Answer
$x=-2$
$y=3$
$z=3$
Work Step by Step
$ \begin{bmatrix}
1 & 1 & 1 & 4\\
-1 & 2 & 3 & 17\\
2 & -1 & 0 & -7
\end{bmatrix} $
Add 1times the 1st row to the 2nd row to produce a new 2nd row.
Add -2 times the 1st row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & 1 & 1 & 4\\
0 & 3 & 4 & 21\\
0 & -3 & -2 & -15
\end{bmatrix} $
Add 1 times the 2nd row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & 1 & 1 & 4\\
0 & 3 & 4 & 21\\
0 & 0 & 2 & 6
\end{bmatrix} $
Divide the last row by 2.
$ \begin{bmatrix}
1 & 1 & 1 & 4\\
0 & 3 & 4 & 21\\
0 & 0 & 1 & 3
\end{bmatrix} $
Use back-substitution to find the solution.
$z=3$
$3y+4z=21 \rightarrow y=3$
$x+y+z=4 \rightarrow x=-2$
