equation 6y + x = 7 for y yields                         line is given to be (0, –6), we conclude that
      the equivalent equation y =      –1
                                        6   x+   7
                                                 6 . The                     the equation of the line is y = 2 x – 6. Now,
                                                                                                             3
      slope of this line is 1 .
                            6                                                observe that substituting the point (–6,–10)
252. c. We must first determine the equation of the                          into the equation yields the true statement
      line. To do so, choose two of the five points                          –10 = 2 (–6) – 6. Therefore, the point (–6,–10)
                                                                                   3
      listed and compute the slope. Using (–4,15)                            lies on this line.
                                           15 – 11             4
      and (–2,11), the slope is m =      (–4) – (–2)     =    –2   =   256. d. The slope of a line containing the points
      –2. Next, use the points, (–4,15), and the slope                       (–3,–1), (0,y), and (3,–9) can be computed
      m = –2 to write the equation of the line using                         using any two pairs of these points. Specifi-
      the point-slope formula y – y1 = m(x – x1),                            cally, using (–3,–1) and (3,–9), we see that


                                                                                                                                 175
                                                ANSWERS & EXPLANATIONS–


                                                                     264. d. The line falls from left to right at a rate of
      the slope is m =        (–1) – (–9)
                                (–3) –3     = – 8 = – 4 . Now, we
                                                6     3
                                                                           one vertical unit down per one horizontal unit
      equate the expression obtained by computing
                                                                           right, and it crosses the y-axis at the point
      the slope of this line using the points (–3,–1)
                                                                           (0,7). So, the slope of the line is –1 and its
      and (0,y) to – 4 , and solve for y:
                     3                                                     y-intercept is (0,7). Its equation is therefore
       y – (–1)
       0 – (–3)
                  = –4
                     3
                                                                           y = –x + 7.
                                                                     265. b. Using the two points (0,5) and (–9,–1) on
       y+1
        3     = –4
                 3
                                                                           the line, we observe that the line rises from left
      y + 1 = –4
                                                                           to right at a rate of six vertical units up per
      y = –5
                                                                           nine horizontal units right. Hence, its slope is
                                                                           6    2
Set 17 (Page 31)                                                           9 = 3 . Also, it crosses the y-axis at (0,5). Thus,
                                                                           the equation of this line is y = 2 x + 5.
                                                                                                              3
257. d. The points on the line y = –3 are of the form                                                         2   1
                                                                     266. c. First, convert the equation 3 y – 2 x = 0 into
      (x,–3), for all real numbers x. This set of points
                                                                           slope-intercept form:
      forms a horizontal line containing the point
                                                                            2
      (0,–3). The correct graph is given by choice d.                       3y   – 1x = 0
                                                                                   2
                                                          0 – (–5)          2
258. b. The slope of this line segment is m = –3 – 0
                                                                            3y   = 1x
                                                                                   2
      = –5.
         3                                                                         3         1
                                                                           y=      2         2x     = 3x
                                                                                                      4
259. b. The y-axis is a vertical line and hence, its
      slope is undefined.                                                  From this, we observe that since the slope is 3 ,
                                                                                                                           4
260. d. The slope of this line segment is                                  the graph of the line rises from left to right at a
                              8
      m=      2 –(–6)
             10 –(–2)
                         =   12   = 2.
                                    3
                                                                           rate of 3 vertical units up per 4 horizontal units
261. c. The slope is 2 (so that the graph of the line
                                                                           right. The y-intercept is the origin, and the
      rises from left to right at a rate of two vertical                   correct graph is shown in choice c.
      units up per one horizontal unit right) and the                267. b. A line with a positive slope rises from left to
      y-intercept is (0,3). The correct graph is shown                     right. The only line that rises from left to right
      in choice c.                                                         is the one in choice b.
262. a. The slope is –2 (so that the graph of the line               268. a. A line with an undefined slope must be ver-
      falls from left to right at the rate of two vertical                 tical. The only graph that satisfies this criterion
      units down per one horizontal unit right) and                        is choice a.
      the y-intercept is (0,9). The correct graph is                 269. d. First, convert the equation into slope-intercept
      shown in choice a.                                                   form as follows:
263. d. The slope is         –5
                           (so that the graph of the line
                              2                                            0.1x – 0.7y = 1.4
      falls from left to right at a rate of five vertical
                                                                           0.1x = 0.7y + 1.4
      units down per two horizontal units right) and
                                                                           0.1x – 1.4 = 0.7y
      the y-intercept is (0,–5). The correct graph is                            0.1         1.4
                                                                           y=    0.7 x   –   0.7   = 1x – 2
                                                                                                     7
      in choice d.




   176
                                               ANSWERS & EXPLANATIONS–


      Since the slope is 1 , the graph of the line rises
                         7                                       275. b. The line provided in choice b is equivalent
      from left to right at a rate of one vertical unit                to y = –2x +6. Since this has the same slope as
      up per seven horizontal units right. The y-                      the given line, namely –2, we conclude that the
      intercept is (0,–2), and the correct graph is                    correct answer is b.
      given by choice d.
270. c. The graph of y = c is a horizontal line that is          Set 18 (Page 42)
      either above or below the x-axis. If it lies above         276. d. Since we want a line perpendicular to a line
      the x-axis, the graph crosses into only Quad-                                                                    1
                                                                       with slope m1 = 3 , we must use m2 = –
                                                                                       4                               3   = –4
                                                                                                                              3
                                                                                                                       4
      rant I and Quadrant II, while if it lies below
                                                                       as the slope. Since the point (–6,4) must be on
      the x-axis, it crosses into only Quadrant III
                                                                       the line, the point-slope formula for the line is
      and Quadrant IV.
271. b. The graph of y = c, where c      0, is a hori-                 y – 4 = – 4 (x + 6). This is equivalent to the
                                                                                 3

      zontal line that lies either above or below the                  equation y = – 4 x – 4.
                                                                                      3
      x-axis, and must cross the y-axis.                         277. b. A line parallel to y = 3x + 8 must have
272. a. For instance, consider the line whose equa-                    slope 3. Using the point-slope formula for a
      tion is y = –x – 1. Its graph is shown here:                     line with the point (4,4), we see that the
                                                                       equation of the line we seek is y – 4 = 3(x – 4),
                                 10
                                                                       which simplifies to y = 3x – 8.
                                  8
                                                                 278. b. The slope of the line passing through the
                                  6                                                                   6–2      4
                                                                       two given points is m = –5 – 4 = – 9 . This is
                                  4
                                                                       actually the slope of the line we seek because
                                  2
                                                                       the line parallel to the one containing the two
                                                             x
       –10   –8   –6   –4   –2         2   4    6   8   10             given points. Using this slope with the point
                                  –2
                                                                       (0,12), we see that the point-slope form of the
                                  –4
                                                                                                           4
                                                                       equation of the lines is y – 12 = – 9 (x – 0),
                                                                                                  4
                                  –6
                                                                       which simplifies to y = – 9 x + 12.
                                  –8
                                                                 279 c. A line perpendicular to the given line must
                                 –10                                                 18
                                                                       have slope    13 . Using   this slope with the point
                                   y
                                                                       (0,0), we see that the point-slope form
      Observe that the graph does indeed cross into
                                                                       of the equation of the line we seek is y – 0 =
      three of the four quadrants.                                     18                                      18
273. e. A line perpendicular to the given line must                    13 (x   – 0), which simplifies to y =   13 x.
                             3                                   280. a. Only vertical lines have undefined slopes.
      have a slope m = 2 . So, the line given in choice
      e is the correct choice.                                         The only vertical line among the choices
                                                                       provided is given by choice a.
274. e. Two lines are parallel if and only if they
      have the same slope. This is true for the line             281. b. Only horizontal lines have zero slopes. The
      provided in choice e since the slopes of both                    only horizontal line among the choices pro-
      this lineand the given one are 6.                                vided is given by choice b.



                                                                                                                       177
                                        ANSWERS & EXPLANATIONS–


282. c. Let x = the length of the first piece. Then, 2x – 1 = the length of the second piece and 3(2x – 1) + 10 =
      the length of the third piece. The sum of the lengths of these three smaller pieces will be the length of
      the original piece of rope. This is represented as the equation x + (2x – 1) + 3 (2x – 1) + 10 = 60. To
      solve this equation, we first simplify the left side to obtain 9x + 6 = 60. Solving this equation gives us x =
      6. Therefore, the length of the first piece is 6 feet, the second piece is 11 feet, and finally, the third piece is
      43 feet long. So, we conclude that the longest piece of rope is 43 feet long.
283. b. Let x = number of canisters of Ace balls. Then, x + 1 = number of canisters of Longline balls. The
      important observation is that multiplying the price of one canister of Ace balls by the number of canis-
      ters of Ace balls results in the portion of the total amount spent on Ace balls. The same reasoning is true
      for the Longline balls. So, we must solve an equation of the form:

                  amount spent on Ace balls + amount spent on Longline balls = total amount spent

      Using the information provided, this equation becomes 3.50x + 2.75(x + 1) = 40.25. Simplifying the left
      side of the equation yields 6.25x + 2.75 = 40.25. Subtracting 2.75 from both sides and then dividing by
      6.25 yields the solution x = 6. So, we conclude that he bought 6 canisters of Ace balls and 7 canisters of
      Longline balls.
284. d. Let x = the number of gallons needed of the 30% nitrogen. Then, since we are supposed to end up
      with 10 gallons, it must be the case that 10 – x = the number of gallons needed of the 90% nitrogen.
      Multiplying the number of gallons of 30% nitrogen by its concentration yields the amount of nitrogen
      contained within the 30% solution. A similar situation holds for the 90% nitrogen, as well as for the
      final 70% solution. We must solve an equation of the following form:

           amount of nitrogen contributed + amount of nitrogen contributed = total amount of nitrogen
              from the 30% solution            from the 90% solution          in the entire 10 gallons

      Using the information provided, this equation becomes 0.30x + 0.90(10 – x) = 0.70(10), which is solved
      as follows:

      0.30x + 0.90(10x – x) = 0.70(10)
      30x + 90(10 – x) = 70(10)
      30x + 900 – 90x) = 700
      –60x + 900 = 700
      –60x = –200
      x = ––26000 = 130

      Thus, rounding to two decimal places, we conclude that she should mix approximately 3.33 gallons of
      the 30% nitrogen solution with 6.67 gallons of the 90% nitrogen solution to obtain the desired mixture.




   178
                                      ANSWERS & EXPLANATIONS–


285. a. The important concept in this problem is how rate, time, and distance interrelate. It is known that
      distance = rate time. We need to determine the amount of time that the girl is bicycling, and at pre-
      cisely what time the girl and the instructor meet and have therefore traveled the exact same distance
      from the starting point. So, we must determine expressions for the distances traveled by both the girl
      and her instructor, then equate them. To this end, let x = number of hours the girl has been bicycling
      when she intercepts her instructor. Then, since the instructor had a 3-hour head start, the amount of
      time that he has been bicycling when the girl catches him must be 3 + x hours.

      Now, write an equation for the girl, and one for the instructor that relates their respective times, rates,
      and distance traveled.

      Let     Rg         =       rate of the girl = 17 mph
              Tg         =       time the girl is bicycling when she meets her instructor = x hours
              Dg         =       distance the girl has biked when she finally intercepts the instructor = 17x
              RI         =       rate of the instructor = 7 mph
              TI         =       time the instructor is bicycling when he meets the girl = 3 + x hours
              DI         =       distance the instructor has biked when he is intercepted by the girl = 7(3 + x)

      Using the information provided, we must solve the equation 17x = 7(3 + x), as follows:

      17x = 2(3 + x)
      17x = 21 + 7x
      10x = 21
      x = 2.1

      Thus, it takes the girl 2.1 hours (or 2 hours 6 minutes) to overtake her instructor.
286. d. Let x = the amount invested at 10% interest. Then, she invested 1,500 + x dollars at 11% interest. The
      amount of interest she earns in one year from the 10% investment is 0.10x, and the amount of interest
      earned in one year from the 11% investment is 0.11(1,500 + x). Since her total yearly interest earned is
      795 dollars, the following equation describes this scenario:

      0.10x + 0.11(1,500 + x) = 795

      This equation is solved as follows:

      0.10x + 0.11(1,500 + x) = 795
      0.10x + 165 + 0.11x = 795
      0.21x = 630
          6
      x = 0.30 = 3,000
            21

      Hence, she invested $3,000 at 10% interest and $4,500 at 11% interest.




                                                                                                            179
                                       ANSWERS & EXPLANATIONS–


287. b. Let x = the number of nickels in the piggy          290. a. The graph of the line is solid, so it is
      bank. Then there are 65 – x dimes in the bank.              included in the solution set and the inequality
      The amount contributed to the total by the                  describing the shaded region must include
      nickels is 0.05x and the amount contributed by              equality (either or ). Next, since the graph
      the dimes is 0.10(65 – x). Since the total in the           of the line rises from left to right at the rate of
      bank is $5.00, we must solve the following                  two vertical units up per one horizontal unit
      equation.                                                   right, its slope is 2. And, since it crosses the y-
                                                                  axis at (0,7), we conclude that the equation of
      0.05x + 0.10(65 – x) = 5.00
                                                                  the line is y = 2x + 7. Finally, since the shaded
      The equation is solved as follows:                          region is below the line y = 2x + 7, the inequal-
                                                                  ity illustrated by this graph is y 2x + 7. This
      0.05x + 0.10(65 – x) = 5.00
                                                                  can be verified by choosing a point in the
      0.05x + 6.5 – 0.10x = 5.00
                                                                  shaded region, say (0,0), and observing that
      –0.05x = –1.5
                                                                  substituting it into the inequality results in the
      x = ––01..055 = 30
                                                                  true statement 0 7.
      Thus, there are 30 nickels and 35 dimes in the        291. b. The graph of the line is dashed, so it is not
      piggy bank.                                                 included in the solution set and the inequality
288. b. Let x = Lisa’s current age (in years). Lori’s             describing the shaded region must not include
      age is 2x. The statement, “In 5 years, Lisa will            equality (it must be either or ). Next, since
      be the same age as her sister was 10 years ago”             the graph of the line falls from left to right at the
      can be expressed symbolically as the equation               rate of four vertical units down per one horizon-
      x + 5 = 2x – 10, which is solved as follows:                tal unit right, its slope is –4. Since it crosses the
                                                                  y-axis at (0,–3), we conclude that the equation of
      x + 5 = 2x – 10
                                                                  the line is y = –4x – 3. Finally, the shaded region
      x = 15
                                                                  is below the line y = –4x – 3, so we conclude that
      Thus, Lisa is currently 15 years old and Lori is            the inequality illustrated by this graph is y
      30 years old.                                               –4x – 3. This can be verified by choosing any
                                                                  point in the shaded region, say (0,–4), and
Set 19 (Page 45)                                                  observing that substituting it into the inequality
289. b. The fact that the graph of the line is solid              results in the true statement –4 –3.
      means that it is included in the solution set,        292. c. The graph of the line is dashed, so it is not
      so the inequality describing the shaded region              included in the solution set and the inequality
      must include equality (that is, it must be either           describing the shaded region must not include
         or ). Next, since the shaded region is below             equality (it must be either or ). Next,
      the horizontal line y = –2, all points in the solu-         since the shaded region is to the left of the ver-
      tion set have a y-value that is less than or equal          tical line x = 8, all points in the solution set
      to –2. Hence, the inequality illustrated by this            have an x-value that is less than or equal to 8.
      graph is y – 2.                                             Hence, the inequality illustrated by this graph
                                                                  is x 8.


   180
                                         ANSWERS & EXPLANATIONS–


                                                                                         1     1
293. c. The graph of the line is solid, so it is                    above the line y = – 6 x – 2 , the inequality illus-
                                                                                                    1    1
      included in the solution set and the inequality               trated by this graph is y – 6 x – 2 . Multiplying
      describing the shaded region must include                     both sides of this inequality by 2 and moving
      equality (it must be either or ). Next,                       the x-term to the left results in the equivalent
                                                                                1
      since the graph of the line rises from left to                inequality 3 x + 2y –1.We can verify this by
      right at the rate of one vertical unit up per one             substituting any point from the shaded region,
      horizontal unit right, its slope is 1. Since it               such as (0,1), into the inequality, which results
      crosses the y-axis at (0,2), the equation of the              in the true statement 2 > –1.
      line is y = x + 2. Finally, since the shaded            296. a. Because the graph of the line is solid, we
      region is above the line y = x + 2, the inequal-              know that it is included in the solution set, so
      ity illustrated by this graph is y x + 2, which               the inequality describing the shaded region
      is equivalent to x – y –2. We can verify this                 must include equality ( it must be either or
      by choosing any point in the shaded region,                   ≤). Next, the graph of the line falls from left to
      such as (0,3), and observing that substituting                right at the rate of three vertical unit down per
      it into the inequality results in the true state-             one horizontal unit right, so its slope is –3.
      ment –3 –2.                                                   And, since it crosses the y-axis at (0, 4), we
294. c The fact that the graph of the line is solid                 conclude that the equation of the line is y =
      means that it is included in the solution set, so             –3x + 4. The shaded region is below the line y
      the inequality describing the shaded region                   = –3x + 4, so the inequality illustrated by this
      must include equality (it must be either or                   graph is y –3x + 4 Multiplying both sides of
      ≤). Next, since the graph of the line rises from              the inequality by 2 and moving the x-term to
      left to right at the rate of one vertical unit up             the left results in the equivalent inequality 2y +
      per one horizontal unit right, its slope is 1.                6x 8. This can be further verified by choosing
      Since it crosses the y-axis at (0,0), the equation            a point from the shaded region, such as (0,0),
      of the line is y = x. The shaded region is above              and observing that substituting it into the
      the line y = x, so we conclude that the inequal-              inequality results in the true statement 0 0.
      ity illustrated by this graph is y x, which is          297. c. The graph of the line is dashed, so it is not
      equivalent to y – x 0. Substituting a point                   included in the solution set and the inequality
      from the shaded region, such as (0,3) into the                describing the shaded region must not include
      inequality results in the true statement 3 0.                 equality (it must be either or ). Next,
295. a. The graph of the line is dashed, so it is not               since the graph of the line rises from left to
      included in the solution set, and the inequality              right at the rate of three vertical units up per
      describing the shaded region must not include                 one horizontal unit right, its slope is 3. And,
      equality (it must be either or ). Next, since                 since it crosses the y-axis at (0,–2), the equa-
      the graph of the line falls from left to right at             tion of the line is y = 3x – 2. Finally, since the
      the rate of one vertical unit down per six hori-              shaded region is above the line y = 3x – 2, we
                                          1
      zontal units right, its slope is – 6 . It crosses the         conclude that the inequality illustrated by this
                      1
      y-axis at (0,– 2 ), so the equation of the line is y          graph is y 3x – 2. Moving the y-term to the
          1     1
      = – 6 x – 2 . Finally, since the shaded region is             right results in the equivalent inequality 3x – y


                                                                                                                181
                                       ANSWERS & EXPLANATIONS–



      – 2 0. This can be verified by choosing an            300. d. Substituting x = 3 and y = –2 into the
      arbitrary point in the shaded region, say (0,0),            inequality 9x – 1 y yields the true statement
      and observing that we can verify this by substi-            26 –2. We can therefore conclude that
      tuting a point from the shaded region, such as              (3,–2) satisfies this inequality.
      (0,0) into the inequality, resulting in the true      301. d. First, since the given inequality does not
      statement –2 0.                                             include equality, the horizontal line y = 4 is
298. a. The graph of the line is solid, so we know                not included in the solution set and should be
      that it is included in the solution set, and that           dashed. Because y 4, any point in the solu-
      the inequality describing the shaded region                 tion set (the shaded region) must have a y-
      must include equality (either or ). Next,                   coordinate that is larger than 4. Such points
      since the graph of the line rises from left to              occur only above the line y = 4. The correct
      right at the rate of three vertical unit up per             graph is given by choice d.
      one horizontal unit right, its slope is 3. It         302. c. Since the given inequality does not include
      crosses the y-axis at (0,1), so the equation of             equality, the vertical line x = 4 is not included
      the line is y = 3x + 1. Finally, since the shaded           in the solution set and should be dashed. Also,
      region is above the line y = 3x + 1, the inequal-           since x 4, any point in the solution set (the
      ity illustrated by this graph is y 3x + 1. This             shaded region) must have an x-coordinate that
      can be verified by choosing any point in the                is larger than 4. Such points occur to the right
      shaded region, such as (2,0), and substituting              of the line x = 4. The correct graph is shown in
      it into the inequality, which results in the true           choice c.
      statement 2 1.                                        303. b. The graph of the line is solid, so it is included
299. d. The fact that the graph of the line is solid              in the solution set, and the inequality describing
      means that it is included in the solution set, so           the shaded region must include equality (it must
      the inequality describing the shaded region                 be either or ≤). Next, since the graph of the
      must include equality (that is, it must be either           line falls from left to right at the rate of one ver-
          or ). Next, since the graph of the line falls           tical unit down per seven horizontal units right,
                                                                                 1
      from left to right at the rate of two vertical              its slope is – 7 . It crosses the y-axis at (0,10), so
                                                                                                       1
      unit down per one horizontal unit right, its                the equation of the line is y = – 7 x + 10. Finally,
                                                                                                                        1
      slope is –2. And, since it crosses the y-axis at            since the shaded region is below the line y = – 7
      (0,4), we conclude that the equation of the line            x + 10, the inequality illustrated by this graph is
                                                                         1
      is y = –2x + 4. Finally, since the shaded region            y – 7 x + 10. Observe that simplifying
      is above the line y = –2x + 4, we conclude that             –28y 2x – 14(y + 10) results in this inequal-
      the inequality illustrated by this graph is y               ity. This can be verified by choosing any point
      –2x + 4. Observe that simplifying 3x – y 7x +               in the shaded region, such as (0,5), and substi-
      y – 8 results in this inequality. This can be veri-         tuting it into the inequality to produce the true
      fied by choosing a point in the shaded region,              statement 5 10.
      such as (0,5), substituting it into the inequality    304. b. The following graph illustrates the inequal-
      to produce the true statement 5 4.                          ity y ( 2x + 7, whose solution set intersects
                                                                  all four quadrants.


   182
                                               ANSWERS & EXPLANATIONS–



                                 10
                                                                 308. d. First, multiply the second equation by 2 to

                                  8
                                                                       obtain y +8x = 24. Then, subtract the first
                                  6
                                                                       equation from this one to obtain 6x = 18,
                                                                       which simplifies to x = 3.
                                  4

                                                                 309. b First, simplify the second equation by sub-
                                  2

                                                             x
                                                                       tracting 9 from both sides of the equation. The
       –10   –8   –6   –4   –2         2   4    6   8   10
                                                                       second equation becomes –2x – 6 = y. Then,
                                  –2
                                                                       multiply the equation by 2 and add it to the
                                  –4
                                                                       first equation to obtain –6 = –y, the solution of
                                  –6
                                                                       which is y = 6. Now, substitute the value of y
                                  –8
                                                                       into the first equation and solve for x:
                                 –10

                                   y                                   4x + 6 = –3(6)
                                                                       4x + 6 = –18
Set 20 (Page 51)                                                           4x = –24
305. c. Adding the two equations together yields                            x = –6
      the equation 10a = –40, the solution of which                                                               –6
                                                                       Since x = –6 and y = 6, the value of x =
                                                                                                            y      6   = –1.
      is a = –4. Now, substitute –4 in for a in the first
                                                                 310. e. First, multiply the first equation by –4 to
      equation and solve for b:
                                                                       obtain 28a – b = –100. Then, add this to the
      5(–4) + 3b = –2                                                  second equation to obtain 29a = –87, a = –3.
      –20 + 3b = –2                                                    To find b, substitute this value into the second
      3b = 18                                                          equation to obtain b + (–3) = 13, so b = 16.
      b=6                                                        311. b. In the first equation, multiply the (m + n)
306. d. Add the two equations together to get the                      term by 2 and add m to obtain
      equation –2y = 8, which simplifies to y = –4.
                                                                        2(m + n) + m = 2m + 2n + m = 3m + 2n
      Next, substitute –4 for y in the second equa-
      tion and solve for x:                                            Now, subtract the second equation from the
                                                                       first equation to obtain the equation 5n = –15,
      x – 5(–4) = –3
                                                                       which simplifies to n = –3.
      x + 20 = – 3
                                                                 312. a. Simplifying the left side of the first equation
      x = – 23
                                                                       results in 14a + 21b = 56. Multiplying the sec-
307. b. In the first equation, multiply the (x + 4)
                                                                       ond equation by –7 yields –7b – 14a = 28.
      term by 3 to obtain 3(x + 4) = 3x + 12. Then,
                                                                       Now, adding these two equations together
      subtract 12 from both sides of the equation, so
                                                                       yields 14b = 84, the solution of which is b = 6.
      that the first equation becomes 3x – 2y = –7.
      Now, add the two equations together to obtain
      –x = 1, or x = –1.




                                                                                                                   183
                                       ANSWERS & EXPLANATIONS–


      Finally, substitute this into the second equa-              Since a = 5 and b = 10, the value of a + b = 5 +
      tion and solve for a:                                       10 = 15.
                                                            316. b. Multiply the first equation to 5 and simplify
      6 + 2a = –4
                                                                  to obtain c – d = 10. Then, subtract the second
      2a = –10
                                                                  equation from this to obtain 5d = 10, or d = 2.
      a = –5
                                                                  Now, substitute this value into the second
313. c. Multiply the first equation by 8 to obtain
                                                                  equation and solve for c:
      the equivalent equation 4x + 48y = 56. Adding
      this to the second equation in the system results           c – 6(2) = 0
      in the equation 33y = 66, which simplifies to               c = 12
      y = 2.                                                                         c        12
                                                                  So, the value of   d   is    2   = 6.
314. d. Divide the second equation by 2 and add it          317. a. Multiply the second equation by 2, then add
      to the first equation to obtain the equation 7a =           to the first equation to obtain –7x = –63, which
      21, the solution of which is a = 3. Now, substi-            simplifies x = 9. Now, substitute this value into
      tute the value of a into the first equation and             the second equation and solve for y:
      solve for b:
                                                                  –9 – y = –6
      4(3) + 6b = 24                                              –y = 3
      12 + 6b = 24                                                y = –3
      6b = 12
                                                                  So, the value of xy is (9)(–3)= –27.
      b=2
                                                            318. e. First, simplify the first equation by multi-
      Since a = 3 and b = 2, the value of a + b =                 plying (x – 1) by 9 to obtain 9(x – 1) = 9x – 9.
      3 + 2 = 5.                                                  Then, add 9 and 4y to both sides of the equa-
315. b. First, simplify the first equation by multi-              tion. The first equation becomes 9x + 4y = 11.
      plying (a + 3) by 1 to obtain the equivalent
                         2                                        Multiply the second equation by –2 and add it
      equation 1 a + 3 – b = –6. Then, subtract 3
                 2    2                          2                to the first equation to obtain –5x = 5 or x = –1.
      from both sides to further obtain 1 a – b = – 125 .
                                         2                        Now, substitute the value of x into the second
      Next, multiply the equation by –6 and                       equation and solve for y:
      add it to the second equation to obtain the
                                                                  2y + 7(–1) = 3
      equation 4b = 40, or b = 10. Now, substitute
                                                                  2y – 7 = 3
      the value of b into the second equation and
                                                                  2y = 10
      solve for a:
                                                                  y= 5
      3a – 2(10) = –5
                                                                  Since y = 5 and x = –1, the value of (y – x)2 =
      3a – 20 = –5
                                                                  (5 – (–1))2 = 62 = 36.
      3a = 15
      a=5




   184
                                       ANSWERS & EXPLANATIONS–


319. c. Multiply the second equation by 3 and add          322. d. Solve the first equation for y in terms of x:
      it to the first equation to obtain 14q = 98, which
                                                                 2x + y = 6
      simplifies to q = 7. Now, substitute the value of
                                                                 y = 6 – 2x
      q into the second equation and solve for p:
                                                                 Substitute this expression for y in the second
      –5p + 2(7) = 24
                                                                 equation and solve for x:
      –5p + 14 = 24
                                                                 6 – 2x
      –5p = 10                                                      2  + 4x = 12
      p = –2                                                     3 – x + 4x = 12
                                                                 3x + 3 = 12
      Since p = –2 and q = 7, the value of (p + q)2 =
                                                                 3x = 9
      (–2 + 7)2 = 52 = 25.
                                                                 x=3
320. b. Multiply the first equation by 2 to obtain
                                                           323. d. Solve the first equation for a in terms of b
      8x – 6y = 20 and the second equation by 3 to
                                                                 by multiplying both sides of the equation by
      obtain 15x + 6y = 3. Then, add these equations
                                                                 2 to obtain a = 2b + 2. Now, substitute this
      to obtain 23x = 23, which simplifies to x = 1.
                                                                 expression for a in the second equation to
      Now, substitute this into the first equation and
                                                                 find b:
      solve for y:
                                                                 3(2b + 2 – b) = –21
      4(1) – 3y = 10
                                                                 3(b + 2) = –21
      –3y = 6
                                                                 3b + 6 = –21
      y = –2
                                                                 3b = –27
      So, the solution of the system is x = 1, y = –2.           b = –9

                                                                 Substitute the value of b into the first equation
Set 21 (Page 53)
                                                                 and solve for a:
321. a. Since the first equation is already solved for
                                                                 a
      x, substitute it directly into the second equation         2 = –9 + 1
                                                                 a
      and solve for y:                                           2 = –8
                                                                 a = –16
      2(–5y) + 2y = 16
      –10y + 2y = 16                                             Since a = –16 and b = –9, the value of
      –8y = 16                                                     a = –16 = 16 = 4
                                                                     b    –9     9   . 3
      y=–2

      Now, substitute this value for y into the first
      equation to find the corresponding value of x:
      x = –5(–2) = 10. Hence, the solution of the
      system is x = 10, y = –2.




                                                                                                           185
                                       ANSWERS & EXPLANATIONS–


324. e. Solve the second equation for a in terms of b:         Substitute the value of d into the second equa-
      b + a = 13                                               tion and solve for c:
      a = 13 – b
                                                               c – 6(2) = 0
      Substitute this expression for a in the first            c – 12 = 0
      equation and solve for b:                                c = 12
               b                                                                                      c       12
      –7a +    4   = 25                                        Since c = 12 and d = 2, the value of   d   =    2   = 6.
                          b
      –7(13 – b) +        4   = 25                       327. a. Solve the second equation for y in terms of x:
                    b
      7b – 91 +     4   = 25                                   –x – y = –6
      29b                                                      –y = x – 6
       4    = 116
                                                               y = –x + 6
      29b = 464
      b = 16                                                   Substitute this expression for y in the first
325. a. Solve the second equation for b in terms of a:         equation and solve for x:
      b + 2a = –4
                                                               –5x + 2(–x + 6) = –51
      b = –2a – 4
                                                               –5x – 2x + 12 = –51
      Substitute this expression for b in the first            –7x + 12 = –51
      equation and solve for a:                                –7x = –63
                                                               x=9
      7(2a + 3(–2a – 4)) = 56
      7(2a – 6a – 12) = 56                                     Substitute the value of x into the second equa-
      7(–4a – 12) = 56                                         tion and solve for y:
      –28a – 84 = 56
                                                               –9 – y = –6
      –28a = 140
                                                               –y = 3
      a = –5
                                                               y = –3
326. b. Solve the second equation for c in terms of d:
      c – 6d = 0                                               Since x = 9 and y = –3, the value of xy = (9)(–3)
      c = 6d                                                   = –27.
                                                         328. e. Solve the second equation for b in terms of a:
      Substitute this expression for c in the first
                                                               b–a=1
      equation and solve for d:
                                                               b=a+1
      c–d
        5 –2=0                                                 Substitute this expression for b in the first
      6d – d
         5 –2=0                                                equation and solve for a:
      5d
       5 –2=0                                                  10(a + 1) – 9a = 6
      d–2=0                                                    10a + 10 – 9a = 6
      d=2                                                      a + 10 = 6
                                                               a = –4




   186
                                       ANSWERS & EXPLANATIONS–


      Substitute the value of a into the second equa-       331. b. Since the two lines intersect in exactly one
      tion and solve for b:                                       point, we conclude that the system of equa-
                                                                  tions represented by the graph has one solution.
      b – (–4) = 1
                                                            332. b. The slope-intercept form of the line y – 3x =
      b+4=1
                                                                  –2 is y = 3x – 2. As such, since the slope of this
      b = –3
                                                                  line, (3) is the same as the slope of the line
      Since a = –4 and b = –3, the value of ab =                  given by the first equation, we conclude that
      (–4)(–3) = 12.                                              the lines are parallel. Their graphs never inter-
329. b. Solve the second equation for y in terms of x:            sect, so the system has no solution.
      2x – y = 9                                            333. b. Solve the first equation for y to obtain y =
      –y = –2x + 9                                                3x – 2. Now, substitute this into the second
      y = 2x – 9                                                  equation and solve for x, as follows:
      Substitute this expression for y in the first               2(3x – 2) – 3x = 8
      equation and solve for x:                                   6x – 4 – 3x = 8
       x + 2x – 9                                                 3x – 4 = 8
           3        =8
                                                                  3x = 12
       3x – 9
         3      =8                                                x=4
      x–3=8
                                                                  Next, substitute this value of x into the first
      x = 11
                                                                  equation to determine that the corresponding
      Substitute the value of x into the second equa-             value of y is y = 3(4) – 2 = 10. Thus, the value
      tion and solve for y:                                            2x        2(4)     4
                                                                  of    y   is    10    = 5.
      2(11) – y = 9                                         334. c. Since the graph consists of a single line, we
      22 – y = 9                                                  conclude that the two equations that make up
      –y = –13                                                    the system are exactly the same, so every point
      y = 13                                                      on the line is a solution of the system. There
                                                                  are infinitely many such points.
      Since x = 11 and y = 13, the value of x – y =
                                                            335. b. Since the two lines are parallel, they never
      11 – 13 = –2.
                                                                  intersect. There are no solutions of this system.
330. c. Two lines are parallel if and only if they
                                                            336. c. Observe that dividing both sides of the sec-
      have the same slope. The slope-intercept form
                                                                  ond equation –3y + 9x = –6 by –3 and rear-
      of the line x – y = 7 is y = x – 7, and the slope-
                                                                  ranging terms results in the first equation.
      intercept form of the line 2 – y = –x is y = x + 2.
                                                                  This means that the equations are identical, so
      The slope of each of these lines is 1, so, they are
                                                                  any point that satisfies the first equation auto-
      parallel.
                                                                  matically satisfies the second. Since there are
                                                                  infinitely many such points, the system has
                                                                  infinitely many solutions.




                                                                                                             187
                                       ANSWERS & EXPLANATIONS–



Set 22 (Page 57)                                            341. d. The slope-intercept form of the line 2y – 3x
                                                                             3
337. c. The graphs of the lines y = 4 and y = x + 2               = –6 is y = 2 x – 3. The graphs of this line and
                                                                          5
      are dashed, so that the inequality signs used in            y = 5 – 2 x are solid, so the inequality signs
      both of the inequalities comprising the system              used in both of the inequalities are either or
      are either or . Next, note that points in the                 . Points in the shaded region lie above (or
      shaded region lie above the line y = 4 and                  on) the line 2y – 3x = – 6 and above (or on)
                                                                                    5
      below the line y = x + 2. This means that the               the line y = 5 – 2 x. This means that the system
      system of linear inequalities for which the                 of linear inequalities for which the shaded
      shaded region is the solution set is given by               region is the solution set is given by 2y – 3x
                                                                               5
      y 4, y x + 2.                                               –6, y 5 – 2 x.
338. a. The graphs of the lines y = 5 and x = 2 are         342. a. Given that the first inequality does not
      solid, which means that the inequality signs                include equality, but the second inequality
      used in both of the inequalities are either or              does, we know that the graph of the line y = 2
         . Next, note that points in the shaded region            is dashed and the graph of the line y = 2x + 1
      lie above (or on) the line y = 5 and to the left of         is solid. Points that satisfy the inequality y 2
      (or on) the line x = 2. Therefore, the system of            must be above the line y = 2, and those satisfy-
      linear inequalities for which the shaded region             ing y 2x + 1 must lie below the line y = 2x +
      is the solution set is given by y 5, x 2.                   1. The intersection of these two regions is given
339. a. First, note that the graphs of the lines y = –x
                                                                  by the illustration in choice a.
      + 4 and y = x + 2 are dashed, which means             343. b. The slope-intercept forms of the lines 5y =
      that the inequality signs used in both of the               8(x + 5) and 12(5 – x) = 5y are, respectively,
                                                                       8                12
      inequalities in the system are either or .                  y = 5 x + 8 and y = – 5 x + 12. The graph of
                                                                                8
      Next, note that points in the shaded region lie             the line y = 5 x + 8 is solid (so that the corre-
      below the line y = x + 2 and below the line y =             sponding inequality should involve one of the
                                                                                                        12
      –x + 4. This implies that the system of linear              signs or ). The graph of y = – 5 x + 12 is
      inequalities for which the shaded region is the             dashed, so that the corresponding inequality
      solution set is given by y x + 2, y –x + 4.                 should involve one of the signs or . Points
                                          1
340. a. First, the graph of the line y = 4 x is dashed,
                                                                  in the shaded region lie below (or on) the line
      so the corresponding inequality should involve              5y = 8(x + 5) and below the line 12(5 – x) = 5y.
      one of the signs or . The graph of y = –4x                  This implies that the system of linear inequali-
      is solid (so the corresponding inequality should            ties for which the shaded region is the solution
      involve one of the signs or ). Points in the                set is given by 5y 8(x + 5), 12(5 – x) 5y.
                                             1              344. d. The graph of the line y = 3x is dashed, so
      shaded region lie above the line y = 4 x and
      below the line y = –4x. Therefore, the system               that the corresponding inequality should
      of linear inequalities for which the shaded                 involve one of the signs < or >. The graph of y
                                                 1                = –5 is solid, so that the corresponding
      region is the solution set is given by y 4 x,
      y –4x.                                                      inequality should involve one of the signs
                                                                  or . Note that points in the shaded region lie



   188
                                       ANSWERS & EXPLANATIONS–


      above the line y = 3x and below (or on) the                lines are solid, which means that both inequal-
      line y = –5. The system of linear inequalities             ity signs are either or . Points in the
      for which the shaded region is the solution set            shaded region lie above (or on) the line y =
                                                                 5
      is given by y 3x, y –5.                                    2 x – 10 and below (or on) the line y = –2x – 3,
345. b. The slope-intercept form of the lines 9(y – 4)           so the system of linear inequalities for which
      = 4x and –9y = 2(x + 9) are, respectively, y =             the shaded region is the solution set is given by
      4                  2                                             5
      9 x + 4 and y = – 9 x – 2. The graphs of both
                                                                 y 2 x – 10, y –2x – 3 This system is equiv-
      lines are dashed, so the inequality signs used in          alent to 5x – 2(y + 10) 0, 2x + y –3, which
      both inequalities are either or . Next, note               can be seen by reversing the simplification
      that points in the shaded region lie below the             process used to obtain the slope-intercept
                4                                2
      line y = 9 x + 4 and above the line y = – 9 x – 2.         forms of the lines in the first step. In doing so,
      This tells us that the system of linear inequali-          remember that multiplying both sides of an
      ties for which the shaded region is the solution           inequality results in a reversing of the inequal-
                           4            2
      set is given by y 9 x + 4, y – 9 x – 2. This sys-          ity sign.
      tem is equivalent to 9(y – 4) 4x, –9y 2(x            348. b. The slope-intercept forms of the lines
                                                                                            1
      + 9), which can be seen by reversing the sim-              7(y – 5) = –5x and –3 = 4 (2x – 3y) are, respec-
                                                                               5               2
      plification process used to obtain the slope-              tively, y = – 7 x + 5 and y = 3 x + 4. The graphs
      intercept forms of the lines in the first step. In         of both lines are dashed, so the inequality
      doing so, remember that multiplying both                   signs used in both of the inequalities compris-
      sides of an inequality results in a switching of           ing the system are either or . Points in the
                                                                                                          5
      the inequality sign.                                       shaded region lie below the line y = – 7 x + 5
                                                                                          2
346. c. The slope-intercept forms of the lines y – x             and below the line y = 3 x + 4. This tells us that
      = 6 and 11y = –2(x + 11) are y = x + 6 and y =             the system of linear inequalities for which the
         2
      – 11 x – 2, respectively. The graphs of both lines         shaded region is the solution set is given by y
                                                                      5             2
      are solid, so the inequality signs used in both               – 7 x + 5, y 3 x + 4. This system is equiva-
                                                                                                 1
      inequalities are either or . Next, points in               lent to 7(y – 5) –5x, –3 4 (2x – 3y), which
      the shaded region lie above (or on) the line y =           can be seen by reversing the simplification
                                               2
      x + 6 and above (or on) the line y = – 11 x – 2.           process used to obtain the slope-intercept
      The system of linear inequalities for which the            forms of the lines in the first step. In doing so,
      shaded region is the solution set is given by              remember that multiplying both sides of an
                         2
      y x + 6, y – 11 x – 2. This system is equiv-               inequality results in a reversing of the inequal-
      alent to y – x 6, 11y –2(x + 11), which                    ity sign.
      can be seen by reversing the simplification          349. d. The solution set for the system in choice a is
      process used to obtain the slope-intercept                 the empty set. The solution set for the system in
      forms of the lines in the first step.                      choice b consists of only the points that lie on
347. c. The slope-intercept forms of the lines 5x –              the line y = 3x + 2, and the solution set of the
      2(y + 10) = 0 and 2x + y = 3 are, respectively, y          system in choice c consists of only the points
        5
      = 2 x – 10 and y = –2x 3. The graphs of both               that lie on the line y = x. So, the solution sets of




                                                                                                             189
                                         ANSWERS & EXPLANATIONS–


      none of these systems span the entire Cartesian               Section 3—Polynomial
      plane. In fact, it is impossible for such a system                 Expressions
      of linear inequalities to have a solution set that
      spans the entire Cartesian plane.
                                                              Set 23 (Page 66)
350. b. Note that the graphs of the lines y = x + 3
                                                              353. d.
      and y = x – 1 are parallel, where the graph of
                                                                   (x2 – 3x + 2) + (x3 – 2x2 + 11) =
      y = x + 3 lies strictly above the graph of y = x – 1.
                                                                   x3 + x2 – 2x2 – 3x + 2 + 11 =
      Using the first inequality specified in the system,
                                                                   x3 – x2 + 3x + 13
      any point that it is in the solution set of the
                                                              354. a.
      system to y x + 3, y x – 1 would necessarily                                               2
                                                                   (3x2 – 5x + 4) – (– 3 x + 5)
      be above the line y = x + 3, and therefore, by
                                                                                             2
      the previous observation, also above the line                = 3x2 – 5x + 4 + 3 x – 5
                                                                                         2
      y = x – 1. However, the second inequality in                 = 3x2 – 5x + 3 x + 4 – 5
      the system requires that the point be below the                         15         2
                                                                   = 3x2 – 3 x + 3 x – 1
      line y = x – 1, which is not possible. Hence, the                       13
                                                                   = 3x2 – 3 x – 1
      solution set of this system is the empty set.
                                                              355. b.
351. d. The boundaries of Quadrant III are the                      1       1        2       2  7      1
                                                                   ( 3 x2 – 5 x – 3 ) – ( 3 x2 –
                                                                                               10 x + 2 )
      x-axis and y-axis; the equations of these axes
                                                                      1      1     2    2      7      1
      are y = 0 and x = 0, respectively. Since points              = 3 x2 – 5 x – 3 – 3 x2 + 10 x – 2
      in the solution set are not to be on either axis,              1      2      1      7      2   1
                                                                   = 3 x2 – 3 x2 – 5 x + 10 x – 3 – 2
      both inequalities comprising the system we                        1      2       7    4 3
                                                                   = – 3 x2 – 10 x + 10 x – 6 – 6
      seek must involve one of the signs or .
                                                                        1       5     7
      Next, note that the sign of both the x- and                  = – 3 x2 + 10 x – 6
      y-coordinate of a point in Quadrant III is neg-                   1      1     7
                                                                   = – 3 x2 + 2 x – 6
      ative. We conclude that the system with this            356. c.
      solution set is given by x 0, y 0.                           (9a2b + 2ab – 5a2) – (–2ab – 3a2 + 4a2b)
352. b. A system of linear inequalities whose solu-                = 9a2b + 2ab – 5a2 + 2ab + 3a2 – 4a2b
      tion set consists of the points on a single line             = 9a2b – 4a2b + 2ab + 2ab – 5a2 + 3a2
      must be of the form y mx + b, y mx + b,                      = 5a2b + 4ab – 2a2
      assuming that the lines are not vertical. Observe       357. a.
      that the first inequality in the system 2y – 6x               1        2                       2          7            1
                                                                   ( 6 x2 + 3 x + 1) + (2x – 3 x2 + 4) –        2   + 3x + 2 x2)
      4, or y 2 + 3x, is equivalent to y 2 + 3x, so                     1        2                   2      7            1
      this system is of the form specified. The solu-              = 6 x2 + 3 x + 1 + 2x – 3 x2 + 4 –       2   – 3x – 2 x2
                                                                        1        2    1       2                              7
      tion set consists of those points on the line                = 6 x2 – 3 x2 –        2
                                                                                      2 x + 3 x + 2x – 3x +         1+4–     2
      y = 3x + 2.                                                       1       4    3      2           7
                                                                   = 6 x2 –       2     2
                                                                                6x –6x + 3x – x + 5 – 2
                                                                                1    3
                                                                   = –x2 –      3x + 2




   190
                                         ANSWERS & EXPLANATIONS–


358. d.                                                             expression given in choice a is a polynomial;
      (2 – 3x3)–  [(3x3 + 1) – (1 –  2x3)]                          the coefficients, not the variable, involve nega-
      = 2 – 3x 3 – [3x3 + 1 – 1 + 2x3]                              tive exponents. The expression in choice b is a
      = 2 – 3x3 – [5x3]                                             polynomial for similar reasons; note that the
      = 2 – 3x3 – 5x3                                               first term is really just a constant since x0 = 1.
      = 2 – 8x3                                               364. d. The statements in choices a, b, and c are all
359. b. The degree of a polynomial is the highest                   true, and follow from the fact that simplifying
      power to which the variable x is raised. For the              such arithmetic combinations of polynomials
      polynomial –5x8 + 9x 4 – 7x 3 – x2, the term                  simply involves adding and subtracting the
      involving the highest power of x is –5x8, so the              coefficients of like terms. Note also that, by
      degree of the polynomial is 8.                                definition, a trinomial is a polynomial with
                                3
360. c. For the polynomial – 2 x + 5x 4 – 2x2 + 12,                 three terms and a binomial is a polynomial
      the term involving the highest power of x is                  with two terms.
      5x 4, so the degree of the polynomial is 4.             365. a. In general, dividing one polynomial by
361. a. A constant polynomial is of the form cx0 =                  another will result in an expression involving a
      c, where c is a constant. By this definition, the             term in which the variable is raised to a nega-
      degree of the constant polynomial 4 is zero.                  tive power. For instance, the quotient of even
                                                                                                              3
362. c. By definition, a polynomial is an expression                the very simple polynomials 3 and x2 is x2 =
      of the form anxn + an–1x n–1 + ... + a1x + a0                 3x–2, which is not a polynomial.
      where a0, a1, ..., an are real numbers and n is a       366. b.
      nonnegative integer. Put simply, once the                     –(–2x0)–3 + 4–2x2 – 3–1x – 2
      expression has been simplified, it cannot                                           1           1
                                                                    = –(–2)–3 +           42 x
                                                                                               2    – 3x – 2
      contain negative powers of the variable x.                             1            1           1
      Therefore, the expression x – 3x–2 is not a                   =–     (–2)3      +   42 x
                                                                                              2     – 3x – 2
      polynomial.                                                   =–
                                                                            1
                                                                                +
                                                                                       1        1
                                                                           –8         16 x2   – 3x – 2
363. c. A polynomial is an expression of the form
                                                                         1        1        15
      anxn + an–1x n–1 + ... + a1x + a0, where a0, a1, ...,         =   16 x
                                                                            2   – 3x –      8

      an are real numbers and n is a nonnegative              367. d.
      integer. That is, once the expression has been
                                                                    –(2 – (1 – 2x2 – (2x2 – 1))) – (3x2 – (1 – 2x2))
      simplified, it cannot contain negative powers
                                                                    = –(2 – (1 – 2x2 – 2x2 + 1)) – (3x2 – 1 + 2x2)
      of the variable x. If we simplify the expression
                                                                    = –(2 – (2 – 4x2)) – (5x2 – 1)
      (–2x)–1 – 2 using the exponent rules, we obtain
        1                                                           = –(2 – 2 + 4x2) – (5x2 – 1)
      – 2 x–1 – 2, which cannot be a polynomial
                               1                                    = –4x2 – 5x2 + 1
      because of the term – 2 x–1. Note that the
                                                                    = –94x2 + 1




                                                                                                               191
                                                                  ANSWERS & EXPLANATIONS–


368. b.                                                                           376. d. Use FOIL to find the product of two bino-
      –22(2–3          –   2–2x2)      +   33(3–2     –   3–3x3)                        mials. Then, add the products:
                  1            1                 1        1
      = –4        23   –       22 x
                                   2   + 27      32   –   33 x
                                                              3                         (x – 6)(x – 6) = x2 – 6x – 6x + 36 = x2 – 12x + 36

                  1        1                 1         1
                                                                                  377. a. To find the product of two binomials, mul-
      = –4        8    – 4 x2 + 27           9   –    27 x
                                                          3
                                                                                        tiply the first term of each binomial, the outside
          1                                                                             terms, the inside terms, and the last terms.
      = – 2 + x2 + 3 – x3
                                                                                        Then, add the products:
                                 5
      = –x3 + x2 +               2
                                                                                        (x – 1)(x + 1) = x2 + x – x – 1 = x2 – 1
Set 24 (Page 67)                                                                  378. e. First, note that (x + c)2 = (x + c)(x + c).
369. a.                                                                                 Then, use FOIL to find the product of the two
      (3x3) (7x2) = (3 7) (x3x2) = 21(x3+2) = 21x5                                      binomials. Finally, add the products:
370. c. 2x(5x2 + 3y) = 2x(5x2) + 2x(3y) = 10x3 + 6xy                                    (x + c)(x + c) = x2 + cx + cx + c2 = x2 + 2cx + c2
371. a. x3 + 6x = x x2 + 6 x = x(x2 + 6)                                          379. b. To find the product of two binomials, mul-
              2                              3             2          2
372. b. 2x (3x + 4xy – 2xy ) = 2x (3x) + 2x (4xy)                                       tiply the first term of each binomial, the out-
          2                3           3         3             3 3
      – 2x (2xy ) = 6x + 8x y – 4x y                                                    side terms, the inside terms, and the last terms.
373. d.                                                                                 Then, add the products:
      7x5(x8 + 2x4 – 7x –9)
                                                                                        (2x + 6)(3x – 9) = 6x2 – 18x + 18x – 54 = 6x2 – 54
      = 7x5(x8) + 7x5(2x4) – 7x5(7x) – 7x5(9)
                                                                                  380. e. Begin by multiplying the first two terms:
      = (7)       (x5x8)
                 + (7 2)                      (x5x4)       – (7 7)   (x 5x)   –         –3x(x + 6) = –3x2 – 18x. Then, multiply the
        (7 9) (x5)                                                                      two binomials, –3x2 – 18x and x – 9:

      = 7x13 + 14x9 – 49x6 – 63x5                                                       (–3x2 – 18x)(x – 9) = –3x3 + 27x2 – 18x2 +
374. c.                                                                                 162x = –3x3 + 9x2 + 162x
      4x2z(3xz3 – 4z2 + 7x5)                                                      381. c.
      = 4x2z(3xz3) + 4x2z(–4z2) + 4x2z(7x5)                                             (x – 4) (3x2 + 7x – 2)
      = 12x3z4 – 16x2z3 + 28x7z                                                         = x(3x2 + 7x – 2) – 4(3x2 + 7x – 2)
375. c. To find the product of two binomials, mul-                                      = x(3x2) + x(7x) – x(2) – 4(3x2) –4(7x) – 4(–2)
      tiply the first term of each binomial, the out-                                   = 3x3 + 7x2 – 2x – 12x2 – 28x + 8
      side terms, the inside terms, and the last terms                                  = 3x3 – 5x2 – 30x + 8
      (FOIL). Then, add the products:                                             382. e. Begin by multiplying the first two terms:

      (x – 3)(x + 7) = x2 + 7x – 3x – 21 = x2 + 4x – 21                                 (x – 6)(x – 3) = x2 – 3x – 6x + 18 = x2 – 9x + 18

                                                                                        Then, multiply (x2 – 9x + 18) by (x – 1):

                                                                                        (x2 – 9x + 18)(x – 1) = x3 – 9x2 + 18x – x2 +
                                                                                        9x – 18 = x3 – 10x2 + 27x – 18




   192
                                            ANSWERS & EXPLANATIONS–


383. c. First, simplify the left side of the equation:     392. c. 73x3 – 72x2 + 7x – 49 = 7(72x3 – 7x2 + x – 7)
                                                                 = 7(49x3 – 7x2 + x – 7)
      (5x + 1)(2y + 2) = 10xy + 2y + 10x + 2
                                                           393. b. 5x(2x + 3) – 7(2x + 3) = (2x + 3)(5x – 7)
      Now, simplify the equation by rearranging and        394. c. 5x(6x – 5) + 7(5 – 6x) = 5x(6x – 5) – 7(6x – 5)
      combining like terms:                                      = (5x – 7)(6x – 5)
      (5x + 1)(2y + 2) = 10xy + 12                         395. a.
      10xy + 2y + 10x + 2 = 10xy + 12                            6(4x + 1) – 3y(1 + 4x) + 7z(4x + 1)
      2y + 10x + 2 = 12                                          = 6(4x + 1) – 3y(4x + 1) + 7z(4x + 1)
      10x + 2y = 10                                              = (6 – 3y + 7z)(4x + 1)
                                                                        2           2                    2
      5x + y = 5                                           396. b. 5x( 3 x + 7) – ( 3 x + 7) = (5x – 1)( 3 x + 7)
384. a.                                                    397. c.
      (2x3 – 2x2 + 1)(6x3 + 7x2 – 5x – 9)                        3x(x + 5)2 – 8y(x + 5)3 + 7z(x + 5)2
                                                                 = (x + 5)2(3x) + (x + 5)2 (–8y(x + 5)) + (x + 5)2(7z)
      = 2x3(6x3 + 7x2 – 5x – 9) – 2x2 (6x3 + 7x2 – 5x –
                                                                 = (x + 5)2(3x – 8y(x + 5) + 7z)
        9) + (6x3 + 7x2 – 5x – 9)
                                                                 = (x + 5)2(3x – 8yx – 40y + 7z)
      = 12x6 + 14x5 – 10x4 – 18x3 – 12x5 – 14x4 +          398. a.
        10x3 + 18x2 + 6x3 + 7x2 – 5x –9                          8x4y2(x – 9)2 – 16x3y5(x – 9)3 + 12x5y3(9 – x)
      = 12x6 + (14x5 – 12x5) + (–10x4 – 14x4) +                  = 8x4y2(x – 9)2 – 16x3y5(x – 9)3 – 12x5y3(x – 9)
        (–18x3 + 10x3 + 6x3) + (18x2 + 7x2) – 5x –9
                                                                 = 4x3y2(x – 9)[2x(x –9)] + 4x3y2(x – 9)[–4y3(x
      = 12x6 + 2x5 – 24x4 – 2x3 + 25x2 – 5x –9                   – 9)2]+ 4x3y2(x – 9)[–3x2y]

Set 25 (Page 69)                                                 = 4x3y2(x – 9)[2x(x –9) – 4y3(x – 9) – 3x2y]
385. b. 15x – 10 = 5(3x) – 5(2) = 5(3x – 2)                      = 4x3y2(x – 9)[2x2 – 18x – 4y3(x2 – 18x + 81)
386. b.                                                          –3x2y]
      9x5 + 24x2 – 6x
                                                                 = 4x3y2(x – 9)[2x2 – 18x – 4y3x2 + 72y3 – 324y3
      = 3x(3x4) + 3x(8x) – 3x(2)
                                                                 –3x2y]
      = 3x(3x4 + 8x – 2)
                                                           399. c.
387. c.
                                                                 8x4y2z(2w – 1)3 – 16x2y4z3(2w – 1)3 +
      36x 4 – 90x3 – 18x
                                                                 12x4y4z(2w–1)4
      = 18x(2x3) + 18x(–5x2) + 18x(–1)
      = 18x(2x3 – 5x2 –1)                                        = 4x2y2z(2w – 1)3[2x2] + 4x2y2z(2w – 1)3[–4y2z2]
388. a. x3 – x = x(x2) + x(–1) = x(x2 –1)                        + 4x2y2z(2w – 1)3[3x2y2(2w – 1)]
389. d. 5x2 + 49 cannot be factored further.                     = 4x2y2z(2w – 1)3[2x2 – 4y2z2 + 3x2y2(2w – 1)]
390. a. 36 –   81x2   = 9(4) –   9(9x2)   = 9(4 –   9x2)
                                                                 = 4x2y2z(2w – 1)3[2x2 – 4y2z2 + 6x2y2w – 3x2y2]
391. c. 125x3 – 405x2 = 5x2(25x) + 5x2(–81) =
      5x2(25x – 81)



                                                                                                              193
                                                ANSWERS & EXPLANATIONS–


400. b.                                                     411. b.
      –22a3bc2(d – 2)3(1 – e)2 + 55a2b2c2(d – 2)2(1 – e)          1 + 2x + x2 = x2 + 2x + 1 = x2 + x + x + 1 =
      – 44a2bc4(d – 2)(1 – e)                                     (x2 + x) + (x + 1)

      = 11a2bc2(d – 2)(1 – e)[–2a(d – 2)2(1– e)] +                = x(x + 1) + (x + 1) = (x + 1)(x + 1) = (x + 1)2
      11a2bc2(d – 2)(1 – e)[5b(d – 2)] + 11a2bc2            412. c.
      (d – 2)(1 – e)[ –4c2]                                       4x2 – 12x + 9 = 4x2 – 6x – 6x + 9 = (4x2 – 6x) –
      = 11a2bc2(d – 2)(1 – e)[ –2a(d – 2)2(1– e) +                (6x – 9)
      5b(d – 2) –4c2]                                             = 2x(2x – 3) – 3(2x – 3) = 2x – 3)(2x – 3) =
                                                                  (2x – 3)2
Set 26 (Page 71)
                                                            413. c.
401. b.
                                                                  75x4 + 30x3 + 3x2 = 3x2[25x2 + 10x + 1] =
      x2   – 36 =   x2   –   62   = (x – 6)(x + 6)
                                                                  3x2[25x2 + 5x + 5x + 1]
402. a.
      144 – y2 = 122 – y2 = (12 – y)(12 + y)                      = 3x2[5x(5x + 1) + (5x + 1)] = 3x2[(5x + 1)
                                                                  (5x + 1)] = 3x2(5x + 1)2
403. d. 4x2 + 1 cannot be factored further.
                                                            414. a.
404. b. 9x2 – 25 = (3x)2 – (5)2 = (3x –5)(3x + 5)
                                                                  9x2(3 + 10x) – 24x(10x + 3) + 16(3 + 10x)
405. a. 121x4 – 49z2 = (11x2)2 – (7z)2 =
                                                                  = 9x2(3 + 10x) – 24x(3 + 10x) + 16(3 + 10x)
      (11x2 – 7z)(11x2 + 7z)
                                                                  = (3 + 10x)(9x2 – 24x + 16)
406. c. 6x2 – 24 = 6(x2) – 6(4) = 6(x2 –4) = 6((x)2 –
                                                                  = (3 + 10x)(9x2 – 12x – 12x + 16)
      (2)2) = 6(x – 2)(x + 2)
                                                                  = (3 + 10x)(3x(3x – 4) – 4(3x –4)
407. c.                                                           = (3 + 10x)(3x – 4)(3x – 4)
      32x5 – 162x = 2x(16x4 – 81) = 2x[(4x2)2 – 92] =             = (3 + 10x)(3x – 4)2
      2x(4x2 – 9)(4x2 + 9)
                                                            415. b.
      = 2x[(2x)2 – 32](4x2 + 9) = 2x(2x – 3)(2x + 3)              1 – 6x2 + 9x 4 = 1 – 3x2 – 3x2 + 9x 4 = (1 – 3x2)
      (4x2 + 9)                                                   – 3x2(1 – 3x2) = (1 – 3x2)(1 – 3x2) = (1 – 3x2)2
408. b.                                                     416. b.
      28x(5 – x) – 7x3(5 – x) = (28x – 7x3)(5 – x) =              8x7 – 24x4 + 18x = 2x(4x6 – 12x3 + 9) =
      7x(4 – x2)(5 – x)                                           2x[4x6 – 6x3 – 6x3 + 9]

      = 7x(22 – x2)(5 – x) = 7x(2 – x)(2 + x)(5 – x)              = 2x[2x3(2x3 – 3) – 3(2x3 – 3)]
409. a.                                                           = [(2x3 – 3)(2x3 – 3)] = 2x(2x3 – 3)2
      x2(3x – 5) + 9(5 – 3x) = x2(3x – 5) – 9(3x – 5)
      = (x2 – 9)(3x – 5) = (x – 3)(x + 3)(3x – 5)           Set 27 (Page 72)
410. a.                                                     417. a. x2 + 2x –8 = x2 + 4x – 2x – 8 = (x2 + 4x) –

      x(x2 + 7x) – 9x3(x2 + 7x) = (x – 9x3)(x2 + 7x) =            (2x + 8) = x(x + 4) – 2(x + 4) = (x + 4)(x – 2)
      [x(1 – 9x2)][x(x + 7)]                                418. a.
                                                                  x2 – 9x + 20 = x2 – 5x – 4x + 20 = (x2 – 5x) –
      = x(1 – (3x)2) x(x + 7) = x(1 – 3x)(1 + 3x)
                                                                  (4x – 20) = x(x – 5) – 4(x – 5) = (x – 4)(x – 5)
      x(x + 7) = x2(1 – 3x)(1 + 3x)(x + 7)
   194
                                         ANSWERS & EXPLANATIONS–


419. c.                                                    428. b.
     6x2 + 11x – 2 =    6x2
                          – x + 12x – 2 =    (6x2– x) +
                                                                27(x – 3) + 6x(x – 3) – x2(x – 3) = (x – 3)(27 +
     (12x – 2) = x(6x – 1) + 2(6x – 1) = (x + 2)(6x – 1)
                                                                6x – x2) = (x – 3)(27 + 6x + x2)
420. b.
     12x2 – 37x – 10 = 12x2 + 3x – 40x – 10 = 3x(4x             = –(x – 3)(x2 – 6x – 27) = –(x – 3)(x2 – 9x + 3x
     + 1) – 10(4x + 1) = (3x – 10)(4x + 1)                      – 27) = –(x – 3)(x(x – 9) + 3(x – 9))
421. c.                                                         = –(x – 3)(x + 3)(x – 9)
     7x2 – 12x + 5 = 7x2 – 5x – 7x + 5 = x(7x – 5) –
                                                           429. c.
     (7x – 5) = (7x – 5)(x – 1)
                                                                (x2 + 4x + 3)x2 + (x2 + 4x + 3)3x + 2(x2 + 4x +
422. a.                                                         3) = (x2 + 4x + 3)[x2 + 3x + 2]
     9 – 7x – 2x2 = 9 + 2x – 9x – 2x2 = 1(9 + 2x) –
     x(9 + 2x) = (9 + 2x)(1 – x)                                = (x2 + 3x + x + 3)[x2 + x + 2x + 2] = (x(x +
                                                                3) + (x + 3))[x(x + 1) + 2(x + 1)]
423. c.
     2x3 + 6x2 + 4x = 2x(x2 + 3x + 2) =                         = ((x + 1)(x + 3))[(x + 2)(x + 1)] = (x + 1)2(x
     2x(x2 + x + 2x + 2)                                        + 2)(x + 3)

     = 2x(x(x + 1) + 2(x + 1) = 2x(x + 2)(x + 1)           430. a.
                                                                18(x2 + 6x + 8) – 2x2(x2 + 6x + 8) = (x2 + 6x +
424. c.
                                                                8)[18 – 2x2] = (x2 + 6x + 8)[2(9 – x2)]
     –4x5 + 24x4 – 20x3 = –4x3(x2 – 6x + 5) =
     –4x3(x2 –x – 5x + 5)                                       = (x2 + 4x + 2x + 8)[2(32 – x2)] = (x(x+ 4)
                                                                +2(x + 4))[2(3 – x)(3 + x)]
     = –4x3(x(x – 1) – 5(x –1)) = –4x3(x – 5)(x –1)
425. b.                                                         = 2(x + 2)(x + 4)(3 – x)(3 + x)
     –27x4   +   27x3–  6x2=  –3x2(9x2   – 9x + 2) =       431. a.
     –3x2(9x2    – 6x – 3x + 2)                                 2x2(16 + x4) + 3x(16 + x4) = (16 + x4) = (16 +
                                                                x4)[2x2 + 3x + 1]
     = –3x2[(9x2 –6x) – (3x – 2)] = –3x2[3x(3x – 2)
     – (3x – 2)] = –3x2(3x – 1)(3x – 2)                         = (16 + x4)[2x2 + 2x + x + 1] = (16 + x4)[2x(x
426. b.                                                         + 1) + (x + 1)] = (16 + x4)(2x + 1)(x + 1)
     x2(x + 1) – 5x(x + 1) + 6(x + 1) = (x + 1)(x2 –       432. a.
     5x + 6) = (x + 1)(x2 – 2x – 3x + 6)                        6x2(1 – x4) + 13x(1 – x4) + 6(1 – x4) = (1 – x4)
                                                                [6x2 + 13x + 6] = (1 – x4)[6x2 + 4x + 9x + 6]
     = (x + 1)[x(x – 2) – 3(x – 2)] = (x + 1)(x – 3)
     (x – 2)                                                    = (12 – (x2)2)[2x(3x + 2) + 3(3x + 2)] =
427. a.                                                         (1 – x2)(1 + x2)(2x + 3)(3x + 2)
     2x2(x2– 4) – x(x2 – 4) + (4 – x2) = 2x2(x2 – 4) –
                                                                = (1 – x)(1 + x)(1 + x2)(2x + 3)(3x + 2)
     x(x2 – 4) – (x2 – 4) = (x2 – 4)[2x2 – x – 1]
     = (x2– 4)[2x2 – 2x + x – 1] = (x2 – 4)[2x(x – 1)
     + (x –1)] = (x2 – 4)(2x + 1)(x – 1)
     = (x– 2)(x + 2)(2x + 1)(x – 1)


                                                                                                           195
                                         ANSWERS & EXPLANATIONS–



Set 28 (Page 74)                                           439. a. Begin by factoring the polynomial:
433. c. First, factor the polynomial:                            28x(5 – x) – 7x3(5 – x) = (28x – 7x3)(5 – x) =
      x2 – 36 = x2 – 62 = (x – 6)(x + 6)                         7x(4 – x2)(5 – x)

      Now, set each of the factors equal to zero and             = 7x(22 – x2)(5 – x) = 7x(2 – x)(2 + x)(5 – x)
      solve for x to conclude that the zeros of the              There are four factors: 7x, 2 – x, 2 + x, and 5 –
      polynomial are –6 and 6.                                   x. Now, set each of these factors equal to zero
434. a. First, factor the polynomial:                            and solve for x. The zeros of the polynomial
      9x2 – 25 = (3x)2 – (5)2 = (3x – 5)(3x + 5)                 are 0, –2, 2, and 5.
                                                           440. c. First, factor the polynomial:
      The factors are 3x – 5 and 3x + 5. Now, set
      each factor equal to zero and solve for x. The             75x4 + 30x3 + 3x2 = 3x2[25x2 + 10x + 1] =
                                    5      5                     3x2[25x2 + 5x + 5x + 1]
      zeros of the polynomial are – 3 and 3 .
435. d. First, note that 5x2 + 49 cannot be factored             = 3x2[5x(5x + 1) + (5x + 1)] = 3x2[(5x + 1)
      further. Since both terms are positive, the sum            (5x + 1)] = 3x2(5x + 1)2
      is positive, so there is no x-value that makes
      the expression equal to zero.                              The factors are 3x2 and (5x + 1)2. Now, set
                                                                 each factor equal to zero and solve for x to
436. b. First, factor the polynomial:
                                                                 conclude that the zeros of the polynomial are
      6x2 – 24 = 6(x2) – 6(4) = 6(x2 – 4) = 6((x)2) =            0 and – 1 .
                                                                         5
      6(x – 2)(x + 2)                                      441. a. First, factor the polynomial:

      The factors are 6, x – 2, and x + 2. Set each fac-         x2 – 9x + 20 = x2 – 5x– 4x + 20 =
      tor equal to zero and solve for x to conclude              (x2 – 5x) – (4x – 20) = x(x – 5) = (x – 4)(x – 5)
      that the zeros of the polynomial are –2 and 2.             Now, set each factor on the right side of the
437. d. Begin by factoring, the polynomial:                      string of equalities equal to zero and solve for
                                                                 x. The zeros of the polynomial are 4 and 5.
      5x(2x + 3) – 7(2x + 3) = (2x + 3)(5x – 7)
                                                           442. c. Begin by factoring the polynomial:
      The factors are 2x + 3 and 5x – 7. Now, set
      each factor equal to zero and solve for x to find          12x2 – 37x – 10 = 12x2 + 3x – 40x – 10 =
                                             3       7
      that the zeros of the polynomial are – 2 and 5 .           3x(4x + 1) – 10(4x + 1) = (3x – 10)(4x + 1)

438. d. First, factor the polynomial:                            The factors are 3x – 10 and 4x + 1. Now, set
                                                                 each factor equal to zero and solve for x to find
      5x( 2 x + 7) – ( 2 x + 7) = (5x –1)( 2 x + 7)
          3            3                   3
                                                                 that the zeros of the polynomial are 130 and – 1 .
                                                                                                                 4
      The factors are 5x – 1 and 2 x + 7. Now, set
                                 3
      each factor equal to zero and solve for x. The
      zeros of the polynomial are 1 and – 221 .
                                   5




   196
                                        ANSWERS & EXPLANATIONS–


443. d. First, factor the polynomial:                      447. d. Begin by factoring the polynomial:

      9 –7x – 2x2 = 9 + 2x – 9x – 2x2 = 1(9 + 2x)                2x2(16 + x4) + 3x(16 + x4) + (16 + x4) =
      –x(9 + 2x) = (9 + 2x)(1 – x)                               (16 + x4)[2x2 + 3x + 1]

      The factors are 9 + 2x and 1 – x. Set each fac-            = (16 + x4)[2x2 + 2x + x + 1] = (16 + x 4)[2x(x +
      tor equal to zero and solve for x. The zeros of            1) + (x + 1)] = (16 + x4)(2x + 1)(x + 1)
      the polynomial are – 9 and 1.
                           2
                                                                 The three factors are 16 + x4, 2x + 1, and x + 1.
444. b. Begin by factoring the polynomial:
                                                                 Now, set each factor equal to zero. Solve for x
      2x3 + 6x2 + 4x = 2x(x2 + 3x + 2) = 2x(x2 + x +             to find that zeros of the polynomial: –1 and – 1 .
                                                                                                                2
      2x + 2)                                              448. b. First, factor the polynomial:

      2x(x(x + 1) + 2(x + 1) = 2x(x + 2)(x + 1)                  18(x2 + 6x + 8) – 2x2(x2 + 6x + 8) = (x2 +
                                                                 6x + 8)[18 – 2x2] = (x2 + 6x + 8)[2(9 – x2)]
      There are three factors: 2x, x + 2, and x + 1.
      Set each factor equal to zero and solve for x to           = (x2 + 4x + 2x + 8)[2(32 – x2)] = (x(x + 4) +
      conclude that the zeros of the polynomial are              2(x + 4))[2(3 – x)(3 + x)]
      –2, –1, and 0.
                                                                 = 2(x + 2)(x + 4)(3 – x)(3 + x)
445. c. First, factor the polynomial:
                                                                 Set each of the four factors equal to zero and
      –4x5 + 24x4 – 20x3 = –4x3(x2 – 6x + 5) =
                                                                 solve for x. The zeros of the polynomial are
      –4x3(x2 – x – 5x + 5)
                                                                 –4, –2, –3, and 3.
      = –4x3(x(x – 1) – 5(x – 1)) = –4x3(x – 5)(x – 1)
                                                           Set 29 (Page 75)
      The three factors are –4x3, x – 5, and x – 1. Now,
                                                           449. b. The strategy is to determine the x-values
      set each factor equal to zero and solve for x to
                                                                 that make the expression on the left side equal
      find that zeros of the polynomial: 0, 1, and 5.
                                                                 to zero. Doing so requires that we first factor
446 a. First, factor the polynomial:                             the polynomial:
      2x2(x2 – 4) – x(x2 – 4) + (4 – x2) = 2x2(x2 – 4)           x2 – 36 = x2 – 62 = (x – 6)(x + 6)
      – x(x2 – 4) – (x2 – 4) = (x2 – 4)[2x2 – x – 1]
                                                                 Next, set each factor equal to zero and solve
      = (x2 – 4)[2x2 – 2x + x – 1] = (x2 – 4)[2x(x – 1)          for x to conclude that the zeros of the polyno-
      + (x – 1] = (x2 – 4) (2x +1)(x – 1)                        mial are –6 and 6. Now, we assess the sign of
      = (x– 2)(x + 2)(2x + 1)(x – 1)                             the expression on the left side on each subin-
                                                                 terval formed using these values. To this end,
      Now, set each of the four factors equal to zero            we form a number line, choose a real number
      and solve for x. The zeros of the polynomial               in each of the subintervals, and record the sign
      are 1, 2, –2, and – 1 .
                          2                                      of the expression above each:

                                                                             +        –        +
                                                                                 –6        6



                                                                                                           197
                                       ANSWERS & EXPLANATIONS–


     Since the inequality does not include “equals,”     452. a. Find the x-values that make the expression
     we do not include those values from the num-              on the left side equal to zero. First, factor the
     ber line that make the polynomial equal to zero.          polynomial:
     Therefore, the solution set is (–∞, –6)∪(6, ∞).
                                                               6x2 – 24 = 6(x2) – 6(4) = 6(x2 –4) =
450. b. Determine the x-values that make the
                                                               6((x)2 – (2)2) = 6(x –2)(x + 2)
     expression on the left side equal to zero. First,
     factor the polynomials:                                   Next, set each factor equal to zero. Solve for x
                                                               to find that the zeros of the polynomial: –2
     9x2 – 25 = (3x)2 – (5)2 = (3x – 5)(3x + 5)
                                                               and 2. Now, assess the sign of the expression
     Next, set each factor equal to zero and solve             on the left side on each subinterval formed
     for x to conclude that the zeros of the polyno-           using these values. Form a number line, choose
     mial are – 5 and 5 . Now, assess the sign of the
                3     3
                                                               a real number in each of the subintervals, and
     expression on the left side on each subinterval           record the sign of the expression above each:
     formed using these values. To this end, form a
                                                                           +        –        +
     number line, choose a real number in each of
     the subintervals, and record the sign of the                              –2        2

     expression above each:                                    Since the inequality includes “equals,” we
                 +        –        +                           include those values from the number line that
                                                               make the polynomial equal to zero. The solu-
                     –5
                      3
                              5
                              3                                tion set is (–∞, –2]∪[2, ∞).
                                                         453. a. The strategy is to determine the x-values
     Since the inequality includes “equals,” include
                                                               that make the expression on the left side equal
     those values from the number line that make
                                                               to zero. Doing so requires that we first factor
     the polynomial equal to zero. The solution set
                                                               the polynomial:
     is [– 5 , 5 ].
           3 3
451. c. Determine the x-values that make the                   5x(2x + 3) – 7(2x + 3) = (2x + 3)(5x + 7)
     expression on the left side equal to zero. Begin          Set each factor equal to zero and solve for x.
     by factoring the polynomial, if possible. How-            The zeros of the polynomial: – 3 and 7 . Next
                                                                                               2      5
     ever, note that 5x2 + 49 cannot be factored fur-          assess the sign of the expression on the left side
     ther. Moreover, since both terms are positive             on each subinterval formed using these values.
     for any value of x, the sum is positive for every         Form a number line, choose a real number in
     value of x. Therefore, the solution set is the            each of the duly formed subintervals, and
     empty set.                                                record the sign of the expression above each:

                                                                           +        –        +

                                                                               –3
                                                                                2
                                                                                         7
                                                                                         5




   198
                                         ANSWERS & EXPLANATIONS–


      Since the inequality does not include “equals,”           each of the subintervals, and record the sign of
      do not include those values from the number               the expression above each:
      line that make the polynomial equal to zero. As
      such, the solution set is(–∞, – 3 )∪( 7 , ∞).
                                      2     5
                                                                   +        –        +       –        +
454. c. Find the x-values that make the expression                     –2       0        2        5

      on the left side equal to zero. First, factor the
      polynomial:                                               Since the inequality includes “equals,” include
      5x( 2 x + 7) – ( 2 + 7) = (5x – 1)( 2 x + 7)              the values from the number line that make the
          3            3                  3
                                                                polynomial equal to zero. The solution set
      Next, set each factor equal to zero and solve             is(–∞, –2)∪[0, 2]∪(5, ∞).
      for x. The zeros of the polynomial are 1 and
                                                 5        456. c. First, determine the x-values that make the
      – 221 . Assess the sign of the expression on the
                                                                expression on the left side equal to zero. Doing
      left side on each subinterval formed using
                                                                so requires that we factor the polynomial:
      these values. Form a number line, choose a
      real number in each of the subintervals, and              75x4 + 30x3 + 3x2 = 3x2[25x2 + 10x + 1] =
      record the sign of the expression above each:             3x2[25x2 + 5x + 5x + 1]

                                                                = 3x2[5x(5x + 1) + (5x + 1)] = 3x2[(5x + 1) +
                  +          –       +
                                                                (5x + 1)] = 3x2(5x + 1)2
                      – 21
                        2
                                 1
                                 5
                                                                Set each factor equal to zero, then solve for x
                                                                to find the zeros of the polynomial: 0 and – 1 .
                                                                                                              5
      The inequality includes “equals,” so we include
                                                                Assess the sign of the expression on the left
      those values from the number line that make
                                                                side on each subinterval formed using these
      the polynomial equal to zero. The solution set
                                                                values: Form a number line, choose a real
      is [– 221 , 1 ].
                  5
                                                                number in each subinterval, and record the
455. c. Determine the x-values that make the
                                                                sign of the expression above each:
      expression on the left side equal to zero. To do
      this, factor the polynomial:                                          +        +       +

      28x(5 – x) – 7x3(5 – x) = (28x – 7x3)(5 – x) =                            –1
                                                                                 5
                                                                                         0

      7x(4 – x2)(5 – x)
                                                                The inequality includes “equals,” so we include
      = 7x(22 – x2)(5 – x) = 7x(2 – x)(2 + x)(5 – x)            those values from the number line that make
      Next, set each factor equal to zero. Solve for x          the polynomial equal to zero. Since every
      to find the zeros of the polynomial, which are            x-value that is not a zero of the polynomial
      0, –2, 2, and 5. Now, assess the sign of the              results in a positive quantity, the solution set
      expression on the left side on each subinterval           consists of only the zeros of the polynomial,
      formed using these values. To this end, we                namely {– 1 , 0}.
                                                                            5

      form a number line, choose a real number in




                                                                                                          199
                                        ANSWERS & EXPLANATIONS–


457. d. Find the x-values that make the expression              Because the inequality does not include
      on the left side equal to zero. Begin by factor-          “equals,” we exclude those values from the
      ing the polynomial:                                       number line that make the polynomial equal
                                                                to zero. The solution set is(– 1 , 130 ).
                                                                                               4
      x2 – 9x + 20 = x2 – 5x – 4x + 20 = (x2 – 5x) –
                                                          459. d. Determine the x-values that make the
      (4x – 20) = x(x – 5) – 4(x – 5) = (x – 4)(x – 5)
                                                                expression on the left side equal to zero. To do
      Set each factor equal to zero, then solve for x           this, we first factor the polynomial:
      to find the zeros of the polynomial, which are
                                                                9 – 7x – 2x2 = 9 + 2x – 9x – 2x2 = 1(9 + 2x) –
      4 and 5. Now, assess the sign of the expression
                                                                x(9 + 2x) = (9 + 2x)(1 –x)
      on the left side on each subinterval formed
      using these values. To this end, form a number            Next, set each factor equal to zero and solve
      line, choose a real number in each subinterval,           for x to find the zeros of the polynomial which
      and record the sign of the expression above               are – 9 and 1. Now, we assess the sign of the
                                                                      2
      each:                                                     expression on the left side on each subinterval
                                                                formed using these values. Form a number line,
                 +         –        +
                                                                choose a real number in each subinterval, and
                       4       5
                                                                record the sign of the expression above each:
      The inequality does not include “equals,” so we                      –        +        –
      exclude those values from the number line that
                                                                                         1
      make the polynomial equal to zero. Therefore,                            –9
                                                                                2
      the solution set is (4,5).
                                                                The inequality does not include “equals,” so we
458. d. First, find the x-values that make the expres-          do not include those values from the number
      sion on the left side equal to zero. Doing so             line that make the polynomial equal to zero.
      requires that we factor the polynomial:                   The solution set is (– 9 ,1).
                                                                                       2
                                                          460. b. The strategy is to determine the x-values
      12x2 –37x – 10 = 12x2 + 3x – 40x – 10 =
                                                                that make the expression on the left side equal
      3x(4x + 1) – 10(4x + 1) = (3x – 10)(4x + 1)
                                                                to zero. First, factor the polynomial:
      Next, set each factor equal to zero and solve
      for x to conclude that the zeros of the polyno-           2x3 + 6x2 + 4x = 2x(x2 + 3x + 2) =
      mial are 130 and – 1 . Now, we assess the sign of
                         4
                                                                2x(x2 + x + 2x + 2)
      the expression on the left side on each subin-
                                                                = 2x(x(x + 1) + 2(x + 1)) = 2x(x + 2)(x + 1)
      terval formed using these values. To this end,
      we form a number line, choose a real number               Next, set each factor equal to zero and solve
      in each subinterval, and record the sign of the           for x to conclude that the zeros of the polyno-
      expression above each, as follows:                        mial are –2, –1, and 0. Now, we assess the sign
                                                                of the expression on the left side on each
                  +        –        +
                                                                subinterval formed using these values. To this
                      –1
                       4
                               10
                                3
                                                                end, we form a number line, choose a real



   200
                                           ANSWERS & EXPLANATIONS–


      number in each subinterval, and record the          462. a. First, determine the x-values that make the
      sign of the expression above each:                        expression on the left side equal to zero. This
                                                                requires that we factor the polynomial:
           –         +        –        +
               –2        –1        0                            2x2(x2 –4) – x(x2 – 4) + (4 – x2) = 2x2(x2 – 4) –
                                                                x(x2 – 4) – (x2 – 4) = (x2 – 4)[2x2 –x –1]
      Since the inequality includes “equals,” we
      include those values from the number line                 = (x2 – 4)[2x2 – 2x + x – 1] = (x2 – 4)[2x(x – 1)
      that make the polynomial equal to zero.                   + (x – 1)] = (x2 – 4)(2x + 1)(x – 1)
      The solution set is[–2, –1]∪[0, ∞).
                                                                = (x – 2)(x + 2)(2x + 1)(x – 1)
461. a. Find the x-values that make the expression
      on the left side equal to zero. First, factor the         Set each factor equal to zero and solve for x to
      polynomial:                                               find the zeros of the polynomial are 1, 2, –2,
                                                                and – 1 . Assess the sign of the expression on
                                                                       2
      –4x5 + 24x4 – 20x3 = 4x3(x2 – 6x + 5) =                   the left side on each subinterval formed using
      –4x3(x2 – x – 5x + 5)                                     these values. To this end, form a number line,
      = –4x3(x(x – 1) – 5(x – 1)) = –4x3(x – 5)(x – 1)          choose a real number in each subinterval, and
                                                                record the sign of the expression above each:
      Next, set each factor equal to zero and solve
      for x. The zeros of the polynomial are 0, 1, and               +        –        +        –       +
      5. Now, we assess the sign of the expression on                     2
                                                                                  –1       1        2
                                                                                   2
      the left side on each subinterval formed using
      these values: We form a number line, choose a             The inequality does not include “equals,” so we
      real number in each subinterval, and record               exclude those values from the number line that
      the sign of the expression above each:                    make the polynomial equal to zero. The solu-
                                                                tion set is (–2, – 1 )∪(1, 2).
                                                                                   2
               +         –        +         –
                                                          463. c. Determine the x-values that make the
                    0         1        5
                                                                expression on the left side equal to zero. First,
      The inequality includes “equals,” so we include           factor the polynomial:
      those values from the number line that make               2x2(16 + x4) + 3x)16 + x4) + (16 + x4) =
      the polynomial equal to zero. The solution set            (16 + x4)[2x2 + 3x + 1]
      is(–∞, 0]∪[1, 5).
                                                                = (16 + x4)[2x2 +2x + x + 1] = (16 + x4)[2x(x +
                                                                1) + (x + 1)] =(16 + x4)(2x + 1)(x + 1)

                                                                Set each factor equal to zero and solve for x.
                                                                The zeros of the polynomial are – 1 and – 1 .
                                                                                                            2
                                                                Assess the sign of the expression on the left
                                                                side on each subinterval formed using these




                                                                                                            201
                                                  ANSWERS & EXPLANATIONS–


      values: Form a number line, choose a real               Section 4—Rational Expressions
      number in each of the subinterval, and record
      the sign of the expression above each, as               Set 30 (Page 78)
      follows:                                                          2z2 – z – 15       (2z + 5)(z – 3)        2z + 5
                                                              465. d. z2 + 2z – 15 = (z + 5)(z – 3) = z + 5
                                                                        25(–x)4         25x4        1
                    +        –                +               466. d.
                                                                        x(5x2)2
                                                                                   =   x 25x4
                                                                                                =   x
                        –1               1                              z3 – 16z       z(z2 – 16)  z(z –4)(z + 4)
                                     –
                                         2
                                                              467. a. 8z – 32 =
                                                                                8(z – 4)
                                                                                         =            8(z – 4)
                                                                                                                         +
                                                                                                                   = z(z 8 4)
                                                                        y2 – 64    (y – 8)(y + 8)    (y – 8)(y + 8)
                                                              468. b. 8 – y =                     = –(y – 8) = –(y +                 8)
      The inequality includes “equals,” so we include                                   8–y
                                                                          2
      those values from the number line that make                       x + 8x          x(x + 8)           x(x + 8)             1
                                                              469. a.              =                =                      =   x–8
                                                                        x3 – 64x       x(x2 – 64)       x(x + 8)(x – 8)
      the polynomial equal to zero. Therefore, the
                                                                            2x2 + 4x              2x(x + 2)
                                                              470. c.                      =                       =
      solution set is [–1, – 1 ].
                             2                                          4x3 – 16x2 – 48x       4x(x2 – 4x – 12)

464. b. Find the x-values that make the expression                         2x(x + 2)              1              1
                                                                        4x(x – 6)(x + 2)
                                                                                           =   2(x – 6)
                                                                                                          =   2x – 12
      on the left side equal to zero. First, factor the
                                                              471. a. A rational expression is undefined at any
      polynomial:
                                                                    value of x that makes the denominator equal
      18(x2 + 6x + 8) – 2x2(x2 + 6x + 8) = (x2 +                    to zero even if the corresponding factor can-
      6x + 8)[18 – 2x2] = (x2 + 6x + 8)[2(9 – x2)]                  cels with one in the numerator. Observe that
                                                                    the denominator factors as 4x3 + 44x2 + 120x
      = (x2 + 4x + 2x + 8)[2(32 – x2)] = (x(x + 4) +
                                                                    = 4x(x2 + 11x + 30) = 4x(x + 5)(x + 6).
      2(x + 4))[2(3 – x)(3 + x)]
                                                                    Setting each factor equal to zero shows that
      = 2(x + 2)(x + 4)(3 – x)(3 + x)
                                                                    the rational expression is undefined at x = 0, –5,
      Set each factor equal to zero and solve for x to              and –6.
      find the zeros of the polynomial, which are –4,         472. c. The domain of a rational expression is the
      –2, –3, and 3. Assess the sign of the expression              set of all real numbers that do not make the
      on the left side on each subinterval formed using             denominator equal to zero. For this function,
      these values. To this end, form a number line,                the values of x that must be excluded from
      choose a real number in each subinterval, and                 the domain are the solutions of the equation
      record the sign of the expression above each:                 x3 – 4x = 0. Factoring the left side yields the
                                                                    equivalent equation
           –        +            –            +        –
               –4       –3               –2        3                x3 – 4x = x(x2 – 4) = x(x – 2)(x + 2) = 0

                                                                    The solutions are x = –2, 0, and 2. Hence, the
      Because the inequality does not include
                                                                    expression is defined for any x in the set
      “equals,” we exclude those values from the
                                                                    (–∞,–2)∪(–2,0) ∪(0,–2) ∪(2,∞).
      number line that make the polynomial equal
      to zero. The solution set is(–4, –3)∪(–2, 3).




   202
                                                        ANSWERS & EXPLANATIONS–



               x2 – 16             (x – 4)(x + 4)
473. d. x3 + x2 – 20x = x(x2 + x –20) = x(x + 5)(x – 4) =
                                                            (x – 4)(x + 4)   Set 31 (Page 79)
            x+4                                                              481. a.
           x2 + 5x                                                                  4x– 45               2x – 9             3x + 1
                                                                                     x–9         +       x–9           –     x–9           =
           x       1        5
474. e.   4x   =   4   =   20 , provided     that x is not zero.                    (4x – 45) + (2x – 9) – (3x + 1)
                                                                                                 x–9                                        =
475. b. Any value of x that makes the denominator
                                                                                    4x – 45 + 2x – 9 – 3x – 1                           3x – 55
       equal to zero, even if it also makes the numer-                                        x–9                               =        x–9
       ator equal to zero, renders a rational expres-
                                                                                       5a                2a            5a + 2a                 7a          7
       sion undefined. For the given expression, both                        482. a.             +              =                       =             =
                                                                                       ab3               ab3             ab3                   ab3         b3
       4 and –4 make the denominator equal to zero.                          483. d.
476. b. Any value of x that makes the denominator                                       3 – 2x                          2–x                          3 – 2x – (2 – x)
                                                                                    (x + 2)(x – 1)
                                                                                                               –   (x – 1)(x + 2)
                                                                                                                                               =      (x – 1)(x + 2)
                                                                                                                                                                           =
       equal to zero, even if it also makes the numer-                                   1–x                           –(x – 1)                            1
                                                                                    ( x – 1)(x + 2)
                                                                                                               =    (x – 1)(x + 2)
                                                                                                                                                =–        x+ 2
       ator equal to zero, renders a rational expres-
       sion undefined. To determine these values for                                    4             2              4s             2r2              4s + 2r2        2(2s + r2)
                                                                             484. d.           +               =            +                  =                 =
                                                                                       sr3           rs2            s2r3            s2r3               s2r3             s2r3
       the given expression, we factor the denomina-                         485. c.
       tor as x3 + 3x2 – 4x = x(x2 + 3x – 4) = x(x + 4)                                2                     5 – 2x                           2             (x – 1)
                                                                                    x(x – 2)
                                                                                                 –       (x – 2)(x – 1)
                                                                                                                                    =      x(x – 2)         (x – 1)
                                                                                                                                                                       –
       (x – 1). Note that the values –4, 0, and 1 all
                                                                                         5 – 2x                    x         2(x – 1) – x(5 – 2x)
       make the given expression undefined.                                          (x – 2)(x – 1)                x    =      x(x – 1)(x – 2)
                                                                                                                                                                 =
                                                                                                               2                2
                                                                                    2x – 1 – 5x + 2x                        2x – 3x – 1
477. b.                                                                                                             =
                                                                                     x(x – 1)(x – 2)                       x(x – 1)(x – 2)
        5x2(x – 1) – 3x(x – 1) – 2(x – 1)
       10x2(x – 1) + 9x(x – 1) + 2(x – 1)           =                                        4              2               4                  2(t = 2)          4 – 2(t + 2)
                                                                             486. b. t(t + 2) – t = t(t + 2) – t(t + 2) =                                          t(t + 2)     =
         (x – 1)(5x2 – 3x – 2)           (x –1)(5x + 2)(x – 1)
        (x – 1)(10x2 + 9x + 2)      =   (x –1)(5x + 2)(2x + 1)       =                –2t
                                                                                                 =        –2
                                                                                    t(t + 2)             t+2
        x–1
       2x + 1                                                                487. b.
                                                2
           6x3 – 12x           6x(x2 – 2)                                              1                        2x                         3            1            (x + 2)
478. b.      24x2          =   4 6 x x      = x 4– 2
                                                 x                                  x(x + 1)         –    (x + 1)(x + 2)             +     x    =    x(x + 1)        (x + 2)
                                                                                             2x                        x        3          (x + 1)(x + 2)
            4ab2 – b2              b2(4a – 1)                b2                    –   (x + 1)(x + 2)                  x   +    x          (x + 1)(x + 2)
                                                                                                                                                                 =
479. a. 8a2 + 2a – 1 = (4a – 1)(2a + 1) = 2a + 1
                                                                                    x + 2 – 2x2 + 3(x + 1)(x + 2)
                                                                                           x(x + 1)(x + 2)
                                                                                                                                         =
           (2x – 5)(x + 4) – (2x – 5)(x + 1)
480. c.                9(2x – 5)
                                                    =
                                                                                    x + 2 – 2x2 + 3x2 + 3x + 6x + 6                                  x2 + 10x + 8
                                                                                            x(x + 1)(x + 2)
                                                                                                                                               =    x(x + 1)(x + 2)
          (2x – 5)((x + 4) – (x + 1))           3       1
                   9(2x – 5)
                                            =   9   =   3
                                                                             488. b.
                                                                                      x                1                     2x2                  x               2x –1
                                                                                    2x + 1       –   2x – 1         +      4x2 – 1       =      2x + 1            2x – 1   –
                                                                                       1             2x + 1                  2x2
                                                                                     2x –1           2x + 1         +      4x2 – 1         =
                                                                                    x(2x – 1) – 1(2x + 1) + 2x2                                2x2 – x – 2x – 1 + 2x2
                                                                                         (2x – 1)(2x + 1)                            =           (2x – 1)(2x + 1)
                                                                                                                                                                           =
                                                                                      4x2 – 3x – 1                       (4x + 1)(x – 1)
                                                                                    (2x – 1)(2x + 1)
                                                                                                                   =    (2x – 1)(2x + 1)




                                                                                                                                                                        203
                                                                           ANSWERS & EXPLANATIONS–


489. c.                                                                                                   495. c.
        3y + 2                  7y –3                        5               3y + 2            y+1              x–1          3x – 4             x–1           3x – 4            x(x – 1)
       (y – 1)2         –   (y – 1)(y + 1)          +       y+1        =    (y – 1)2           y+1              x– 2     –   x2 – 2x      =     x–2     –    x(x – 2)       =   x(x – 2)    –
                                                                                       2                                                                               2
                  7y – 3                  (y – 1)             5               (y – 1)                            3x – 4          x(x – 1) – (3x – 4)                 x – x – 3x + 4
      –       (y – 1)(y + 1)              (y – 1)      +     y+1              (y – 1)2     =                    x(x – 2)
                                                                                                                             =        x(x – 2)
                                                                                                                                                               =        x(x – 2)             =
                                                                               2                                 2                              2
       (3y + 2)(y + 1) – ( 7y – 3)(y – 1) + 5(y – 1)                                                            x – 4x + 4             (x – 2)           x–2
                      (y – 1)2(y + 1)
                                                                                   =                                             =                  –     x
                                                                                                                 x(x – 2)              x(x – 2)
       3y2 + 5y + 2 – (7y2 – 10y + 3) + 5(y2 – 2y + 1
                                                                                   =                      496. c.
                       (y – 1)2(y + 1)
                                                                                                                       x–1            3x – 3                  x–1             3x – 3
       3y2 + 5y + 2 – 7y2 + 10y – 3 + 5y2 – 10y + 5
                                                                                   =                            1+      x        –    x2 + 3x    =1+           x        –    x(x + 3)    =
                      (y – 1)2(y +1)
                                                                                                                x(x + 3)             (x – 1)(x + 3            3x – 3
         y2 + 5y + 4
                                    =    (y + 4)(y + 1)
                                                                  =       y+4
                                                                                                                x(x + 3)     +         x(x + 3)         –    x(x + 3)       =
       (y – 1)2(y + 1)                   (y – 1)2(y + 1)                 (y – 1)2
                                                                                                                x(x + 3) + (x – 1)(x + 3) – (3x – 3)
                6z + 12              2z – 6
                                                   –1
                                                                 6z + 12 + 2z – 6
                                                                                            –1                                x(x + 3)                              =
490. a.          4z + 3         +    4z + 3            =              4z + 3                   =
                                                                                                                x2 + 3x + x2 + 2x – 3 – 3x + 3
                                                                                                                           x(x + 3)                          =
                                –1                           –1
                8z + 6                     2(4z + 3)                     –1        1
                4z + 3              =        4z + 3               =2 =             2                            2x2 + 2x             2x(x + 1)           2(x + 1)
                                                                                                                x(x + 3)     =       x(x + 3)       =      x+3
                4           x+5                4                 x+5                   4         x+5
491. a. x –3 + 3 – x = x – 3 + –(x – 3) = x – 3 – x – 3
                                                                                                          Set 32 (Page 81)
              4 –(x + 5)            x–1                    x+1
      =          x–3            = – x–3 =                  x–3
                                                                                                                    3 2 y3z4  4x3y5z4 2y5z
                                                                                                          497. c. 4x 3y 2x5 = 2x5z3 = x2
                                                                                                                   z
492. a.
                                                                                                          498. d.
             x                        3                            x                        3                    8a4         5a2 + 13a – 6                     8a4
       x2 –10x + 24             –    x–6      +1=            (x – 6)(x – 4)            –   x–6       +1         9 – a2        24a – 60a2
                                                                                                                                                    =    (3 – a)(3 + a)
                    x                         3(x – 4)                   (x – 6)(x – 4)                         (5a – 2)(a + 3)                     8a4                (5a –2)(a + 3)
      =       (x – 6)(x – 4)
                                     –     (x – 6)(x – 4)
                                                                   +     (x – 6)(x – 4)
                                                                                                 =                                      =                                                    =
                                                                                                                  12a(2 – 5a)                 (3 – a)(3 + a)            12a(5a – 2)
       x – 3(x – 4) + (x – 6)(x – 4)                         x – 3x + 12 + x2 – 10x + 24                           2a3
              (x – 6)(x – 4)
                                                        =           (x – 6)(x – 4)                              –3(3 – a)

              x2 – 12x + 36                (x – 6)(x – 6)                  x –6                           499. b.
      =       (x – 6)(x – 4)
                                     =     (x – 6)(x + 4)
                                                                     =     x+4                                  9x – 2         10 – 5x           9x – 2             5(2 – x)            5
                                                                                                                8 – 4x         2 – 9x       =   4(2 – x)           –(9x – 2)     =     –4
493. d.                                                                                                              12x2y           –24xy2          (12)(–24)x3y3               2x2
                                                                                                          500. a.                               =                           =    7y
                                                                                                                     –18xy            56y3           (–18)(56)xy4
       –x2 + 5x                 x+1           –x(x – 5)                x+1
       (x – 5)2
                        +       x+5
                                          =    (x – 5)2
                                                                 +     x+5         =                      501. a.
        –x              x+1                 –x(x + 5)                    (x + 1)(x – 5)                         x2 – x – 12                                                 x2 x – 12
       x–5          +   x+5          =    (x – 5)(x + 5)          +       (x + 5)(x 5)         =                3x2 – x – 2            (3x2 – 10x – 8) =                   3x2 – x – 2
       –x(x + 5) + (x + 1)(x – 5)                          –x2 – 5x + x2 – 4x – 5
             (x – 5)(x + 5)                         =          (x – 5)(x + 5)                  =                      1
                                                                                                                                      =    (x – 4)(x + 3)                   1
                                                                                                                                                                                            =
                                                                                                                3x2 – 10x – 8             (3x + 2)(x – 1)            (3x + 2)(x – 4)
                2
           x – 9x                           x(x – 9)
       (x – 5)(x + 5)            =       (x – 5)(x + 5)                                                              x+3
                                                                                                                (3x + 2)2(x – 1)
494. b.
                                                                                                                     x–3             x2 – 3x      x–3         4x         2
                                                                                                          502. c.                               =                    = x3
        2x2               1                  1                    2x2                        1                        2x3              4x         2x3      x(x – 3)
       x4 –1
                    –   x2 –1
                                     +    x2 + 1
                                                       =    (x2 –1)(x2 + 1)
                                                                                       –   x2 –1
                                                                                                     +
                                                                                                                     x2 – 64            6x2 + 48x     x2 – 64       2x – 6
                                                                                                          503. d.     x2 –9              2x – 6 = x2 –9           6x2 + 48x                 =
         1                    2x2                              x2 + 1
          2
       x +1
                    =       2
                        (x – 1)(x2 + 1)
                                                    –        2
                                                           (x – 1)(x2 + 1)
                                                                                       +
                                                                                                                (x – 8)(x + 8) 2(x – 3)                       x –8
                                                                                                                (x – 3)(x + 3) 6x(x +8)
                                                                                                                                                     =      3x(x + 3)
            x2 – 1                        2x2 – x2 –1 + x2 – 1
       (x2 –1)(x2 + 1)
                                     =      (x2 – 1)(x2 + 1)
                                                                           =
                                                                                                                     2(x 6)2           –(5 + x)             –(x – 6)
            2x –2   2                              2
                                             2(x – 1)                       2                             504. a.     x+5              4(x – 6)     =          2
       (x2 – 1)(x2 + 1)
                                     =    (x2 –1)(x2 +1)
                                                                   =     x2 + 1
                                                                                                                     9x2y3            21y       10x            (9)(21)(10)x3y4              3x
                                                                                                          505. b.                                       =                              =    4y
                                                                                                                      14x            15xy2      12y3          (14)(15)(12)x2y5

   204
                                                                ANSWERS & EXPLANATIONS–


506. b.                                                                                   511. b.
      4x2 + 4x + 1             2x2 + 3x + 1              4x2 + 4x + 1                              x2 – x            10xy2                       3x2 + 3x          x2 – x           10xy2
        4x2 – 4x                 2x2 – 2x          =       4x2 – 4x                                  4y              2x – 2                       15x2y2     =       4y             2x – 2
        2x2 – 2x              (2x + 1)2             2x(2x – 2)
      2x2 + 3x + 1      =     4x(x – 1)           (2x + 1)(x + 1)         =                     15x2y2
                                                                                                                    =    x(x – 1)                 10xy2           15x2y2
                                                                                                                                                                                    =
                                                                                               3x2 + 3x                     4y                   2(x – 1)        3x(x + 1)
      (2x + 1)(x – 2)
      2(x – 1)(x + 1)                                                                           25x4y4                         25 x3y3
                                                                                               4xy(x +1)             =        4(x + 1)
507. b.                                                                                   512. d.
                                                                                                  x+2                         2x2 + 7x + 3              6x2 + 5x + 1                 x2 – 4
       2
      x –1           2x +               2
                                    x +x–2
                                                   =   (x – 1)(x + 1)                          x2 + 5x + 6                    4x2 + 4x + 1                3x2 + x                   x2 + 2x   =
      x2 + x         1 – x2           x2 – x              x(x + 1)
                                                                                                   x+2                               (2x + 1)(x + 3)               (3x + 1)(2x + 1)
        2(x + 1)              (x +2)(x – 1)              (x – 1)         2                     (x + 2)(x + 3)                       (2x + 1)(2x + 1)                  x(3x + 1)
      (1 –x)(1 + x)              x(x –1)           =        x         –(x – 1)
                                                                                               (x – 2)(x + 2)                          x–2
                                                                                                  x(x + 2)                    =         x2
      (x + 2)         –2(x + 2)
         x       =       x2

508. c.
                                                                                          Set 33 (Page 82)
                                                                                          513. b.
                                            –(x – 3)      2x2 – 3x – 5
      (4x2 – 8x – 5)                         x+1             x–3              =                          3
                                                                                                                    1
                                                                                                                                                   2
                                                                                               1–        4
                                                                                                          9         4    +          5
                                                                                                                                    2    –   1
                                                                                                                                             4         = 1 – (3
                                                                                                                                                              4
                                                                                                                                                                                9
                                                                                                                                                                               16   )
                                               –(x –3)          (2x – 5)(x + 1)                          16
      (2x + 1)(2x – 5)                          x+1                  x–3              =                                      2
                                                                                               1
                                                                                               4    +       10 – 1
                                                                                                              4                    = 1 – (3
                                                                                                                                          4
                                                                                                                                                         9
                                                                                                                                                        16 )
                                                                                                                                                                   1
                                                                                                                                                                   4   +       81
                                                                                                                                                                               16   =
      (2x + 1)(2x – 5)                      [–(2x – 5)] = (2x + 1)
                                                                                                            3       16              1        81                        4        1       81
                           1                                                                   1–           4        9              4    +   16   =1– 1                3        4   +   16    =
      (2x + 5)          –(2x – 5)           = –(2x + 1)

509. a.                                                                                        1–       1
                                                                                                            +       81
                                                                                                                         =         48 – 16 + 243
                                                                                                                                                        =    275
          a2 – b2              2a2 – 7ab + 3b2                  ab – 3b2                                3           16                   48                   48
      2a2 – 3ab + b2               a2 +ab                    a2 + 2ab + b2        =
                                                                                                     2          3             8          9        17
                                                                                                     3   +      4            12    +    12                       17
                                                                                          514. b.    3          1
                                                                                                                    =          3        2
                                                                                                                                             =    12
                                                                                                                                                        4=        3
           2     2                  2              2        2         2       2                                                                   1
          a –b                 2a – 7ab + 3b               a + 2ab + b                               4   –      2              4   –    4
      2a2 – 3ab + b2              a2 +ab                     ab – 3b2             =                                                               4


                                                                                          515. b.
      (a – b)(a + b)               (2a – b)(a – 3b)         (a + b)(a + b)
      (2a – b)(a – b)                  a(a + b)                b(a – 3b)          =
                                                                                                   3x2 + 6x                   2+x                  3x – 1
                                                                                                    x–5              +        5–x                  25 – x2     =
      (a + b)         (2a – b)(a – 3b)             (a + b)(a + b)             (a – 3b)
      (2a – b)            a(a + b)                    b(a – 3b)           =       a                3x(x + 2)                   x+2                (5 – x)(5 + x)
                                                                                                     x–5                 –     x–5                    3x – 1           =
      (a + b)(a + b)               (a + b)2
         b(a – 3b)        =           ab                                                           3x(x + 2) – (x + 2)                            –(5 – x)(5 + x)
                                                                                                          x–5                                         3x – 1               =
                               2
                              x + 3x – 18                                x
510. a. (x – 3)                    x              = (x – 3)           2
                                                                     x + 3x –18       =
                             x                     x                                           (3x – 1)(x + 2)                          –(5 – x)(5 + x)
      (x – 3)         (x – 3)(x + 6)          =   x+6                                               x–5                                     3x – 1
                                                                                               = –(x + 2)(x + 5)




                                                                                                                                                                                        205
                                                                                 ANSWERS & EXPLANATIONS–


516. d.                                                                                                    521. d.
                                                                                                                  a–2        a+2
             1                       1                                                                                   –
          (x + h)2              –    x2                    h=                                                     a+2
                                                                                                                  a–2
                                                                                                                             a–2
                                                                                                                             a+2
                                                                                                                                       =
                                                                                                                  a+2    +   a–2

             1                        x2               1          (x + h)2                                        (a – 2)(a – 2)
                                                                                                                                   –    (a + 2)(a + 2)
          (x + h)2              –     x2      –        x2         (x + h)2            h=                          (a + 2)(a – 2)        (a + 2)(a – 2)
                                                                                                                                                              =
                                                                                                                  (a – 2)(a – 2)        (a + 2)(a + 2)
                                                                                                                  (a + 2)(a – 2)   +    (a + 2)(a – 2)

             x2 – (x + h)2                                        x2 – x2 – 2hx – h2
              x2(x + h)2                               h=             x2(x +h)2                   h=              (a – 2)(a – 2) – (a + 2)(a – 2)
                                                                                                                          (a + 2)(a – 2)
                                                                                                                                                                        =
      –h(2x + h)                        1              –(2x + h)                                                  (a – 2)(a – 2) + (a + 2)(a + 2)
      x2(x + h)2                        h    =         x2(x + h)2                                                         (a + 2)(a – 2)
                       1                ab         1
              a+       b                 b   +     b            ab + 1        a           a                       a2 – 4a + 4 – (a2 + 4a + 4)
517. d.                1
                                =       ba         1
                                                           =      b         ba + 1    =   b                              (a + 2)(a – 2)
              b+       a                 a   +     a                                                                                                                    =
518. a.                                                                                                           a2 – 4a + 4 + (a2 + 4a + 4)
                                    4x       3    –    1                                                                 (a + 2)(a – 2)                           =
         3        1                          x         2
         x   –    2
      5            1
                           =                 5         1
                                                                =                                                 a2 – 4a + 4 – (a2 + 4a + 4)                                (a + 2)(a – 2)
      4x     –    2x
                                  4x         4x   –    2x                                                                (a + 2)(a – 2)                               a2 – 4a + 4 + (a2 + 4a + 4)

         4x       3
                  x    – 4x         1
                                    2                                                                            = – 2a8a 8 =
                                                                                                                       2
                                                                                                                         +
                                                                                                                                             4a
                                                                                                                                           a2 + 4
                                                        12 – 2x            2(6 – x)
                                                 =       5–2          =       3
      4x      5
              4x       – 4x         1
                                    2x
                                                                                                           522. a.
                                                                                                                        4
519. c.                                                                                                               4–x2 – 1
                                                                                                                      1      1
                                                                                                                                        =
                                                                                                                     x+2 + x–2
         5                    2
      (x – 1)3        –    (x – 1)2               (x – 1)4
         2                    5                   (x – 1)4        =                                                            4         4 – x2
                                                                                                                                    – 4 – x2
      (x – 1)3        –    (x – 1)4                                                                                            – x2
                                                                                                                                                                  =
                                                                                                                          x–2                x+2
        5                                        2                                                                   (x + 2)( x – 2) + (x + 2)( x – 2)
     (x – 1)3              (x – 1)4 –         (x – 1)3          (x – 1)4
                           –       5
                                             (x – 1)4                      =
                                (x – 1)4                                                                               4 – 4 + x2
                                                                                                                         4 – x2
                                                                                                                      x – 2 + x +2      =
      5(x – 1) – 2(x – 1)2                                     5x – 5 – 2x2 + 4x – 2                                 (x + 2)(x – 2)
          2(x – 1) –5                                  =              2x – 7              =
                                                                                                                        x2             (x + 2)(x – 2)                   x2           (x + 2)(x – 2)
                                                                                                                 =    4 – x2                 2x                   =   4 – x2               2x         =
              2                                            2
      –2x + 9x – 7                                –(2x –9x + 7)                –(2x – 7)(x – 1)
         2x – 7                         =             x–7                  =        2x – 7             =
                                                                                                                    x2             x2 – 4                 x                x2 – 4          x
                                                                                                                  4 – x2             2x         =     –(x2 – 4)              2        =–   2
      – (x – 1)
                                                                                                                                                                      –1                  –1
                                                                                                                                                          1       1             b+a                 ab
520. a.                                                                                                    523. a. (a–1 + b–1)–1= a + b                                    =     ab            =   b+a
                       x                            x
                       5                                               x          5
     1–                     x
                                 =1–                5
                                                   5+x
                                                                = 1 – (5        x + 5)        =                        x–1 – y–1
                                                                                                                                            1
                                                                                                                                            x     –   1
                                                                                                                                                      y
                                                                                                                                                                  y–x
                                                                                                                                                                   xy          y–x       xy        y–x
                 1+         5                       5                                                      524. b. x–1 + y–1 =              1         1
                                                                                                                                                              =   y+x
                                                                                                                                                                        =       xy      y+x    =   y+x
                  x                 x+5                     x             5                                                                 x     +   y
     1–          x+5            =   x+5            –       x+5      =    x+5
                                                                                                                                                                   xy




   206
                                                                 ANSWERS & EXPLANATIONS–


525. b.                                                                                 Set 34 (Page 84)
       x2 + 4x – 5              2x + 3            2                                     529. a. First, clear the fractions from all terms in the
       2x2 + x – 3              x+1         –    x+2      =
                                                                                              equation by multiplying both sides by the least
           (x = 5)(x – 1)                2x + 3            2        x+5        2              common denominator (LCD). Then, solve the
          (2x + 3)(x – 1)                 x+1         –   x+2   =   x+1   –   x+2   =
                                                                                              resulting equation using factoring techniques:
      (x + 5)(x + 2)                   2(x + 1)
      (x + 1)(x + 2)            –   (x + 2)(x + 1)        =                                    3
                                                                                                   =2+x
                                                                                               x
                                                                                                       3
      (x + 5)(x + 2) – 2(x + 1)
                                                =     x2 + 7x + 10 – 2x – 2
                                                                              =               x        x   = x (2 + x)
            (x +1)(x +2)                                  (x + 1)(x + 2)
                                                                                              3 = 2x + x2
       x2 + 5x + 8
      (x + 1)(x + 2)                                                                          x2 + 2x – 3= 0
526. c.                                                                                       (x + 3)(x – 1) = 0
       x+5
                  –x                 1
                                           –       x+5
                                                          –   x(x – 3)         1              x + 3 = 0 or x – 1 = 0
       x–3                          x–3            x–3         x–3            x–3
                                                                                              x = –3 or x = 1
                       2
           x + 5 – x + 3x
     =          x–3                      (x – 3)                                              Since neither of these values makes any of the
                  2
      = – (x – 4x – 5) – (x – 5)(x + 1                                                        expressions in the original equation, or any
527. c.                                                                                       subsequent step of the solution, undefined, we
      3+       1                 x+3              +
                                          = [ 3(xx+ 33) +         1       x+3                 conclude that both of them are solutions to
              x+3                x–2                            x + 3]    x –2    =
                                                                                              the original equation.
     3x + 10          x+3                3x + 10                                        530. d. First, clear the fractions from all terms in the
      x+3             x–2           =     x+3
528. d.                                                                                       equation by multiplying both sides by the least
     1–       2
                  –        3
                                –   1
                                               =                                              common denominator. Then, solve the result-
              x            2x       6x
                                                                                              ing equation using factoring techniques:
              2       3             1                                                          2       3       1
     1–       x   –   2x    +       6x    =                                                        –       =
                                                                                               3       x       2
                                                                                               2               3          1
                                                                                               3   6x –        x   6x =   2   6x
     1–        (6
            ( 26x)    –     3(3)
                             6x      +    1
                                          6x )    =
                                                                                              4x – 18 = 3x
            12 – 9 + 1              6x – 4          3x – 2
     1–         6x              =     6x       =      3x                                      x = 18

                                                                                              This value does not make any of the expres-
                                                                                              sions in the original equation, or any subse-
                                                                                              quent step of the solution, undefined, so we
                                                                                              conclude that it is indeed a solution of the
                                                                                              original equation.




                                                                                                                                        207
                                                        ANSWERS & EXPLANATIONS–


531. c. First, clear the fractions from all terms in                          533. c. First, clear the fractions from all terms in
      the equation by multiplying both sides by the                                 the equation by multiplying both sides by the
      least common denominator. Then, solve the                                     least common denominator. Then, solve the
      resulting equation using factoring techniques:                                resulting equation using factoring techniques:
                                  2t       1                                         x         2        3
      (t – 7)(t – 1)             t –7   +t – 1     = 2 (t – 7)(t – 1)               x–3    +   x   –   x–3
                                                                                     x                        2                 3
       2t                                   1                                       x–3     x(x – 3) +        x   x(x – 3) =   x–3   x(x – 3)
      t–7         (t – 7)(t – 1) +       (t – 1)    (t – 7)(t – 1) =
                                                                                     2
                                                                                    x + 2(x – 3) = 3x
      2(t – 7)(t – 1) = 2t(t – 1) + (t – 7) = 2(t – 7)(t – 1)
                                                                                    x2 – x – 6 = 0
      2t2 – 2t + t – 7 = 2t2 – 16 + 14                                              (x – 3)( x + 2) = 0
      –t – 7 = –16t + 14                                                            x = 3, –2

      15t =21                                                                       Because x = 3 makes some of the terms in the
             21       7                                                             original equation undefined, it cannot be a
      t=     15   =   5
                                                                                    solution of the equation. Thus, we conclude
      Since this value does not make any of the                                     that the only solution of the equation is x = –2.
      expressions in the original equation, or any                            534. a. First, clear the fractions from all terms in
      subsequent step of the solution, undefined, it                                the equation by multiplying both sides by the
      is indeed a solution of the original equation.                                least common denominator. Then, solve the
532. a. First, clear the fractions from all terms in                                resulting equation using factoring techniques:
      the equation by multiplying both sides by the                                  3                        6
                                                                                    x+2    +1=         (2 – x)(2 + x)
      least common denominator. Then, solve the
                                                                                      3
      resulting equation using factoring techniques:                                 x+2       (2 – x)(2 + x) + 1 (2 – x )(2 + x) =
                          x+8                         12                  2                 6
      x(x + 2)            x+2   + x(x + 2)         x2 + 2x   = x(x + 2)   x          (2 – x)(2 + x)      (2 – x)(2 + x)
      x(x + 8) + 12 = 2(x + 2)                                                      3(2 – x) + 4 – x2 = 6
      x2 + 8 + 12 = 2x + 4
                                                                                    10 – 3x – x2 = 6
         2
      x + 6x + 8 = 0
                                                                                    x2 + 3x – 4 = 0
      (x + 4)(x + 2) = 0
      x = –4 or x = – 2                                                             (x + 4)(x – 1) = 0

      Note that x = –2 makes some of the terms in                                   x = –4, 1
      the original equation undefined, so it cannot                                 Neither of these values makes any of the
      be a solution of the equation. Thus, we con-                                  expressions in the original equation, or any
      clude that the only solution of the equation is                               subsequent step of the solution, undefined, so
      x = –4.                                                                       we conclude that both of them are solutions to
                                                                                    the original equation.




   208
                                           ANSWERS & EXPLANATIONS–


535. b. First, clear the fractions from all terms in        537. d. First, clear the fractions from all terms in
      the equation by multiplying both sides by the               the equation by multiplying both sides by the
      least common denominator. Then, solve the                   least common denominator. Then, solve the
      resulting equation using factoring techniques:              resulting equation using factoring techniques,
          10                       3                              as follows:
       (2x – 1)2   =4+           2x – 1
                                                                   x–1        4
          10                                         3                   =   x–5
       (2x – 1)2       (2x – 1)2 = 4 (2x – 1)2 +   2x – 1
                                                                   x–5
                                                                              x–1                        4
                                                                  (x – 5)     x–5     = (x – 5)         x–5
      (2x –      1)2
                                                                  x–1=4
                                 2
      10 = 4(2x – 1) + 3(2x – 1)                                  x=5
      10 = 16x2 – 16x + 4 + 6x – 3                                Because this value of x makes the expressions
      10 = 16x2 – 10x + 1                                         in the original equation undefined, we con-
                                                                  clude that the equation has no solution.
      16x2 – 10x – 9 = 0
                                                            538. d. First, clear the fractions from all terms in
      (2x + 1)(8x – 9) = 0                                        the equation by multiplying both sides by the
      x=    –1 9                                                  least common denominator. Then, solve the
             2,8
                                                                  resulting equation using factoring techniques:
      Since neither of these values makes any of the
                                                                        22              3           2
      expressions in the original equation, or any                 2p2 – 9p – 5   –   2p + 1   =   p–5

      subsequent step of the solution, undefined,                        22                 3           2
                                                                   (2p + 1)(p – 5)    –   2p + 1   =   p–5
      both of them are solutions to the original                                                  22
                                                                  (2p +1)(p – 5)            (2p + 1)(p – 5)   –
      equation.
                                                                                                 3                              2
536. b.                                                               (2p + 1)(p – 5)          2p + 1    = (2p + 1)(p – 5)    p –5
      1                     1        1
      f   = (k – 1)         pq   +   q                            22 – 3(p – 5) = 2(2p + 1)
           1           1         1
       f(k – 1)    =   pq   +    q                                22 – 3p + 15 = 4p + 2
          1
      f(k – 1)     f(k – 1)pq = [ p1q + 1 ] f(k – 1)pq
                                        q                         –3p + 37 = 4p + 2
      pq = f(k – 1) + f)k – 1)p                                   35 = 7p
            f(k –1) + f(k – 1)p                                   p=5
      q=             p
            f(k –1)(1 + p)                                        This value of p makes the expressions in the
      q=          p
                                                                  original equation undefined, so the equation
                                                                  has no solution.




                                                                                                                        209
                                                       ANSWERS & EXPLANATIONS–


539. b. First, clear the fractions from all terms in                                subsequent step of the solution, undefined.
      the equation by multiplying both sides by the                                 Therefore, we conclude that both of them are
      least common denominator. Then, solve the                                     solutions to the original equation.
      resulting equation using factoring techniques:                          541. c. First, clear the fractions from all terms in
        x+1               1                   1                                     the equation by multiplying both sides by the
       x2 – 9x   –   2x2+ x – 21   =   2x2 + 13x + 21
                                                                                    least common denominator. Then, solve the
            x+1                     1                     1
       x(x – 3)(x + 3)   –   (2x + 7)(x – 3)   =   (2x + 7)(x + 3)                  resulting equation using factoring techniques:
                                            x+1                    1                       2             4
      x(x– 3)(x+ 3)(2x+ 7)             (x – 3)(x + 3)   –   (2x + 7)(x – 3)
                                                                                    1+    x–3   =   x2 – 4x + 3
                                                                                           2              4
                                                      1                             1+    x–3   =   (x – 3)(x – 1)
      = x(x – 3)(x + 3)(2x + 7)                 (2x + 7)(x – 3    )
                                                                                                                   2
      (x + 1)(2x + 7) – x(x + 3) = x(x – 3)                                         (x – 3)(x – 1)         1+     x–3   =

      2x2 + 9x + 7 – x2 – 3x = x2 – 3x
                                                                                                                 4
                                                                                    (x – 3)(x – 1)         (x – 3)(x – 1)
      x2 + 6x + 7 = x2 – 3x
      6x + 7 = –3x                                                                  (x – 3)(x – 1) + 2(x – 1) = 4
      9x –7
                                                                                    x2 – 4x + 3 + 2x – 2 = 4
      x = –7
           9
                                                                                    x2 – 2x + 3 = 0
      Since this value does not make any of the
      expressions in the original equation, or any                                  (x – 3)(x + 1) = 0
      subsequent step of the solution, undefined, it                                x = 3 or x = –1
      is indeed a solution of the original equation.
                                                                                    Because x = 3 makes some of the terms in the
540. c. First, clear the fractions from all terms in
                                                                                    original equation undefined, it cannot be a
      the equation by multiplying both sides by the
                                                                                    solution of the equation. Thus, we conclude
      least common denominator (LCD). Then,
                                                                                    that the only solution of the equation is x = –1.
      solve the resulting equation using factoring
                                                                              542. d. First, clear the fractions from all terms in
      techniques:
                                                                                    the equation by multiplying both sides by the
       x          3            3
      x+1   –    x+4   =   x2 +5x + 4                                               least common denominator (LCD). Then,
       x
            –     3
                       =          3                                                 solve the resulting equation using factoring
      x+1        x+4       (x + 1)( x + 4)
                         x                  3
                                                                                    techniques:
      (x + 1)(x + 4) [ x + 1 –            x + 4]   =
                                                                                     3        x–3
                             3
      (x + 1)(x + 4) [ (x + 1)(x + 4) ]                                             x+2   =   x–2
                                                                                                     3                      x–3
      x(x + 4) – 3(x + 1) = 3                                                       (x + 2)(x – 2) x + 2 = (x + 2)(x – 2)   x–2
                                                                                    3(x – 2) = (x + 2)(x – 3)
      x2 + 4x – 3x – 3 = 3
                                                                                    3x – 6 = x2 –x – 6
      x2 + x – 6 = 0
                                                                                    x2 –4x = 0
      (x + 3)(x – 2) = 0                                                            x(x – 4) = 0
      x = –3 or x = 2                                                               x = 0 or x = 4
      Neither of these values makes any of the
      expressions in the original equation, or any
   210
                                                           ANSWERS & EXPLANATIONS–


      Since neither of these values makes any of the                                 Set 35 (Page 86)
      expressions in the original equation, or any                                   545. b. First, determine the x-values that make the
      subsequent step of the solution, undefined, we                                       expression on the left side equal to zero or
      conclude that both of them are solutions to                                          undefined. Then, we assess the sign of the
      the original equation.                                                               expression on the left side on each subinterval
543. c. First, clear the fractions from all terms in                                       formed using these values. To this end, observe
      the equation by multiplying both sides by the                                        that these values are x = –3, –2, and 1. Now, we
      least common denominator. Then, solve the                                            form a number line, choose a real number in
      resulting equation using factoring techniques:                                       each of the subintervals, and record the sign of
      t+1                  4                                                               the expression above each:
      t–1      =        t2 – 1
      t+1
      t–1      =           4
                    (t – 1)(t + 1)
                                                                                                             +          +          –       +
                                                                                                                 –3           –2       1
                            1
      (t – 1)(t + 1) [ tt + 1 ] = (t – 1)(t + 1)
                          –
            4                                                                              Since the inequality includes “equals,” we include
      (t + 1)(t – 1)
                                                                                           those values from the number line that make
      (t + 1)(t + 1) = 4                                                                   the numerator equal to zero. The solution set
      t2 + 2t + 1 = 4                                                                      is [–2, 1].
                                                                                     546. c. First, we must make certain that the numer-
      t2 + 2t – 3 = 0
                                                                                           ator and denominator are both completely fac-
      (t + 3)(t – 1) = 0                                                                   tored and that all common terms are canceled:
      t = –3 or t = 1                                                                        x2 + 9              x2 + 9
                                                                                           x2 – 2x – 3   =   (x – 3)(x + 1)
      Note that t = 1 makes some of the terms in the
                                                                                           Next, determine the x-values that make this
      original equation undefined, so it cannot be a
                                                                                           expression equal to zero or undefined. Then,
      solution of the equation. Thus, we conclude
                                                                                           we assess the sign of the expression on the left
      that the only solution of the equation is t = –3.
                                                                                           side on each subinterval formed using these
544. a.
               v1 + v2                                                                     values. To this end, observe that these values
      v=                v1v2
              1+         c2                                                                are x = –1 and 3. Now, we form a number line,
                   v1v2
                                                                                           choose a real number in each subinterval, and
      v[1 +         c2 ]       = v1 + v2                                                   record the sign of the expression above each:
              vv1v2
      v+        c2       = v1 + v2                                                                           +          –          +
      vv1v2
        c2     –v1 = v2 –v                                                                                       –1            3

          vv
      v1( c 2 – 1) = v2 – v                                                                Since the inequality does not include “equals,”
              2
                  v2 – v           v2 – v                    c2          2
                                                                         c (v2 –v)
      v1 =        vv2          =              = (v2 –v)              =                     we do not include those values from the num-
                    2   –1
                                   vv2 – c2               vv2 – c2        vv2 – c2
                   c                  c2                                                   ber line that make the numerator equal to zero.
                                                                                           Therefore, the solution set is (–∞, – 1)∪(3, ∞).




                                                                                                                                               211
                                                      ANSWERS & EXPLANATIONS–


547. a. Determine the solution set for the inequal-                     548. a. Determine the solution set for the inequal-
               –x2 – 1
      ity   6x4 – x3 – 2x2   ≥ 0.                                                   2
                                                                                    x2
                                                                                           1
                                                                                        – x–1
                                                                              ity     1      4
                                                                                                        0.
                                                                                    x + 3 – x2
      First, we must make certain that the numera-
      tor and denominator are both completely fac-                            First, we must simplify the complex fraction
      tored and that all common terms are canceled,                           on the left side of the inequality:
                                                                                2      1
      as follows:                                                              x2
                                                                                    – x–1
                                                                                  1      4    =
         –x2 – 1                –(x2 + 1)               –(x2 + 1)              x + 3 – x2
      6x4 – x3 – 2x2   =     x2(6x2 – x – 2)   =    x2(2x +1)(3x – 2)
                                                                              2(x – 1) – x2
                                                                                x2(x – 1)
      Next determine the x-values that make this                              x2 – 4(x + 3)   =
                                                                                x2(x + 3)
      expression equal to zero or undefined. Then,
      we assess the sign of the expression on the left                        –(x2 – 2x + 1)             x2(x + 3)
                                                                                                                       =
                                                                                x2(x – 1)               x2 – 4x – 12
      side on each subinterval formed using these
                                                                              –(x – 1)2             x2(x + 3)
      values. To this end, observe that these values                          (x2(x –1)           (x – 6)(x + 2)   =
      are x = – 1 , 0, and 2 . Now, we form a number
                2          3                                                      (x –1)(x + 3)
                                                                              –   (x – 6)(x + 2)
      line, choose a real number in each subinterval,
      and record the sign of the expression above                             So, the original inequality can be written as
      each:
                                                                                  x –1
                                                                              – ((x – 6))((x + 3)) ≥ 0, or equivalently (upon multi-
                                                                                           x+2
                                                                                                                       x –1
                                                                              plication by –1 on both sides), ((x – 6))((x + 3)) 0.
                                                                                                                            x+2
               –             +          +             –
                                                                              Next, we determine the x-values that make this
                                    0
                   –3
                    2
                                               –2
                                                3                             expression equal to zero or undefined, includ-
                                                                              ing the values that make any factors common
      Since the inequality includes “equals,” we                              to both numerator and denominator equal to
      include those values from the number line that                          zero. Then, we assess the sign of the expression
      make the numerator equal to zero. Since none                            on the left side on each subinterval formed
      of these values make the numerator equal to                             using these values. To this end, observe that
      zero, we conclude that the solution set is                              these values are x = –3, –2, 0, 1, and 6. Now,
      (– 1 , 0)∪(0, 2 ).
         2          3                                                         we form a number line, choose a real number
                                                                              in each subinterval, and record the sign of the
                                                                              expression above each:

                                                                              +               –           +            +       –       +
                                                                                  –3               –2              0       1       6


                                                                              The inequality includes “equals,” so we include
                                                                              those values from the number line that make
                                                                              the numerator equal to zero. The solution set
                                                                              is [–3, –2)∪[1, 6).




   212
                                                     ANSWERS & EXPLANATIONS–


549. c. First, we must make certain that the numer-                    Since, the inequality includes “equals,” so include
      ator and denominator are both completely fac-                    those values from the number line that make
      tored and that all common terms are canceled:                    the numerator equal to zero. Since none of
      2z2 – z – 15        (2z + 5)(z – 3)        (2z + 5)              these values make the numerator equal to zero,
      z2 + 2 z – 15   =    (z + 5)(z – 3)    =     z+5
                                                                       we conclude that the solution set is (–∞, 0).
      Now, the strategy is to determine the z-values             551. d. First, we must make certain that the numer-
      that make this expression equal to zero or                       ator and denominator are both completely fac-
      undefined. Then, we assess the sign of the                       tored and that all common terms are canceled:
      expression on the left side on each subinterval                  z3 – 16z       z(z2 – 16)       z(z – 4)(z + 4)       z(z + 4)
                                                                       8z – 32    =    8(z – 4)    =       8(z – 4)      =       8
      formed using these values. To this end, observe
      that these values are z = –5, – 5 . Next, we form
                                      2                                Next, determine the z-values that make this
      a number line, choose a real number in each                      expression equal to zero or undefined. Then,
      subinterval, and record the sign of the expres-                  we assess the sign of the expression on the left
      sion above each:                                                 side on each subinterval formed using these
                                                                       values. To this end, observe that these values
                      +             –            +                     are z = –4, 0, 4. Now, we form a number line,
                           –5           –5
                                         2
                                                                       choose a real number in each of the duly
                                                                       formed subintervals, and record the sign of the
      Since, the inequality includes “equals,” we                      expression above each:
      include those values from the number line that
      make the numerator equal to zero. Therefore,                                +           –             +            +
      the solution set is (–∞, –5)∪[– 5 ,∞).
                                      2
                                                                                      –4               0          4

550. d. Our first step is to make certain that the
                                                                       Since the inequality does not include “equals,”
      numerator and denominator are both com-
                                                                       we do not include those values from the num-
      pletely factored and that all common terms are
                                                                       ber line that make the numerator equal to
      canceled:
                                                                       zero. The solution set is (–4, 0).
       25(–x)4         25x4         1
       x(5x2)2
                 =    x 25x4
                                =   x

      Now, determine the x-values that make this
      expression equal to zero or undefined. Then,
      we assess the sign of the expression on the left
      side on each subinterval formed using these
      values. To this end, observe that the only value
      for which this is true is x = 0. Next, we form a
      number line, choose a real number in each
      subinterval, and record the sign of the expres-
      sion above each, as follows:

                            –           +
                                    0


                                                                                                                                        213
                                                      ANSWERS & EXPLANATIONS–


552. b. To begin, we must make certain that the                              Since the inequality does not include “equals,”
      numerator and denominator are both com-                                we do not include those values from the num-
      pletely factored and that all common terms are                         ber line that make the numerator equal to
      canceled:                                                              zero. Therefore, the solution set is (8, ∞).
      y2 – 64       (y – 8)(y + 8)       (y – 8)(y + 8)                554. a. To begin, we must make certain that the
       8–y      =       (8 – y)      =      –(y – 8)      = –(y + 8)
                                                                             numerator and denominator are both com-
      Now, determine the y-values that make this                             pletely factored and that all common terms
      expression equal to zero or undefined. Then,                           are canceled:
      we assess the sign of the expression on the left                       5x2(x – 1) – 3x(x –1) – 2(x – 1)          (x – 1)(5x2 – 3x – 2)
                                                                             10x2(x – 1) + 9x(x – 1 +2(x – 1)     =   (x – 1)(10x2 + 9x +2)    =
      side on each subinterval formed using these
                                                                             (x – 1)(5x + 2)(x – 1)         x–1
      values. To this end, observe that these values                         (x – 1)(5x +2)(2x + 1)
                                                                                                      =    2x + 1
      are y = –8, 8. Next, we form a number line,
                                                                             Next determine the x-values that make this
      choose a real number in each subinterval, and
                                                                             expression equal to zero or undefined. Then,
      record the sign of the expression above each,
                                                                             we assess the sign of the expression on the left
      as follows:
                                                                             side on each subinterval formed using these
                         +           –           –                           values. To this end, observe that these values
                                                                             are x = – 1 , – 2 , 1. Now, we form a number line,
                              –8           8                                           2     5
                                                                             choose a real number in each subinterval, and
      The inequality includes “equals,” we include                           record the sign of the expression above each:
      those values from the number line that make
      the numerator equal to zero. We conclude that                                   +          –            –            +
      the solution set is [–8, 8)∪(8, ∞).                                                 –1
                                                                                           2
                                                                                                      –2
                                                                                                       5
                                                                                                                      1

553. a. First, we must make certain that the numer-
      ator and denominator are both completely fac-                          The inequality includes “equals,” we include
      tored and that all common terms are canceled:                          those values from the number line that make
                                                                             the numerator equal to zero. We conclude that
      x2 + 8x            x(x + 8)           x(x + 8)           1
      x2 – 64x      =   x(x2 – 64)   =   x(x + 8)(x– 8)   =   x–8            the solution set is (– 1 , – 2 )∪(– 2 , 1].
                                                                                                    2     5      5

      Next, determine the x-values that make this
      expression equal to zero or undefined. Then,
      we assess the sign of the expression on the left
      side on each subinterval formed using these
      values. To this end, observe that these values
      are x = –8, 8. Now, we form a number line,
      choose a real number in each subinterval, and
      record the sign of the expression above each:

                          –          –            +
                              –8             8




   214
                                                        ANSWERS & EXPLANATIONS–


555. c. First, we must make certain that the numer-                                                      –               +
      ator and denominator are both completely fac-                                                              5
      tored and that all common terms are canceled:                                                              2

      6x3 – 24x       6x(x2 – 4)       x2 – 4          (x – 2)(x + 2)
        24x2      =   4 6 x x      =     4x        =         4x
                                                                              Since the inequality does not include
                                                                              “equals,”so we do not include those values
      Now, the strategy is to determine the x-values
                                                                              from the number line that make the numera-
      that make this expression equal to zero or
                                                                              tor equal to zero. There are no such values, and
      undefined. Then, we assess the sign of the
                                                                              furthermore, the expression is always positive.
      expression on the left side on each subinterval
                                                                              Therefore, the solution set is the empty set.
      formed using these values, which are x = –2, 0,
                                                                        557. c. First, make certain that the numerator and
      2. Now, we form a number line, choose a real
                                                                              denominator are both completely factored and
      number in each subinterval, and record the
                                                                              that all common terms are canceled, as
      sign of the expression above each:
                                                                              follows:
              –            +               –             +                        3 – 2x                2–x              3 – 2x – (2 – x)
                                                                              (x + 2)(x – 1)   –   (x – 1)(x + 2)    =    (x – 1)(x + 2)    =
                      –2           0               2
                                                                                   1–x                –(x – 1)            1
                                                                              (x – 1)(x + 2)   =   (x – 1)(x + 2)    = –x + 2
      Since the inequality includes “equals,” we
      include those values from the number line that                          Next, determine the x-values that make this
      make the numerator equal to zero. The solu-                             expression equal to zero or undefined. Then,
      tion set is [–2, 0)∪(2, ∞).                                             we assess the sign of the expression on the left
556. c. First, we must make certain that the numer-                           side on each subinterval formed using these
      ator and denominator are both completely fac-                           values, which are x = –2 and 1. Now, we form a
      tored and that all common terms are canceled:                           number line, choose a real number in each
                                                                              subinterval, and record the sign of the expres-
      (2x – 5)(x + 4) – (2x – 5)(x + 1)
                  9(2x – 5)
                                               =                              sion above each:
      (2x – 5)((x + 4) – (x + 1))          3       1
                                       =   9   =   3
               9(2x – 5)                                                                       +             –               –
      Now, determine the x-values that make this                                                    –2               1
      expression equal to zero or undefined. Then,
      we assess the sign of the expression on the left                        Since the inequality includes “equals,” so we
      side on each subinterval formed using these                             include those values from the number line that
      values. To this end, observe that the only value                        make the numerator equal to zero. The solu-
      for which this is true is x = 5 . Next, we form a                       tion set is (–∞, –2).
                                    2
      number line, choose a real number in each of
      the duly formed subintervals, and record the
      sign of the expression above each:




                                                                                                                                            215
                                                           ANSWERS & EXPLANATIONS–


558. b. To begin, we must make certain that the                                          undefined. Then, we assess the sign of the
      numerator and denominator are both com-                                            expression on the left side on each subinterval
      pletely factored and that all common terms are                                     formed using these values. To this end, observe
      canceled:                                                                          that these values are x = – 1 , – 1 , 1 , and 1. Next,
                                                                                                                     2     4 2
                                                                                         form a number line, choose a real number in
            5
      [ x + 3 – x]
        x–
                             1
                            x–3     = [x + 3 –
                                       x–
                                           5         x(x – 3)
                                                      x–3 ]
                                                                       1
                                                                      x–3     =
                                                                                         each subinterval, and record the sign of the
       x + 5 – x2 + 3x
            x–3             (x – 3) = – (x2 – 4x – 5) =                                  expression above each:
      –(x –5)(x + 1)
                                                                                             +              –             +             +              –
      Determine the x-values that make this expres-                                                 –1             1             1             1
                                                                                                     2             4             2

      sion equal to zero or undefined. Then, we assess
      the sign of the expression on the left side on                                     Since the inequality includes “equals,” we
      each subinterval formed using these values. To                                     include those values from the number line that
      this end, observe that these values are x = –1,                                    make the numerator equal to zero. The solution
      3, 5. Now, we form a number line, choose a                                         set is (– 1 , – 1 ]∪( 1 , 1].
                                                                                                   2     4     2
      real number in each subinterval, and record                                  560. d. First, we must make certain that the numer-
      the sign of the expression above each:                                             ator and denominator are both completely fac-
                                                                                         tored and that all common terms are canceled,
                   –            +             +            –                             as follows:
                       –1                 3           5
                                                                                          3y + 2             7y – 3              5             3y + 2
                                                                                         (y – 1)2   –    (y – 1)(y + 1)   +   (y + 1)    =    (y – 1)2
      Since the inequality does not include “equals,”
      we would not include those values from the                                         y +1           7y – 3            y –1        5         (y –1)2
                                                                                         y+1    –   (y – 1)(y + 1)        y–1    +   y+1        (y – 1)2     =
      number line that make the numerator equal to
      zero. As such, we conclude that the solution set                                   (3y + 2)(y + 1) – (7y –3)(y – 1) + 5(y –1)2
                                                                                                       (y – 1)2(y +1)                           =
      is [–1, 3)∪(3, 5).
                                                                                          3y2 + 5y + 2 – (7y2 – 10y + 3) + 5(y2 – 2y + 1)
559. d. Our first step is to make certain that the                                                        (y – 1)2(y + 1)                                =
      numerator and denominator are both com-                                            3y2 + 5y + 2 – 7y2 + 10y – 3 + 5y2 – 10y + 5
                                                                                                        (y – 1)2(y + 1)                            =
      pletely factored and that all common terms
      are canceled:                                                                        y2 + 5y + 4             (y +4)(y + 1)             y+4
                                                                                         (y – 1)2(y + 1)
                                                                                                              =   (y – 1)2(y + 1)
                                                                                                                                     =      (y – 1)2
        x            1                2
                                   2x               x        2x – 1         1
      2x + 1   –   2x – 1   +    4x2 – 1
                                              =   2x + 1     2x – 1   –   2x – 1         Now, the strategy is to determine the y-values
         2x + 1          2x2            x(2x – 1) – 1(2x + 1) + 2x2                      that make this expression equal to zero or
         2x + 1    +   4x2 – 1    =          (2x – 1)(2x + 1)             =
       2x2 – x – 2x – 1 + 2x2               4x2 – 3x – 1
                                                                                         undefined. Then, we assess the sign of the
         (2x – 1)(2x + 1)           =     (2x – 1)(2x + 1)     =
                                                                                         expression on the left side on each subinterval
        (4x + 1)(x – 1)                                                                  formed using these values. To this end, observe
       (2x – 1)(2x + 1)
                                                                                         that these values are y = –4, –1, 1. Next, we
      Now, the strategy is to determine the x-values
                                                                                         form a number line, choose a real number in
      that make this expression equal to zero or




   216
                                                       ANSWERS & EXPLANATIONS–


      each subinterval, and record the sign of the                         572. c. We break up the fractional exponent into
                                                                                                                                                                   3                 1
      expression above each:                                                      two separate exponents to obtain 32 5 = (32 5 )3
                                                                                     5
                                                                                  = ( 32)3 = (2)3 = 8.
                  –                +           +           +
                                                                                           8     2             2        2                2               3             32        9
                          –4           –1          1
                                                                           573. c. ( 27 )– 3 = ( 3 )3                  –3
                                                                                                                             = ( 3 )–2 = ( 2 )2 =                      22   = 4.
                                                                                                 –3
                                                                                                     1                      1                                 1             1
                                                                           574. a. (–64)                 = [(–4)3]– 3 = (–4)–1 =                             –4    = –4
      The inequality does not include “equals,” so we
                                                                                                                                 –1
      do not include those values from the number                                                 –2
                                                                                                     1                      –2                                –1        1        x2
                                                                           575. c. (4x–4)                = (2x–2)                    2       = (2x–2) =                2x–2     =2
      line that make the numerator equal to zero.
      The solution set is (–∞, –4).                                        576. b. 4           x144 = 4 (x72)2= 4x72

                                                                           Set 37 (Page 91)
     Section 5—Radical                                                     577. b.
                                                                                       3
                                                                                         9     –3 =
                                                                                                     3             3
                                                                                                                       (9)(–3) =
                                                                                                                                                     3
                                                                                                                                                         –27 =
  Expressions and Quadratic                                                        3
                                                                                       (–3)3 = –3
          Equations                                                                         x5            x5            1                    1   2       1
                                                                           578. b.          x7   =        x7   =        x2       =           x       =   x

                                                                           579. a. a3            a3 = a3 a2a = a3a a = a4 a
Set 36 (Page 90)
                                                                           580. a. Factor      4g into two radicals. 4 is a perfect
561. c. –125 since (–5)3 = –125.
                                                                                  square, so factor 4g into 4 g = 2 g.
562. c. –7 and 7 are both second roots (square
                                                                                  Simplify the fraction by dividing the numera-
      roots) since (–7)2 = 49 and (7)2 = 49.
                                                                                  tor by the and denominator. Cancel the g
563. a. Note that 625 = 54. So, the principal root of
                                                                                  terms from the numerator and denominator.
      625 is 5.                                                                                4
                                                       5
                                                                                  That leaves 2 = 2.
564. d. Since(–2)5 = –32, we write                         –32 = –2.
                                                                           581. a. The cube root of 27y3 = 3y, since (3y)(3y)(3y)
565. a. Since     43      = 64, b = 64 satisfies the equation.                    = 27y3. Factor the denominator into two radi-
          4               4
566. a.       312 =            (33)4 = 33 = 27                                    cals: 27y2. = 9y2            3. The square root of
          5               5
567. c.       515 =            (53)5 = 53 = 125                                      2
                                                                                  9y = 3y, since (3y)(3y) = 9y2. The expression
                      4                                                                                 y
568. b. Since             (2b)4 = 2b, b = 3 satisfies the                         is now equal to 3y 3 . Cancel the 3y terms
      equation.                                                                   from the numeratorand denominator, leaving
              1                1                                                    1
569. b. 64 6 = (26) 6 = 2                                                             3
                                                                                        . Simplify the fraction by multiplying the
                                                                                        numeratorand denominatorby : 3 :
570. d. We break up the fractional exponent into                                      1      3        3
                                                                       1
                                                               5
                                                                                  ( 3 )( 3 ) = 3 .
      two separate exponents to obtain 49 2 = (49 2 )5
      = 75 = 16,807.                                                       582. c. Factor each term in the numerator:                                                           a2b =
571. a. We break up the fractional exponent into                                       a2            b = a b and                         ab2 =               a          b2 =
                                                               3                  b     a. Next, multiply the two radicals. Multiply
      two separate exponents to obtain 81– 4 =
          1                        1    1
                                                                                  the coefficients of each radical and multiply
      (81 4 )–3 = 3–3 =            33 = 27 .                                      the radicands of each radical: (a b)(b a) =
                                                                                  ab ab. The expression is now                                           ab     ab     . Cancel
                                                                                                                                                               ab




                                                                                                                                                                            217
                                                                        ANSWERS & EXPLANATIONS–



      the             ab terms from the numerator and                                                                  Set 38 (Page 93)
      denominator, leaving ab.                                                                                         593. b.         –25 = 25 (–1) =                           52            i2 =
583. c. First, cube the 4g2 term. Cube the constant 4                                                                              5 2   i2 = 5i
                                                                                                           3
      and multiply the exponent of g by 3: (4g2) =                                                                     594. a.   –32 =                 (32(–1) =               32          –1 = (4√2)(i) =
      64g6. Next, multiply 64g6 by g4. Add the expo-                                                                         4i 2
      nents of the g terms. (64g6)(g4) = 64g.10 Finally,                                                               595. a.
      taking the square root of 64g10 yields 8g5, since                                                                      – 48 + 2 27 – 75 = – 42 3 + 2 32 3
      (8g5)(8g5) = 64g.10                                                                                                    – 52 3 = –4 3 + 6 3 – 5 3 =
584. e. First, find the square root of 9pr.                                                   9pr =                          (–4 + 6 – 5) 3 = –3 3
                                                                                              3
      9       pr = 3 pr. The denominator (pr) has a                                           2                        596. d. 3           3+4 5–8                     3 = (3 – 8)                 3+ 4 5 =
      negative exponent, so it can be rewritten in the                                                                       –5           3+4 5
      numerator with a positive exponent. The                                                                          597. d. First, simplify each radical expression. Then,
                                                                                                       1
                                                                                                                             because the variable/radical parts are alike, we
      expression                  pr can be written as ( pr)                                           2

                                                                                          1                                  can add the coefficients:
      since a value raised to the exponent                                                2   is
      another way of representing the square root of                                                                         xy 8xy2 + 3y2 18x3 =xy(2y) 2x + 3y2(3x)
                                                                                              3                1
      the value. The expression is now 3(pr) (pr) .                                           2                2               2x = 2xy2 2x + 9xy2 2x = 11xy2 2x

                                                                                                                                                  }
                                                                                                                                                  }
                                                                                                                   1
      To multiply the pr terms, add the exponents. 2                                                                                              Same                  Same
          3           4                              3          1
      +   2       =   2   = 2, so =3(pr) (pr) = 3(pr)2 = 3p2r2.
                                                     2          2                                                      598. c. We first simplify each fraction. Then, we
                                                          20 + 5            20                                               find the LCD and add.
585. e. Substitute 20 for n:                                20          (2            5) =
          25                                5                                                                                     18              32           18           32         3       2       4       2
          20      (10         5) =      2       5   (10        5). Cancel the                                  5                  25      +        9   =       25   +        9   =         5       +       3       =
                                                                                                       5                      3       2       3            2    5       9     2 + 20       2           29 2
      terms and multiply the fraction by 10:                                                      2        5                      5           3   +4       3    5   =           15                 =     15
                              5(10)         50
      (10 5) =                          =           = 25                                                               599. a. (5 –            3)(7 + 3) = 5(7) + 5( 3) – 7(                                           3)
                                2            2
                  125             125               52 5            5       5
                                                                                                                             –            32 = 35 + (5 – 7) 3 – 3 = 32 – 2 3
586. c.            9      =        9    =             32    =           3                                              600. b. (4 +       6)(6 – 15) = 24 – 4 15 + 6 6 –
              4
                   243    4 243                 4          4
587. d.           4    = 3 =                        81=        34 = 3                                                              90 = 24 –4 15 + 6 6 – 3 10
                    3
588. d.            x2 + 4x + 4 =                    (x + 2)2 = x + 2                                                                  –10 + –25                –10 +     25      –1            –10 + 5i
                                                                                                                       601. a.             5               =            5              =          5            =
              4                   4                                     4             4            4
589. d.    32x8 =                     24 2 (x2)4 =                          24            2                (x2)4                  5(–2 + i)
                                                                                                                                                   = –2 +i
      = 2x2 4√4                                                                                                                      5

590. b.
              4
                  x21 =
                              4
                                  (x5)4 x =
                                                     4
                                                         (x5)4
                                                                            4
                                                                                 x = x5
                                                                                                       4
                                                                                                           x           602. c. (4 + 2i)(4 –2i) = 16 – (2i)2 = 16 – 22i2 =
              3                   3                                               3                                          16 – (4)(–1) = 16 + 4= 20
591. b.           54x5 =              2 33 x3 x2 = 3x                                     2x2
592. a.     x3 + 40x2 + 400x =                              x(x2 + 40x + 400) =                                        603. d. (4 + 2i)2 = 16 +(4)(2i) + (2i)(4) + (2i)2 =
          x(x + 20)2 = (x + 20)                             x                                                                16 + 16i + 22i2 = 16 +16i –4 = 12 + 16i




   218
                                                    ANSWERS & EXPLANATIONS–



                        3        7
                                                                                 Substituting x = 4 into the original equation
                                               3 3+ 7 7
604. b.       21(       7   +    3   ) = 21                        =             yields the true statement 7 = 7, but substitut-
                                                  7 3
                 3+7
           21(          ) = 10                                                   ing x = –2 into the original equation results in
                  21
605. d.(2 +          3x)2 = 4 + (2)( 3x) + (                3x)(2) +             the false statement 5 = –5. So, only x = 4 is a
      (     3x)2   = 4+4 3x + 3x                                                 solution to the original equation.
                                                                                      4      2
                                                                           611. e. a 3 = (a 3 )2 = 62 = 36.
606. a.
                                                                                                       1             1
      ( 3 + 7)(2 3 – 5 7) = (                       3)( 2 3) +             612. d. Observe that (– 3 )–2 = (–1)–2 ( 3 )–2 =
      ( 7)(2 3) – ( 3)(5 7) – (                      7)(5 7) =                            3 2
                                                                                 1(–1)2   1      = 9. We must solve the equation
      2( 3)2 = 2 7 3 – 5 3 7 –                      5( 7)2 = 2 3                                                         1
                                                                                 ( p)4 = 9 for p. Since, ( p)4 = (p 2 )4 =
      + 2 21 – 5 21 – 5 7 = 6 – 3                     21 – 35 =
      –29 – 3 21                                                                 p2, this equation is equivalent to p2 = 9,

              1               1          3+5    2        3+5 2                   the solutions of which are –3 and 3.
607. d.                 =                           =   2        2     =
            3–5 2           3–5 2        3+5    2       3 – (5 2)          613. d. To eliminate the radical term, raise both
           3+5 2            3+5 2             3+5 2
          9 – 52( 2)2
                        =   9 – 25(2)    =–     41
                                                                                 sides to the third power and solve for x:
                                                                                  3
               2x             2x         2+3    x                                  5x – 8 = 3
608. d.                 =                           =
            2–3 x           2–3 x        2+3    x                                5x – 8 = 33 = 27
      2(     2x) + (3 x)( 2x)            2 2x + 3(2x)       2    2x + 6x         5x = 35
              22 – (3 x)2
                                     =    4 – 32( x)2
                                                        =       4 – 9x
                                                                                 x=7
Set 39 (Page 94)
                                                                                 Substituting this value into the original equa-
609. a. Square both sides of the equation and then                               tion yields the true statement 3 = 3, so it is
      solve for x:                                                               indeed a solution.
        7 + 3x = 4                                                         614. b. To eliminate the radical term, raise both
      ( 7 + 3x)2 = (4)2                                                          sides to the third power and solve for x:
      7 + 3x = 16                                                                 3
                                                                                    7 3x = –2
      3x = 9
                                                                                 7 – 3x = (–2)3 = –8
      x=3
                                                                                 –3x = –15
      Substituting this value into the original                                  x=5
      equation yields the true statement 4 = 4,
                                                                                 Substituting this value into the original equa-
      so we know that it is indeed a solution.
                                                                                 tion yields the true statement –2 = –2, so it is
610. a. Square both sides of the equation and then                               indeed a solution.
      solve for x:
                                                                           615. a. Take the square root of both sides and solve
         4x + 33 = 2x – 1                                                        for x:
      ( 4x + 33)2 = (2x – 1)2
                                                                                 (x – 3)2 = –28
      4x + 33 = 4x2 – 4x + 1
                                                                                 (x – 3)2 = ± –28
      0 = 4x2 – 8x – 32
                                                                                 x – 3 = ±2i 7
      0 = 4(x2 – 2x – 8)
                                                                                 x = 3 ±2i 7
      0 = 4(x – 4)(x + 2)
      x = 4, –2
                                                                                                                              219
                                      ANSWERS & EXPLANATIONS–


616. c. Square both sides of the equation and then        619. b.
      solve for x:                                             x3 = – 27
                                                                         3
                                                               3 x3 =      –27
         10 – 3x = x – 2                                            3
                                                               x = (–3)3
      ( 10 – 3x)2 = (x – 2)2
                                                               x=–3
      10 – 3x = x2 – 4x + 4
                                                          620. c.
      0 = x2 – x – 6
                                                               x2 = 225
      0 =(x – 3)( x + 2)
                                                                  x2= ± 225
      x = 3, –2
                                                               x = ± 15
      Substituting x = 3 into the original equation       621. a.
      yields the true statement 1 = 1, but substitut-          x3 = –125
      ing x = –2 into the original equation results in          3     3
                                                                  x3= –125
      the false statement 4 = –4. Only x = 3 is a solu-        x = –5
      tion to the original equation.
                                                          622. c.
617. d. Square both sides of the equation and then             (x + 4)2 = 81
      solve for x:                                                (x + 4)2 = ±   81
         3x + 4 +x = 8                                         x + 4 = ±9
         3x + 4 = 8 – x                                        x = – 4 ±9
      ( 3x + 4)2 = (8 – x)2                                    x = 5, –13
      3x + 4 = 64 – 16x + x2                              623. d.
      0 = x2 – 19x + 60                                        x2 + 1 = 0
      0 = (x – 4)(x – 15)                                      x2 = – 1
      x = 4, 15                                                   x2 = ± – 1
                                                               x = ± i2
      Substituting x = 4 into the original equation
                                                                x=±i
      yields the true statement 8 = 8, but substitut-
                                                          624. b.
      ing x = 15 into the original equation results in
                                                               x2 + 81 = 0
      the false statement 22 = 8. Therefore, only x =
                                                               x2 = –81
      4 is a solution to the original equation.
                                                                  x2 = ± –81
618. b. Isolate the squared expression on one side
                                                               x = ±9i
      and then, take the square root of both sides
      and solve for x:

      (x – 1)2 + 16 = 0
      (x – 1)2 = – 16
         (x – 1)2 = ± –16
      x – 1 = ± 4i
      x = 1 ± 4i




   220
                                                               ANSWERS & EXPLANATIONS–



Set 40 (Page 95)                                                                  631. b. Apply the quadratic formula with a = 3, b = 5,
                                                                                        and c = 2 to obtain:
625. d. Apply the quadratic formula with a = 1, b = 0,
                                                                                                –b ±    b2 –4ac           –(5) ±        (5)2 –4(3)(2)
      and c = –7 to obtain:                                                             x=             2a           =                   2(3)               =
                –b ±     b2 – 4ac          –(0) ±         (0)2 – 4(1)(–7)               –5 ± 1             –5 ± 1                  2
      x=                2a          =                      2(1)               =            6           =     6      = –1, – 3
            28             2 7                                                    632. a. Apply the quadratic formula with a = 5,
      ±     2      =±       2     =±             7
                                                                                        b = 0, and c = –24 to obtain:
626. a. Apply the quadratic formula with a = 2, b = 0,
                                                                                                –b ±    b2 –4ac          –(0) ±        (0)2 –4(5)(–24)
      and c = –1 to obtain:                                                             x=             2a           =                   2(5)               =
                                                                                              480           4     30           2       30
                –b ±    b2 – 4ac           –(0) ±         (0)2 – 4(2)(–1)               ±    10        =±       10      =±         5
      x=               2a           =                      2(2)               =
              8         2 2                  2
                                                                                  633. a. First, put the equation into standard form
      ±   4       =±     4       =±      2                                              by moving all terms to the left side of the
627. a. Apply the quadratic formula with a = 4, b = 3,                                  equation to obtain the equivalent equation
      and c = 0 to obtain:                                                              2x2 +5x + 4 = 0. Now, apply the quadratic for-
                –b ±    b2 – 4ac         –(3)±           (3)2 – 4(4)(0)
                                                                                        mula with a = 2, b = 5, and c = 4 to obtain:
      x=               2a           =                    2(4)
                                                                                              –b ±      b2 –4ac          –(5)±         (5)2 –4(2)(4)       –5        –7
          –3 ± 9               –3 ± 3                3                                  x=                          =                                  =
      =      8             =     8      = 0, – 4                                                       2a                              2(2)                     4
                                                                                             –5 ±i 7
628. b. Apply the quadratic formula with a = –5,                                        =        4
      b = 20, and c = 0 to obtain:                                                634. a. Apply the quadratic formula with a = 1,

                –b ±    b2 – 4ac         –(20)±           (20)2 –4(–5)(0)
                                                                                        b = –2         2 , and c = 3 to obtain:
      x=               2a           =                     2(–5)               =
                                                                                                –b ±    b2 –4ac
       –20 ± 202                –20 ± 20                                                x=                          =
          –10              =      –10            = 0, 4                                                2a

629. c. Apply the quadratic formula with a = 1,                                          –(–2     2) ±      (–2     2)2–4(1)(3)              2    2 ± 8 –12
                                                                                                           2(1)                         =            2          =
      b = 4, and c = 4 to obtain:
                                                                                         2   2 ± –4             2      2 ±2i
                                                                                               2           =           2       =            2±i
                –b ±    b2 – 4ac         –(4)±           (4)2 –4(1)(4)
      x=               2a           =                    2(1)             =       635. b. First, put the equation into standard form
       –4 ± 0
          2            = –2 (repeated solution)                                         by moving all terms to the left side of the
630. c. Apply the quadratic formula with a = 1,                                         equation to obtain the equivalent equation
      b = –5, and c = –6 to obtain:                                                     x2 + 2x = 0. Now, apply the quadratic formula
                                                                                        with a = 1, b = 2, and c = 0 to obtain:
                –b ±    b2 – 4ac         –(–5)             (–5)2 – 4(1)(–6)
      x=               2a           =                       2(1)              =                 –b ±    b2 –4ac           –(2) ±        (2)2 –4(1)(0)
       5±         49       5±7
                                                                                        x=             2a           =                   2(1)               =
            2          =    2     = –1, 6                                                –2 ± 4            –2 ±2
                                                                                            2          =     2      = –2, 0




                                                                                                                                                               221
                                                     ANSWERS & EXPLANATIONS–


636. c. First, put the equation into standard form                         639. b. The simplification process will be easier if
      by expanding the expression on the left side,                              we first eliminate the fractions by multiplying
                                                                                                             1     5
      and then moving all terms to the left side of                              both sides of the equation 6 x2 – 3 x + 1 = 0 by 6.
      the equation:                                                              Doing so yields the equivalent equation x2 – 10x
                                                                                 + 6 = 0. Now, apply the quadratic formula
      (3x – 8)2 = 45
                                                                                 with a = 1, b = –10, and c = 6 to obtain:
      9x2 – 48x + 64 = 45
      9x2 – 48x + 19 = 0                                                                –b ±     b2 –4ac           –(–10) ±    (–10)2 –4(1)(6)
                                                                                 x=             2a             =               2(1)              =
      Now, apply the quadratic formula with a = 9,                                10 ± 76           10 ± 2 19
                                                                                      2         =        2           = 5 ± 19
      b = –48, and c = 19 to obtain:
                                                                           640. b. First, put the equation into standard form
               –b ±    b2 –4ac        –(–48) ±    (–48)2 –4(9)(19)               by expanding the expression on the left side,
      x=              2a          =               2(9)                =
                                                                                 and then moving all terms to the left side of
      –48 ± 1620              –48 ±18 5          –8 ±3 5
           18             =       18        =        3                           the equation:
637. d. We first multiply both sides of the equation
                                                                                 (x – 3)(2x + 1) =x(x – 4)
      by 100, then divide both sides by 20 in order
                                                                                 2x2 – 6x +x – 3 =x2 – 4x
      to make the coefficients integers; this will help
                                                                                 x2 –x – 3 = 0
      with the simplification process. Doing so
      yields the equivalent equation x2 – 11x + 10 =                             Now, apply the quadratic formula with a = 1,
      0. Now, apply the quadratic formula with a = 1,                            b = –1, and c = –3 to obtain:
      b = –11, and c = 10 to obtain:
                                                                                        –b ±     b2 –4ac           –(–1) ±     (–1)2 –4(1)(–3)
                                                                                 x=             2a             =               2(1)              =
               –b ±    b2 –4ac        –(–11) ±    (–11)2 – 4(1)(–10)
      x=              2a          =                2(1)                =          1 ± 13
                                                                                     2
      11 ± 81             11 ±9
          2           =     2     = 1, 10
638. d. Apply the quadratic formula with a = 1,                            Set 41 (Page 97)
      b = –3, and c = –3 to obtain:                                        641. a. Isolate the squared expression on one side,
                                                                                 take the square root of both sides, and solve
               –b ±    b2 –4ac        –(–3) ±    (–3)2 –4(1)(3)
      x=              2a          =              2(1)             =              for x, as follows:
      3±        21
           2                                                                     4x2 = 3
                                                                                        3
                                                                                 x2 =   4
                                                                                            3              3               3
                                                                                 x=±        4
                                                                                                =±         4
                                                                                                               =±      2




   222
                                                     ANSWERS & EXPLANATIONS–


642. c. Isolate the squared expression on one side,              647. a. To solve the given equation graphically, let y1
      take the square root of both sides, and solve                    = 5x2 – 24, y2 = 0. Graph these on the same set
      for x:                                                           of axes and identify the points of intersection:

      –3x2 = –9                                                                                             y1
      x2 = 3                                                                                      9
      x=± 3                                                                                       6
                                                                                                  3
643. b. Take the square root of both sides, and solve
                                                                                                                      y2
                                                                                   –4 –3 –2 –1         1 2 3 4
      for x:                                                                                 –3
                                                                                             –6
      (4x + 5)x2 = –49                                                                       –9
                                                                                             –12
      4x + 5 = ± –49 = ±7i                                                                   –15
                                                                                             –18
      4x = –5 ± 7i                                                                           –21
           –5 ±7i                                                                            –24
      x=     4
644. c. Take the square root of both sides, and solve
                                                                       The x-coordinates of the points of intersection
      for x:
                                                                       are the solutions of the original equation. We
      (3x – 8)2 = 45                                                   conclude that the solutions are approximately
      3x – 8 = ± 45                                                    ±2.191.
      3x = 8 ± 45                                                648. d. To solve the equation graphically, let y1 = 2x2,
           8±       45       8 ±3 5                                    y2 = –5x – 4. Graph these on the same set of
      x=        3        =      3
                                                                       axes and identify the points of intersection:
645. c. Isolate the squared expression on one side,
                                                                                  y2                                 y1
      take the square root of both sides, and solve
                                                                                                 10
      for x:
                                                                                                  8

      (–2x +    1)2
                 – 50 = 0                                                                         6
               2
      (–2x + 1) = 50
                                                                                                  4
      –2x + 1 = ± 50
                                                                                                  2
      –2x = –1 ± 50
           –1 ± 50           1±       50       1 ±5 2                   –5   –4   –3   –2   –1          1   2    3        4     5
      x=      –2         =        2        =      2                                               –2
646. b. Isolate the squared expression on one side,
                                                                                                  –4
      take the square root of both sides, and solve
                                                                                                  –6
      for x:
                                                                                                  –8

      –(1 –    4x)2
                 –121 = 0                                                                        –10
      (1 – 4x)2 = –121

      1 – 4x = ± –121                                                  The x-coordinates of the points of intersection
      –4x = –1 ± –121                                                  are the solutions of the original equation. Since
           –1 ± –121             1±    –121        1 ±11i
      x=                     =                 =                       the curves do not intersect, the solutions are
               –4                     4               4
                                                                       imaginary.
                                                                                                                              223
                                                            ANSWERS & EXPLANATIONS–



649. a. To solve the given equation graphically, let y1 = 4x2, y2 = 20x – 24. Graph these on the same set of axes
       and identify the points of intersection:

                                          y1 y2                           y1                                                                          y2
60                                                                                                           60
54                                                                                                           54
48                                                                                                           48
42                                                                                                           42
36                                                                                                           36
30                                                                                                           30
24                                                                                                           24
18                                                                                                           18
12                                                                                                           12
 6                                                                                                            6

      0.5   1   1.5   2   2.5   3   3.5       4   4.5   5        –5       –4        –3      –2         –1             1       2       3       4                5
–6                                                                                                            –6

                                                                                                             –12



       The x-coordinates of the points of intersection are the solutions of the original equation. The solutions
       are x = 2, 3.

650. c. To solve the given equation graphically, let                                     651. c. To solve the equation graphically, let y1 =
                            2
       y1 = 12x – 15x , y2 = 0. Graph these on the                                               (3x – 8)2, y2 = 45. Graph these on the same set
       same set of axes and identify the points of                                               of axes and identify the points of intersection:
       intersection:                                                                                                                              y1
                                                                                                                    60
                                      4
                                                                                                                    50
                                      3                                                                                                                        y2
                                                                                                                    40
                                      2
                                                                                                                    30
                                          1
                                                                                                                    20
                                                                               y2
                –2         –1                               1         2
                                                                                                                    10
                                     –1

                                     –2                                                           –4    –3    –2   –1     1       2   3   4       5        6
                                                                                                                    –10
                                     –3
                                     –4                                                          The x-coordinates of the points of intersection
                                     –5                                                          are the solutions of the original equation. We
                                     –6                                                          conclude that the solutions are approximately
                                     –7                                                          x = 3.875, 4.903.
                                     –8

                                     –9

                                    –10
                                                                y1


       The x-coordinates of the points of intersection
       are the solutions of the original equation, so
       the solutions are x = 0, 1.25.

     224
                                                     ANSWERS & EXPLANATIONS–



652. b. To solve the given equation graphically, let y1 = 0.20x2 – 2.20x + 2, y2 = 0. Graph these on the same
        set of axes and identify the points of intersection:
                                                                                                                    y1
                                        10


                                            8



                                            6



                                            4



                                            2


                                                                                                                                       y2
 –9               –6         –3                                3               6          9          12              15         18
                                            –2


                                            –4



                                            –6


        The x-coordinates of the points of intersection are the solutions of the original equation. We conclude
        that the solutions are x = 1, 10.

653. a. To solve the equation graphically, let y1 =                            654. b. To solve the given equation graphically, let y1
        x2 – 3x – 3, y2 = 0. Graph these on the same set                             =x2 and y2 = –2x. Graph these on the same set
        of axes and identify the points of intersection:                             of axes and identify the points of intersection:
                                                      y1
                                                                                                                           y1
                                                                                                            10
                                  10


                                   8                                                                            8

                                   6
                                                                                                                6
                                   4
                                                                                                                4
                                   2

                                                                                                                2
                                                                        y 2x
      –10    –8    –6   –4   –2         2        4     6   8       10

                                   –2
                                                                                               –5 –4 –3 –2 –1        1 2 3 4 5
                                   –4
                                                                                                            –2
                                   –6
                                                                                                            –4
                                   –8
                                                                                                                          y2
                                  –10

                                                                                     The x-coordinates of the points of intersection
        The x-coordinates of the points of intersection                              are the solutions of the original equation: x =
        are the solutions of the original equation:                                  –2, 0.
        approximately x = –0.791, 3.791.
                                                                                                                                     225
                                           ANSWERS & EXPLANATIONS–


                                                       1      5
655. b. To solve the equation graphically, let y1 = 6 x2 – 3 x + 1, y2 = 0. Graph these on the same set of axes and
        identify the points of intersection:
                                                                                            y1
                                     10


                                       8



                                       6



                                       4



                                       2


                                                                                                                  y2
   –9           –6         –3                      3              6         9       12           15        18


                                      –2


                                      –4


        The x-coordinates of the points of intersection are the solutions of the original equation. We conclude
        that the solutions are approximately x = 0.641, 9.359.
656. c. To solve the equation graphically, let y1 = (2x + 1)2 – 2(2x + 1) – 3, y2 = 0. Graph these on the same set
        of axes and identify the points of intersection:
                                                                      y1
                                                           10


                                                              8


                                                              6


                                                              4

                                                              2


                                                                            y2
                                                –4 –3 –2 –1       1 2 3 4

                                                           –2


                                                           –4

        The x-coordinates of the points of intersection are the solutions of the original equation. The solutions
        are x = –1, 1.




   226
                                             ANSWERS & EXPLANATIONS–



Set 42 (Page 98)                                            659. d. Note that 4b4 + 20b2 +25 = 0 can be written
657. b. Observe that b4 – 7b2 + 12 = 0 can be written             as 4(b2)2 + 20(b)2 + 25 = 0. Let u= b2 . Rewrit-
      as (b2)2 – 7(b)2 + 12 = 0. Let u= b2. Then, rewrit-         ing the original equation yields the equation,
      ing the previous equation yields the equation               4u2 + 20u+25 = 0, which is quadratic. Factor-
      u2 – 7u+ 12 = 0, which is quadratic. Factoring              ing the left side results in the equivalent equa-
      the left side results in the equivalent equation            tion (2u + 5)2 = 0. Solving this equation for u
                                                                                              5
      (u– 4)(u– 3) = 0. Solving this equation for u               yields the solution u = – 2 . Solving the original
      yields the solutions u = 4 or u = 3. In order to            equation requires that we go back to the sub-
      solve the original equation, we must go back                stitution and write u in terms of the original
      to the substitution and write u in terms of the             variable b:
      original variable b:                                                 5                                 5
                                                                  u= – 2 is the same as b2 = – 2 , which gives us
      u = 4 is the same as b2 = 4, which gives us                 b=±          – 5 = ±i    5   = ±i(          10   )
                                                                                   2       2                 2
      b = ±2
                                                                  Therefore, the solutions of the original equa-
      u = 3 is the same as b2 = 3, which gives us                 tion are b = ±i( 10 ).
                                                                                           2
      b=± 3
                                                            660. b. Observe that 16b4 – 1 = 0 can be written as
      The solutions of the original equation are
                                                                  16(b2)2 – 1 = 0. Let u = b2. Rewriting the origi-
      b = ±2, ± 3.
                                                                  nal equation yields the equation 16u2 – 1 = 0,
658. a. Let u = b2. Observe that (3b2 – 1)(1– 2b2) = 0
                                                                  which is quadratic. Factoring the left side results
      can be written as (3u– 1)(1– 2u) = 0, which is
                                                                  in the equivalent equation (4u– 1)(4u+ 1) = 0.
      quadratic. Solving this equation for u yields
                        1       1                                 Solving this equation for u yields the solution
      the solutions u = 3 or u= 2 . In order to solve                      1
                                                                  u= ± 4 . In order to solve the original equation,
      the original equation, we must go back to the
                                                                  we must go back to the substitution and write
      substitution and write u in terms of the origi-
      nal variable b:                                             u in terms of the original variable b:
                                                                           1                                 1
      u=
           1                             1
               is the same as b2 = 3 , which gives us             u= – 4 is the same as b2 = – 4 , which gives us
           3
                                                                                                   1
      b=±
                 1
                        =±
                                     3                            b = ± – 1 = ±i           1   ±i( 2 )
                 3               3                                                 4       4
                                                                       1                                 1
      u=
           1                             1
               is the same as b2 = 2 , which gives us             u=   4   is the same as b2 = 4 , which gives us
           2
                                                                                       1
                 1                   2                            b=±          1   = ±2
      b=±        2      =±       2                                             4

                                                                  The solutions of the original equation are
      The solutions of the original equation are                          1      1
                    2            3                                b = ±i( 2 ), ± 2 .
      b=±       2       ,±   3       .




                                                                                                                       227
                                            ANSWERS & EXPLANATIONS–


                                    1
661. c. Note that, x + 21 = 10x 2 or equivalently, x –       663. a. Note that x –     x = 6, or equivalently x –
          1                                  1        1
      10x2 + 21 = 0 can be written as (x2)2 – 10(x2) +                x – 6 = 0, can be written as ( x)2 – ( x) –
                      1
      21 = 0. Let u=x2.Then, rewriting the original                6 = 0. Let u = x. Then, rewriting the above
      equation yields the equation u2 – 10u+ 21 = 0,               equation yields the equation u2 – u– 6 = 0,
      which is quadratic. Factoring the left side results          which is quadratic. Factoring yields the equiv-
      in the equivalent equation (u– 3)(u– 7) = 0.                 alent equation (u– 3)(u+ 2) = 0. Solving this
      Solving this equation for u yields the solution              equation for u yields the solutions u = –2 or
      u = 3 or u = 7. In order to solve the original               u= 3. In order to solve the original equation,
      equation, we go back to the substitution and                 we must go back to the substitution and write
      write u in terms of the original variable x:                 u in terms of the original variable x:
                             1
      u = 3 is the same as x2 = 3, which gives x = 32 = 9          u = –2 is the same as x = –2, which does not
                            1
      u = 7 is the same as x2 = 7, which gives x = 72 = 49         have a solution
                                                                   u = 3 is the same as x = 3, so that x = 9
      The solutions of the original equation are
      x = 9, 49.                                                   Therefore, the solution of the original equa-
662. a. Observe that 16 – 56      x + 49x = 0 can be               tion is x = 9.
      written as 16 – 56 x+ 49( x)2 = 0. Let u=              664. c. We must first write the equation in the cor-
         x. Rewriting the original equation yields the             rect form:
      equation 16 – 56u+ 49u2 = 0, which is quadratic.               1   1
                                                                   2x6 –x3 – 1 = 0
      Factoring the left side results in the equivalent             1      1
                                                                   x3 – 2x6 + 1 = 0
      equation (4 – 7u)2 = 0. Solving this equation                   1       1
                                    4                              (x6)2 – 2(x6) + 1 = 0
      for u yields the solution u= 7 . In order to solve
                                                                                  1
      the original equation, we go back to the substi-             Next, let u = x6. Then, we must solve the equa-
      tution and write u in terms of the original                  tion u2 – 2u+ 1 = 0. Observe that factoring
      variable x:                                                  this equation yields (u– 1)2 = 0. Consequently,
          4                             4                          u = 1. Next, we must go back to the actual sub-
      u = 7 is the same as       x = 7 , which gives us
            4      16                                              stitution and solve the new equations obtained
      x = ( 7 )2 = 49
                                                                   by substituting in this value of u. Specifically,
                                                                                    1
      Therefore, solution of the original equation is              we must solve x6 = 1. This is easily solved by
          16
      x = 49 .                                                     raising both sides to the power 6. The result is
                                                                   that x = 1.




   228
                                                                ANSWERS & EXPLANATIONS–


665. b. We must first rewrite the equation in a nicer form. Observe that
                 –1     –1
      3 +x 4 –x 2 = 0
         –1    –1
      –x 2 +x 4 = 0
            –1      –1
      – (x 4 )2 + (x 4 ) + 3 = 0
                        –1
      Let u = x 4 . Then, solve the quadratic equation
      –u2 + u+ 3 = 0. Using the quadratic formula yields
                 –1 ±       1 – 4(–1)(3)         –1 ± 13            1±        13
      u=                    2(–1)            =      –2          =         2           .

      Now, we must go back to the actual substitution and solve the following equations involving the original
      variable x:
       –1         1+         13                                     –1        1–          13
      x4 =              2                                       x4 =               2
            –1                1+       13 –4                         –1                   1–           13 –4
      (x 4 ) –4 = (                2     )                      (x 4 )–4 = (                   2         )
                        2                       16                                 2                              16
      x = ( 1+               13
                                  )4 =   (1 +        13)4
                                                                x = ( 1–                  13
                                                                                               )4 =        (1 –        13)4

                                                                                                                         16                     16
      So, the two solutions to the original equation are x =                                                      (1 +        13)4
                                                                                                                                     ,   (1 –        13)4
                                                                                                                                                            .

666. d. Let u = x3 + 5. Then, the equation (x3 + 5)2 – 5(x3 + 5) + 6 = 0 can be written equivalently as u2 – 5u+
      6 = 0. This factors as (u– 3)(u– 2) = 0, so we conclude that u= 3, 2. Next, we must solve the following
      equations obtained by going back to the actual substitution:

      x3 + 5 = 3                          x3 + 5 = 2
      x3 = –2                             x3 = –3
          1       1                           1       1
      (x3)3 = (–2)3                       (x3)3 = (–3)3
            3                                   3
      x = –2                              x = –3
                                                                                                                  3            3
      So, the two solutions to the original equation are x =                                                           –2,         –3.
                                                                                                                                                                2
                                         6                  3                                          6
667. c. Observe that 4x + 1 = 5x , or equivalently 4x – 5x + 1 = 0, can be written as 4(x3) – 5(x3) + 1 = 0.      3

      Let u = x3. Rewriting the original equation yields the equation 4|u|2 – 5|u| + 1 = 0, which is quadratic.
      Factoring yields the equivalent equation (4u – 1)(u – 1) = 0. Solving this equation for u yields the solu-
                1
      tions u = 4 or u = 1. Solving the original equation requires that we go back to the substitution and write
      u in terms of the original variable x:
             1                                        1                       3   1
      u=     4    is the same as x3 = 4 , so that x =                             4
      u = 1 is the same as x3 = 1, so that x = 1
                                                                                                   3   1
      The solutions of the original equation are x =                                                   4
                                                                                                           , 1.




                                                                                                                                                                    229
                                       ANSWERS & EXPLANATIONS–


                                                                                  3                                  3   3
668. b. Let u=x2 +x. Observe that (x2 +x)2 + 12 =         670. a. Let u = r– r . Observe that (r– r )2 – (r– r ) –
      8(x2 +x)or equivalently (x2 +x)2 – 8(x2 +x) +             6 = 0 can be written as u2 – u– 6 = 0, which is
      12 = 0, can be written as u2 – 8u + 12 = 0,               quadratic. Factoring yields the equivalent
      which is quadratic. Factoring yields the equiv-           equation (u – 3)(u + 2) = 0. Solving this equa-
      alent equation (u – 6)(u – 2) = 0. Solving this           tion for u yields the solutions u = – 2 or u= 3.
      equation for u yields the solutions u = 2 or u=           In order to solve the original equation, we
      6. To solve the original equation, we go back to          must go back to the substitution and write u in
      the substitution and write u in terms of the              terms of the original variable r. Doing so yields
      original variable x. Doing so yields two more             two more equations involving rational expres-
      quadratic equations, this time in x, that must            sions, this time in r, that must be solved. First,
                                                                                           3
      be solved: First, u = 2 is the same as x2 +x = 2,         u = –2 is the same as r– r = –2. Multiply both
      or equivalently x2 +x – 2 = 0. Factoring yields           sides by r and solve for r:
      the equation (x + 2)(x – 1) = 0, so that x = – 2
                                                                   3
      or 1. Similarly, u = 6 is the same as x2 + x = 6,         r– r = – 2
      or equivalently x2 + x – 6 = 0. Factoring yields          r2 – 3 = – 2r
      the equation (x + 3)(x – 2) = 0, so that x = –3           r2 + 2r– 3 = 0
      or 2. Therefore, the solutions of the original            (r + 3)(r– 1) = 0
      equation are x = –3, –2, 1, or 2.                         r = –3, 1
                                                      2                                                              3
                                                                Similarly, u = 3 is the same as r– r = 3. Multi-
669. a. Let u = 1 +       w. Observe that 2 1 +   w
                                                                ply both sides by r and solve for r:
      = 13 1 +     w – 6, or equivalently                          3
                                                                r– r = 3
                   2
      2 1+     w       – 13 1 +   w + 6 = 0, can be             r2 – 3 = 3r
                                                                r2 – 3 r – 3 = 0
      written as 2u2 – 13u + 6 = 0, which is qua-
      dratic. Factoring yields the equivalent equa-             Using the quadratic formula then yields r =
      tion (2u– 1)(u– 6) = 0. Solving this equation              –(–3) ±   (–3)2 –4(1)(–3)         3±       21
                                     1                                     2(1)                =        2        .
      for u yields the solutions u= 2 or u= 6. Solving
      the original equation requires that we go back            The solutions of the original equation are
      to the substitution and write u in terms of the                        3±       21
      original variable w. Doing so yields two more             r = –3, 1,        2        .
      radical equations, this time in w, that must be
                        1                          1
      solved. First, u= 2 is the same as 1 + w = 2 .
                                                1
      Isolating the radical term yields w = – 2 ,
      which has no solution. Similarly, u = 6 is the
      same as 1 + w = 6. Isolating the radical term
      yields w = 5, so that w = 25. The solution of
      the original equation is w = 25.




   230
                                                 ANSWERS & EXPLANATIONS–


                                         4
671. b. Observe that 6     x – 13 x + 6 = 0 can be                               Section 6—Elementary
                     4            4
      written as 6( x )2 – 13( x ) + 6 = 0. Let u =                                    Functions
       4
         x . Rewriting the original equation yields
      the equation 6u2 – 13u + 6 = 0, which is qua-
                                                                           Set 43 (Page 102)
      dratic. Factoring yields the equivalent equa-
                                                                           673. b. Draw a horizontal line across the coordinate
      tion (2u – 3)(3u – 2) = 0. Solving this equation
                                     2       3                                   plane where f(x) = 3. This line touches the
      for u yields the solutions u = 3 or u= 2 . Solv-
                                                                                 graph of f(x) in exactly one place. Therefore,
      ing the original equation requires that we go
                                                                                 there is one value for which f(x) = 3.
      back to the substitution and write u in terms
                                                                           674. d. The x-axis is the graph of the line f(x) = 0,
      of the original variable x:
                                                                                 so every time the graph touches the x-axis.
          2                 4        2                       2        16
      u= 3 is the same as       x = 3 , so that x = ( 3 )4 = 81 .                The graph of f(x) touches the x-axis in 5 places.
          3                 4        3                       3        81         Therefore, there are 5 values for which f(x) = 0.
      u= 2 is the same as       x = 2 , so that x = ( 2 )4 = 16 .
                                                                           675. b. Draw a horizontal line across the coordinate
      Therefore, the solution of the original equation
               16 81                                                             plane where f(x) = 10. The arrowheads on the
      is x =   81 , 16 .                                                         ends of the curve imply that the graph extends
                      1                      2           1
672. c. Let u = a 3 . Observe that 2a 3 – 11a 3 + 12 = 0                         upward, without bound, as x tends toward
                                1                1                               both positive and negative infinity. This line
      can be written as 2(a 3 )2 – 11(a 3 ) + 12 = 0.
                                                                                 touches the graph of f(x) in 2 places. There-
      Rewriting the original equation yields the
                                                                                 fore, there are 2 values for which f(x) = 10.
      equation 2u2 – 11u + 12 = 0, which is qua-
                                                                           676. e. The domain of a real-valued function is
      dratic. Factoring yields the equivalent equa-
      tion (2u – 3)(u– 4) = 0. Solving this equation                             the set of all values that, when substituted
                                     3
      for u yields the solutions u = 2 or u= 4. In                               for the variable, produce a meaningful output,
      order to solve the original equation, we go                                while the range of a function is the set of all
      back to the substitution and write u in terms                              possible outputs. All real numbers can be
      of the original variable a:                                                substituted for x in the function f(x) = x2 – 4,
                                1
                                                                                 so the domain of the function is the set of all
           3                         3               3           27
      u = 2 is the same as a 3 = 2 , so x = ( 2 )3 =             8               real numbers. Since the x term is squared, the
                                1
      u= 4 is the same as a 3 = 4, so x = (4)3 = 64                              smallest value that this term can equal is 0
                                                                                 (when x = 0). Therefore, the smallest value that
      The solutions of the original equation are                                 f(x) can attain occurs when x = 0. Observe that
              27
      a = 64, 8 .                                                                f(0) = 02 – 4 = –4. The range of f(x) is the set of
                                                                                 all real numbers greater than or equal to –4.




                                                                                                                            231
                                       ANSWERS & EXPLANATIONS–


677. b. The domain of the function is the set of all       680. d. Using the graphs yields f(0) = 0, f(2) = –1,
      real numbers, so any real number can be sub-               and g(4) = –4. Substituting these values into
      stituted for x. The range of a function is the set         the given expression yields
      of all possible outputs of the function. Since
                                                                 2 f(0) + [f(2) g(4)]2 = 2(0) + [(–1)( –4)]2 =
      the x term is squared, then made negative, the
                                                                 0 + 42 = 16
      largest value that this term can equal is 0 (when
                                                           681. b. The zeros of a polynomial are precisely its x-
      x = 0). Every other x value will result in a neg-
                                                                 intercepts, which are –3, 1, and 3.
      ative value for f(x). As such, the range of f(x)
      is the set of all real numbers less than or equal    682. c. The lowest point on the graph of y = p(x)

      to 0.                                                      occurs at (2, –1), so the smallest possible y-
                                                                 value attained is –1. Further, every real num-
678. c. You must identify all possible y-values that
                                                                 ber greater than –1 is also attained at some
      are attained within the graph of f. The graph
                                                                 x-value. Hence, the range is [–1, ∞).
      of f is comprised of three distinct components,
      each of which contributes an interval of values      683. b. The domain of any polynomial function is

      to the range of f. The set of y-values correspond-            because any real number can be substituted
      ing to the bottommost segment is (–2,–1];                  in for x in p(x) and yield another real number.
      note that –2 is excluded due to the open hole        684. d. We must identify the x-values of the por-
      at (5,–2) on the graph, and there is no other x-           tion of the graph of y = p(x) that lies between
      value in [–5,5] whose functional value is –2.              the horizontal lines y = –1 and y = 0 (i.e.,
      Next, the portion of the range corresponding               the x-axis). Once this is done, we exclude
      to the middle segment is [0,2); note that 2 is             the x-values of the points where the graph
      excluded from the range for the same reason                of y = p(x) intersects the line y = –1 (because
      –2 is excluded. Finally, the horizontal segment            of the strict inequality), and we include those
      contributes the singleton {3} to the range; even           x-values of the points where the graph of y =
      though there is a hole in the graph at (0,3),              p(x) intersects the line y = 0. This yields the set
      there are infinitely many other x-values in                [1, 2)∪(2, 3]∪{–3}.
      [–5,5] whose functional value is 3, thereby          685. b. The graph of f has a vertical asymptote at
      requiring that it be included in the range.                x = 1 and a horizontal asymptote at y = –2.
      Thus, the range is (–2, –1]∪[0, 2)∪{3}.                    Since the graph follows the vertical asymptote
679. c. The graph of g is steadily decreasing from               up to positive infinity as x approaches x = 1
      left to right, beginning at the point (–5,–4)              from the left and down to negative infinity as x
      and ending at (5,–4), with the only gap occur-             approaches x = 1 from the right, and it does
      ring in the form of a hole at (0,1). Since there           not cross the horizontal asymptote, we con-
      is no x-value in [–5,5] whose functional value             clude that the graph attains all y-values except
      is 1, this value must be excluded from the                 –2. Hence, the range is (– ∞, – 2)∪(–2, ∞).
      range. All other values in the interval [–4,4]
      do belong to the range. Thus, the range is
      [–4,1)∪(1, 4].




   232
                                                  ANSWERS & EXPLANATIONS–



          9f(x)      9[ –(2x – (–1 – x2))]       –9(x2 + 2x + 1)               A is not a function. It fails the vertical line test
686. b. g(x) =             3(1 + x)          =      3(x + 1)         =
                                                                               for all x-values where –2 x 2. The equa-
      –9(x + 1) 2
       3(x + 1)     = –3(x + 1) = –g(x)                                        tions graphed in diagrams B and D are func-
                                                      6(x + 1)
687. d. Since the function 2g(x)h(x) = x2 + 1 is a                             tions whose ranges contain negative values.
      rational function, its domain is the set of all                    691. e. The equation of the graph in diagram B is
      those x-values for which the denominator is                              y = |x| – 3. Any real number can be substituted
      not equal to zero. There is no real number x                             into this equation. There are no x-values that
      that satisfies the equation x2 + 1 = 0. There-                           will generate an undefined or imaginary y-value.
      fore, the domain is the set of all real numbers.                         The equation of the graph in diagram E is y =
688. b.
                                                                               (x – 3)2 + 1.With this equation as well, any real
                                                                               number can be substituted for x—there are no
                              1
      3f(x) – 2xg(x) –       h(x)                                              x-values that will generate an undefined or
                                                                 1
                                                                 1             imaginary y-value. The equation of the graph
      = 3[– (2x – (–1 – x2))] – 2x[3(x + 1)] –               x2 + 1
                                                                               in diagram D is y = 1 . If x = 0, this function
                                                                                                      x
      = –3(x2 + 2x + 1) – 6x(x + 1) – (x2 + 1)                                 will be undefined. Therefore, the domain of
      = –3x2 – 6x – 3 – 6x2 – 6x – x2– 1                                       this function is all real numbers excluding 0.
      = – 10x2 – 12x – 4                                                       Only the functions in diagrams B and E have a
      = – 2(5x2 + 6x + 2)                                                      domain of all real numbers with no exclusions.
                                                                         692. b. The equation of the graph in diagram C is y
Set 44 (Page 105)                                                              = x. Since the square root of a negative
689. b. The graph of the equation in diagram A is                              number is imaginary, the domain of this equa-
      not a function. A function is an equation in                             tion is all real numbers greater than or equal
      which each unique input yields no more than                              to 0. The square roots of real numbers greater
      one output. The equation in diagram A fails                              than or equal to 0 are also real numbers that
      the vertical line test for all x-values where                            are greater than or equal to 0. Therefore, the
      –2 x 2. For each of these x-values                                       range of the equation y = x is all real num-
      (inputs), there are two y-values (outputs).                              bers greater than 0, and the domain and range
690. d. The range of a function is the set of possible                         of the equation are the same. The equation of
      outputs of the function. In each of the five                             the graph in diagram D is y = 1 . If x = 0, this
                                                                                                                x
      equations, the set of possible y-values that can                         function will be undefined. Therefore, the
      be generated for the equation is the range of                            domain of this function is all real numbers
      the equation. Find the coordinate planes that                            excluding 0. Dividing 1 by a real number
      show a graph that extends below the x-axis.                              (excluding 0) will yield real numbers, exclud-
      These equations have negative y-values, which                            ing 0. Therefore, the range of the equation
      means that the range of the equation contains                            y = 1 is all real numbers excluding 0, and the
                                                                                    x
      negative values. The graphs of the equations in                          domain and range of the equation are the
      diagrams A, B, and D extend below the x-axis.                            same. The equation of the graph in diagram B
      However, the graph of the equation in diagram                            is y = |x| – 3. Any real number can be substituted




                                                                                                                            233
                                       ANSWERS & EXPLANATIONS–


      into this equation. There are no x-values that              excluding 0. The equation of the graph in
      will generate an undefined or imaginary y                   diagram E is y = (x – 3)2 + 1. Any x-value
      value. However, it is impossible to generate a              greater than or less than 3 will generate a y
      y-value that is less than –3. Any x-value greater           value that is greater than 1; no values less than
      than or less than 3 will generate a y value that            1 can be generated by this equation. Therefore,
      is greater than –3. Therefore, the range of the             the range of the equation y = (x – 3)2 + 1 is all
      equation y = |x| –3 is all real numbers greater             real numbers greater than or equal to 1. Of the
      than or equal to –3. The domain and range of                four equations that are functions, the equation
      y = |x| –3 are not the same. The equation of                y = (x – 3)2 + 1 (E), has the smallest range
      the graph in diagram E is y = (x – 3)2 + 1.                 (fewest elements), since the set of real num-
      With this equation as well, any real number                 bers that are greater than or equal to 1 is
      can be substituted for x—there are no x-values              smaller than the set of all real numbers greater
      that will generate an undefined or imaginary                than or equal to –3 (B), smaller than the set of
      y-value. However, it is impossible to generate a            all real numbers greater than or equal to 0 (C),
      y-value that is less than 1. Any x-value greater            and smaller than the set of all real numbers
      than or less than 3 will generate a y-value that            excluding 0 (D).
      is greater than 1. Therefore, the range of the        694. b. Substitute the expression 2y – 1 for every
      equation y = (x – 3)2 + 1 is all real numbers               occurrence of x in the definition of the func-
      greater than or equal to 1. The domain and                  tion f(x), and then simplify:
      range of y = (x – 3)2 + 1 are not the same.
                                                                  f(2y – 1) =
693. e. The graph of the equation in diagram A is
                                                                  (2y – 1)2 + 3(2y – 1) – 2 =
      not a function. It fails the vertical line test for
                                                                  4y2 – 4y + 1 + 6y – 3 – 2 =
      all x-values where –2 x 2. The equation of
                                                                  4y2 + 2y – 4
      the graph in diagram B is y = |x| – 3. Any x-
                                                            695. c. Simplifying f(x + h) requires we substitute
      value greater than or less than 3 will generate a
                                                                  the expression x + h for every occurrence of x
      y-value that is greater than –3; no values less
                                                                  in the definition of the function f(x), and then
      than –3 can be generated by this equation.
                                                                  simplify:
      Therefore, the range of the equation y = |x| – 3
      is all real numbers greater than or equal to –3.            f(x + h) =
      The equation of the graph in diagram C is y =               –((x + h) –1)2 + 3 =
          x. Since the square roots of negative num-              –[(x + h)2 – 2(x + h)+ 1] + 3 =
      bers are imaginary, the range of this equation              –[x2 + 2hx + h2 – 2x – 2h + 1] + 3 =
      is all real numbers greater than or equal to 0.             –x2 – 2hx – h2+ 2x + 2h – 1 + 3 =
      The equation of the graph in diagram D is y =               –x2 – 2hx – h2+ 2x + 2h + 2
      1
      x . Dividing 1 by a real number (excluding 0)
                                                                  Next, in anticipation of simplifying f(x + h) – f(x),
      will yield real numbers, excluding 0. There is
                                                                  we simplify the expression for f(x) = –(x – 1)2 + 3
      no value for x that can make y = 0. Therefore,
                                     1                            in order to facilitate combining like terms:
      the range of the equation y = x is all real numbers




   234
                                       ANSWERS & EXPLANATIONS–


      f(x) =                                                      Finally, substitute this result for x in f(x):
      –(x – 1)2 + 3 =
                                                                  f(5) = 2(5) + 1 = 10 + 1 = 11
      –(x2 – 2x + 1) + 3 =
      –x2 + 2x – 1 + 3 =                                          Thus, f(g(f(3))) = 11.
      –x2 – 2x + 2                                          700. c. Begin with the innermost function. You are
                                                                  given the value of f(x): f(x) = 6x + 4. Substitute
      Finally, simplify the original expression
                                                                  this expression for x in the equation g(x), and
      f(x + h) – f(x):
                                                                  then simplify:
      f(x + h) – f(x) =
                                                                  g(6x + 4) =
      (–x2 – 2hx – h2 + 2x + 2h + 2) – (–x2 + 2x + 2) =
                                                                  (6x + 4)2 – 1 =
      –x2 – 2hx – h2 + 2x + 2h + 2 + x2 – 2x –2 =
                                                                  36x2 + 24x + 24x + 16 – 1 =
      –2hx – h2 + 2h =
                                                                  36x2 + 48x + 15
      h(h – 2h + 2)
696. c. By definition, (g ˚ h)(4) = g(h(4)). Observe              Therefore, g(f(x)) = 36x2 + 48x + 15.
      that h(4) = 4 – 2 4 = 4 – 2(2) = 0, so g(h(4))        701. b. Since g(0) = 2 and f(2) = –1, we have
      = g(0) = 2(0)2 – 0 – 1 = –1. Thus, we conclude
                                                                  (f ˚ g)(0) = f(g)(0)) = f(2) = –1.
      that (g ˚ h)(4) = –1.
                                                            702. b. Since f(5) = 0 and f(0) = 0, we work from
697. d. By definition, (f ˚ f ˚ f)(2x) = f(f(f(2x))).
                                                                  the inside outward to obtain
      Working from the inside outward, we first
      note that f(2x) = –2(x)2 = –4x2. Then, f(f(2x))             f(f(f(f(5)))) = f(f(f(0))) = f(f(0)) = f(0) = 0
      = f(–4x2) = – (–4x2)2 = –16x4. Finally, f(f(f(2x)))   703. c. Simplify the given expression:
      = f(–16x4) = – (–16x4)2 = –256x8. Thus, we
      conclude that (f ˚ f ˚ f)(2x) = –256x8.                     f(x + 2) =

698. b. Begin with the innermost function: find                     (x + 2)2 – 4(x + 2) =
      f(–2) by substituting –2 for x in the function
                                                                    x2 + 4x + 4 – 4x – 8 =
      f(x):
                                                                    x2 – 4
      f(–2) = 3(–2) + 2 = –6 + 2 = –4
                                                            704. b. The domain of g ˚ f consists of only those
                                                                  values of x for which the quantity f(x) is
      Then, substitute the result for x in g(x).
                                                                  defined (that is, x belongs to the domain of f)
      g(–4) = 2(–4) – 3 = –8 – 3 = –11                            and for which f(x) belongs to the domain of g.
                                                                  For the present scenario, the domain of f con-
      Thus, g(f(–2)) = –11
                                                                  sists of only those x-values for which –3x 0,
699. e. Begin with the innermost function: Find
                                                                  which is equivalent to x 0. Since the domain
      f(3) by substituting 3 for x in the function f(x):
                                                                  of g(x) = 2x2 + 18 is the set of all real num-
      f(3) = 2(3) + 1 = 6 + 1 = 7                                 bers, it follows that all x-values in the interval
                                                                  (–∞, 0] are permissible inputs in the composi-
      Next, substitute that result for x in g(x).
                                                                  tion function (g ˚ f)(x), and that, in fact, these
      g(7) = 7 – 2 = 5                                            are the only permissible inputs. Therefore, the
                                                                  domain of g ˚ f is (–∞, 0].
                                                                                                               235
                                       ANSWERS & EXPLANATIONS–



Set 45 (Page 107)                                           709. b. The intersection of the graph of f(x) = x3
705. b. The radicand of an even-indexed radical                   and the graph of the horizontal line y = a can
      term (e.g., a square root) must be nonnegative              be found by solving the equation x3 = a. Taking
      if in the numerator of a fraction and strictly              the cube root of both sides yields the solution
                                                                       3
      positive if in the denominator of a fraction.               x = a, which is meaningful for any real
      For the present function, this restriction takes            number a.
                                                                                                  1
      the form of the inequality –x 0, which upon           710. c. The graph of f(x) = x , in fact, decreasing on
      multiplication on both sides by –1, is equiva-              its entire domain, not just (0, ∞). Its graph is
      lent to x 0. Hence, the domain of the func-                 given here:
      tion f(x) = –x is (–∞, 0].                                                            10

706. d. There is no restriction on the radicand of an                                        8

      odd-indexed radical term (e.g., a cube root) if                                        6

      it is in the numerator of a fraction, whereas                                          4

      the radicand of such a radical term must be                                            2

      nonzero if it occurs in the denominator of a
                                                                  –10   –8   –6   –4   –2             2   4   6   8   10
      fraction. For the present function, this restric-                                      –2
      tion takes the form of the statement –1 – x ≠ 0,
                                                                                             –4
      which is equivalent to x ≠ –1. Hence, the
                                                                                             –6
                                          1
      domain of the function g(x) =     3       is
                                         –1 – x                                              –8
      (–∞,–1)∪(–1, ∞).
                                                                                            –10
707. b. The equation y = 2 is the equation of hori-
      zontal line that crosses the y-axis at (0, 2).        711. c. The square root of a negative value is imagi-
      Horizontal lines have a slope of 0. This line is a          nary, so the value of 4x – 1 must be greater than
      function, since it passes the vertical line test: A         or equal to 0. Symbolically, we have:
      vertical line can be drawn through the graph
                                                                  4x – 1 0
      of y = 2 at any point and will cross the graphed
                                                                  4x 1
      function in only one place. The domain of the
                                                                  x 1  4
      function is infinite, but all x-values yield the
      same y-value: 2. Therefore, the range of y = 2
                                                                  Hence, the domain of f(x) is the set of all real
      is 2.
708. b. The graph of f(x) = |x| has its lowest point              numbers greater than or equal to 1 . The small-
                                                                                                   4
      at the origin, which is both an x-intercept and             est value of f(x) occurs at x = 1 , and its value is
                                                                                                  4
      a y-intercept. Since f(x) 0 or any nonzero real
      number x, it cannot have another x-intercept.               √4( 14 ) – 1 =       0 = 0. So, the range of the
      Moreover, a function can have only one y-                   function is the set of all real numbers greater
      intercept, since if it had more than one, it
                                                                  than or equal to 0.
      would not pass the vertical line test.




   236
                                       ANSWERS & EXPLANATIONS–


712. c. The minimum value for both functions               716. d. The x-values of the points of intersection of
      occurs at their vertex, which occurs at (0, 0).            the graphs of f(x) = 2x and g(x) = 4x3must sat-
      Also, for any positive real number a, the                  isfy the equation 4x3 = 2x. This equation is
      graphs of both f and g intersect the horizontal            solved as follows:
      line y = a twice. Therefore, the range of both
                                                                 4x3 = 2x
      functions is [0, ∞).
713. c. The radicand of an odd-indexed radical                   4x3 – 2x = 0
      term (e.g., a fifth root) must be nonzero if it            2x(x2 – 1) = 0
      occurs in the denominator of a fraction, which
      is presently the case. As such, the restriction            4x(x2 – 1 ) = 0
                                                                         2

      takes the form of the statement 2 – x ≠ 0, which           4x(x – √ 1 )(x + √ 1 ) = 0
                                                                                2              2
      is equivalent to x ≠ 2. Thus, the domain is
      (–∞,2) (2, ∞).                                             x =0, ±√ 1 )
                                                                                2
714. c. The x-intercepts of f are those values of x
                                                                 So, the points of intersection are (0,0),
      satisfying the equation 1 – |2x – 1| = 0, which
                                                                 (√ 1 , 2√ 1 ), and (–√ 1 , –2√ 1 ). There are
      is equivalent to |2x – 1| = 1. Using the fact that            2           2                  2         2
      |a| = b if and only if a = ±b, we solve the two            more than two points of intersection.
      equations 2x – 1 = ±b separately:                    717. b. The x-values of the points of intersection of

      2x – 1 = – 1       2x – 1 = 1                              the graphs of f(x) = 3 x2 and g(x) =
                                                                                      4
                                                                                                                    5 2
                                                                                                                   16 x   must
      2x = 0             2x = 2                                  satisfy the equation 3 x2 =
                                                                                      4
                                                                                                         5
                                                                                                        16   x2. This equation
      x =0               x=1                                     is solved as follows:
      Thus, there are two x-intercepts of the given              3 2        5
                                                                                x2
                                                                 4x    =   16
      function.
                                                                 3 2        5
                                                                 4x    –   16   x2 = 0
715. d. The x-values of the points of intersection of
      the graphs of f(x) = x2and g(x) = x4must sat-              (3–
                                                                  4
                                                                         5 2
                                                                        16 )x    =0
      isfy the equation x4 = x2. This equation is
                                                                 x2 = 0
      solved as follows:
                                                                 x=0
      x4 = x2
      x4 – x2 = 0                                                Hence, there is only one point of intersection,
      x2(x2 – 1) = 0                                             (0,0).
      x2(x – 1)(x + 1) = 0
                                                           718. b. The y-intercept for a function y = f(x) is the
      x = –1, 0, 1
                                                                                                                   –2 –|2 – 3(0)|
                                                                 point (0, f(0)). Observe that f(0) =              4 – 2(0)2|–0|
      The points of intersection are (–1,1), (0,0),
                                                                                    –4
      and (1,1): there are more than two points of               = –2––02 =
                                                                    4                4   = –1. So, the y-intercept is (0, –1).
      intersection.




                                                                                                                          237
                                                  ANSWERS & EXPLANATIONS–


719. b. If there is no x-value that satisfies the equa-       723. b. We must identify the intervals in the domain
      tion, f(x) = 0 then the graph of y = f(x) does                of p(x) on which the graph of y = p(x) rises
      not cross the horizontal line y = 0, which is the             from left to right. This happens on the intervals
      x-axis.                                                       (–3, 0)∪( 2, ∞).
720. b. If there is no x-value that satisfies the equa-       724. b. The graph of f passes the horizontal line test
      tion, f(x) = 3 then there is no point on the                  on this interval, so, it has an inverse.
      graph of y = f(x) when y = 3. Therefore, 3 is           725. c. Determining the inverse function for f
      not in the range of f.                                        requires that we solve for x in the expression
                                                                          x–1
                                                                    y=             :
Set 46 (Page 109)                                                        5x + 2
                                                                          x–1
721. c. The domain of a rational function is the set                y=   5x + 2
      of all real numbers that do not make the                      y(5x + 2) = x – 1
      denominator equal to zero. For this function,
                                                                    5xy + 2y = x – 1
      the values of x that must be excluded from the
                                                                    5xy – x = –2y – 1
      domain are the solutions of the equation x3 – 4x
      = 0. Factoring the left side yields the equivalent            x(5y – 1) = –2y – 1
      equation                                                           –2y – 1
                                                                    x=    5y – 1

      x3 – 4x = x(x2 – 4) = x(x – 2)(x + 2) = 0                     Now, we conclude that the function f –1(y) =
                                                                    –2y –1       1
      The solutions of this equation are x = –2, 0,                 5y – 1 , y ≠ 5 is the inverse function of f.

      and 2. Hence, the domain is (–∞,–2)∪( –2,               726. d. Remember that the domain of f is equal to
      0)∪(0, 2)∪(2, ∞).                                             the range of f –1. As such, since 0 does not
722. d. First, simplify the expression for f(x) as
                                                                    belong to the range of f –1, it does not belong to
      follows:                                                      the domain of f, so f(0) is undefined. Also, the
                                                                    fact that (1, 4) is on the graph of y = f(x) means
       (x – 3)(x2 – 16)       (x – 3)(x – 4)(x + 4)
       (x2 + 9)(x – 4)
                          =      (x2 + 9)(x – 4)
                                                      =             that f(1) = 4; this is equivalent to saying that
       (x – 3)(x + 4)
                        =   x2 + x – 12)                            (4, 1) is on the graph of f –1, or that f –1(4) = 1.
          (x2 + 9)             x2 + 9
                                                                    All three statements are true.
      While there is a hole in the graph of f at x = 4,
                                                              727. b. Determining the inverse function for f
      there is no x-value that makes the denomina-
                                                                    requires that we solve for x in the expression
      tor of the simplified expression equal to zero.
                                                                    y = x3 + 2:
      Hence, there is no vertical asymptote. But, since
      the degrees of the numerator and denomina-                    y = x3 + 2
      tor are equal, there is a horizontal asymptote                y – 2 = x3
                                                                         3
      given by y = 1 (since the quotient of the coeffi-             x= y–2
      cients of the terms of highest degree in the                                                    3
                                                                    Hence, the inverse is f –1(y) =       y – 2.
      numerator and denominator is 1 1 = 1).




   238
                                                    ANSWERS & EXPLANATIONS–



728. b. First, note that x2 + 1 does not factor, so x = 1                        Next, since the degrees of the numerator and
      is a vertical asymptote for the graph of f. Since                          denominator are equal, there is a horizontal
      the degree of the numerator of the fraction is                             asymptote given by y = 1 (since the quotient of
      exactly one more than that of the denominator,                             the coefficients of the terms of highest degree
      we can conclude that the graph has no horizon-                             in the numerator and denominator is 1 1 =
      tal asymptote, but does have an oblique asymp-                             1). Because x = 0 makes the denominator equal
      tote. Hence, II is a characteristic of the graph                           to zero, but does not make the numerator equal
      of f.                                                                      to zero, it is a vertical asymptote. So, statement
                                                                                 II holds.
      Next, while there is a y-intercept, (0,3), there is
      no x-intercept. To see this, we must consider                              Finally, since x = 0 is a vertical asymptote, the
      the equation, f(x) = 0 which is equivalent to                              graph of f cannot intersect it, and there is no y-
           x2 + 1        2(x – 1) – (x2 + 1)       –(x2 – 2x + 3)
                                                                                 intercept. So, statement III does not hold.
      2–   x–1      =           x–1            =        x–1         = 0.                                          1
                                                                           730. c. The y-values of f(x) = x get smaller as x-
      The x-values that satisfy such an equation are                             values move from left to right. The graph is as
      those that make the numerator equal to zero                                follows:
      and do not make the denominator equal to                                                              10

      zero. Since the numerator does not factor, we                                                          8

      know that (x – 1) is not a factor of it, so we
                                                                                                             6
      need only solve the equation. x2 – 2x + 3 = 0.
                                                                                                             4
      Using the quadratic formula yields
                                                                                                             2

            –(–2) ±      (–2)2 –4(1)(3)        2± –8
      x=                 2(1)             =      2      =1±i 2                    –10   –8   –6   –4   –2             2   4   6   8   10

                                                                                                             –2
      Since the solutions are imaginary, we conclude
                                                                                                             –4
      that there are no x-intercepts. Hence, III is not
                                                                                                             –6
      a characteristic of the graph of f.
                                                                                                             –8
729. b. The expression for f can be simplified as
                                                                                                            –10
      follows:
       (2 – x)2(x + 3)       (–(x – 2))2(x + 3)
          x(x – 2)2
                         =       x(x – 2)2
                                                   =
       (x – 2)2(x + 3)       x+3
          x(x – 2)2
                         =    x

      Since x = 2 makes both the numerator and
      denominator of the unsimplified expression
      equal to zero, there is a hole in the graph of f at
      this value. So, statement I holds.




                                                                                                                                      239
                                      ANSWERS & EXPLANATIONS–


731. d. The y-values on all three graphs increase as                                        10

      the x-values move from left to right in the                                               8
      interval (0, ∞). Their graphs are as follows:
                                                                                                6


                                                                                                4

                                                                                                2

                                                                                                                                          x
                                              f(x) = x3     –10    –8    –6    –4    –2                 2       4       6       8    10

                                                                                            –2


                                                                                            –4


                                                                                            –6


                                                                                            –8

                                                                                           –10



                                                                                          10


                                                                                           8


                                                                                           6


                                                                                           4

                                                                                           2

                                                                                                                                          x
                                                           –10    –8    –6    –4    –2              2       4       6       8       10
                                          f(x) = 2x + 5
                                                                                           –2


                                                                                           –4


                                                                                           –6

                                                                                           –8

                                                                                          –10



                                                                                            10


                                                                                                8


                                                                                                6


                                                                                                4

                                                                                                2

                                                                                                                                          x
                                                            –10    –8    –6    –4    –2                 2       4       6       8    10
                                              f(x) = |x|
                                                                                            –2


                                                                                            –4

                                                                                            –6


                                                                                            –8


                                                                                           –10

   240
                                       ANSWERS & EXPLANATIONS–


                              x2 + 1
732. c. The function f(x) = x2 + 3 has no vertical asymptote since no value of x makes the denominator equal to
      zero, and has the horizontal asymptote y = 1.
733. d. The function f(x) = x3 is an example that shows that both statements a and b are true. And the func-
      tion g(x) = –1 – x2 illustrates the truth of statement c. Their graphs are as follows:

                                                                                       10


                                                                                        8


                                                                                        6


                                                                                        4

                                                                                        2


                                               f(x) = x3     –10   –8   –6   –4   –2           2   4   6   8    10

                                                                                        –2


                                                                                        –4

                                                                                        –6


                                                                                        –8

                                                                                       –10



                                                                                        10


                                                                                         8


                                                                                         6


                                                                                         4

                                                                                         2



                                         g(x) = –1 – x2      –10   –8   –6   –4   –2           2   4   6   8    10

                                                                                        –2