Answer
$\{(4,0,-2)\}$
Work Step by Step
The given first equation can be re-written as:
$x=2-z~~~(1) $
Now, $-x+2y=-4 \implies 2y+z=-2 ~~~~(2)$
and $2y-z+2y+z=2-2 \implies y=0$
Plug $y=0$ in equation $2y-z=2$ to obtain: $z=-2$
Back substitution for $z=-2$ and $y=0$ to solve for $x$: $x-2=2 \implies x=4$
Solution set: $\{(4,0,-2)\}$
