1. Chapter 6: Systems of Equations and Inequalities
Algebra I

2. Table of Contents 6.1 - Solving Systems by Graphing
6.2 - Solving Systems by Substitution
6.3 - Solving Systems by Elimination
6.4 - Solving Special Systems
6.5 - Solving Linear Inequalities
6.6 - Solving Systems of Linear Inequalities

3. 6.1 - Solving Systems by Graphing
Algebra I

4. 6.1
Algebra 1 (bell work)
Just Read
A system of linear equations is a set of two or more linear equations containing two or more variables.
A solution of a system of linear equations with two variables is an ordered pair that satisfies each equation in the system.
-So, if an ordered pair is a solution, it will make both equations true.

5.   3x – y = 13 6.1 Example 1 Identifying Solutions of Systems
Tell whether the ordered pair is a solution of the given system.
(5, 2);
3x – y = 13
2 – 
3(5) –
15 – 
3x – y 13
The ordered pair (5, 2) makes both equations true.
(5, 2) is the solution of the system.

6. 6.1
Tell whether the ordered pair is a solution of the given system.
(1, 3);
2x + y = 5
–2x + y = 1
2x + y = 5
2(1) 
–2x + y = 1
–2(1) – 
The ordered pair (1, 3) makes both equations true.
(1, 3) is the solution of the system.

7. 6.1
Tell whether the ordered pair is a solution of the given system.
Optional
x – 2y = 4
(2, –1);
3x + y = 6
x – 2y = 4
2 – 2(–1) 4 
3x + y = 6
3(2) + (–1) 6
6 –
The ordered pair (2, –1) makes one equation true, but not the other.
(2, –1) is not a solution of the system.

8. 6.1
Math Joke
Teacher: What is the name of the formula that describes the phases of the moon?
Student: The lunear (linear) equation

9. 6.1
Just Read
All solutions of a linear equation are on its graph.
To find a solution of a system of linear equations, you need a point that each line has in common.
In other words, you need their point of intersection.
y = 2x – 1
y = –x + 5
The point (2, 3) is where the two lines intersect and is a solution of both equations, so (2, 3) is the solution of the systems.

10. 6.1
Example 2
Solving a System of Linear Equations by Graphing
Solve the system by graphing. Check your answer.
y = x
y = –2x – 3
Check
Substitute (–1, –1) into the system.
y = x
(–1) (–1)
–1 –1 
y = –2x – 3
(–1) –2(–1) –3
– – 3
–1 – 1 

11. 6.1
Solve the system by graphing. Check your answer.
y = x – 6
y x = –1
y + x = –1
− x − x
y =

12. 6.1
Solve the system by graphing. Check your answer.
Day 2
y = –2x – 1
y = x + 5
Check Substitute (–2, 3) into the system.
y = –2x – 1
–2(–2) – 1
– 1 
y = x + 5
–2 + 5 

13. 2x + y = 4 6.1 Solve the system by graphing. Check your answer.
–2x – 2x
y = –2x + 4

14. Wren and Jenni are reading the same book.
6.1
Example 3
Problem Solving Application
Wren and Jenni are reading the same book.
Wren is on page 14 and reads 2 pages every night.
Jenni is on page 6 and reads 3 pages every night.
After how many nights will they have read the same number of pages? How many pages will that be?
Total
pages is
number read
every
night
plus
already read.
Wren y = 2
 x + 14
Jenni y = 3
 x + 6

15. The lines appear to intersect at (8, 30).
6.1
Method 1 (Graph)
Method 2 (Substitution) 
(8, 30)
Nights
Graph y = 2x + 14 and y = 3x + 6.
The lines appear to intersect at (8, 30).
So, the number of pages read will be the same at 8 nights with a total of 30 pages.

16. Video club A charges $10 for membership and $3 per movie rental.
6.1
Video club A charges $10 for membership and $3 per movie rental.
Video club B charges $15 for membership and $2 per movie rental.
For how many movie rentals will the cost be the same at both video clubs? What is that cost?
Optional
Total
cost is
price
for each
rental
plus
member-
ship fee.
Club A y = 3
 x + 10
Club B y = 2
 x + 15

17. 6.1
Method 1 (Graph)
Method 2 (Substitution)
Graph y = 3x + 10 and y = 2x The lines appear to intersect at (5, 25). So, the cost will be the same for 5 rentals and the total cost will be $25.

18. HW pg. 400
6.1-
Day 1: 1-7, 27, 33-42
Day 2: 12-15, 8, 16, 17, 25, 26, 28
For the story problems you can use method 1 or 2
Ch: 18, 23, 24, 31, 32
HW Guidelines or ½ off
Start a new HW sheet for Ch 6

19. 6.2 – Solving Systems by Substitution
Algebra I

20. Steps for solving by Substitution
6.2
Algebra 1 (bell work)
Steps for solving by Substitution
Solve one equation for x= or y=
Substitute that equation into the other equation
Use the new equation to solve for the value of x or y
Plug the value back in to find other variable (Back Substition)

21. 6.2
Example 1
Solving a System of Linear Equations by Substitution
Solve the system by substitution.
y = 3x
Step 4
y = 3x
y = x – 2
y = 3(–1)
y = –3
Step 1 y = 3x
Step 5
(–1, –3)
y = x – 2
Solution is always an ordered point or pair
Step 2 y = x – 2
3x = x – 2
Step 3
–x –x
2x = –2
2x = –2
x = –1

22. y = x + 1 4x + y = 6 6.2 Solve the system by substitution. Step 4
Step 1 y = x + 1
Step 5
(1, 2)
Step 2 4x + y = 6
4x + (x + 1) = 6
5x + 1 = 6
Step 3
–1 –1
5x = 5
x = 1
5x = 5

23. 6.2
Math Joke
Q: Why did the chef throw the unsolved system of equations out of the restaurant?
A: The menu said, “No substitution.”

24. x + 2y = –1 x – y = 5 6.2 Optional Solve the system by substitution.
Step 3
–3y – 1 = 5
x – y = 5
+1 +1
–3y = 6
Step 1 x + 2y = –1
–3y = 6
–3 –3
y = –2
−2y −2y
x = –2y – 1
Step 4
x – y = 5
Step 2 x – y = 5
(–2y – 1) – y = 5
x – (–2) = 5
x + 2 = 5
–3y – 1 = 5
–2 –2
x = 3
Step 5 (3, –2)

25. y + 6x = 11 3x + 2y = –5 6.2 Example 2 Using the Distributive Property
Solve the system by substitution.
y + 6x = 11
Step 3
3x + 2(–6x) + 2(11) = –5
3x + 2y = –5
3x – 12x + 22 = –5
–9x + 22 = –5
Step 1 y + 6x = 11
– 6x – 6x
y = –6x + 11
–9x = –27
– 22 –22
–9x = –27
– –9
3x + 2(–6x + 11) = –5
3x + 2y = –5
Step 2
x = 3
Step 4
y + 6x = 11
y + 6(3) = 11
3x + 2(–6x + 11) = –5
y + 18 = 11
–18 –18
y = –7
Step 5
(3, –7)

26. –2x + y = 8 3x + 2y = 9 6.2 Day 2 Solve the system by substitution.
Step 3
3x + 2(2x) + 2(8) = 9
3x + 2y = 9
3x + 4x = 9
7x = 9
Step 1 –2x + y = 8
7x = –7
–16 –16
+ 2x x
y = 2x + 8
7x = –7
3x + 2(2x + 8) = 9
3x + 2y = 9
Step 2
x = –1
Step 4
–2x + y = 8
3x + 2(2x + 8) = 9
–2(–1) + y = 8
y + 2 = 8
–2 –2
y = 6
Step 5
(–1, 6)

27. Jenna is deciding between two cell-phone plans.
6.2
Example 3
Application
Jenna is deciding between two cell-phone plans.
The first plan has a $50 sign-up fee and costs $20 per month.
The second plan has a $30 sign-up fee and costs $25 per month.
After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.
Total paid
sign-up fee
payment
amount
for each
month. is
plus
Option 1 t =
$50 +
$20 m
Option 2 t =
$30 +
$25 m
Step 1 t = m
t = m
Step 2
m = m

28. Total paid sign-up fee payment amount for each month. is plus Option 1
6.2
Total paid
sign-up fee
payment
amount
for each
month. is
plus
Option 1 t =
$50 +
$20 m
Option 2 t =
$30 +
$25 m
Step 3
m = m
Step 4
t = m
–20m – 20m
t = (4)
50 = m
t =
t = 130
– –30
= m
m = 4
20 = 5m

29. 6.2
After how many months will the total costs be the same? What will the costs be?
If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.
Step 5
(4, 130)
In 4 months, the total cost for each option would be the same $130.
If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.
Option 1: t = (12) = 290
Option 2: t = (12) = 330
Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330.

30. 6.2
One cable television provider has a $60 setup fee and $80 per month, and the second has a $160 equipment fee and $70 per month.
Optional
a. In how many months will the cost be the same? What will that cost be.
Total paid
payment
amount
for each
month. is
fee
plus
Option 1 t =
$60 +
$80 m
Option 2 t =
$160 +
$70 m
Step 1 t = m
t = m
Step 2
m = m

31. Step 3 60 + 80m = 160 + 70m –70m –70m 60 + 10m = 160 –60 –60 10m = 100
6.2
Step 3
m = m
–70m –70m
m = 160
– –60
10m = 100
m = 10
Step 4
t = m
t = (10)
t =
t = 860

32. In 10 months, the total cost for each option would be the same, $860.
6.2
Step 5
(10, 860)
In 10 months, the total cost for each option would be the same, $860.
b. If you plan to move in 6 months, which is the cheaper option? Explain.
Option 1: t = (6) = 540
Option 2: t = (6) = 580
The first option is cheaper for the first six months.

33. HW pg. 408 6.2- Day 1: 1-5 (Odd), 9-15 (Odd), 45-47
Ch: 24, 25, 27
HW Guidelines or ½ off

34. 6.3 - Solving Systems by Elimination
Algebra I

35. Steps for solving by Elimination
6.3
Algebra 1 (bell work)
Steps for solving by Elimination
Line x and y up vertically
Eliminate one variable and solve for the other
Take solved x or y and use back substitution
Check your answer

36. 3x – 4y = 10 Solve by elimination. x + 4y = –2 6.3 Example 1
Elimination Using Addition
3x – 4y = 10
Solve by elimination.
x + 4y = –2
Step 1
3x – 4y = 10
Step 3
x + 4y = –2
x + 4y = –2
Step 2
4x = 8
2 + 4y = –2
4x = 8
– –2
4y = –4
4x = 8
x = 2
4y –4
y = –1
Step 4 (2, –1)

37. y + 3x = –2 Solve by elimination. 2y – 3x = 14 6.3 Step 1 2y – 3x = 14
– –4
3x = –6
3y = 12
3x = –6
x = –2
y = 4
Step 4
(–2, 4)

38. 2x + y = –5 Solve by elimination. 2x – 5y = 13 6.3 Example 2
Elimination Using Subtraction
2x + y = –5
Solve by elimination.
2x – 5y = 13
Step 3
2x + y = –5
2x + y = –5
Step 1
2x + (–3) = –5
–(2x – 5y = 13)
2x – 3 = –5
2x + y = –5
–2x + 5y = –13
0 + 6y = –18
Step 2
2x = –2
6y = –18
y = –3
x = –1
Step 4
(–1, –3)

39. Math Joke Teacher: What’s your solution of the system of equations?
6.3
Math Joke
Teacher: What’s your solution of the system of equations?
Student: Nothing – I used elimination and got rid of the whole thing!

40. x + 2y = 11 –3x + y = –5 6.3 Solve the system by elimination. Step 3
– –3
2y = 8
x + 2y = 11
+(6x –2y = +10)
y = 4
7x = 21
Step 4
(3, 4)
Step 2
7x = 21
x = 3

41. 6.3
Example 3
Elimination Using Multiplication First
Day 2
Solve the system by elimination.
–5x + 2y = 32
2x + 3y = 10
Step 3
2x + 3y = 10
Step 1
2(–5x + 2y = 32)
5(2x + 3y = 10)
2x + 3(6) = 10
2x + 18 = 10
–10x + 4y = 64
+(10x + 15y = 50)
–18 –18
2x = –8
Step 2
19y = 114
x = –4
y = 6
Step 4
(–4, 6)

42. 2x + 5y = 26 –3x – 4y = –25 6.3 Solve the system by elimination.
Step 1
3(2x + 5y = 26)
Step 3
2x + 5y = 26
+(2)(–3x – 4y = –25)
2x + 5(4) = 26
6x + 15y = 78
2x + 20 = 26
–20 –20
2X = 6
+(–6x – 8y = –50)
y = 28
Step 2
y = 4
x = 3
Step 4
(3, 4)

43. 1.25l + 0.90t = 14.85 l + t = 13 6.3 Example 4 Application
Sally spent $14.85 to buy 13 flowers.
She bought lilies, which cost $1.25 each, and tulips, which cost $0.90 each.
How many of each flower did Sally buy?
Write a system. Use l for the number of lilies and t for the number of tulips.
1.25l t = 14.85
l t = 13

44. 6.3
Step 1
1.25l + .90t = 14.85
+ (–.90)(l + t) = 13
1.25l t = 14.85
+ (–0.90l – 0.90t = –11.70)
0.35l = 3.15
Step 2
Step 4
(9, 4)
l = 9
Step 3
l + t = 13
Sally bought 9 lilies and 4 tulips.
9 + t = 13
– –9
t = 4

45. HW pg. 415
6.3-
Day 1: 1-6, 15, 25, 27, (Odd)
Day 2: 7-10, 17, 19, 20
Ch: 21, 30, 31
HW Guidelines or ½ off

46. 6.4 – Solving Special Systems
Algebra I

47. Systems with at least one solution are called consistent.
6.4
Algebra 1 (bell work)
Write down the following definitions
Systems with at least one solution are called consistent.
A system that has no solution is an inconsistent system. (Parallel Lines)
An independent system has exactly one solution.
A dependent system has infinitely many solutions. (Same Line, which means same slope and y-intercept)

48. One solution, consistent, independent All real, consistent, dependent.
6.4
One solution, consistent, independent
All real, consistent, dependent.
No solution, inconsistent
Pg. 421

49. Method 1 Compare slopes and y-intercepts.
6.4
y = x – 4
Solve
–x + y = 3
Method 1 Compare slopes and y-intercepts.
y = x – y = 1x – 4
Write both equations in slope-intercept form.
–x + y = y = 1x + 3
The lines are parallel because they have the same slope and different y-intercepts.
This system has no solution so it is an inconsistent system.

50.  y = x – 4 Solve –x + y = 3 Method 2 Solve the system algebraically.
6.4
y = x – 4
Solve
–x + y = 3
Method 2 Solve the system algebraically.
–x + (x – 4) = 3
–4 = 3 
This system has no solution so it is an inconsistent system.

51. Method 1 Compare slopes and y-intercepts.
6.4
y = –2x + 5
Solve
2x + y = 1
Method 1 Compare slopes and y-intercepts.
y = –2x y = –2x + 5
2x + y = y = –2x + 1
Write both equations in slope-intercept form.
The lines are parallel because they have the same slope and different y-intercepts.
This system has no solution so it is an inconsistent system.

52. 6.4
y = –2x + 5
Solve
2x + y = 1
Method 2 Solve the system algebraically.
2x + (–2x + 5) = 1
5 = 1 
This system has no solution so it is an inconsistent system.

53. Method 1 Compare slopes and y-intercepts.
6.4
y = 3x + 2
Solve
3x – y + 2= 0
Method 1 Compare slopes and y-intercepts.
Write both equations in slope-intercept form. The lines have the same slope and the same y-intercept.
y = 3x y = 3x + 2
3x – y + 2= y = 3x + 2
If this system were graphed, the graphs would be the same line. There are infinitely many solutions, consistent, dependent.

54. 6.4
y = 3x + 2
Solve
3x – y + 2= 0
Method 2 Solve the system algebraically.
y = 3x y − 3x = 2
3x − y + 2= −y + 3x = −2
0 = 0 
There are infinitely many solutions, consistent, dependent.

55. 6.4
Classify the system. Give the number of solutions.
Day 2
3y = x + 3
Solve
x + y = 1
The system is consistent and dependent. It has infinitely many solutions.

56. The system is inconsistent. It has no solutions.
6.4
Classify the system. Give the number of solutions.
x + y = 5
Solve
4 + y = –x
The system is inconsistent. It has no solutions.

57. The system is consistent and independent. It has one solution.
6.4
Classify the system. Give the number of solutions.
y = 4(x + 1)
Solve
y – 3 = x
The system is consistent and independent. It has one solution.

58. 6.4
Example 4
Application
Jared and David both started a savings account in January.
If the pattern of savings in the table continues, when will the amount in Jared’s account equal the amount in David’s account?

59. Total saved amount saved for each month. start amount is plus Jared y
6.4
Total saved
amount
saved
for each
month.
start
amount is
plus
Jared y =
$25 + $5 x
David y =
$40 + $5 x
y = 5x + 25
y = 5x + 40
y = 5x + 25
y = 5x + 40
The graphs of the two equations are parallel lines, so there is no solution.
If the patterns continue, the amount in Jared’s account will never be equal to the amount in David’s account.

60. 6.4
Optional
Matt has $100 in a checking account and deposits $20 per month.
Ben has $80 in a checking account and deposits $30 per month.
Will the accounts ever have the same balance? Explain.
y = 20x + 100
y = 30x + 80
y = 20x + 100
y = 30x + 80
The accounts will have the same balance.
The graphs of the two equations have different slopes so they intersect.

61. HW pg. 423
6.4-
Day 1: 2-7, 15, 19, (Odd)
Day 2: 8-11, 21, 23, 24, 29, 31
Ch: 28, 30
HW Guidelines or ½ off

62. 6.5 - Solving Linear Inequalities
Algebra I

63. 6.5
Algebra 1 (bell work)
Copy Definitions
A linear inequality is similar to a linear equation, but the equal sign is replaced with an inequality symbol.
A solution of a linear inequality is any ordered pair that makes the inequality true.

64.   (–2, 4); y < 2x + 1 (3, 1); y > x – 4 y > x − 4 1 3 – 4
6.5
Identifying Solutions of Inequalities
Tell whether the ordered pair is a solution of the inequality.
(–2, 4); y < 2x + 1
(3, 1); y > x – 4
y > x − 4
– 4
– 1
>
y < 2x + 1
(–2) + 1
–4 + 1 –3
< 
> 
<
<
(–2, 4) is not a solution.
(3, 1) is a solution.

65. 6.5
Don’t Copy Down
A linear inequality describes a region of a coordinate plane called a half-plane.
All points in the region are solutions of the linear inequality.
The boundary line of the region is the graph of the related equation.

66. Don’t Copy Down
6.5

67. Graphing Linear Inequalities
6.5
Steps to Graphing Linear Inequalities
Graphing Linear Inequalities
Step 1
Solve the inequality for y (slope-intercept form).
Step 2
Graph the boundary line. Use a solid line for ≤ or ≥. Use a dashed line for < or >.
Step 3
Shade the half-plane above the line for y > or ≥. Shade the half-plane below the line for y < or y ≤. Check your answer.

68. Graph the solutions of the linear inequality.
6.5
Example 2
Graphing Linear Inequalities in Two Variables
Graph the solutions of the linear inequality.
y  2x – 3
Check ( 0, 0 )

69. Math Joke Q: Why did the math student name her boundary line Hope?
6.5
Math Joke
Q: Why did the math student name her boundary line Hope?
A: Because it was dashed.

70. Graph the solutions of the linear inequality.
6.5
Graph the solutions of the linear inequality.
5x + 2y > –8
Check ( 0, 0 )

71. Graph the solutions of the linear inequality.
6.5
Graph the solutions of the linear inequality.
4x – 3y > 12
Check ( 0, 0 )

72. Graph the solutions of the linear inequality.
6.5
Optional
Graph the solutions of the linear inequality.
2x – y – 4 > 0
Check ( 0, 0 )

73. Graph the solutions of the linear inequality.
6.5
Day 2
Graph the solutions of the linear inequality.
4x – y + 2 ≤ 0
Check ( 0, 0 )

74. Necklace beads bracelet
6.5
Example 4
Application
Ada has at most 285 beads to make jewelry. A necklace requires 40 beads, and a bracelet requires 15 beads.
Write a linear inequality to describe the situation.
Let x represent the number of necklaces and y the number of bracelets.
Write an inequality. Use ≤ for “at most.”
Necklace
beads
bracelet
plus
is at
most
285
beads.
40x +
15y ≤
Solve the inequality for y.
40x + 15y ≤ 285
–40x –40x
15y ≤ –40x + 285

75. Graph the boundary line
6.5
b. Graph the solutions.
Step 1 Since Ada cannot make a negative amount of jewelry, the system is graphed only in Quadrant I.
Graph the boundary line
Use a solid line for ≤.

76. 6.5
b. Graph the solutions.
Step 2 Shade below the line. Ada can only make whole numbers of jewelry. All points on or below the line with whole number coordinates are the different combinations of bracelets and necklaces that Ada can make.

77. 6.5
c. Give two combinations of necklaces and bracelets that Ada could make.
Two different combinations of jewelry that Ada could make with 285 beads could be 2 necklaces and 8 bracelets or 5 necklaces and 3 bracelets.
(2, 8)
(5, 3) 

78. Green olives cost $2 per pound and black olives cost $2.50 per pound.
6.5
Example 4
Application
Dirk is going to bring two types of olives to the Honor Society induction and can spend no more than $6.
Green olives cost $2 per pound and black olives cost $2.50 per pound.
a. Write a linear inequality to describe the situation.
b. Graph the solutions.
c. Give two combinations of olives that Dirk could buy.

79. Write an inequality. Use ≤ for “no more than.”
6.5
Write an inequality. Use ≤ for “no more than.”
Green
olives
black
plus
is no more than
total
cost. 2x +
2.50y ≤ 6
Solve the inequality for y.
2x y ≤ 6
–2x –2x
2.50y ≤ –2x + 6
2.50y ≤ –2x + 6
2.50

80. y ≤ –0.80x + 2.4 b. Graph the solutions.
6.5
y ≤ –0.80x + 2.4
Green Olives
Black Olives
b. Graph the solutions.
Step 1 Since Dirk cannot buy negative amounts of olive, the system is graphed only in Quadrant I. Graph the boundary line for y = –0.80x Use a solid line for≤.

81. c. Give two combinations of olives that Dirk could buy.
6.5
c. Give two combinations of olives that Dirk could buy.
Two different combinations of olives that Dirk could purchase with $6 could be 1 pound of green olives and 1 pound of black olives or 0.5 pound of green olives and 2 pounds of black olives.
Black Olives 
(1, 1)
(0.5, 2)
Green Olives

82. Write an inequality to represent the graph.
6.5
Example 4
Writing an Inequality from a Graph
Write an inequality to represent the graph.
y-intercept: 0 slope: –1
Write an equation in slope-intercept form.
y = mx + b y= – 1x
The graph is shaded below a dashed boundary line.
Replace = with < to write the inequality y < –x.

83. Write an inequality to represent the graph.
6.5
Write an inequality to represent the graph.
y-intercept: 1; slope:
Write an equation in slope-intercept form.
The graph is shaded above a dashed boundary line.
Replace = with > to write the inequality

84. HW pg. 432
6.5-
Day 1: 2-8, 15-18, (Odd)
Day 2: 9-11, 19-21, 37 (Equation Only), 38
Ch: 22, 27, 28, 40-42

85. 6.6 - Solving Systems of Linear Inequalities
Algebra I

86. 6.6
Algebra 1 (bell work)
A system of linear inequalities is a set of two or more linear inequalities containing two or more variables.
The solutions of a system of linear inequalities consists of all the ordered pairs that satisfy all the linear inequalities in the system.

87.   y ≤ –3x + 1 (–1, –3); y < 2x + 2 (–1, –3) (–1, –3) y ≤ –3x + 1
6.6
Example 1
Identifying Solutions of Systems of Linear Inequalities
Tell whether the ordered pair is a solution of the given system.
y ≤ –3x + 1
(–1, –3);
y < 2x + 2
(–1, –3)
(–1, –3)
y ≤ –3x + 1
y < 2x + 2
–3 –3(–1) + 1 – – ≤ 
–3 –2 + 2 –
< 
– (–1) + 2
(–1, –3) is a solution to the system because it satisfies both inequalities.

88.   y < –2x – 1 (–1, 5); y ≥ x + 3 (–1, 5) (–1, 5) y < –2x – 1
6.6
Optional
Tell whether the ordered pair is a solution of the given system.
y < –2x – 1
(–1, 5);
y ≥ x + 3
(–1, 5)
(–1, 5)
y < –2x – 1 ≥ 
5 –1 + 3
y ≥ x + 3
5 –2(–1) – 1
– 1
< 
(–1, 5) is not a solution to the system because it does not satisfy both inequalities.

89. 6.5
Example 2
Solving a System of Linear Inequalities by Graphing
y ≤ 3
y > –x + 5

90. Math Joke Q: Why did the math student wear two pairs of sunglasses?
A: He wanted to have overlapping shades

91. 6.5
y > x – 7
3x + 6y ≤ 12

92. 6.5
–3x + 2y ≥ 2
y < 4x + 3

93. y ≤ –2x – 4 y > –2x + 5 6.5 Example 3
Graphing Systems with Parallel Boundary Lines
Day 2
Graph and Describe the
Solutions
y ≤ –2x – 4
y > –2x + 5

94. 6.5
Graph and Describe the
Solutions
y > 3x – 2
y < 3x + 6

95. Step 1 Write a system of inequalities.
6.6
Example 4
Application
In one week, Ed can mow at most 9 times and rake at most 7 times. He charges $20 for mowing and $10 for raking. He needs to make more than $125 in one week.
Show and describe all the possible combinations of mowing and raking that Ed can do to meet his goal. List two possible combinations.
Earnings per Job ($)
Mowing 20
Raking 10
Step 1 Write a system of inequalities.
Let x represent the number of mowing jobs and y represent the number of raking jobs.

96. 6.6
In one week, Ed can mow at most 9 times and rake at most 7 times. He charges $20 for mowing and $10 for raking. He needs to make more than $125 in one week. Show and describe all the possible combinations of mowing and raking that Ed can do to meet his goal. List two possible combinations.
Step 1 Write a system of inequalities.
Let x represent the number of mowing jobs and y represent the number of raking jobs.
x ≤ 9
He can do at most 9 mowing jobs.
He can do at most 7 raking jobs.
y ≤ 7
He wants to earn more than $125.
20x + 10y > 125

97. 6.6
Step 2 Graph the system.
Solutions
The graph should be in only the first quadrant because the number of jobs cannot be negative.

98. 6.6
Step 3 Describe all possible combinations.
All possible combinations represented by ordered pairs of whole numbers in the solution region will meet Ed’s requirement of mowing, raking, and earning more than $125 in one week. Answers must be whole numbers because he cannot work a portion of a job.
Step 4 List the two possible combinations.
Two possible combinations are:
7 mowing and 4 raking jobs
8 mowing and 1 raking jobs
Solutions

99. 6.6
At her party, Alice is serving pepper jack cheese and cheddar cheese. She wants to have at least 2 pounds of each. Alice wants to spend at most $20 on cheese. Show and describe all possible combinations of the two cheeses Alice could buy. List two possible combinations.
Price per Pound ($)
Pepper Jack 4
Cheddar 2
Step 1 Write a system of inequalities.
Let x represent the pounds of pepper jack
and y represent the pounds of cheddar.
x ≥ 2
She wants at least 2 pounds of pepper jack.
She wants at least 2 pounds of cheddar.
y ≥ 2
She wants to spend no more than $20.
4x + 2y ≤ 20

100. 6.6
Step 2 Graph the system.
Solutions
The graph should be in only the first quadrant because the amount of cheese cannot be negative.

101. Step 3 Describe all possible combinations.
6.6
Step 3 Describe all possible combinations.
All possible combinations within the gray region will meet Alice’s requirement of at most $20 for cheese and no less than 2 pounds of either type of cheese. Answers need not be whole numbers as she can buy fractions of a pound of cheese.
Step 4 Two possible combinations are (3, 2) and (2.5, 4).
3 pepper jack, 2 cheddar or 2.5 pepper jack, 4 cheddar.

102. HW pg. 438
6.6-
Day 1: 2-8, 19, 21, 55, 57
Day 2: 9-15, 25, (Equations Only)
Ch: 38-42