# 3 variable systems

1) $x+y-z = 6$

2) $x+3y-2z = 14$

3) $3x - 2y + z = -5$

I multiply 1 by -1 and add to 2, I multiply 2 by -3 and add to 3. I do not get anyone close to a proper answer.

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This gives you a new smaller system $2y-z=8$ and $-11y+7z=-47$. Solve this for $y$ and $z$ as usual by substitution as you mentioned you were familiar with, and then $x$ follows by plugging those values for $y$ and $z$ back in. –  yunone Apr 16 '11 at 22:01
I did and I got y=33.6 which I know is wrong. –  Adam Apr 16 '11 at 22:09

From the comments, you have $2y-z=8$ implies $z=2y-8$, and so $-11y+7(2y-8)=-47$, that is, $3y-56=-47$, or $3y=9$, which implies $y=3$, not $33.6$. Now you can plug back into $z=2y-8$ to find $z$, and $x$ should be easily calculable as well.
@Adam don't swear and despair... Add $56$ to both sides of $3y - 56 = -47$, so $3y = - 47 + 56 = 9$. –  t.b. Apr 16 '11 at 22:30