1. 3-2 Warm Up Lesson Presentation Lesson Quiz Using Algebraic Methods
to Solve Linear Systems
3-2
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 2
Holt Algebra 2

2. Warm Up
Determine if the given ordered pair is an element of the solution set of
2x – y = 5 .
3y + x = 6
2. (–1, 1) no
1. (3, 1)
yes
Solve each equation for y.
3. x + 3y = 2x + 4y – 4
y = –x + 4
4. 6x y = 3y + 2x – 1
y = 2x + 3

3. Objectives Solve systems of equations by substitution.
Solve systems of equations by elimination.

4. Vocabulary
substitution
elimination

5. The graph shows a system of linear equations
The graph shows a system of linear equations. As you can see, without the use of technology, determining the solution from the graph is not easy. You can use the substitution method to find an exact solution. In substitution, you solve one equation for one variable and then substitute this expression into the other equation.

6. Example 1A: Solving Linear Systems by Substitution
Use substitution to solve the system of equations.
y = x – 1
x + y = 7
Step 1 Solve one equation for one variable. The first equation is already solved for y: y = x – 1.
Step 2 Substitute the expression into the other equation.
x + y = 7
x + (x – 1) = 7
Substitute (x –1) for y in the other equation.
2x – 1 = 7
Combine like terms.
2x = 8
x = 4

7. Example 1A Continued
Step 3 Substitute the x-value into one of the original equations to solve for y.
y = x – 1
y = (4) – 1
Substitute x = 4.
y = 3
The solution is the ordered pair (4, 3).

8. Example 1A Continued
Check A graph or table supports your answer.

9. Example 1B: Solving Linear Systems by Substitution
Use substitution to solve the system of equations.
2y + x = 4
3x – 4y = 7
Method 2 Isolate x.
Method 1 Isolate y.
2y + x = 4
2y + x = 4
First equation.
y =
x = 4 – 2y
Isolate one variable.
3x – 4y= 7
3x – 4y = 7
Second equation.
3x – = 7
3(4 – 2y)– 4y = 7
Substitute the expression into
the second equation.
12 – 6y – 4y = 7
3x + 2x – 8 = 7
12 – 10y = 7
5x – 8 = 7
Combine like terms.
–10y = –5
5x = 15
x = 3
First part of the solution

10. By either method, the solution is .
Example 1B Continued
Substitute the value into one of the original equations to solve for the other variable.
Method 1
Method 2
2y + (3) = 4
Substitute the value to solve for the other variable.
+ x = 4
1 + x = 4
2y = 1
x = 3
Second part of the solution
By either method, the solution is

11. Check It Out! Example 1a
Use substitution to solve the system of equations.
y = 2x – 1
3x + 2y = 26
Step 1 Solve one equation for one variable. The first equation is already solved for y: y = 2x – 1.
Step 2 Substitute the expression into the other equation.
3x + 2y = 26
Substitute (2x –1) for y in the other equation.
3x + 2(2x–1) = 26
3x + 4x – 2 = 26
Combine like terms.
7x = 28
x = 4

12. Check It Out! Example 1a Continued
Step 3 Substitute the x-value into one of the original equations to solve for y.
y = 2x – 1
y = 2(4) – 1
Substitute x = 4.
y = 7
The solution is the ordered pair (4, 7).

13. Check It Out! Example 1a Continued
Check A graph or table supports your answer.

14. Use substitution to solve the system of equations. 5x + 6y = –9
Check It Out! Example 1b
Use substitution to solve the system of equations.
5x + 6y = –9
2x – 2 = –y
Method 1 Isolate y.
Method 2 Isolate x.
2x – 2 = –y
2x – 2 = –y
First equation.
y = –2x + 2
x = 1 – y
Isolate one variable.
5x + 6y = –9
5x + 6y = –9
Second equation.
Substitute the expression into the second equation.
5(1 – y)+ 6y = –9
5x + 6(–2x + 2) = –9

15. Check It Out! Example 1b Continued
Method 1 Isolate y.
Method 2 Isolate x.
5x + 6(–2x + 2) = –9
Combine like terms.
5x – 12x + 12 = –9
10 – 5y + 12y = –18
–7x = –21
10 + 7y = –18
7y = –28
x = 3
y = –4
First part of the solution.

16. By either method, the solution is (3, –4).
Example 1b Continued
Substitute the value into one of the original equations to solve for the other variable.
5(3) + 6y = –9
5x + 6(–4) = –9
Substitute the value to solve for the other variable.
5x + (–24) = –9
15 + 6y = –9
6y = –24
5x = 15
y = –4
x = 3
Second part of the solution
By either method, the solution is (3, –4).

17. You can also solve systems of equations with the elimination method
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

18. Example 2A: Solving Linear Systems by Elimination
Use elimination to solve the system of equations.
3x + 2y = 4
4x – 2y = –18
Step 1 Find the value of one variable.
3x + 2y = 4
The y-terms have opposite coefficients.
+ 4x – 2y = –18
7x = –14
Add the equations to eliminate y.
x = –2
First part of the solution

19. Example 2A Continued
Step 2 Substitute the x-value into one of the original equations to solve for y.
3(–2) + 2y = 4
2y = 10
Second part of the solution
y = 5
The solution to the system is (–2, 5).

20. Example 2B: Solving Linear Systems by Elimination
Use elimination to solve the system of equations.
3x + 5y = –16
2x + 3y = –9
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
Add the equations.
y = –5
First part of the solution

21. Example 2B Continued
Step 2 Substitute the y-value into one of the original equations to solve for x.
3x + 5(–5) = –16
3x = 9
3x – 25 = –16
x = 3
Second part of the solution
The solution for the system is (3, –5).

22. Example 2B: Solving Linear Systems by Elimination
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 

23. Check It Out! Example 2a
Use elimination to solve the system of equations.
4x + 7y = –25
–12x –7y = 19
Step 1 Find the value of one variable.
4x + 7y = –25
– 12x – 7y = 19
The y-terms have opposite coefficients.
–8x = –6
Add the equations to eliminate y.
x =
First part of the solution

24. Check It Out! Example 2a Continued
Step 2 Substitute the x-value into one of the original equations to solve for y.
4( ) + 7y = –25
3 + 7y = –25
7y = –28
Second part of the solution
y = –4
The solution to the system is ( , –4).

25. Use elimination to solve the system of equations.
Check It Out! Example 2b
Use elimination to solve the system of equations.
5x – 3y = 42
8x + 5y = 28
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
Add the equations.
49y = –196
First part of the solution
y = –4

26. Check It Out! Example 2b
Step 2 Substitute the y-value into one of the original equations to solve for x.
5x – 3(–4) = 42
5x = 30
5x + 12 = 42
x = 6
Second part of the solution
The solution for the system is (6,–4).

27. Check It Out! Example 2b
Check Substitute 6 for x and –4 for y in each equation.
5x – 3y = 42
8x + 5y = 28 42
5(6) – 3(–4)
8(6) + 5(–4) 28 

28. In Lesson 3–1, you learned that systems may have infinitely many or no solutions. When you try to solve these systems algebraically, the result will be an identity or a contradiction.
An identity, such as 0 = 0, is always true and indicates infinitely many solutions. A contradiction, such as 1 = 3, is never true and indicates no solution.
Remember!

29. Example 3: Solving Systems with Infinitely Many or No Solutions
Classify the system and determine the number of solutions.
3x + y = 1
2y + 6x = –18
Because isolating y is straightforward, use substitution.
3x + y = 1
y = 1 –3x
Solve the first equation for y.
2(1 – 3x) + 6x = –18
Substitute (1–3x) for y in the second equation.
2 – 6x + 6x = –18
Distribute.
2 = –18 x
Simplify.
Because 2 is never equal to –18, the equation is a contradiction. Therefore, the system is inconsistent and has no solution.

30. Check It Out! Example 3a
Classify the system and determine the number of solutions.
56x + 8y = –32
7x + y = –4
Because isolating y is straightforward, use substitution.
7x + y = –4
y = –4 – 7x
Solve the second equation for y.
56x + 8(–4 – 7x) = –32
Substitute (–4 –7x) for y in the first equation.
56x – 32 – 56x = –32
Distribute.
–32 = –32 
Simplify.
Because –32 is equal to –32, the equation is an identity. The system is consistent, dependent and has infinite number of solutions.

31. Check It Out! Example 3b
Classify the system and determine the number of solutions.
6x + 3y = –12
2x + y = –6
Because isolating y is straightforward, use substitution.
2x + y = –6
y = –6 – 2x
Solve the second equation.
6x + 3(–6 – 2x)= –12
Substitute (–6 – 2x) for y in the first equation.
6x –18 – 6x = –12
Distribute.
–18 = –12 x
Simplify.
Because –18 is never equal to –12, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.

32. Example 4: Zoology 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.

33. Example 4 Continued
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) + =

34. Example 4 Continued
x + y = 60
0.18x +0.09y = 9
Solve the system.
x + y = 60
First equation
y = 60 – x
Solve the first equation for y.
0.18x (60 – x) = 9
Substitute (60 – x) for y.
0.18x – 0.09x = 9
Distribute.
0.09x = 3.6
Simplify.
x = 40

35. Example 4 Continued
Substitute x into one of the original equations to solve for y.
Substitute the value of x into one equation.
40 + y = 60
y = 20
Solve for y.
The mixture will contain 40 lb of the beef mix and 20 lb of the bacon mix.

36. Check It Out! Example 4
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.

37. Check It Out! Example 4 Continued
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) + =

38. Check It Out! Example 4 Continued
x + y = 50
5x + 13y = 500
Solve the system.
x + y = 50
First equation
y = 50 – x
Solve the first equation for y.
5x + 13(50 – x) = 500
Substitute (50 – x) for y.
5x – 13x = 500
Distribute.
–8x = –150
Simplify.
x = 18.75

39. Check It Out! Example 4 Continued
Substitute x into one of the original equations to solve for y.
Substitute the value of x into one equation.
y = 50
y = 31.25
Solve for y.
The mixture will contain lb of the Sumatra beans and lb of the Kona beans.

40. Lesson Quiz
Use substitution or elimination to solve each system of equations.
3x + y = 1
5x – 4y = 10 1. 2.
y = x + 9
3x – 4y = –2
(–2, 7)
(6, 5)
3. The Miller and Benson families went to a theme park. The Millers bought 6 adult and 15 children tickets for $423. The Bensons bought 5 adult and 9 children tickets for $293. Find the cost of each ticket.
adult: $28; children’s: $17