# Solving Systems of Linear Equations By Elimination

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```					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)
to equation (2)   7x + 0y = 42
7x = 42  By eliminating y, we
can now solve for x
x=6
Example 1
Substitute x= 6 into               Check your solution x = 6
equation (1) to solve for y        and y = -1 in equation (2)
2x – 3y = 15                         5x + 3y = 27
2(6) – 3y = 15                     5(6) + 3(-1) = 27
12 – 3y = 15                          30 – 3 = 27
– 3y = 15 – 12                         27 = 27
– 3y = 3                               LS = RS
y = -1
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)     When you change the
(2) x -2   -6x – 20y = 26 (4)     equations you need to
Add (3) & (4)  19x = -114         renumber them.

x = -6
Example 2
Substitute x = -6 into equation (1)
5x + 4y = -28         Check your answer x = -6 and
5(-6) + 4y = -28         y = ½ into equation (2)

-30 + 4y = -28            3(-6) + 10(½) = -13
4y = -28 +30                 -18 + 5 = -13
4y = 2                          -13 = -13
2
y                             LS = RS
4
1
y        Therefore, the solution set = {(-6, ½)}
2
Questions?

   Any Questions?

   Homework: 1.4 # 5 – 12

   Complete Homework for Monday!

Have a Great Weekend 

```
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