#### Answer

$(x,y)=(\dfrac{3}{2},1)$

#### 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}
2 & 3\\ 1 & -1 \end{array}\right]$ and$\left[\begin{array}{l}
p\\q \end{array}\right]=\left[\begin{array}{l} 6\\ \dfrac{1}{2} \end{array}\right]$
$\begin{array}{cccccc} \triangle =& & \triangle_{1} =& & \triangle_{2} = \\\left|\begin{array}{ll} 2 & 3\\ 1 & -1
\end{array}\right|= & & \left|\begin{array}{ll} 6 & 3\\
\dfrac{1}{2} & -1 \end{array}\right|= & & \left|\begin{array}{ll}
2 & 6\\ 1 & \dfrac{1}{2} \end{array}\right|=\\
=-2-3 & & =-6-\dfrac{3}{2}& & =1-6 \\=-5 (\ne 0) & & =\dfrac{-15}{2} & & =-5\\ & & & & \end{array}$
So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{-15/2}{-5}=\dfrac{3}{2}$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{-5}{-5}=1$
Thus, our solution is: $(x,y)=(\dfrac{3}{2},1)$