Answer
Solution set: $\{ ( -2t+3$ , $-2t+1$ , $t )$ , $t\in \mathbb{R}\}$
Work Step by Step
Augmented matrix:
$\left[\begin{array}{lllll}
3 & -1 & 4 & | & 8\\
0 & 1 & 2 & | & 1
\end{array}\right]\left\{\begin{array}{l}
\div 3\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & -1/3 & 4/3 & | & 8/3\\
0 & 1 & 2 & | & 1
\end{array}\right]$
The system is consistent and does not have a unique solution.
Let $z=t, t\in \mathbb{R}$.
Substituting into row 2,
$y+2t=1$
$y=-2t+1$
Substituting into row 1,
$x-\displaystyle \frac{1}{3}(-2t+1)+\frac{4}{3}t=\frac{8}{3}$
$x+\displaystyle \frac{6}{3}t-\frac{1}{3}=\frac{8}{3}$
$x=-2t+\displaystyle \frac{8}{3}+\frac{1}{3}$
$x=-2t+3$
Solution set: $\{ ( -2t+3$ , $-2t+1$ , $t )$ , $t\in \mathbb{R}\}$
