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