Answer
$x=4, \ y=-2, \ z=1$
or $(4,-2,1)$
Work Step by Step
Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$.
Here, $A=\left[\begin{array}{rrr}{1}&{0}&{2}\\{-1}&{2}&{3}\\{1}&{-1}&{0}\end{array}\right]$, as in exercise $38$.
There, we found $A^{-1}=\left[\begin{array}{rrr}
{3}&{-2}&{-4}\\
{3}&{-2}&{-5}\\
{-1}&{1}&{2}\end{array}\right]$
With $B=\left[\begin{array}{l}
6\\
-5\\
6
\end{array}\right]$, the solution, $X=A^{-1}B$ equals
$X=\left[\begin{array}{l}
18+10-24\\
18+10-30\\
-6-5+12
\end{array}\right]=\left[\begin{array}{l}
4\\
-2\\
1
\end{array}\right]$
$x=4, \ y=2, \ z=1$
or $(4,-2,1)$
