Answer
$(x,y)=(2,3)$
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}
2 & 4\\ 3 & -5\end{array}\right],\quad \left[\begin{array}{l}
p\\q \end{array}\right]=\left[\begin{array}{l}16\\-9\end{array}\right]$
$\begin{array}{cccccc} \triangle =& & \triangle_{1} =& & \triangle_{2} = \\\left|\begin{array}{ll}
2 & 4\\ 3 & -5 \end{array}\right|= & & \left|\begin{array}{ll}
16 & 4\\-9 &-5\end{array}\right|= & & \left|\begin{array}{ll}
2 & 16\\3 & -9 \end{array}\right|=\\ =-10-12 & & =-80+36 & & =-18-48 \\ =-22 (\ne0) & & =-44 & & =-66\\ & & & & \end{array}$
So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{-44}{-22}=2$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{-66}{-22}=3$
Thus, our solution is: $(x,y)=(2,3)$
