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