TITLE: 3 variable systems
QUESTION [7 upvotes]: $$\begin{align*}
x + 2y - z &= -3\\
2x - 4y + z &= -7\\
-2x + 2y - 3z &= 4
\end{align*}$$
I multiply (1) by $-2$ and get $-2x-4y+2z=6$ and then add it to (2) which gives me $-8y+3z = -1$.
I add (2) and (3) and get $-2y-2z=-3$
I take the first result and get 
$$y=\frac{-1-3z}{8}$$
putting that into $-2y-2z=-3$ I get
$$\frac{-2(-1-3z)}{8-2z}=-3$$
which gives me $2+6z-16z = -24$, which gives me $10z=-26$ which I know isn't right already. I messed up the math somewhere, but I have done this problem about a dozen times and always get the wrong answer no matter what.

REPLY [3 votes]: You should get $y=\frac{1+3z}{8}$, not $y=\frac{-1-3z}{8}$. 
The answer should be x=-3, y=1/2, z=1.

REPLY [3 votes]: Your mistake is when you solve for $y$ from the new (2). You have
$$-8y + 3z = -1$$
from which you get
$$-8y = -1 -3z.$$
If you now divide by $-8$, you get
$$y = \frac{-1-3z}{-8} = \frac{-(1+3z)}{-8} = \frac{1+3z}{8}.$$
I should also point out that you need to be careful with the order. After you add a multiple of the first equation to the second, your system now looks like
$$\begin{align*}
x + 2y - z &= -3\\
 - 8y + 3z &= -1\\
-2x + 2y - 3z &= 4
\end{align*}$$
so you are not adding the second equation to the third to eliminate $-2x$. Rather, you first add equation (2) to equation (3), and then you add -2 times the first equation to the second.