Answer
$$
x=-\frac{13}{2} \quad y=\frac{7}{4}
$$
Work Step by Step
The system can be written as
$$
\underbrace{\left[\begin{array}{cc}
-1 & -2 \\
2 & 8
\end{array}\right]}_{\text {Coefficient Matrix }} \cdot\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
3 \\
1
\end{array}\right]
$$
Multiply both sides from the left by the inverse of the coefficient matrix. Therefore, the solution is
$$
\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{cc}
-1 & -2 \\
2 & 8
\end{array}\right]^{-1} \cdot\left[\begin{array}{l}
3 \\
1
\end{array}\right]
$$
The inverse of the matrix
$$
\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]
$$
is
$$
\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]
$$
Here $a=-1, b=-2, c=2, d=8$, and
$$
\begin{aligned}
\left[\begin{array}{cc}
-1 & -2 \\
2 & 8
\end{array}\right]^{-1} &=\frac{1}{(-1)(8)-(-2)(2)}\left[\begin{array}{cc}
8 & 2 \\
-2 & -1
\end{array}\right] \\
&=\frac{1}{-8-(-4)}\left[\begin{array}{cc}
8 & 2 \\
-2 & -1
\end{array}\right] \\
&=\frac{1}{-8+4}\left[\begin{array}{cc}
8 & 2 \\
-2 & -1
\end{array}\right] \\
&=-\frac{1}{4}\left[\begin{array}{cc}
8 & 2 \\
-2 & -1
\end{array}\right]
\end{aligned}
$$
$\mathrm{So}$
$$
\left[\begin{array}{l}
x \\
y
\end{array}\right]=-\frac{1}{4}\left[\begin{array}{cc}
8 & 2 \\
-2 & -1
\end{array}\right] \cdot\left[\begin{array}{l}
3 \\
1
\end{array}\right]=-\frac{1}{4}\left[\begin{array}{c}
(8)(3)+(2)(1) \\
(-2)(3)+(-1)(1)
\end{array}\right]=-\frac{1}{4}\left[\begin{array}{l}
24+2 \\
-6-1
\end{array}\right]=-\frac{1}{4}\left[\begin{array}{l}
26 \\
-7
\end{array}\right]=\left[\begin{array}{c}
-\frac{26}{4} \\
\frac{7}{4}
\end{array}\right]=\left[\begin{array}{c}
-\frac{13}{2} \\
\frac{7}{4}
\end{array}\right]
$$
To write it clearly,
$$
\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
-\frac{13}{2} \\
\frac{7}{4}
\end{array}\right]
$$
This yields:
$$
x=-\frac{13}{2} \quad y=\frac{7}{4}
$$
