Answer
$(2,3)$
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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{array}{l}{2 x+4 y=16}\\{3x-5y=-9}\end{array}\right.\Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
2 & 4\\
3 & -5
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
16\\
-9
\end{array}\right]$
$\begin{array}{lllll}
D & ... & D_{x} & ... & D_{y}\\
\left|\begin{array}{ll}
2 & 4\\
3 & -5
\end{array}\right|= & & \left|\begin{array}{ll}
16 & 4\\
-9 & -5
\end{array}\right|= & & \left|\begin{array}{ll}
2 & 16\\
3 & -9
\end{array}\right|=\\
=-10-12 & & =-80+36 & & =-18-48\\
=-22\neq 0 & & =-44 & & =-66\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{-44}{-22}=2$
$y=\displaystyle \frac{D_{y}}{D}=\frac{-66}{-22}=3$
Solution: $(x,y)=(2,3)$