$\left[\begin{array}{cc}-2 & 5 \\ 3 & -1\end{array}\right]_{2 x \times 2}\left[\begin{array}{l}x \\ y\end{array}\right]_{2 \times 1}=\left[\begin{array}{cc}1 & 2 \\ 3 & 4\end{array}\right]_{2 \times 2}\left[\begin{array}{c}3 \\ -1\end{array}\right]_{2 \times 1}$ $\Rightarrow\left[\begin{array}{c}-2 x+5 y \\ 3 x-y\end{array}\right]=\left[\begin{array}{l}1 \\ 5\end{array}\right]$
$\Rightarrow-2 x+5 y=1,3 x-y=5$
or $y=3 x-5 \quad \text{...(i)}$
$\Rightarrow \quad-2 x+5(3 x-5)=1$
$\Rightarrow \quad-2 x+15 x-25=1$
$\Rightarrow \quad 13 x=26 \Rightarrow x=2$
Substituting $x=2$ in Eq. (i), we get
$$
y=6-5=1
$$
Hence, $(x, y)=(2,1)$