Answer
$x_1=2-37a$
$x_2=16a$
$x_3=1-7a$
$x_4=a$
Work Step by Step
Build the augmented matrix of the system of equations;
$\begin{bmatrix}1&4&3&-6&|&5\\1&3&1&-4&|&3\\2&8&7&-5&|&11\\2&5&0&-6&|&4\end{bmatrix}$
Bring the matrix to the row reduced echelon form:
Add $-R_1$ to $R_2$, $-2R_1$ to $R_3$ and $R_4$:
$\begin{bmatrix}1&4&3&-6&|&5\\0&-1&-2&2&|&-2\\0&0&1&7&|&1\\0&-3&-6&6&|&-6\end{bmatrix}$
Multiply $R_2$ by -1:
$\begin{bmatrix}1&4&3&-6&|&5\\0&1&2&-2&|&2\\0&0&1&7&|&1\\0&-3&-6&6&|&-6\end{bmatrix}$
Add $-4R_2$ to $R_1$ and $3R_2$ to $R_4$:
$\begin{bmatrix}1&0&-5&2&|&-3\\0&1&2&-2&|&2\\0&0&1&7&|&1\\0&0&0&0&|&0\end{bmatrix}$
Add $5R_3$ to $R_1$ and $-2R_3$ to $R_2$:
$\begin{bmatrix}1&0&0&37&|&2\\0&1&0&-16&|&0\\0&0&1&7&|&1\\0&0&0&0&|&0\end{bmatrix}$
As Row 4 has only zeros, it means the variable $x_4$ is independent. Note $x_4=a$.
$\begin{cases}
x_1+37x_4=2\\
x_2-16x_4=0\\
x_3+7x_4=1
\end{cases}$
$\begin{cases}
x_1+37a=2\\
x_2-16a=0\\
x_3+7a=1
\end{cases}$
The solution is:
$x_1=2-37a$
$x_2=16a$
$x_3=1-7a$
$x_4=a$
