## Precalculus (10th Edition)

$\{(2,-1)\}$
We are given the system of equations: $\begin{cases} x-2y=4\\ 3x+2y=4 \end{cases}$ Compute $D,D_x,D_y$: $D=\begin{vmatrix}1&-2\\3&2\end{vmatrix}=1(2)-3(-2)=8$ $D_x=\begin{vmatrix}4&-2\\4&2\end{vmatrix}=4(2)-4(-2)=16$ $D_y=\begin{vmatrix}1&4\\3&4\end{vmatrix}=1(4)-3(4)=-8$ Determine $x$: $x=\dfrac{D_x}{D}=\dfrac{16}{8}=2$ Determine $y$: $y=\dfrac{D_y}{D}=\dfrac{-8}{8}=-1$ The solution set is: $\{(2,-1)\}$