Answer
$x=\frac{1}{3},$
$y=-\frac{1}{3},$
$z=\frac{2}{3}$
Work Step by Step
$\begin{cases}
y-z=-1\\
6x+2y+z=2\\
-x-y-3z=-2
\end{cases}$
Multiplying Equation 3 by 6 and adding it to Equation 2, results in new Equation 3.
$\begin{cases}
6x+2y+z=2\\
-6x-6y-18z=-12\\
-- -- -- --\\
-4y-17z=-10
\end{cases}$
Multiplying Equation 1 by 4 and adding it to new Equation 3.
$\begin{cases}
4y-4z=-4\\
-4y-17z=-10\\
-- -- --\\
-21z=-14
\end{cases}$
Thus, $z=\frac{2}{3}$. Substituting it back into Equation 1, $y-\frac{2}{3}=-1, y=-\frac{1}{3}$. Substituting it back into Equation 3, $-x+\frac{1}{3}-2=-2, x=\frac{1}{3}$
