Answer
$(\displaystyle \frac{1}{2},\frac{3}{4})$
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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{array}{l}{2 x-4 y=-2}\\{3x+2y=3}\end{array}\right.\Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
2 & -4\\
3 & 2
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
-2\\
3
\end{array}\right]$
$\begin{array}{lllll}
D=\left|\begin{array}{ll}
2 & -4\\
3 & 2
\end{array}\right|= & & D_{x}=\left|\begin{array}{ll}
-2 & -4\\
3 & 2
\end{array}\right|= & & D_{y}=\left|\begin{array}{ll}
2 & -2\\
3 & 3
\end{array}\right|=\\
=4+12 & & =-4+12 & & =6+6\\
=16\neq 0 & & =8 & & =12\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{8}{16}=\frac{1}{2}$
$y=\displaystyle \frac{D_{y}}{D}=\frac{12}{16}=\frac{3}{4}$
Solution: $(x,y)=(\displaystyle \frac{1}{2},\frac{3}{4})$