1. **State the problem:** Solve the system of equations: $$\frac{y}{2} - \frac{y+1}{3} = 0$$ $$2x = \frac{y}{2}$$ 2. **Solve the first equation for $y$:** Start with: $$\frac{y}{2} - \frac{y+1}{3} = 0$$ Multiply both sides by 6 (the least common multiple of 2 and 3) to clear denominators: $$6 \times \left(\frac{y}{2} - \frac{y+1}{3}\right) = 6 \times 0$$ $$3y - 2(y+1) = 0$$ 3. **Simplify the equation:** $$3y - 2y - 2 = 0$$ $$y - 2 = 0$$ 4. **Solve for $y$:** $$y = 2$$ 5. **Use the second equation to find $x$:** Given: $$2x = \frac{y}{2}$$ Substitute $y=2$: $$2x = \frac{2}{2}$$ $$2x = 1$$ 6. **Solve for $x$:** $$\cancel{2}x = \frac{1}{\cancel{2}}$$ $$x = \frac{1}{2}$$ **Final answer:** $$x = \frac{1}{2}, \quad y = 2$$