#### Answer

$x_3$ and $x_5$ are free
$$x_1=3x_5+5$$
$$x_2=4x_5+1$$
$$x_4=-9x_5+4$$

#### Work Step by Step

We're given the augmented matrix
$$
\begin{bmatrix}
1 & -3 & 0 & -1 & 0 & -2 \\
0 & 1 & 0 & 0 & -4 & 1 \\
0 & 0 & 0 & 1 & 9 & 4 \\
0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
First, add $3$ times the second row to the first
$$
\begin{bmatrix}
1 & 0 & 0 & -1 & -12 & 1 \\
0 & 1 & 0 & 0 & -4 & 1 \\
0 & 0 & 0 & 1 & 9 & 4 \\
0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
And add the third row to the first.
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & -3 & 5 \\
0 & 1 & 0 & 0 & -4 & 1 \\
0 & 0 & 0 & 1 & 9 & 4 \\
0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
The matrix is now in reduced row echelon form, so we can see the solution to the system:
$x_3$ and $x_5$ are free because their corresponding columns have no pivot entries.
From the first three rows, we can clearly see that
$$x_1=3x_5+5$$
$$x_2=4x_5+1$$
$$x_4=-9x_5+4$$