Answer
Solution set = $\displaystyle \{(x,y,z)\ | \ x=\frac{5}{13}z-\frac{12}{13},\ y=-\frac{2}{13}z+\frac{10}{13}, z\in \mathbb{R}\}$
Work Step by Step
Method: Gauss-Jordan. Row reduce the augmented matrix:
$\left[\begin{array}{rrr|r}
-4 & 3 & 2 & 6 \\
3 & 1 & -1 & -2 \\
1 & 9 & 1 & 6 \end{array}\right]\qquad\left\{\begin{array}{l}
R_{1}=r_{1}.\\
.\\
R_{3}=r_{1}.
\end{array}\right\}\rightarrow$
$\rightarrow\left[\begin{array}{rrr|r}
1 & 9 & 1 & 6 \\
3 & 1 & -1 & -2 \\
-4 & 3 & 2 & 6 \end{array}\right]\qquad\left\{\begin{array}{l}
.\\
R_{2}=r_{2}-3r_{1}.\\
R_{3}=r_{3}+4r_{1}.
\end{array}\right\}\rightarrow$
$\rightarrow\left[\begin{array}{rrr|r}
1 & 9 & 1 & 6 \\
0 & -26 & -4 & -20 \\
0 & 39 & 6 & 30 \end{array}\right]\qquad\left\{\begin{array}{l}
.\\
R_{2}=r_{2}\div(-26).\\
R_{3}=r_{3}-\frac{3}{2}r_{2}.
\end{array}\right\}\rightarrow$
$\rightarrow\left[\begin{array}{rrr|r}
1 & 9 & 1 & 6 \\
0 & 1 & 2/13 & 10/13 \\
0 & 0 & 0 & 0 \end{array}\right]\qquad\left\{\begin{array}{l}
R_{1}=r_{1}-9r_{2}.\\
..\\
.
\end{array}\right\}\rightarrow$
$\left[\begin{array}{ccc|c}
{1}&{0}&{-\displaystyle \frac{5}{13}}&{-\displaystyle \frac{12}{13}}\\
{0}&{1}&{\displaystyle \frac{2}{13}}&{\displaystyle \frac{10}{13}}\\
{0}&{0}&{0}&{0}\end{array}\right]$
The system is consistent (the last row represents 0=0, which is always satisfied).
Taking $z\in \mathbb{R}$ as a parameter, we have:
Eq.2 $\Rightarrow y=-\displaystyle \frac{2}{13}z+\frac{10}{13},$
Eq.$1 \displaystyle \Rightarrow x=\frac{5}{13}z-\frac{12}{13}$
Solution set = $\displaystyle \{(x,y,z)\ | \ x=\frac{5}{13}z-\frac{12}{13},\ y=-\frac{2}{13}z+\frac{10}{13}, z\in \mathbb{R}\}$
