Answer
The solution is $(-2,2,-1)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
1& 2 &1 |1\\
3& 5 &1| 3\\
2 & 6 & 7|1
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
1& 2 &1 |1\\
3& 5 &1| 3\\
2 & 6 & 7|1
\end{bmatrix} \approx^1\begin{bmatrix}
1& 2 &1 |1\\
0& -1 &-2| 0\\
0 & 2 & 5|-1
\end{bmatrix} \approx^2\begin{bmatrix}
1& 2 &1 |1\\
0& 1 &2| 0\\
0 & 2 & 5|-1
\end{bmatrix} \approx^3 \begin{bmatrix}
1& 2 &1 |1\\
0& 1 &2| 0\\
0 & 0 & 1|-1
\end{bmatrix}$
The equivalent system is:
$x_1+2x_2+x_3=5$
$x_2+2x_3=0$
$x_3=-1$
hence here
$x_2+2(-1)=0 \rightarrow x_2=2$
$x_3=-1$
$x_1+2.2+(-1)=5 \rightarrow x_1=-2$
The solution is $(-2,2,-1)$
