#### Answer

$(x,y)=(4,-2)$

#### 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}
3 & -6\\ 5 & 4\end{array}\right],\quad \left[\begin{array}{l}
p\\q \end{array}\right]=\left[\begin{array}{l}24\\12\end{array}\right]$
$\begin{array}{cccccc} \triangle =& & \triangle_{1} =& & \triangle_{2} = \\\left|\begin{array}{ll}
3 & -6\\ 5 & 4 \end{array}\right|= & & \left|\begin{array}{ll}
24 & -6\\12 & 4\end{array}\right|= & & \left|\begin{array}{ll}
3 & 24\\5 & 1 \end{array}\right|=\\ =12+30 & & =96+72 & & =36-120 \\ =42 (\ne0) & & =168 & & =-84\\ & & & & \end{array}$
So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{168}{42}=4$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{-84}{42}=-2$
Thus, our solution is: $(x,y)=(4,-2)$