Answer
$x_1=\frac{3}{2}$
$x_2=-2$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
4& -1 |8\\
2 & 1 | 1
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
4& -1 |8\\
2 & 1 | 1
\end{bmatrix} \approx^1 \begin{bmatrix}
1& -\frac{1}{4} |2\\
2 & 1 | 1
\end{bmatrix} \approx^2\begin{bmatrix}
1& -\frac{1}{4} |2\\
0 & \frac{3}{2} | -3
\end{bmatrix} \approx^3 \begin{bmatrix}
1& -\frac{1}{4} |2\\
0 & 1 | -2
\end{bmatrix} $
The equivalent system is:
$x_1-\frac{1}{4}x_2=2$
$x_2=-2$
hence here
$x_1-\frac{1}{4}(-2)=2 \rightarrow x_1=\frac{3}{2}$
$x_2=-2$
