Answer
$\{(\frac{7}{4},−\frac{25}{8})\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& -2y &=&8 \\
3x& +2y &=&-1
\end{matrix}\right.$
Determinant $D$ consists of the $x−$ and $y−$ coefficients.
$D=\begin{vmatrix}
1& -2 \\
3&2
\end{vmatrix}=(1)(2)-(3)(-2)=2+6=8$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
8& -2 \\
-1&2
\end{vmatrix}=(8)(2)-(-1)(-2)=16-2=14$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& 8 \\
3&-1
\end{vmatrix}=(1)(-1)-(3)(8)=-1-24=-25$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{14}{8}=\frac{7}{4}$
And
$y=\frac{D_y}{D}=\frac{-25}{8}$
Hence, the solution set is $\{(x,y)\}=\{(\frac{7}{4},−\frac{25}{8})\}$.
