Unit 4 L-8 Math 8
Aim: To solve systems of two linear equations in two variables – Algebraically (8.EE.8b)
Solving a System of Equations by Elimination
Systems of Equations - Elimination Method
We are looking to eliminate one of the variables through addition.
Here’s what it looks like:
Solve the following system by elimination.
3x – 4y = -16
-1x + 4y = 24
1.
Rewrite both equations in the same order.
• We’re lucky both the variables in both equations already line up.
2.
Look to see if one of the variables would equal zero if we added the two equations
together. This would mean that the coefficients on the same variables would be
additive inverses (opposites) of each other.
• In our example the y variable will cancel out when we add the equations.
3.
Add the equations together and solve for one variable.
3x – 4y = -16
+ -1x + 4y = 24
2x
=8
2
2
x= 4
4.
Pick one equation and solve for y
-1x + 4y = 24
-1(4) + 4y = 24
-4 + 4y = 24
+4
+4
4y = 28
4 4
y=7
The solution to the system is (4, 7).
Now that we have the solution, we can check it in the other equation.
3x – 4y = -16
3(4) – 4(7) = -16
12 - 28 = -16
-16 = -16
2014-2015
Unit 4 L-8 Math 8
Solve each system below by elimination. Show all work
1.
3x – 5y = -42
5x + 5y = -30
SOLUTION ___________
2.
-5x + 3y = 12
-2x – 3y = -33
SOLUTION ___________
3.
-3x + y = -2
y + 3x = -2
SOLUTION ___________
2014-2015
Unit 4 L-8 Math 8
4.
2y – 5x = -13
5x + 1y = 1
SOLUTION ___________
5.
7y = 3x - 56
3x – 7y = 1
SOLUTION ___________
6.
y – 3x = -20
3x + y = 40
SOLUTION ___________
2014-2015
Unit 4 L-8 Math 8
7.
y = 2x - 2
y – 2x = -2
SOLUTION ___________
8.
2x + 2y = 22
13 = 3y – 2x
SOLUTION ___________
9.
3x + y = 0
-2x – y = 0
SOLUTION ___________
10.
x + 5y = 10
-x + 5y = 0
SOLUTION ___________
11.
2x – 5y = 29
4x + 5y = 13
SOLUTION ___________
12.
-3x – 3y = -12
3x + 3y = 6
SOLUTION ___________
13.
71 = 5x + 4y
-1x – 4y = -43
SOLUTION ___________
14.
6x – 5y = 21
2x + 5y = -5
SOLUTION ___________
2014-2015