Answer
The solution set is $\{(7t-5\ ,\ 8t-4,\ t) \ |\ \ t\in \mathbb{R}\}$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
-4 & -1 & 36 & 24\\
1 & -2 & 9 & 3\\
-2 & 1 & 6 & 6
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftrightarrow R_{2}.\\
.\\
+2R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & -2 & 9 & 3\\
-4 & -1 & 36 & 24\\
0 & -3 & 24 & 12
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
+4R_{1}.\\
\div(-3).
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & -2 & 9 & 3\\
0 & -9 & 72 & 36\\
0 & -3 & 24 & 12
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
\div(-9).\\
\div(-3).
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & -2 & 9 & 3\\
0 & 1 & -8 & -4\\
0 & 1 & -8 & -4
\end{array}\right]\rightarrow\left(\begin{array}{l}
+2R_{2}.\\
.\\
-R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & -7 & -5\\
0 & 1 & -8 & -4\\
0 & 0 & 0 & 0
\end{array}\right]$
The system is consistent (has solutions) and dependent.
$z$ has no leading 1 in its corresponding column.
Let $z=t\in \mathbb{R}$, an arbitrary number,
Row $2 \Rightarrow y=8t-4$
Row $1 \Rightarrow x=7t-5$
The solution set is $\{(7t-5\ ,\ 8t-4,\ t) \ |\ \ t\in \mathbb{R}\}$
