1. **State the problem:** Given the equation $$2025^{-2} = \frac{3^{\frac{25}{x}} \cdot 5^{\frac{25}{y}}}{5^{\frac{25}{x}} \cdot 45^{\frac{5}{y}}}$$ we need to find the value of $x - y$. 2. **Rewrite the bases in prime factorization:** Note that: $$2025 = 45^2 = (3^2 \cdot 5)^2 = 3^4 \cdot 5^2$$ and $$45 = 3^2 \cdot 5$$ 3. **Express the left side:** $$2025^{-2} = (3^4 \cdot 5^2)^{-2} = 3^{-8} \cdot 5^{-4}$$ 4. **Rewrite the right side:** $$\frac{3^{\frac{25}{x}} \cdot 5^{\frac{25}{y}}}{5^{\frac{25}{x}} \cdot 45^{\frac{5}{y}}} = 3^{\frac{25}{x}} \cdot 5^{\frac{25}{y}} \cdot 5^{-\frac{25}{x}} \cdot (3^2 \cdot 5)^{-\frac{5}{y}}$$ 5. **Simplify the right side by combining exponents:** $$= 3^{\frac{25}{x}} \cdot 5^{\frac{25}{y}} \cdot 5^{-\frac{25}{x}} \cdot 3^{-\frac{10}{y}} \cdot 5^{-\frac{5}{y}}$$ $$= 3^{\frac{25}{x} - \frac{10}{y}} \cdot 5^{\frac{25}{y} - \frac{25}{x} - \frac{5}{y}}$$ 6. **Simplify the exponent of 5:** $$\frac{25}{y} - \frac{25}{x} - \frac{5}{y} = \frac{20}{y} - \frac{25}{x}$$ 7. **Equate the powers of 3 and 5 on both sides:** For base 3: $$-8 = \frac{25}{x} - \frac{10}{y}$$ For base 5: $$-4 = \frac{20}{y} - \frac{25}{x}$$ 8. **Rewrite the system:** $$\frac{25}{x} - \frac{10}{y} = -8$$ $$-\frac{25}{x} + \frac{20}{y} = -4$$ 9. **Add the two equations to eliminate $\frac{25}{x}$:** $$\left(\frac{25}{x} - \frac{10}{y}\right) + \left(-\frac{25}{x} + \frac{20}{y}\right) = -8 + (-4)$$ $$-\frac{10}{y} + \frac{20}{y} = -12$$ $$\frac{10}{y} = -12$$ $$y = \frac{10}{-12} = -\frac{5}{6}$$ 10. **Substitute $y = -\frac{5}{6}$ into the first equation:** $$\frac{25}{x} - \frac{10}{-\frac{5}{6}} = -8$$ $$\frac{25}{x} + 12 = -8$$ $$\frac{25}{x} = -20$$ $$x = \frac{25}{-20} = -\frac{5}{4}$$ 11. **Find $x - y$:** $$x - y = -\frac{5}{4} - \left(-\frac{5}{6}\right) = -\frac{5}{4} + \frac{5}{6} = \frac{-15 + 10}{12} = -\frac{5}{12}$$ **Final answer:** $$x - y = -\frac{5}{12}$$