1. **State the problem:** Solve the system of three equations with three variables: $$\begin{cases} x + 2y + 2z = 8 \\ 3x - 2y + z = -9 \\ 2x + y - z = -4 \end{cases}$$ 2. **Use substitution or elimination method:** We will use elimination to find $x$, $y$, and $z$. 3. **Add equations (1) and (2) to eliminate $y$:** $$ (x + 2y + 2z) + (3x - 2y + z) = 8 + (-9) $$ $$ x + 3x + 2y - 2y + 2z + z = -1 $$ $$ 4x + 3z = -1 $$ 4. **Label this as equation (4):** $$ 4x + 3z = -1 $$ 5. **Add equations (2) and (3) to eliminate $z$:** $$ (3x - 2y + z) + (2x + y - z) = -9 + (-4) $$ $$ 3x + 2x - 2y + y + z - z = -13 $$ $$ 5x - y = -13 $$ 6. **Label this as equation (5):** $$ 5x - y = -13 $$ 7. **From equation (5), express $y$ in terms of $x$:** $$ y = 5x + 13 $$ 8. **Substitute $y = 5x + 13$ into equation (1):** $$ x + 2(5x + 13) + 2z = 8 $$ $$ x + 10x + 26 + 2z = 8 $$ $$ 11x + 2z = 8 - 26 $$ $$ 11x + 2z = -18 $$ 9. **Label this as equation (6):** $$ 11x + 2z = -18 $$ 10. **Now solve equations (4) and (6) for $x$ and $z$:** Equation (4): $$4x + 3z = -1$$ Equation (6): $$11x + 2z = -18$$ 11. **Multiply equation (4) by 2 and equation (6) by 3 to align $z$ coefficients:** $$ 2(4x + 3z) = 2(-1) \Rightarrow 8x + 6z = -2 $$ $$ 3(11x + 2z) = 3(-18) \Rightarrow 33x + 6z = -54 $$ 12. **Subtract the first from the second to eliminate $z$:** $$ (33x + 6z) - (8x + 6z) = -54 - (-2) $$ $$ 33x - 8x + 6z - 6z = -54 + 2 $$ $$ 25x = -52 $$ 13. **Solve for $x$:** $$ x = \frac{-52}{25} $$ 14. **Substitute $x$ back into equation (4) to find $z$:** $$ 4\left(\frac{-52}{25}\right) + 3z = -1 $$ $$ \frac{-208}{25} + 3z = -1 $$ $$ 3z = -1 + \frac{208}{25} = \frac{-25}{25} + \frac{208}{25} = \frac{183}{25} $$ $$ z = \frac{183}{75} = \frac{61}{25} $$ 15. **Substitute $x$ into $y = 5x + 13$ to find $y$:** $$ y = 5\left(\frac{-52}{25}\right) + 13 = \frac{-260}{25} + 13 = \frac{-260}{25} + \frac{325}{25} = \frac{65}{25} = \frac{13}{5} $$ 16. **Final solution:** $$ x = \frac{-52}{25}, \quad y = \frac{13}{5}, \quad z = \frac{61}{25} $$