1. Unit 3 – Chapter 7

2. Unit 3 Section 7.1 – Solve Linear Equations by graphing
Section 7.2 – Solve Linear Equations by substitution.
Section 7.3 – Solve Linear Equations by Adding or Subtracting
Section 7.4 – Solve Linear Systems by multiplying first
Section 7.5 – Solve Special Types of Linear Systems
Section 7.6 – Solve systems of linear inequalities

3. Warm-Up – 7.1

4. Lesson 7.1, For use with pages 426-434
1. Graph the equation –2x + y = 1.
ANSWER
2. It takes 3 hours to mow a lawn and 2 hours to trim hedges. You spend 16 hours doing yard work. What are 2 possible numbers of lawns you mowed and hedges you trimmed?
ANSWER
2 lawns and 5 hedges, or
4 lawns and 2 hedges

5. Vocabulary – 7.1 System of Linear Equations
2 or more linear equations with the same variables
Solution of a Linear Equation
The solution set that MAKES ALL THE EQUATIONS TRUE AT THE SAME TIME!!
Usually a single point!
Consistent Independent System
A linear system that has EXACTLY one solution.

6. Notes – 7.1 – Solving Linear Systems by Graphing
Remember the steps to convert English to Mathlish:
Read and highlight key words
DEFINE THE VARIABLES (MOST CRITICAL STEP!!)
Write Mathlish sentence left to right (careful with subtraction and division!)
3 Step Process To Solve Linear Eqns by graphing
Write equations in slope-intercept form
Graph them on calculator
Find intersection of lines
CHECK ANSWERS!

7. Examples 7.1

8. 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.

9. 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

10. 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.

11. 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
Put eqns in slope-
Intercept form
y = 5x
5x + y = -5x + 10
The Intersection point is at (1,5).

12. 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. •
Write a system of equations to model these situations and find out where they are equal.

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. EXAMPLE 3
Standardized Test Practice
EQUATION 2
y = x
ANSWER
The correct answer is y = 13x and y = 4x + 90.

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.
SOLUTION
Let the number of sessions be x
So, x = x
13x = x
9x = 90
x = 10
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?
SOLUTION
Let the number of sessions be x
So, x = x
13x = x
9x =
x = 15
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. 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.
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 =20
Equation for number of rentals.
15x + 30y = 420
Equation for money collected from rentals.

21. EXAMPLE 4
Solve a multi-step problem
GUIDED PRACTICE
for Example 4
STEP 2
Graph both equations.
STEP 3
Estimate the point of intersection. The two lines appear to intersect at (12, 8).
STEP 4
Check whether (12, 8) is a solution.
x + y 20 =
15x + 30y =
= 20
15(12) + 30(8) =
20 = 20
420 = 420

22. EXAMPLE 4
EXAMPLE 4
Solve a multi-step problem
GUIDED PRACTICE
for Example 4
ANSWER
The business rented number of bicycle is 8.

23. Warm-Up – 7.2

24. Lesson 7.2, For use with pages 435-441
Solve the equation.
1. 6a – 3 + 2a = 13
ANSWER
a = 2
2. 4(n + 2) – n = 11
ANSWER
n = 1

25. Lesson 7.2, For use with pages 435-441
Solve the equation.
3. You burned 8 calories per minute on a treadmill and 10 calories per minute on an elliptical trainer for a total of 560 calories in 60 minutes. How many minutes did you spend on each machine?
HINT
Define the variables
Create two equations (one for total minutes
And one for total calories)
ANSWER
EQN 1 – 8x + 10y = 560
EQN 2 – x + y = 60
treadmill: 20 min,
elliptical trainer: 40 min

26. Do the Dance!!!! Lesson 7.2, For use with pages 435-441
Solve the system of equations BUT YOU CAN’T USE THE
CALCULATOR OR GRAPH IT!!! WORK WITH YOUR
GROUP!!.
2x + 3y = 40
y = x + 5
HINT!
Do the Dance!!!!
ANSWER
X = 5
Y = 10 26

27. Vocabulary – 7.2
None!!

28. Notes – 7.2 – Solving Systems w/Substitution
If it’s EASY to get one of the variables in an equation by itself, substitution may be the easiest way to solve the system of equations.
To Solve Linear Systems with Substitution
Get ONE of the variables in ONE of the equations by itself.
Substitute that variable into the OTHER equation.
Solve the equation from #2.
Plug the answer back into the original equation.

29. Examples 7.2

30. 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

31. 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.

32. 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).

33. 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

34. Use the substitution method for Examples 1 and 2
GUIDED PRACTICE
Use the substitution method
for Examples 1 and 2
Solve the linear system using the substitution method.
y = 2x + 5 1.
Equation 1
3x + y = 10
Equation 2
SOLUTION
STEP 1
Solve for y. Equation 1 is already solved for y.

35. Use the substitution method for Examples 1 and 2
GUIDED PRACTICE
Use the substitution method
for Examples 1 and 2
STEP 2
Substitute 2x + 5 for y in Equation 2 and solve for x.
3x + y = 10
Write Equation 2.
3x + (2x + 5) = 10
Substitute 2x+5 for x.
5x = 10
Simplify.
5x = 5
x = 2

36. EXAMPLE 2
GUIDED PRACTICE
for Examples 1 and 2
STEP 3
Substitute 1 for x in the revised Equation 1 to find the value of y.
y = 2x + 5 = 2(1) + 5 = 7
ANSWER
The solution is ( 1, 7).

37. Use the substitution method for Examples 1 and 2 for Example 1
GUIDED PRACTICE
GUIDED PRACTICE
Use the substitution method
for Examples 1 and 2
for Example 1
CHECK
Substitute 1 for x and 7 for y in each of the original
equations.
Equation 1
Equation 2
y = 2x + 5
3x + y = 10
7 = 2(1) + 5 ?
3 ( 1 ) + 7 = 11 ?
7 = 7
10 = 10

38. Use the substitution method for Examples 1 and 2
GUIDED PRACTICE
Use the substitution method
for Examples 1 and 2
x – y = 3 2.
Equation 1
x + 2y = – 6
Equation 2
SOLUTION
STEP 1
Solve Equation 1 for x.
x – y = 3
Write original Equation 1.
x = y + 3
Revised Equation 1

39. Use the substitution method for Examples 1 and 2
GUIDED PRACTICE
Use the substitution method
for Examples 1 and 2
STEP 2
Substitute y + 3 for x in Equation 2 and solve for y.
x + 2y = – 6
Write Equation 2.
( y + 3) + 2y = – 6
Substitute y + 3 for x.
3y = – 6
Simplify.
3y = – 9
Add 3 to each side.
y = – 3
Divide each side by 3.

40. Use the substitution method for Examples 1and 2
GUIDED PRACTICE
Use the substitution method
for Examples 1and 2
STEP 3
Substitute – 3 for y in the revised Equation 1 to find the value of x.
x = y + 3
Revised Equation 1
x = – 3 + 3
Substitute – 3 for y.
x = 0
Simplify.
ANSWER
The solution is ( 0, – 3 ).

41. Use the substitution method GUIDED PRACTICE for Examples 1 and 2
CHECK
Substitute 0 for x and – 3 for y in each of the original
equations.
Equation 1
Equation 2
x + 2y = – 6
x – y = 3
0 – (–3) = 3 ?
(–3 ) = – 6 ?
3 = 3
– 6 = – 6

42. 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.

43. 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

44. 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

45. 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

46. GUIDED PRACTICE
for Example 3 4.
In Example 3, what is the total cost for website hosting for each company after 20 months?
SOLUTION
Let y be the total cost.
y = =

47. GUIDED PRACTICE
for Example 3
The total cost for website hosting for each company after 20 months is $ 449.
ANSWER

48. 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?
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.

49. 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

50. 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.
Write Equation 1.
x + y = 11
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.

51. 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.

52. 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.
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.

53. GUIDED PRACTICE
for Example 4
Equation 1: Total number of quarts
x + y = 16
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 = (16)
x + 0.5y = 11.2

54. The system of equations is: x + y =16
GUIDED PRACTICE
for Example 4
The system of equations is:
x + y =16
Equation 1
x + 0.5y = 11.2
Equation 2.
STEP 2
Solve Equation 1 for x.
Write Equation 1.
x + y = 16
x = 16 – y
Revised Equation 1

55. Substitute 16 – y for x in Equation 1 and solve for x.
GUIDED PRACTICE
for Example 4
STEP 3
Substitute 16 – y for x in Equation 1 and solve
for x.
x + 0.5y = 7.7
Write Equation 2.
(16 – y) + 0.5y = 7.7
Substitute 16 – y for x.
Solve for y.
y = 9.6

56. GUIDED PRACTICE
for Example 4
STEP 4
Substitute 9.6 for y in the revised Equation 1 to
find the value of x.
x = 16 – y = 16 – 9.6 = 6.4
ANSWER
Mix 6.4 quarts of 100% antifreeze and 9.6 quarts of a 50%
Antifreeze.

57. Warm-Up – 7.3

58. Lesson 7.2, For use with pages 435-441
Solve the linear systems by GRAPHING!!!
x + y = -2
-x + y = 6
ANSWER
x = -4
y = 2
x – y = 0
5x + 2y = -7
ANSWER
x = -1
y = -1

59. Lesson 7.2, For use with pages 435-441
Solve the linear systems by SUBSTITUTION!!!!!!
y = x – 4
-2x + y = 18
ANSWER
x = -22
y = -26
5x – 4y = 27
-2x + y = 3
ANSWER
x= -13
y = -23

60. Lesson 7.3, For use with pages 443-450
1. Solve the linear system using substitution.
2x + y = 12
3x – 2y = 11
ANSWER
(5, 2)
One auto repair shop charges $30 for a diagnosis and $25 per hour for labor. Another auto repair shop charges $35 per hour for labor. For how many hours are the total charges for both of the shops the same?
HINT: FIND EQUATIONS FOR TOTAL AND SUBSTITUTE!
ANSWER
3 h

61. Lesson 7.3, For use with pages 443-450
1. Add the two equations together (combine like terms) and solve for x and y.
2x + 3y = 11
-2x + 5y = 13
ANSWER
(1,3)

62. Vocabulary – 7.3 - REVIEW System of Linear Equations
2 or more linear equations with the same variables
Solution of a Linear Equation
The solution set that MAKES ALL THE EQUATIONS TRUE AT THE SAME TIME!!
Consistent Independent System
A linear system that has EXACTLY one solution.

63. Notes – 7.3 – Solving systems with elimination
If we have two equations and two variables, how many solutions should we USUALLY have??
What’s the goal of solving every algebra eqn. you will ever see?
You can eliminate variables from some systems by adding or subtracting equations to eliminate variables.
RULES/HINTS TO MAKE PROCESS EASIER!
Remember the goal: You are trying to eliminate one variable!
Line up like terms under each other.
NEVER subtract. Add the negative instead!

64. Examples 7.3

65. 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
STEP 3
Substitute 3 for y in either equation and
Solve for x.

66. Use addition to eliminate a variable
EXAMPLE 1
Use addition to eliminate a variable
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).

67. 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
STEP 3
Substitute 4 for x in either equation and solve
for y.

68. Use subtraction to eliminate a variable
EXAMPLE 2
Use subtraction to eliminate a variable
4x + 3y = 2
Write Equation 1.
4(– 4) + 3y = 2
Substitute – 4 for x.
y = 2
Solve for y.
ANSWER
The solution is (– 4, 6).

69. 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
STEP 4
Substitute 2 for x in either equation and
solve for y.

70. EXAMPLE 3 Arrange like terms 4y = 3x + 14 4y = 3(2) + 14 y = 5 ANSWER
Write Equation 2.
4y = 3(2) + 14
Substitute 2 for x.
y = 5
Solve for y.
ANSWER
The solution is (2, 5).

71. Solve the linear system:
GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: ` 1.
4x – 3y = 5
Equation 1
– 2x + 3y = – 7
Equation 2
SOLUTION
STEP 1
Add the equations to
eliminate one variable.
4x – 3y = 5
– 2x +3y = – 7
STEP 2
Solve for x.
2x = – 2
x = – 1
STEP 3
Substitute – 1 for y in either equation and
Solve for x.

72. GUIDED PRACTICE for Example 1,2 and 3 4x – 3y = 5 2(– 1) – 3y = 5
Write Equation 1.
2(– 1) – 3y = 5
Substitute – 1 for x.
y = – 3
Solve for x.
ANSWER
The solution is (– 1, – 3).

73. GUIDED PRACTICE
for Example 1,2 and 3
CHECK Substitute 1 for x and 3 for y in each of the.
original equation
4x – 3y = 5
– 2x + 3y = – 7
4(– 1) – 3(– 3) = 5 ?
– 2(– 1) + 5(– 3) – 7 ? =
5 = 5
– 7 = – 7

74. Solve the linear system: 4. 7x – 2y = 5 7x – 3y = 4
GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 4.
7x – 2y = 5
Equation 1
7x – 3y = 4
Equation 2
SOLUTION
STEP 1
Subtract the equations to
eliminate one variable.
7x – 2y = 5
7x – 3y = 4
STEP 2
y = 1
Solve for y.
STEP 3
Substitute 1 for y in either and solve for x.

75. GUIDED PRACTICE for Example 1,2 and 3 7x – 2y = 5 7x – 2(1) = 5 x = 1
Write Equation 1.
7x – 2(1) = 5
Substitute 1 for y.
x = 1
Solve for x.
ANSWER
The solution is (1, 1).

76. Solve the linear system: 5. 3x + 4y = – 6 = 3x + 6 2y
GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 5.
3x + 4y = – 6
Equation 1 =
3x + 6 2y
Equation 2
SOLUTION
STEP 1
Rewrite Equation 2 so that the like terms are
arranged in columns.
3x + 4y = – 6
3x – 2y = – 6
3x + 4y = – 6 = 3x 2y
STEP 2
Subtract the equations.
6y = 0
STEP 3
Solve for y.
y = 0
STEP 4
Substitute 0 for y in either equation and
solve for x.

77. GUIDED PRACTICE for Example 1,2 and 3 2y = 3x + 6 2(0) = 3x + 6
Write Equation 2. =
3x + 6
2(0)
Substitute 0 for y.
x = – 2
Solve for x .
ANSWER
The solution is (– 2, 0).

78. Solve the linear system: 6. 2x + 5y = 12 = 4x + 6 5y
GUIDED PRACTICE
for Example 1,2 and 3
Solve the linear system: 6.
2x + 5y = 12
Equation 1 =
4x + 6 5y
Equation 2
SOLUTION
STEP 1
Rewrite Equation 2 so that the like terms are
arranged in columns.
2x + 5y = 12
– 4x + 5y = 6
2x + 5y = 12 =
4x + 6 5y
STEP 2
Subtract the equations.
6x = 6
STEP 3
Solve for x.
x = 1
STEP 4
Substitute 1 for x in either equation and
solve for y.

79. GUIDED PRACTICE for Example 1,2 and 3 2x + 5y = 12 2(1) + 5y = 6 x = 2
Write Equation 2. +
5y = 6
2(1)
Substitute 1 for x.
x = 2
Solve for y .
ANSWER
The solution is (1, 2).

80. 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.

81. 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

82. 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 = –

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

84. 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.

85. 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.

86. 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.
SOLUTION
STEP 1
Write a system of equations. First find the speed of the kayak going downstream.

87. GUIDED PRACTICE
for Example 4
Upstream:
d = rt
Downstream:
d = rt
10 = r 5
10 = r 2
2 = r
5 = r
Use the speeds to write a linear system. Let x be the average speed of the kayakar in still water, and let y be the speed of the current.

88. GUIDED PRACTICE
for Example 4
Going upstream
Equation 1: x y 2 = –
Equation 2:
Going downstream x y 5 = +

89. Solve the system of equations.
GUIDED PRACTICE
for Example 4
STEP 2
Solve the system of equations.
x – y= 2
Equation 1.
x + y = 5
Equation 2.
2x = 7
Add equations.
x = 3.5
Solve for x.
Substitute 3.5 for x in Equation 2 and solve for y.

90. GUIDED PRACTICE for Example 4 3.5 + y = 6 y = 1.5 ANSWER
Substitute 3.5 for x in Equation 2.
y = 1.5
Subtract 3.5 from each side.
ANSWER
The average speed of the kayakar in still water is 3.5 miles per hour, and the speed of the current is 1.5 mile per hour.

91. Warm-Up – 7.4

92. Lesson 7.4, For use with pages 451-457
Solve the linear system.
1. 4x – 3y = 15
2x – 3y = 9
ANSWER
(3, – 1)
2. –2x + y = – 8
2x – 2y = 8
ANSWER
(4, 0)

93. Lesson 7.4, For use with pages 451-457
Solve the linear system.
3. You can row a canoe 10 miles upstream in 2.5 hours and 10 miles downstream in 2 hours. What is the average speed of the canoe in still water?
ANSWER
4.5 mi/h
Multiply the second equation by -2 and rewrite it. Then use it to solve the system.
8x – 6y = 30
2x – 3y = 9
ANSWER
(3, – 1)

94. Vocabulary – 7.4 Least Common Multiple
Smallest POSITIVE number that is a multiple of two or more factors

95. Notes – 7.4 – Solving systems by multiplying first.
In order to add equations and eliminate a variable, two of the coefficients must be opposite signs.
Learned 3 ways to solve systems of linear eqns:
Graphing
Easiest when I can get y by itself and have a calculator!
Substitution
Easiest when I can get one variable by itself.
Elimination
Easiest when I can get opposite coefficients.
There is a 4th way - multiply and then eliminate.
To get coefficients with opposite signs, you can multiply one or more equations by constants.
May need to identify LCM of two coefficients.

96. Examples 7.4

97. 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

98. 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.
ANSWER
The solution is (4, –1).

99. 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.

100. 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
ANSWER
The solution is (5, 3).

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

102. GUIDED PRACTICE for Examples 1 and 2 2x + 5y = 3 2. 3x + 10y = –3
Equation 1
3x + 10y = –3
Equation 2
SOLUTION
ANSWER
The solution is (9, –3).

103. GUIDED PRACTICE for Examples 1 and 2 3. 3x – 7y = 5 9y = 5x +5
Equation 1
9y = 5x +5
Equation 2
SOLUTION
ANSWER
The solution is (–10, –5).

104. 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. Write a 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?

105. 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

106. 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
The correct answer is x = 10 and y = 6.

107. 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.
SOLUTION
Write a system of equations where x is the cost of soccer ball and y is the cost of soccer ball bag.

108. GUIDED PRACTICE
for Example 3
Equation 1:
Cost of soccer ball
Total Cost
Cost of soccer bag + =
10x + 2y =
155

109. The system of equations is:
GUIDED PRACTICE
for Example 3
Equation 2:
Cost of soccer ball
Total Cost
Cost of soccer bag + =
12x + 3y =
189
The system of equations is:
10x +2y = 155
Equation 1
12x +3y = 189
Equation 2

110. GUIDED PRACTICE
for Example 3
STEP 1
Multiply equation 1 by – 3 and equation 2 by 2 so that the coefficient of y in each equation is the least common multiple of – 3 and 2 .
STEP 2
10x +2y = 155
– 30x – 6y = –465
12x +3y = 189
24x + 6y = 378
STEP 3
Add the equation
6x = 87
STEP 4
Solve for x
x = 14.50

111. Cost of soccer ball is $ 14.50 and soccer ball bag is $5.
GUIDED PRACTICE
for Example 3
STEP 5
Substitute for x in either of the original equations and solve for y.
10x + 2y = 155
Write Equation 1.
10(14.50) + 2y = 155
Substitute for x.
y = 5
solve for y
ANSWER
Cost of soccer ball is $ and soccer ball bag is $5.

112. Warm-Up – 7.5

113. Lesson 7.5, For use with pages 459-465
1. Solve the linear system.
2x + 3y = – 9
x – 2y = 6
ANSWER
(0, – 3)
2. You buy 8 pencils for $8 at the bookstore. Standard pencils cost $.85 and specialty pencils cost $1.25.
How many specialty pencils did you buy?
ANSWER
3 specialty pencils

114. Lesson 7.5, For use with pages 459-465
1. Solve the linear system by GRAPHING. Describe the lines on your whiteboard.
x + y = -2
y = -x+5
1. Solve the linear system by SUBSTITUTION. What do you get?
x + y = -2
y = -x+5
1. Solve the linear system by ELIMINATION. What do you get?
x + y = -2
y = -x+5
ANSWER
No solution.

115. Vocabulary – 7.5 Consistent Independent System
System of equations with ONE solution
Inconsistent System
System of equations with NO solution.
Consistent Dependent System
System of equations with INFINITE solutions.

116. Notes – 7.5–Special Types of Systems
Systems can have one soln, no soln, or infinite.
Easiest ways to check for solutions:
Graph them (put them in slope-intercept form)
Intersect = 1 solution
Parallel = No solution
Same line = Infinite solutions
Check if equations are multiples of each other.
Yes = infinite solutions
No = Check some more!
Eliminate the variables (using Add. or Mult.)
Always False statement = No solutions
Always True statement = Infinite solutions

117. Notes – 7.5–Special Types of Systems

118. Examples 7.4

119. A linear system with no solution
EXAMPLE 1
A linear system with no solution
Solve the linear system by graphing and by elimination!
3x + 2y = 10
Equation 1
3x + 2y = 2
Equation 2
SOLUTION
METHOD 1 Graphing
Graph the linear system.
Answer: NO SOLUTION.

120. A linear system with no solution
EXAMPLE 1
A linear system with no solution
METHOD 2 Elimination
Subtract the equation.
3x + 2y = 10
3x + 2y = 2
0 = 8
This is a false statement.
ANSWER
The variables are eliminated and you are left with a false statement regardless of the values of x and y. This tells you that the system has no solution.

121. A linear system with infinitely many solutions
EXAMPLE 2
A linear system with infinitely many solutions
Solve the system by graphing and substitution.
x – 2y = – 4
Equation 1
y = x + 2 1 2
Equation 2
SOLUTION
Graphing
METHOD 1
Graph the linear system.

122. A linear system with infinitely many solutions
EXAMPLE 2
A linear system with infinitely many solutions
METHOD 2 Substitution
Substitute x + 2 for y in Equation 1 and solve for x. 1 2
x – 2y =
– 4
Write Equation 1
x – x + 2 = 1 2
– 4 2
Substitute x + 2 for y. 1
– 4 =
– 4
Simplify.
The variables are eliminated and you are left with a true statement regardless of the values of x and y. This tells you that the system has infinite solutions.
ANSWER

123. Subtract the equations.
GUIDED PRACTICE
for Examples 1 and 2
Tell whether the linear system has no solution or infinitely many solutions. Explain.
x + 3y = 6
Equation 1
– 5x – 3y = 3
Equation 2
METHOD 2 Elimination
Subtract the equations.
5x + 3y = 6
– 5x – 3y = 3
0 = 9
This is a false statement.

124. GUIDED PRACTICE
for Examples 1 and 2
ANSWER
The variables are eliminated and you are left with a false statement regardless of the values of x and y. This tells you that the system has no solution.

125. Substitute 2x – 4 for y in Equation 2 and solve for x.
GUIDED PRACTICE
for Examples 1 and 2
2. y = 2x – 4
Equation 1
– 6x + 3y = – 12
Equation 2
METHOD 2 Elimination
Substitute 2x – 4 for y in Equation 2 and solve for x.
– 6x + 3y = – 12
Write Equation 2
– 6x + 3(2x – 4) = – 12
Substitute (2x – 4) for y.
– 12 = – 12
Simplify.

126. GUIDED PRACTICE
for Examples 1 and 2
ANSWER
The variables are eliminated and you are left with a true statement regardless of the values of x and y. This tells you that the system has infinitely many solution.

127. Warm-Up – 7.6

128. Lesson 7.6, For use with pages 466-472
1. Graph y < x – 1. 2 3
ANSWER
1. Solve -3x >= 12
ANSWER: x <= -4

129. Lesson 7.6, For use with pages 466-472
2. You are running one ad that costs $6 per day and another that costs $8 per day. You can spend no more than $120. Graph this inequality. HINT: WRITE THE EQUATION FIRST!
Graph the following on a number line:
x <= 5 and x >= 0
ANSWER:

130. Vocabulary – 7.6 System of Linear Inequalities
Two or more linear inequalities in the same variables.
Solution of a system of linear inequalities
An ordered pair that makes ALL the inequalities true at the same time.
Graph of a system of linear inequalitites
Graph of all the solutions of the system.

131. Notes–7.6–Solve Systems of linear inequalities.
REVIEW
Graphing inequalities - similar to graphing lin.eqns
Play the pretend game and let equation be =.
Dotted line is < or >
Solid line is <= or >=
Pick a point NOT ON THE LINE, check the answer, and shade the correct side of the line.
TO GRAPH SYSTEMS OF INEQUALITIES
Graph each inequality
Find the AREA where solutions intersect.
Pick a point and check your solution.

132. Examples 7.6

133. 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.

134. 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

135. 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.

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

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

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

139. 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.

140. 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

141. 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.

142. 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.

143. 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.

144. 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

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

146. 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.
x y ≤ 8, x ≥ 26, x ≤ 34, y ≥ 0
ANSWER

147. Review – Ch. 7 – PUT HW QUIZZES HERE

148. Daily Homework Quiz
For use after Lesson 7.1
Use the graphing to solve the linear system 1.
3x – y = 5
– x + 3y = 5
ANSWER
(2.5,2.5)

149. Daily Homework Quiz
For use after Lesson 7.1
2. Solve the linear system by graphing.
2x + y = – 3
– 6x + 3y = 3
ANSWER
(–1, –1)

150. Daily Homework Quiz
For use after Lesson 7.1
A pet store sells angel fish for $6 each and clown loaches for $4 each . If the pet store sold 8 fish for $36, how many of each type of fish did it sell? 3.
ANSWER
2 angel fish and 6 clown loaches

151. Daily Homework Quiz
For use after Lesson 7.2
Solve the linear system using substitution
1. –5x – y = 12
3x – 5y = 4
ANSWER
(–2, –2)
x + 9y = –4
x – 2y = 11
ANSWER
(7, –2 )

152. Daily Homework Quiz
For use after Lesson 7.2 3.
You are making 6 quarts of fruit punch for a party. You want the punch to contain 80% fruit juice. You have bottles of 100% fruit juice and 20% fruit juice. How many quarts of 20% fruit juice should you mix to make 6 quarts of 80% fruit juice?
ANSWER
4.5 quarts of 100% fruit juice and 1.5 quarts of 20% fruit juice

153. Daily Homework Quiz
For use after Lesson 7.3
Solve the linear system using elimination.
1. –5x +y = 18
3x – y = –10
ANSWER
(–4, –2)
x + 2y = 14
4x – 3y = –11
ANSWER
(1, 5)
x – y = –14
y = 3x + 6
ANSWER
(8, 30)

154. Daily Homework Quiz
For use after Lesson 7.3
4. x + 4y = 15
2y = x – 9
ANSWER
(11, 1)
A business center charges a flat fee to send faxes plus a fee per page.You send one fax with 4 pages for $5.36 and another fax with 7 pages for $7.88. Find the flat fee and the cost per page to send a fax.
HINT: Find two points and then utilize the slope-intercept form of a line. 5.
ANSWER
flat fee: $2, price per page: $.84

155. Daily Homework Quiz
For use after Lesson 7.4
Solve the linear system using elimination.
x + 3y = 12
– 2x + y = 4
ANSWER
(0,4)
2. – 3x + 2y = 7
5x – 4y = – 15
ANSWER
(1,5)
3. – 7x – 3y = 11
4x – 2y = 16
ANSWER
(1, – 6)

156. Daily Homework Quiz
For use after Lesson 7.4
A recreation center charges nonmembers $3 to use the pool and $5 to use the basketball courts. A person pays $42 to use the recreation facilities 12 times. How many times did the person use the pool. 4.
ANSWER
9 times
(-4,-2)

157. Daily Homework Quiz
For use after Lesson 7.5
Without solving the linear system, tell whether the linear system has one solution, no solution, or infinitely many solutions.
x + 2y = 12
y = –2x + 8
ANSWER
no solution
2. – 2x + 5y = 5
y = x + 1 2 5
ANSWER
infinitely many solutions

158. Daily Homework Quiz
For use after Lesson 7.3
A group of 12 students and 3 teachers pays $57 for admission to a primate research center. Another group of 14 students and 4 teachers pays $69. Find the cost of one student ticket. 3.
ANSWER
$3.50

159. Daily Homework Quiz
For use after Lesson 7.6
Write a system of inequalities for the shaded region. 1.
ANSWER
x < 2, y > x =1

160. Daily Homework Quiz
For use after Lesson 7.6
A bibliography can refer to at most 8 articles, at most 4 books, and at most 8 references in all. Write and graph a system of inequalities that models the situation. 2.
ANSWER
x = articles, y = books; x 8, y 4, x + y 8, x 0, and y 0
>
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161. Warm-Up – X.X

162. Vocabulary – X.X
Holder
Holder 2
Holder 3
Holder 4

163. Notes – X.X – LESSON TITLE.
Holder

164. Examples X.X