Answer
$(x,y)=(3,2)$
Work Step by Step
Cramer's rule states that
$a x+b y=p \\ cx+dy=q$
$\triangle=\left|\begin{array}{ll}{a}&{b}\\{c}&{d}\end{array}\right|, \triangle_{1}=\left|\begin{array}{ll}{p}&{b}\\{q}&{d}\end{array}\right|, \triangle_{2}=\left|\begin{array}{ll}{a}&{p}\\{c}&{q}\end{array}\right|$; $ x=\dfrac{\triangle_1}{\triangle}; y=\dfrac{\triangle_2}{\triangle} (D\displaystyle \neq 0)$
From the given system of equations, we have:
$ \left[\begin{array}{ll}a & b\\c & d \end{array}\right]=\left[\begin{array}{ll}
5 & -1\\ 2 & 3\end{array}\right],\quad \left[\begin{array}{l}
p\\q \end{array}\right]=\left[\begin{array}{l}
13\\12\end{array}\right]$
$\begin{array}{cccccc} \triangle =& & \triangle_{1} =& & \triangle_{2} = \\\left|\begin{array}{ll}
5 & -1\\ 2 &3 \end{array}\right|= & & \left|\begin{array}{ll}
13 & -1\\12 & 3\end{array}\right|= & & \left|\begin{array}{ll}
5 &13\\2 & 12 \end{array}\right|=\\ =15+2 & & =39+12 & & =60-26\\ =17(\ne0) & & =51 & & =34\\ & & & &
\end{array}$
So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{51}{17}=3$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{34}{17}=2$
Thus, our solution is: $(x,y)=(3,2)$
