Elementary algebra graphs and authentic applications 2nd edition lehmann solutions manual

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Chapter 6 Systems of Linear Equations Homework 6.1 2. To be a solution of the system, an ordered pair must satisfy both equations in the system.

Check (3, 5) : y  x  4 y  3 x  2 5 ⱨ   3  4

5 ⱨ 3  3  2

5 ⱨ 1 false 5 ⱨ 11 false Since the ordered pair  3,5  did not satisfy

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

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

Check (2, 5) : 4x  5 y  17

3 x  2 y  16

4  2   5  5  ⱨ 17

3  2   2  5  ⱨ 16

17 ⱨ 17 16 ⱨ 16 true true Since the ordered pair  2, 5  satisfies both

equations, it is a solution of the system.

6. y  4 x  2 y  2 x  4 To begin, we graph both equations in the same coordinate system. y

3 ⱨ 3  7   2

y  4x  2

5

3 ⱨ 3 true 3 ⱨ 23 false Since the ordered pair  7,3 did not satisfy

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

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

7 ⱨ 3  3  2

7 ⱨ 7 true 7 ⱨ 7 true Since the ordered pair  3, 7  satisfies both

equations, 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.

Check (3  1) : 4x  5 y  17 3x  2 y  16 4  3  5  1 ⱨ 17

3  3  2  1 ⱨ 16

17 ⱨ 17 7 ⱨ 16 true false Since the ordered pair  3, 1 did not satisfy

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

Check (4, 2) : 4x  5 y  17 3x  2 y  16 4  4   5  2  ⱨ 17

3  4   2  2  ⱨ 16

6 ⱨ 17 16 ⱨ 16 false true Since the ordered pair  4, 2  did not satisfy

(1, 2) x

5



y  2 x  4

5

The intersection point is 1, 2  . So, the solution is the ordered pair 1, 2  .

2 8. y   x  4 3 5 y  x3 3 To begin, we graph both equations in the same coordinate system. y y

5

2 y  x4 3

(3, 2)



5

5 x 3 3

x

5

The intersection point is  3, 2  . So, the solution is the ordered pair  3, 2  .

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

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Chapter 6: Systems of Linear Equations y 5 x  3 y  9

10. y  x y  x To begin, we graph both equations in the same coordinate system. y

5 3y  2x

5



5

y

x

(3, 2)

x

5 (0, 0)



y

5

The intersection point is  3, 2  . So, the

x

5

solution is the ordered pair  3, 2  .

x

The intersection point is  0, 0  . So, the solution is the ordered pair  0, 0  .

12. 4 x  y  9 2 x  y  3 To begin, we rewrite each equation in slope–intercept form.

4 x  y  9 2 x  y  3 y  2 x  3 y  4x  9 Next, we graph both equations in the same coordinate system. y

5

16. x  2 y  4 3 x  2 y  4 To begin, we rewrite each equation in slope–intercept form. x  2 y  4 3x  2 y  4 1 3 y   x2 y  x2 2 2 Next, we graph both equations in the same coordinate system. y

5

3 x  2 y  4

(2, 1)

4 x  y  9

5



x  2 y  4

5

( 2, 1)



5 5

x

x

The intersection point is  2, 1 . So, the solution is the ordered pair  2, 1 .

2 x  y  3

The intersection point is  2,1 . So, the solution is the ordered pair  2,1 .

14. 3 y  2 x 5 x  3 y  9 To begin, we rewrite each equation in slope–intercept form.

3y  2x 5 x  3 y  9 2 5 y x y  x3 3 3 Next, we graph both equations in the same coordinate system.

18. 6 y  10 x  18 3y  x  9 To begin, we rewrite each equation in slope–intercept form.

6 y  10 x  18 3y  x  9 5 1 y   x 3 y  x3 3 3 Next, we graph both equations in the same coordinate system. y

5 (3, 2)



3y  x  9

5 5

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x

6 y  10 x  18

163

164

ISM: Elementary Algebra The intersection point is  3, 2  . So, the

26. We use “intersect” to find the solution  3, 4  .

solution is the ordered pair  3, 2  .

20. x  0 y0 To begin, we graph both equations in the same coordinate system. y

5 y0 x

(0, 0)

5

 5

28. We first write both equations in slope–intercept form.

2 7 y  3x  7 y   x 5 3 3 y  x 1 7 We use “intersect” to find the approximate solution  5.48, 1.35  .

x0

The intersection point is  0, 0  . So, the solution is the ordered pair  0, 0  .

22. y  x  2 1 y 2 To begin, we rewrite each equation in slope–intercept form.

1 y  x2 x y 2 y  2x Next, we graph both equations in the same coordinate system. x

y

5

x

1 y 2 y  x2

5

 ( 2, 4)

30. We first write both equations in slope–intercept form.

12 x  9 y  18  x  4 y  20 4 1 y   x2 y   x5 3 4 We use “intersect” to find the approximate solution  6.46, 6.62  .

3 x 1 5 3x  5 y  5 We first write both equations in slope–intercept form.

3 3x  5 y  5 y  x 1 5 3 y  x 1 5 The two equations are equivalent, so the graph 3 of y  x  1 and the graph of 3 x  5 y  5 5 are the same line. The solution set of the system is the infinite set of ordered pairs 3 represented by points on the line y  x  1 .

5 The system is dependent.

32. y 

x

5

The intersection point is  2, 4  . So, the solution is the ordered pair  2, 4  .

24. We use “intersect” to find the approximate solution  2.23, 3.51 .

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Chapter 6: Systems of Linear Equations y  3x  2

34.

3x  2 y  2 We first write both equations in slope–intercept form. y  3x  2 3x  2 y  2 3 y  x 1 2 Next, we graph both equations in the same coordinate system. y y  3x  2

5

3x  2 y  2

5



165

38. n  16.98t  395.66 n  29.87t  1369.89

The models predict that the number of daily morning papers equaled the number of daily evening papers (749 of each) in the year 1980  21  2001 .

40. a. Using the regression feature of a graphing calculator, we obtain the model r  0.16t  54.17 .

x

( 2, 4)

5

The intersection point is  2, 4  . So, the solution is the ordered pair  2, 4  .

2 x2 3 2x  3y  9 We first write both equations in slope–intercept form. y  23 x  2 2x  3y  9

36. y 

y  23 x  3

b. Using the regression feature of a graphinc calculator, we obtain the model r  0.10t  45.48 .

c. r  0.16t  54.17 r  0.10t  45.48

Next, we graph both equations in the same coordinate system. y

5 y



2 x2 3 2x  3 y  9

5

x

5

Since the distinct lines have equal slopes, the lines are parallel. Parallel lines do not intersect, so there are no ordered pairs that satisfy both equations. The solution set is the empty set. The system is inconsistent.

The models predict that the women’s record time will equal the men’s record time (31.00 seconds) in the year 1900  145  2045 .

42. The solution of the system is the ordered pair that corresponds to the intersection point of the graphs. The graphs appear to intersect at approximately (–1.8, 2.6), so (–1.8, 2.6) is the approximate solution of the system.

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166

ISM: Elementary Algebra

44. First estimate the slope of the lines.

Blue: The line appears to pass through the points  3,3 and  0,5 . The slope is m

53 2  .

0   3 3

Red: The line appears to pass through the points  3, 3 and  0, 4  . The slope is m

4   3 0   3



1 1  .

3 3

Next, use the slopes to find additional points on each line.

Blue  6, 1  9, 1  12, 3  15, 5 

Red  6,  2   9, 1  12, 0   15, 1

Since the point  9, 1 lies on both lines, the ordered pair  9, 1 is the solution of the system.

46. The input 2 leads to the output 11 for both equations. Therefore, the ordered pair  2,11

50. Answers may vary. Example: y  2 x  10 3x  5 y  22 52. No. Answers may vary. Example: Two lines can intersect in at most one point, so a system of linear equations can have at most one solution.

54. Answers may vary. Example: An ordered pair solution of an equation in two variables is a point on the graph of the equation. A solution of a system of linear equations in two variables is an ordered pair that satisfies each equation in the system.

That is, an ordered pair that represents a point that is on the graph of each equation in the system. Such an ordered pair is an intersection point of the graphs.

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

3 3x  7 y  14 y  x2 7 3 y  x2 7 The lines have the same slope and the same y-intercepts. Therefore, the lines are not parallel; they are the same line.

is the solution of the system.

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

48. a. Answers may vary. Any point on the graph of y  4 x  5 except 1, 1 will work.

c. Since the lines have the same slope and the same y-intercepts, there are an infinite number of intersection points. Therefore, the system has an infinite number of solutions. Any point on the graph of 3 y  x  2 is a solution of the system.

7

b. Answers may vary. Any point on the graph of y  3 x  2 except 1, 1 will work. c. Solve the system by graphing. y

5



y  4x  5

(1, 1) 5

5

58. Answers may vary. Example: 2 x  y  6  x  3 y  2

x

60.

y  3 x  2

 2 x  5    3x  5  2 x  5  3x  5  2 x  3x  5  5

There is only one intersection point, so the only ordered pair that satisfies both equations is 1, 1 .

 5 x  10 This is a linear expression in one variable.

d. Answers may vary.

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Chapter 6: Systems of Linear Equations 62. The equation y  3 x  5 is in slope–intercept form, so the slope is m  3 and the y-intercept is  0, 5  . We first plot  0, 5  .

From this point we move 1 unit to the left and 3 units up, plotting the point  1, 2  . We then draw the line that connects the two points, extending in both directions. y  3x  5 y

5

x

4 x  3x  12  13 x  12  13

The solution is 1, 3 .

(1, 2)

5 (0, 5)

This is a linear equation in two variables.

Homework 6.2 2.

6. 4 x  3 y  13 y  x4 Substitute x  4 for y in the first equation and solve for x.

4 x  3  x  4   13

x 1 Substitute 1 for x in the equation y  x  4 and solve for y. y  1  4  3

5



167

y  4x 2 x  y  18 Substitute 4x for y in the second equation and solve for x.

2 x   4 x   18 6 x  18 x3 Substitute 3 for x in the equation y  4 x and solve for y. y  4  3  12

The solution is  3,12  .

4. x  2 y  4 x  3y 1 Substitute 3 y  1 for x in the first equation and solve for y.  3 y  1  2 y  4 3y 1  2 y  4 5 y 1  4 5y  5 y 1 Substitute 1 for y in the equation x  3 y  1 and solve for x. x  3 1  1  3  1  2

8. 3x  2 y  9 x  1 2 y Substitute 1  2 y for x in the first equation and solve for y.

3 1  2 y   2 y  9 3  6 y  2 y  9 3  4 y  9 4 y  12 y3 Substitute 3 for y in the equation x  1  2 y and solve for x. x  1  2  3  1  6  5

The solution is  5,3 .

10. 4 x  3 y  2  0 y  1  3x Substitute 1  3x for y in the first equation and solve for x.

4 x  3 1  3 x   2  0 4x  3  9x  2  0 5 x  5  0 5 x  5 x 1 Substitute 1 for x in the equation y  1  3 x and solve for y. y  1  3 1  1  3  2

The solution is 1, 2  .

The solution is  2,1 .

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168

12.

ISM: Elementary Algebra x  2 y  5 7 x  2 y  13  0 Substitute 2 y  5 for x in the second equation and solve for y.

7  2 y  5   2 y  13  0 14 y  35  2 y  13  0 12 y  48  0 12 y  48 y4 Substitute 4 for y in the equation x  2 y  5 and solve for x. x  2  4   5  8  5  3

The solution is  3, 4  .

14. y  4 x  2 y  3 x  5 Substitute 3x  5 for y in the first equation and solve for x. 3x  5  4 x  2 7 x  5  2 7 x  7 x  1 Substitute 1 for x in the equation y  3 x  5 and solve for y. y  3  1  5  3  5  2

The solution is  1, 2  .

16. y  x y  x Substitute  x for y in the first equation and solve for x. x  x 2 x  0 x0 Substitute 0 for x in the equation y   x and solve for y. y   0  0

The solution is  0, 0  .

18. y  5  x  1 y  2  x  2

Substitute 5  x  1 for y in the second

5  x  1  2  x  2  5x  5  2x  4 3x  5  4 3x  9 x3 Substitute 3 for x in the equation y  5  x  1

and solve for y. y  5  3  1  5  2   10 The solution  3,10  .

20. y  1.45 x  6.18 y  2.63 x  2.73 Substitute 2.63 x  2.73 for y in the first equation and solve for x.

2.63x  2.73  1.45 x  6.18 4.08 x  2.73  6.18 4.08 x  3.45 3.45 x  0.846 4.08 Substitute 0.846 for x in the equation y  2.63 x  2.73 and solve for y. y  2.63  0.846   2.73  4.955

The approximate solution is  0.85, 4.96  .

22. y  0.51x  2.64 y  2.79 x  5.94 Substitute 2.79 x  5.94 for y in the first equation and solve for x. 2.79 x  5.94  0.51x  2.64 2.28 x  5.94  2.64 2.28 x  8.58 8.58 x  3.763 2.28 Substitute 3.763 for x in the equation y  2.79 x  5.94 and solve for y. y  2.79  3.763  5.94  4.559

The approximate solution is  3.76, 4.56  .

24. y  2.49 x  6.17 y  4.81x  3.45 Substitute 4.81x  3.45 for y in the first equation and solve for x.

equation and solve for x.

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Chapter 6: Systems of Linear Equations 4.81x  3.45  2.49 x  6.17 7.3 x  3.45  6.17 7.3 x  9.62 9.62  1.318 7.3 Substitute 1.318 for x in the equation y  4.81x  3.45 and solve for y. x

y  4.81 1.318   3.45  2.890

The approximate solution is  1.32, 2.89  .

26. 3x  y  14 2 x  7 y  22 Solve the first equation for y. 3x  y  14 y  3 x  14 Substitute 3x  14 for y in the second equation and solve for x.

2 x  7  3 x  14   22 2 x  21x  98  22 19 x  98  22 19 x  76 x4 Substitute 4 for x in the equation y  3 x  14 and solve for y. y  3  4   14  12  14  2

The solution is  4, 2  .

28. 2 x  3 y  1 x  3 y  5 Solve the second equation for x. x  3 y  5 x  3 y  5 Substitute 3 y  5 for x in the first equation and solve for y.

2  3 y  5   3 y  1 6 y  10  3 y  1 3 y  10  1 3 y  9 y  3 Substitute 3 for y in the equation x  3 y  5 and solve for x. x  3  3  5  9  5  4

169

30. 6 x  5 y  8 x  3 y  14 Solve the second equation for x. x  3 y  14 x  3 y  14 Substitute 3 y  14 for x in the first equation and solve for y.

6  3 y  14   5 y  8 18 y  84  5 y  8 23 y  84  8 23 y  92 y4 Substitute 4 for y in the equation x  3 y  14 and solve for x. x  3  4   14  12  14  2

The solution is  2, 4  .

32. 4 x  3 y  7 3 x  y  9 Solve the second equation for y.

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

4 x  3  3 x  9   7 4 x  9 x  27  7 5 x  27  7 5 x  20 x  4 Substitute 4 for x in the equation y  3 x  9 and solve for y. y  3  4   9  12  9  3

The solution is  4, 3 .

34. 4 x  5 y  2 3x  y  7 Solve the second equation for y.

3x  y  7 y  3x  7 Substitute 3x  7 for y in the first equation and solve for x.

The solution is  4, 3 .

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170

ISM: Elementary Algebra 4 x  5  3 x  7   2 4 x  15 x  35  2 11x  35  2 11x  33 x  3 Substitute 3 for x in the equation y  3 x  7 and solve for y. y  3  3  7  9  7  2

42. 3x  12 y  6 x  4 y  2 Substitute 4 y  2 for x in the first equation and solve for y.

3  4 y  2   12 y  6 12 y  6  12 y  6 6  6 true The result is a true statement, so the system is dependent. The solution is the infinite set of ordered pairs that satisfy the equation x  4 y  2 .

The solution is  3, 2  .

36.

x  2 y 1 4 x  8 y  4 Substitute 2 y  1 for x in the second equation and solve for y.

4  2 y  1  8 y  4 8 y  4  8 y  4 4  4 true The result is a true statement, so the system is dependent. The solution is the infinite set of ordered pairs that satisfy the equation x  2 y 1.

38. 3x  2 y  10 y  4  2x Substitute 4  2x for y in the first equation and solve for x. 3x  2  4  2 x   10 3x  8  4 x  10 x  8  10 x  2 Substitute 2 for x in the equation y  4  2 x and solve for y. y  4  2  2   4  4  8

The solution is  2,8  .

40.

y  2x  5 4 x  2 y  6 Substitute 2 x  5 for y in the second equation and solve for x. 4 x  2  2 x  5   6 4 x  4 x  10  6 10  6 false The result is a false statement, so the system is inconsistent. The solution is the empty set,  .

y  4x 1

44.

8 x  2 y  5 Substitue 4 x  1 for y in the second equation and solve for x. 8 x  2  4 x  1  5 8 x  8 x  2  5 2  5 false The result is a false statement, so the system is inconsistent. The solution is the empty set,  .

46. We begin by finding the equations for each table.

Table 15: The input x  0 yields the output y  61 , so

the y-intercept is  0, 61 . m

67  73 6   6 , so the slope is 6 . 1   2  1

The first equation is y  6 x  61 .

Table 16: The input x  0 yields the output y  9 , so the y-intercept is  0, 9  . m

13   17  1   2 



4  4 , so the slope is 4.

1

The second equation is y  4 x  9 .

Now we solve the system: y  6 x  61 y  4x  9 Substitute 4 x  9 for y in the first equation and solve for x.

4 x  9  6 x  61 10 x  9  61 10 x  70 x7 Substitute 7 for x in the equation y  4 x  9 and solve for y.

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Chapter 6: Systems of Linear Equations y  4  7   9  28  9  19

171

Substitute 2 for y in the equation x  3 y  9 and solve for x.

The solution is  7,19  .

x  3  2   9  6  9  3

48. A: This point is the origin so it has coordinates  0, 0  .

B: This point is the y-intercept for line l2 . Let x  0 in the equation and solve for y.

2 y   0  6 2y  6

The solution is  3, 2  . b. Solve the first equation for y.

2x  y  8 y  2 x  8 Substitute 2 x  8 for y in the second equation and solve for x. x  3  2 x  8   9 x  6 x  24  9

y3

The coordinates of the point are  0,3 .

C: This point is the intersection of the two lines.

Therefore, it is the solution of the system: y  2 x  8 2y  x  6 Substitute 2 x  8 for y in the second equation and solve for x.

2  2 x  8   x  6 4 x  16  x  6 5 x  16  6 5 x  10 x2 Substitute 2 for x in the equation y  2 x  8 and solve for y. y  2  2   8  4  8  4

The coordinates of the point are  2, 4  .

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

0  2 x  8 x4 The coordinates of the point are  4, 0  .

50. a. Solve the second equation for x. x  3y  9 x  3 y  9 Substitute 3 y  9 for x in the first equation and solve for y.

2  3 y  9   y  8 6 y  18  y  8 5 y  18  8

5 x  24  9 5 x  15 x3 Substitute 3 for x in the equation y  2 x  8 and solve for y. y  2  3  8  6  8  2

The solution is  3, 2  . 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. 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.

56. a. y  y1  m  x  x1  y  4  3  x  3 y  4  3x  9 y  3x  5

5 y  10 y2

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172

ISM: Elementary Algebra b. y  y1  m  x  x1 

Homework 6.3

y  4  2  x  3  y  4  2 x  6 y  2 x  10

c. Our system of equations is y  3x  5 y  2 x  10 Substitute 2 x  10 for y in the first equation and solve for x. 2 x  10  3 x  5 5 x  10  5 5 x  15 x3 Substitute 3 for x in the equation y  2 x  10 and solve for y.

2. 4 x  5 y  11 Equation (1) 4 x  3 y  13 Equation (2) 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. 4 x  5 y  11 4 x  3 y  13 8 y  24 y3 Substitute 3 for y in equation (2) and solve for x.

4 x  3  3  13 4 x  9  13 4x  4

y  2  3  10  6  10  4

x 1 The solution is 1,3 .

The solution is  3, 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 in the system.

58.

4. 3 x  5 y  19 Equation (1) 2 x  5 y  4 Equation (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.

3 x  5 y  19 2 x  5 y  4 5x  15 x3 Substitute 3 for x in equation (1) and solve for y.

3  3  5 y  19

7  3 x  5 7  5  3 x  5  5 2  3 x 2 3 x  3 3 2  x 3

9  5 y  19 5 y  10 y2

2 The solution is  .

3 This is a linear equation in one variable.

60. 9 x  2 y  3 x  5  9 x  3x  2 y  5   9  3 x  2 y  5  6x  2 y  5 This is a linear expression in two variables.

The solution is  3, 2  .

6.

x  4y  5

Equation (1)

5 x  2 y  11 Equation (2) To eliminate the y terms, we multiply both sides of equation (2) by 2 , yielding the system: x  4y  5 10 x  4 y  22

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Chapter 6: Systems of Linear Equations 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. x  4y  5 10 x  4 y  22

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. 8 x  14 y  50 8 x  3 y  1

9 x  27

17 y  51

x  3 Substitute 3 for x in equation (1) and solve for y.  3  4 y  5

8.

y  3 Substitute 3 for y in equation (2) and solve for x.

8 x  3  3  1

4 y  8

8 x  9  1

y  2

8x  8

The solution is  3, 2  .

x 1 The solution is 1, 3 .

3 x  4 y  5

173

Equation (1)

5 x  2 y  17 Equation (2) To eliminate the y terms, we multiply both sides of equation (2) by 2 , yielding the system: 3x  4 y  5 10 x  4 y  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.

3 x  4 y  5 10 x  4 y  34

12.

6 x  5 y  14

Equation (1)

4 x  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: 12 x  10 y  28 12 x  21 y  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.

12 x  10 y  28 12 x  21y  6

13x  39 x  3 Substitute 3 for x in equation (2) and solve for y. 5  3  2 y  17 15  2 y  17

11y  22 y2 Substitute 2 for y in equation (1) and solve for x.

6 x  5  2   14 6 x  10  14

2y  2

6 x  24

y 1

x  4 The solution is  4, 2  .

The solution is  3,1 .

10. 4 x  7 y  25 Equation (1) 8 x  3 y  1 Equation (2) To eliminate the x terms, we multiply both sides of equation (1) by 2 , yielding the system: 8 x  14 y  50 8 x  3 y  1

14. 5 x  2 y  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: 15 x  6 y  24 4 x  6 y  2

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174

ISM: Elementary Algebra Substitute 1 for y in equation (2) and solve for x.

5 x  5 1  10  0

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.

15 x  6 y  24 4 x  6 y  2

5 x  15  0 5 x  15 x3 The solution is  3,1 .

11x  22 x2 Substitute 2 for x in equation (1) and solve for y.

5  2  2 y  8

20.

10  2 y  8 2 y  2 y 1

The solution is  2,1 .

16. y  2 x  7  2 x  y  7 Equation (1) y   x  8  x  y  8 Equation (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.

2 x  y  7 x  y  8

6x  8 y  8 23 y  46 y  2 Substitute 2 for y in equation (1) and solve for x.

2 x  5  2   18

3x  15 x  5 Substitute 5 for x in equation (1) and solve for y. y  2  5   7  10  7  3

2 x  10  18 2x  8 x4 The solution is  4, 2  .

The solution is  5, 3 .

18. 3  2 x  9 y  2 x  9 y  3 Eq. (1) 5 x  5 y  10  0  5 x  5 y  10 Eq. (2) 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: 10 x  45 y  15 10 x  10 y  20 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. 10 x  45 y  15 10 x  10 y  20 35 y  35

1 5 9 x  y   2 x  5 y  18 Equation (1) 2 4 2 3 1 1 x  y   3 x  4 y  4 Equation (2) 8 2 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: 6 x  15 y  54 6x  8 y  8 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. 6 x  15 y  54

22.

4 x  6 y  10

Equation (1)

6 x  9 y  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: 12 x  18 y  30 12 x  18 y  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.

12 x  18 y  30 12 x  18 y  30 0  0 true

y 1

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Chapter 6: Systems of Linear Equations 30 x  36 y  15

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 4 x  6 y  10 .

24.

3x  2 y  5

0  3 false The result is a false statement. Therefore, the system is inconsistent. The solution set is the empty set,  .

26.

8 x  3 y  2

Equation (1)

5 x  12 y  29 Equation (2) To eliminate the y terms, we multiply both sides of equation (1) by 4, yielding the system: 32 x  12 y  8 5 x  12 y  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.

32 x  12 y  8 5 x  12 y  29 37x  37 x  1 Substitute 1 for x in equation (1) and solve for y.

8  1  3 y  2 8  3 y  2 3 y  6 y  2

The solution is  1, 2  .

28.

10 x  12 y  5

30 x  36 y  16 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.

30 x  36 y  15 30 x  36 y  16

Equation (1)

12 x  8 y  17 Equation (2) To eliminate the x terms, we multiply both sides of equation (1) by 4, yielding the system: 12 x  8 y  20 12 x  8 y  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.

12 x  8 y  20 12 x  8 y  17

Equation (1)

15 x  18 y  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:

175

0  1 false The result is a false statement. Therefore, the system is inconsistent. The solution set is the empty set,  .

30.

9 x  6 y  15

Equation (1)

12 x  8 y  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: 36 x  24 y  60 36 x  24 y  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.

36 x  24 y  60 36 x  24 y  60 0  0 true 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 9 x  6 y  15 .

32. y  1.28 x  2.05  1.28 x  y  2.05 Eq. (1) y  3.94 x  8.83  3.94 x  y  8.33 Eq. (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. 1.28 x  y  2.05 3.94 x  y  8.33 2.66 x  10.38 10.38  3.902 x 2.66 Substitute 3.902 for x in equation (1) and solve for y. y  1.28  3.902   2.05  7.04

The approximate solution is  3.90, 7.04  .

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176

ISM: Elementary Algebra

34. y  3.28 x  1.43  3.28 x  y  1.43 Eq. (1) y  0.56 x  6.72 0.56 x  y  6.72 Eq. (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.

3.28 x  y  1.43 0.56 x  y  6.72 2.72 x  5.29 5.29 x  1.945 2.72 Substitute 1.945 for x in equation (1) and solve for y. y  3.28 1.945   1.43  7.810

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

6  3  y  13 18  y  13 y5

The solution is  3,5  .

40. 4 x  5 y  23 y  1 2x Substitute 1  2x for y in the first equation and solve for x.

4 x  5 1  2 x   23 4 x  5  10 x  23 14 x  5  23 14 x  28 x2 Substitute 2 for x in the equation y  1  2 x and solve for y. y  1  2  2   1  4  3

The approximate solution is 1.95, 7.81 .

36. y  1.64 x  5.07  1.64 x  y  5.07 Eq. (1) y  2.57 x  3.39  2.57 x  y  3.39 Eq. (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.

1.64 x  y  5.07 2.57 x  y  3.39 4.21x  8.46 8.46 x  2.010 4.21 Substitute 2.010 for x in equation (1) and solve for y. y  1.64  2.010   5.07  8.366

The approximate solution is  2.01,8.37  .

38.

6 x  y  13 Equation (1) 3x  2 y  19 Equation (2) To eliminate the y terms, we multiply both sides of equation (1) by 2, yielding the system: 12 x  2 y  26 3x  2 y  19 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.

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

The solution is  2, 3 .

42.

5 x  2 y  7 Equation (1) 4 x  3 y  7 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: 15 x  6 y  21 8 x  6 y  14 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.

15 x  6 y  21 8 x  6 y  14 7x  7 x 1 Substitute 1 for x in equation (1) and solve for y.

5 1  2 y  7 5 2y  7 2 y  2 y  1

The solution is 1, 1 .

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Chapter 6: Systems of Linear Equations 44.

3 x  8 y  4 Equation (1) x  5 y  6 Equation (2) To eliminate the x terms, we multiply both sides of equation (2) by 3 , yielding the system: 3x  8 y  4 3 x  15 y  18 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.

3x  8 y  4 3 x  15 y  18 7 y  14 y2 Substitute 2 for y in equation (1) and solve for x.

3x  8  2   4 3 x  16  4 3 x  12 x4 The solution is  4, 2  .

46. y  4 x  9 y  2x  3 Substitute 2 x  3 for y in the first equation and solve for x.

2 x  3  4 x  9 6 x  3  9 6 x  12 x  2 Substitute 2 for x in the equation y  2 x  3 and solve for y. y  2  2   3  4  3  1

The solution is  2, 1 .

48. 3x  2 y  19  3 x  2 y  19 Eq. (1) 5 y  4 x  10  4 x  5 y  10 Eq. (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: 12 x  8 y  76 12 x  15 y  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.

177

12 x  8 y  76 12 x  15 y  30 23 y  46 y2 Substitute 2 for y in equation (1) and solve for x. 3x  2  2   19 3 x  4  19 3x  15 x  5 The solution is  5, 2  .

50. 4  2 x  7   3 y  7

7 x  5  2 y  3  3 Begin by writing each equation in standard form.

4 2x  7  3y  7 7 x  5  2 y  3  3 8 x  28  3 y  7 7 x  10 y  15  3 8 x  3 y  35 7 x  10 y  18 This yields the system: 8 x  3 y  35 Equation (1) 7 x  10 y  18 Equation (2) To eliminate the y terms, we multiply both sides of equation (1) by 10 and both sides of equation (2) by 3, yielding the system: 80 x  30 y  350 21x  30 y  54 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.

80 x  30 y  350 21x  30 y  54 101x  404 x4 Substitute 4 for x in equation (1) and solve for y.

8  4   3 y  35 32  3 y  35 3y  3 y 1

The solution is  4,1 .

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178

52.

ISM: Elementary Algebra y  2 x  7 5x  2 y  4 Graphing by hand: Begin by writing each equation in slope–intercept form. y  2 x  7 5x  2 y  4 2 y  5 x  4 5 y  x2 2 Next we graph both equations in the same coordinate system.

4 x  2 y  14 5x  2 y  4 9x  18 x2 Substitute 2 for x in equation (1) and solve for y.

2  2  y  7 y3

The solution is  2,3 .

Preference may vary.

y

5

5x  2 y  4

54.

(2, 3)



5 5

x y  2 x  7

The intersection point is  2,3 . So, the ordered pair  2,3 is the solution.

Substitution: y  2 x  7 5x  2 y  4 Substitute 2 x  7 for y in the second equation and solve for x.

5 x  2  2 x  7   4 5 x  4 x  14  4 9 x  14  4 9 x  18 x2 Substitute 2 for x in the equation y  2 x  7 and solve for y. y   2  2   7  4  7  3

The solution is  2,3 .

Elimination: 2 x  y  7 Equation (1) 5 x  2 y  4 Equation (2) To eliminate the y terms, we multiply both sides of equation (1) by 2, yielding the system: 4 x  2 y  14 5x  2 y  4 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.

4 x  7 y  15 5x  3 y  7

Equation (1) 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: 20 x  35 y  75 20 x  12 y  28 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.

20 x  35 y  75 20 x  12 y  28  47 y  47 y  1 Substitute 1 for y in equation (1) and solve for x.

4 x  7  1  15 4 x  7  15 4x  8 x2 The solution is  2, 1 .

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: 12 x  21 y  45 35 x  21 y  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.

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Chapter 6: Systems of Linear Equations 12 x  21 y  45 35 x  21 y  49 47 x  94 x2 Substitute 2 for x in equation (2) and solve for y.

5  2  3 y  7 10  3 y  7 3 y  3 y  1

c. The results are the same.

56. We begin by finding the equations for each table.

Table 19: The input x  0 yields the output y  76 , so

the y-intercept is  0, 76  . 80   84  1   2 



4  4 , so the slope is 4.

1

The first equation is y  4 x  76 .

Table 20: The input x  0 yields the output y  85 , so the y-intercept is  0,85  . m

88  91 3   3 , so the slope is 3 . 1   2  1

The second equation is y  3 x  85 .

Now we solve the system: y  4 x  76 y  3 x  85 Writing in standard form, we get: 4 x  y  76 Equation (1) 3 x  y  85 Equation (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.

4 x  y  76 3x  y  85 7x  161 x  23

Substitute 23 for x in equation (2) and solve for y.

3  23  y  85 69  y  85 y  16

The solution is  23,16  .

58. A: This point is the origin so it has coordinates  0, 0  .

B: This point is the y-intercept for line l1 . Let x  0 in the equation and solve for y. 2  0   5 y  10

The solution is  2, 1 .

m

179

5 y  10 y2

The coordinates of the point are  0, 2  .

C: This point is the intersection of the two lines.

Therefore, it is the solution of the system: 2 x  5 y  10 Equation (1) 7 x  3 y  47 Equation (2) To eliminate the x terms, we multiply both sides of equation (1) by 7 and both sides of equation (2) by 2, yielding the system: 14 x  35 y  70 14 x  6 y  94 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. 14 x  35 y  70 14 x  6 y  94 41 y  164 y4 Substitute 4 for y in equation (1) and solve for x. 2 x  5  4   10 2 x  20  10 2 x  10 x5 The coordinates of the point are  5, 4  .

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180

ISM: Elementary Algebra D: This point is the x-intercept of line l2 . Let y  0 in the equation and solve for x.

7 x  3  0   47

d. The equation of the line is y  3x  2 . e. Graph the equation and check if the points  1,5  and  2, 4  are on the graph.

7 x  47 47 x 7  47  The coordinates of the point are  , 0  .  7 

60. Answers may vary. Example: If elimination results in 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.

66. 5 x  4 y  8

x-intercept: 5 x  4  0   8 5x  8 x

 

The x-intercept is 85 , 0 .

y-intercept: 5  0   4 y  8

62. Answers may vary 64. a.

4y  8 y2

y  mx  b

The y-intercept is  0, 2  .

5  m  1  b

We plot the intercepts and connect them with a line, extending the line in both directions.

m  b  5

b.

8 5

y  mx  b

y

4  m  2   b

5

2m  b  4

(0, 2)

c. m  b  5 Equation (1) 2m  b  4 Equation (2) To eliminate the b terms, we multiply both sides of equation (1) by 1 , yielding the system: m  b  5 2m  b  4 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. m  b  5 2m  b  4 3m  9 m  3 Substitute 3 for m in equation (1) and solve for b.   3  b  5 3b  5 b2 The solution of the system is  m, b   3, 2 .



8 ,0 5

5

x

5

This is a linear equation in two variables.

68. 5 x  4 y  8 Equation (1) 3 x  2 y  2 Equation (2) To eliminate the y terms, we multiply both sides of equation (2) by 2 , yielding the system: 5x  4 y  8 6 x  4 y  4 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.

5x  4 y  8 6 x  4 y  4 x  4 x  4

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Chapter 6: Systems of Linear Equations Substitute 4 for x in equation (2) and solve for y.

3  4   2 y  2

181

6. a. Start by creating a scatter diagram of the data.

12  2 y  2 2 y  14 y7

The solution is  4, 7  .

This is a system of linear equations in two variables.

The data appear fairly linear so we use the regression feature of a graphing utility to obtain the line of best fit.

Homework 6.4 2. n  16.98t  395.66 n  29.87t  1369.89 Substitute 29.87t  1369.89 for n in the first equation and solve for t. 29.87t  1369.89  16.98t  395.66 46.85t  1369.89  395.66 46.85t  974.23 974.23 t  20.79 46.85 Substitute 20.79 for t in the equation n  29.87t  1369.89 and solve for n. n  29.87  20.79   1369.89  748.89

The approximate solution of the system is  20.79, 748.89  .

According to the models, there were the same number of morning newspapers as evening newspapers (749 each) in 2001.

4. r  0.16t  54.17 r  0.10t  45.48 Substitute 0.10t  45.48 for r in the first equation and solve for t. 0.10t  45.48  0.16t  54.17 0.06t  45.48  54.17 0.06t  8.69 8.69 t  144.83 0.06 Substitute 144.83 for t in the equation r  0.10t  45.48 and solve for r. r  0.10 144.83  45.48  31.00

The solution of the system is 144.83, 31.00  .

According to the models, the women’s winning time will equal the men’s winning time (31.00 seconds) in 2045.

A reasonable model for the data is V  0.66t  28.7 . b. Start by creating a scatter diagram of the data.

The data appear fairly linear so we use the regression feature of a graphing utility to obtain the line of best fit.

A reasonable model for the data is V  1.24t  31.22 . c. To estimate when the number of stores will be equal, we solve the system: V  0.66t  28.7 V  1.24t  31.22 Substitute 1.24t  31.22 for V in the first equation and solve for t. 1.24t  31.22  0.66t  28.7 1.90t  31.22  28.7 1.90t  2.52 2.52  1.33 t 1.90 Substitute 1.33 for t in the equation V  1.24t  31.22 and solve for V. V  1.24 1.33   31.22  29.57

The approximate solution of the system is 1.33, 29.57  .

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182

ISM: Elementary Algebra According to the models, the average weekly viewing time of ad-supported cable was equal to the average weekly viewing time of broadcast television (29.6 hours each) in 2001.

8. a. Start by creating a scatter diagram of the data.

Substitute 17.99 for t in the equation c  4.65t  28.91 and solve for c. c  4.65 17.99   28.91  112.56 The approximate solution of the system is 17.99, 112.56  .

According to the models, North America’s energy consumption was equal to energy consumption in Asia and Oceania (112.56 quadrillion Btu) in 1998.

10. a. Start by creating a scatter diagram of the data.

The data appear fairly linear so we use the regression feature of a graphing utility to obtain the line of best fit.

The data appear fairly linear so we use the regression feature of a graphing utility to obtain the line of best fit.

A reasonable model for the data is c  1.5t  85.57 . b. Start by creating a scatter diagram of the data.

A reasonable model for the data is n  0.21t  11.58 . b. Start by creating a scatter diagram of the data.

The data appear fairly linear so we use the regression feature of a graphing utility to obtain the line of best fit.

The data appear fairly linear so we use the regression feature of a graphing utility to obtain the line of best fit.

A reasonable model for the data is c  4.65t  28.91 . c. To estimate when the Btu consumption will be equal, we solve the system: c  1.5t  85.57 c  4.65t  28.91 Substitute 4.65t  28.91 for c in the first equation and solve for t.

4.65t  28.91  1.5t  85.57 3.15t  28.91  85.57 3.15t  56.66 56.66 t  17.99 3.15

A reasonable model for the data is n  0.24t  6.75 . c. The slopes are very close in value. This means that the number of men who live alone changes at about the same rate as the number of women who live alone.

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Chapter 6: Systems of Linear Equations d. Answers may vary. Example: Although the two slopes are close, they are not equivalent. Therefore, the graphs of the two equations will intersect. This means that, at some point in time, the number of men who live alone will equal the number of women who live alone. Because the difference in slopes is very small compared to the difference in the y-intercepts, this event will not happen for a very long time.

12. a. Since the number of students who earned a bachelor’s degree in computer science decreases 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 d-intercept is  0, 38.0  . So, a reasonable model is d  4.3t  38.0 .

b. By the same reasoning as in part (a), the model for bachelor’s degrees earned in mathematics and statistics is d  0.5t  15.5 . c. d  4.3t  38.0 d  0.5t  15.5 Substitute 0.5t  15.5 for d 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 22.5  4.69 t 4.8 Substitute 4.69 for t in the equation d  0.5t  15.5 and solve for d. d  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.

14. a. Since the Miami Marlins’ payroll has increased by $22.5 million each year, we can model the situation by a linear equation.

The slope is $22.5 million per year. Since the payroll in 2010 was $57 million, the p-intercept is  0, 57  . So, a reasonable

model is p  22.5t  57 .

183

b. By the same reasoning as in part (a), the model for the New York Mets payroll is p  18.7t  134.4 . c. p  22.5t  57 p  18.7t  134.4 Substitute 18.7t  134.4 for p in the first equation and solve for t. 18.7t  134.4  22.5t  57 41.2t  134.4  57 41.2t  77.4 77.4 t  1.88 41.2 Substitute 1.88 for t in the equation p  18.7t  134.4 and solve for p. p  18.7 1.88   134.4  99.24

The approximate solution of the system is 1.88, 99.24  .

According to the models, the Marlins’ and Mets’ payrolls ($99.24 million each) were equal in 2012.

16. a. Since a one-year-old Ford Crown Victoria depreciates about $1774 per year, we can model the situation by a linear equation.

The slope is 1774 dollars per year. Since a one-year-old Ford Crown Victoria was worth $18.389 in 2012, the V-intercept is  0, 18,389  . So, a reasonable model is V  1774t  18,389 .

b. By the same reasoning as in part (a), the model for the Dodge Avenger is V  940t  14,205 . c. V  1774t  18,389 V  940t  14,205 Substitute 940t  14,205 for V in the first equation and solve for t. 940t  14,205  1774t  18,389 834t  14,205  18,389 834t  4184 4184 t  5.02 834 Substitute 5.02 for t in the equation V  940t  14,205 and solve for V. V  940  5.02   14,205  9486.20 .

The approximate solution of the system is  5.02, 9486.20  .

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184

ISM: Elementary Algebra According to the models, the two cars will have the same value ($9486) in 2017.

18. 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

22. 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.

Broadband Internet: When t  0 , we have p  21 , so the

p-intercept is  0, 21 .

84  21 63   7.875 10  2 8 Therefore, the linear equation for households with broadband Internet access is p  7.875t  21 .

Dial-up Internet: When t  0 , we have p  78 , so the m

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 .

p-intercept is  0, 78  .

13  78 65   8.125 10  2 8 Therefore, the linear equation for households with dial-up Internet access is p  8.125t  78 .

The two equations yield the system: p  7.875t  21 p  8.125t  78 Substitute 8.125t  78 for p in the first equation and solve for t. 8.125t  78  7.875t  21 16t  78  21 16t  57 57  3.56 t 16 Substitute 3.56 for t in the equation p  8.125t  78 and solve for p. m

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: p  0.9t  60.5 p  0.8t  31.8 Substitute 0.8t  31.8 for p in the first equation and solve for t.

0.8t  31.8  0.9t  60.5 1.7t  31.8  60.5 1.7t  28.7 28.7 t  16.88 1.7 Substitute 16.88 for t in the equation p  0.8t  31.8 and solve for p. p  0.8 16.88   31.8  45.30

p  8.125  3.56   78  49.08

The approximate solution of the system is 16.88, 45.30  .

The approximate solution of the system is  3.56, 49.08 . According to the models, the

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.

20. Answers may vary.

percentage of households with broadband access was equal to the percentage of households with dial-up access (49%) in 2006.

24. 2 x  6  2 1  6  2  6  4

This is a linear expression in one variable.

26.

 2 x  6    3x  1  2 x  6  3x  1  2 x  3x  6  1  x7 This is a linear expression in one variable.

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Chapter 6: Systems of Linear Equations Homework 6.5 2. Let L be the length and W be the width.

P  2 L  2W L  1.62W 50  2 L  2W 25  L  W To determine the dimensions of the floor, we solve the system: L  1.62W 25  L  W Substitute 1.62W for L in the second equation and solve for w.

25  1.62W   W 25  2.62W 25  9.542 2.62 Substitute 9.542 for W in the equation L  1.62W and solve for L. L  1.62  9.542   15.458 W

The width of the floor is approximately 9.54 feet and the length is approximately 15.46 feet.

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

P  2 L  2W L  2.25W 200  2 L  2W 100  L  W To determine the dimensions of the football field, we solve the system: L  2.25W 100  L  W Substitute 2.25W for L in the second equation and solve for W. 100   2.25W   W 100  3.25W 100  30.769 3.25 Substitute 30.769 for W in the equation L  2.25W and solve for L. L  2.25  30.769   69.230 W

The width of the field is about 30.77 yards and the length is about 69.23 yards.

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

P  2 L  2W L  2W  1 28  2 L  2W 14  L  W To determine the dimensions of the table, we solve the system: L  2W  1 14  L  W

185

Substitute 2W  1 for L in the second equation and solve for W. 14   2W  1  W 14  3W  1 15  3W 5 W Substitute 5 for W in the equation L  2W  1 and solve for L. L  2  5   1  10  1  9

The width of the table is 5 feet and the length is 9 feet.

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

P  2 L  2W L  2W  5 112  2 L  2W 56  L  W To determine the dimensions of the rectangle, we solve the system: L  2W  5 56  L  W Substitute 2W  5 for L in the second equation and solve for W. 56   2W  5   W 56  3W  5 51  3W 17  W Substitute 17 for W in the equation L  2W  5 and solve for L. L  2 17   5  34  5  39

The width of the rectangle is 17 inches and the length is 39 inches.

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

P  2 L  2W L  3W  2 148  2 L  2W 74  L  W To determine the dimensions of the rectangle, we solve the system: L  3W  2 74  L  W Substitute 3W  2 for L in the second equation and solve for W. 74   3W  2   W 74  4W  2 76  4W 19  W Substitute 19 for W in the equation L  3W  2 and solve for L. L  3 19   2  57  2  55

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186

ISM: Elementary Algebra The width of the rectangle is 19 yards and the length is 55 yards.

12. Let x be the number of $45 tickets and y be the number of $65 tickets.

There are a total of 10,000 tickets, so our first equation is x  y  10, 000 .

The total revenue is $560,000. The revenue for each ticket type is obtained by multiplying the ticket price by the number of tickets sold at that price. Therefore, our second equation is 45 x  65 y  560, 000 The system is: x  y  10, 000 Equation (1) 45 x  65 y  560, 000 Equation (2) To eliminate the x terms, multiply both sides of equation (1) by 45 , yielding the system: 45 x  45 y  450, 000 45 x  65 y  560, 000 Add the left sides and right sides and solve for y. 45 x  45 y  450, 000 45 x  65 y  560, 000 20y  110, 000 y  5,500 Substitute 5,500 for y in equation (1) and solve for x. x   5,500   10, 000 x  4,500 The theater should sell 4500 of the $45 tickets and 5500 of the $65 tickets.

14. Let x be the number of the Blood Pressures albums sold and y be the number of The Kills EPs sold.

A total of 373 recordings were sold, so our first equation is x  y  373 .

The total revenue was $2647.71. The revenue for each recording is obtained by multiplying the recording price by the number of recordings sold at that price. Therefore, our second equation is 9.99 x  4.95 y  2647.71 The system is: x  y  373 9.99 x  4.95 y  2647.71 Solve the first equation for y. x  y  373 y   x  373 Substitute  x  373 for y in the second equation and solve for x.

9.99 x  4.95   x  373  2647.71 9.99 x  4.95 x  1846.35  2647.71 5.04 x  1846.35  2647.71 5.04 x  801.36 x  159 Substitute 159 for x in the equation y   x  373 and solve for y. y  159  373  214 159 Blood Pressures albums and 214 The Kills EPs were sold.

16. Let x be the number of hardcovers and y be the number of paperbacks.

A total of 800 books were sold, so our first equation is x  y  800 .

The total revenue was $8792. The revenue for each product is obtained by multiplying the product price by the number sold of the product. Therefore, our second equation is 16.99 x  8.99 y  8792 .

The system is: x  y  800 Equation (1) 16.99 x  8.99 y  8792 Equation (2) To eliminate the y terms, multiply both sides of equation (1) by 8.99 , yielding the system: 8.99 x  8.99 y  7192 16.99 x  8.99 y  8792 Add the left sides and right sides and solve for x. 8.99 x  8.99 y  7192 16.99 x  8.99 y  8792 8x  1600 x  200 Substitute 200 for x in equation (1) and solve for y.

200  y  800 y  600 The bookstore sold 200 hardcovers and 66 paperbacks of You Killed Wesley Payne.

18. Let x be the price of balcony seats and y be the price of main-level seats.

Tickets for the balcony are $15 less than tickets for the main-level, so our first equation is x  y  15 .

The total revenue is $100,750. To obtain the revenue for each seat type, we multiply the price of each seat type by the number of seats at that price. Therefore, our second equation is 450 x  1700 y  100, 750 .

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Chapter 6: Systems of Linear Equations The system is: x  y  15 450 x  1700 y  100, 750 Substitute y  15 for x in the second equation and solve for y.

450  y  15   1700 y  100, 750 450 y  6750  1700 y  100, 750 2150 y  6750  100, 750 2150 y  107,500 y  50 Substitute 50 for y in the equation x  y  15 and solve for x. x  50  15  35 Each balcony seat should cost $35 and each main-level seat should cost $50.

20. Let x be the price of general seats and y be the price of reserved seats.

Tickets for the general are $17 less than tickets for the reserved, so our first equation is x  y  17 .

The total revenue is $435,400. To obtain the revenue for each seat type, we multiply the price of each seat type by the number of seats at that price. Therefore, our second equation is 9500 x  3200 y  435, 400 .

The system is: x  y  17 9500 x  3200 y  435, 400 Substitute y  17 for x in the second equation and solve for y.

9500  y  17   3200 y  435, 400 9500 y  161,500  3200 y  435, 400 12, 700 y  161,500  435, 400 12, 700 y  596,900 y  47 Substitute 47 for y in the equation x  y  17 and solve for x. x  47  17  30 Each general seat should cost $30 and each reserved seat should cost $47.

22. a. We find 5% of 4000.  0.05  4000   200

The interest in one year would be $200.

187

c. We find 5% of d.  0.05  d   0.05d

The interest in one year would be 0.05d dollars.

24. a. 0.04  3000   0.12  6000   120  720  840 The total interest would be $840.

b. 0.04  4000   0.12  8000   160  960  1120 The total interest would be $1120.

c. 0.04  x   0.12  2 x   0.04 x  0.24 x  0.28 x The total interest would be 0.28x .

26. Let x be the amount invested in the cdbank.com account and y be the amount invested in the Turner Small Cap Value account.

The total amount invested is $3500 so our first equation is x  y  3500 .

The total interest after one year is $200.

Therefore, our second equation is 0.03 x  0.16 y  200 .

The system is: x  y  3500 0.03 x  0.16 y  200 Solve the first equation for y. x  y  3500 y   x  3500 Substitute  x  3500 for y in the second equation and solve for x.

0.03 x  0.16   x  3500   200 0.03 x  0.16 x  560  200 0.13x  560  200 0.13 x  360 x  2769.23 Substitute 2769.23 for x in the equation y   x  3500 and solve for y. y    2769.23  3500  730.77

He should invest $2769.23 in the cdbank.com account and $730.77 in the Turner Small Cap Value account.

b. We find 5% of 5000.  0.05  5000   250

The interest in one year would be $250.

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28. Let x be the amount invested in the Connecticut Bank CD account and y be the amount invested in the Artisan International account.

The total amount invested is $6500, so our first equation is x  y  6500 .

The total interest after one year is $400, so our second equation is 0.03 x  0.14 y  400 .

The system is: x  y  6500 Equation (1) 0.03 x  0.14 y  400 Equation (2) To eliminate the x terms, multiply both sides of equation (1) by 0.03 , yielding the system: 0.03x  0.03 y  195 0.03x  0.14 y  400 Add the left sides and the right sides, and solve for y. 0.03 x  0.03 y  195 0.03 x  0.14 y  400 0.11y  205 205 y  1863.64 0.11 Substitute 1863.64 for y in equation (1) and solve for x. x  1863.64  6500 x  4636.36 She should invest $4636.36 in the Connecticut Bank CD account and $1863.64 in the Artisan International account.

30. Let x be the amount invested in the BankUSA CD account and y be the amount invested in the Fidelity Select Health Care account.

The amount invested in the BankUSA CD account will be two times as much as the amount in the Fidelity Select Health Care account, so our first equation is x  2 y .

The total interest for one year is $700, so our second equation is 0.035 x  0.17 y  700 .

The system is: x  2y 0.035 x  0.17 y  700 Substitute 2 y for x in the second equation and solve for y.

0.035  2 y   0.17 y  700 0.07 y  0.17 y  700 0.24 y  700 y  2916.667 Substitute 2916.667 for y in the equation x  2 y and solve for x.

x  2  2916.667   5833.334

The person should invest $5833.33 in the BankUSA CD account and $2916.67 in the Fidelity Select Health Care account.

32. Let x be the amount invested in the LaSalle Bank CD account and y be the amount invested in the Franklin Stategic Mortgage account.

The amount invested in the LaSalle Bank CD account will be two times as much as the amount in the Franklin Stategic Mortgage account, so our first equation is x  2 y .

The total interest for one year is $650, so our second equation is 0.025 x  0.08 y  650 .

The system is: x  2y 0.025 x  0.08 y  650 Substitute 2 y for x in the second equation and solve for x.

0.025  2 y   0.08 y  650 0.05 y  0.08 y  650 0.13 y  650 y  5000 Substitute 5000 for y in the equation x  2 y and solve for x. x  2  5000   10, 000

The person should invest $10,000 in the LaSalle Bank CD account and $5000 in the Franklin Strategic Mortgage account.

34. a. We compute 2% of 4.

0.02  4   0.08

There are 0.08 cups of butterfat in the solution. b. We compute 2% of 8.

0.02  8   0.16

There are 0.16 cups of butterfat in the solution. c. We compute 2% of M. 0.02  M   0.02M

There are 0.02M cups of butterfat in the solution.

36. a. 0.24  2   0.08  5   0.48  0.4  0.88 There are 0.88 liters of pure alcohol in the mixture.

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Chapter 6: Systems of Linear Equations b. Let x be the number of liters of the 24% solution and y be the number of liters of the 8% solution.

The total number of liters is 8, so our first equation is x  y  8 .

The total amount of acid between the two solutions must be the same as the total amount of acid in the mixture, so our second equation is 0.24 x  0.08 y  0.20  8  .

The system is: x y 8 0.24 x  0.08 y  0.20  8  Solve the first equation for y. x y 8 y  x  8 Substitute  x  8 for y in the second equation and solve for x.

0.24 x  0.08   x  8   0.20  8  0.24 x  0.08 x  0.64  1.60 0.16 x  0.64  1.60 0.16 x  0.96 x6 Substitute 6 for x in the equation y   x  8 and solve for y. y   6  8  2

The mixture should contain 6 liters of the 24% solution and 2 liters of the 8% solution.

38. Let x be the number of quarts of the 30% solution and y be the number of quarts of the 50% solution.

The total number of quarts is 10, so our first equation is x  y  10 .

The total amount of acid between the two solutions must be the same as the total amount of acid in the mixture, so our second equation is 0.30 x  0.50 y  0.42 10  .

The system is: x  y  10 0.30 x  0.50 y  0.42 10  Solve the first equation for y. x  y  10 y   x  10 Substitute  x  10 for y in the second equation and solve for x.

189

0.30 x  0.50   x  10   0.42 10  0.30 x  0.50 x  5.0  4.2 0.2 x  5.0  4.2 0.2 x  0.8 x4 Substitute 4 for x in the equation y   x  10 and solve for y. y    4   10  6

The mixture should contain 4 quarts of the 30% solution and 6 quarts of the 50% solution.

40. Let x be the number of liters of the 25% solution and y be the number of liters of 40% solution.

The total number of liters is 6, so our first equation is x  y  6 .

The total amount of pure antifreeze between the two solutions must be the same as the total amount of pure antifreeze in the mixture, so our second equation is 0.25 x  0.40 y  0.30  6  .

The system is: x y 6 0.25 x  0.40 y  0.30  6  Solve the first equation for y. x y 6 y  x  6 Substitute  x  6 for y in the second equation and solve for x.

0.25 x  0.40   x  6   0.30  6  0.25 x  0.40 x  2.40  1.80 0.15 x  2.40  1.80 0.15 x  0.60 x4 Substitute 4 for x in the equation y   x  6 and solve for y. y    4  6  2

The mixture should contain 4 liters of the 25% antifreeze solution and 2 liters of the 40% antifreeze solution.

42. Let x be the number of quarts of the 5% solution and y be the number of quarts of the 20% solution.

The total number of quarts is 6, so our first equation is x  y  6 .

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ISM: Elementary Algebra The total amount of acid between the two solutions must be the same as the total amount of acid in the mixture, so our second equation is 0.05 x  0.20 y  0.10  6  .

The system is: x y 6 0.05 x  0.20 y  0.10  6 

Solve the first equation for y. x y 6 y  x  6 Substitute  x  6 for y in the second equation and solve for x.

0.05 x  0.20   x  6   0.10  6  0.05 x  0.20 x  1.2  0.6 0.15 x  1.2  0.6

The mixture should contain 2 quarts of the 10% acid solution and 3 quarts of the 45% acid solution.

46. Let x be the number of liters of the 30% alcohol solution and y be the number of liters of pure water.

The total number of liters is 6, so our first equation is x  y  6 .

The total amount of alcohol in the 25% solution must be the same as the total amount of alcohol in the mixture since there is no alcohol in pure water. Therefore, our second equation is 0.30 x  0.25  6  .

The system is: x y 6 0.30 x  0.25  6 

0.15 x  0.6 x4 Substitute 4 for x in the equation y   x  6 and solve for y. y    4  6  2

The mixture should contain 4 quarts of the 5% solution and 2 quarts of the 20% solution.

44. Let x be the number of quarts of the 10% acid solution and y be the number of quarts of the 45% acid solution.

The total number of quarts is 5, so our first equation is x  y  5 .

The total amount of acid between the two solutions must be the same as the total amount of acid in the mixture, so our second equation is 0.10 x  0.45 y  0.31 5  .

The system is: x y 5

Solve the second equation for x.

0.30 x  0.25  6  0.30 x  1.5 x5 Substitute 5 for x in the equation x  y  6 and solve for y.

5 y  6 y 1 The mixture should contain 5 liters of the 30% solution and 1 liter of pure water.

48. a.

The interest would be $150. ii. 3000  0.10   300

The interest would be $300. b.

0.10 x  0.45 y  0.31 5 

Solve the first equation for y. x y 5 y  x  5 Substitute  x  5 for y in the second equation and solve for x.

0.10 x  0.45   x  5   0.31 5  0.10 x  0.45 x  2.25  1.55 0.35 x  2.25  1.55 0.35 x  0.70 x2 Substitute 2 for x in the equation y   x  5 and solve for y. y    2  5  3

i. 3000  0.05   150

i. No; from part (a), the minimum amount of interest she will earn is $150. ii. No; from part (a), the maximum amount of interest she will earn is $300. iii. Yes; $180 is between the maximum and minimum amount of interest she can earn.

Let x be the amount invested at 5% and y be the amount invested at 10%.

The total amount invested is $3000, so our first equation is x  y  3000 .

The total interest after one year is $180, so our second equation is 0.05 x  0.10 y  180 .

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Chapter 6: Systems of Linear Equations The system is: x  y  3000 0.05 x  0.10 y  180 Solve the first equation for y: x  y  3000 y   x  3000 Substitute  x  3000 for y in the second equation and solve for x.

0.05 x  0.10   x  3000   180

54. Descriptions may vary. x y 8  y  x  8

0.35 x  0.55 y  0.40  8  Substitute  x  8 for y in the second equation and solve for x.

0.35 x  0.55   x  8   0.40  8  0.35 x  0.55 x  4.4  3.2 0.2 x  4.4  3.2 0.2 x  1.2

0.05 x  0.10 x  300  180

x6 Substitute 6 for x in the equation y   x  8 and solve for y. y   6  8  2 .

0.05 x  300  180 0.05 x  120 x  2400 Substitute 2400 for x in the equation x  y  3000 and solve for y.

2400  y  3000 y  600 She should invest $2400 at 5% and $600 at 10%.

50. 7% Answers may vary. Example: Let x be the amount invested in each account.

The interest earned after one year is 0.05 x  0.09 x  0.14 x .

Since the total amount invested is 2x dollars, the total interest earned as a percentage of the total money invested will be: 0.14 x  0.07 or 7% 2x 52. Descriptions may vary. x  y  500  y   x  500 15 x  25 y  8300 Substitute  x  500 for y in the second equation and solve for x.

15 x  25   x  500   8300 15 x  25 x  12,500  8300 10 x  4200 x  420 Substitute 420 for x in the equation y   x  500 and solve for y.

The solution is  420,80  .

The solution is  6, 2  .

56.

P  2L W  P  2 L  2W P  2 L  2W P  2L P  2L  W or W  2 2

58.

A  P 1  rt  P 1  rt  A  1  rt 1  rt  A A  P or P  1  rt 1  rt

60. 3x  2 y  6 y  2 x  11 Substitute 2 x  11 for y in the first equation and solve for x.

3x  2  2 x  11  6 3x  4 x  22  6 7 x  22  6 7 x  28

10 x  12,500  8300

y    420   500  80

191

x4 Substitute 4 for x in the equation y  2 x  11 and solve for y. y  2  4   11  8  11  3

The solution of the system is  4, 3 .

This is a system of linear equations in two variables.

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62. 3x  2 y  6

x-intercept: 3x  2  0   6

4. Check (2,1) : 3 y  4 x

3x  6

3 1   4  2  ?

x2 The x-intercept is  2, 0  .

?

3  8 false

The ordered pair  2,1 does not satisfy the

y-intercept: 3  0   2 y  6 2y  6 y3

inequality.

Check (3, 4) : 3 y  4 x

The y-intercept is  0,3 .

3  4    4  3  ?

We plot the intercepts and connect them with a line, extending the line in both directions. y

The ordered pair  3, 4  satisfies the inequality.

Check (2, 1) : 3 y  4 x

5 (0, 3) (2, 0)



?

12  12 true

5

3  1   4  2  ?

x

?

3  8 true 5

The ordered pair  2, 1 satisfies the

This is a linear equation in two variables.

inequality.

6. The graph of y  2 x  4 is the region below the line y  2 x  4 . We use a dashed line along the border to indicate that the points on the line y  2 x  4 are not solutions of y  2x  4 .

Homework 6.6 2. Check (4, 1) : y  3x  2 1  3  4   2 ?

y

?

1  12  2

5

?

1  10 true

The ordered pair  4, 1 satisfies the inequality.

Check (4, 3) : y  3 x  2



5

x

5

3  3  4   2 ? ?

3  12  2 ?

3  14 false

The ordered pair  4,3 does not satisfy the inequality.

Check (2, 5) : y  3 x  2

8. The graph of y  3x  4 is the line y  3 x  4 and the region above that line.

We use a solid line along the border to indicate that the points on the line y  3 x  4 are solutions of y  3 x  4 . y

5

5   3  2   2 ? ?

5  6  2



?

5

5  8 false

The ordered pair  2, 5  does not satisfy the

5

inequality.

Copyright © 2015 Pearson Education, Inc.

x

Chapter 6: Systems of Linear Equations 10. The graph of y 

2 x  5 is the region above 3

2 x  5 . We use a dashed line 3 along the border to indicate that the points on 2 the line y  x  5 are not solutions of 3 2 y  x 5.

3

the line y 

193

16. First, we get y alone on one side of the inequality. y  3x  0 y  3 x The graph of y  3x is the line y  3x and the region below the line. We use a solid line along the border to indicate that the points on the line y  3 x are solutions of y  3 x  0 . y

5

y

5  

5

5

x

x

5

5

1 12. The graph of y   x  3 is the line 2 1 y   x  3 and the region below that line.

2 We use a solid line along the border to indicate 1 that the points on the line y   x  3 are 2 1 solutions of y   x  3 .

2

18. First, we get y alone on one side of the inequality.

2x  y  3 y  2 x  3 The graph of y  2 x  3 is the line y  2 x  3 and the region above that line.

We use a solid line along the border to indicate that the points on the line are solutions to 2x  y  3 . y

5

y

5  

5

5

14. The graph of y   x is the region below the line y   x . We use a dashed line along the border to indicate that the points on the line y   x are not solutions of y   x . y

5

5

x

x

5



5

20. First, we get y alone on one side of the inequality. x y 3  y  x  3 y  x3 The graph of y  x  3 is the region above the line y  x  3 . We use a dashed line along the border to indicate that the points on the line are not solutions to x  y  3 . y

x

5

5 

5 5

Copyright © 2015 Pearson Education, Inc.

x

194

ISM: Elementary Algebra

22. First, we get y alone on one side of the inequality.

5x  2 y  0 2 y  5 x 5 y x 2 5 5 The graph of y  x is the line y  x and 2 2 the region above that line. We use a solid line along the border to indicate that the points on the line are solutions to 5 x  2 y  0 . y

5

along the border to indicate that the points on the line are not solutions of 3x  4 y  12 . y

5



5

x

5

28. The graph of y  2 is the region below the horizontal line y  2 . We use a dashed line along the border to indicate that the points on the line are not solutions to y  2 . y



5

x

5

5

24. First, we get y alone on one side of the inequality.

2x  4 y  4 4 y  2 x  4 1 y   x 1 2 1 The graph of y   x  1 is the region above 2 1 the line y   x  1 . We use a dashed line 2 along the border to indicate that the points on the line are not solutions of 2 x  4 y  4 . y



x

5

30. The graph of x  4 is the vertical line x  4 and the region to the left of that line. We use a solid line along the border to indicate that the points on the line are solutions to x  4 . y

5



5



5

5

x

5

5

32. The graph of y  1 is the region above the horizontal line y  1 . We use a dashed line along the border to indicate that the points on the line are not solutions to y  1 .

x

5

y

26. First, we get y alone on one side of the inequality.

3x  4 y  12 4 y  3 x  12 3 y  x3 4 3 The graph of y  x  3 is the region below 4 3 the line y  x  3 . We use a dashed line 4

5



5 5

Copyright © 2015 Pearson Education, Inc.

x

Chapter 6: Systems of Linear Equations y

34. The graph of x  0 is the vertical line x  0 and the region to the right of that line. We use a solid line along the border to indicate that the points on the line are solutions to x  0 .

5

y



5

195

5

x

5 

5

x

4 x  2 and 3 the graph of y   x  4 . The graph of

40. First we sketch the graph of y 

5

36. First we sketch the graph of y 

2 x  1 and 3

4 the graph of y   x  2 . The graph of 3 2 2 y  x  1 is the line y  x  1 (graph the 3 3 line with a solid line) and the region below the 4 line. The graph of y   x  2 is the line 3 4 y   x  2 (graph the line with a solid line) 3 and the region above that line. The graph of the solution set of the system is the intersection of the graphs of the inequalities.

4 4 x  2 is the line y  x  2 (graph the 3 3 line with a solid line) and the region below that line. The graph of y   x  4 is the region above the line y   x  4 (graph the line with a dashed line). The graph of the solution set of the system is the intersection of the graphs of the inequalities. y

y

5



y

5

x

5

5

42. First we sketch the graph of y  3 x  3 and the 

5

x

5

38. First we sketch the graph of y  3x  5 and 2 x  3 . The graph of 5 y  3 x  5 is the region below the line y  3 x  5 (graph the line with a dashed

the graph of y 

line). The graph of y 

2 x  3 is the region 5

2 below the line y  x  3 (graph the line with 5 a dashed line). The graph of the solution set of the system is the intersection of the graphs of the inequalities.

1 graph of y   x  1 . The graph of 4 y  3x  3 is the line y  3x  3 (graph the line with a solid line) and the region below 1 that line. The graph of y   x  1 is the 4 1 region below the line y   x  1 (graph the 4 line with a dashed line). The graph of the solution set of the system is the intersection of the graphs of the inequalities. y

5



5 5

Copyright © 2015 Pearson Education, Inc.

x

196

ISM: Elementary Algebra

44. First we sketch the graph of y  4 x  5 and the graph of y   x  1 . The graph of y  4 x  5 is the region above the line y  4 x  5 (graph the line with a dashed line). The graph of y   x  1 is the line y   x  1 (graph the line with a solid line) and the region below that line. The graph of the solution set of the system is the intersection of the graphs of the inequalities. y

5

1 x  3 and 2 1 the graph of y  2 x . The graph of y  x  3 2 1 is the region above the line y  x  3 (graph 2 the line with a dashed line). The graph of y  2 x is the line y  2 x (graph the line with a solid line), and the region above that line.

The graph of the solution set of the system is the intersection of the graphs of the inequalities.

Next we sketch the graph of y 

y



5

5

x

5



1 46. First we sketch the graph of y   x and the 2 4 1 graph of y  x . The graph of y   x is 3 2 1 the line y   x (graph the line with a solid 2 line) and the region below that line. The graph 4 of y  x is the region below the line 3 4 y  x (graph the line with a dashed line).

3 The graph of the solution set of the system is the intersection of the graphs of the inequalities. y

5

5

x

5

50. First we get y alone on one side of each inequality. x  y 1 3x  y  2 y  x 1  y  3x  2 y  3x  2 Next we sketch the graph of y   x  1 and the graph of y  3 x  2 . The graph of y   x  1 is the region below the line y   x  1 (graph the line with a dashed line).

The graph of y  3x  2 is the line y  3x  2 (graph the line with a solid line) and the region below the line. The graph of the solution set of the system is the intersection of the graphs of the inequalities. y



5

5

x

5



48. First we get y alone on one side of each inequality.

3y  6x x  2y  6 y  2x 2 y   x  6 1 y  x3 2

5

x

5

52. First we get y alone on one side of each inequality. 3x  4 y  4 5x  2 y  2 4 y  3x  4 2 y  5 x  2 3 5 y  x 1 y  x 1 4 2

Copyright © 2015 Pearson Education, Inc.

Chapter 6: Systems of Linear Equations Next we sketch the graph of y 

3 x  1 and 4

5 x  1 . The graph of 2 3 3 y  x  1 is the line y  x  1 (graph the 4 4 line with a solid line) and the region below 5 that line. The graph of y  x  1 is the region 2 5 above the line y  x  1 (graph the line with 2 a dashed line). The graph of the solution set of the system is the intersection of the graphs of the inequalities.

the graph of y 

y

5

5



x

5

54. First we sketch the graph of y  2 and the graph of x  1 . The graph of y  2 is the horizontal line y  2 and the region above that line (graph the line with a solid line). The graph of x  1 is the region to the left of the vertical line x  1 (graph the line with a dashed line). The graph of the solution set of the system is the intersection of the graphs of the inequalities. y

5



5

x

5

The graph of x  5 is the vertical line x  5 (graph the line with a solid line) and the region to the right of that line. The graph of the solution set of the system is the intersection of the graphs of the inequalities. y

5

5



x

5

58. First we get y alone on one side of each of the first two inequalities.

3x  4 y  16 3x  4 y  8 4 y  3 x  16 4 y  3 x  8 3 3 y   x4 y   x2 4 4 3 Next, we sketch the graph of y   x  4 , 4 3 the graph of y   x  2 , the graph of x  1 , 4 and the graph of x  4 . The graph of 3 3 y   x  4 is the line y   x  4 (graph 4 4 the line with a solid line) and the region below 3 that line. The graph of y   x  2 is the line 4 3 y   x  2 (graph the line with a solid line) 4 and the region above that line. The graph of x  1 is the vertical line x  1 (graph the line with a solid line) and the region to the right of that line. The graph of x  4 is the vertical line x  4 (graph the line with a solid line) and the region to the left of that line. The graph of the solution set of the system is the intersection of the graphs of the inequalities. y

56. First we sketch the graph of y  1 , the graph of y  4 , the graph of x  2 , and the graph of x  5 . The graph of y  1 is the horizontal line y  1 (graph the line with a solid line) and the region below that line. The graph of y  4 is the horizontal line y  4 (graph the line with a solid line) and the region above that line. The graph of x  2 is the vertical line x  2 (graph the line with a solid line) and the region to the left of that line.

197

5



5 5

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x

198

ISM: Elementary Algebra

60. a. We are given equations for the upper and lower limits on H and we know that the the number of dog years is between 10 and 20, inclusive. The system is: H  5.4d  12.3 H  4.5d  15.5 d  10 d  20

Human Years

b. First we sketch the graph of H  5.4d  12.3 , the graph of H  4.5d  15.3 , the graph of d  10 , and the graph of d  20 . The graph of H  5.4d  12.3 is the line H  5.4d  12.3 (graph the line with a solid line) and the region below that line.

The graph of H  4.5d  15.3 is the line H  4.5d  15.3 (graph the line with a solid line) and the region above that line.

The graph of d  10 is the vertical line d  10 (graph the line with a solid line) and the region to the right of that line. The graph of d  20 is the vertical line d  20 (graph the line with a solid line) and the region to the left of that line. The graph of the solution set of the system is the intersection of the graphs of the inequalities.

From the graph, we can see the y-intercept is  0,1 , the x-intercept is 1, 0  , and so the slope is 1  1 . Therefore, the equation of the 1 border line is y   x  1 . The shaded region is below the dashed line and above the solid line.

Therefore, the system of inequalities is: y2 m

y  x 1

64. The student is incorrect. The graph of y  3 x  1 is the line y  3x  1 and the region below the line y  3x  1 .

66. a. Answers may vary. b. Answers may vary. c. Answers may vary. d. Answers may vary.

68. Answers may vary. Examples: Solutions: 1,3 ,  3,5  ,  0, 0 

Non solutions:  4, 2  ,  4, 4  ,  1, 7 

H 160

70. Answers may vary. Example: The line represents the “equal to” part of the “greater than or equal to” inequality.

80

72. Answers may vary.

74. 3x  4 y  4 20 10 Dog Years

d

x-intercept: 3x  4  0   4 3x  4

c. Lower limit: H  4.5 15   15.3  67.5  15.3  82.8

Upper limit: w  5.4 15   12.3  93.3 The numbers of human years are between 82.8 years and 93.3 years, inclusive.

62. First we find the equation of the dashed line.

From the graph we see the line is horizontal and passes through the point  0, 2  . Therefore,

the equation of the line is y  2 . Next we find the equation of the solid line.

x

 

4 3

The x-intercept is 43 , 0 .

y-intercept: 3  0   4 y  4 4 y  4 y  1

The y-intercept is  0, 1 .

We plot the intercepts and connect them with a line, extending the line in both directions.

Copyright © 2015 Pearson Education, Inc.

Chapter 6: Systems of Linear Equations y

199

The graph of y   12 x  4 is the region below

5

the line y   12 x  4 (graph the line with a



5

dashed line). The graph of the solution set of the system is the intersection of the graphs of the inequalities.

x

y

5

5

76. 3x  4 y  4 4 y  3 x  4 3 y  x 1 4 The graph of 3x  4 y  4 is the region below the line 3 x  4 y  4 (see problem 72). We use a dashed line to indicate that the points on the line are not solutions of 3 x  4 y  4 . y

5



5

x

5

78. 3x  4  4 3x  8



5

x

5

82. 3x  4 y  4 Equation (1) x  2 y  8 Equation (2) To eliminate the y terms, we multiply both sides of equation (2) by 2, yielding the system: 3x  4 y  4 2 x  4 y  16 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  4 2 x  4 y  16 5 x  20

x  83





In interval notation, the solution set is 83 ,  . 

8 3

2y  4

80. First we get y alone on one side of each inequality.

3x  4 y  4 x  2y  8 4 y  3 x  4 2 y  x  8 y  34 x  1

x4 Substitute 4 for x in equation (2) and solve for y.  4  2 y  8 y2

The solution is  4, 2  .

84. Answers may vary.

y   12 x  4

86. Answers may vary.

Next we sketch the graph of y  34 x  1 and

88. Answers may vary.

the graph of y   12 x  4 . The graph of y  34 x  1 is the region below the line y  34 x  1 (graph the line with a dashed line).

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200

ISM: Elementary Algebra

Chapter 6 Review 1. y  2 x  3 y  3 x  7 To begin, we graph both equations in the same coordinate system. y y  2x  3

5

4. 4 x  y  3  3 x  y  4 To begin, we rewrite each equation in slope–intercept form.

4x  y  3 3x  y  4 y  4 x  3 y  3x  4 Next, we graph both equations in the same coordinate system. y

(2, 1)

5



5

x

y  3 x  7

5



The intersection point is  2,1 . So, the solution is the ordered pair  2,1 .

3 2. y  x  4 2 1 y  x4 2 To begin, we graph both equations in the same coordinate system. y

5

y

 (4, 2)

3 x4 2

5

x

2 3. y  x 5 y  2 x To begin, we graph both equations in the same coordinate system. y

(0, 0)



5 5

The intersection point is 1, 1 . So, the solution is the ordered pair 1, 1 .

5.  x  y  4 2 x  y  5 To begin, we rewrite each equation in slope–intercept form. x  y  4 2 x  y  5 y  x4 y  2 x  5 Next, we graph both equations in the same coordinate system.

5

solution is the ordered pair  4, 2  .

y

4x  y  3

y

The intersection point is  4, 2  . So, the

5

x

(1, 1) 5

5

1 y   x4 2

5

3x  y  4

2 x 5 x

y  2 x

The intersection point is  0, 0  . So, the solution is the ordered pair  0, 0  .

x  y  4

( 3, 1)



5 5

x

2 x  y  5

The intersection point is  3,1 . So, the solution is the ordered pair  3,1 .

6. x  3 y  3 2 x  3 y  12 To begin, we rewrite each equation in slope–intercept form.

2 x  3 y  12 x  3y  3 3 y   x  3 3 y  2 x  12 1 2 y  x 1 y  x4 3 3 Next, we graph both equations in the same coordinate system.

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Chapter 6: Systems of Linear Equations y

Substitute 0 for x in the equation y  2 x and solve for y. y  2  0  0

5 x  3y  3

5

 (3, 2)

5

201

The solution is  0, 0  .

x

10. y  3  x  2  y  4  x  5

2 x  3 y  12

The intersection point is  3, 2  . So, the

Substitute 4  x  5  for y in the first equation

solution is the ordered pair  3, 2  .

and solve for x.

4  x  5   3  x  2  4 x  20  3x  6 7 x  20  6 7 x  14 x2 Substitute 2 for x in the equation y  4  x  5 

7. 3x  2 y  11 y  5 x  16 Substitute 5 x  16 for y in the first equation and solve for x.

3x  2  5 x  16   11 3 x  10 x  32  11 7 x  32  11 7 x  21 x3 Substitute 3 for x in the equation y  5 x  16 and solve for y. y  5  3  16  1

and solve for y. y  4  2  5   4  3  12 The solution is  2, 12  .

11.

The solution is  3, 1 .

8. 4 x  3 y  5  0 x  4  2y Substitute 4  2 y for x in the first equation and solve for y.

4 4  2 y  3y  5  0 16  8 y  3 y  5  0 11 y  11  0 11y  11 y 1 Substitute 1 for y in the equation x  4  2 y and solve for x. x  4  2 1  4  2  2

The solution is  2,1 .

9. y  5 x y  2x Substitute 2x for y in the first equation and solve for x.

2 x  5 x 7x  0 x0

x  y  1 2x  y  4 Solve the first equation for y. x  y  1 y  x 1 Substitute  x  1 for y in the second equation and solve for x.

2 x    x  1  4 2x  x  1  4 3x  1  4 3x  3 x 1 Substitute 1 for x in the equation y   x  1 and solve for y. y   1  1  2

The solution is 1, 2  .

12.

x  2y  5 4 x  2 y  4 Solve the first equation for x. x  2y  5 x  2 y  5 Substitute 2 y  5 for x in the second equation and solve for y.

4  2 y  5   2 y  4 8 y  20  2 y  4 6 y  20  4 6 y  24 y4

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ISM: Elementary Algebra Substitute 4 for y in the equation x  2 y  5 and solve for x. x  2  4   5  8  5  3 The solution is  3, 4  .

13. y  2.19 x  3.51 y  1.54 x  6.22 Substitute 1.54 x  6.22 for y in the first equation and solve for x.

1.54 x  6.22  2.19 x  3.51 3.73x  6.22  3.51 3.73 x  9.73 9.73  2.609 x 3.73 Substitute 2.609 for x in the equation y  1.54 x  6.22 and solve for y. y  1.54  2.609   6.22  2.202

The approximate solution is  2.61, 2.20  .

14. y  4.98 x  1.18 y  0.57 x  4.08 Substitute 0.57 x  4.08 for y in the first equation and solve for x. 0.57 x  4.08  4.98 x  1.18 4.41x  4.08  1.18 4.41x  5.26 5.26 x  1.193 4.41 Substitute 1.193 for x in the equation y  0.57 x  4.08 and solve for y. y  0.57  1.193  4.08  4.760

The approximate solution is  1.19, 4.76  .

15.

x  2 y  1 Equation (1) 3x  5 y  19 Equation (2) To eliminate the x terms, we multiply both sides of equation (1) by 3 , yielding the system: 3x  6 y  3 3 x  5 y  19 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. 3x  6 y  3 3 x  5 y  19 11 y  22 y2

Substitute 2 for y in equation (1) and solve for x. x  2  2   1 x  4  1 x3 The solution is  3, 2  .

16. 2 x  5 y  3 Equation (1) 4 x  3 y  19 Equation (2) To eliminate the x terms, we multiply both sides of equation (1) by 2 , yielding the system: 4 x  10 y  6 4 x  3 y  19 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. 4 x  10 y  6 4 x  3 y  19 13 y  13 y  1 Substitute 1 for y in equation (1) and solve for x.

2 x  5  1  3 2 x  5  3 2 x  8 x  4 The solution is  4, 1 .

17. 3 x  8 y  2 Equation (1) 5 x  2 y  12 Equation (2) To eliminate the y terms, we multiply both sides of equation (2) by 4, yielding the system: 3x  8 y  2 20 x  8 y  48 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  8 y  2 20 x  8 y  48 23x  46 x  2 Substitute 2 for x in equation (2) and solve for y.

5  2   2 y  12 10  2 y  12 2 y  2 y 1

The solution is  2,1 .

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Chapter 6: Systems of Linear Equations 18.

4 x  3 y  6 Equation (1) 7 x  5 y  11 Equation (2) To eliminate the y terms, we multiply both sides of equation (1) by 5 and both sides of equation (2) by 3, yielding the system: 20 x  15 y  30 21x  15 y  33 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.

20 x  15 y  30 21x  15 y  33 x  3 x  3 Substitute 3 for x in equation (1) and solve for y.

4  3  3 y  6 12  3 y  6 3 y  6 y  2

The solution is  3, 2  .

19. 2 x  5 y  2 Equation (1) 3 x  6 y  0 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: 6 x  15 y  6 6 x  12 y  0 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. 6 x  15 y  6 6 x  12 y  0  3y  6 y  2 Substitute 2 for y in equation (2) and solve for x.

3x  6  2   0 3 x  12  0 3 x  12 x4 The solution is  4, 2  .

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20. y  3x  5  3x  y  5 Equation (1) y  2 x  5  2 x  y  5 Equation (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.

3x  y  5 2x  y  5 5 x  10 x2 Substitute 2 for x in equation (2) and solve for y. y  2  2   5  4  5  1

The solution is  2,1 .

21. 2  x  3  y  6 x  3  y  2   1 Begin by writing each equation in standard form.

2  x  3  y  6 x  3  y  2   1 2x  6  y  6 x  3 y  6  1 2x  y  0 x  3 y  7 This yields the system: 2x  y  0 Equation (1) x  3 y  7 Equation (2) To eliminate the y terms, we multiply both sides of equation (1) by 3, yielding the system: 6x  3y  0 x  3 y  7 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.

6x  3y  0 x  3 y  7 7 x  7 x  1 Substitute 1 for x in equation (1) and solve for y.

2  1  y  0 2  y  0 y2

The solution is  1, 2  .

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22.

ISM: Elementary Algebra 1 2 5 x y  Equation (1) 2 3 3 1 3 13 x y Equation (2) 3 2 6 To eliminate the x terms (and clear fractions), we multiply both sides of equation (1) by 12 and both sides of equation (2) by 18, yielding the system: 6 x  8 y  20 6 x  27 y  39 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. 6 x  8 y  20 6 x  27 y  39 19 y  19 y 1 Substitute 1 for y in equation (2) and solve for y.

1 3 13 x  1   3 2 6 1 3 13 x  3 2 6 1 13 9 x  3 6 6 1 4 x 3 6 4 x    2 2 The solution is  2,1 .

23. y  4.59 x  1.25  4.59 x  y  1.25 Eq. (1) y  0.52 x  4.39 0.52 x  y  4.39 Eq. (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.

4.59 x  y  1.25 0.52 x  y  4.39 4.07 x  3.14 3.14 x  0.771 4.07 Substitute 0.771 for x in equation (1) and solve for y. y  4.59  0.771  1.25  4.789

The approximate solution is  0.77, 4.79  .

24. y  0.91x  3.57  0.91x  y  3.57 Eq. (1) y  3.58 x  6.05  3.58 x  y  6.05 Eq. (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.

0.91x  y  3.57 3.58 x  y  6.05 4.49 x  9.62 9.62 x  2.143 4.49 Substitute 2.143 for x in equation (1) and solve for y. y  0.91 2.143  3.57  1.620

The approximate solution is  2.14, 1.62  .

25. 2 x  7 y  13 Equation (1) 5 x  3 y  12 Equation (2) 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: 10 x  35 y  65 10 x  6 y  24 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.

10 x  35 y  65 10 x  6 y  24  41 y  41 y 1 Substitute 1 for y in equation (1) and solve for x.

2 x  7 1  13 2 x  7  13 2 x  6 x  3 The solution is  3,1 .

26. 4 x  7 y  8 x  3  2y Substitute 3  2 y for x in the first equation and solve for y.

4 3  2 y   7 y  8 12  8 y  7 y  8 12  y  8  y  4 y4

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