Answer
$x=1,$
$y=-3,$
$z=2$
Work Step by Step
$\begin{cases}
x+y+z=0\\
-x+2y+5z=3\\
3x-y+0z=6
\end{cases}$
Adding Equation 1 and Equation 2, results in new Equation 2.
$\begin{cases}
x+y+z=0\\
-x+2y+5z=3\\
-- -- -- -- --\\
3y+6z=3
\end{cases}$
Multiplying Equation 1 by -3 and Adding it to Equation 3, results in new Equation 3.
$\begin{cases}
-3x-3y-3z=0\\
3x-y+0z=6\\
-- -- -- -- --\\
-4y-3z=6
\end{cases}$
Multiplying new Equation 3 by 2 and Adding it to new Equation 2.
$\begin{cases}
-8y-6z=12\\
3y+6z=3\\
-- -- -- --\\
-5y=15
\end{cases}$
Thus, $y=-3$. Substituting back into Equation 3, $3x+3=6, x=1$.
Substituting back into Equation 1, $1-3+z=0, z=2$.
