Answer
$\{(x,y,z)|x=5z-2,y=4z-3,z\text{ is any real number}\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x-y-z=1\\
-x+2y-3z=-4\\
3x-2y-7z=0
\end{cases}$
Use the elimination method. Add the first equation to the second. Then multiply the second equation by 3 and add it to the third equation to eliminate $x$:
$\begin{cases}
x-y-z-x+2y-3z=1-4\\
3x-2y-7z+3(-x+2y-3z)=0+3(-4)
\end{cases}$
$\begin{cases}
y-4z=-3\\
3x-2y-7z-3x+6y-9z=-12
\end{cases}$
$\begin{cases}
y-4z=-3\\
4y-16z=-12
\end{cases}$
$\begin{cases}
y-4z=-3\\
y-4z=-3
\end{cases}$
Multiply the first equation by -1, and add it to the second equation to eliminate $y$ and determine $x$:
$\begin{cases}
-y+4z=3\\
y-4z=-3
\end{cases}$
$-y+4z+y-4z=3-3$
$0=0$
As we got an identity, the system has infinitely many solutions.
Solve the equation $y-4z=-3$ in terms of $z$:
$y=4z-3$
Determine $x$:
$x-y-z=1$
$x=y+z+1$
$x=4z-3+z+1$
$x=5z-2$
The solution set of the system is:
$\{(x,y,z)|x=5z-2,y=4z-3,z\text{ is any real number}\}$
