Answer
$(\displaystyle \frac{11}{3},\frac{2}{3})$
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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{array}{l}{x+2 y=5}\\{x-y=3}\end{array}\right..\Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
1 & 2\\
1 & -1
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
5\\
3
\end{array}\right]$
$\begin{array}{ccccc}
D & ... & D_{x} & ... & D_{y}\\
\left|\begin{array}{ll}
1 & 2\\
1 & -1
\end{array}\right|= & & \left|\begin{array}{ll}
5 & 2\\
3 & -1
\end{array}\right|= & & \left|\begin{array}{ll}
1 & 5\\
1 & 3
\end{array}\right|=\\
=-1-2 & & =-5-6 & & =3-5\\
=-3\neq 0 & & =-11 & & =-2\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{-11}{-3}=\frac{11}{3}$
$y=\displaystyle \frac{D_{y}}{D}==\frac{-2}{-3}=\frac{2}{3}$
Solution: $(x,y)=(\displaystyle \frac{11}{3},\frac{2}{3})$
