The reduced row echelon form of the augmented matrix of a system of linear equations...

${\displaystyle x_1-s_3+3x_4=9}$
${\displaystyle x_2+2x_3-5x_4=8}$
${\displaystyle 0=0}$ The augmented matrix does not contain a row in which the only nonzero entry appears in the last column. Therefore, this system of equations must be consistent. Convert the augmented matrix into a system of equations. ${\displaystyle x_1=9+x_3-3x_4}$
${\displaystyle x_2=8-2x_3+5x_4}$
${\displaystyle x_3,\mathfrak{e}e}$
${\displaystyle x_4,\mathfrak{e}e}$ Solve for the leading entry for each individual equation. Determine the free variables, if any. ${\displaystyle x_1=9+s-3t}$
${\displaystyle x_2=8-2s+5t}$
${\displaystyle x_3=s}$
${\displaystyle x_4=t}$ Parameterize the free variables. $x=\left[\begin{array}{c}9\\ 8\\ 0\\ 0\end{array}\right]+s\left[\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}-3\\ 5\\ 0\\ 1\end{array}\right]$ And write the solution in vector form.