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