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10 01 SKILL BUILDERS PRACTICE Boost Your Score! FREE Access to Instantly Scored Online Practice Questions 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 987654321 For more information or to place an order, contact LearningExpress at: 2 Rector Street 26th Floor New York, NY 10006 Or visit us at: www.learnatest.com 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