Answer
$x=1, \ y=-1, \ z=1/2$
or $(1,-1,1/2)$
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 $B=\left[\begin{array}{l}
2\\
-3/2\\
2
\end{array}\right]$, the solution, $X=A^{-1}B$ equals
$X=\left[\begin{array}{l}
6+3-8\\
6+3-10\\
-2-3/2+4
\end{array}\right]=\left[\begin{array}{l}
1\\
-1\\
1/2
\end{array}\right]$
$x=1, \ y=-1, \ z=1/2$
or $(1,-1,1/2)$
