Answer
$x=0$
$y=4$
$z=2$
Work Step by Step
First we have to compute the determinants $D,D_x,D_y,D_z$:
$D=\begin{vmatrix}3&0&2\\5&-1&0\\0&4&3\end{vmatrix}=3(-3-0)-0(15-0)+2(20+0)=31$
$D_x=\begin{vmatrix}4&0&2\\-4&-1&0\\22&4&3\end{vmatrix}=4(-3-0)-0(-12-0)+2(-16+22)=0$
$D_y=\begin{vmatrix}3&4&2\\5&-4&0\\0&22&3\end{vmatrix}=3(-12-0)-4(15-0)+2(110+0)=124$
$D_z=\begin{vmatrix}3&0&4\\5&-1&-4\\0&4&22\end{vmatrix}=3(-22+16)-0(110+0)+4(20+0)=62$
We use Cramer's Rule to determine the solutions of the system:
$x=\dfrac{D_x}{D}=\dfrac{0}{31}=0$
$y=\dfrac{D_y}{D}=\dfrac{124}{31}=4$
$z=\dfrac{D_z}{D}=\dfrac{62}{31}=2$
The solution is:
$x=0$
$y=4$
$z=2$
