Question

Algebra 1 solve by elimination 6x-3y=30 6x+y=18

Answer 1

Slimination

the teacher probably wants you to eliminate the x terms since there are an equal number of them

so multiply first equation by -1
-6x+3y=-30
add to second equation
6x-6x+3y+y=18-30
0x+4y=-12
4y=-12
divide by 4
y=-3
subsitute
6x+-3=18
add 3
6x=21
divide both sides by 6
x=21/6=7/2



x=7/2
y=-3

Answer 2

First, we need the coefficient of one of our variables to be the same.
In this case, it's already done for us. We have 6x in both equations.

Now, we just subtract the equations.
$\ \ 6x-3y=30\\\underline{-6x+y=18}\\\overline{} \ \ \ \ -\ 4y=12\\\overline{} \ \ \ \ \ \ \ \ \ \ y=-3$

Note that the subtraction distributes to each term of that equation.

(you can also do something similar by adding equations--as long as you get rid of of that variable...and if your coefficients aren't the same to begin with, just multiply both sides of one of those equations to get them equal.)

Now, we just use that y = -3 to find x.

$6x+y=18\\6x-3=18\\6x=21\\\boxed{x=3.5}$