1. **State the problem:** Solve the system of equations: $$-x_2 - x_3 - 4 = 0$$ $$0 = -1 + x_3 - x_1$$ $$x_2 = x_3 + x_1$$ 2. **Rewrite the equations for clarity:** $$-x_2 - x_3 = 4$$ $$x_3 - x_1 = 1$$ $$x_2 = x_3 + x_1$$ 3. **Use the third equation to substitute $x_2$ in the first equation:** $$-(x_3 + x_1) - x_3 = 4$$ 4. **Simplify the above:** $$-x_3 - x_1 - x_3 = 4$$ $$-2x_3 - x_1 = 4$$ 5. **From the second equation, express $x_3$ in terms of $x_1$:** $$x_3 = 1 + x_1$$ 6. **Substitute $x_3 = 1 + x_1$ into the simplified first equation:** $$-2(1 + x_1) - x_1 = 4$$ $$-2 - 2x_1 - x_1 = 4$$ $$-2 - 3x_1 = 4$$ 7. **Isolate $x_1$:** $$-3x_1 = 4 + 2$$ $$-3x_1 = 6$$ 8. **Divide both sides by $-3$:** $$x_1 = \frac{6}{\cancel{-3}} \times \cancel{-1} = -2$$ 9. **Find $x_3$ using $x_3 = 1 + x_1$:** $$x_3 = 1 + (-2) = -1$$ 10. **Find $x_2$ using $x_2 = x_3 + x_1$:** $$x_2 = -1 + (-2) = -3$$ **Final answer:** $$x_1 = -2, \quad x_2 = -3, \quad x_3 = -1$$