1. **State the problem:** Solve the simultaneous equations: $$\begin{cases} 2x + 3y - z = -4 \\ 3x + 2y + 2z = 14 \\ x - 3z = -13 \end{cases}$$ 2. **Use substitution or elimination:** From the third equation, express $x$ in terms of $z$: $$x - 3z = -13 \implies x = -13 + 3z$$ 3. **Substitute $x$ into the first two equations:** First equation: $$2(-13 + 3z) + 3y - z = -4$$ Simplify: $$-26 + 6z + 3y - z = -4$$ $$3y + 5z - 26 = -4$$ $$3y + 5z = 22$$ Second equation: $$3(-13 + 3z) + 2y + 2z = 14$$ Simplify: $$-39 + 9z + 2y + 2z = 14$$ $$2y + 11z - 39 = 14$$ $$2y + 11z = 53$$ 4. **Solve the system for $y$ and $z$:** $$\begin{cases} 3y + 5z = 22 \\ 2y + 11z = 53 \end{cases}$$ Multiply the first equation by 2 and the second by 3 to eliminate $y$: $$\begin{cases} 6y + 10z = 44 \\ 6y + 33z = 159 \end{cases}$$ Subtract the first from the second: $$6y + 33z - (6y + 10z) = 159 - 44$$ $$23z = 115$$ $$z = \frac{115}{23} = 5$$ 5. **Find $y$ using $z=5$ in $3y + 5z = 22$:** $$3y + 5(5) = 22$$ $$3y + 25 = 22$$ $$3y = 22 - 25 = -3$$ $$y = \frac{-3}{3} = -1$$ 6. **Find $x$ using $x = -13 + 3z$ and $z=5$:** $$x = -13 + 3(5) = -13 + 15 = 2$$ **Final solution:** $$\boxed{(x, y, z) = (2, -1, 5)}$$