1. **State the problem:** Rationalize the expression $$\frac{3\sqrt{3}+\sqrt{5}}{3\sqrt{3}-\sqrt{5}} \times \frac{3\sqrt{3}+\sqrt{5}}{3\sqrt{3}+\sqrt{5}}$$ 2. **Formula and rule:** To rationalize a denominator with surds, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $$a - b$$ is $$a + b$$ and vice versa. 3. **Multiply the numerators:** $$\left(3\sqrt{3}+\sqrt{5}\right) \times \left(3\sqrt{3}+\sqrt{5}\right) = \left(3\sqrt{3}\right)^2 + 2 \times 3\sqrt{3} \times \sqrt{5} + \left(\sqrt{5}\right)^2$$ 4. **Calculate each term:** $$\left(3\sqrt{3}\right)^2 = 9 \times 3 = 27$$ $$2 \times 3\sqrt{3} \times \sqrt{5} = 6\sqrt{15}$$ $$\left(\sqrt{5}\right)^2 = 5$$ 5. **Sum the numerator:** $$27 + 6\sqrt{15} + 5 = 32 + 6\sqrt{15}$$ 6. **Multiply the denominators (difference of squares):** $$\left(3\sqrt{3}\right)^2 - \left(\sqrt{5}\right)^2 = 27 - 5 = 22$$ 7. **Write the rationalized expression:** $$\frac{32 + 6\sqrt{15}}{22}$$ 8. **Simplify by dividing numerator and denominator by 2:** $$\frac{\cancel{2}(16 + 3\sqrt{15})}{\cancel{2}11} = \frac{16 + 3\sqrt{15}}{11}$$ **Final answer:** $$\frac{16 + 3\sqrt{15}}{11}$$