Answer
$(-1,2)$
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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{array}{l}{x+3 y=5}\\{2x-3y=-8}\end{array}\right. \Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
1 & 3\\
2 & -3
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
5\\
-8
\end{array}\right]$
$\begin{array}{lllll}
D & ... & D_{x} & ... & D_{y}\\
\left|\begin{array}{ll}
1 & 3\\
2 & -3
\end{array}\right|= & & \left|\begin{array}{ll}
5 & 3\\
-8 & -3
\end{array}\right|= & & \left|\begin{array}{ll}
1 & 5\\
2 & -8
\end{array}\right|=\\
=-3-6 & & =-15+24 & & =8-10\\
=-9\neq 0 & & =9 & & =-18\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{9}{-9}=-1$
$y=\displaystyle \frac{D_{y}}{D}=\frac{-18}{-9}=2$
Solution: $(x,y)=(-1,2)$