Answer
The solution is $(1-t,2-3t,t)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
3& -1 &0 |1\\
2& 1 &5| 4\\
7 & -5 & -8|-3
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
3& -1 &0 |1\\
2& 1 &5| 4\\
7 & -5 & -8|-3
\end{bmatrix} \approx^1 \begin{bmatrix}
1& -2 &-5 |-3\\
2& 1 &5| 4\\
7 & -5 & -8|-3
\end{bmatrix} \approx^2\begin{bmatrix}
1& -2 &-5 |-3\\
0& 5 &15| 10\\
0 & 9 & 27|18
\end{bmatrix} \approx^3 \begin{bmatrix}
1& -2 &-5 |-3\\
0& 1 &3| 2\\
0 & 9 & 27|18
\end{bmatrix} \approx^4 \begin{bmatrix}
1& 0 &1 |1\\
0& 1 &3| 2\\
0 & 0 & 0|0
\end{bmatrix} $
The equivalent system is:
$x_1+x_3=1$
$x_2+3x_3=2$
$x_3=t$
There is one free variable, which we take to be $x_3 = t$, where t can assume any complex value. Applying back substitution yields:
$x_1=1-t$
$x_2=2-3t$
$x_3=t$
The solution is $(1-t,2-3t,t)$
