Answer
$(x,y)=(\dfrac{11}{3},\dfrac{2}{3})$
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}
1 & 2\\1 & -1
\end{array}\right],\quad \left[\begin{array}{l}
p\\q \end{array}\right]=\left[\begin{array}{l}
5\\3 \end{array}\right]$
$\begin{array}{cccccc}
\triangle =& & \triangle_{1} =& & \triangle_{2} = \\\left|\begin{array}{ll} 1 & 1 \\ 1 & -1
\end{array}\right|= & & \left|\begin{array}{ll} 5 & 2\\
3 & -1 \end{array}\right|= & & \left|\begin{array}{ll}
1 & 5\\ 1 & 3 \end{array}\right|=\\
=-1-2 & & =-5-6 & & =3-5\\
=-3(\ne 0) & & =-11 & & =-2\\
& & & &
\end{array}$
So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{-11}{-3}=\dfrac{11}{3}$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{-2}{-3}=\dfrac{2}{3}$
Thus, our solution is: $(x,y)=(\dfrac{11}{3},\dfrac{2}{3})$
