1. **State the problem:** Solve the equation $$\frac{5}{2(y-3)} = \frac{10 - y}{(y-3)^2}$$ for $y$. 2. **Identify the formula and rules:** We have a proportion between two fractions. To solve, we can cross-multiply, remembering that $y \neq 3$ because the denominators contain $(y-3)$ and division by zero is undefined. 3. **Cross-multiply:** $$5 \cdot (y-3)^2 = (10 - y) \cdot 2(y-3)$$ 4. **Simplify both sides:** Left side: $$5(y-3)^2 = 5(y-3)(y-3)$$ Right side: $$2(10 - y)(y-3)$$ 5. **Expand both sides:** Left side: $$5(y^2 - 6y + 9) = 5y^2 - 30y + 45$$ Right side: $$2(10y - 30 - y^2 + 3y) = 2(-y^2 + 13y - 30) = -2y^2 + 26y - 60$$ 6. **Set the equation:** $$5y^2 - 30y + 45 = -2y^2 + 26y - 60$$ 7. **Bring all terms to one side:** $$5y^2 + 2y^2 - 30y - 26y + 45 + 60 = 0$$ $$7y^2 - 56y + 105 = 0$$ 8. **Simplify by dividing all terms by 7:** $$y^2 - 8y + 15 = 0$$ 9. **Factor the quadratic:** $$(y - 3)(y - 5) = 0$$ 10. **Solve for $y$:** $$y = 3 \quad \text{or} \quad y = 5$$ 11. **Check for restrictions:** Since $y \neq 3$ (denominator zero), discard $y=3$. 12. **Final solution:** $$\boxed{y = 5}$$