1. Lesson Objectives: I will be able to …
Solve systems of linear equations in two
variables by elimination
Compare and choose an appropriate method
for solving systems of linear equations
Language Objective: I will be able to …
Read, write, and listen about vocabulary, key
concepts, and examples

2. Remember that an equation stays balanced if you add equal amounts to both sides.
Consider equations: x - 2y = -19 and
5x + 2y = 1
When you use the elimination method to solve a system of linear equations, align all like terms in the equations. Then determine whether any like terms can be eliminated because they have opposite coefficients.

3. Solving Systems of Equations by Elimination
Page 11
Solving Systems of Equations by Elimination
Step 1
Write the system so that like terms are aligned.
Eliminate one of the variables and solve for the other variable.
Step 2
Substitute the value of the variable into one of the original equations and solve for the other variable.
Step 3
Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check.
Step 4

4. Example 1: Elimination Using Addition
Page 13
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
– –2
4y = –4
4x = 8
4x = 8
x = 2
4y –4
y = –1
Step 4 (2, –1)

5. Your Turn 1 y + 3x = –2 Solve by elimination. 2y – 3x = 14 Step 1
Page 14
y + 3x = –2
Solve by elimination.
2y – 3x = 14
Step 1
2y – 3x = 14
y + 3x = –2
Step 3 y + 3x = –2
4 + 3x = –2
Step 2
3y = 12
– –4
3x = –6
3y = 12
y = 4
3x = –6
x = –2
Step 4
(–2, 4)

6. Example 2: Elimination Using Subtraction
Page 15
2x + y = –5
Solve by elimination.
2x – 5y = 13
2x + y = –5
Step 1
Step 3
2x + y = –5
–(2x – 5y = 13)
2x + (–3) = –5
2x + y = –5
–2x + 5y = –13
2x – 3 = –5
0 + 6y = –18
Step 2
2x = –2
6y = –18
y = –3
x = –1
Step 4
(–1, –3)

7. Your Turn 2 3x + 3y = 15 Solve by elimination. –2x + 3y = –5
Page 16
3x + 3y = 15
Solve by elimination.
–2x + 3y = –5
3x + 3y = 15
–(–2x + 3y = –5)
Step 1
Step 3
3x + 3y = 15
3(4) + 3y = 15
3x + 3y = 15
+ 2x – 3y = +5
12 + 3y = 15
Step 2
5x = 20
– –12
3y = 3
5x = 20
y = 1
x = 4
(4, 1)
Step 4

8. In some cases, you will first need to multiply one or both of the equations by a number so that one variable has opposite coefficients. This will be the new Step 1.

9. Example 3: Elimination Using Multiplication First
Page 17
Solve the system by elimination.
x + 2y = 11
–3x + y = –5
x + 2y = 11
Step 1
–2(–3x + y = –5)
Step 3
x + 2y = 11
3 + 2y = 11
x + 2y = 11
+(6x –2y = +10)
– –3
2y = 8
7x = 21
y = 4
Step 2
7x = 21
Step 4
(3, 4)
x = 3

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

11. Solve the system by elimination.
Your Turn 4
Page 19
Solve the system by elimination.
2x + 5y = 26
–3x – 4y = –25
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
+(–6x – 8y = –50)
–20 –20
2X = 6
y = 28
Step 2
x = 3
y = 4
Step 4
(3, 4)

12. Page 12

13. Homework
Assignment #32
Holt 6-3 #11, 12, 14, 15, 17-20