Algebra ( Period 5)

Using Addition With Multiplication to Solve a System 6-4

This is still  second Algebraic method to solve a system but to get the additive inverses of one variable—you will need to multiply ONE or BOTH equations by a factor.

EXAMPLE 1

Multiplying just ONE equation

5x + 6y = -8
2x + 3y = -5

Multiply the bottom by -2 to eliminate y

5x + 6y = -8
-4x -6y = 10

x = 2

EXAMPLE 2

Multiply BOTH Equations

4x + 2y = 8
3x + 3y = 9

To eliminate x, you would need to multiply the top by 3 and the bottom by -4
so you would get 12x and – 12x
OR

Multiply the top by 3 and the bottom by -2 so that you will get 6y and -6y

It is your choice…
 [and you could actual divide the second equation by 3 to start out with
x + y = 3  Making it even easier… but if you did not see that—use the other two…]

I think keeping the numbers as SMALL as possible is usually easier so I will eliminate y

3(4x + 2y) =  3(8)
-2(3x + 3y) = -2(9)

12x + 6y = 24
-6x -6y = -18

6x = 6
x= 1