College Mathematics for Business, Economics, Life Sciences & Social Sciences – 14th Edition (Instructor’s Solutions Manual) | Barnett, Byleen & Stocker | ISBN 9780134676081

INSTRUCTOR’S SOLUTIONS MANUAL

COLLEGE MATHEMATICS FOR BUSINESS, ECONOMICS, LIFE SCIENCES
& SOCIAL SCIENCES
14TH EDITION

CHAPTER 1: LINEAR EQUATIONS AND GRAPHS
EXERCISE 1-1

2. 7 x  6  5 x  24 4. 3( x  6)  5  2( x  1)
3x  18  5  2 x  2
2 x  18
5 x  15
x  9
x  3

2 x  1 5x
6.   4 (multiply both sides by 6) 8. 3  x  5
3 2
2(2 x  1)  15 x  24
4 x  2  15 x  24
11x  22
x  2

10. 6  x  1 12. x  9 or 9  x  

14. [1,5) 16. [4, )

18. 8  4 x  12 (divide the inequalities by – 4 and reverse the directions)
2  x  3 or  3  x  2
[3, 2)

m 2 x 5
20. 4 22. 
3 3 4 6
Multiply both sides of the equation by 3 to obtain: Multiply both sides by (−4) which will
m − 12 = 2 result in changing the direction of
m = 14 the inequality as well.
20 10
x and simplified we have x   .
6 3
24. 3 y  9  y  13  8 y 26. −3(4 − x) = 5 − (x + 1)
2 y  9  13  8 y −12 + 3x = 5 − x − 1
6y  4 −12 + 3x = 4 − x
4 2 12 − 12 + 3x = 12 + 4 − x
y 
6 3 3x = 16 − x
4x = 16
x=4

28. x − 2 ≥ 2(x − 5) y y 1
30. − =
x − 2 ≥ 2x − 10 4 3 2
x − 2 + 2 ≥ 2x − 10 + 2 Multiply both sides by 12:
x ≥ 2x − 8 3y − 4y = 6
 x  8 −y = 6
x≤8 y = −6

,1-2 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS



u 2 u 34. −4 ≤ 5x + 6 < 21
32. − < +2
2 3 3 −6 − 4 ≤ 5x < 21 − 6
u u 2 −10 ≤ 5x < 15
− <2+
2 3 3 −2 ≤ x < 3 or [−2, 3)
u 8
<
6 3
u < 16

2 2
36. −1 ≤ t + 5 ≤ 11 38. y=− x+8
3 3
2 2
−5 − 1 ≤ t ≤ 11 − 5 y–8=− x+8−8
3 3
2 2
−6 ≤ t ≤ 6 − x=y−8
3 3
−18 ≤ 2t ≤ 18 −2x = 3y − 24
−9 ≤ t ≤ 9 or [9,9]. 3 y  24 3
x= = − y + 12
2 2

40. y = mx + b 5
42. C= ( F  32)
y − b = mx + b − b 9
mx = y − b 9
C = F − 32
yb 5
m=
x 9
32 + C = F
5
9
F = C + 32
5

44. 10  8  3u  6
18  3u  14
18  3u  14
14
6 u
3
14
 u  6 or [, 6]
3

b
46. If a and b are negative and > 1, then multiplying both sides by the negative number a we obtain b < a
a
and hence a − b > 0.

48. Let x = number of quarters in the meter. Then
100 − x = number of dimes in the meter.

Now, 0.25x + 0.10(100 − x) = 14.50 or
0.25x + 10 − 0.10x = 14.50
0.15x = 4.50
4.50
x= = 30
0.15
Thus, there will be 30 quarters and 70 dimes.

, EXERCISE 1-1 1-3


50. Let x be the amount invested in “Fund A” and (500,000 – x) the amount invested in “Fund B”. Then 0.052x
+ 0.077(500,000 – x) = 30,000.
Solving for x:
(0.077)(500,000) – 30,000 = (0.077 – 0.052)x
8,500 = 0.025x
8,500
x= = $340,000
0.025
So, $340,000 should be invested in Fund A and $160,000 in Fund B.

52. Let x be the price of the house in 1960. Then
x 29.6
 (refer to Table 2, Example 10)
200,000 229.6
29.6
x  200,000 ≈ $25,784
229.6

To the nearest dollar, the house would be valued $25,784 in 1960.

54. (A) It is 60 – 0.15(60) = $51

(B) Let x be the retail price. Then
136 = x – 0.15x = 0.85x
136
So, x = = $160.
0.85

56. Let x be the number of times you must clean the living room carpet to make buying cheaper than renting.
Then
(20 + 2(16))x = 300 + 3(9)x
Solving for x
52x = 300 + 27x
25x = 300
300
x= = 12
25

58. Let x be the amount of the second employee’s sales during the month. Then
(A) 3,000 + 0.05x = 4,000
4, 000  3, 000
or x = = $20,000
0.05

(B) In view of Problem 57 we have:
2,000 + 0.08(x – 7,000) = 3,000 + 0.05x
Solving for x:
2,000 – (0.08)7,000 – 3,000 = 0.05x – 0.08x
−1,560 = −0.03x
1,560
x= = $52,000
0.03

(C) An employee who chooses (A) will earn more than he or she would with the other option until
$52,000 in sales is achieved, after which the other option would earn more.

, 1-4 CHAPTER 1: LINEAR EQUATIONS AND GRAPHS



60. Let x = number of books produced. Then

Costs: C = 2.10x + 92,000
Revenue: R = 15x

To find the break-even point, set R = C:

15x = 2.10x + 92,000
12.9x = 92,000
92, 000
x= ≈ 7,132
12.9

Thus, 7,132 books will have to be sold for the publisher to break even.

62. Let x = number of books produced.
Costs: C(x) = 92,000 + 2.70x
Revenue: R(x) = 15x

(A) The obvious strategy is to raise the price of the book.

(B) To find the break-even point, set R(x) = C(x):
15x = 92,000 + 2.70x
12.30x = 92,000
x = 7,480
The company must sell more than 7,480 books to make a profit.

(C) From Problem 60, the production level at the break-even point is:
7,132 books. At this production level, the costs are
C(7,132) = 92,000 + 2.70(7,132) = $111,256.40

If p is the new price of the book, then we need
7,132p = 111,256.40
and p ≈ $15.60

The company should sell the book for at least $15.60.

64. −49 ≤ F ≤ 14 MA
66. Note that IQ =  100
9 CA
−49 ≤ C + 32 ≤ 14
5 (see problem 65). Thus
9
−32 – 49 ≤ C ≤ 14 – 32 80 < IQ < 140
5
9 MA
−81 ≤ C ≤ −18 80 <  100 < 140
5 12
(80)(12) (140)(12)
(−81) · 5 ≤ 9C ≤ (−18) · 5 or < MA <
(81)  5 (18)  5 100 100
≤C≤ or 9.6 < MA < 16.8
9 9
−45 ≤ C ≤ −10