# Thinking Mathematically by Robert Blitzer - PowerPoint 4

Document Sample

```					  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

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

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

38. 7x2 - 10x2
(7 - 10)x2 distributive law
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.

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 42 posted: 8/13/2011 language: English pages: 37