Chapter 1 Linear Equations · 1 Chapter 1 Linear Equations 1.1 Lines 1. True 2. 2610 24 2 x x x 3. True 4. The equation of a vertical line with x-intercept at (–3, 0) is x = –3

1

Chapter 1 Linear Equations

1.1 Lines 1. True

2. 2 6 102 4

2

xxx

+ ===

3. True

4. The equation of a vertical line with x-intercept at (–3, 0) is x = –3.

5. If the slope of a line is undefined, the line is vertical.

6. The line y = –4x + 6 has slope m = –4 and y-intercept (0, 6).

7. The point-slope form of the equation of a line with slope m containing the point ( )1 1,x y is

( )1 1 .y y m x x− = −

8. If the graph of a line slants downward from left to right, its slope m is negative.

9. A = (4, 2); B = (6, 2); C = (5, 3); D = (–2, 1) E = (–2, –3); F = (3, –2); G = (6, –2) H = (5, 0)

10. a. 2nd quadrant

b. x-axis

c. 3rd quadrant

d. 1st quadrant

e. y-axis

f. 4th quadrant

11. The set of points of the form, (2, y), where y is a real number, is a vertical line passing through 2 on the x-axis. The equation of the line is x = 2.

12. The set of points, (x, 3), where x is a real number, is a horizontal line passing through 3 on the y-axis. The equation of the line is y = 3.

13. y = 2x + 4

0 2 2 4 4

0

x

y

− − −−

2

4 8 0 12 4

14. y = –3x + 6

0 2 2 4 4

0

x

y

− −−

2

6 0 12 6 18

Solution Manual for Finite Mathematics 11th Edition by Sullivan

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2 Chapter 1 Linear Equations

15. 2x – y = 6

0 2 2 4 4

0

x

y

− −− − − −

3

6 2 10 2 14

16. x + 2y = 8

0 2 2 4 4

0

x

y

− −8

4 3 5 2 6

17. a. The vertical line containing the point (2, –3) is x = 2.

b. The horizontal line containing the point (2, –3) is y = –3.

c. ( )3 5 23 5 10

13 5

y xy x

x y

+ = −+ = −

= −

The line with a slope of 5 containing the point (2, –3) is 5x – y = 13.

18. a. The vertical line containing the point (5, 4) is x = 5.

b. The horizontal line containing the point (5, 4) is y = 4.

c. ( )4 5 54 5 25

21 5

y xy x

x y

− = −− = −

= −

The line with a slope of 5 containing the point (5, 4) is 5x – y = 21.

19. a. The vertical line containing the point (–4, 1) is x = –4.

b. The horizontal line containing the point (–4, 1) is y = 1.

c. ( )1 5 41 5 20

21 5

y xy x

x y

− = +− = +− = −

The line with a slope of 5 containing the point (–4, 1) is 5x – y = –21.

20. a. The vertical line containing the point (–6, –3) is x = –6.

b. The horizontal line containing the point (–6, –3) is y = –3.

c. ( )3 5 63 5 30

27 5

y xy x

x y

+ = ++ = +− = −

The line with a slope of 5 containing the point (–6, –3) is 5x – y = –27.

21. a. The vertical line containing the point (0, 3) is x = 0.

b. The horizontal line containing the point (0, 3) is y = 3.

c. 3 53 5

y xx y

− =− = −

The line with a slope of 5 containing the point (0, 3) is 5x – y = –3.

22. a. The vertical line containing the point (–6, 0) is x = –6.

b. The horizontal line containing the point (–6, 0) is y = 0.

c. ( )0 5 65 30

30 5

y xy x

x y

− = += +

− = −

The line with a slope of 5 containing the point (–6, 0) is 5x – y = –30.

23. 2 1

2 1

1 0 1

2 0 2

y ym

x x

− −= = =− −

We interpret the slope to mean that for every 2 unit change in x, y changes 1 unit. That is, for every 2 units x increases, y increases by 1 unit.

24. 2 1

2 1

1 0 1

( 2) 0 2

y ym

x x

− −= = = −− − −

We interpret the slope to mean that for every 2 unit change in x, y changes –1 unit. That is, for every 2 units x increases, y decreases by 1 unit.






1.1 Lines 3

25. 2 1

2 1

3 11

1 1

y ym

x x

− −= = = −− − −

We interpret the slope to mean that for every 1 unit change in x, y changes by –1 unit. That is, for every 1 unit increase in x, y decreases by 1 unit.

26. ( )2 1

2 1

2 1 1

2 1 3

y ym

x x

− −= = =− − −

We interpret the slope to mean that for every 3 unit change in x, y changes by 1 unit. That is, for every 3 units x increases, y increases by 1 unit.

27. 2 1

2 1

3 0 33

2 1 1

y ym

x x

− −= = = =− −

A slope of 3 means that for every 1 unit change in x, y will change 3 units.

28. 2 1

2 1

4 2 21

3 1 2

y ym

x x

− −= = = =− −

A slope of 1 means that for every 1 unit change in x, y will change 1 unit.

29. 2 1

2 1

1 3 2 1

2 ( 2) 4 2

y ym

x x

− − −= = = = −− − −

A slope of 1

2− means that for every 2 unit

increase in x, y will decrease –1 unit.

30. 2 1

2 1

3 1 2

2 ( 1) 3

y ym

x x

− −= = =− − −

A slope of 2

3 means that for every 3 unit

increase in x, y will increase 2 units.

31. 2 1

2 1

( 1) ( 1) 00

2 ( 3) 5

y ym

x x

− − − −= = = =− − −

A slope of zero indicates that regardless of how x changes, y remains constant.

32. 2 1

2 1

2 2 00

( 5) 4 9

y ym

x x

− −= = = =− − − −

A slope of zero indicates that regardless of how x changes, y remains constant.

33. 2 1

2 1

( 2) 2 4

( 1) ( 1) 0

y ym

x x

− − − −= = =− − − −

The slope is not defined.







34. 2 1

2 1

2 0 2

2 2 0

y ym

x x

− −= = =− −

The slope is not defined.

35.

36.

37.

38.

39.

40.

41.

42.

43. Use the points (0, 0) and (2, 1) to compute the slope of the line:

2 1

2 1

1 0 1

2 0 2

y ym

x x

− −= = =− −

Since the y-intercept, (0, 0), is given, use the slope-intercept form of the equation of the line:

1 10

2 2y x y x= + ⇒ =

Then write the general form of the equation: x – 2y = 0.






1.1 Lines 5

44. Use the points (0, 0) and (–2, 1) to compute the slope of the line:

( )2 1

2 1

1 0 1

2 0 2

y ym

x x

− −= = = −− − −

Since the y-intercept (0, 0) is given, use the slope-intercept form of the equation of the line:

10

212

22 0

y mx b

y x

y x

y xx y

= +

= − +

= −

= −+ =

This is the general form of the equation.

45. Use the points (1, 1) and (–1, 3) to compute the slope of the line:

2 1

2 1

3 1 21

( 1) 1 2

y ym

x x

− −= = = = −− − − −

Now use the point (1, 1) and the slope m = –1 to write the point-slope form of the equation of the line:

1 1( )1 ( 1)( 1)1 1

2

y y m x xy xy x

x y

− = −− = − −− = − ++ =


46. Use the points (–1, 1) and (2, 2) to compute the slope of the line:

( )2 1

2 1

2 1 1

2 1 3

y ym

x x

− −= = =− − −

Next use the point (–1, 1) and the slope 1

3m =

to write the point-slope form of the equation of the line.

1 1( )1

1 ( ( 1))31

1 ( 1)3

3 3 13 4

y y m x x

y x

y x

y xx y

− = −

− = − −

− = +

− = +− = −


47. Since the slope and a point are given, use the point-slope form of the line:

( )( )( )

1 1

1 2 41 2 8

2 9

y y m x x

y xy x

x y

− = −− = − −− = +− = −



( )( )( )

1 1

4 3 34 3 9

3 13

y y m x x

y xy xx y

− = −− = − −− = +− = −



( )( ) ( )

( )

1 12

1 13

3 3 2 13 3 2 2

2 3 1

y y m x x

y x

y xy x

x y

− = −

− − = − −

+ = − −+ = − +

+ = −



( )( )

1 11

1 32

2 2 32 1

y y m x x

y x

y xx y

− = −

− = −

− = −− =


51. Since we are given two points, (1, 3) and (–1, 2), first find the slope.

( )3 2 1

1 1 2m

−= =− −

Then use the slope, one of the points, (1, 3), and the point-slope form of the line:

( )1 1( )

13 1

22 6 1

2 5

y y m x x

y x

y xx y

− = −

− = −

− = −− = −


52. Since we are given two points, (–3, 4) and (2, 5), first find the slope.

( )4 5 1

3 2 5m

−= =− −

Then use the slope, one of the points, (–3, 4), and the point-slope form of the line:

( )( )1 1( )

14 3

55 20 3

5 23

y y m x x

y x

y xx y

− = −

− = − −

− = +− = −








53. Since we are given the slope m = –2 and the y-intercept (0, 3), we use the slope-intercept form of the line:

2 32 3

y mx by x

x y

= += − +

+ =


54. Since we are given the slope m = –3 and the y-intercept (0, –2), we use the slope-intercept form of the line:

3 23 2

y mx by x

x y

= += − −

+ = −


55. We are given the slope m =3 and the x-intercept (–4, 0), so we use the point-slope form of the line:

( )( )1 1( )0 3 4

3 123 12

y y m x x

y xy x

x y

− = −− = − −

= +− = −


56. We are given the slope m = –4 and the x-intercept (2, 0). So we use the point-slope form of the line:

( )1 1 ( )0 4 2

4 84 8

y y m x xy x

y xx y

− = −− = − −

= − ++ =


57. We are given the slope 4

5m = and the point

(0, 0), which is the y-intercept. So, we use the slope-intercept form of the line:

40

55 4

4 5 0

y mx b

y x

y xx y

= +

= +

=− =


58. Since we are given the slope 7

3m = and the

point (0, 0), which is the y-intercept, we use the slope-intercept form of the line:

70

33 7

7 3 0

y mx b

y x

y xx y

= +

= +

=− =


59. We are given two points, the x-intercept (2, 0) and the y-intercept (0, –1), so we need to find the slope and then use the slope-intercept form of the line to get the equation.

( )0 1 1slope

2 0 2

11

22 22 2

y mx b

y x

y xx y

− −= =

−= +

= −

= −− =


60. We are given two points, the x-intercept (–4, 0) and the y-intercept (0, 4), so we need to find the slope and then use the slope-intercept form of the line to get the equation.

4 0slope 1

0 ( 4)

44

y mx by x

x y

−= =− −

= += +

− = −


61. Since the slope is undefined, the line is vertical. The equation of the vertical line containing the point (1, 4) is x = 1.

62. Since the slope is undefined, the line is vertical. The equation of the vertical line containing the point (2, 1) is x = 2.

63. Since the slope = 0, the line is horizontal. The equation of the horizontal line containing the point (1, 4) is y = 4.

64. Since the slope = 0, the line is horizontal. The equation of the horizontal line containing the point (2, 1) is y = 1.

65. y = 2x + 3 slope: m = 2; y-intercept: (0, 3)






1.1 Lines 7

66. y = –3x + 4 slope: m = –3; y-intercept: (0, 4)

67. To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y. 1

12

2 2

y x

y x

= −

= −

slope: m = 2; y-intercept: (0, –2)

68. To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y. 1

23

12

3

x y

y x

+ =

= − +

slope: 1

3m = − ; y-intercept: (0, 2)

69. To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y. 2 3 6

22

3

x y

y x

− =

= −

slope: 2

3m = ; y-intercept: (0, –2)


33

2

x y

y x

+ =

= − +

slope: 3

2m = − ; y-intercept: (0, 3)

71. To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y.

11

x yy x

+ == − +

slope: m = –1; y-intercept: (0, 1)








22

x yy x

− == −

slope: m = 1; y-intercept: (0, –2)

73. The slope is not defined; there is no y- intercept. So the graph is a vertical line.

74. slope: m = 0; y-intercept: (0, –1)

75. slope: m = 0; y-intercept: (0, 5)

76. The slope is not defined; there is no y-intercept. So the graph is a vertical line.


0y xy x

− ==

slope: m = 1; y-intercept: (0, 0)


0x yy x

+ == −

slope: m = –1; y-intercept: (0, 0)


2 332

y xy x

y x

− ==

=

slope: 3

2m = ; y-intercept: (0, 0)


2 332

x yy x

y x

+ == −

= −

slope: 3

2m = − ; y-intercept: (0, 0)

(continued on next page)






1.1 Lines 9

(continued)

81. To graph an equation on a graphing utility, first solve the equation for y. 1.2 0.8 2

0.8 1.2 21.5 2.5

x yy xy x

+ == − += − +

Window: Xmin = –10; Xmax = 10 Ymin = –10; Ymax = 10

The x-intercept is (1.67, 0).

The y-intercept is (0, 2.50).

82. To graph an equation on a graphing utility, first solve the equation for y.

1.3 2.7 82.7 1.3 8

13 8027 27

x yy x

y x

− + == +

= +


The x-intercept is (–6.15, 0).


83. To graph an equation on a graphing utility, first solve the equation for y. 21 15 53

15 21 5321 53 7 5315 15 5 15

x yy x

y x x

− == −

= − = −



The y-intercept is (0, –3.53).


3 5 825 823 3

x yy x

y x

− == −

= −





17 23 36 4 2

23 17 323 4 2 46 23

6 17 3 51 9

x y

y x

y x x

+ =

= − +

⎛ ⎞= − + = − +⎜ ⎟⎝ ⎠











14 8 73 9 28 14 7

8 9 2 36 163 14 7 21 2112 16

7 21

x y

y x

y x x

y x

− =

= −

⎛ ⎞= − = −⎜ ⎟⎝ ⎠

= −





3 6

3 6

23

x y

y x

y x

πππ

− == −

= −





15

15

15

x y

y x

xy

ππ

π π

+ == − +−= +




89. The graph passes through the points (0, 0) and (4, 8). We use the points to find the slope of the line:

2 1

2 1

8 0 82

4 0 4

y ym

x x

− −= = = =− −

The y-intercept (0, 0) is given, so we use the y-intercept and the slope m = 2, to obtain the slope-intercept form of the line.

2 0 2

y mx by x y x= += + ⇒ =

This is choice (b).


2 1

2 1

4 0 4 1

8 0 8 2

y ym

x x

− −= = = =− −

The y-intercept (0, 0) is given, so we use the

y-intercept and the slope m = 1

2, to obtain the

slope-intercept form of the line.

1 10

2 2

y mx b

y x y x

= +

= + ⇒ =

This is choice (c).


2 1

2 1

8 0 84

2 0 2

y ym

x x

− −= = = =− −


4 0 4y mx by x y x= += + ⇒ =

This is choice (d).


2 1

2 1

2 0 21

2 0 2

y ym

x x

− −= = = =− −


1 0y mx by x y x= += + ⇒ =

This is choice (a).






1.1 Lines 11

93. Using the intercepts (–2, 0) and (0, 2),

2 1

2 1

y ym

x x

−=− ( )

2 0 21

0 2 2

−= = =− −

Slope-intercept form: y = x + 2 General form: x – y = –2

94. Using the intercepts (1, 0) and (0, 1),

2 1

2 1

y ym

x x

−=−

1 0 11

0 1 1

−= = = −− −

Slope-intercept form: y = –x + 1 General form: x + y = 1

95. Using the intercepts (3, 0) and (0, 1),

2 1

2 1

y ym

x x

−=−

1 0 1 1

0 3 3 3

−= = = −− −

Slope-intercept form: 1

13

y x= − +

General form: 1

3 3 13

3 33 3

y x

y xx y

⎛ ⎞= − +⎜ ⎟⎝ ⎠= − +

+ =

96. Using the intercepts (–2, 0) and (0, –1),

2 1

2 1

y ym

x x

−=−

( )( )

1 0 1 10 2 2 2

− − −= = = −− −

Slope-intercept form: 1

12

y x= − −

General form: 1

2 2 12

2 22 2

y x

y xx y

⎛ ⎞= − −⎜ ⎟⎝ ⎠= − −

+ = −

97. a. The equation is C = 0.54x, where x is the number of miles the car is driven.

b. x = 15,000 0.54(15,000) 8100C = =

It costs $8100 to drive the car 15,000 miles.

c.

d. For every unit increase in x, the number of miles driven, C, the cost, increases by 0.54x. This represents the additional cost of driving a standard-sized car another mile.

98. a. Each week it costs $224 to rent the truck and an additional $0.52 per mile for each mile the truck is driven. So the total cost for a weekly rental is given by the equation C = 0.52x + 224, where x is the number of miles the truck is driven.

b. x = 500 0.52(500) 224 484C = + =

It costs $484 to rent the truck if you drive it 500 miles.

c.

d. The cost of renting a truck increases by $0.52 for every mile driven.

e. The y-intercept $224 represents the fixed cost of renting the truck.

99. a. The fixed cost of electricity for the month is $8.23. In addition, the electricity costs $0.10438 (10.438 cents) for every kilowatt-hour (KWH) used. If x represents the number of KWH of electricity used in a month, the total monthly charge is represented by the equation

0.10438 8.23, 0 400.C x x= + ≤ ≤

b.

c. The charge for using 100 KWH of electricity is found by substituting 100 for x in part (a):

( )0.10438 100 8.2310.438 8.23 18.668 $18.67

C = += + = ≈

d. The charge for using 300 KWH of electricity is found by substituting 300 for x in part (a):

( )0.10438 300 8.2331.314 8.23 39.544 $39.54

C = += + = ≈







e. The slope of the line, m = 0.10438, indicates that for every extra KWH used (up to 400 KWH), the electric bill increases by 10.438 cents.

100. a. The fixed monthly cost of electricity is $5.69. In addition, the electricity costs $0.08735 (8.735 cents) for every kilowatt-hour (KWH) used. If x represents the number of KWH of electricity used in a month, the total monthly charge is represented by the equation

0.08735 5.69, 0 1000.C x x= + ≤ ≤

b.

c. If 200 KWH of electricity are used the cost is

( )0.08735 200 5.69 $23.16.C = + =

d. If 500 KWH of electricity are used the cost is

( )0.08735 500 5.69 49.365 $49.37.C = + = ≈

e. The slope of the line, m = 0.08735, indicates that for every extra KWH used (up to 1000 KWH), the electric bill increases by 8.735 cents.

101. a. Since we are told the relationship is linear, we will use the two points to get the slope of the line:

2 1

2 1

100 0 100 5

212 32 180 9

C Cm

F F

° − ° −= = = =° − ° −

We use the point (0, 32) and the fact that the

slope is 5

9 to obtain the point-slope form of

the equation.

( )( )

( )

1 15

0 3295

329

C C m F F

C F

C F

° − ° = ° − °

° − = ° −

° = ° −

b. To find the Celsius measure of 68 ºF, substitute 68 for F in the equation and simplify:

( )568 32 20

9C° = − = °

102. a. K = C° + 273

b. Since we only have relations between Kelvin and Celsius, and Celsius and Fahrenheit, we first use the relationship between Celsius and

Fahrenheit, and then substitute ( )532

9F° −

for C in the equation from (a):

( )

( )

532

9273

532 273

95

255.229

C F

K C

K F

K F

° = ° −

= ° +

= ° − +

= ° +

103. a. If t = 0 represents December 21, then January 20 is represented by t = 30. Use the points (0, 102.7) and (30, 104.1) to find the slope of the equation.

104.1 102.7 1.40.0467

30 0 30m

−= = =−

We will use the slope and the point (0, 102.7) to write the point-slope form of the equation.

( )102.7 0.0467 00.0467 102.7

A tA t

− = −= +

b. A = 0.0467(10) + 102.7 = 103.167. On December 31, 2009 there were 103.167 billion gallons of water in the reservoir.

c. The slope indicates that the reservoir gains 0.0467 billion (or 46.7 million) gallons of water a day.

d. A = 0.0467(41) + 102.7 = 104.6147. The model predicts that there will be 104.6147 billion gallons of water in the reservoir on January 31, 2010.

e. The reservoir will be full when A = 265.5. 0.0467 102.7 265.5

0.0467 162.83486.1

ttt

+ ==≈

The reservoir will be full after day 3486, or about 9.55 years.

104. a. Since we are given two points and are told that sales are linearly related to advertising costs, we compute the slope of the line:

2 1

2 1

300,000 100,000100

6000 4000

N Nm

A A

− −= = =− −







1.1 Lines 13

(continued)

We use the point (4000, 100,000) and the slope 100 to obtain the point-slope form of the equation of the line:

( )( )

1 1

100,000 100 4000100 400,000 100,000100 300,000

N N m A A

N AN AN A

− = −− = −

= − += −

b. To determine how much advertising is needed to sell 200,000 boxes of cereal, we will let N be 200,000 in the equation from part (a) and solve for A. 200,000 100 300,000500,000 100

5000

AA

A

= −==

So, the company would need to spend $5000 on advertising to sell 200,000 boxes of cereal.

c. The firm will sell 100 extra boxes of cereal for every additional dollar spent on advertising.

105. a. When x = 0, that is in 2006,

( )5000 0 80,000 $80,000S = + =

b. When x = 3, that is in 2009,

( )5000 3 80,000 $95,000S = + =

c. If the trend continues, sales in 2012 should be equal to S when x = 6,

( )5000 6 80,000$110,000

SS= +=

d. If the trend continues, sales in 2015 should be equal to S when x = 9.

( )5000 9 80,000$125,000

SS= +=

106. We are given two points and are told that the number of diseased mice is linearly related to the number of days since exposure. In this problem we will let n represent the number of diseased mice in the cage and let t represent the number of days after the first exposure. We first compute the slope of the line:

2 1

2 1

14 8 63

6 4 2

n nm

t t

− −= = = =− −

We now use the point (4, 8) and the fact that the slope m = 3 to obtain the point-slope form of the equation of the line.

( )( )

1 1

8 3 4n n m t t

n t

− = −− = −

Writing the equation in slope-intercept form, will give the number of diseased mice after any given number of days.

8 3 12 3 4n t n t− = − ⇒ = − To determine how long it will take for all 40 of the mice in the cage to become infected, we will substitute 40 for n and solve the equation for t: 40 3 444 3443

tt

t

= −=

=

So, after 2

14 3

t = days all 40 of the mice will

be infected.

107. a. Writing 5% as a decimal, 5% = 0.05, we form the equation S = 0.05x + 400.

b. If x = $4000.00, then Dan’s earnings will be S = 0.05(4000) + 400 = 200 + 400 = $600.

c. Let S equal the median earnings, and solve for x. 857.31 0.05 400457.31 0.059146.2

xx

x

= +==

Dan would need to have sales that generate $9146.20 in profit to earn the median amount.

108. a. Since the relationship between time and depletion is linear, we use the points to find the slope of the line.

2 1

2 1

3.5 4.9 1.40.14

2009 1999 10

A Am

t t

− − −= = = = −− −

Then we choose one point, say (2009, 3.5), and write the point-slope form of the line.

( )( )

1 1

3.5 0.14 20090.14 281.26 3.50.14 284.76

A A m t t

A tA tA t

− = −− = − −

= − + += − +

b. To determine the year when the fields run dry, let A = 0 and solve for t.

0 0.14 284.760.14 284.76

284.762034

0.14

tt

t

= − +=

= =

The fields are predicted to run dry in 2034.

c. The slope represents the annual depletion rate. These fields decrease by 140 million barrels per year.

d. Divide the current reserves in the Jack Field by the annual depletion rate (that is, the slope).

15 billion 15107.14

140 million 0.14= ≈

The Jack Field will last about 107 years.







109. a. Since the rate of increase is constant, we use the points to find the slope of the line.

2 1

2 1

496 475 212.1

2009 1999 10

S Sm

t t

− −= = = =− −

Then we choose a point, say (1999, 475), and write the point-slope form of the line.

( )( )( )

1 1

475 2.1 19992.1 1999 4752.1 3722.9

S S m t t

S tS tS t

− = −− = −

= − += −

b. If the trend continues, in 2011,

( )2.1 2011 3722.9 500.2S = − =

The projected SAT mathematics score is about 500.

110. a. Since we assume the rate of increase is constant, we use the points to find the slope of a line.

2 1

2 1

1.81 1.450.36

2 1

I Im

t t

− −= = =− −

Now we choose a point, say (1, 1.45), and write the point-slope form of the line.

( )( )

1 1

1.45 0.36 10.36 0.36 1.450.36 1.09

I I m t t

I tI tI t

− = −− = −

= − += +

b. To find the projected net income for the fourth quarter, let t = 4. I = 0.36(4) + 1.09 = 2.53 AT&T is projected to have a fourth quarter 2006 income of $2.53 billion.

c. m = 0.36(1,000,000,000) = 360,000,000 = 360 million The slope indicates that the net income will increase by $360 million each quarter.

111. a. Since we assume the rate of increase is constant, we use the points to find the slope of a line.

2 1

2 1

29.4 24.4 50.5

2008 1998 10

P Pm

t t

− −= = = =− −

Now we choose a point, say (24.4, 1998), and write the point-slope form of the line.

( )( )( )

1 1

24.4 0.5 19980.5 1998 24.40.5 974.6

P P m t t

P tP tP t

− = −− = −

= − += −

b. To find the predicted percentage of people who will have a bachelor’s degree, let t = 2011. P = 0.5(2011) – 974.6 = 30.9 By 2011 it is predicted that 30.9% of people over 25 years of age will have a bachelor’s degree or higher.

c. The slope is the annual average increase in the percentage of people over 25 years of age who have a bachelor’s degree or higher. For every unit change in the year t, P increases by 0.5.

112. a. Since we assume the relationship between time and number of degrees conferred is linear, we use the points to find the slope of the line.

2 1

2 1

1,524,092 1, 237,875

2007 2000286, 217

40,888.17

N Nm

t t

− −= =− −

= ≈

Now choose a point, say (2000, 1237875), and write the point-slope form of the line.

( )( )

1 1

1, 237,875 40,888.1 20001, 237,875 40,888.1 81,776, 200

40,888.1 80,538,325

N N m t t

N tN t

N t

− = −− = −− = −

= −

b. N = 40,888.1(2011) – 80,538,325 = 1,687,644.1 The model predicts that 1,687,644 bachelor’s degrees will be awarded in 2011.

c. The slope is the average annual increase in the number of bachelor’s degrees awarded.

113. a. Since the cost of the houses is linear, we first use the points to find the slope of the line.

2 1

2 1

88, 280 94,8236543

2009 2008

C Cm

t t

− −= = = −− −

Next, choose a point, say (2008, 94823), and write the point slope form of the line.

( )( )( )

1 1

94,823 6543 20086543 2008 94,8236543 13, 233,167

C C m t t

C tC tC t

− = −− = − −

= − − += − +

b. The projected average cost of a house in 2011 is found by letting t = 2011 in the equation.

( )6543 2011 13, 233,167 $75,194C = − + =






1.1 Lines 15

114. Two points are given, (h1, w1) = (67, 139) and (h2, w2) = (70, 151), and we are told they are linearly related. So we will first compute the slope of the line:

2 1

2 1

151 139 124

70 67 3

w wm

h h

− −= = = =− −

We use the point (151, 70) and the fact that the slope 4m = to get the point-slope form of the equation of the line.

( )151 4 70151 4 280

4 129

w hw h

w h

− = −− = −

= −

115. a. First we find the slope of the line.

2 1

2 1

264.908 214.09150.817

2008 2007

S Sm

t t

− −= = =− −

Then we use the point (2007, 214.091) to write the point-slope form of the equation of the line.

( )( )( )

1 1

214.091 50.817 200750.817 2007 214.09150.817 101,775.628

S S m t t

S tS tS t

− = −− = −

= − += −

b. ( )50.817 2011 101,775.628 417.359S = − =

The model predicts that the total sales and other operating income for Chevron Corporation will be $417.359 billion in 2011.

116. a. We first find the slope.

2 1

2 1

61.1 61.130.03

1 0

R Rm

t t

− −= = = −− −

Then we use the point (0, 61.1) to write the point-slope form of the equation of the line.

( )( )

0 0

61.1 0.03 10.03 61.13

R R m t t

R tR t

− = −− = − −

= − +

b. ( )0.03 3 61.13 61.04R = − + =

The model predicts that the net revenue for Dell, Inc. will be $61.04 billion in 2011.

117. a. If the Smiths drive x miles in a year, and their car averages 17 miles per gallon of gasoline,

then they will use about 17

x gallons of

gasoline per year. In 2010, the Smith’s annual fuel cost is given by

2.7392.739 0.1611 .

17 17

xC x x

⎛ ⎞= = ≈⎜ ⎟⎝ ⎠

b. In 2009, the Smiths annual fuel cost is given

by 1.847

1.847 0.1086 .17 17

xC x x

⎛ ⎞= = ≈⎜ ⎟⎝ ⎠

c. Assuming that the Smiths drove 15,000 miles, their fuel cost in 2010 was

15,0002.739 $2417.

17C

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

d. Assuming that the Smiths drove 15,000 miles, their fuel cost in 2009 was

15,0001.847 $1630.

17C

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

e. The difference in the annual costs is about $2417 – $1630 = $787. The Smiths spend $787 more at the 2010 price than at the 2009 price.

118. a. If the Jones’s drive x miles in a year, and their car averages 39 miles per gallon of gasoline,

then they will use about 39

xgallons of

gasoline per year. In 2010, the Jones’s annual fuel cost is given by

2.739 0.0702 .39

xC x

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

b. In 2009, the Jones’s annual fuel cost is given

by 1.847 0.0474 .39

xC x

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

c. Assuming that the Jones’s drove 15,000 miles, their fuel cost in 2010 was

15,0002.739 $1053.

39C

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

d. Assuming that the Jones’s drove 15,000 miles, their fuel cost in 2009 was

15,0001.847 $710.

39C

⎛ ⎞= ≈⎜ ⎟⎝ ⎠

e. The difference in the annual costs is about $1053 – $710 = $343.

119. a. First we compute the slope of the line.

2 1

2 1

1147.5 954.7

3 0192.8

64.33

N Nm

t t

− −= =− −

= ≈

Then we use the slope and the point (0, 954.7) to write the point-slope form of the line.

( )( )

1 1

954.7 64.3 064.3 954.7

N N m t t

N tN t

− = −− = −

= +







b. The slope indicates that credit and debit cards in force are increasing at an average rate of 64.3 million cards per year.

c. In 2011, t = 5, and N = 64.3(5) + 954.7 = 1276.2 There will be an estimated 1.276 billion credit cards in force at the end of the first quarter of 2011.

d. To estimate the year that the number of credit cards will first exceed 1.5 billion, let N = 1500 (since the equation is written in millions), and solve for t. 1500 64.3 954.7545.3 64.3

545.38.5

64.3

tt

t

= +=

= ≈

Since t = 0 represented the year 2006, t = 8.5 is in the year 2014. The number of credit and debit cards will surpass 1.5 billion in 2014.

120. From the graph we can see that the line has a positive slope and a y-intercept of the form (0, b) where b is a positive number. Put each of the equations into slope-intercept form and choose those with both positive slope and positive y-intercept.

2 3(b) 2 (c) 3

3 4(e) 1 (g) 2 3

y x y x

y x y x

= + = +

= + = +

121. From the graph we can see that the line has a negative slope and a y-intercept of the form (0, b) where b is a positive number. Put each of the equations into slope-intercept form and choose those with negative slope and positive y-intercept.

2 3(a) 2 (c) 3

3 41

(f) 2 1 (g) 102

y x y x

y x y x

= − + = − +

= − + = − +

122. Answers will vary.

123. A vertical line cannot be written in slope-intercept form since its slope is not defined.

124. Not every line has two distinct intercepts. A line passing through the origin has the point (0, 0) as both its x- and y-intercept. Usually vertical lines have only an x-intercept and horizontal lines have only a y-intercept.

125. Two lines that have equal slopes and equal y-intercepts have equivalent equations and identical graphs.

126. If two lines have the same x-intercept and the same y-intercept, and the x-intercept is not (0, 0), then the two lines have equal slopes. Lines that have equal slopes and equal y-intercepts have equivalent equations and identical graphs.

127. If two lines have the same slope, but different x-intercepts, they cannot have the same y-intercept. If Line 1 has x-intercept (a, 0) and Line 2 has x-intercept (c, 0), but both have the same slope m, write the equation of each line using the point-slope form then change them to slope-intercept form and compare the y-intercepts: Line 1: ( )0y m x a

y mx ma− = −= −

y-intercept is (0, –ma) Line 2: ( )0y m x c

y mx mc− = −= −

y-intercept is (0, –mc)

128. Two lines can have the same x-intercept but different slopes only if their y-intercept is the point (0, 0).

129. The line y = 0 has infinitely many x-intercepts, so yes, a line can have two distinct x-intercepts or a line can have infinitely many x-intercepts.

130. Yes, a line can have no x-intercepts. For example, the line y = 2 has no x-intercepts. No, a line cannot have neither an x-intercept nor a y-intercept. It must have one or the other or both.

131. monter (t.v.) to go up; to ascend, to mount; to climb; to embark; to rise, to slope up, to be uphill; to grow up; to shoot; to increase. (source: Cassell’s French Dictionary). We use m to represent slope. The French verb monter means to rise, to climb or to slope up.

1.2 Pairs of Lines 1. parallel

2. intersect






1.2 Pairs of Lines 17

3. To determine whether the pair of lines is parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their y-intercepts.

: 10 10L x y

y x+ =

= − +

slope: m= –1; y-intercept: (0, 10) : 3 3 6

3 3 62

M x yy xy x

+ == − += − +

slope: m = –1; y-intercept: (0, 2) The slopes of the two lines are the same, but the y-intercepts are different, so the lines are parallel.


: 5 5L x y y x− = ⇒ = −

slope: m = 1; y-intercept: (0, –5) : 2 2 8

2 2 84

M x yy xy x

− + == += +

slope: m = 1; y-intercept: (0, 4) The slopes of the two lines are the same, but the y-intercepts are different, so the lines are parallel.


: 2 4 2 4L x y y x+ = ⇒ = − +

slope: m = –2; y-intercept: (0, 4) : 2 2 8

2 2 8 4M x y

y x y x− =− = − + ⇒ = −

slope: m = 1; y-intercept: (0, –4) The slopes of the two lines are the different, so the lines intersect.


: 2 8 2 8L x y y x+ = ⇒ = − +

slope: m = –2; y-intercept: (0, 8) : 2 4

2 42 4

M x yy xy x

− = −− = − −

= +

slope: m = 2; y-intercept: (0, 4) The slopes of the two lines are the different, so the lines intersect.


: 2 2L x y y x− + = ⇒ = +

slope: m = 1; y-intercept: (0, 2) : 2 2 4

2 2 42

M x yy xy x

− = −− = − −

= +

slope: m = 1; y-intercept: (0, 2) Since both the slopes and the y-intercepts of the two lines are the same, the lines are coincident.


: 44

L x yy x

+ = −= − −

slope: m = –1; y-intercept: (0, –4) : 3 3 12

3 3 124

M x yy xy x

+ = −= − −= − −

slope: m = –1; y-intercept: (0, –4) Since both the slopes and the y-intercepts of the two lines are the same, the lines are coincident.


: 2 3 8 3 2 8

2 8

3 3

L x yy x

y x

− = −− = − −

= +

slope: m = 2

3; y-intercept:

80,

3⎛ ⎞⎜ ⎟⎝ ⎠

: 6 9 2

9 6 22 2

3 9

M x yy x

y x

− = −− = − −

= +

slope: m = 2

3; y-intercept:

20,

9⎛ ⎞⎜ ⎟⎝ ⎠

The slopes of the two lines are the same, but the y-intercepts are different, so the lines are parallel.








: 4 2 7 2 4 7

7 2

2

L x yy x

y x

− = −− = − −

= +

slope: m = 2; y-intercept: 7

0,2

⎛ ⎞⎜ ⎟⎝ ⎠

: 2 2 2 2M x y

y x− + = −

= −

slope: m = 2; y-intercept: (0, –2) The slopes of the two lines are the same, but the y-intercepts are different, so the lines are parallel.


: 3 4 14 3 1

3 14 4

L x yy x

y x

− =− = − +

= −

slope: m = 3

4; y-intercept:

10,

4⎛ ⎞−⎜ ⎟⎝ ⎠

: 2 4

2 41

22

M x yy x

y x

− = −− = − −

= +

slope: m = 1

2; y-intercept: (0, 2)

The slopes of the two lines are the different, so the lines intersect.


: 4 3 2 3 4 2

4 2

3 3

L x yy x

y x

+ == − +

= − +

slope: m = 4

3− ; y-intercept:

20,

3⎛ ⎞⎜ ⎟⎝ ⎠

: 2 1 2 1 2 1

M x yy x y x

− = −− = − − ⇒ = +

slope: m = 2; y-intercept: (0, 1) The slopes of the two lines are the different, so the lines intersect.

13. L: x = 3 slope: not defined; no y-intercept M: y = –2 slope: m = 0, y-intercept: (0, –2) Since the slopes of the two lines are different, the lines intersect.

14. L: x = 4 slope: not defined; no y-intercept M: x = –2 slope: not defined; no y-intercept These are two vertical lines. Since they have different x-intercepts, they are parallel.

15. To find the point of intersection of two lines, first put the lines in slope-intercept form.

: 55

L x yy x

+ == − +

: 3 73 7

M x yy x

− == −

Since the point of intersection, (x0, y0), must be on both L and M, we set the two equations equal to each other and solve for x0. Then we substitute the value of x0 into the equation of one of the lines to find y0.

0 0

0 0

5 3 712 4 3

x xx x

− + = −= ⇒ =

0 (3) 5 2y = − + =

The point of intersection is (3, 2).








: 2 72 7

L x yy x

+ == − +

: 44

M x yy x

− = −= +


0 0

0

0

2 7 43 3

1

x xx

x

− + = +==

0 1 4 5y = + =



: 22

L x yy x

− == −

: 2 72 7

M x yy x

+ == − +


0 0

0

0

2 2 73 9

3

x xxx

− = − +==

0 3 2 1y = − =



: 2 12 1

L x yy x

− = −= +

: 44

M x yy x

+ == − +


0 0

0

0

2 1 43 3

1

x xxx

+ = − +==

0 2(1) 1 3y = + =



: 4 2 42 2

L x yy x

+ == − +

: 4 2 42 2

M x yy x

− == −


0 0

0

0

2 2 2 24 4

1

x xx

x

− + = −==

0 2(1) 2 0y = − + =









: 4 2 82 4

L x yy x

− == −

: 6 3 02

M x yy x

+ == −


0 0

0

0

2 4 24 4

1

x xxx

− = −==

0 2(1) 2y = − = −

The point of intersection is (1, −2).


: 3 4 23 1

4 2

L x y

y x

− =

= − : 2 4

12

2

M x y

y x

+ =

= − +


0 0

0 0

0

0

3 1 12

4 2 23 2 2 8

5 102

x x

x xxx

− = − +

− = − +==

01

(2) 2 12

y = − + =



: 4 3 24 2

3 3

L x y

y x

+ =

= − +

: 2 12 1

M x yy x

− == −


0 0

0

0 0

4 22 1

3 34 2 6 3

110 5

2

x x

x x

x x

− + = −

− = − +

= ⇒ =

01

2 1 02

y⎛ ⎞= − =⎜ ⎟⎝ ⎠

The point of intersection is 1

, 02

⎛ ⎞⎜ ⎟⎝ ⎠

.


: 3 2 53 5

2 2

L x y

y x

− = −

= + : 3 2

3 2M x y

y x+ = −

= − −


0 0

0 0

0 0

3 53 2

2 23 5 6 4

9 9 1

x x

x xx x

+ = − −

+ = − −= − ⇒ = −

( )0 3 1 2 1y = − − − =

The point of intersection is (–1, 1).








: 4 64 6

L x yy x

+ == − +

: 4 2 02

M x yy x

− ==


0 0

0

0

4 6 26 6

1

x xx

x

− + ===

( )0 2 1 2y = =


25. L is the vertical line on which the x-value is always 4. M is the horizontal line on which y-value is always –2. The point of intersection is (4, –2).

26. L is the vertical line on which the x-value is always 0. It is the equation of the y-axis.M is the horizontal line on which y-value is always 0. It is the equation of the x-axis. The point of intersection is (0, 0), which is the origin.

27. L is parallel to y = 2x, so the slope of L is m = 2. We are given the point (3, 3) on line L. Use the point-slope form of the line.

( )( )

1 1

3 2 33 2 6

y y m x x

y xy x

− = −− = −− = −

slope-intercept form: 2 3y x= −

general form: 2 3x y− =

28. L is parallel to y = –x, so the slope of L is m = –1. We are given the point (1, 2) on line L. Use the point-slope form of the line.

( )( )

1 1

2 1 12 1

y y m x x

y xy x

− = −− = − −− = − +

slope-intercept form: 3y x= − +

general form: 3x y+ =

29. We want a line parallel to y = 4x. So our line will have slope m = 4. It must also contain the point (−1, 2). Use the point slope form of the equation of a line:

( )( )

1 1

2 4 12 4 4

y y m x x

y xy x

− = −− = +− = +

slope-intercept form: 4 6y x= +

general form: 4 6x y− = −

30. We want a line parallel to y = –3x. So our line will have slope m = –3. It must also contain the point (–1, 2). Use the point slope form of the equation of a line:

( )( )

1 1

2 3 12 3 3

y y m x x

y xy x

− = −− = − +− = − −

slope-intercept form: 3 1y x= − −

general form: 3 1x y+ = −

31. We want a line parallel to 2 2.x y− = − Find the

slope of the line and use the given point, (0, 0) to obtain the equation. Since the y-intercept (0, 0) is given, use the slope-intercept form of the equation of a line. Original line: 2 2

2 22

x yy xm

− = −= +=

Parallel line:

2 02

y mx by xy x

= += +=








32. We want a line parallel to 2 5.x y− = − Find the

slope of the line and use the given point, (0, 0) to obtain the equation. Since the y-intercept (0, 0) is given, use the slope-intercept form of the equation of a line.

Original line: 2 5

2 51 52 2

x yy x

y x

− = −= +

= +

Parallel line: 1

2

12

m

y mx b

y x

=

= +

=


33. We want a line parallel to the line x = 3. This is a vertical line so the slope is not defined. A parallel line will also be vertical, and it must contain the point (4, 2). The parallel line will have the equation x = 4.

34. We want a line parallel to y = 3. This is a horizontal line; so it has a slope m = 0. The parallel line will also have a slope m = 0 but will contain the point (4, 2). Use the point-slope form of the equation of a line, or recognize the fact that horizontal lines have an equation of the form y = b. The equation of the line is y = 2.

35. To find the equation of the line, we must first find the slope of the line containing the points (–2, 9) and (3, –10):

( )( )

2 1

2 1

9 10 19 192 3 5 5

y ym

x x

− −−= = = = −− − − −

The slope of a line parallel to the line containing

these points is also 19

5− . Use the slope and the

point (–2, –5) to write the point-slope form of the parallel line.

( )( )

1 119

5 25

y y m x x

y x

− = −

+ = − +

Solve for y to get the slope-intercept form: 19 38 19 63

55 5 5 5

y x y x= − − − ⇒ = − −

Multiply both sides by 5 and rearrange terms to obtain the general form of the equation: 19 5 63x y+ = −

36. To find the equation of the line parallel to the line containing the points (–4, 5) and (2, –1), first find the slope of the line containing the two

points: ( )

( )2 1

2 1

5 1 61

4 2 6

y ym

x x

− −−= = = = −− − − −

The slope of a line parallel to the line containing these points is also –1. Use the slope and the point (–2, –5) to write the point-slope form of the parallel line.

( )( )

1 1

5 1 25 2

y y m x x

y xy x

− = −+ = − ++ = − −

Solve for y to get the slope-intercept form: 7y x= − −

Rearrange terms to obtain the general form of the equation: x + y = –7.

37. We will let x = the number of caramels the box of candy, and y = the number of creams in the box of candy. Since there are a total of 50 pieces of candy in a box, we have x + y = 50, or y = 50 – x. Each caramel costs $0.10 to make, and each cream costs $0.20 to make. So, the cost of making a box of candy is given by the equation:

0.1 0.20.1 0.2(50 )

C x yx x

= += + −

The box of candy sells for $8.00. So to break even, we need

8 0.1 0.2(50 )8 0.1 10 0.28 10 0.1

0.1 22050 50 20 30

R Cx xx x

xxxy x

== + −= + −= −=== − = − =

To break even, put 20 caramels and 30 creams into each box. If the candy shop owner increases the number of caramels to more than 20 (and decreases the number of creams) the owner will obtain a profit since the caramels cost less to produce than the creams.

38. Let x denote the number of pounds of cashews in the mixture, and let y denote the number of pounds of pecans in the mixture. Use a table to organize the data.








(continued)

Number of Pounds

Cost per Pound

Total Value

Cashews x $6.50 6.5x

Pecans y = 60 – x $7.50 7.5(60 – x)

Peanuts 40 $2.00 2(40)

Mixture x + y + 40 = 100 $4.89 4.89(100)

The last column gives the information needed to solve the problem since the sum of the values of the ingredients must equal the total value of the mixture. 6.5 7.5(60 ) 80 4896.5 450 7.5 80 489

530 1.0 48941

60 41 19

x xx x

xx

y

+ − + =+ − + =

− === − =

We need 41 pounds of cashews and 19 pounds of pecans to be mixed with the 40 pounds of peanuts to make 100 pounds of mixture worth $4.89 per pound.

39. Investment problems are simply mixture problems involving money. We will use a table to organize the information. Let x denote the amount Mr. Nicholson invests in AA bonds, and y denote the amount he invests in S & L Certificates.

Investment Amount Invested

Interest Rate

Interest Earned

AA Bonds x 0.10 0.10x

S&L Certificates

y = 150,000 – x 0.05 0.05(150,000 – x)

Total x + y = 150,000 10,000

The last column gives the information we need to set up the equation solve since the sum of the interest earned on the two investments must equal the total interest earned. 0.1 0.05(150,000 ) 10,000

0.1 7500 0.05 10,0000.05 7500 10,000

0.05 250050,000150,000100,000

x xx x

xxxy x

+ − =+ − =

+ ==== −=

Mr. Nicholson should invest $50,000 in AA Bonds and $100,000 in Savings and Loan Certificates in order to earn $10,000 per year.

40. The only difference between this problem and Problem 41 is that the total interest earned must equal $12,000. The equation which will give Mr. Nicholson’s distribution of funds is: 0.1 0.05(150,000 ) 12,000

0.1 7500 0.05 12,0000.05 7500 12,000

0.05 4,50090,000150,000 90,00060,000

x xx x

xxxy

+ − =+ − =

+ ==== −=

To earn the extra $2000 in interest, Mr. Nicholson should increase his investment in the higher yielding AA Bonds to $90,000 and reduce his investment in the Savings and Loan Certificates to $60,000.

41. Let x denote the amount of Kona coffee and y denote the amount of Columbian coffee in the mix. We will use the hint and assume that the total weight of the blend is 100 pounds.

Coffee Amount Mixed

Price per Pound

Total Value

Kona x 22.95 22.95x

Columbian y = 100 – x 6.75 6.75(100 – x)

Mixture x + y = 100 10.80 10.80(100)

The last column gives the information necessary to write the equation, since the sum of the values of each of the two individual coffees must equal the total value of the mixture.

( ) ( )22.95 6.75 100 10.80 10022.95 675 6.75 1080

16.2 40525100 25 75

x xx x

xxy

+ − =+ − =

=== − =

Mix 25 pounds of Kona coffee with 75 pounds of Columbian coffee to obtain a blend worth $10.80 per pound.

42. Let x denote the amount of corn meal and y denote the amount of soybean meal in the mix. We will use a table to organize the information.

Amount Mixed

Protein Content

Amount of Protein

Corn meal x 0.22 0.22x

Soybean y = 300 – x 0.44 0.44(300 – x)

Mixture x + y = 300 0.30 0.30(300)








(continued)

The last column gives the information necessary to write the equation, since the sum of the protein amounts of each of the two individual meals must equal the total value of the mixture.

( ) ( )0.22 0.44 300 0.30 3000.22 132 0.44 90

0.22 42191300 191 109

x xx x

xxy

+ − =+ − =

− = −≈= − =

The farmer should mix about 191 pounds of corn meal with about 109 pounds of soybean meal to obtain a blend of 300 pounds that is 30% protein.

43. Let x represent the amount of Acid A used and let y represent the amount of Acid B used in the solution. We will use a table to organize the information. We will use a table to organize the information.

Amount Mixed

Protein Content

Amount of Protein

Acid A x 0.15 0.15x

Acid B y = 100 – x 0.05 0.05(100 – x)

Mixture x + y = 100 0.08 0.08(100)

The last column gives the information necessary to write the equation, since the sum of the two individual solutions must equal the total solution. 0.15 0.05(100 ) 8

0.15 5 0.05 80.1 3

30100 30 70

x xx x

xxy

+ − =+ − =

=== − =

We should mix 30 cubic centimeters of 15% solution with 70 cubic centimeters of 5% solution to obtain a 100 cubic centimeters of 8% solution.

44. Investment problems are simply mixture problems involving money. We will use a table to organize the information. Let x be amount of loaned at 8%, and let y be the amount loaned at 12%.

Amount Invested

Interest Rate

Interest Earned

Loan A x 0.08 0.08x

Loan B y = 10,000 – x 0.12 0.12(10,000 – x)

Total x + y = 10,000 1000

The last column gives the information necessary to write the equation that will be used to solve the problem, since the sum of the interest paid on each of the two separate loans must equal the total interest paid. 0.08 0.12(10,000 ) 1000

0.08 1200 0.12 10001200 0.04 1000

200 0.045000

10,000 5000 5000

x xx x

xx

xy

+ − =+ − =

− ==== − =

The bank lent $5000 at 8% interest and $5000 at 12% interest for a total loan of $10,000.

45. a. The realized gain, y, is the difference between the 2006 value and the 2010 value of the gold. y = 1112.30x – 549.86x = 562.44x.

b. The point of intersection is the point (x0, yo) that satisfies both equation y = 562.44x and equation y = 10,000. 10,000 562.44

10,00017.8

562.44

x

x

=

= ≈

The individual would have had to bought and sold about 17.8 ounces of gold to realize a gain of $10,000.00.

46. a. Let x be the number of miles driven in the Honda Civic. The amount of gasoline used is

given by .42

xy = The annual cost of the

gasoline used is 2.775 .42Hx

C⎛ ⎞= ⎜ ⎟⎝ ⎠

b. Let x be the number of miles driven in the Ford Fusion. The amount of gasoline used is

given by .39

xy = The annual cost of the

gasoline used is 2.775 .39Fx

C⎛ ⎞= ⎜ ⎟⎝ ⎠

c. 15,000

2.775 991.0742

15,0002.775 1067.31

39

H

F

C

C

⎛ ⎞= ≈⎜ ⎟⎝ ⎠⎛ ⎞= ≈⎜ ⎟⎝ ⎠






1.3 Applications to Business and Economics 25

d. The distance between the two points of intersection is the difference in annual fuel cost between driving a Honda Civic Hybrid 15,000 miles and driving a Ford Fusion Hybrid 15,000 miles.

47. a. Assuming that the rate of growth is constant, we find the slope of the line passing through the points (0, 1615) and (365, 1877).

2 1

2 1

1877 1615 262

365 0 365

y ym

x x

− −= = =− −

Next use the slope and the point (0, 1615) to write the point-slope form of the line.

( )( )

1 1262

1615 0365262

1615365

y y m x x

y x

y x

− = −

− = −

= +

b. 262

2000 1615365262

385365536.4

x

x

x

= +

=

≈

There were 2000 HD radio stations 537 days after February 1, 2008 or on July 22, 2009.

48. The two lines in the graph are parallel. Parallel lines have equal slopes, so to determine which set of equations is parallel we must compare their slopes. Do this by writing each equation in slope-intercept form. Also notice that the slopes of the graphed equations are positive, and one line has a positive y-intercept and the other has a negative y-intercept. So choice (c) is the only possible answer.

2 21 1

x y y xx y y x− = − ⇒ = +− = ⇒ = −

Both equations have slope m = 1. The first has y-intercept (0, 2) and the second has y-intercept (0, –1), so these equations might have the graph illustrated.

1.3 Applications to Business and Economics

1. False. The break even point is the intersection of the revenue graph and the cost graph.

2. True

3. The break-even point is the point where the revenue and the cost are equal. Setting R = C, we find 30 10 60020 600

30

x xxx

= +==

That is, 30 units must be sold to break even. The break-even point is x = 30, R = 30(30) = 900 or (30, 900).

4. The break-even point is the point where the revenue and the cost are equal. Setting R = C, we find 8 5 2003 200

66.667

x xxx

= +==

That is, 66.667 units must be sold to break even. Since R = 8x, R = 8(66.667) = 533.336. Break-even point: (66.67, 533.34)

5. The break-even point is the point where the revenue and the cost are equal. Setting R = C, we find 0.30 0.20 500.10 50

500

x xxx

= +==

That is, 500 units must be sold to break even. Break-even point: (500, 150)







6. The break-even point is the point where the revenue and the cost are equal. Setting R = C, we find 2500 1800 3000

700 30004.29

x xxx

= +==

That is, 4.29 units must be sold to break even. Break-even point: (4.29, 10725)

7. The market price is the price at which the supply and the demand are equal.

1 32 2

1

S Dp p

pp

=+ = −

==

At a price of $1.00 supply and demand are equal, so $1.00 is the market price.


2 3 63 3

1

S Dp p

pp

=+ = −

==



20 500 1000 3050 500

10

S Dp p

pp

=+ = −

==



40 300 1000 3070 700

10

S Dp p

pp

=+ = −

==


11. The break-even point is the point where the revenue and the cost are equal. Cost is given by the variable cost of producing x pennants at $0.75 per pennant, plus the fixed operational overhead of $300 per day.

$0.75 $300C x= + Revenue is the product of price of each pennant ($1) and the number of pennants sold.

$1R x= Setting R = C, we find

1 0.75 3000.25 300

1200

x xxx

= +==

1200 pennants must be sold each day to break even.

12. The break-even point is the point where the revenue and the cost are equal. Cost is given by the variable cost of producing x units at $0.65 per unit, plus the fixed operational overhead of $350 per day.

$0.65 $350C x= + Revenue is the product of price of each pennant ($1) and the number of pennants sold.

$1R x= Setting R = C, we find

1 0.65 3500.35 350

1000

x xxx

= +==

1000 pennants must be sold each day to break even. Since these are fewer units than are needed to break-even with the original cost of production, the company will start to realize a profit sooner. It would be advantageous to change processes.






1.3 Applications to Business and Economics 27

13. a. The market price is the price at which the supply and the demand are equal.

0.7 0.4 0.5 1.61.2 1.2

1

S Dp p

pp

=+ = − +

==

The market price is $1.00 per pound.

b. To find the quantity supplied at market price, let p = 1 and solve for S:

0.7(1) 0.4 1.1S = + =

So 1.1 million pounds are demanded at $1.00.

c.

d. The point of intersection called the market equilibrium. It is the price where the quantity supplied equals the quantity demanded.

14. To find the supply equation use the market price and quantity (5, 14000) and the fact that at a price of $1.00 no units are produced. First, determine the slope of the line passing through the points.

2 1

2 1

14,000 0 14,0003500

5 1 4

S Sm

p p

− −= = = =− −

Then use the point (1, 0), the slope m = 3500, and the point-slope form of the equation:

( )( )

1 1

0 3500 13500 3500

S S m p p

S pS p

− = −− = −

= −

To find the demand equation we use the market price and quantity (5, 14000) and the fact that at a price of $19.00 no units are purchased. First, we find the slope of the line passing through the points.

2 1

2 1

0 14,000

19 514,000

100014

D Dm

p p

− −= =− −

−= = −

Then we use the point (19, 0), the slope m = –1000, and the point-slope form of an linear equation:

( )( )

1 1

0 1000 191000 19,000

D D m p p

D pD p

− = −− = − −

= − +

15. At the market price of $3.00, S = 2(3) + 5 = 11, so 11 units of the commodity are supplied. Since at the market price the supply and demand are equal, the point (3, 11) satisfies the demand equation. In addition, we are told that at a price of $1.00, 19 units are demanded. To find the demand equation, use the points (3, 11) and (1, 19) to find the slope.

2 1

2 1

19 11 84

1 3 2

D Dm

p p

− −= = = = −− − −

We then use the point (1, 19), the slope m = –4, and the point-slope form of the equation.

( )( )

1 1

19 4 14 23

D D m p p

D pD p

− = −− = − −

= − +

16. At the market price of $4.00, D = –3(4) + 20 = 8, so 8 units of the commodity are demanded at the market price. Since at the market price the supply and demand are equal, the point (4, 8) satisfies the supply equation. In addition, we are told that at $1.00, four units are supplied. To find the supply equation use the points (1, 4) and (4, 8) to determine the slope of the equation.

2 1

2 1

8 4 4

4 1 3

S Sm

p p

− −= = =− −

Use the point (1, 4), the slope 4

3m = , and the

point-slope form of the equation.

( )( )

1 14

4 134 4

43 3

4 83 3

S S m p p

S p

S p

S p

− = −

− = −

− = −

= − +

17. Let x denote the number of DVDs purchased. The cost of purchasing x DVDs from the club is

0.49(4) 17.95( 4) 2.311.96 17.95 71.8 2.3120.26 69.84

C x xx x

x

= + − += + − += −

The cost of purchasing x DVDs from the discount retailer is D = 14.95(1.07)x = 16.00x. We want to buy as many DVDs as possible from the club, while keeping the cost below the retailers. So we set C = D and solve for x. 20.26 69.84 16.00

4.26 69.8416.39

x xxx

− ===

At most sixteen DVDs can be ordered from the club to keep the price lower than that of the discount retailer.







18. a. Profit = Revenue – Cost

2004 2007

Radio 3.897 3.636

Television 7.287 7.205

b. Radio: Use the two points (2004, 3.897) and (2007, 3.636) to find the equation that describes radio profits.

2 1

2 1

3.636 3.897

2007 20040.261

0.0873

P Pm

t t

− −= =− −

−= = −

( )( )

1 1

3.897 0.087 20040.087 178.245

R

R

R

P P m t t

P tP t

− = −− = − −

= − +

Television: Use the two points (2004, 7.287) and (2007, 7.205) to find the equation that describes radio profits.

2 1

2 1

7.205 7.287

2007 20040.082

0.0273

P Pm

t t

− −= =− −

−= ≈ −

( )( )

1 1

7.287 0.027 20040.027 61.395

T

T

T

P P m t t

P tP t

− = −− = − −

= − +

c. Set the two equations equal and solve for t.

0.087 178.245 0.027 61.3950.6 116.85

1947.5

R TP Pt t

tt

=− + = − +

− = −=

The two media had the same profit in 1947.

1.4 Scatter Diagrams; Linear Curve Fitting

1. True

2. The mathematical measure of the goodness of fit of a line is called the correlation coefficient.

3. A relation exists, and it appears to be linear.

4. A relation exists, but it is not linear.



7. No relation exists.

8. A relation exists, but it is not linear.

9. a.

b. We select points (3, 4) and (9, 16). The slope of the line containing these points is:

2 1

2 1

16 4 122

9 3 6

y ym

x x

− −= = = =− −

The equation of the line is:

( )( )

1 1

4 2 34 2 6

2 2

y y m x x

y xy x

y x

− = −− = −− = −

= −

c. The line on the scatter diagram will vary depending on the choice of points in part (b).

d.

[–2, 10] by [–3, 20]

e.

y = 2.0357x – 2.357

f.

[–2, 10] by [–3, 20]






1.4 Scatter Diagrams; Linear Curve Fitting 29

10. a.


2 1

2 1

11 0 11

13 3 10

y ym

x x

− −= = =− −


( )( )

1 111

0 31011 3310 10

y y m x x

y x

y x

− = −

− = −

= −


d.

[–2, 14] by [–4, 15]

e.

y = 1.12857x – 3.86190

f.

[–2, 14] by [–4, 15]

11. a.

b. We select points (–2, –4) and (2, 5). The slope of the line containing these points is:

( )( )

2 1

2 1

5 4 92 2 4

y ym

x x

− −−= = =− − −


( )( )

1 19

5 249 9

54 29 14 2

y y m x x

y x

y x

y x

− = −

− = −

− = −

= +


d.

[–6, 6] by [–6, 7]

e.

y = 2.2x + 1.2







f.

[–6, 6] by [–6, 7]

12. a.

b. We select points (–2, 7) and (2, 0). The slope of the line containing these points is:

( )2 1

2 1

7 0 7 7

2 2 4 4

y ym

x x

− −= = = = −− − − −


( )( )

1 17

0 247 7

4 2

y y m x x

y x

y x

− = −

− = − −

= − +


d.

[–6, 6] by [–2, 10]

e.

y = −1.8x + 3.6

f.

[–6, 6] by [–2, 10]

13. a.


2 1

2 1

70 100 30 3

60 20 40 4

y ym

x x

− − −= = = = −− −


( )( )

1 13

100 2043

100 1543

1154

y y m x x

y x

y x

y x

− = −

− = − −

− = − +

= − +


d.

[0, 100] by [1, 120]

e.

y = –0.72x + 116.6







f.

[0, 100] by [1, 120]

14. a.


2 1

2 1

11 2 9

25 5 20

y ym

x x

− −= = =− −


( )( )

1 19

2 5209 9

220 49 120 4

y y m x x

y x

y x

y x

− = −

− = −

− = −

= −


d.

[–10, 35] by [–10, 35]

e.

y = 0.46x – 0.3

f.

[–10, 35] by [–10, 35]

15. a.

b. We select points (–20, 100) and (–10, 140). The slope of the line containing these points is:

( ) ( )2 1

2 1

140 100 404

10 20 10

y ym

x x

− −= = = =− − − −


( )( )( )

1 1

100 4 20100 4 80

4 180

y y m x x

y xy x

y x

− = −− = − −− = +

= +


d.

[–30, 10] by [0, 160]

e.

y = 3.86131x + 180.29197







f.

[–30, 10] by [0, 160]

16. a.

b. We select points (−30, 10) and (−14, 18). The slope of the line containing these points is:

( ) ( )2 1

2 1

18 10 8 1

14 30 16 2

y ym

x x

− −= = = =− − − −


( )( )( )

1 11

10 3021

10 1521

252

y y m x x

y x

y x

y x

− = −

− = − −

− = +

= +


d.

[–40, 10] by [–10, 35]

e.

y = 0.44207x + 23.45592

f.

[–40, 10] by [–10, 35]

17. a.

b. Answers will vary. We select points (20, 16) and (50, 39) with numbers in thousands. The slope of the line containing these points is:

2 1

2 1

39 16 23

50 20 30

C Cm

I I

− −= = =− −


( )( )

1 123

16 203023 46

1630 323 230 3

C C m I I

C I

C I

C I

− = −

− = −

− = −

= +

c. The slope of this line indicates that a family will spend $23 of every extra $30 of disposable income.

d. To find the consumption of a family whose disposable income is $42,000, substitute 42 for x in the equation from part (b).

( )23 242

30 3473

32.8666715

C = +

= ≈

The family will spend about $32,867

e.

y = 0.75489x + 0.62663







18. a. From the chart in Problem 17 we construct the savings chart and the scatter diagram.

Disposable Income I (thousands)

Savings (thousands)S = I – C

20 4

20 2

18 5

27 6

36 9

37 11

45 9

50 11

b. Answers will vary. We select points (20, 4) and (50, 11). The slope of the line containing these points is:

2 1

2 1

11 4 7

50 20 30

S Sm

I I

− −= = =− −


( )( )

1 17

4 20307 14

430 37 2

30 3

S S m I I

S I

S I

S I

− = −

− = −

− = −

= −

c. The slope indicates that a family of three will save $7.00 for every additional $30 of disposable income.

d. To find the predicted amount of savings of a family with $42,000 of disposable income, evaluate S with I = 42.

( )7 2 13742 9.1333

30 3 15S = − = =

A family of three with $42,000 of disposable income will save about $9,133.33.

e.

y = 0.24511x – 0.62663

19. a.

[–10, 110] by [50, 70]

b.

y = 0.07818x + 59.0909

c.

[–10, 110] by [50, 70]

d. The slope indicates the apparent change in temperature in a 65ºF room for every percent increase in relative humidity.

e. To determine the apparent temperature when the relative humidity is 75%, evaluate the equation of the line of best fit when x = 75.

( )0.07818 75 59.0909 64.95y = + =

When the relative humidity is 75%, the temperature of the room will appear to be 65ºF.

20. a.

[–10, 110] by [60, 90]

b.

y = 0.10818x + 68.68182







c.

[–10, 110] by [60, 90]

d. The slope indicates the apparent change in temperature in a 75ºF room for every percent increase in relative humidity.

e. To determine the apparent temperature when the relative humidity is 75%, evaluate the equation of the line of best fit when x = 75.

( )0.10818 75 68.68182 76.80y = + =

When the relative humidity is 75%, the temperature of the room will appear to be 77ºF.

21. a. Using the LinReg function, the line of best fit is y = 0.0651t + 10.6049.

The predicted energy use in 2015, t = 20, is

y = 0.0651(20) + 10.6049 ≈ 11.91 quadrillion BTU. The predicted energy use in 2030, t = 35, is y = 0.0651(35) + 10.6049 ≈ 12.88 quadrillion BTU. The predicted energy use in 2035, t = 40, is y = 0.0651(40) + 10.6049 ≈ 13.21 quadrillion BTU.

b. When we graph the scatterplot and the line of best fit, we see that the points are not strictly linear. Therefore, estimates based on the line of best fit are approximations. The line of best fit may differ depending on which points are chosen or a linear model may not have been used.

[0, 35] by [9, 13]

Chapter 1 Review Exercises

1.

2.

3.

4.

5. a. 2 1

2 1

4 2 2 1

( 3) 1 4 2

y ym

x x

− −= = = = −− − − −

A slope of 1


change in x, y will change (−1) unit. That is, for every 2 units x moves to the right, y will move down 1 unit.

b. Use the point (1, 2) and the slope to get the point-slope form of the equation of the line:

( )( )

1 11

2 12

y y m x x

y x

− = −

− = − −







Chapter 1 Review Exercises 35

(continued)

Simplifying and solving for y gives the slope-intercept form:

1 12

2 21 52 2

y x

y x

− = − +

= − +

Rearranging terms gives the general form of the equation: 2 5x y+ =

c.

6. a. 2 1

2 1

3 1 21

( 1) 1 2

y ym

x x

− −= = = = −− − − −

A slope of –1 means that for every one unit change in x, y changes (–1) unit. That is, for every 1 unit x moves to the right, y moves down 1 unit.


( )( )

1 1

1 1 1y y m x x

y x

− = −− = − −


1 12

y xy x

− = − += − +

Rearranging terms gives the general form of the equation: 2x y+ =

c.

7. a. ( )2 1

2 1

5 3 22

( 1) 2 1

y ym

x x

− −= = = =− − − −

A slope of 2 means that for every one unit change in x, y changes by 2 units. That is, for every 1 unit x moves to the right, y moves up 2 units.

b. Use the point (–1, 5) and the slope to get the point-slope form of the equation of the line:

( )( )

1 1

5 2 ( 1)y y m x x

y x

− = −− = − −


5 2 22 7

y xy x

− = += +

Rearranging terms gives the general form of the equation: 2 7x y− = −

c.

8. a. 2 1

2 1

3 0 3 3

( 2) 0 2 2

y ym

x x

− −= = = = −− − − −

A slope of 3


change in x, y will change (–3) units. That is, for every 2 units x moves to the right, y will move down 3 units.


3 30

2 2

y mx b

y x x

= +

= − + = −

Rearranging terms gives the general form of the equation: 3 2 0x y+ =

c.







9. Since we are given the slope m = –3 and a point, we get the point-slope equation of the line:

( )( )

1 1

( 1) 3 2y y m x x

y x

− = −− − = − −

Solving for y puts the equation into the slope-intercept form:

1 3 63 5

y xy x

+ = − += − +

Rearranging terms gives the general form of the equation: 3 5x y+ =

10. Since we are given the slope m = 4 and a point, we get the point-slope equation of the line:

( )( )

1 1

( 3) 4 ( 1)y y m x x

y x

− = −− − = − −

Solving for y puts the equation into the slope-intercept form:

3 4 44 1

y xy x

+ = += +

Rearranging terms gives the general form of the equation: 4 1x y− = −

11. Since we are given the slope m = 0 and a point on the line. We either use the point-slope formula or recognize that this is a horizontal line and the equation of a horizontal line y = b. The general form of the equation is y = 4.

12. Since we are told the slope is undefined, we know the line is vertical. We use the point (–3, 4) to write the equation of a vertical line x = –3. This is the general form of the equation of a vertical line.

13. We are told that the line is vertical, so the slope is not defined. We also know the line contains the point (8, 5). The general equation of a vertical line is x = a, so the general form of this equation is x = 8.

14. We are told that the line is horizontal, so the slope is 0. We also know the line contains the point (5, 8). The general equation of a horizontal line is y = b, so the general form of this equation is y = 8.







15. We are given the x-intercept and a point. First we find the slope of the line containing the two

points. 2 1

2 1

( 5) 0 5 5

4 2 2 2

y ym

x x

− − − −= = = = −− −

We then use the point (2, 0) and the slope to get the point-slope form of the equation of the line.

( )( )

1 15

0 22

y y m x x

y x

− = −

− = − −

To get the slope-intercept form, solve for y: 5

52

y x= − +

Rearrange the terms for the general form of the equation: 5 2 10x y+ = .

16. We are given two points, the y-intercept and another point. First use the two points to find the slope of the line containing them.

2 1

2 1

( 3) ( 2) 1 1

5 0 5 5

y ym

x x

− − − − −= = = = −− −

Since one of the points is the y-intercept, use it and the slope to get the slope-intercept form of

the equation: 1

25

y x= − −

Rearrange the terms for the general form of the equation: 5 10x y+ = − .

17. We are given the x-intercept and the y-intercept. Use the two points to find the slope of the line.

2 1

2 1

( 4) 0 4 4

0 ( 3) 3 3

y ym

x x

− − − −= = = = −− − −

Since one of the points is the y-intercept, use it and the slope to write the slope-intercept form of

the equation: 4

43

y x= − −

Rearrange the terms for the general form of the equation: 4 3 12x y+ = − .

18. We are given two points. Use them to find the slope of the line:

( )2 1

2 1

1 4 55

2 3 1

y ym

x x

− −−= = = = −− − −

Use the point (2, 1) and the slope to write the point-slope form of the equation:

( )( )

1 1

1 5 2y y m x x

y x

− = −− = − −

Simplifying and solving for y, gives the slope-intercept form of the equation.

1 5 105 11

y xy x

− = − += − +

Rearrange the terms for the general form of the equation: 5 11x y+ = .







19. Since the line we are seeking is parallel to 2 3 4x y+ = − , the slope of the two lines are the

same. Find the slope of the given line by putting it into slope-intercept form: 2 3 4

3 2 42 43 3

x yy x

y x

+ = −= − −

= − −

The slopes of the two lines are 2

3m = − . Use the

slope and the point (–5, 3) to write the point-slope form of the equation of the parallel line.

( )( )( )

( )

1 12

3 532

3 53

y y m x x

y x

y x

− = −

− = − − −

− = − +

To put the equation into slope-intercept form, solve for y.

2 103

3 32 1

3 3

y x

y x

− = − −

= − −

Rearrange the terms to obtain the general form of the equation: 2 3 1x y+ = −

20. The line we are seeking is parallel to 2x y+ = ,

so the slopes of both lines are the same. To find the slope put the equation into slope-intercept form.

22

x yy x

+ == − +

Use the slope m = –1 and the point (1, –3) to write the point-slope form of the equation of the parallel line.

( )( )

1 1

( 3) 1 1y y m x x

y x

− = −− − = − −

Solve for y to get the slope-intercept form: 2y x= − −

Rearrange the terms to get the general form of the equation: 2x y+ = −

21. To find the slope and y-intercept of the line, put the equation into the slope-intercept form. 9 2 18

2 9 189

92

x yy x

y x

+ == − +

= − +

The slope is 9

2− , and the y-intercept is (0, 9).


5 4 204

45

x yy x

y x

+ == − +

= − +

The slope is 4

5− , and the y-intercept is (0, 4).


2 4 99

22

x yy x

y x

+ == − +

= − +

The slope is –2, and the y-intercept is 9

0, 2

⎛ ⎞⎜ ⎟⎝ ⎠

. (continued on next page)







(continued)


2 3 83

42

x yy x

y x

+ == − +

= − +

The slope is 3

,2

− and the y-intercept is (0, 4).

25. To find the slope and y-intercept of the line, put

the equation into the slope-intercept form. 1 1 12 3 6

1 1 1

3 2 63 12 2

x y

y x

y x

+ =

= − +

= − +

The slope is 3

2− , and the y-intercept is

10,

2⎛ ⎞⎜ ⎟⎝ ⎠

.

26. To find the slope and y-intercept of the line, put the equation into the slope-intercept form. 1 1 54 3 12

1 1 5

3 4 123 54 4

x y

y x

y x

− =

− = − +

= −

The slope is 3

4, and the y-intercept is

50,

4⎛ ⎞−⎜ ⎟⎝ ⎠

.

27. Put each equation into slope-intercept form: 3 4 12 6 8 9

4 3 12 8 6 93 6 9

34 8 8

3 9

4 8

x y x yy x y x

y x y x

y x

− = − − = −− = − − − = − −

= + = +

= +

Since both lines have the same slope but different y-intercepts, the lines are parallel.

28. Put each equation into slope-intercept form: 2 3 5 4 6 10

3 2 5 6 4 102 5 4 10

3 3 6 62 5

3 3

x y x yy x y x

y x y x

y x

+ = − + = −= − − = − −

= − − = − −

= − −

Since both lines have the same slope and the same y-intercept, the lines are coincident.

29. Put each equation into slope-intercept form: 2

22

x yy xy x

− = −− = − −

= +

3 4 124 3 12

33

4

x yy x

y x

− = −− = − −

= +

Since the lines have different slopes, they intersect.

30. Put each equation into slope-intercept form: 2 3 5

3 2 52 53 3

x yy x

y x

+ == − +

= − +

1212

x yy x

+ == − +

Since the lines have different slopes, they intersect.







31. Put each equation into slope-intercept form: 4 6 12

6 4 122

23

x yy x

y x

+ = −= − −

= − −

2 3 63 2 6

22

3

x yy x

y x

+ = −= − −

= − −

Since both lines have the same slope and the same y-intercept, the lines are coincident.

32. Put each equation into slope-intercept form: 3 0

3x y

y x− + =

= 6 2 5

2 6 55

32

x yy x

y x

− = −− = − −

= +

Since both lines have the same slope but different y-intercepts, the lines are parallel.

33. To find the point of intersection of two lines, first put the lines in slope-intercept form:

: 44

L x yy x

− == −

: 2 71 72 2

M x y

y x

+ =

= − +


0 0

0 0

0

0

1 74

2 22 8 7

3 155

x x

x xxx

− = − +

− = − +==

0 0

0

0

45 41

y xyy

= −= −=


: 44

L x yy x

+ == − +

: 2 11 12 2

M x y

y x

− =

= −


0 0

0 0

0

0

1 14

2 22 8 1

3 93

x x

x xxx

− + = −

− + = −==

0 0

0

0

43 4

1

y xyy

= − += − +=



: 22

L x yy x

− = −= +

: 2 71 72 2

M x y

y x

+ =

= − +


0 0

0 0

0

0

1 72

2 22 4 7

3 31

x x

x xxx

+ = − +

+ = − +==

0 0

0

0

21 23

y xyy

= += +=



: 2 4 44 2 4

11

2

L x yy x

y x

+ == − +

= − +

: 2 4 84 2 8

12

2

M x yy x

y x

− =− = − +

= −


0 0

0 0

0

0

1 11 2

2 22 4

2 63

x x

x xxx

− + = −

− + = −− = −

=

0 0

0

0

12

21

3 22

1

2

y x

y

y

= −

= ⋅ −

= −

The point of intersection is 1

3,2

⎛ ⎞−⎜ ⎟⎝ ⎠.








: 2 4 84 2 8

12

2

L x yy x

y x

− = −− = − −

= +

: 3 6 06 3

12

M x yy x

y x

+ == −

= −


0 0

0 0

0

0

1 12

2 24

2 42

x x

x xxx

+ = −

+ = −= −= −

0 0

0

0

1

21

( 2)2

1

y x

y

y

= −

= − ⋅ −

=

The point of intersection is (–2, 1).


: 3 4 24 3 2

3 14 2

L x yy x

y x

+ == − +

= − +

: 2 12 1

1 12 2

M x yy x

y x

− =− = − +

= −


0 0

0 0

0

0

3 1 1 1

4 2 2 23 2 2 2

5 44

5

x x

x xx

x

− + = −

− + = −− = −

=

0 0

0

0

1 1

2 21 4 1

2 5 22 1 1

5 2 10

y x

y

y

= −

= ⋅ −

= − = −

The point of intersection is 4 1

,5 10⎛ ⎞−⎜ ⎟⎝ ⎠

.

39. Use a table to organize the information.

Amount Invested

Interest Rate

Interest Earned

B-Bonds x 0.12 0.12x

Bank y = 90,000 – x 0.05 0.05(90,000 – x)

Total x + y = 90,000 10,000 The last column gives the information needed

for the equation since the sum of the interest earned on the individual investments must equal the total interest earned.

( )0.12 0.05 90,000 10,0000.12 4500 0.05 10,000

0.07 5,50078,571.4390,000 78,571.4311, 428.57

x xx x

xxy

+ − =+ − =

=≈= −=

Karen should invest $78,571.43 in B-rated bonds and $11,428.57 in the well-known bank in order to achieve their investment goals.


Amount Mixed

Percent Acid Total Acid

Solution 1 x 0.20 0.20x

Solution 2 y = 100 – x 0.12 0.12(100 – x)

Mixture x + y = 100 0.15 0.15(100)

The last column contains the information needed to solve the problem because the sum of the concentrations of the individual solutions must equal the concentration in the mixture.

( )0.20 0.12 100 150.20 12 0.12 15

0.08 337.5100 37.5 62.5

x xx x

xxy

+ − =+ − =

=== − =








(continued)

So, 37.5 cubic centimeters of 20% HCL acid should be mixed with 62.5 cubic centimeters of 12% HCL acid to obtain 100 cubic centimeters of a solution with a 15% HCL acid.

41. a. The break-even point is the point where the cost equals the revenue, or when the profit is zero. Before we can find the break-even point we need the equation that describes cost. We are told the fixed costs, the band and the advertising, and the variable costs. If we let x denote the number of tickets sold, the cost of the dance is described by the equation:

500 100 5 5 600.C x x= + + = + The revenue is given by the equation, R = 10x, since each ticket costs $10. Setting C = R, and solving for x, will tell how many tickets must be sold to break even. 5 600 10

600 5120

x xx

x

+ ===

So, 120 tickets must be sold for the group to break even.

b. Profit is the difference between the revenue and the cost. To determine the number of tickets that need to be sold to clear a profit of $900, we will solve the equation:

( )900 10 5 600900 5 600

1500 5300

P R Cx x

xx

x

= −= − += −==

The church group must sell 300 tickets to realize a profit of $900.

c. If tickets cost $12, the break-even point will come from the equation

12 5 6007 600

85.71

R Cx xxx

== +==

To break even, 86 tickets must be sold. To find the number of ticket sales needed to have a $900 profit, solve the equation:

( )900 12 5 600900 7 600

1500 7214.29

P R Cx x

xx

x

= −= − += −==

So the church group needs to sell 215 tickets at $12 each to realize a profit of $900.


Type of Coffee

Amount of Coffee

Price per Pound

Total Cost of Coffee

Type 1 x 5 5x

Type 2 y = 100 – x 7.5 7.5(100 – x)

Blend x + y = 100 6 6(100)

The last column provides the information necessary to solve the problem, since the sum of the costs of the individual coffees must equal the cost of the blend.

( )5 7.5 100 6005 750 7.5 600

2.5 15060

x xx x

xx

+ − =+ − =

− = −=

The manufacturer should mix 60 pounds of type 1 coffee and 40 pounds of type 2 coffee to get 100 pounds of a blend worth $6.00 per pound.

43.

This relation does not appear to be linear.

44.

This relation appears to be linear.

45. a. The market price is the price for which supply equals demand. To find market price set S = D and solve for p. 0.8 0.2 0.4 1.8

1.2 1.61.33

p ppp

+ = − +==

The market price is $1.33 per bushel.

b. When p = 1.33, S = 0.8(1.33) + 0.2 = 1.264. There will be 1.264 million bushels of corn supplied at the market price of $1.33.







c.

d. At a price of $1.33 per bushel the supply and the demand for corn are equal.

46. a.

b. 2 1

2 1

4.03 5.47

1998 19921.44

0.246

y ym

x x

− −= =− −

−= = −

c. The slope indicates the average annual decrease in concentration of carbon monoxide in the air between 1992 and 1998.

d. 2 1

2 1

2.57 3.51

2008 20000.94

0.11758

y ym

x x

− −= =− −

−= = −

e. The slope indicates the average annual decrease in concentration of carbon monoxide in the air between 2000 and 2008.

f.

The slope of the line of best fit is a = –0.2377.

g. The slope indicates the average annual decrease in the concentration of carbon monoxide in the air between 1992 and 2008.

h. The average annual decrease in the concentration of carbon monoxide is not constant. The slope depends on the points used to calculate it.

i. The average level of carbon monoxide is decreasing.

47. a.

b. 2 1

2 1

228,700 181,900

2002 199846,800

11,7004

p pm

x x

− −= =− −

= =

c. The mean price of houses sold in the United States increased by an average of $11,700 per year from 1998 through 2002.

d. 2 1

2 1

292,600 228,700

2008 200263,900

10,6506

p pm

x x

− −= =− −

= =

e. The mean price of houses sold in the United States increased by an average of $10,650 per year from 2002 through 2008.

f.

The slope of the line of best fit is a = 13,960.

g. The mean price of houses sold in the United States increased by an average of $13,960 per year from 1998 through 2008.

h. The average annual increase in mean price of houses sold in the United States is not constant. The slope depends on the points used to calculate it.

i. The average price of houses sold in the United States increased from 1998 though 2007 and then started to decline.







48. a.

b. Yes, the data are roughly linear.

c. 2 1

2 1

135.15 102.67

2007 200332.48

8.124

v vm

x x

− −= =− −

= =

d. The slope is the average annual rate of change in the value of a share of the Vanguard 500 Index Fund. The positive value indicates that the value of a share is increasing.

e.

The slope of the line of best fit is a = 8.391.

f. Evaluate the line of best fit at x = 2008.

( )8.391 16,7058.391 2008 16,705 144.13

y xy= −= − ≈

According to the model, the value of a share of Vanguard 500 Index Fund in 2008 was worth 144.13.

49. a.

b. The data do not appear to be linearly related.

c. Since the data are not linearly related, the prediction in Problem 48 is not valid.

d. Answers will vary.

50. Answers will vary.

51. a. This is the equation of a vertical line that includes the point (0, 0). Its graph is the y-axis.

b. The graph of y = 0 is the x-axis. It is a horizontal line and has a slope of zero.

c. This is an equation of a line with a slope of –1 and a y-intercept of (0, 0). Its graph is a negatively sloped diagonal line through the origin.

Chapter 1 Project 1. Since Metro PCS charges a flat rate that does

not depend on the number of minutes, x, the equation is that of a horizontal line. M = 60

2. If 450,x ≤ the cost of the Verizon plan is constant, so V = 39.99. If x > 450, then V = 0.45(450 – x) + 39.99.

3, 4.

Set V (for x ≥ 450) = M to find the intersection of the two graphs.

( )0.45 450 39.99 600.45 162.51 60

0.45 222.51494.5

xx

xx

− + =− =

=≈

5. From the graph, we can see that if you talk more than 494.5 minutes per month, the Metro PCS plan is more economical.

6. If 650,x ≤ the cost of the T-Mobile plan is constant, so T = 49.99. If x > 650, then T = 0.50(650 – x) + 49.99.






Chapter 1 Mathematical Questions from Professional Exams 45

7.

Set T (for x ≥ 650) = M to find the intersection

of the two graphs. ( )0.50 650 49.99 60

0.50 275.01 600.50 335.01

670

xx

xx

− + =− =

=≈

Set V (for x ≥ 450) = T = 49.99 to find the intersection of the two graphs.

( )0.45 450 39.99 49.990.45 162.51 49.99

0.45 212.5472.2

xx

xx

− + =− =

=≈

8. If you talk 480 minutes a month, then the T-Mobile plan is the cheapest.

9. T-Mobile is cheapest if you talk between 472 and 670 minutes per month. Metro PCS is cheapest if you talk more than 670 minutes per month.

10. ( )0.50 650 49.99 550.50 275.01 55

0.50 330.01660

xx

xx

− + =− =

=≈

If you want to spend only $55 a month using T-Mobile, then you are limited to about 660 minutes per month.

11. The monthly charge for Metro PCS is $70 including the phone.

( )0.45 450 39.99 700.45 162.51 70

0.45 232.51516.7

xx

xx

− + =− =

=≈

Verizon is the better deal if you talk less than 517 minutes per month.

12. T-Mobile: 2000 minutes:

( )0.50 2000 650 49.99 724.99− + =

2400 minutes: ( )0.50 2400 650 49.99 924.99− + =

Verizon: 2000 minutes:

( )0.45 2000 450 39.99 737.49− + =

2400 minutes:

( )0.45 2400 450 39.99 917.49− + =

Find when the costs of the two plans are equal. ( ) ( )0.50 650 49.99 0.45 450 39.99

0.50 275.01 0.45 162.510.05 112.5

2250

x xx x

xx

− + = − +− = −

==

If you talk between 2000 and 2250 minutes, then the T-Mobile plan is less expensive. If you talk more than 2250 minutes, then the Verizon plan is less expensive. Both plans are expensive.

Mathematical Questions from Professional Exams 1. The break-even point is the value of x for which

the revenue equals cost. If x units are sold at price of $2.00 each, the revenue is R = 2x. Cost is the total of the fixed and variable costs. We are told that the fixed costs are $6000, and that the variable cost per item is 40% of the price. So the cost is given by the equation:

(0.40)(2) 6000 0.8 6000.C x x= + = +

Setting R = C and solving for x yields

2 0.8 60001.2 6000

5000

R Cx xxx

== +==

The answer is (b).

2. Profit is defined as revenue minus cost. If x rodaks are sold for $6.00 each, the revenue equation is R = 6x. We are told that to manufacture rodaks costs $2.00 per unit and $37,500 in fixed costs. So the cost equation for producing x rodaks is C = 2x + 37,500. The Breiden Company wants to realize a before tax profit equal to 15% of sales, which is P = 0.15R = (0.15)(6x) = 0.9x.








(continued)

To find the number of rodaks that need to be sold to meet Breiden’s goal, we solve the equation

0.9 6 (2 37,500)0.9 4 37,5003.1 37,500

12,096.77

P R Cx x xx xxx

= −= − += −==

The answer is (d).

3. The break-even point is the number of units that must be sold for revenue to equal cost. Using the notation given, we have R = SPx and C = VCx + FC. To find the sales level necessary to break even, we set R = C and solve for x.

( )

R CSPx VCx FC

SPx VCx FCSP VC x FC

FCx

SP VC

== +

− =− =

=−

The answer is (d)

4. At the break-even point of x = 400 units sold, cost equals revenue. We are told cost, C = $400 + 200 = 600, so revenue R = $600. Since R = price · quantity, the price of each item

is $600

$1.50.400

= The variable cost per unit is

$1.00, so the 401st unit contributes $1.50 – $1.00 = $0.50 to profits. The answer is (b).

5. Since straight-line depreciation remains the same over the life of the property, its expense over time will be a horizontal line. Sum-of-year’s-digits depreciation expense decreases as time increases. The answer is (c).

6. The answer is (c). Y = $1000 + $2X is a linear relationship.

7. The answer is (b). Y is an estimate of total factory overhead.

8. The answer is (b). In the equation $2 is the estimate of variable cost per direct labor hour.






Documents

Chapter 1 Linear Equations · 1 Chapter 1 Linear Equations 1.1 Lines 1. True 2. 2610 24 2 x x x 3. True 4. The equation of a vertical line with x-intercept at (–3, 0) is x = –3