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