Answer
$(8,-4)$
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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{array}{r}{3 x=24}\\{x+2y=0}\end{array}\right.\Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
3 & 0\\
1 & 2
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
24\\
0
\end{array}\right]$
$\begin{array}{lllll}
D & ... & D_{x} & ... & D_{y}\\
\left|\begin{array}{ll}
3 & 0\\
1 & 2
\end{array}\right|= & & \left|\begin{array}{ll}
24 & 0\\
0 & 2
\end{array}\right|= & & \left|\begin{array}{ll}
3 & 24\\
1 & 0
\end{array}\right|=\\
=6-0 & & =48-0 & & =0-24\\
=6\neq 0 & & =48 & & =-24\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{48}{6}=8$
$y=\displaystyle \frac{D_{y}}{D}=\frac{-24}{6}=-4$
Solution: $(x,y)=(8,-4)$