Answer
$\{(2,3)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2x+3y-13=0\\
3x-2y=0
\end{cases}$
Bring the system to standard form:
$\begin{cases}
2x+3y=13\\
3x-2y=0
\end{cases}$
Compute $D,D_x,D_y$:
$D=\begin{vmatrix}2&3\\3&-2\end{vmatrix}=2(-2)-3(3)=-13$
$D_x=\begin{vmatrix}13&3\\0&-2\end{vmatrix}=13(-2)-0(3)=-26$
$D_y=\begin{vmatrix}2&13\\3&0\end{vmatrix}=2(0)-3(13)=-39$
Determine $x$:
$x=\dfrac{D_x}{D}=\dfrac{-26}{-13}=2$
Determine $y$:
$y=\dfrac{D_y}{D}=\dfrac{-39}{-13}=3$
The solution set is:
$\{(2,3)\}$
