1. Examples and Guided Practice come from the Algebra 1 PowerPoint Presentations available at

2. Lesson 7.1

3. Check the intersection point
EXAMPLE 1
Check the intersection point
Use the graph to solve the system. Then check your solution algebraically.
x + 2y = 7
Equation 1
3x – 2y = 5
Equation 2
SOLUTION
The lines appear to intersect at the point (3, 2).
CHECK
Substitute 3 for x and 2 for y in each equation.
x + 2y = 7
3 + 2(2) = ? 7
7 = 7

4. Check the intersection point
EXAMPLE 1
Check the intersection point
3x – 2y = 5
3(3) – 2(2) 5 = ?
5 = 5
ANSWER
Because the ordered pair (3, 2) is a solution of each equation, it is a solution of the system.

5. Use the graph-and-check method
EXAMPLE 2
Use the graph-and-check method
Solve the linear system:
–x + y = –7
Equation 1
x + 4y = –8
Equation 2
SOLUTION
STEP 1
Graph both equations.

6. Use the graph-and-check method
EXAMPLE 2
Use the graph-and-check method
STEP 2
Estimate the point of intersection. The two lines appear to intersect at (4, – 3).
STEP 3
Check whether (4, –3) is a solution by substituting 4 for x and –3 for y in each of the original equations.
Equation 1
Equation 2
–x + y = –7
x + 4y = –8
–(4) + (–3) –7 = ?
4 + 4(–3) –8 = ?
–7 = –7
–8 = –8

7. EXAMPLE 2
Use the graph-and-check method
ANSWER
Because (4, –3) is a solution of each equation, it is a solution of the linear system.

8. EXAMPLE 2
GUIDED PRACTICE
Use the graph-and-check method
for Examples 1 and 2
Solve the linear system by graphing. Check your solution.
–5x + y = 0 1.
5x + y = 10
ANSWER
(1, 5)

9. EXAMPLE 2
GUIDED PRACTICE
Use the graph-and-check method
for Examples 1 and 2
Solve the linear system by graphing. Check your solution.
2x + y = 4
–x + 2y = 3 2.
ANSWER
(1, 2)

10. EXAMPLE 2
GUIDED PRACTICE
Use the graph-and-check method
for Examples 1 and 2
Solve the linear system by graphing. Check your solution.
3x + y = 3
x – y = 5 3.
ANSWER
(2, 3)

11. EXAMPLE 3
Standardized Test Practice
The parks and recreation department in your town offers a season pass for $90. •
As a season pass holder, you pay $4 per session to use the town’s tennis courts. •
Without the season pass, you pay $13 per session to use the tennis courts.

12. EXAMPLE 3
Standardized Test Practice
Which system of equations can be used to find the number x of sessions of tennis after which the total cost y with a season pass, including the cost of the pass, is the same as the total cost without a season pass?
y = 13x
y = 4x A
y = 4x
y = x B
y = 13x
y = x C
y = x
y = x D

13. EXAMPLE 3
Standardized Test Practice
SOLUTION
Write a system of equations where y is the total cost (in dollars) for x sessions.
EQUATION 1
y = x

14. Standardized Test Practice
EXAMPLE 3
Standardized Test Practice
EQUATION 2
y = x
ANSWER
The correct answer is C. A C B D

15. GUIDED PRACTICE
for Example 3
4. Solve the linear system in Example 3 to find the number of sessions after which the total cost with a season pass, including the cost of the pass, is the same as the total cost without a season pass.
ANSWER
10 sessions

16. GUIDED PRACTICE
for Example 3
5. WHAT IF? In Example 3, suppose a season pass costs $135. After how many sessions is the total cost with a season pass, including the cost of the pass, the same as the total cost without a season pass?
ANSWER
15 sessions

17. EXAMPLE 4
Solve a multi-step problem
RENTAL BUSINESS
A business rents in-line skates and bicycles. During one day, the business has a total of 25 rentals and collects $450 for the rentals. Find the number of pairs of skates rented and the number of bicycles rented.

18. Solve a multi-step problem
EXAMPLE 4
Solve a multi-step problem
SOLUTION
STEP 1
Write a linear system. Let x be the number of pairs of skates rented, and let y be the number of bicycles rented.
x + y = 25
Equation for number of rentals
15x + 30y = 450
Equation for money collected from rentals
STEP 2
Graph both equations.

19. Solve a multi-step problem
EXAMPLE 4
Solve a multi-step problem
STEP 3
Estimate the point of intersection. The two lines appear to intersect at (20, 5).
STEP 4
Check whether (20, 5) is a solution. = ?
15(20) + 30(5) = ?
25 = 25
450 = 450
ANSWER
The business rented 20 pairs of skates and 5 bicycles.

20. EXAMPLE 4
Solve a multi-step problem
GUIDED PRACTICE
for Example 4
In Example 4, suppose the business has a total of 20 rentals and collects $420. Find the number of bicycles rented. 6.
ANSWER
8 bicycles

21. Lesson 7.2

22. Use the substitution method
EXAMPLE 1
Use the substitution method
Solve the linear system:
y = 3x + 2
Equation 1
x + 2y = 11
Equation 2
SOLUTION
STEP 1
Solve for y. Equation 1 is already solved for y.

23. Use the substitution method
EXAMPLE 1
Use the substitution method
STEP 2
Substitute 3x + 2 for y in Equation 2 and solve for x.
x + 2y = 11
Write Equation 2.
x + 2(3x + 2) = 11
Substitute 3x + 2 for y.
7x + 4 = 11
Simplify.
7x = 7
Subtract 4 from each side.
x = 1
Divide each side by 7.

24. EXAMPLE 1
Use the substitution method
STEP 3
Substitute 1 for x in the original Equation 1 to find the value of y.
y = 3x + 2 = 3(1) + 2 = = 5
ANSWER
The solution is (1, 5).

25. Use the substitution method
EXAMPLE 1
GUIDED PRACTICE
Use the substitution method
CHECK
Substitute 1 for x and 5 for y in each of the original
equations.
y = 3x + 2
x + 2y = 11
5 = 3(1) + 2 ?
1 + 2 (5) = 11 ?
5 = 5
11 = 11

26. Use the substitution method
EXAMPLE 2
Use the substitution method
Solve the linear system:
x – 2y = –6
Equation 1
4x + 6y = 4
Equation 2
SOLUTION
STEP 1
Solve Equation 1 for x.
x – 2y = –6
Write original Equation 1.
x = 2y – 6
Revised Equation 1

27. Use the substitution method
EXAMPLE 2
Use the substitution method
STEP 2
Substitute 2y – 6 for x in Equation 2 and solve for y.
4x + 6y = 4
Write Equation 2.
4(2y – 6) + 6y = 4
Substitute 2y – 6 for x.
8y – y = 4
Distributive property
14y – 24 = 4
Simplify.
14y = 28
Add 24 to each side.
y = 2
Divide each side by 14.

28. Use the substitution method
EXAMPLE 2
Use the substitution method
STEP 3
Substitute 2 for y in the revised Equation 1 to find the value of x.
x = 2y – 6
Revised Equation 1
x = 2(2) – 6
Substitute 2 for y.
x = –2
Simplify.
ANSWER
The solution is (–2, 2).

29. Use the substitution method
EXAMPLE 2
GUIDED PRACTICE
Use the substitution method
CHECK
Substitute –2 for x and 2 for y in each of the original
equations.
Equation 1
Equation 2
4x + 6y = 4
x – 2y = –6
–2 – 2(2) = –6 ?
4(–2) + 6 (2) = 4 ?
–6 = –6
4 = 4

30. EXAMPLE 1
GUIDED PRACTICE
Use the substitution method
for Examples 1 and 2
Solve the linear system using the substitution method.
y = 2x + 5 1.
3x + y = 10
ANSWER
(1, 7)

31. EXAMPLE 2
GUIDED PRACTICE
Use the substitution method
for Examples 1 and 2
Solve the linear system using the substitution method.
x – y = 3 2.
x + 2y = –6
ANSWER
(0, –3)

32. EXAMPLE 2
GUIDED PRACTICE
Use the substitution method
for Examples 1 and 2
Solve the linear system using the substitution method.
3x + y = –7 3.
–2x + 4y = 0
ANSWER
(–2, –1)

33. EXAMPLE 3
Solve a multi-step problem
WEBSITES
Many businesses pay website hosting companies to store and maintain the computer files that make up their websites. Internet service providers also offer website hosting. The costs for website hosting offered by a website hosting company and an Internet service provider are shown in the table. Find the number of months after which the total cost for website hosting will be the same for both companies.

34. EXAMPLE 3
Solve a multi-step problem
SOLUTION
STEP 1
Write a system of equations. Let y be the total
cost after x months.
Equation 1: Internet service provider
y = x

35. Solve a multi-step problem
EXAMPLE 3
Solve a multi-step problem
Equation 2: Website hosting company
y = x
The system of equations is:
y = x
Equation 1
y = 22.45x
Equation 2

36. Solve a multi-step problem
EXAMPLE 3
Solve a multi-step problem
STEP 2
Substitute 22.45x for y in Equation 1 and solve
for x.
y = x
Write Equation 1.
22.45x = x
Substitute 22.45x for y.
0.5x = 10
Subtract 21.95x from each side.
x = 20
Divide each side by 0.5.
The total cost will be the same for both companies after 20 months.
ANSWER

37. GUIDED PRACTICE
for Example 3 4.
In Example 3, what is the total cost for website hosting for each company after 20 months?
$449
ANSWER

38. GUIDED PRACTICE
for Example 3 5.
WHAT IF? In Example 3, suppose the Internet service provider offers $5 off the set-up fee. After how many months will the total cost for website hosting be the same for both companies?
10 mo
ANSWER

39. EXAMPLE 4
Solve a mixture problem
ANTIFREEZE
For extremely cold temperatures, an automobile manufacturer recommends that a 70% antifreeze and 30% water mix be used in the cooling system of a car. How many quarts of pure (100%) antifreeze and a 50% antifreeze and 50% water mix should be combined to make 11 quarts of a 70% antifreeze and 30% water mix?

40. EXAMPLE 4
Solve a mixture problem
SOLUTION
STEP 1
Write an equation for the total number of quarts and an equation for the number of quarts of antifreeze. Let x be the number of quarts of 100% antifreeze, and let y be the number of quarts of a 50% antifreeze and 50% water mix.

41. EXAMPLE 4
Solve a mixture problem
Equation 1: Total number of quarts
x + y = 11
Equation 2: Number of quarts of antifreeze
x quarts of
100% antifreeze
y quarts of
50%–50% mix
11 quarts of
70%–30% mix
1 x y = (11)
x + 0.5y = 7.7

42. Solve a mixture problem
EXAMPLE 4
Solve a mixture problem
The system of equations is:
x + y =11
Equation 1
x + 0.5y = 7.7
Equation 2
STEP 2
Solve Equation 1 for x.
x + y = 11
Write Equation 1
x = 11 – y
Revised Equation 1
STEP 3
Substitute 11 – y for x in Equation 2 and solve
for y.
x + 0.5y = 7.7
Write Equation 2.

43. Solve a mixture problem
EXAMPLE 4
Solve a mixture problem
(11 – y) + 0.5y = 7.7
Substitute 11 – y for x.
Solve for y.
y = 6.6
STEP 4
Substitute 6.6 for y in the revised Equation 1 to
find the value of x.
x = 11 – y = 11 – 6.6 = 4.4
ANSWER
Mix 4.4 quarts of 100% antifreeze and 6.6 quarts of a 50%
antifreeze and 50% water mix to get 11 quarts of a 70%
antifreeze and 30% water mix.

44. GUIDED PRACTICE
for Example 4
WHAT IF? How many quarts of 100% antifreeze and a 50% antifreeze and 50% water mix should be combined to make 16 quarts of a 70% antifreeze and 30% water mix? 6.
ANSWER
6.4 quarts of 100% antifreeze and 9.6 quarts of a 50%
antifreeze and 50% water mix

45. Lesson 7.3

46. Use addition to eliminate a variable
EXAMPLE 1
Use addition to eliminate a variable
Solve the linear system:
2x + 3y = 11
Equation 1
–2x + 5y = 13
Equation 2
SOLUTION
STEP 1
Add the equations to
eliminate one variable.
2x + 3y = 11
–2x + 5y = 13
STEP 2
Solve for y.
8y = 24
y = 3

47. Use addition to eliminate a variable
EXAMPLE 1
Use addition to eliminate a variable
STEP 3
Substitute 3 for y in either equation and
solve for x.
2x + 3y = 11
Write Equation 1
2x + 3(3) = 11
Substitute 3 for y.
x = 1
Solve for x.
ANSWER
The solution is (1, 3).

48. Use addition to eliminate a variable
EXAMPLE 1
Use addition to eliminate a variable
CHECK
Substitute 1 for x and 3 for y in each of the original equations.
2x + 3y = 11
2x + 5y = 13
2(1) + 3(3) = 11 ?
2(1) + 5(3) = 13 ?
11 = 11
13 = 13

49. Use subtraction to eliminate a variable
EXAMPLE 2
Use subtraction to eliminate a variable
Solve the linear system:
4x + 3y = 2
Equation 1
5x + 3y = –2
Equation 2
SOLUTION
STEP 1
Subtract the equations to
eliminate one variable.
4x + 3y = 2
5x + 3y = –2
STEP 2
Solve for x.
– x = 4
x = 4

50. Use subtraction to eliminate a variable
EXAMPLE 2
Use subtraction to eliminate a variable
STEP 3
Substitute 4 for x in either equation and solve
for y.
4x + 3y = 2
Write Equation 1.
4(–4) + 3y = 2
Substitute –4 for x.
y = 6
Solve for y.
ANSWER
The solution is (–4, 6).

51. Solve the linear system: 8x – 4y = –4 4y = 3x + 14
EXAMPLE 3
Arrange like terms
Solve the linear system:
8x – 4y = –4
Equation 1
4y = 3x + 14
Equation 2
SOLUTION
STEP 1
Rewrite Equation 2 so that the like terms are
arranged in columns.
8x – 4y = –4
8x – 4y = –4
4y = 3x + 14
3x + 4y = 14
STEP 2
Add the equations.
5x = 10
STEP 3
Solve for x.
x = 2

52. Substitute 2 for x in either equation and solve for y.
EXAMPLE 3
Arrange like terms
STEP 4
Substitute 2 for x in either equation and
solve for y.
4y = 3x + 14
Write Equation 2.
4y = 3(2) + 14
Substitute 2 for x.
y = 5
Solve for y.
ANSWER
The solution is (2, 5).

53. GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 1.
4x – 3y = 5 `
–2x + 3y = –7
ANSWER
(–1, –3)

54. GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 2.
5x – 6y = 8 –
5x + 2y = 4
ANSWER
(2, –3)

55. GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 3.
6x – 4y = 14
3x + 4y = 1 –
ANSWER
(5, 4)

56. GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 4.
7x – 2y = 5
7x – 3y = 4
ANSWER
(1, 1)

57. GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 5.
3x + 4y = –6 =
3x + 6 2y
ANSWER
(–2, 0)

58. GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 6.
2x + 5y = 12 =
4x + 6 5y
ANSWER
(1, 2)

59. EXAMPLE 4
Write and solve a linear system
KAYAKING
During a kayaking trip, a kayaker travels 12 miles upstream (against the current) and 12 miles downstream (with the current), as shown. The speed of the current remained constant during the trip. Find the average speed of the kayak in still water and the speed of the current.

60. EXAMPLE 4
Write and solve a linear system
STEP 1
Write a system of equations. First find the speed of the kayak going upstream and the speed of the kayak going downstream.
Upstream:
d = rt
Downstream:
d = rt
12 = r 3
12 = r 2
4 = r
6 = r

61. EXAMPLE 4
Write and solve a linear system
Use the speeds to write a linear system. Let x be the average speed of the kayak in still water, and let y be the speed of the current.
Equation 1:
Going upstream x y 4 = –

62. EXAMPLE 4
Write and solve a linear system
Equation 2:
Going downstream x y 6 = +

63. Write and solve a linear system
EXAMPLE 4
Write and solve a linear system
STEP 2
Solve the system of equations.
x – y = 4
Write Equation 1.
x + y = 6
Write Equation 2.
2x = 10
Add equations.
x = 5
Solve for x.
Substitute 5 for x in Equation 2 and solve for y.

64. Write and solve a linear system
EXAMPLE 4
Write and solve a linear system
5 + y = 6
Substitute 5 for x in Equation 2.
y = 1
Subtract 5 from each side.

65. GUIDED PRACTICE
for Example 4
7. WHAT IF? In Example 4, suppose it takes the kayaker 5 hours to travel 10 miles upstream and 2 hours to travel 10 miles downstream. The speed of the current remains constant during the trip. Find the average speed of the kayak in still water and the speed of the current.
ANSWER
average speed of the kayak: 3.5 mi/h, speed of the current 1.5 mi/h

66. Lesson 7.4

67. Multiply one equation, then add
EXAMPLE 1
Multiply one equation, then add
Solve the linear system:
6x + 5y = 19
Equation 1
2x + 3y = 5
Equation 2
SOLUTION
STEP 1
Multiply Equation 2 by –3 so that the coefficients of x are opposites.
6x + 5y = 19
6x + 5y = 19
2x + 3y = 5
–6x – 9y = –15
STEP 2
Add the equations.
–4y = 4

68. Multiply one equation, then add
EXAMPLE 1
Multiply one equation, then add
STEP 3
Solve for y.
y = –1
STEP 4
Substitute –1 for y in either of the original equations and solve for x.
2x + 3y = 5
Write Equation 2.
2x + 3(–1) = 5
Substitute –1 for y.
2x + (–3) = 5
Multiply.
2x = 8
Subtract –3 from each side.
x = 4
Divide each side by 2.

69. Multiply one equation, then add
EXAMPLE 1
Multiply one equation, then add
ANSWER
The solution is (4, –1).
CHECK
Substitute 4 for x and –1 for y in each of the original equations.
Equation 1
Equation 2
6x + 5y = 19
2x + 3y = 5
6(4) + 5(–1) = 19 ?
2(4) + 3(–1) = 5 ?
19 = 19
5 = 5

70. Multiply both equations, then subtract
EXAMPLE 2
Multiply both equations, then subtract
Solve the linear system:
4x + 5y = 35
Equation 1
2y = 3x – 9
Equation 2
SOLUTION
STEP 1
Arrange the equations so that like terms are in columns.
4x + 5y = 35
Write Equation 1.
–3x + 2y = –9
Rewrite Equation 2.

71. EXAMPLE 2
Multiply both equations, then subtract
STEP 2
Multiply Equation 1 by 2 and Equation 2 by 5 so that the coefficient of y in each equation is the least common multiple of 5 and 2, or 10.
4x + 5y = 35
8x + 10y = 70
–3x + 2y = –9
–15x +10y = –45
STEP 3
Subtract: the equations.
23x = 115
STEP 4
Solve: for x.
x = 5

72. Multiply both equations, then subtract
EXAMPLE 2
Multiply both equations, then subtract
STEP 5
Substitute 5 for x in either of the original equations and solve for y.
4x + 5y = 35
Write Equation 1.
4(5) + 5y = 35
Substitute 5 for x.
y = 3
Solve for y.
ANSWER
The solution is (5, 3).

73. Multiply both equations, then subtract
EXAMPLE 2
Multiply both equations, then subtract
CHECK
Substitute 5 for x and 3 for y in each of the original equations.
Equation 1
Equation 2
4x + 5y = 35
2y = 3x – 9
4(5) + 5(3) = 35 ?
2(3) = 3(5) – 9 ?
35 = 35
6 = 6
ANSWER
The solution is (5, 3).

74. GUIDED PRACTICE
for Examples 1 and 2
Solve the linear system using elimination.
6x – 2y = 1 1.
–2x + 3y = –5
ANSWER
The solution is (–0.5, –2).

75. GUIDED PRACTICE
for Examples 1 and 2
Solve the linear system using elimination.
2x + 5y = 3 2.
3x + 10y = –3
ANSWER
The solution is (9, –3).

76. GUIDED PRACTICE
for Examples 1 and 2
Solve the linear system using elimination.
3x – 7y = 5 3.
9y = 5x + 5
ANSWER
The solution is (–10, –5).

77. EXAMPLE 3
Standardized Test Practice
Darlene is making a quilt that has alternating stripes of regular quilting fabric and sateen fabric. She spends $76 on a total of 16 yards of the two fabrics at a fabric store. Which system of equations can be used to find the amount x (in yards) of regular quilting fabric and the amount y (in yards) of sateen fabric she purchased?
x + y = 16 A
x + y = 16 B
x + y = 76
4x + 6y = 76
x + y = 16 D
x + y = 76 C
4x + 6y = 16
6x + 4y = 76

78. EXAMPLE 3
Standardized Test Practice
SOLUTION
Write a system of equations where x is the number of yards of regular quilting fabric purchased and y is the number of yards of sateen fabric purchased.
Equation 1: Amount of fabric x + y = 16

79. Standardized Test Practice
EXAMPLE 3
Standardized Test Practice
Equation 2: Cost of fabric 4 76 6 + = y x
The system of equations is:
x + y = 16
Equation 1
4x + 6y = 76
Equation 2
ANSWER A D C B
The correct answer is B.

80. GUIDED PRACTICE
for Example 3
SOCCER A sports equipment store is having a sale on soccer balls. A soccer coach purchases 10 soccer balls and 2 soccer ball bags for $155. Another soccer coach purchases 12 soccer balls and 3 soccer ball bags for $189. Find the cost of a soccer ball and the cost of a soccer ball bag. 4.
ANSWER
soccer ball $14.50, soccer ball bag: $5

81. Lesson 7.6

82. Graph a system of two linear inequalities
EXAMPLE 1
Graph a system of two linear inequalities
Graph the system of inequalities.
y > –x – 2
y  3x + 6
Inequality 1
Inequality 2
SOLUTION
Graph both inequalities in the same coordinate plane. The graph of the system is the intersection of the two half-planes, which is shown as the darker shade of blue.

83. Graph a system of two linear inequalities
EXAMPLE 1
Graph a system of two linear inequalities
CHECK
Choose a point in the dark blue region, such as (0, 1). To check this solution, substitute 0 for x and 1 for y into each inequality.
1 > 0 – 2 ?
1  0 + 6 ?
1 > –2
1  6

84. Graph a system of three linear inequalities
EXAMPLE 2
Graph a system of three linear inequalities
Graph the system of inequalities.
y > –1
x > –2
Inequality 1
Inequality 2
x + 2y  4
Inequality 3
SOLUTION
Graph all three inequalities in the same coordinate plane. The graph of the system is the triangular region shown.

85. GUIDED PRACTICE
for Examples 1 and 2
Graph the system of linear inequalities. 1.
y < x – 4
y  –x + 3
ANSWER

86. GUIDED PRACTICE
for Examples 1 and 2
Graph the system of linear inequalities. 2.
y  –x + 2
y < 4
x < 3
ANSWER

87. GUIDED PRACTICE
for Examples 1 and 2
Graph the system of linear inequalities. 3.
y > –x
y  x – 4
y < 5
ANSWER

88. EXAMPLE 3
Write a system of linear inequalities
Write a system of inequalities for the shaded region.
SOLUTION
INEQUALITY 1: One boundary line for the shaded region is y = 3. Because the shaded region is above the solid line, the inequality is y  3.
INEQUALITY 2: Another boundary line for the shaded region has a slope of 2 and a y-intercept of 1. So, its equation is y = 2x + 1. Because the shaded region is above the dashed line, the inequality is y > 2x + 1.

89. Write a system of linear inequalities
EXAMPLE 3
Write a system of linear inequalities
ANSWER
The system of inequalities for the shaded region is:
y  3
y > 2x + 1
Inequality 1
Inequality 2

90. EXAMPLE 4
Write and solve a system of linear inequalities
BASEBALL
The National Collegiate Athletic Association (NCAA) regulates the lengths of aluminum baseball bats used by college baseball teams. The NCAA states that the length (in inches) of the bat minus the weight (in ounces) of the bat cannot exceed 3. Bats can be purchased at lengths from 26 to 34 inches.
a. Write and graph a system of linear inequalities that describes the information given above.
b. A sporting goods store sells an aluminum bat that is 31 inches long and weighs 25 ounces. Use the graph to determine if this bat can be used by a player on an NCAA team.

91. Write and solve a system of linear inequalities
EXAMPLE 4
Write and solve a system of linear inequalities
SOLUTION
a. Let x be the length (in inches) of the bat, and let y be the weight (in ounces) of the bat. From the given information, you can write the following inequalities:
x – y  3
The difference of the bat’s length and weight can be at most 3.
x ≥ 26
The length of the bat must be at least 26 inches.
x ≤ 34
The length of the bat can be at most 34 inches.
y ≥ 0
The weight of the bat cannot be a negative number.
Graph each inequality in the system. Then identify the region that is common to all of the graphs of the inequalities. This region is shaded in the graph shown.

92. EXAMPLE 4
Write and solve a system of linear inequalities
b. Graph the point that represents a bat that is 31 inches long and weighs 25 ounces.
ANSWER
Because the point falls outside the solution region, the bat cannot be used by a player on an NCAA team.

93. GUIDED PRACTICE
for Examples 3 and 4
Write a system of inequalities that defines the shaded region. 4.
ANSWER
x ≤ 3, y > x 1 3 2

94. GUIDED PRACTICE
for Examples 3 and 4
Write a system of inequalities that defines the shaded region. 5.
ANSWER
y ≤ 4, x < 2

95. GUIDED PRACTICE
for Examples 3 and 4
6. WHAT IF? In Example 4, suppose a Senior League (ages 10–14) player wants to buy the bat described in part (b). In Senior League, the length (in inches) of the bat minus the weight (in ounces) of the bat cannot exceed 8. Write and graph a system of inequalities to determine whether the described bat can be used by the Senior League player.
ANSWER
x y ≤ 8, x ≥ 26, x ≤ 34, y ≥ 0
The bat can be used.