Answer
$x=1/3, \ y=1, \ z=2/3$
or $(1/3, \ 1, \ 2/3)$
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 $B=\left[\begin{array}{l}
2\\
7/3\\
10/3
\end{array}\right]$, the solution, $X=A^{-1}B$ equals
$X=\left[\begin{array}{l}
\frac{1}{7}(-10+\frac{7}{3}+10)\\
\frac{1}{7}(18+7/3-40/3)\\
\frac{6}{7}-\frac{2}{3}+\frac{10}{21}
\end{array}\right]=\left[\begin{array}{l}
1/3\\
1\\
\frac{18-14+10}{21}
\end{array}\right]=\left[\begin{array}{l}
1/3\\
1\\
\frac{14}{21}
\end{array}\right]=\left[\begin{array}{l}
1/3\\
1\\
2/3
\end{array}\right]$
$x=1/3, \ y=1, \ z=2/3$
or $(1/3, \ 1, \ 2/3)$
