#### Answer

$x_3$ and $x_4$ free
$$x_1=-7x_3-9$$
$$x_2=6x_3+3x_4+2$$
$$x_4=0$$

#### Work Step by Step

We're given the augmented matrix:
$$
\begin{bmatrix}
1 & 2 & -5 & -6 & 0 & -5 \\
0 & 1 & -6 & -3 & 0 & 2 \\
0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
To put it into reduced row echelon form, add $-2$ times the second row to the first row:
$$
\begin{bmatrix}
1 & 0 & 7 & 0 & 0 & -9 \\
0 & 1 & -6 & -3 & 0 & 2 \\
0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
Now, we can see that $x_3$ and $x_4$ are free. From the matrix, we have:
$$x_1=-7x_3-9$$
$$x_2=6x_3+3x_4+2$$
$$x_4=0$$