1. Chapter 3 - Linear Systems
Algebra II

2. Table of Contents
3.1 - Using Graphs and Tables to Solve Linear Systems
3.2 - Using Algebraic Methods to Solve Linear Systems
3.3 - Solving Systems of Linear Inequalities
3.4 - Linear Programming (Skip 15_16)
3.5 - Linear Equations in Three Dimensions (Skip 15_16)
3.6 - Solving Linear Systems in Three Variables (Notes online)

3. 3.1 - Using Graphs and Tables to Solve Linear Systems
Algebra II

4. 3.1
Algebra 2 (bell work)
A consistent system is a set of equations or inequalities that has at least one solution
An inconsistent system will have no solutions.
An independent system has equations with different slopes.
A dependent system has equations with equal slopes and equal y-intercepts.
Pg. 184

5. x + 2y = 10 (4) + 2(3) 10  9 3(4) – (3) 3x – y = 9  3.1 Example 1
Verifying Solutions of Linear Systems
Use substitution to determine if the given ordered pair is an element of the solution set for the system of equations.
(4, 3);
x + 2y = 10
3x – y = 9
x + 2y = 10
(4) + 2(3) 10  9
3(4) – (3)
3x – y = 9 
Because the point is a solution for both equations, it is a solution of the system

6. x + 6 = 4y (–4) + 6 2 2x + 8y = 1 2(–4) + 1 –4  x 3.1 Optional
Use substitution to determine if the given ordered pair is an element of the solution set for the system of equations.
x + 6 = 4y
(–4, );
2x + 8y = 1
x + 6 = 4y
(–4) + 6 2
2x + 8y = 1
2(–4) + 1 –4  x
Because the point is not a solution for both equations, it is not a solution of the system.

7. 3.1
Example 2
Solving Linear Systems by Using Graphs and Tables
Use a graph and a table to solve the system. Check your answer.
2x – 3y = 3
y + 2 = x
Solve each equation for y.
y= x – 2
y= x – 1

8. 3.1
Making tables is optional
Just Watch
y= x – 1
y= x – 2
Make a table of values for each equation.
Notice that when x = 3, the y-value for both equations is 1. 1 3 2 –1 y x x y –2 1
– 1 2 3
The solution to the system is (3, 1).

9. 3.1
Example 3
Classifying Linear Systems
Classify the system and determine the number of solutions.
x = 2y + 6
3x – 6y = 18
Solve each equation for y.
y = x – 3
The system is consistent and
dependent with infinitely many solutions.

10. 3.1
Classify the system and determine the number of solutions.
4x + y = 1
y + 1 = –4x
Solve each equation for y.
y = –4x + 1
y = –4x – 1
The system is inconsistent and
has no solution.

11. 3.1
Classify the system and determine the number of solutions.
Optional
7x – y = –11
3y = 21x + 33
Solve each equation for y.
y = 7x + 11
The system is consistent and
dependent with infinitely many solutions.

12. 3.1
Classify each system and determine the number of solutions.
Optional
x + 4 = y
5y = 5x + 35
Solve each equation for y.
y = x + 4
y = x + 7
The system is inconsistent with no solution.

13. 3.1
Example 4
Application
Day 2
City Park Golf Course charges $20 to rent golf clubs plus $55 per hour for golf cart rental.
Sea Vista Golf Course charges $35 to rent clubs plus $45 per hour to rent a cart.
For what number of hours is the cost of renting clubs and a cart the same for each course?
Step 1 Write an equation for the cost of renting clubs and a cart at each golf course.
Let x represent the number of hours and y represent the total cost in dollars.
City Park Golf Course: y = 55x + 20
Sea Vista Golf Course: y = 45x + 35
Because the slopes are different, the system is independent and has exactly one solution.

14. 3.1
City Park Golf Course: y = 55x + 20
Sea Vista Golf Course: y = 45x + 35
Step 2 Solve the system by using a table of values. Or set equal to each other and solve
Use increments of to represent 30 min.
y = 55x + 20
y = 45x + 35 x y 20
47.5 1 75
102.5 2
120 x y 35
57.5 1 80
102.5 2
125
When x = , the y-values
are both The cost of renting clubs and renting a
cart for hours is $102.50
at either company. So the cost is the same at each golf
course for hours.

15. 3.1
Ravi is comparing the costs of long distance calling cards.
To use card A, it costs $0.50 to connect and then $0.05 per minute.
To use card B, it costs $0.20 to connect and then $0.08 per minute.
For what number of minutes does it cost the same amount to use each card for a single call?
Step 1 Write an equation for the cost for each of the different long distance calling cards.
Let x represent the number of minutes and y represent the total cost in dollars.
Card A: y = 0.05x
Card B: y = 0.08x

16. 3.1
Step 2 Solve the system by using a table of values. Or by setting them equal to each other
y = 0.05x
y = 0.08x
When x = 10 , the y-values are both The cost of using the phone cards of 10 minutes is $1.00 for either cards.
So the cost is the same for each phone card at 10 minutes. x y 1
0.55 5
0.75 10
1.00 15
1.25 x y 1
0.28 5
0.60 10
1.00 15
1.40

17. HW pg.186 3.1- Day1: 3-9 (Odd, No Tables), 10-12, 57-63 (Odd)
Ch: 32-34, 42, 43, 47, 49, 50, 55, 56
Follow All HW Guidelines or ½ off

18. 3.2 - Using Algebraic Methods to Solve Linear Systems
Algebra II

19. Solving Linear Systems by Substitution
3.2
Example 1
Solving Linear Systems by Substitution
Use substitution to solve the system of equations.
y = x – 1
x + y = 7 1 2
Step 1 Solve one equation for one variable.
The first equation is already solved for y: y = x – 1.
Step 3 Substitute the x-value into one of the original equations to solve for y. 1 1
y = x – 1
Step 2 Substitute the expression into the other equation.
y = (4) – 1
y = 3 2
x + y = 7
x + (x – 1) = 7
The solution is the ordered pair (4, 3).
2x – 1 = 7
2x = 8
x = 4

20. 3.2
Use substitution to solve the system of equations.
Method 2 Isolate x.
Method 1 Isolate y.
2y + x = 4
3x – 4y = 7 1 1
2y + x = 4 1
2y + x = 4 2
x = 4 – 2y
y =
3x – 4y= 7 2
3x – 4y = 7 2
3(4 – 2y)– 4y = 7
3x – = 7
12 – 6y – 4y = 7
12 – 10y = 7
3x + 2x – 8 = 7
–10y = –5
5x – 8 = 7
5x = 15
x = 3
Plug Back In
Plug Back In 1
+ x = 4 1
2y + (3) = 4
1 + x = 4
2y = 1
x = 3

21. 3.2
Use substitution to solve the system of equations.
Method 1 Isolate y.
Method 2 Isolate x.
5x + 6y = –9
2x – 2 = –y 1 2
2x – 2 = –y 2
2x – 2 = –y 2
x = 1 – y
y = –2x + 2 1
5x + 6y = –9
5x + 6y = –9
5x + 6(–2x + 2) = –9 1
5(1 – y)+ 6y = –9
5x – 12x + 12 = –9
–7x = –21
x = 3
Plug Back In
Plug Back In 1
5x + 6(–4) = –9
10 – 5y + 12y = –18 1
5(3) + 6y = –9
5x + (–24) = –9
10 + 7y = –18
15 + 6y = –9
5x = 15
7y = –28
6y = –24
x = 3
y = –4
y = –4
By either method, the solution is (3, –4).

22. 3.2
Day 2
Just Read
You can also solve systems of equations with the elimination method.
With elimination, you get rid of one of the variables by adding or subtracting equations.
You may have to multiply one or both equations by a number to create variable terms that can be eliminated.
The elimination method is sometimes called the addition method or linear combination.
Reading Math

23. 3.2
Example 2
Solving Linear Systems by Elimination
Day 2
Use elimination to solve the system of equations.
3x + 2y = 4
4x – 2y = –18 1
Step 1 Find the value of one variable. 2 1
3x + 2y = 4 2
+ 4x – 2y = –18
7x = –14
x = –2
Step 2 Substitute the x-value into one of the original equations to solve for y. 1
3(–2) + 2y = 4
2y = 10
y = 5
The solution to the system is (–2, 5).

24. 3.2
Use elimination to solve the system of equations.
3x + 5y = –16
2x + 3y = –9 1 2
Step 1 To eliminate x, multiply both sides of the first equation by 2 and both sides of the second equation by –3.
2(3x + 5y) = 2(–16)
–3(2x + 3y) = –3(–9)
6x + 10y = –32
–6x – 9y = 27
y = –5
Step 2 Substitute the y-value into one of the original equations to solve for x. 1
3x + 5(–5) = –16
3x = 9
3x – 25 = –16
x = 3
The solution for the system is (3, –5).

25.  3.2 Check Substitute 3 for x and –5 for y in each equation.
3x + 5y = –16
2x + 3y = –9
–16
3(3) + 5(–5)
2(3) + 3(–5) –9 

26. 3.2
Example 2
Solving Linear Systems by Elimination
Use elimination to solve the system of equations.
5x – 3y = 42
8x + 5y = 28 1 2
Step 1 To eliminate x, multiply both sides of the first equation by –8 and both sides of the second equation by 5.
–8(5x – 3y) = –8(42)
5(8x + 5y) = 5(28)
–40x + 24y = –336
40x + 25y = 140
49y = –196
y = –4 1
5x – 3(–4) = 42
5x = 30
5x + 12 = 42
x = 6
The solution for the system is (6,–4).

27. 3.2
Practice
Use elimination to solve the system of equations.
4x + 7y = –25 1
Step 1 Find the value of one variable.
4x + 7y = –25
– 12x – 7y = 19
–12x –7y = 19 2
–8x = –6
x =
Step 2 Substitute the x-value into one of the original equations to solve for y.
4( ) y = –25 1
3 + 7y = –25
7y = –28
y = –4
The solution to the system is ( , –4).

28. 3.2
Example 3
Classifying Systems with Infinitely Many or No Solutions
Day 3
Classify the system and determine the number of solutions.
3x + y = 1 1
2y + 6x = –18 2
Because isolating y is straightforward, use substitution.
3x + y = 1
y = 1 –3x 2
2(1 – 3x) + 6x = –18
2 – 6x + 6x = –18
2 = –18 x
Because 2 is never equal to –18, the equation is a contradiction.
Therefore, the system is inconsistent and has no solution.

29.  3.2 Classify the system and determine the number of solutions.
56x + 8y = –32 1
Because isolating y is straightforward, use substitution.
7x + y = –4 2
7x + y = –4
y = –4 – 7x
56x + 8(–4 – 7x) = –32 1
56x – 32 – 56x = –32
–32 = –32 
Because –32 is equal to –32, the equation is an identity.
The system is consistent, dependent and has infinite number of solutions.

30. 3.2
Example 4
Application
A veterinarian needs 60 pounds of dog food that is 15% protein.
He will combine a beef mix that is 18% protein with a bacon mix that is 9% protein.
How many pounds of each does he need to make the 15% protein mixture?
Let x present the amount of beef mix in the mixture. Let y present the amount of bacon mix in the mixture.
Write one equation based on the amount of dog food:
Amount of beef mix
plus
amount of bacon mix
equals x y
60. 60 + =
Write another equation based on the amount of protein:
Protein of beef mix
plus
protein of bacon mix
equals
0.18x
0.09y
protein in mixture.
0.15(60) + =

31. Substitute x into one of the original equations to solve for y.
3.2
Solve the system. 1
x + y = 60
x + y = 60
0.18x +0.09y = 9 1
y = 60 – x 2
0.18x (60 – x) = 9 2
0.18x – 0.09x = 9
0.09x = 3.6
x = 40
Substitute x into one of the original equations to solve for y. 1
40 + y = 60
y = 20
The mixture will contain 40 lb of the beef mix and 20 lb of the bacon mix.

32. 3.2
A coffee blend contains Sumatra beans which cost $5/lb, and Kona beans, which cost $13/lb.
If the blend costs $10/lb, how much of each type of coffee is in 50 lb of the blend?
Let x represent the amount of the Sumatra beans in the blend.
Let y represent the amount of the Kona beans in the blend.
Write one equation based on the amount of each bean:
Amount of Sumatra beans
plus
amount of Kona beans
equals x y
50. 50 + =
Write another equation based on cost of the beans:
Cost of Sumatra beans
plus
cost of Kona beans
equals 5x
13y
cost of beans.
10(50) + =

33. 3.2
Solve the system. 1
x + y = 50
y = 50 – x
x + y = 50
5x + 13y = 500 1 2
5x + 13(50 – x) = 500 2
5x – 13x = 500
–8x = –150
x = 18.75
Substitute x into one of the original equations to solve for y. 1
y = 50
y = 31.25
The mixture will contain lb of the Sumatra beans and lb of the Kona beans.

34. HW pg. 194 3.2- Day 1: 2-5, 48-51 (Just Simplify)
Ch: 32, 33, 34, 36, 37
Follow All HW Guidelines or ½ off

35. 3.3 - Solving Systems of Linear Inequalities
Algebra II

36. y < – 3 y ≥ –x + 2 3.3 Example 1 Graphing Systems of Inequalities
Graph the system of inequalities.
y ≥ –x + 2
y < – 3

37. Graph the system of inequalities.
3.3
Graph the system of inequalities.
y ≥ –x + 2
y < – 3
The overlapping region is the solution region.

38. x – 3y < 6 2x + y > 1.5 Graph the system of inequalities. 3.3

39. Graph the system of inequalities.
3.3
Graph the system of inequalities.
x – 3y < 6
2x + y > 1.5

40. 3.3
Graph the system of inequalities.
Practice
y < –3x + 2
y ≥ –1

41. Graph each system of inequalities.
3.3
Graph each system of inequalities.
y < –3x + 2
y ≥ –1

42. 3.3
Math Joke
Q: Why didn’t the chicken cross to the other side of the inequality?
A: It couldn’t get past the boundary line

43. 3.3 Example 3 Geometry Application x ≥ –2 x ≤ 3 y ≥ –x + 1 y ≤ 4
Graph the system of inequalities, and classify the figure created by the solution region.
x ≥ –2
x ≤ 3
y ≥ –x + 1
y ≤ 4

44. 3.3
Trapezoid

45. Example 2 Application Day 2 3.3
Lauren wants to paint no more than 70 plates for the art show.
It costs her at least $50 plus $2 per item to produce red plates and $3 per item to produce gold plates.
She wants to spend no more than $215.
Write and graph a system of inequalities that can be used to determine the number of each plate that Lauren can make.
Let x represent the number of red plates, and let y represent the number of gold plates.
The total number of plates Lauren is willing to paint can be modeled by the inequality x + y ≤ 70.
The amount of money that Lauren is willing to spend can be modeled by
50 + 2x + 3y ≤ 215.
x  0
y  0
x + y ≤ 70
The system of inequalities is
50 + 2x + 3y ≤ 215

46. Graph the solid boundary line x + y = 70, and shade below it.
3.3
Graph the solid boundary line x + y = 70, and shade below it.
Graph the solid boundary line x + 3y ≤ 215, and shade below it. The overlapping region is the solution region.

47. 3.3
Check Test the point (20, 20) in both inequalities. This point represents painting 20 red and 20 gold plates.
x + y ≤ 70
50 + 2x + 3y ≤ 215
≤ 70
50 + 2(20) + 3(20) ≤ 215
40 ≤ 70 
150 ≤ 215 

48. 3.3
Leyla is selling hot dogs and spicy sausages at the fair.
She has only 40 buns, so she can sell no more than a total of 40 hot dogs and spicy sausages.
Each hot dog sells for $2, and each sausage sells for $2.50.
Leyla needs at least $90 in sales to meet her goal. Write and graph a system of inequalities that models this situation.
Let d represent the number of hot dogs, and let s represent the number of sausages.
The total number of buns Leyla has can be modeled by the inequality d + s ≤ 40.
The amount of money that Leyla needs to meet her goal can be modeled by 2d + 2.5s ≥ 90.
d  0
s  0
The system of inequalities is
d +s ≤ 40
2d + 2.5s ≥ 90

49. Graph the solid boundary line d + s = 40, and shade below it.
3.3
Graph the solid boundary line d + s = 40, and shade below it.
Graph the solid boundary line 2d + 2.5s ≥ 90, and shade above it. The overlapping region is the solution region.

50. 3.3
Check Test the point (5, 32) in both inequalities. This point represents selling 5 hot dogs and 32 sausages.
2d + 2.5s ≥ 90
d + s ≤ 40
≤ 40
2(5) + 2.5(32) ≥ 90
37 ≤ 40 
90 ≥ 90 

51. 3.3
Graph the system of inequalities, and classify the figure created by the solution region.
Practice
x ≤ 6
y ≤ x + 1
y ≥ –2x + 4

52. 3.3

53. 3.3
Practice
Graph the system of inequalities, and classify the figure created by the solution region.
y ≤ 4
y ≥–1
y ≤ –x + 8
y ≤ 2x + 2

54. 3.3

55. HW pg. 202
3.3 -
Day 1: 3-7 (Odd), 14, 19
Day 2: 6, 15 (Equation Only), 29, 32, 33, 43-49
HW Guidelines or ½ off
Always staple Day 1&2 Together
Put assignment in planner or HW Sheet