Download Solving Systems of Linear Equations By Elimination

Solving Systems of Linear
Equations By Elimination
What is Elimination?

To eliminate means to get rid of or
remove.

You solve equations by eliminating one
of the variables (x or y) using addition
and or subtraction.
Example 1
Solve the following system of linear
equations by elimination.
2x – 3y = 15 (1)
5x + 3y = 27 (2)
Add equation (1)
7x + 0y = 42

to equation (2)
7x = 42  By eliminating y, we
x=6
can now solve for x
Example 1
Substitute x= 6 into
equation (1) to solve for y
2x – 3y = 15
2(6) – 3y = 15
12 – 3y = 15
– 3y = 15 – 12
– 3y = 3
y = -1
Check your solution x = 6
and y = -1 in equation (2)
5x + 3y = 27
5(6) + 3(-1) = 27
30 – 3 = 27
27 = 27
LS = RS
Therefore, the solution set = {(6,-1)}
Example 2
5x + 4y = -28 (1)
3x + 10y = -13 (2)
 If we were to add these equations we would obtain
8x + 14y = -41
 Even though we have only one equation now, we still
have 2 variables.
 We need to multiply the equations by terms that will
allow us to eliminate either x or y.
Example 2
5x + 4y = -28 (1)
3x + 10y = -13 (2)
 If we multiply equation (1) by 5 and equation (2) by -2,
we be able to eliminate y.
(1) x 5
25x + 20y = -140 (3)
(2) x -2
-6x – 20y = 26 (4)
Add (3) & (4)  19x = -114
x = -6
 When you change the
equations you need to
renumber them.
Example 2
Substitute x = -6 into equation (1)
5x + 4y = -28
Check your answer x = -6 and
y = ½ into equation (2)
5(-6) + 4y = -28
-30 + 4y = -28
3(-6) + 10(½) = -13
4y = -28 +30
4y = 2
2
y
4
1
y
2
-18 + 5 = -13
-13 = -13
LS = RS
Therefore, the solution set = {(-6, ½)}
Questions?

Any Questions?

Homework: 1.4 # 5 – 12

Complete Homework for Monday!
Have a Great Weekend 