Answer
$\{(x,y,z) | x=5z-2, y=4z-3,\ z\in \mathbb{R}\}$
Work Step by Step
$\left\{\begin{array}{lllll}
x & -y & -z & =1 & \\
-x & +2y & -3z & =-4 & /R_{2}=r_{2}+r_{1}\\
3x & -2y & -7z & =0 & /R_{3}=r_{3}-3r_{1}
\end{array}\right.$
... Eliminate $x$ in eq. 2 and 3
$\left\{\begin{array}{lllll}
x & -y & -z & =1 & \\
& y & -4z & =-3 & \\
& y & -4z & =-3 & /R_{3}=r_{3}-r_{2}
\end{array}\right.$
... Eliminate y in equation 3
$\left\{\begin{array}{lllll}
x & -y & -z & =1 & (1)\\
& y & -4z & =-3 & (2)\\
& & 0 & =0 & (3)
\end{array}\right.$
Equation 3 is 0=0, which is always satisfied.
The system is consistent and dependent.
Let z be the parameter, $z\in \mathbb{R}.$
$(2) \Rightarrow y=4z-3$
$(1)\Rightarrow x=y+z+1\Rightarrow x=5z-2$
Solution set: $\{(x,y,z) | x=5z-2, y=4z-3,\ z\in \mathbb{R}\}$
