Answer
$(x,y) =\left(\frac{3}{2},1\right)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
2x& +&3y&=&6\\
x& -&y & =&\frac{1}{2}
\end{matrix}\right.$
Determinant $D$ consists of the $x$ and $y$ coefficients.
$D=\begin{vmatrix}
2&3 \\
1& -1
\end{vmatrix}=(2)(-1)-(1)(3)=-2-3=-5$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
6&3 \\
\frac{1}{2}& -1
\end{vmatrix}=(6)(-1)-(\frac{1}{2})(3)=-6-\frac{3}{2}=-\frac{15}{2}$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
2&6 \\
1& \frac{1}{2}
\end{vmatrix}=(2)(\frac{1}{2})-(1)(6)=1-6=-5$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{-\frac{15}{2}}{-5}=\dfrac{3}{2}$
and
$y=\dfrac{D_y}{D}=\dfrac{-5}{-5}=1$
Hence, the solution set is $(x,y) =\left(\frac{3}{2},1\right)$.
