1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x - 3y + 3z = -10 \\ x - 2y + 4z = -3 \\ x - 2y + 2z = -7 \end{cases}$$ 2. **Use elimination or substitution:** We want to find values of $x$, $y$, and $z$ that satisfy all three equations simultaneously. 3. **Subtract the second equation from the first:** $$ (x - 3y + 3z) - (x - 2y + 4z) = -10 - (-3) $$ $$ \cancel{x} - 3y + 3z - \cancel{x} + 2y - 4z = -10 + 3 $$ $$ (-3y + 2y) + (3z - 4z) = -7 $$ $$ -y - z = -7 $$ 4. **Subtract the third equation from the second:** $$ (x - 2y + 4z) - (x - 2y + 2z) = -3 - (-7) $$ $$ \cancel{x} - 2y + 4z - \cancel{x} + 2y - 2z = 4 $$ $$ ( -2y + 2y ) + (4z - 2z) = 4 $$ $$ 2z = 4 $$ 5. **Solve for $z$ from step 4:** $$ z = \frac{4}{2} = 2 $$ 6. **Substitute $z=2$ into the equation from step 3:** $$ -y - 2 = -7 $$ $$ -y = -7 + 2 $$ $$ -y = -5 $$ $$ y = 5 $$ 7. **Substitute $y=5$ and $z=2$ into the first original equation to find $x$:** $$ x - 3(5) + 3(2) = -10 $$ $$ x - 15 + 6 = -10 $$ $$ x - 9 = -10 $$ $$ x = -10 + 9 $$ $$ x = -1 $$ **Final answer:** $$ x = -1, \quad y = 5, \quad z = 2 $$