Chapter 4
Systems of Linear Equations
4.1 Check Points
1. a. To determine if (1,2) is a solution to the system,
replace x with 1 and y with 2 in both equations.
23 4
2(1) 3(2) 4
xy

b. To determine if (7,6) is a solution to the system,
replace x with 7 and y with 6 in both equations.
23 4
2(7) 3(6) 4
14 18 4
xy


2. Graph 26xy by using intercepts.
x-intercept (Set 0.y)
26
206
26
3
xy
x
x
x


y-intercept (Set 0.x)
y-intercept (Set 0.x)
22
2(0) 2
2
2
xy
y
y
y



3. Graph 6yx  by using the y-intercept of 6 and
the slope of –1.
Graph 36yx
by using the y-intercept of –6 and
the slope of 3.
4. Graph 32yx
by using the y-intercept of –2 and
the slope of 3.
Graph 31yx
by using the y-intercept of 1 and
the slope of 3.
Chapter 4 Systems of Linear Equations
5. Graph 3xy by using intercepts.
x-intercept (Set 0.y)
3
03
3
xy
x
x


3
x
y-intercept (Set
0.x
)
22 6
2(0) 2 6
26
3
xy
y
y
y


Both lines have the same x-intercept and the same y–
Any ordered pair that is a solution to one equation is
a solution to the other, and, consequently, a solution
6. a. Graph 2yx by using the y-intercept of 0 and
the slope of 2.
Graph 10yx by using the y-intercept of 10
and the slope of 1.
b. If the bridge is used 10 times in a month, the
total monthly cost without the discount pass is
the same as the monthly cost with the discount
pass, namely $20.
4.1 Exercise Set
1. To determine if (2,−3) is a solution to the system,
replace x with 2 and y with −3 in both equations.
23 5
22 3 3 5
49 5
xy

 
2.
62 2
xy
311
32 5 11
65 11
11 11, true
xy

 

Section 4.1 Solving Systems of Linear Equations by Graphing
3. 31
21
31
39
211
xy





4.
431
71
43 1
25 25
28 3 1
25 25
xy
 
 
 
 

The ordered pair satisfies both equations, so it is a
solution to the system.
5.
53 2
55 39 2
25 27 2
22,true
xy
 
 
410 157 13
40 105 13
145 13, false


The ordered pair does not satisfy both equations, so
8.
4100
4 200 700 100
800 700 100
100 100, true
xy

 

Chapter 4 Systems of Linear Equations
9.
54 20
58 45 20
40 20 20
xy


10.
43 26
45 32 26
20 6 26
xy


11. Graph both equations on the same axes.
6:xy
x-intercept = 6; y-intercept = 6
2:xy
x-intercept = 2; y-intercept = −2
12. Graph both equations on the same axes.
2:xy
x-intercept = 2; y-intercept = 2
4:xy
x-intercept = 4; y-intercept = −4
y
13. Graph both equations on the same axes.
1:xy
x-intercept = 1; y-intercept = 1
14. Graph both equations on the same axes.
4:xy
The solution set is
(0,4) .
15. Graph both equations.
-4
-2
The solution set is
(3,0) .
Section 4.1 Solving Systems of Linear Equations by Graphing
16. Graph both equations
22xy
:
x-intercept = 2: y-intercept = 1
2xy:
17. Graph both equations.
44xy:
x-intercept = 1: y-intercept = 4
33xy:
18. Graph both equations.
510xy :
x-intercept = 2: y-intercept = −10
24xy:
19. Graph both equations.
5yx
:
Slope = 1; y-intercept = 5
3yx  :
20. Graph both equations.
1yx
:
Slope = 1; y-intercept = 1
31yx
:
Slope = 3; y-intercept = −1
21. Graph both equations.
2yx:
slope =2; y–intercept = 0
6yx  :
Chapter 4 Systems of Linear Equations
22. Graph both equations.
21yx
:
slope =2; y–intercept = 1
23. Graph both equations.
23yx :
slope = −2; y-intercept = 3
1yx :
24. Graph both equations.
34yx
:
slope = 3; y–intercept = −4
21yx :
25. Graph both equations.
21yx
:
Slope = 2; y-intercept = −1
26. Graph both equations.
31yx
:
Slope = 3; y-intercept = −1
32yx
:
Slope = 3; y-intercept = 2
27. Graph each equation.
4xy:
x-intercept = 4; y-intercept = 4
2x :
Section 4.1 Solving Systems of Linear Equations by Graphing
28. Graph both lines.
6xy:
x-intercept = 6; y-intercept = 6
3y :
horizontal line with y-intercept −3
y
29. Graph each equation.
24xy
:
x-intercept = 4; y-intercept = −2
24 8xy
:
30. Graph both lines.
23 6xy
:
x-intercept = 3; y-intercept = 2
46 12xy
:
x-intercept = 3; y-intercept = 2
31. Graph both lines.
21yx
:
slope = 2; y–intercept = −1
24xy
:
x-intercept = −4; y–intercept = 2
y
32. Graph both lines.
24yx :
slope = 2; y–intercept = −4
42 8xy
:
x-intercept = 2; y-intercept = −4
y
x-intercept = 5; y-intercept = 5
22 12xy:
x-intercept = 6; y-intercept = 6
x
y
-8 -4 4 8
-4
4
8
0
Chapter 4 Systems of Linear Equations
34. Graph both lines.
2xy:
x-intercept = 2; y-intercept = −2
33 6xy
:
35. 0xy
yx
y
2
4
36. 20xy
2yx
5
y
37. x = 2
y = 4
y
4
38. x = 3
y = 5
5
y
x
y
-4 -2 2 4
-2
2
4
0
5
5
y
x
x
-4 -2 2 4
-4
-2
2
0
Section 4.1 Solving Systems of Linear Equations by Graphing
42. y = 0
y = 5
5
y
43. 13
2
yx:
slope = 1
44. 32
4
yx:
slope = 3
4, y-intercept = −2
45. 14
2
yx
slope = 1
, y-intercept = 4
46.
13
4
yx 
slope =
1
4
, y-intercept = 3
43
xy

47.
36
36
36
xy
yx
yx

 

slope = 3, y-intercept = −6
y
y-intercepts, the graphs will coincide and there are
an infinite number of solutions.
48.
24
24
xy
yx

 
slope = 2, y-intercept = −4
Since the lines have the same slopes and
y-intercepts, the graphs will coincide and there are
an infinite number of solutions.
Chapter 4 Systems of Linear Equations
50.
20
2
xy
yx


51. a. The x-coordinate of the intersection point is 40.
Both companies charge the same for 40 miles
driven.
b. The y-coordinate of the intersection point is
about 55.
52. a. The solution is the ordered pair (2,200).
53. a. The solution is the ordered pair (5,20).
54. – 61. Answers will vary.
62. makes sense
65. makes sense
66. false; Changes to make the statement true will vary.
A sample change is: A linear system could have
graphs of equal slopes that coincide.
system that has one solution must have different
slopes.
70. One linear system whose solution is (5, 1) is
6
4
xy
xy


72. Answers will vary depending on exercises chosen.
73.
22
26
yx
yx

 
The solution set is
(1, 4) .
Section 4.1 Solving Systems of Linear Equations by Graphing
75.
24
4
xy
xy


In order to enter the equations into a graphing
calculator, each of them must be solved for y.
24
xy

76.
2310
43 20
xy
xy


In order to enter the equations into a graphing
calculator, each of them must be solved for y.
77.
35
52 10
xy
xy

 
Solve each equation for y.
35
xy

78.
23 7
327
27
33
xy
yx
yx



80.
12
2
37
4
yx
yx
 

7324
721
3
x
x
x

The solution set is
3.
Chapter 4 Systems of Linear Equations
86.
(5 1) 1 5 5
51155
xx
xx
 
 
4.2 Check Points
1.
513
2312
yx
xy


Substitute
513x
for y in the second equation.
23(513)12
xx

2.
32 1
3
xy
xy


Solve the second equation for x.
3
3
xy
xy


Substitute 3y for x in the first equation.
3.
35
33
xy
yx

 
Substitute
33x
for y in the first equation.
35
y
xy


The solution set is
.
4.
34
9312
yx
xy


Substitute
34x
for y in the second equation.
9312
xy

5. a.
30 1800
30
px
px
 
Substitute
30x
for p in the first equation.
30 1800
30 30 1800
p
px
xx
 
 
offer 30,000 apartments for rent.
4.2 Concept and Vocabulary Check
Section 4.2 Solving Systems of Linear Equations by the Substitution Method
3.
4.
(, )2 6 8 or (, ) 3 4xy x y xy x y 
4.2 Exercise Set
1.
4
3
xy
yx

Substitute 3x for y in the first equation.
4
xy

2.
6
2
xy
yx

Step 1 The second equation is already solved for y.
Step 2 Substitute 2x for y in the first equation.
6
xy

3.
38
29
xy
yx


Substitute
29x
for y in the first equation and solve
for x.
Back-substitute 5 for x into the second equation and
solve for y.
29
yx

4.
23 13
27
xy
yx


Substitute
27x
for y in the first equation and
solve for x.
23 13
2327 13
xy
xx


5.
35
4513
xy
xy


Solve the first equation for x.
35
53
xy
xy


Back-substitute 1 for y in the equation 53xy
and solve for x.
53
531 2
xy
x

 
Chapter 4 Systems of Linear Equations
6.
25
215
xy
xy


Solve the first equation for x.
Back-substitute 5 for y in the equation 52xy
and solve for x.
52
525 5
xy
x

 
The solution set is
5, 5 .
7.
25
514
xy
xy


Solve the second equation for x.
514
xy

8.
2311
40
xy
xy


Solve the second equation for x.
Substitute
4y for x in the first equation.
2311
24 3 11
xy
yy


52 10
xy

Solve the first equation for y.
23
23
23
xy
yx
yx

 

Substitute
23x
for y in the second equation.
522310
54610
610
4
xx
xx
x
x



23 10 8 6
6208 6
14 20 6
yy
yy
y



11. 31
24
xy
xy


Solve the second equation for x.
24
24
xy
xy


Substitute 24y for x in the first equation.
12. 411
23 5
xy
xy


Solve the first equation for y.
411
411
xy
yx


Substitute 411
x
for y in the second equation.
Substitute
14
5 for x in the equation 4 11.yx
411
14
411
5
yx
y





62 7
6223 7
646 7
47,false
xy
xx
xx


 
The false statement 4 = 7 indicates that the system
is inconsistent and has no solution.
The solution set is
.
Chapter 4 Systems of Linear Equations
16. 9312
34
xy
yx


Substitute 3x−4 for y in the second equation.
9312
xy

17. 52 0
30
xy
xy


Solve the second equation for x.
30
3
xy
xy

18. 43 0
20
xy
xy


Solve the second equation for x.
20
2
xy
yx


19. 26
32 5
xy
xy


Solve the first equation for y.
26
xy

717
17
7
x
x
Back-substitute to find y.
17 8
262 6
yx

 

Substitute 24
x
for y in the second equation and
solve for x.
35242
310 202
7202
718
xx
xx
x
x


 

21.
21 3
23
xy
yx


Substitute
23x
for y in the first equation.
22.
12
31
xy yx
yx
 

Simplify the first equation.
23.
29
710
xy
xy


Substitute 7y + 10 for x in the first equation.
29
71029
xy
yy

 
24.
53
84
xy
xy


5384
735944
5 3
3333
yy
 




The solution set is
44 7
,.
33







0.19 38
200
x
x

Back-substitute to find y.
4 100
4 200 100
yx

Chapter 4 Systems of Linear Equations
26.
6 8000
0.3 0.6 0
xy
xy


Multiply second equation by 10 to eliminate
decimal places and solve for x.
36 0
xy

27.
12
33
52
7
yx
yx


First, clear both equations of fractions. Multiply the
first equation by the LCD, 3.
Solve the first of these equations for x.
32
32
yx
yx


Substitute 32y for x in the second equation of the
new system.
7514
yx

28.
12
2
37
4
yx
yx
 

First, clear both equations of fractions. Multiply the
first equation by 2.
4328
yx

Solve the first of these equations for x.
24
24
42
yx
yx
xy
 


424 4
x
 
The solution set is
4, 4 .
Section 4.2 Solving Systems of Linear Equations by the Substitution Method
29.
1
623
xy

23xy
Clear the first equation of fractions by multiplying 6.
1
66
62 3
32
xy
xy





Solve this equation for x.
30.
1
44
xy

49xy
Clear the first equation of fractions by multiplying 4.
441
44
4
xy
xy





Solve this equation for x.
4xy
31.
23 82
34 314
xy x
xyxy


Simplify the first equation.
Simplify the second equation.
34 314
34 3 314 3
214
xyxy
xyxyxy xy
xy



Solve the last equation for y.
214
14 2
xy
yx


Substitute
14 2
x
for y in the equation 43 8.xy
The solution set is
5, 4 .
32.
34 4
26 54
xyxy
xyy
 

Simplify the first equation.
34 4
23 4
xyxy
xy
 

Simplify the second equation.
26 54
24
xyy
xy


21 4 2
y
  
The solution set is
1, 2 .
Chapter 4 Systems of Linear Equations
33.
81
41
xy
xy


Substitute y + 41 for x in the first equation.
81
xy

34.
62
12
xy
xy


Substitute y + 12 for x in the first equation.
62
12 62
xy
yy


35.
5
46
xy
xy

Solve the first equation for x.
5
5
xy
xy


Substitute 5y for x in the second equation.
46
xy
36.
25
212
xy
xy

Solve the first equation for x.
25
xy

25 5 25 30xy  
The numbers are 5 and 30.
37.
1
27
xy
xy


Solve the first equation for x.
36
2
y
y
Back-substitute.
1213xy
The numbers are 2 and 3.
38.
5
214
xy
xy

