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Example 1:
-5x – 6y = 8
5x + 2y = 4
-4y = 12
y = -3
(2, -3)
Example 5:
-5x – 6y = 8
-5x – 6(-3) = 8
-5x + 18 = 8
-5x = -10
x=2
Example 2:
6x – 4y = 14
-3x + 4y = 1
3x = 15
x=5
6x – 4y = 14
6(5) – 4y = 14
30 – 4y = 14
-4y = -16
y=4
3x – 2y = 3
(-x + y = 1 ) x
3x – 2y = 3
3x – 2(6) = 3
3x – 12 = 3
3x = 15
x=5
Example 6:
(2x + 5y = 3 )x-2
3x + 10y = -3
2x + 5y = 3
2(9) + 5y = 3
18 + 5y = 3
5y = -15
y = -3
(5,4)
Example 3:
4x + 3y = 2
5x + 3y = -2
-1x = 4
x = -4
4(-4) + 3y = 2
-16 + 3y = 2
3y = 18
y=6
(4x + 5y = 35 )x3
(-3x + 2y = -9 )x4
4x + 5y = 35
4x + 5(3) = 35
4x + 15 = 35
4x = 20
x=5
Example 4:
6x = 24
x=4
(4,5)
1.
2.
3.
4.
5.
5x + 6y = 50
5(4) + 6y = 50
20 + 6y = 50
6y = 30
y=5
y=6
(5,6)
-4x -10y = -6
3x + 10y = -3
-1x = -9
x=9
(9,-3)
Example 7:
(-4,6)
5x + 6y = 50
-x + 6y = 26
3
3x – 2y = 3
-3x + 3y = 3
12x + 15y = 105
-12x + 8y = -36
23y = 69
y=3
(5,3)
Summary:
Write both equations in Standard Form (Ax + By = C)
If necessary, multiply one (or both) of the equations by an integer, so one of the variables
will eliminate.
Add the columns (like terms)
Solve the equation for the variable that remains
Substitute the variable (from step 4) into the other equation to solve for the other variable.
Solving Linear Systems using
Elimination
© Lisa Davenport 2014
Directions
Print pages 1 & 2 (3 & 4 for the answer key). On my printer, I use the option to print
double sided and to flip along the short edge. Once photocopied have students
fold down the top portion so the rectangular tabs are fully visible. They should lie
just above the title “Solving Linear Functions using Graphing”.
The final product should look like this: