1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 2x - y + 4z = 8 \\ x - 3y + z = -3 \\ -3x + 2y - z = -1 \end{cases}$$ 2. **Use substitution or elimination method.** Here, we use elimination. 3. From the second equation, express $x$: $$x = 3y - z - 3$$ 4. Substitute $x$ into the first and third equations: First equation: $$2(3y - z - 3) - y + 4z = 8$$ Simplify: $$6y - 2z - 6 - y + 4z = 8$$ $$5y + 2z - 6 = 8$$ $$5y + 2z = 14$$ Third equation: $$-3(3y - z - 3) + 2y - z = -1$$ Simplify: $$-9y + 3z + 9 + 2y - z = -1$$ $$-7y + 2z + 9 = -1$$ $$-7y + 2z = -10$$ 5. Now solve the system: $$\begin{cases} 5y + 2z = 14 \\ -7y + 2z = -10 \end{cases}$$ 6. Subtract the second equation from the first: $$\cancel{5y} + 2z - (\cancel{-7y} + 2z) = 14 - (-10)$$ $$5y + 2z + 7y - 2z = 24$$ $$12y = 24$$ $$y = 2$$ 7. Substitute $y=2$ into the first equation: $$5(2) + 2z = 14$$ $$10 + 2z = 14$$ $$2z = 4$$ $$z = 2$$ 8. Substitute $y=2$ and $z=2$ into $x = 3y - z - 3$: $$x = 3(2) - 2 - 3 = 6 - 2 - 3 = 1$$ **Final solution:** $$\boxed{(x, y, z) = (1, 2, 2)}$$