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Thinking Mathematically by Robert Blitzer - PowerPoint 4

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					  Thinking
Mathematically
      MATH1101
Comprehensive Mathematics
   Homework Solutions
       Chapter 6
             Chapter 6, Section 1



1.   5x + 7 2. 9x + 6     3. -7x – 5    4. -6x – 13      5. x2 + 4
     54 + 7   95 + 6    (-7)(-4) – 5 (-6)(-3) – 13   55 + 4
     20+7      45 + 6      28 – 5         18 – 13        25 + 4
     27        51            23             5              29
6.   x2 + 9 7. x2 - 6      8. x2 - 11   9. x2 + 4x       10. x2 + 6x
     33 + 9 (-2)(-2) - 6 (-3)(-3)-11 1010 + 410      99 + 69
      9+9      4-6           9-11          100+40           81 + 54
       18        -2           -2              140             135
         Chapter 6, Section 1


35. 7x + 10x
    (7+10)x distributive law
    17x      addition

36. 5x + 13x
    (5+13)x distributive law
    18x      addition
          Chapter 6, Section 1


37. 5x2 - 8x2
    (5 - 8)x2   distributive law
    -3x2        addition

38. 7x2 - 10x2
    (7 - 10)x2 distributive law
    -3x2       addition
         Chapter 6, Section 1

39. 3(x + 5)
    3x +35    distributive law
    3x + 15    multiplication

40. 4(x + 6)
    4x +46    distributive law
    4x + 24    multiplication
         Chapter 6, Section 1


41. 4(2x - 3)
    42x - 43   distributive law
    8x - 12      multiplication

42. 3(4x - 5)
    34x - 35   distributive law
    12x - 15     multiplication
          Chapter 6, Section 1




61. The difference between a and b: a – b
    5x – 2x which is equal to 3x.
62. 6x – (-2x) which equals 6x + 2x = 8x.
63. 8x – (3x + 6) which equals 8x + (-3x – 6) = 5x – 6.
64. 8 – 3(x + 6) which equals 8 + (-3x – 18) = -3x - 10
               Chapter 6, Section 2


1.   x–7=3             since 7 is subtracted from x     check: 10 – 7 = 3
     x – 7 + 7 = 3 + 7 add 7 to both sides                      3=3
     x + 0 = 10        perform additions
     x = 10


2.   x – 3 = -17         since 3 is subtracted from x   check: -14 – 3 = -17
     x – 3 + 3 = -17 + 3 add 3 to both sides                    -17 = -17 
     x + 0 = -14        perform additions
     x = -14
               Chapter 6, Section 2


3.   x + 5 = -12               since 5 is added to x     check:-17 + 5 = -12
     x + 5 - 5 = -12 - 5      subtract 5 from both sides        12 = 12 
     x + 0 = -17              perform subtraction
     x = -17


4.   x + 12 = -14             since 12 is added to x     check:-26 + 12 = -14
     x + 12 - 12 = -14 - 12   subtract 5 from both sides        -14 = -14 
     x + 0 = -26              perform subtraction
     x = -26
               Chapter 6, Section 2


      x                                                 12
5.      =4          since x is divided by 3    check:      =4
      3                                                  3
     x
       × 3= 4 × 3   multiply both sides by 3            4=4
     3
      x = 12        perform multiplication

       x                                                15
6.     5
         =3         since x is divided by 5    check:      =3
                                                         5
     x
     5
       × 5= 3 × 5   multiply both sides by 5            3=3
      x = 15        perform multiplication
              Chapter 6, Section 2


7.   5x = 45       since x is multiplied by 5   check: 5 × 9= 45
     5x/5 = 45/5   divide both sides by 5              45 = 45 
     x =9          perform division


8.   6x = 18       since x is multiplied by 6   check: 6 × 3= 18
     6x/6 = 18/6   divide both sides by 6              18 = 18 
     x =3          perform division
              Chapter 6, Section 2


9.   8x = -24       since x is multiplied by 8   check: 8 × -3= -24
     8x/8 = -24/8   divide both sides by 8              -24 = -24 
     x = -3         perform division


10. 5x = -25        since x is multiplied by 5   check: 5 × -5 = -25
    5x/5 = -25/5    divide both sides by 5              -25 = -25 
    x = -5          perform division
                Chapter 6, Section 2

      x x 5                                             check:
43.              Get rid of the 3 denominator by
      3 2 6                                             1 1 2 3 23 5
                   multiplying by 3                                 
        x x      5                                    3 2 6 6  6   6
      3    3 
       3 2       6
          3 x 15      Get rid of the 2 denominator by
      x      
           2    6     multiplying by 2
            3x       15
      2 x        2
             2 
                       6
                 30
      2x  3x      5
                  6
      5x  5
      x 1
                Chapter 6, Section 2

      x        x                                               check:
44.      1           Get rid of the 4 denominator by
      4        5                                                        20      20
                       multiplying by 4                                    1 
        x       4x                                                    4       5
      4   1 
        4       5                                                      51  4     
               4x       Get rid of the 5 denominator by                   44
      x4
                5       multiplying by 5
                       4x
      5  x  4  5 
                        5
      5 x  20  4 x      Add 20 to both sides
      5 x  4 x  20
      5 x  4 x  4 x  20  4 x Subtract 4x from both sides
      x  20
               Chapter 6, Section 2

      x          x Get rid of the 2 denominator by      check: 24  20  24
45.       20 
      2          3 multiplying by 2                            2          3
         x           x                                    12  20  8
      2    2  20  
        2            3                                    12  12     
                 2x
      x  40         Get rid of the 3 denominator by
                  3 multiplying by 3
                    2x 
      3 x  3  40 
                     3 
                        
      3 x  120  2 x Add 2x to both sides
      3 x  2 x  120  2 x  2 x
      5 x  120        Divide both sides by 5
      x  24
               Chapter 6, Section 2

      x 1 x Get rid of the 5 denominator by
46.                                                     check:   15 1 15
      5 2 6 multiplying by 5                                           
         x 1  5x                                                  5 2 6
      5                                                        15 1 5
         5 2 6                                                       
           5 5 x Get rid of the 6 denominator by                    5 2 2
      x                                                              1 5
           2 6 multiplying by 6                                    3 
                                                                       2 2
             5
      6 x    5x                                                6 1 5
             2                                                      
                                                                   2 2 2
            30
      6x        6 x  15  5 x Add 15 to both sides              5 5
             2                                                       
                                                                   2 2
                                                                         
      6 x  5 x  15
      6 x  5 x  5 x  15  5 x Subtract 5x from both sides
      x  15
       Chapter 6, Section 2


61. A= ½(a + b)    Multiply both sides by 2
    2A = a + b     Subtract b from both sides
    2A – b = a     to isolate a               This is the formula
                                              for the average of
62. A= ½(a + b)    Multiply both sides by 2   two numbers.
    2A = a + b  Subtract a from both sides
    2A – a = b  to isolate b
       Chapter 6, Section 2


63. S = P + Prt   Subtract P from both sides
    S – P = Prt   Divide both sides by P
   SP
        rt       Divide both sides by t
    P
   SP                                     This is the formula for
       r                                  compound interest that
    Pt                                     we will discover in
                                           Chapter 8.
       Chapter 6, Section 2


64. S = P + Prt   Subtract P from both sides
    S – P = Prt   Divide both sides by P
   SP
        rt       Divide both sides by r
    P
   SP                                     This is the formula for
       t                                  compound interest that
    Pr                                     we will discover in
                                           Chapter 8.
           Chapter 6, Section 3


1. Five times a number: 5x
   decreased by 4: - 4
   5x – 4 = 26
   5x = 26 + 4 = 30
   x=6

   When five times 6 is decreased by 4, 30 is decreased
   by 4, which IS 26.
          Chapter 6, Section 3


2. Two times a number: 2x
   decreased by 3: - 3
   2x – 3 = 11
   2x = 11 + 3 = 14
   x=7

   When twp times 7 is decreased by 3, 14 is decreased
   by 3, which IS 11.
           Chapter 6, Section 3

3. One number exceeds another by 26: x = y + 26
   The sum of the numbers is 64: x + y = 64
   x + y = 64
   (y + 26) + y = 64
   2y + 26 = 64
   2y = 38
   y = 19
   x = y + 26 = 19 + 26 = 45

   45 does exceed 19 by 26, and the sum of 19 and 45 is 64.
           Chapter 6, Section 3

4. One number exceeds another by 24: x = y + 24
   The sum of the numbers is 58: x + y = 58
   x + y = 58
   (y + 24) + y = 58
   2y + 24 = 58
   2y = 34
   y = 17
   x = y + 24 = 17 + 24 = 41

   41 does exceed 17 by 24, and the sum of 17 and 41 is 58.
            Chapter 6, Section 3



15. Plan A: 40 + 25m, where m is the number of months.
    Plan B: 15 + 30m, where m is the number of months.
    These will be equal when 40 + 25m = 15 + 30m.
    40 + 25m = 15 + 30m
    40 – 15 = 30m – 25m
    25 = 5m
    m = 5. Plans even out at 5 months (for 40 + 25×5 = $165)
            Note that 15 + 30×5 = 15 + 150 = $165 as well.
            Chapter 6, Section 3



16. Store A: 9r where r is the number of rentals
    Plan B: 4r + 50. (Since they charge a membership of $50)
    These will be equal when 9r = 4r + 50.
    9r = 4r + 50
    9r – 4r = 50
    5r = 50
    r = 10. Cost evens out after 10 rentals (for 9×10 = $90)
            Note that 4×10 + 50 = 40 + 50 = $90 as well.
              Chapter 6, Section 3



17. With Coupon Book:
      15 + .75x, where x is the number of times used.
  Without Coupon Book:
       1.25x, where x is the number of times used.
  These will be equal when 15 + .75x = 1.25x.
  15 + .75x = 1.25x
  15 = 1.25x – .75x
  15 = 0.5x
  x = 30. Plans even out at 30 uses (for 15 + .75×30 = $15 + 22.50 = $37.50)
           Note that 1.25×30 = $37.50 as well.
               Chapter 6, Section 3



18. With Coupon Book:
        30 + 3.50x, where x is the number of times used.
    Without Coupon Book:
         5.00x, where x is the number of times used.
    These will be equal when 30 + 3.50x = 5.00x.
    30 + 3.50x = 5.00x
    30 = 5.00x – 3.50x
    30 = 1.50x
    x = 20. Plans even out at 20 uses (for 30 + 3.50×20 = $30 + $70 = $100)
             Note that 5.00×20 = $100 as well.
              Chapter 6, Section 3



21. Customer was charged $448. This consisted of parts and labor. Parts
    were $63. Labor was $35x where x is the number of hours.
    So $448 = $63 + $35x.
    448 = 63 + 35x
    448 – 63 = 385 = 63 + 35x – 63
    385 = 35x
    x = 385/35 = 11. Answer: 11 hours of labor.
              Chapter 6, Section 3



22. Customer was charged $1603. This consisted of parts and labor. Parts
    were $532. Labor was $63x where x is the number of hours.
    So $1603 = $532 + $63x.
    1603 = 532 + 63x
    1603 – 532 = 1071 = 532 + 63x – 532
    1071 = 63x
    x = 1071/63 = 17. Answer: 17 hours of labor.
               Chapter 6, Section 3



23. The job pays $33,150. Part of this is a holiday bonus and the rest is
    salary, which is paid 2 times a month. So in a year, there are 24
    paychecks. Each paycheck is 1/24th of the annual salary (minus the
    bonus)
    $33,150 = $750 + 24x
    33150 = 750 + 24x
    32400 = 24x
    1350 = x. Each paycheck, therefore, is for $1,350.
    (Multiply 1350 by 24 and add 750 and see if that equals 33150!)
               Chapter 6, Section 3




24. There are three vertical pieces and four horizontal pieces (even though
    two of these will have to be cut in half.) If x is the height, the length is
    3x. So 3x + 4(3x) = 60
    3x + 12x = 60.
    15x = 60
    x = 4, so the height is four feet. The length (three times the height) is
    twelve feet.)
              Chapter 6, Section 4


1.   24 12
        
      x   7
     12x = 247 = 168
     x = 168  12 = 14
2.   56 8
       
      x 7
     8x = 567 = 392
     x = 392  8 = 49
              Chapter 6, Section 4


3.    x 18
        
      6   4
     4x = 186 = 108
     x = 108  4 = 27
4.    x   3
        
     32 24
     24x = 323 = 96
     x = 96  24 = 4
              Chapter 6, Section 4


5. x   3 , rewrite as x  3   be careful with the minus sign
    3     4            3    4    here. A negative fraction means
   4x = -33 = -9                that only the numerator is
   x = -9  4 = -2¼              negative.
6. x     1            x 1
        , rewrite as 
    2    5            2  5
   5x = -12 = -2
   x = -2  5 = -2/5
          Chapter 6, Section 4



23.      $725        x
                 
        $65 , 000 $100 , 000

      65000x = 725  100000 = 72500000
      x = 72500000  65000
      x =$1115.38
            Chapter 6, Section 4



26.     50 27
           
         x 108
      27x = 50  108 = 5400
      x = 5400  27
      x = 200
There are approximately 200 bass in the lake.

				
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