1. **State the problem:** Solve the system of equations: $$-x + 3y = 6$$ $$x + 3y = 12$$ 2. **Add the two equations to eliminate $x$: ** $$(-x + 3y) + (x + 3y) = 6 + 12$$ $$\cancel{-x} + 3y + \cancel{x} + 3y = 18$$ $$6y = 18$$ 3. **Solve for $y$: ** $$y = \frac{18}{6} = 3$$ 4. **Substitute $y=3$ into the first equation to find $x$: ** $$-x + 3(3) = 6$$ $$-x + 9 = 6$$ 5. **Isolate $x$: ** $$-x = 6 - 9$$ $$-x = -3$$ 6. **Multiply both sides by $-1$: ** $$\cancel{-1} \times -x = \cancel{-1} \times -3$$ $$x = 3$$ **Final answer:** $$x = 3, \quad y = 3$$