Answer
$(x,y) =\left(\frac{1}{2},\frac{3}{4}\right)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
2x& -&4y&=&-2\\
3x& +&2y & =&3
\end{matrix}\right.$
Determinant $D$ consists of the $x$ and $y$ coefficients.
$D=\begin{vmatrix}
2&-4 \\
3& 2
\end{vmatrix}=(2)(2)-(3)(-4)=4+12=16$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
-2&-4 \\
3& 2
\end{vmatrix}=(-2)(2)-(3)(-4)=-4+12=8$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
2&-2 \\
3& 3
\end{vmatrix}=(2)(3)-(3)(-2)=6+6=12$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{8}{16}=\dfrac{1}{2}$
and
$y=\dfrac{D_y}{D}=\dfrac{12}{16}=\dfrac{3}{4}$
Hence, the solution is $(x,y) =\left(\frac{1}{2},\frac{3}{4}\right)$.
