1. **State the problem:** Solve the system of equations: $$\frac{x+1}{3} - \frac{y-1}{2} = 1$$ $$7x - 4(x + y) = 4$$ 2. **Rewrite the first equation to clear denominators:** Multiply both sides by 6 (the least common multiple of 3 and 2): $$6 \times \left(\frac{x+1}{3} - \frac{y-1}{2}\right) = 6 \times 1$$ This gives: $$2(x+1) - 3(y-1) = 6$$ 3. **Expand and simplify:** $$2x + 2 - 3y + 3 = 6$$ $$2x - 3y + 5 = 6$$ $$2x - 3y = 1$$ 4. **Simplify the second equation:** $$7x - 4(x + y) = 4$$ Distribute: $$7x - 4x - 4y = 4$$ $$3x - 4y = 4$$ 5. **Now solve the system:** $$\begin{cases} 2x - 3y = 1 \\ 3x - 4y = 4 \end{cases}$$ 6. **Use elimination:** Multiply the first equation by 3 and the second by 2: $$6x - 9y = 3$$ $$6x - 8y = 8$$ 7. **Subtract the second from the first:** $$(6x - 9y) - (6x - 8y) = 3 - 8$$ $$-9y + 8y = -5$$ $$-y = -5$$ $$y = 5$$ 8. **Substitute $y=5$ into $2x - 3y = 1$:** $$2x - 3(5) = 1$$ $$2x - 15 = 1$$ $$2x = 16$$ $$x = 8$$ **Final answer:** $$\boxed{(x, y) = (8, 5)}$$