Answer
$(6-2t, -9+\frac{5}{2}t, -9+\frac{9}{2}t, t)$
Work Step by Step
Step 1. Establish the augmented matrix of the system and use the Gauss Eliminations method:
$\begin{vmatrix} 0 & 1 & -1 & 2 & 0 \\3 & 2 & 0 & 1 & 0\\2 & 0 & 0 & 4 & 12 \\-2 & 0 & -2 & 5 & 6 \end{vmatrix} \begin{array}(R_3/2\leftrightarrow R_1\\ \\ \\ \\ \end{array}$
Step 2. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 0 & 0 & 2 & 6 \\3 & 2 & 0 & 1 & 0\\0 & 1 & -1 & 2 & 0 \\-2 & 0 & -2 & 5 & 6 \end{vmatrix} \begin{array}( \\R_2-3R_1\to R_2 \\ \\R_4+2R_1\to R_4 \\ \end{array}$
Step 3. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 0 & 0 & 2 & 6 \\0 & 2 & 0 & -5 & -18\\0 & 1 & -1 & 2 & 0 \\0 & 0 & -2 & 9 & 18 \end{vmatrix} \begin{array}( \\R_3\leftrightarrow R_2 \\ \\ \\ \end{array}$
Step 4. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 0 & 0 & 2 & 6 \\0 & 1 & -1 & 2 & 0 \\ 0 & 2 & 0 & -5 & -18 \\0 & 0 & -2 & 9 & 18 \end{vmatrix} \begin{array}( \\ \\R_3-2R_2\to R_3 \\-R_4\to R_4 \\ \end{array}$
Step 5. Do the operations given on the right side of the matrix.
$\begin{vmatrix} 1 & 0 & 0 & 2 & 6 \\0 & 1 & -1 & 2 & 0 \\ 0 & 0 & 2 & -9 & -18 \\0 & 0 & 2 & -9 & -18 \end{vmatrix} \begin{array}( \\ \\ \\ \\ \end{array}$
Step 6. The last two rows are the same. Let $w=t$, row 3 gives $2z-9t=-18$ and $z=-9+\frac{9}{2}t$. From row 2, $y-z+2t=0$ gives $y=-9+\frac{5}{2}t$. Row 1 gives $x+2t=6$ and $x=6-2t$
Step 7. We can write the final results as $(6-2t, -9+\frac{5}{2}t, -9+\frac{9}{2}t, t)$
