Answer
$x=-1$
$y=4$
$z=0$
Work Step by Step
$ \begin{bmatrix}
1 & 1 & 6& 3\\
1 & 1 & 3 & 3\\
1 & 2 & 4 & 7
\end{bmatrix} $
Add -1 times the 1st row to the 2nd row to produce a new 2nd row.
Add -1 times the 1st row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & 1 & 6& 3\\
0 & 0 & -3 & 0\\
0& 1 & -2 & 4
\end{bmatrix} $
From the second row we can express $-3z=0 \rightarrow z=0$
Then use back-substitution to find the solution.
$z=0$
$y-2z=4 \rightarrow y=4$
$x+y+6z=3 \rightarrow x=-1$
