Answer
$x=-1$
$y=3$
$z=2$
Work Step by Step
First we have to compute the determinants $D,D_x,D_y,D_z$:
$D=\begin{vmatrix}2&2&3\\4&-1&1\\5&-2&6\end{vmatrix}=2(-6+2)-2(24-5)+3(-8+5)=-55$
$D_x=\begin{vmatrix}10&2&3\\-5&-1&1\\1&-2&6\end{vmatrix}=10(-6+2)-2(-30-1)+3(10+1)=55$
$D_y=\begin{vmatrix}2&10&3\\4&-5&1\\5&1&6\end{vmatrix}=2(-30-1)-10(24-5)+3(4+25)=-165$
$D_z=\begin{vmatrix}2&2&10\\4&-1&-5\\5&-2&1\end{vmatrix}=2(-1-10)-2(4+25)+10(-8+5)=-110$
We use Cramer's Rule to determine the solutions of the system:
$x=\dfrac{D_x}{D}=\dfrac{55}{-55}=-1$
$y=\dfrac{D_y}{D}=\dfrac{-165}{-55}=3$
$z=\dfrac{D_z}{D}=\dfrac{-110}{-55}=2$
The solution is:
$x=-1$
$y=3$
$z=2$
