Answer
$(x,y)=(6,2)$
Work Step by Step
Cramer's rule states that
$a x+b y=p \\ cx+dy=q$
$\triangle=\left|\begin{array}{ll}{a}&{b}\\{c}&{d}\end{array}\right|, \triangle_{1}=\left|\begin{array}{ll}{p}&{b}\\{q}&{d}\end{array}\right|, \triangle_{2}=\left|\begin{array}{ll}{a}&{p}\\{c}&{q}\end{array}\right|$; $ x=\dfrac{\triangle_1}{\triangle}; y=\dfrac{\triangle_2}{\triangle} (D\displaystyle \neq 0)$
From the given system of equations, we have:
$ \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}
p\\
q
\end{array}\right]=\left[\begin{array}{l}
8\\
4
\end{array}\right]$
$\begin{array}{cccccc}
\triangle =& & \triangle_{1} =& & \triangle_{2} = \\
\left|\begin{array}{ll}
1 & 1\\
1 & -1
\end{array}\right|= & & \left|\begin{array}{ll}
8 & 1\\
4 & -1
\end{array}\right|= & & \left|\begin{array}{ll}
1 & 8\\
1 & 4
\end{array}\right|=\\
=-1-1 & & =-8-4 & & =4-8\\
=-2 & & =-12 & & =-4\\
& & & &
\end{array}$
So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{-12}{-2}=6$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{-4}{-2}=2$
Thus, our solution is: $(x,y)=(6,2)$