Answer
$(3,2)$
Work Step by Step
Cramer's rule
$\left\{\begin{array}{l}{a x+b y=s}\\{cx+dy=t}\end{array}\right.$
$D=\left|\begin{array}{ll}{a}&{b}\\{c}&{d}\end{array}\right|,D_{x}=\left|\begin{array}{ll}{s}&{b}\\{t}&{d}\end{array}\right|,D_{y}=\left|\begin{array}{ll}{a}&{s}\\{c}&{t}\end{array}\right|,$
If $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{aligned}5x-y&=13\\2x+3y&=12\end{aligned}\right. \Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
5 & -1\\
2 & 3
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
13\\
12
\end{array}\right]$
$\begin{array}{ccccc}
D & ... & D_{x} & ... & D_{y}\\
\left|\begin{array}{ll}
5 & -1\\
2 & 3
\end{array}\right|= & & \left|\begin{array}{ll}
13 & -1\\
12 & 3
\end{array}\right|= & & \left|\begin{array}{ll}
5 & 13\\
2 & 12
\end{array}\right|=\\
=15+2 & & =39+12 & & =60-26\\
=17\neq 0 & & =51 & & =34\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{51}{17}=3$
$y=\displaystyle \frac{D_{y}}{D}=\frac{34}{17}=2$
Solution: $(x,y)=(3,2)$