Evaluate a linear system of three equations Solve for $x, y\ \text{and}\ z\ $:
$x-3z=10\\
-x+y+2z=7\\
2x+2y-5z=-8$
My working:
$$\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
-1 & 1 & 2 & 7 \\
2 & 2 & -5 & -8
 \end{array}\right) =
\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17  \\
0 & 3 & 0 & -11
 \end{array}\right) \begin{array}{l} 
\\ 
R_1 + R_2 \\ 
R_3 + R_2 - R_1 
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17  \\
0 & 0 & 3 & -62
 \end{array}\right) \begin{array}{l} 
\\ 
\\ 
R_3 - 3R_2 
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & 0 & -52 \\
0 & 1 & 0 & -7  \\
0 & 0 & 1 & -24
 \end{array}\right) \begin{array}{l} 
R_1 + R_3 \\ 
R_2 + R_3/3 \\ 
R_3/3 
\end{array}$$
Yet plugging these solutions into the original equation does not work: 
$-52+0-3 \times (-24)=20 \ne 10$ and 
$-(-52)-7+2\times(-24)=-3\ne7$
What did I do wrong?

The last line should read
$$\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17  \\
0 & 0 & 3 & -62
 \end{array}\right) \begin{array}{l} 
\\ 
\\ 
R_3 - 3R_2 
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & 0 & -52 \\
0 & 1 & 0 & -11/3  \\
0 & 0 & 1 & -62/3
 \end{array}\right) \begin{array}{l} 
R_1 + R_3 \\ 
R_2 + R_3/3 \\ 
R_3/3 
\end{array}$$
