1. The problem asks to solve the system of three linear equations with three variables: $$\begin{cases} 2x + 3y - z = 7 \\ x + y + z = 8 \\ x + y - z = 0 \end{cases}$$ 2. We will use substitution or elimination to find $x$, $y$, and $z$. 3. Add the last two equations: $$ (x + y + z) + (x + y - z) = 8 + 0 $$ $$ 2x + 2y = 8 $$ 4. Simplify: $$ \cancel{2}x + \cancel{2}y = \cancel{8}4 $$ $$ x + y = 4 $$ 5. From the third equation: $$ x + y - z = 0 \implies z = x + y $$ 6. Substitute $z = x + y$ into the first equation: $$ 2x + 3y - (x + y) = 7 $$ $$ 2x + 3y - x - y = 7 $$ $$ (2x - x) + (3y - y) = 7 $$ $$ x + 2y = 7 $$ 7. Now we have two equations: $$ \begin{cases} x + y = 4 \\ x + 2y = 7 \end{cases}$$ 8. Subtract the first from the second: $$ (x + 2y) - (x + y) = 7 - 4 $$ $$ x + 2y - x - y = 3 $$ $$ y = 3 $$ 9. Substitute $y=3$ into $x + y = 4$: $$ x + 3 = 4 $$ $$ x = 1 $$ 10. Find $z$ using $z = x + y$: $$ z = 1 + 3 = 4 $$ Final answer: $$ (x, y, z) = (1, 3, 4) $$