Does this general solution look right? Here is the starting system:
\begin{cases}
-x+y+4z= -1 \\
3x-y+2z=2\\
2x-2y-8z=2
\end{cases}
I started by performing $2E1+E3\to E3$.  This showed that they cancelled each other out and I was left with $0=0$ for $E3$ (this left me assigning $z$ to $t$).  Then I performed $E1+E2\to E2$ and got $2x+6z=1$ and solved for $x$, getting $x=-3t+1/2$.  After this, I simply back solved for $y$ and got $y=-3t-1/2$.  Does all this look right?
 A: I get:
$$y = \dfrac{-1}{2} - 7z$$
$$x = \dfrac{1}{2} - 3z$$
Of course  $z$ is a free variable.
It looks like you may have just copied your $y$ incorrectly.
A: I would do this in a more systematic way:
\begin{align}
\begin{bmatrix}
-1 & 1 & 4 & -1 \\
3 & -1 & 2 & 2 \\
2 & -2 & -8 & 2
\end{bmatrix}
&\xrightarrow{\substack{E_{31}(-2)\\E_{21}(-3)\\E_1(-1)}}
\begin{bmatrix}
1 & -1 & -4 & 1 \\
0 & 2 & 14 & -1 \\
0 & 0 & 0 & 0
\end{bmatrix}
\\&\xrightarrow{E_2(1/2)}
\begin{bmatrix}
1 & -1 & -4 & 1 \\
0 & 1 & 7 & -1/2 \\
0 & 0 & 0 & 0
\end{bmatrix}
\\&\xrightarrow{E_{12}(1)}
\begin{bmatrix}
1 & 0 & 3 & 1/2 \\
0 & 1 & 7 & -1/2 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
This reduces the system to
\begin{cases}
x+3z=1/2\\
y+7z=-1/2
\end{cases}
and so, setting $z=t$, you get
\begin{cases}
x=\dfrac{1}{2}-3t\\
y=-\dfrac{1}{2}-7t\\[1.5ex]
z=t
\end{cases}
With $E_1(-1)$ I denote “multiply the first row by $-1$”; with $E_{21}(-3)$ I denote “sum to the second row the first multiplied by $-3$” and similarly for the others. The operations are bottom up in the order they are performed.
Doing the work with matrices should reduce the chances of making errors due to wrong reading. In any case, you have a method for checking your solution: substitute the values in one of the equations and look whether you really get $0$.
A: Yes, Amzoti's results look right to me. 
$$\begin{cases}
-x+y+4z= -1 \\
3x-y+2z=2\\
2x-2y-8z=2
\end{cases}$$Note that the first and last equations are equivalent so we don't have a unique solution. 
