# 3 variable systems

\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.

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I have done this so many times, how did you come up with that? I just can't get it, I am getting confused on what to do with a negative times a series of number divided by a number. – Adam Apr 17 '11 at 19:27
From $-8y+3y=-1$ you get first $-8y=-1-3y$. Then $y=\frac{-1-3y}{-8}=\frac{1+3y}{8}$, not $y=\frac{-1-3y}{8}$, because you have to divide $-8y=-1-3y$ by $-8$ and not by $8$. – Américo Tavares Apr 17 '11 at 19:33

You should get $y=\frac{1+3z}{8}$, not $y=\frac{-1-3z}{8}$.
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.