Answer
$x=0,$
$y=1,$
$z=2$
Work Step by Step
$\begin{cases}
2x+4y-z=2\\
x+2y-3z=-4\\
3x-y+z=1
\end{cases}$
Multiplying Equation 2 by -2 and adding it to Equation 1.
$\begin{cases}
2x+4y-z=2\\
-2x-4y+6z=8\\
-- -- -- -- --\\
5z=10
\end{cases}$
Thus, $z=2$.
Substituting back into the Equation...
Multiplying Equation 2 by 2 and adding it to Equation 3.
$\begin{cases}
6x-2y=-2\\
x+2y=2\\
-- -- --\\
7x=0
\end{cases}$
Thus, $x=0$. Substituting back into Equation 1, $4y-2=2, y=1$.
