Answer
$(x,y)=(\dfrac{1}{3},\dfrac{2}{3})$
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}
3 & 3\\ 4 & 2 \end{array}\right]$ and$\left[\begin{array}{l}
p\\q \end{array}\right]=\left[\begin{array}{l}3\\\dfrac{8}{3} \end{array}\right]$
$\begin{array}{cccccc} \triangle =& & \triangle_{1} =& & \triangle_{2} = \\\left|\begin{array}{ll} 3 & 3\\ 4 & 2
\end{array}\right|= & & \left|\begin{array}{ll} 3 & 3\\
\dfrac{8}{3} & 2 \end{array}\right|= & & \left|\begin{array}{ll}
3 & 3\\ 4 & \dfrac{8}{3} \end{array}\right|=\\
=6-12 & & =6-8 & & =8-12 \\=-6(\ne 0) & & =-2 & & =-4\\
& & & & \end{array}$
So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{-2}{-6}=\dfrac{1}{3}$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{-4}{-6}=\dfrac{2}{3}$
Thus, our solution is: $(x,y)=(\dfrac{1}{3},\dfrac{2}{3})$