Solving a 3 equation system using elimination Solve the system using elimination:

X=?
Y=?
Z=?
I'm trying to solve this problem by putting the system into the form of an augmented matrix and using gaussian elimination but I can't find a way to do it without ending up with a bunch of fractions or hitting a dead end. Please help!
 A: $$\begin{align} 
-4x + 5y + 5z &= -20 \\ 
5x - 2y - 4z &= -1 \\
2x + 5y - 6z &= -6
\end{align}$$
For this system we use the following matrix
$$\left[\begin{array}{cccc} 
-4 & 5 & 5 & -20 \\ 
5 & -2 & -4 & -1 \\
2 & 5 & -6 & -6 \\
\end{array}\right]$$
First we want the first column to be
$\left[\begin{array}{c} 
1 \\ 
0 \\
0 \\
\end{array}\right]$. We can do this by multiply the first row by $-1/4$, then subtracting $5$ of the resulting row to the second row and subtracting 2 of the same row from the third row. This gives us 
$$\left[\begin{array}{cccc} 
1 & -5/4 & -5/4 & 5 \\ 
0 & 17/4 & 9/4 & -26 \\
0 & 30/4 & -14/4 & -16 \\
\end{array}\right]$$
We know want to do something similar with column 2. Except now we want 
$\left[\begin{array}{c} 
0 \\ 
1 \\
0 \\
\end{array}\right]$. Notice that adding the second or third row to the first will not alter the first column. Now we want to divide the second row by $17/4$. Then add $5/4$ths of the resultant row to the first row and $-30/4$ of the same row to the third row.
$$\left[\begin{array}{cccc} 
1 & 0 & -10/17 & -45/17 \\ 
0 & 1 & 9/17 & -104/17 \\
0 & 0 & -127/17 & 508/17 \\
\end{array}\right]$$
Now for the final row we want $\left[\begin{array}{c} 
0 \\ 
0 \\
1 \\
\end{array}\right]$. We get this by dividing the row by $-127/17$, then subtracting $9/17$ of this row to the second row, and adding $10/17$ of the same row to the first row.
$$\left[\begin{array}{cccc} 
1 & 0 & 0 & -5 \\ 
0 & 1 & 0 & -4 \\
0 & 0 & 1 & -4 \\
\end{array}\right]$$
Now we have the system 
$$\begin{align} 
x &= -5 \\ 
y &= -4 \\
z &= -4
\end{align}$$
Obviously it is easy to get lost in these calculations. I would usually have scratch work all over the side. I think the main idea here is that we can view a system as a matrix and that row operations do not alter the solution space. It is also worth noting how this idea can pretty easily be implemented as a computer program.
