#### Answer

$x_1 = 7x_2 - 6 x_4 + 5$
$x_2$ free
$x_3 = 2x_4 + 3$
$x_4$ free

#### Work Step by Step

We're given the augmented matrix
$$
\begin{bmatrix}
1 & -7 & 0 & 6 & 5 \\
0 & 0 & 1 & -2 & -3 \\
-1 & 7 & -4 & 2 & 7
\end{bmatrix}
$$
First, add the first row to the third row
$$
\begin{bmatrix}
1 & -7 & 0 & 6 & 5 \\
0 & 0 & 1 & -2 & -3 \\
0 & 0 & -4 & 8 & 12
\end{bmatrix}
$$
And add the second row to the third row
$$
\begin{bmatrix}
1 & -7 & 0 & 6 & 5 \\
0 & 0 & 1 & -2 & -3 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
This leaves us with the general solution to the system:
$x_2$ and $x_4$ are free because they have no pivot columns.
From the equations represented by the first two rows of the matrix, we have:
$x_1 = 7x_2 - 6 x_4 + 5$
$x_3 = 2x_4 + 3$