Solve the equations $x+y+4z=6$, $3x+2y-2z=9$, $5x+y+2z=13$ by using Cramer's Rule.

Given equations are $x+y+4z=6,3x+2y-2z=9,5x+y+2z=13$

Therefore, $D=\left|\begin{array}{ccc}1& 1& 4\\ 3& 2& -2\\ 5& 1& 2\end{array}\right|=1(4+2)-1(6+10)+4(3-10)$

$=6-16-28=-38$

$D_1=\left|\begin{array}{ccc}6& 1& 4\\ 9& 2& -2\\ 13& 1& 2\end{array}\right|=6(4+2)-1(18+26)+4(9-26)$

$=6\left(6\right)-44+4(-17)=36-44-68=-76$

$D_2=\left|\begin{array}{ccc}1& 6& 4\\ 3& 9& -2\\ 5& 13& 2\end{array}\right|=1(18+26)-6(6+10)+4(39-45)$

$=44-6\left(16\right)+4(-6)=44-96-24=-76$

$D_3=\left|\begin{array}{ccc}1& 1& 6\\ 3& 2& 9\\ 5& 1& 13\end{array}\right|=1(26-9)-1(39-45)+6(3-10)$

$=17+6-42=-19$

Thus, $x=\frac{D_1}{D}=\frac{-76}{-38}=2$

$y=\frac{D_2}{D}=\frac{-76}{-38}=2$

$z=\frac{D_3}{D}=\frac{-19}{-38}=\frac{1}{2}=0.5$

Therefore, the solution for he system of equation is $(2,2,0.5)$.