Answer
$(x,y) =(2,-3)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
\frac{1}{2}x& +&y&=&-2\\
x& -&2y & =&8
\end{matrix}\right.$
Determinant $D$ consists of the $x$ and $y$ coefficients.
$D=\begin{vmatrix}
\frac{1}{2}&1 \\
1& -2
\end{vmatrix}=(\frac{1}{2})(-2)-(1)(1)=-1-1=-2$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
-2&1 \\
8& -2
\end{vmatrix}=(-2)(-2)-(8)(1)=4-8=-4$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
\frac{1}{2}&-2 \\
1& 8
\end{vmatrix}=(\frac{1}{2})(8)-(1)(-2)=4+2=6$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{-4}{-2}=2$
and
$y=\dfrac{D_y}{D}=\dfrac{6}{-2}=-3$
Hence, the solution set is $(x,y) =(2,-3)$.
