1. **State the problem:** Solve the system of equations using the substitution method: $$\begin{cases} 4x - 3y + z = -10 \\ 2x + y + 3z = 0 \\ -x + 2y - 5z = 17 \end{cases}$$ 2. **Choose one equation to express one variable in terms of the others.** From the first equation: $$4x - 3y + z = -10 \implies z = -10 - 4x + 3y$$ 3. **Substitute this expression for $z$ into the other two equations:** Second equation: $$2x + y + 3(-10 - 4x + 3y) = 0$$ $$2x + y - 30 - 12x + 9y = 0$$ $$\cancel{2x} - 12x + \cancel{y} + 9y - 30 = 0$$ $$-10x + 10y - 30 = 0$$ $$-10x + 10y = 30$$ $$-\cancel{10}x + \cancel{10}y = \cancel{10}3$$ $$-x + y = 3$$ Fourth step: Third equation: $$-x + 2y - 5(-10 - 4x + 3y) = 17$$ $$-x + 2y + 50 + 20x - 15y = 17$$ $$-x + 20x + 2y - 15y + 50 = 17$$ $$19x - 13y + 50 = 17$$ $$19x - 13y = 17 - 50$$ $$19x - 13y = -33$$ 4. **Now solve the simpler system:** $$\begin{cases} -x + y = 3 \\ 19x - 13y = -33 \end{cases}$$ From the first equation: $$y = x + 3$$ Substitute into the second: $$19x - 13(x + 3) = -33$$ $$19x - 13x - 39 = -33$$ $$6x - 39 = -33$$ $$6x = -33 + 39$$ $$6x = 6$$ $$x = \frac{6}{6} = 1$$ 5. **Find $y$:** $$y = 1 + 3 = 4$$ 6. **Find $z$ using the expression from step 2:** $$z = -10 - 4(1) + 3(4) = -10 - 4 + 12 = -2$$ **Final answer:** $$\boxed{(x, y, z) = (1, 4, -2)}$$