1. **State the problem:** Solve the equation $$\frac{1}{3}(x - 3) - \frac{x+1}{3} + \frac{3+x}{3} = \frac{1}{3} - \frac{2-x}{3} + \frac{y}{3} + 1$$ for $x$ and $y$. 2. **Rewrite the equation:** Multiply each term by 3 to clear denominators: $$3 \times \left(\frac{1}{3}(x - 3) - \frac{x+1}{3} + \frac{3+x}{3}\right) = 3 \times \left(\frac{1}{3} - \frac{2-x}{3} + \frac{y}{3} + 1\right)$$ which simplifies to: $$(x - 3) - (x + 1) + (3 + x) = 1 - (2 - x) + y + 3$$ 3. **Simplify both sides:** Left side: $$x - 3 - x - 1 + 3 + x = x - 1$$ Right side: $$1 - 2 + x + y + 3 = x + y + 2$$ 4. **Set simplified expressions equal:** $$x - 1 = x + y + 2$$ 5. **Isolate $y$:** Subtract $x$ from both sides: $$\cancel{x} - 1 = \cancel{x} + y + 2$$ which gives: $$-1 = y + 2$$ 6. **Solve for $y$:** $$y = -1 - 2 = -3$$ **Final answer:** $$y = -3$$ Note: The equation does not restrict $x$, so $x$ can be any real number.