Answer
$$x_1=1 ,\quad x_2= -1 .$$
Work Step by Step
We have the system
$$\left[ \begin {array}{ccc} 3&2\\ 1& 4 \end {array} \right]
\left[\begin{array}{rrr}{x_1}\\ {x_2} \end{array}\right]=\left[\begin{array}{rrr}{1} \\ {-3} \end{array}\right].$$
The coefficients matrix $A= \left[ \begin {array}{ccc} 3&2\\ 1& 4 \end {array} \right]
$ has the inverse
$$A^{-1}=\left[ \begin {array}{cc} \frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{10}&\frac{3}{10}
\end {array} \right]
.
$$
The system has the solution
$$\left[\begin{array}{rrr}{x_1}\\ {x_2} \end{array}\right]=\left[ \begin {array}{cc} \frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{10}&\frac{3}{10}
\end {array} \right]\left[\begin{array}{rrr}{1} \\ {-3} \end{array}\right]=\left[ \begin {array}{c} {1 }\\ -1
\end {array} \right]
.$$
That is
$$x_1=1 ,\quad x_2= -1 .$$