Answer
$(-\displaystyle \frac{13}{4},2)$
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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{array}{r}{4 x+5 y=-3}\\{-2y=-4}\end{array}\right.\Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
4 & 5\\
0 & -2
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
-3\\
-4
\end{array}\right]$
$\begin{array}{lllll}
D & ... & D_{x} & ... & D_{y}\\
\left|\begin{array}{ll}
4 & 5\\
0 & -2
\end{array}\right|= & & \left|\begin{array}{ll}
-3 & 5\\
-4 & -2
\end{array}\right|= & & \left|\begin{array}{ll}
4 & -3\\
0 & -4
\end{array}\right|=\\
=-8-0 & & =6+20 & & =-16-0\\
=-8\neq 0 & & =26 & & =-16\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{26}{-8}=-\frac{13}{4}$
$y=\displaystyle \frac{D_{y}}{D}=\frac{-16}{-8}=2$
Solution: $(x,y)=(-\displaystyle \frac{13}{4},2)$
