Answer
$x=4; y=1; z=2$
Work Step by Step
Add the first equation to equation $2$, then subtract equation 3 from equation $2$.
The new system of equations is as follows:
$x+y+z=7 \\ 3x+2z=16 \\3z=6$
The third equation yields $z=2$
Plug the value of $z$ into the second equation, giving: $x=4$
Now, we will plug the value of $x$ and $z$ into the first equation to get the value of $y$.
$4+y+2=7 \\ \implies y=1$
Thus, $x=4; y=1; z=2$
