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Solution Manual for Elementary and Intermediate Algebra Functions and Authentic Applications 2nd Edition by Lehmann ISBN 0321922727

9780321922724

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Chapter 6 Systems of Linear Equations and Systems of Linear Inqualities

Homework 6.1

2. To be a solution of the system, an ordered pair must satisfy both equations in the system.

The ordered pair (–3, 5) does not satisfy both

equations, so it is not a solution of the system.

The ordered pair (–7, 3) does not satisfy both equations, so it is not a solution of the system.

The ordered pair (3,

7) satisfies both equations, so it is a solution of the system.

4. To be a solution of the system, an ordered pair must satisfy both equations in the system.

that (4, 2) satisfies both equations

Check ( 3,5): y  x 4 5  3 4 5 1 false y  3x  2 5 3 3  2 5 11 false

that (3, –2) satisfies both equations.

Check ( 7,3): y  x 5 2 3 5 y  2x  4 2 23  4 y  x 4 3  7 4 3 3 true y  3x  2 3 3 7  2 3 23 false 2 2 true 8. 2 6  4 2 2 true

Verify

Check (3, 7): y  x 4 7 3 4 7 7 true y  3x  2 7 33  2 7 7 true

–

Verify

Check (3 1): y  1 x  3 4 2 1 4  3 y  1 x 2 2 1 4 4x 5y  17 43 5 1 17 17 17 true 3x  2y  16 33  2  1 16 7 16 false 4 2 1  3 2 2 true 10. y  x 2 2 2 true y  2x  4     y  x  3

The ordered pair (3,

1) does not satisfy both equations, so it is not a solution of the system. Check ( 4, 2):

To begin, we graph both equations in the same coordinate system.

the ordered pair (–4, –2) does not satisfy both equations, so it is not a solution of the system. Check ( 2, 5):

The intersection point is (0, 0). So, the solution is the ordered pair (0, 0).

The ordered pair (–2, –5) satisfies both equations, so it is a solution of the system.

–

y  x

4x 5y  17 4  4 5 2 17 6 17 false 3x  2y  16 3 4  2  2 16 16 16 true

4x 5y 

4  2 5 5 17

true 3x  2y  16 3 2  2  5 16 16 16 true x

17 17

x (0, 0)  5

12. Write

in slope

intercept form.

satisfies both equations.

14. Write both equations in slope–intercept form.

To begin, we graph both equations in the same coordinate system.

The system is dependent. The solution set is the infinite set of ordered pairs that satisfy the equation

16. Write both equations in slope–intercept form.

The intersection point is (0, 0). So, the solution is the ordered pair (0, 0).

22. Write the second equation in slope–intercept form.

The system is dependent. The solution set is the infinite set of ordered pairs that satisfy the This system is an inconsistent system, since the lines are parallel. The lines are parallel equation y  2x  1 since they have the same slope.

18. Write 2x 3y

6 in slope–intercept form.

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 171 y  2 x 3

–

y  2x 6

that ( 3,0)

Verify that (2, –

2x 3y  6 2( 3) 3(0) 6 x  3y  3 3  0 3 y  3x  4 2 3(2)  4 2 6  4 y  2 x 3 2 22 3 2 2 true 20. x  0 y  0 6 6 true 3 3 true 2 2 true

Verify

satisfies both equations.

9x 3y  21 3y  21 9x 3y  9x  21 y  3x 7 y  3x 7

y  3x 7

3y 1  23x  1 20x 8y  16 8y  16 20x y  2  5 x 2 y  5 x 2 15x  6y  18 6 y  15x  18 y  15 x  3 6 y  5 x  3 3y 1  6x  2 3y  6x  3 y  2x  1



2x 3y  6 3y  2x 6 y  2 x  2 3 Write x  3y  3in slope–intercept form. x  3y  3 3y  x 3 (0, 0) y  0 x  0

24. Write both equations in slope–intercept form.

y  1 x 1 3

3(y 2)  2(3x 1)  7 3y 6  6x  2  7 3y 6  6x  9 3y  6x  15 y  2x  5 y  7 4(x  2) y  7 4x 8 y  4x 1

30. Use “intersect” on a graphing calculator. Verify that ( 1,3) satisfies both equations.

The solution is (3, –4).

26. Write both equations in slope–intercept form.

We first write both equations inslope–intercept form.

Use “intersect” on a graphing calculator. Verify that ( 6, 2) satisfies both systems of equations. The approximate solution is (–5.48, –1.35).

28. Use “intersect” on a graphing calculator.

34. We first write both equations inslope

intercept form. The approximate solution is (–2.23, –3.51).

172 ISM: Elementary and Intermediate Algebra

y  2x  5 3 2( 1)  5 3 2  5 3 3 true y  4x 1 3 4( 1) 1 3 4 1 3 3 true

Check: y  7 x  3 3 4 7 (3)  3 y  2 x 2 3 4 2(3) 2

3 3 1 x  1 y  3 3 2 1 x 1 y  7 2 3 3 4 7  3 4 4 true 4 2 2 4 4 true 1 y  1 x 3 2 3 1 y  1 x 7 3 2 3 32.

y  2 x 6 3 y  3 x 7 2 y  3 x  7 2 y  2 x 5 3 7y 3x  7 y  3 x  1 7

y  2 3 x 6 y  3 2 x  7 Check: y  2 x 5 3 2 2( 6) 6 3 2 12 6 2 3( 6)  7 2 2 18  7 1.35 2( 5.48) 5 3 1.35 3.653 5 3 2 2 7 1.35  1.346 2 2 true 2 2 true 7 y 3x  7

7( 1.35) 3( 5.48) 7 9.45 16.44 7 6.99  7

–

Check: y  4.95



12x  9 y  18 y  4 3 x  2 x 4y  20 y  1 4 x 5

7.51

Copyright © 2015 Pearson Education, Inc. 3.51 4.95( 2.23)  7.51 3.51 11.0385  7.51 3.51  3.5285 y  0.84x 5.38 3.51 0.84( 2.23) 5.38 3.51 1.8732 5.38 3.51  3.5068 Use “intersect” on a graphing calculator. The approximate solution is (6.46, –6.62).

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities

b. The absolute value of the slope of C is more than the absolute value of the slope

of R. This shows that the consumption of chicken is increases at a greater rate than the consumption of red meat decreases.

c. Answers may vary. Example: Since the consumption of chicken is increasing and the consumption of red meat is decreasing, the consumption of chicken will be equal to the consumption of red meat at some point in time.

Use “intersect” on a graphing calculator.

The consumption of chicken will equal that of red meat in 1970  51  2021 The consumption will be 96.93 pounds per person for each type of meat.

Answers may vary. Example: This is not a confident prediction. There is no guarantee that the trend will continue that many years after 1970.

The solution is approximately (6.474, 5.842).

40. a. Start by plotting the data. Then find the regression lines for the data. Fixed telephone lines: F

t  0.51t  22.38

Cell-phone subscriptions: C

 8.9t 12.38

38. a. C(41)  1.13(41)  39.29  85.62

R(41)  0.94(41)  144.88  106.34

The consumption of chicken estimate is 85.62 pounds per person and the consumption of red meat estimate is 106.34 pounds per person. The error in the consumption of chicken estimate is 85.62 84.2  1.42 pounds per person and the error in the consumption of red meat

estimate is 106.34 104.3  2.04 pounds per person.

173

12x  9 y  18

Check:

12(6.46)  9( 6.62) 18 77.52 59.58 18 17.94  18 x 4y  20 (6.46) 4( 6.62) 20 6.46  26.48 20 20.02  20 36. 4 x 7 y  5 3 3 1 x  3 y  6 4 4 Write both equations in slope–intercept form.

4 x 7 y  5 3 3 7 y  4 x 5 3 3 y  4 x  15 7 7 1 x  3 y  6 4 4 3 y  1 x  6 4 4 y  1 x  8 3

Check: 4 x 7 y  5 3 3 4 (6.474) 7 (5.842) 5 3 3 8.632 13.631 5 5 5 true

1 x  3 y  6 4 4 1 (6.474)  3(5.842) 6 4 4 1.6185  4.3815 6

(rounded)

6 6 true





he number of cell-phone subscriptions per 100 people in 2000  3.69  2004

b. T h e n u m b e r o f f i x e d t e l e p h o n e l i n e s p e r 1 0 0 p e o p l e w a s e q u a l t o t

c. In 2017, t  17.

17

0.5117  22.38  13.71 C

 8.917 12.38  138.92

In 2017, the total number of fixed telephone lines and cell-phone subscriptions per 100 people will be 13.71 138.92  152.6. The result is greater than 100. Answers may vary. Example: It is possible for the total to be more than 100 because some people will have both fixed telephone lines and cell-phone subscriptions.

42. a. Start by plotting the data. Then find the regression lines for the data. Women who are married:

46. The slope of the blue line is The slope of 5 the red line is 1 . We must examine the 5 values on the left of the graph to find the intersection point, since on the right side of the graph the equations are diverging from one another. The red graph in this case will go down 2 units every 5 units we go to the left. The blue graph will rise 1 unit every 5 units we go to the left. Moving 10 units to the left will have the two graphs meet. Thus, the intersection point will be ( 16, 1) .

48. Since the graphs have the same value at x  1(3), the solution of the system will be (1,3)

50. The slope of

is 4 and the f-intercept is

. The solution of the system is

58. a. Find the intersection point via graphing calculator.

The time at which the percentages will be equal is t = 61, or 1990 + 61 = 2051. The percentage will be 47.54%. Answers may vary.

c. Use t = 25 for 2015.

165.1

0.5435  89.73 million women

160.4 0.5690  91.27 million men

The total number of people who will be married in 2015 is 89.73

91.27  181 million.

The intersection point of the two graphs is (2, 1) . Answers may vary. Example: Choose a point on y

3x 7which uses

44. Answers may vary. Example: The solution of the system appears to be approximately (–4.8, 1.4).

174 ISM:

Elementary and Intermediate Algebra











0,99 . Therefore, f  x  4x  99.

Men who are married: 52.

slope of g  x is 2

g-intercept

0,3 . Therefore, g  x  2x  3

16,35 g  4  3 M (t)  0.29t  64.15 b. 54. 56. g(x)  1 when x = 2. g  x  0 when x  1 2

W (t)  0.22t  59.85

The

and the

W 25  0.2225  59.85  54.35



M 25  0.2925 



64.15

56.90





an x that is anything other than 2, such as3.

b. Answers may vary. Example: Choose a point on y  2x  3 that is

y  3x 7  3(3) 7  9 7  2 (3,2)

other than 2, such as 3.

c. (2, 1)satisfies both equations.

d. Answers may vary. Example:

68. Answers may vary. Example: The three types of systems are one-solution systems, inconsistent systems, and dependent systems

70. a. Write the equations in slope–intercept form.

(0,0)is a point that lies on neither line and is an acceptable answer.

60. Graph the two equations and find the intersection point.

The lines have the same slope and different y-intercepts. Therefore, the lines are parallel.

b. Parallel lines do not cross so they do not have an intersection point.

c. Since the lines have the same slope and different y-intercepts, there are no intersection points. Therefore, the system has no solutions.

62. Graph all three equations. 72. The graph shows that y

. So the solution of the given equation is x = –5.

The three equations do not meet at a common point. The system is inconsistent. There are no 76. The graph shows that y  1 x  5 3 3 and

solutions that satisfy all the equations.

64. a. Since the system of linear equations has morethan one solution,the systemmust be dependent. The equation of the line that contains the two given points is

1 x  7 . A third solution would be 3 3 anyother point that liesonthe graphofthis

equation. One possible answer is  0,7 

y = x – 1 intersect at the point (4,3) . So the solution of the given system of equations is (4, 3).

80. The equation y  3x 2 is in slope–intercept

b. There are an infinite number of solutions.

and

Inqualities 175

y  2x  3  2(3)  3  6 



(3, 3)

Systems

Linear Equations

Systems

Linear

anything

3x 7y  14 y  3 x 2 7 y  3 x 4 7

x  3 The solution is  3,2 .

 1 x  5 3 3

y y  2x 5  y  0.6x 1 y = –3x – 5

( 2,1)

the

is x = –2.   x 74. The

that y  1 x  5 3 3 and y = 0 y  1.2x  5  intersect at

5,0)

and

intersect at the point

solution of the given equation

graph shows

the point (

y 

2x  8  3x 2  2x  8  3x  2  2x  3x  8  2  5x  10 This

78.

is a linear expression in one variable.



 3 



66. Answers may vary. Example: An ordered pair (a, b) is a solution of a system of two equations in two variables if it satisfies both equations.

form, so the slope is m  3 and the y-intercept is 0, 2 We first plot 0, 2 From this point we move 1 unit to the left and 3 units up, plotting the point  1,1 We then draw the line that connects the two points, extending in both directions.

This is a linear equation in two variables

3 for y in the equation x

3y 1 for x in the first equation and

Substitute 1 for x in the equation

Substitute 2y  5 for x in the second equation and solve for y

Substitute x 4 for y in the first equation and solve for x

Substitute 4 for y in the equation x  2 y  5 and solve for x.

equation and

first

. Homework 6.2 31 2 y  2 y  9 3 6y  2y  9 3 4y  9 4y  12 y  3 2. y  4x 2x  y  18 Substitute 4x

solve for x. 2x  (4x)  18 6x  18 x  3 Substitute



solve for x. x  1 2(3)  1 6  5 The solution is (–5, 3). 10. 4x  3y  2  0 y  1 3x Substitute 3 for x

y  4x and solve for y. Substitute 1 3x

and solve for x y  4(3)  12 The solution

(3, 12). 4. x  2y  4 x  3y 1 Substitute

solve for y (3y 1)  2y  4 3y 1  2y  4 5y 1  4 5y  5 y  1 12. 4x  31 3x  2  0 4x  3 9x  2  0 5x  5  0 5x  5 x  1

and solve for y. y  1 3(1)  1 3  2 The solution is (1, –2). x  2y  5 7x  2y 13  0 y  1 3x

and

for x. x  3(1) 1 



6. 4x

y 

y  x

for y in the second equation and

1 2y and

in the equation

for y in the first equation

Substitute 1 for y in the equation x  3y 1

solve

3 1

2 The solution is (2, 1).

4x 3(x 4) 

4x 3x  12 

x  12  13 x  1

7 2 y  5  2 y 13  0 14y  35  2y 13  0 12y  48  0 12y  48 y  4

x 







(-1,1) 2 4 -2 (0,-2) -4 -4 -2

2(4)

5  3 The solution is (–3, 4).

14. 4x 7y  3 x  2 y  4 3 2 20. y  5 x 1 y  2 x  2 Substitute 5 x 1 for y in the second Substitute y  4 for x in the first equation. 3 equation and solve for x 4x 7y  3 4  2 y  4 7y  3 5 x 1  2 x  2 5x 5  2x  4  3    8 y  16 7y  3 3 13 y 16  3 3 13 y  13 3 y  3 3x 5  4 3x  9 x  3 Substitute 3 for x in the equation and solve for y y  53 1  52  10 The solution (3, 10). y  5 x 1 Solve for x when x  2 y  4 3  2 3  4 3  6 y  3 22. y  1.45x 6.18 y  2.63x 2.73 Substitute 2.63x 2.73 for y in the first equation and solve for x. 2.63x 2.73  1.45x 6.18 16. The solution is (6, 3). y  4x  2 y  3x 5 Substitute 3x 5 for y in the first equation and solve for x. 3x 5  4x  2 7x 5  2 7x  7 x  1 Substitute 1 for x in the equation y  3x 5 and solve for y y  3( 1) 5  3 5  2 The solution is (–1, –2). 24. 4.08x 2.73  6.18 4.08x  3.45 x  3.45  0.846 4.08 Substitute 0.846 for x in the equation y  2.63x 2.73 and solve for y y  2.63 0.846 2.73  4.955 The approximate solution is ( 0.85, 4.96) y  0.51x 2.64 y  2.79x  5.94 Substitute 2.79x  5.94 for y in the first equation and solve for x. 2.79x  5.94  0.51x 2.64 18. y  x y  x Substitute

and solve for x x  x 2x  0 2.28x  5.94  2.64 2.28x  8.58 x  8.58  3.763 2.28

3.763

y  2.79x  5.94 and solve for y. x  0



solve for y. y  (0)  0

solution

(0,

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 177

x for y in the first equation

Substitute

for x in the equation

Substitute 0 for x in the equation y

x and

The

0).

Solve the first equation for y

Substitute 3x 14 for y in the second equation and solve for

Substitute 4 for y in the equation x  3y 14 and solve for x. x

3(4) 14  12 14  2 The solution is (2, 4).

32. 4x 3y  7

3x y  9 Solve the second equation for y 3x y

3x 9

Substitute 3x  9 for y in the first equation and solve for x.

Substitute 4 for x in the equation and solve for y

12 14

The solution is (4, 2) 28.

Solve the second equation for x.

Substitute 3y 5 for x in the first equation and solve for

Substitute 3 for y in the equation

y 5 and solve for x.

14 Solve the second equation for x.

14 for x in the first equation

2015 Pearson



Education, Inc. 178 ISM: Elementary and Intermediate Algebra 26. 3x

y  14 2x 7 y  22

3x  y  14 y  3x

x 2x 7 3x 14  22 2x 21x  98  22 19x  98  22 19x  76 x  4



y 

 9

 9



3(4) 14

 2

2x  3y  1 x  3y  5

x  3y  5 x  3y 5 y  3x 14 4x 33x  9  7 4x 9x 27  7 5x 27  7 5x  20 x  4 Substitute 4 for x in the equation and solve for y. y  3( 4)  9  12  9  3 The solution is ( 4, 3) y  3x  9

2 3y 5  3y  1 6y 10  3y  1 3y 10  1 34. 4x 5y  2 3x  y 7 Solve the second equation for y. 3x  y 7 y  3x  7 3y  9 y  3 Substitute 3x  7 for y in the first equation and solve for x

x  3

x  3( 3) 5 



30. 6



x 



x 



x  3



9 5

4 The solution is (4, 3)

x 5y



and

36. 4x 53x  7  2 4x 15x 35  2 11x 35  2 11x  33 x  3 Substitute 3 for

and solve for y. y  3( 3)  7  9  7  2 The solution is ( 3, 2) x  2 y 1 4x 8y  4 y  3x  7 6 3y 14 5 y  8 18y  84 5y  8 23y  84  8 23y  92

Substitute 3y

solve for y.

the equation

The result is a true statement. The system is dependent. The solution set is the infinite set of ordered pairs that satisfy the equation



in the first



substitution.

in the second equation and solve for

Substitute

and

solve for y y  4 2( 2)  4  4  8 The solution is (–2, 8). y  2x 5 4x  2y  6 y  4 2x 2x  195 x  97.5 Solve for y when x  97.5 y  397.5  201  493.5 The solution is 97.5,493.5 . 48. A:

This point is the origin so it has coordinates (0, 0).



false

for x in the first equation

The coordinates of the point are (0, 3).

y  2x  8 2y x  6

Substitute

solve for

6 true

and solve for

solve for

Inqualities 179

6: Systems of Linear Equations and Systems of Linear

x  2 y 1. 3x 2y  10 y  4 2x Substitute 4 2x for y

38. equation and solve for x. 3x 24 2x  10 3x 8  4x  10 x 8  10 x  2

f  x 

g  x 

y  3x  201 y  5x  6 Solve using

x y  5x  6 3x  201  5x  6

46. The function f has a slope of 3 and an f-intercept of 0,201 Therefore,

201. The function g has a slope of 5 and a g-intercept of 0,6 Therefore,

6 The system that describes the functions

and g is:

Substitute 3x  201 for

for x 4x  22x 5  6 4x  4x 10  6 10  6

40.

–2 for x in the equation

Substitute

2x 5 for y in the second equation and solve

 . 42. 3x 12y  6 x  4 y  2 Substitute 4y  2

and solve for y. 3 4 y  2 12 y  6 12 y  6 12 y  6 6 

The result is a false statement, so the system is inconsistent. The solution is the empty set,

x  4 y  2

The result is a true statement. The system is dependent. The solution set is the infinite set of ordered pairs that satisfy the equation

2 y 0  6 2 y  6 y  3

This point is the y-intercept for line l2 Let

0 in the equation and solve for y

2x  8

and

x 2 2x  8 x  6 4x 16 x  6 5x 16  6 5x  10 44. y  4x 1 8x  2y  5

This point is the intersection of the two lines. Therefore, it is the solution of the system:

for y in the second equation

x. x  2

y y  22  8  4  8  4 y  2x  8 8x  24x 1  5 8x  8x 2  5 2  5 false

Substitute 4x 1 for y in the second equation and

Substitute 2 for x in the

equation

The result is a false statement, so the system is inconsistent. The solution is the empty set, 

The coordinates of the point are (2, 4).

This point is the x-intercept of line l1 Let y  0 in the equation and solve for x

The coordinates of the point are (4, 0).

50. a. Solve the second equation for x

Substitute 3y  9 for x in the first equation and solve for y.

Substitute 2 for y in the equation

9 and solve for x

b. Answers may vary. Example: If the result of applying substitution to a linear system of two equations is a true statement that can be put into the form a  a , then the system is dependent; that is, the two equations represent the same line and the solution set is the infinite set of ordered pairs represented by every point on the line.

54. Answers may vary. Example: Substitution may tend to be more reliable since the exact values can be obtained rather than estimating intersection points.

b. Solve the first equation for y

c. The results are the same. The solution of the system is (3, 2).

52. a. Answers may vary. Example:

If the result of applying substitution to a system of equations is a false statement, then the system is inconsistent; that is, the solution set is the empty set.

4).

d. Answers may vary. Example: The solution of the system is the common point that was used to find the equations in parts (a) and (b). A solution of a system of linear equations in two variables is an ordered pair that lies on the graph of all equations in the system. We know the point 3,4 is a solution of our system since it was on the graph of both equations

0 

2x  8 x  4

x 



9 x

3y  9

2 3y  9 



6y 





5y 



5y 

y 

x  3y 

x  32  9  6  9  3

The solution is (3, 2).

2x  y  8 56. a. b. y y1  mx x1 y 4  3 x 3 y 4  3x 9 y  3x 5 y y1  m x x1 y 4  2 x 3 y 4  2x  6 y  2x 10 y  2x  8 Substitute 2x  8 for y in

equation and solve for x

y  3x 5 y  2x 10 x  3 2x  8  9 Substitute 2x 10 for y in the first x 6x  24  9 5x  24 

5x 

x  3 Substitute 3 for x in the equation y  2x  8 and solve for y. y  2 3  8  6  8  2 The solution is (3,

the second

c. Our system of equations is

2).

equation and solve for x 2x 10  3x 5 5x 10  5 5x  15 x  3 Substitute 3 for x in the equation y  2x 10 and solve for

. y  2 3 10  6 10 

The solution

is (3,

in the system.

2015 Pearson Education, Inc.

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y

(3, 2). 6.

To eliminate the y terms, we multiply both sides of equation (2) by –2, yielding the system:

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for x

58. 60. 62. 64. The solution of the

x = –1.29. The solution of the

x = 3.30. The solution of the

x = –6.69. 7  3x  5 7 5  3x  5 5 2  3x 4x  5y  11 4x  3y  13 8y  24 y  3 Substitute 3 for y in

(2) and solve for x. 4x  33  13 4x  9  13 4x  4 x  1 The solution is (1, 3). 4. 3x  5 y  19 Equation (1) 2x 5y 

Equation

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 181

equation is approximately

equation

(2)

and solve for x 3x  5 y  19 2x 5 y  4 5x  15 x  3 Substitute 3 for x in equation (1) and solve for y 2  3x 3 3 33  5y  19 9  5y  19 66. 2  x 3 The solution is 2 3 This is a linear equation in one variable. 9x  2y 3x  5  9x 3x  2y  5  9 3 x  2y  5  6x  2y  5 This is a linear expression in two variables. 5y  10 y  2

solution is

x 4y 

Equation

x 2y 

Equation

The

(1) 5

(2)

x 4y  5 10x  4y  22 Homework 6.3 2. 4x  5y 

4x  3y 

11 Equation (1)

13 Equation (2)

x 4 y  5 10x  4y  22 9x  27 x  3

Substitute –3 for x in equation (1) and solve for y

The solution is (–3, –2).

8. 3x  4 y  5 Equation (1)

5x  2y  17 Equation (2)

To eliminate the y terms, we multiply both sides of equation (2) by –2, yielding the system:

3x  4y  5 10x 4y  34

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for x 3x  4 y  5 10x 4y  34

The solution is (1, –3).

12. 6x  5y  14 Equation (1)

4x 7 y  2 Equation (2)

To eliminate the x terms, we multiply both sides of equation (1) by 2 and both sides of equation (2) by 3, yielding the system:

12x 10y  28

12x 21y  6

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y

12x 10y  28

12x 21y  6

Substitute2 for y inequation(1)andsolve for x

Substitute –3 for x in equation (2) and solve for y

The solution is (–3, 1).

10. 4x 7y  25 Equation (1)

8x  3y  1 Equation (2)

To eliminate the x terms, we multiply both sides of equation (1) by –2, yielding the system:

8x 14y  50

8x  3y  1

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y.

8x 14y  50 8x  3y  1 17y  51 y

Substitute –3 for y in equation (2) and solve for x

The solution is (–4, 2).

14. 5x 2y  8 Equation (1)

2x 3y  1 Equation(2)

To eliminate the y terms, we multiply both sides of equation (1) by 3 and both sides of equation (2) by 2 , yielding the system:

15x 6y  24

4x  6y  2

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for x.

15x 6y  24

6y  2 11x  22

Substitute2 for x inequation(1)andsolve for y.



3 4 y  5 4y  8

 2

13x  39 x  3





15 



y 

3

2 y  17

2 y



8x  3 3  1 8x 9  1

8x  8 x  1

11y  22

 2

x  52 

6x 10 

6x 

x 



x  2

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations

To eliminate the x terms, we multiply both sides of equation (1) by 5 and both sides of equation (2) by 2, yielding the system:

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y

Multiply the first equation by 2 and add the equations.

Substitute1for y inequation (2)and solvefor x

Multiply the first equation by 3 and add the equations.

Multiply the first equation by 4 and the second equation by 30, then add the equations.

183 52 2y  8 10 2y  8 2y  2 y  1 The solution is (2, 1). 0.6x 1.5y  0.6 (0.8x  1.5y  6.2) 1.4x  5.6 x  4 Solve for y when x  4 0.2x 0.5y  0.2 0.2 4 0.5 y  0.2 16. y  2x  7  2x y  7 Equation (1) y  x 8  x  y  8 Equation (2) 0.8 0.5y  0.2 0.5y  1

and solve for x 2x y  7 x  y  8 3x  15 x  5 Substitute 5 for x in equation (1) and solve for y y  2 5  7  10  7  3 The solution is ( 5, 3) . 18. 3  2x  9y  2x  9 y  3 Eq.(1) 22. y  2 The solution is( 4, 2) 2  x 3 3 y  1  5 4 x 2  5 y  3  13 Simplify both equations. 2 x 3 3 y  1  5 2x 6 3y 3  5 2x 3y  4 4 x 2  5 y  3  13 4x  8  5y  15  13 4x  5 y  10 Solve the system: 2x 3y  4 5x 5y 10  0  5x 5y  10 Eq. (2) 4x  5 y  10

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities

10x  45y  15 10x 10y  20

10x 



10x 10y 

35 y 

y  1

45y

5x 51 10  0 5x 15  0 5x 

x  3

0.2

0.5y  0.2

 1.5y  6.2

The solution is (3, 1). 20.

0.8x

4x 6y  8 ( 4x  5y  10) y  2 y  2 Solve for x when y  2 2 x 3 32  1  5 2x 6 33  5 2x 6 9  5 2x 15  5 2x  10 x  5 The solution is (5, 2) 24. 1 x  5 y  2 4 2 5 x 1 y  2 6 3

x  10 y  8  (25x 10y  60) 26x  52

To eliminate the x terms, we multiply both sides of equation (1) by 3 and both sides of equation (2) by 2, yielding the system:

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve

The result is a true statement of the form

Therefore, the system is dependent.

The solution set is the infinite set of ordered pairs that satisfy the equation 4x

30. 3x  2y  5 Equation (1)

12x 8y  17 Equation (2)

To eliminate the x terms, we multiply both

sides of equation (1) by 4, yielding the system:

12x  8y  20

12x 8y  17

The coefficients of the x terms are equal in

absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y.

12x  8y  20

12x 8y  17

0  3 false

The result is a false statement. Therefore, the system is inconsistent. The solution set is the empty set, 

32. 8x 3y  2 Equation (1)

5x 12y  29 Equation (2)

To eliminate the y terms, we multiply both sides of equation (1) by 4, yielding the system:

Substitute 2 for y in equation (1) and solve for

28. 4x  6y  10 Equation (1)

6x 9y  15 Equation (2)

To eliminate the x terms, we multiply both sides of equation (1) by 3 and both sides of equation (2) by 2, yielding the system:

12x 18y  30

12x 18y  30

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y.

32x 12y  8

5x 12y  29

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for x

32x 12y  8

5x  12y  29

37x  37 x  1

Substitute 1 for x in equation (1) and solve for y

8 1 3y  2

8 3y  2

3y  6 y  2

The solution is ( 1, 2)

Pearson Education, Inc. 184 ISM: Elementary and Intermediate Algebra Solve for y when x  2 1 x  5 y  2 4 2 1  2  5 y  2





2015

12x

18y

30 12x 18y

30 0

0 true

4 1 2



 5 y  2 2 2 5 y  5 2 2 y  1 The solution is ( 2,1)



y 

26. 1 x 5 y  9 2 4 2 3 x  1 y  1 8 2 2  2x 5y  18 Equation (1)  3x  4y  4 Equation (2)

x 15y  54 6x  8y  8

for y 6x  15y  54 6x  8y  8 23y  46 y  2

x 2x 5 2 

2x 10 

2x 

x 

4 The solution is (4, 2)

34. 10x 12y  5 Equation (1)

15x 18y  8 Equation (2)

To eliminate the x terms, we multiply both sides of equation (1) by 3 and both sides of equation (2) by 2, yielding the system:

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for

The result is a false statement. Therefore, the system is inconsistent. The solution set is the empty set, 

36. 9x 6 y  15 Equation (1)12x 8y  20 Equation (2)

To eliminate the x terms, we multiply both sides of equation (1) by 4 and both sides of equation (2) by 3, yielding the system:

36x 24y  60

36x  24y  60

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y 36x 24y  60 36x  24y  60

The result is a true statement of the form a  a . Therefore, the system is dependent. The solution set is the infinite set of ordered pairs that satisfy the equation 9x 6y  15 .

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve

Substitute

for x in equation (1) and solve for

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve

To eliminate the y terms, we multiply both sides of equation (1) by 2, yielding the system:

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 185

y y  1.283.902  2.05  7.04

approximate solution

(3.90,7.04) 30x 36y  15 40. y  3.28x 1.43  3.28x y  1.43 Eq. (1) 30x  36y 

3.902

The

30x

y 

30x  36y 

0 

1 false

0  0 true

38. y  1.28x  2.05  1.28x  y  2.05 Eq. (1) y  0.56x  6.720.56x  y  6.72 Eq. (2)

for x. 3.28x y  1.43 0.56x  y  6.72 2.72x  5.29 x  5.29  1.945 2.72

1.945

equation

and solve for y y  3.281.945 1.43  7.810 The approximate solution is (1.95,7.81)

6x  y 

Equation

3x 2y  19 Equation

Substitute

for x in

(1)

42.

(1)

(2)

12x  2y  26 3x 2y  19

x. 12x  2y  26 3x 2y  19 15x  45 x  3

y  3.94x 8.83  3.94x y  8.33 Eq. (2) 6 3  y  13

Substitute 3 for x

equation (1) and solve for y.

x 1.28x  y  2.05 3.94x y  8.33 2.66x  10.38 x  10.38  3.902 2.66

for

Substitute 2 for x in the equation

To eliminate the y terms, we multiply both sides of equation (1) by 3 and both sides of equation (2) by 2 , yielding the system:

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations

Substitute1 for x inequation(1)andsolve for y

4 Equation (1)

6 Equation (2)

To eliminate the x terms, we multiply both sides of equation (2) by 3 , yielding the system:

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for

equation

x 3x 82  4 3x 16  4 3x  12 x  4

and solve for y. y  1 22  1 4  3 The solution

(2,

. 46. 5x 2y 

4x 3y 

Equation

y  1 2x 50. The solution is (4, 2). y  4x 9 y  2x  3 Substitute 2x  3 for

and solve for x 2x  3  4x 9 6x  3  9

Substitute2 for y in

(1)andsolve for

7 Equation (1)

(2)

the first equation

15x 6y 

8x  6y  14

6x  12 x  2 Substitute 2 for x in the equation and solve for y y  2 2  3  4  3  1 The solution is ( 2, 1) . y  2x  3 and solve for x 52. 3x  2y  19  3x  2y  19 Eq. (1) 15x 6y  21 5y  4x 10  4x  5y  10 Eq. (2) 8x  6y  14 7x  7 x  1

51 2y  7 5 2 y  7 2y  2 y  1 The solution is (1,



1) 48. 3x 8y







3x 8y

4 3x

15y

y. 3x 8y  4 3 x  1 5 y  1 8 7 y  1 4 y  2

on (1) by4 and both sides ofequation (2) by 3, yielding the system:

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y

Substitute2 for y inequation(1)andsolve for x

Begin by writing each equation in standard form.

x  8

 76

x 15y 

x  8 y  76

x 15

 30



y  2

23y

3x  22  19 3x  4  19 3x  15 x  5

solution is ( 5,2) . 42x 7  3y  7 7x 52 y  3  3

The

To eliminate the y terms, we multiply both sides of equation (1) by 10 and both sides of

by 3, yielding the system:

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and

Substitute4 for x inequation(1)andsolve for y

First, we need to simplify the equations. Multiply each of the equations by 12 and simplify so that the variables are on the same side of the equation for the two.

Multiply the first equation by –3 and rewrite the equations so that the variables are on one side.

This is a dependent system – the lines are the same. The solution set is the infinite set of ordered pairs that satisfy the equation

3 and add them

Multiply the second equation by

eliminate the y variable and find

187 42x 7  3y  7 8x 28  3y  7 8x  3y  35 This yields the system: 7x 52 y  3  3 7x 10 y 15  3 7x 10y  18 2x 3y  16 ( 2x  y  12) 2y  4 y  2 Solve for x when y  2 8x 



7x 10y 

1 x 3 y  4

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities

35 Equation (1)

18 Equation (2)

2 4 1 x 3 2  4 equation (2)

2 1 4 3 80x  30y  350 21x 30y  54

solve for x 80x  30y  350 21x 30y  54 101x  404 x  4

. 84  3y  35 32  3y  35 3y  3 x  4 2 2 1 x  5 2 2 x  5 The solution is (–5, 2). 60. 2x  y 3x y  1 3 6 x  y  2x y  31 3 4 12

 2x  y 3x y 12  112  3 6  56. y  1 The solution is (4, 1). y  1 x  4   4(2x  y) 2(3x y)  12 8x  4y 6x  2y  12 2x  6y  12 x  3y  6 3 x  3y   x  y 2x y  31 12  3  4  12  12 12

3y  x 12 x 3y  12 3y  x  12 x  3y  12

1   4(x  y)  3(2x y)  31 4x  4y  6x 3y  31 10x  y  31

–

x. x  3y  6 30x  3y  93 29x  87 x  3

3 for x and find y y  3 1 3 x  4. x  3y  6 3  3y  6 3y  3

Substitute

Multiply the first equation by 4 and the second equation by 3, then add the equations.

The solution is (3,1)

Pearson Education, Inc. 58. x y  4 2 4 2 x 1 y  4 3 3

2015

y  1

Graphing by hand: Begin by writing each equation in slope–intercept form.

Substitute2 for x inequation (1)and solve for y.

The solution is (2, 3).

Next we graph both equations in the same coordinate system.

Preference may vary.

The intersection point is (2, 3). So, the ordered pair (2, 3) is the solution.

Substitute 2x  7 for y in the second equation and solve for

To eliminate the y terms, we multiply both sides of equation (1) by 2, yielding the system:

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for x

64. 4x 7 y  15 Equation (1) 5

3y  7 Equation (2)

a. To eliminate the x terms, we multiply both

sides of equation (1) by 5 and both sides of equation (2) by 4, yielding the system:

The coefficients of the x terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for y

Substitute 1 for y in equation (1) and solve for

b. To eliminate the y terms, we multiply both sides of equation (1) by 3 and both sides of equation (2) by 7, yielding the system:

12x 21y  45

35x  21y  49

The coefficients of the y terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for x.

12x 21y  45

35x  21y  49

Education, Inc.

2015 Pearson

188 ISM: Elementary and Intermediate Algebra





62. y

5x 2 y  4

4x  2y  14 5x 2y  4 9x  18 x  2 y  2x  7 5x 2y  4 2y  5x  4 y  5 x 2 2

22  y  7 y  3

y 5 5

2y  4 (2, 3)



 5 5 x y  2x  7

20x



20x 12y 

35y

y  2x  7 5x 2y  4

Substitution:

x 35y  75

x 20x 12y  28 47 y  47 5x 2 2x  7  4 5x  4x 14  4 9x 14  4 9x  18 x  2

2 for x in the equation and solve for y y  22  7  4  7  3

solution is (2, 3).

2x  y  7 Equation

5x 2y  4 Equation

y  2x  7 y  1

Substitute

The

Elimination:

(1)

(2)

. 4x 7 1  15 4x  7  15 4x  8 x  2

solution is (2, 1)

The

x 



14 5x 2y 

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 189

Substitute 2 for x in equation (2) and solve for

Multiply the second equation by 2 and add the equations.

solution is (2, 1)

c. The results are the same.

66. The function f has a slope of 4 and an f-intercept of (0, 76) . Therefore,

x  4x 76 The function g has a slope of

3 and a g-intercept of (0, 85). Therefore, g

x  3x  85 Write the system that describes the functions f and g

4x 76

3x  85

Solve using substitution.

The solution is (4, 7) so the coordinates of D are (4, 7). The coordinates for E are the same as the coordinates of the point of intersection of

3 and �4 . Solve the following system.

3 Solve the following system

The solution is (23, 16).

68. The coordinates for A are (0, 0) since it lies at the origin. The coordinates for B are the same as the coordinates of the y-intercept of �1, which is (0, 2). The coordinates for C are the same as the point of intersection of �1 and �2 Solve the following system.

2 for y in the second equation.

into the first equation and solve

Substitute

The solution is (2, 8) so the coordinates of C are (2, 8). The coordinates for D are the same as the coordinates of the point of intersection

y 52 

 7 10  3y  7 3y  3 y  1

The



y 

�2 :2y  x  18 �3 : y  4x 

–





x 





y 2y  

 



y  7

y 8x  46

4 Substitute x

4 into 2y

x  18 and solve for

�

�3

y  4



Substitute

solve for x 4x

 3x 

7x 

y  3x 

�4 : y  3x 12 Substitute 3x 12 in for y in 3x 12  4x  23 7x  35 x  5 y  4x  23 x  23

x  23

for y y 

Substitute x  5 into for x. y  35 12  3 y  3x 12 and solve y  423 76  92 76 

4x 76 for y in

161

85 and

Substitute

into

4x 76 and solve

�1 : y  3x  2

2 :2 y  x 

�



23x  2  x  18 6x  4  x  18 7x  14 x  2

Substitute 3x

x  2

for y. y  32  2  6  2  8

�2

and �

The solution is (5, 3) so the coordinates of E are (5, 3). The coordinates of F are the same asthe coordinates of the x-intercept of �4

0 in �4 and solve for x.

The coordinates of F are (4, 0).

70. The student is incorrect. He has found the value for x, but he needs to also find the valueof y by substituting 2 for x in one of the equations and solving for y

Let y

3x 

x 



 3x 120  3x 12

4 Equation(2)

To eliminate the b terms, we multiply both sides of equation (1) by 1 , yielding the system:

The coefficients of the b terms are equal in absolute value and opposite in sign. Add the left sides and the right sides of the equations and solve for m

76. Notice that when x = 3, both y

and y

Substitute 3 for m in equation (1) and solve for

3,2 .

1 x

7 2 2 are equal to 5. So the solution

78. Notice that when x = 1, both y

4 are equal to 4. So the solution to the given equation is x = 1.

80. Notice that when x = 5, both y  1 x

7 2 2 and

y  4 x  2 are equal to 6. So the solution to 5 the given system of equations is (5, 6).

82. 5x  4y  8 x-intercept: 5x  4(0)  8

The x-intercept is 8 ,0 y-intercept: 5(0)  4y  8

The y-intercept is (0, 2). We plot the intercepts and connect them with a line, extending the line in both directions. This is a linear equation in two variables.

,0 (0,

d. The equation of the line is y  3x  2 . 84. 5x  4y  8 Equation (1)

e. Graph the equation and check if the points ( 1,5) and (2, 4) are on the graph.

 11 x  17

6 y  3 3  y

10 10





The solution is (2, 3). 72. Answers may vary to the given equation

x = 3.



74. a. y  mx  b 5  m 1  b m  b  5



2 and

b. y  mx  b





4  m2  b 2m  b  4 c. m  b 

m  b 

5 Equation (1) 2

m b  5 2m  b 

m b  5 2m  b  4 3m  9 m  3

b.  3  b  5 3  b  5 b  2

m,b

The solution of the system is

 

5x  8 x  8 5

4 y  8

 2



3x  2y  2

Equation (2)

To eliminate the y terms, we multiply bothsides of equation (2) by 2 , yielding the system:

5x  4y  8

6x 4y  4

The coefficients of the y terms are equal in absolute value and opposite in sign. Add theleft sides and the right sides of the equationsand solve for x

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 191

6. a. Find the regression lines for the data.

Substitute 4 for x in equation (2) and solve

is a system of linear equations in two variables.

I t  1.26t  22.46

Substitute D t  0.71t  25.78 for I(t) in the second equation and solve for t. 0.71t  25.78  1.26t  22.46 3.32

The fuel efficiency of domestic cars was equal to the fuel efficiency of imported cars in the year 2000

According to the models, the average annual consumption of chicken and red meat will be equal in the year 1970 + 51.01  2021. This is not a confident prediction. There is no guarantee that the trend will continue that many years after 1970.

22.38 for n in the second equation and solve for t

According to the models, domestic passenger cars will not meet this standard, but imported passenger cars will meet it.

According to the models, the number of fixed telephone lines per 100 people was equal to the number of cell-phone subscriptions per 100 people in the year2000  3.69  2004

Answers may vary. Example:

According to the models, in 2016, the fuel efficiency of domestic cars will be less than the fuel efficiency of imported cars. If more domestic cars than imported cars are sold, the average fuel efficiency of all cars will likely be lower than the result.

5x  4y  8 6x 4y  4 x  4 x  4

for y. 3 4  2 y  2 12  2y  2 2y  14 y  7

This

Homework 6.4

The solution is ( 4,7)

Domestic: D t  0.71t  25.78 Imported: I t  1.26t  22.46 b. Solve the system D t  0.71t  25.78 2. y  C t  1.13t  39.29 y  R t  0.94t  144.88 Substitute 1.13t  39.29

equation and solve for t. 1.13t  39.29  0.94t 144.88 2.07t  105.59 t  51.01 Solve for y when t  51.01 y  1.1351.01  39.29  96.93

for y in the second

4. n  F t  0.51t 

n  C t  

0.51





34.76 

3.69 

Solve

y 





22.38

8.9t 12.38 Substitute



22.38

8.9t 12.38

9.41t

for y when t  51.01.

1.13

51.01

39.29  96.93

 0.55t 6.04  t



D 6.04  0.716.04  25.78  30.07 The fuel efficiency was 30.07

c. D 16  0.7116  25.78  37.14 I 16  1.2616  22.46  42.62

6.04  2006

miles per gallon.

d. D 16  I 16  37.14  42.62 2 2  79.76 2  39.88

8. a. Find the regression lines for the data. Gamble online:

I a  0.58a 6.94 Gamble at traditional casinos:

W (t)  1.22t  338.47

Men:

C a  0.83a  58.01

b. Solve the system

I a  0.58a 6.94

C a  0.83a  58.01

Substitute I(a)  0.58a 6.94 for C(a) in the second equation and solve for a.

0.58a 6.94  0.83a  58.01

1.41a  64.95 a  46.06

The percentage of Americans who gamble online is equal to the percentage of Americans who gamble at traditional casinos at age 46.

I 46.06  0.5846.06 6.94  19.77

The percentage is about 19.8%.

c. Find the intersection point using a graphing utility.

M (t)  0.36t  240.44

b. Solve the system

y  1.22t  338.47

 0.36t  240.44

Substitute 1.22t  338.47 for y in the second equation and solve for t

 113.99

Substitute this result into the first equation and solve for y

 1.22(113.99)  338.47  199.40

According to the models, the record times for men and women will both be approximately 199.40 seconds in 2014.

This confirms our original solution of 46 years old and 19.8%.

10. a. Start by plotting the data, then compute the regression line for each data set.

Women:

c. Women: W (t)  1.01t  323.57

Men:

M (t)  0.36t  240.73

Pearson Education, Inc. 192 ISM: Elementary and Intermediate Algebra

2015

1.22

t  338.47  0.36t  240.44 0.86t  98.03

323.57 for y in the second equation and solve for t.

Substitute 1.01t

Substitute this result into the first equation and solve for y y

194.85 Now the models predict that the record times for men and women will be the same in 2027.

d. For the most part the data appear to be linear. However, the data value for women in 1927 seems to deviate somewhat from the linear pattern. Removing this value makes the linear model a better fit.

12. a. Since a 2011 Honda Accord’s value decreases by a constant $1231 each year, the function H is linear and its slope is 1231 The H-intercept is (0,15,905) since a 2011 Honda Accord was worth $15,905 at year t  0. So, an equation for H is: H(t)  1231t  15,905

Similar work for the student’s savings gives the equation:

500

b. The student will be able to buy a 2011 Honda Accord when the amount saved is equal to the value of the car. Solve the system:

y  1231t 15,905

y  1700t  500

Substitute 1231t  15,905 for y in the second equation and solve for t

1231t 15,905  1700t  500

1231t 15,405  1700t

15,405  2931t

5.26  t

According to the models, the student will save enough money to buy the 2011 Honda Accord in approximately 5.26 years, or in 2012  5.26  2017 .

c. The graph intersection confirms the calculations.

d. The calculation is an overestimate. If interest is being made, then the amount of money in the student’s account would increase at a faster rate over time. Therefore, the amount of time it would take to save enough money would be less than the prediction in part (b).

14. a. Since the monthly fee for Option 1 is $29.99, the cost of Option 1 increases by a constant $29.99 per month. The function is linear and its slope is 29.99. The intercept is (0,149.99) , since at t = 0 the cost is an initial investment of $149.99. The function for the cost of this is therefore

. Similar work with the Option 2 information gives us

 dollars dollars  dollars

The unit analysis for the other equation is identical.

c. The cost of each option will be equal when the functions are equal. Thus, solve the system

y  29.99t 149.99

y  39.99t  49.99

Substitute 29.99t 149.99 for y in the second equation and solve for t.

t 149.99  39.99t  49.99

t  100

In 10 months the cost of the two options will be identical.

Education, Inc. Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 193

2015 Pearson



y  1.01t

323.57 y  0.36t  240.73

1.01t  323.57  0.36t 

0.65t 

t 



240.73

82.84

127.45

 1.01127.45 

323.57 

f (t)  29.99t 

g(t)  39.99t  49.99

b. f (t)  29.99t 149.99 dollars

149.99

dollars  month dollars  months  dollars

S(t)  1700t 

dollars

29.99

 10

16.

The graph intersection confirms the calculations.

e. Since the total cost of each payment option is identical at month 10, it does not matter which option she chose.

a. Since the number of students who earned a bachelor’s degree in computer science decreased by 4.3 thousand each year, we can model the situation by a linear equation. The slope is 4.3 thousand degrees per year. Since the number of computer science degrees earned in 2009 was 38.0 thousand, the C-intercept is (0,38.0). So, a reasonable model for C is

C(t)  4.3t  38.0 .

Since the number of students who earned a bachelor’s degree in mathematics and statistics increased by 0.5 thousand each year, we can model the situation by a linear equation. The slope is 0.5 thousand degrees per year. Since the number of mathematics and statistics degrees earned in 2009 was 15.5 thousand, the M-intercept is (0,15.5) So, a reasonable model for M is M (t)  0.5t  15.5 .

b. C(t)  4.3t  38.0

M (t)  0.5t  15.5

Substitute 0.5t  15.5 for C (t ) in the first equation and solve for t

0.5t 15.5  4.3t  38.0

4.8t 15.5  38.0 4.8t  22.5

t  22.5  4.69 4.8

Substitute 4.69 for t in the equation

M (t)  0.5t  15.5 and solve for M (t)

M (t)  0.5(4.69)  15.5  17.85

The approximate solution of the system is (4.69,17.85)

According to the models, the same number of bachelor’s degrees (17.9 thousand each) will be earned in 2014.

c. Find the intersection point using a graphing utility.

18. Find equations to model the ad spending information. Since the rate for each is a constant, these can be modeled by a linear equation. For the spending on Internet ads, the rate of increase per year was $2 million, making the slope 2. The intercept is (0,21) , since we are going to define time in years since 2007. I(t)  2t  21

Follow the same method for the spending on print newspaper ads.

N(t)  6t  42

The cost of spending on each type of ad was equal when the functions are equal. Thus, solve the system

I(t)  2t  21

N(t)  6t  42

Substitute 2t  21 for N(t) in the second equation and solve for t

2t  21  6t  42

8t  21

t  2.625

We estimate that the total spending on Internet ads was equal to the total spending on newspaper ads in the year

2007  (2.625)  2010

I(2.625)  2(2.625)  21  26.25

The total spending on each type of ad was $26.3 million.

20. Let p be percentage of 30–34-year-olds who are married at t years since 2010.

Since the percentage of 30–34-year-olds who are married has decreased by 0.9 percentage point each year, we can model the situation by a linear equation. The slope is 0.9 percentage point per year. Since the percentage of 30–34-year-olds who are married was 60.5% in 2010, the p-intercept is 0, 60.5 So, a reasonable model is p  0.9t  60.5

By the same reasoning as for the percentage of 30–34-year-olds who are married, the model for percentage of 30–34-year-olds who have never married is p  0.8t  31.8

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 195

To predict when the percentage of 30–34-yearolds who are married will equal the percentage of 30–34-year-olds who have never married, we solve the system:

Substitute 0.8t  31.8 for p in the first equation and solve for t

Substitute 3.56 for t in the equation p

78 and solve for p

Substitute 16.88 for t in the equation

0.8t  31.8 and solve for p.

The approximate solution of the system is (3.56, 49.08) According to the models, the percentage of households with broadband access was equal to the percentage of households with dial-up access (49%) in 2006.

 0.8(16.88)  31.8  45.30

The approximate solution of the system is

(16.88, 45.30)

According to the models, the percentage of 30–34-year-olds who are married will equal the percentage of 30–34-year-olds who have never married (45.3%) in 2027.

22. Answers may vary.

24. We are given that the changes in percentage of

households with broadband Internet access and percentage of households with dial-up Internet accessare approximately linear.

Let t be the number of years since 2002, and p be the percentage of households.

2. Let L be the length and W be the width.

To determine the dimensions of the floor, we solve the system:

0, we have p-intercept is (0, 21). 84 21 63

When t

21, so the

for L in the second equation m

10 2 8  7.875

Therefore, the linear equation for households with broadband Internet access is

7.875t  21.

Dial-up Internet:

solve for w

When t  0, we have p  78, so the Substitute 9.542 for W in the equation

-intercept is 0, 78 .

13 78  10 2 65 8  8.125

1.62(9.542)  15.458

and solve for L m

The width of the floor is approximately 9.54

 0.9t  60.5



0.8t  31.8

0.8

1.7t 

1.7

t 

1.7

t  31.8  0.9t  60.5

31.8  60.5

t  28.7

28.7  16.88



26. 8.125t 



16t  78 

16t 

t 



7.875t  21

3.56



p  8.1253.56  78  49.08

 8.125t

2x  6  21  6  2  6  4 This is a linear expression in one variable.

28.  2x  6  3x 1  2x  6  3x 1  2x  3x  6 1  x  7 This is a linear

expression

one variable. Homework 6.5

L  1.62W P  2L  2W 50  2L  2W 25  L W

Broadband

Internet:

L  1.62

p 

25  L W

1.62

 



Substitute

and

p 

25  1.62W   W 25  2.62W W  25  9.542 2.62

L  1.62W



Elementary and intermediate algebra functions and authentic applications 2nd edition jay lehmann sol

Solution Manual for Elementary and Intermediate Algebra Functions and Authentic Applications 2nd Edition by Lehmann ISBN 0321922727

Chapter 6

Systems of Linear Equations and Systems of Linear Inqualities

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 189

Equation (2)

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 191

Chapter 6: Systems of Linear Equations and Systems of Linear Inqualities 195

(16.88, 45.30)