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.
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.
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.