Algebra (Period 1)

Using Addition w/ Multiplication to Solve a System 6-4

This is the same method as Chapter 6-3, but to get ADDITIVE INVERSES of one variable you’ll need to multiply one or both equations by a factor.

MULTIPLYING JUST ONE EQUATION:

5x + 6y = -8

2x + 3y = -5

--------------------

Multiply either the bottom by -2 to eliminate y:

 5x + 6y = -8

-4x - 6y = 10

--------------------

 x +  0  =  2

x = 2

MULTIPLYING BOTH EQUATIONS:

4x + 2y = 8

3x + 3y = 9

--------------------

To eliminate x, you’d need to multiply the top by 3 and the bottom by -4 so that you’d get 12x and -12x

OR

Multiply the top by 3 and the bottom by -2 so that you’d get 6y and -6y.

It’s your choice!

I think keeping the numbers as small as possible is usually easier, so I’ll choose eliminating y.

3(4x + 2y) =  3(8)

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

--------------------

12x + 6y = 24

-6x - 6y = -18

--------------------

 6x +  0  =  6

x = 1