Answer
$x=-34/7, \ y=85/7, \ z=12/7$
or $(-34/7, \ 85/7, \ 12/7)$
Work Step by Step
Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$.
Here, $A=\left[\begin{array}{lll}{1}&{1}&{1}\\{3}&{2}&{-1}\\{3}&{1}&{2}\end{array}\right]$, as in exercise $39$.
There, we found $A^{-1}=\left[\begin{array}{rrr}
{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{1}{7}}&{\displaystyle \frac{3}{7}}\\
{\displaystyle \frac{9}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{4}{7}}\\
{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{2}{7}}&{\displaystyle \frac{1}{7}}\end{array}\right]$
With $B=\left[\begin{array}{l}
9\\
8\\
1
\end{array}\right]$, the solution, $X=A^{-1}B$ equals
$X=\left[\begin{array}{l}
\frac{1}{7}(-45+8+3)\\
\frac{1}{7}(81+8-4)\\
\frac{1}{7}(27-16+1)
\end{array}\right]=\left[\begin{array}{l}
-34/7\\
85/7\\
12/7
\end{array}\right]$
$x=-34/7, \ y=85/7, \ z=12/7$
or $(-34/7, \ 85/7, \ 12/7)$
