1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x + 4y + z = 9 \\ -x + 2y + 2z = 0 \\ 2x + 2y - z = 9 \end{cases}$$ 2. **Method:** We will use the elimination method to find $x$, $y$, and $z$. 3. **Add the first two equations to eliminate $x$:** $$ (x + 4y + z) + (-x + 2y + 2z) = 9 + 0 $$ Simplifies to: $$ 6y + 3z = 9 $$ Divide both sides by 3: $$ 2y + z = 3 \quad (4) $$ 4. **Use equation (4) and the third original equation:** Third equation: $$ 2x + 2y - z = 9 \quad (3) $$ From (4), express $z$: $$ z = 3 - 2y $$ 5. **Substitute $z$ into equation (3):** $$ 2x + 2y - (3 - 2y) = 9 $$ Simplify: $$ 2x + 2y - 3 + 2y = 9 $$ $$ 2x + 4y - 3 = 9 $$ $$ 2x + 4y = 12 $$ Divide both sides by 2: $$ x + 2y = 6 \quad (5) $$ 6. **Use equation (1) to substitute $z$:** $$ x + 4y + z = 9 $$ Substitute $z = 3 - 2y$: $$ x + 4y + 3 - 2y = 9 $$ Simplify: $$ x + 2y + 3 = 9 $$ $$ x + 2y = 6 $$ This matches equation (5), confirming consistency. 7. **Express $x$ from (5):** $$ x = 6 - 2y $$ 8. **Substitute $x$ and $z$ into equation (2):** Equation (2): $$ -x + 2y + 2z = 0 $$ Substitute $x = 6 - 2y$ and $z = 3 - 2y$: $$ -(6 - 2y) + 2y + 2(3 - 2y) = 0 $$ Simplify: $$ -6 + 2y + 2y + 6 - 4y = 0 $$ $$ (2y + 2y - 4y) + (-6 + 6) = 0 $$ $$ 0 = 0 $$ This is always true, so $y$ is a free variable. 9. **Parametric solution:** Let $y = t$ (a parameter). Then: $$ x = 6 - 2t $$ $$ z = 3 - 2t $$ 10. **Final answer:** $$ (x, y, z) = (6 - 2t, t, 3 - 2t) \quad \text{for any real number } t $$ This means the system has infinitely many solutions along a line parameterized by $t$.