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