SOLUTION: Solve By The Addition Method ? x+3y+z=6 3x+y-z=-2 2x+2y-z=1

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Question 266793: Solve By The Addition Method ?
x+3y+z=6
3x+y-z=-2
2x+2y-z=1

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

system%28x%2B3y%2Bz=6%2C%0D%0A3x%2By-z=-2%2C%0D%0A2x%2B2y-z=1%29

Pick a letter and pick two equations to 
eliminate it from:

I'll pick z and the first and second equations
to eliminate it from:

system%28x%2B3y%2Bz=6%2C%0D%0A3x%2By-z=-2%29

Just add them term by term as they are and the
z's will cancel out.

4x%2B4y=4

Since you can divide that one through by 4, we will
go ahead and do that:

x%2By=1

Now pick another pair (you have to use one of the
same equations you used before, but make sure
that you use the third one, the one you 
haven't used) and eliminate the SAME letter from
them.

I'll pick the first and third:

system%28x%2B3y%2Bz=6%2C%0D%0A2x%2B2y-z=1%29

Just add them term by term as they are and the
z's will also cancel out.

3x%2B5y=7

So now we have reduced it from three equations in three
variables, to just two equations in two variables:

system%28x%2By=1%2C3x%2B5y=7%29

We will eliminate the y's by multiplying the first equation
through by -5

system%28-5x-5y=-5%2C3x%2B5y=7%29

Add them term by term and get:

-2x=2

Divide both sides by -2

x=-2%2F2

x=-1

Substitute in 

x%2By=1
%28-1%29%2By=1
y=2

Substitute in any one of the original equations:

x%2B3y%2Bz=6
%28-1%29%2B3%282%29%2Bz=6
-1%2B6%2Bz=6
5%2Bz=6
z=1

So (x,y,z) = (-1,2,1)

Edwin