1. **State the problem:** Solve the system of linear equations: $$-x + 3y = 5$$ $$-3y = 1 + x$$ 2. **Rewrite the second equation:** $$-3y = 1 + x \implies -3y - x = 1$$ or equivalently: $$-x - 3y = 1$$ 3. **Observe the system:** $$-x + 3y = 5$$ $$-x - 3y = 1$$ 4. **Subtract the second equation from the first:** $$(-x + 3y) - (-x - 3y) = 5 - 1$$ Simplify the left side: $$-x + 3y + x + 3y = 4$$ which simplifies to: $$6y = 4$$ 5. **Solve for $y$:** $$y = \frac{4}{6} = \frac{2}{3}$$ 6. **Substitute $y = \frac{2}{3}$ into the first equation:** $$-x + 3 \times \frac{2}{3} = 5$$ Simplify: $$-x + 2 = 5$$ 7. **Solve for $x$:** $$-x = 5 - 2 = 3$$ $$x = -3$$ **Final answer:** $$x = -3, \quad y = \frac{2}{3}$$