Answer
$(4,-2)$
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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{array}{l}{3 x-6 y=24}\\{5x+4y=12}\end{array}\right.\Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
3 & -6\\
5 & 4
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
24\\
12
\end{array}\right]$
$\begin{array}{lllll}
D & ... & D_{x} & ... & D_{y}\\
\left|\begin{array}{ll}
3 & -6\\
5 & 4
\end{array}\right|= & & \left|\begin{array}{ll}
24 & -6\\
12 & 4
\end{array}\right|= & & \left|\begin{array}{ll}
3 & 24\\
5 & 1
\end{array}\right|=\\
=12+30 & & =96+72 & & =36-120\\
=42\neq 0 & & =168 & & =-84\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{168}{42}=4$
$y=\displaystyle \frac{D_{y}}{D}=\frac{-84}{42}=-2$
Solution: $(x,y)=(4,-2)$
