Answer
$\{(1,-2)\}$.
Work Step by Step
Solve the given system using Cramer's Method. Rewrite the given system of equations in standard form:
$\left\{\begin{matrix}
4x & +y & = &2 \\
2x &-3y& = &8
\end{matrix}\right.$
The determinant $D$ consists of the $x-$ and $y-$ coefficients. We determine its value:
$D=\begin{vmatrix}
4& 1\\
2 & -3
\end{vmatrix}=(4)(-3)-(2)(1)=-12-2=-14$.
For determinant $D_x$ replace the $x-$ coefficients with the constants.
$D_x=\begin{vmatrix}
2& 1\\
8 & -3
\end{vmatrix}=(2)(-3)-(8)(1)=-6-8=-14$.
For determinant $D_y$ replace the $y-$ coefficients with the constants.
$D_y=\begin{vmatrix}
4& 2\\
2 & 8
\end{vmatrix}=(4)(8)-(2)(2)=32-4=28$.
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{-14}{-14}=1$
And
$x=\frac{D_y}{D}=\frac{28}{-14}=-2$
Hence, the solution set is $\{(x,y)\}=\{(1,-2)\}$.
