1. **State the problem:** Rationalise the denominator of the expression $$\frac{6 + 5\sqrt{2}}{3\sqrt{2} + 4}$$ and express the answer in the form $$a + b\sqrt{2}$$. 2. **Formula and rule:** To rationalise a denominator with a surd, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $$3\sqrt{2} + 4$$ is $$4 - 3\sqrt{2}$$. 3. **Multiply numerator and denominator by the conjugate:** $$\frac{6 + 5\sqrt{2}}{3\sqrt{2} + 4} \times \frac{4 - 3\sqrt{2}}{4 - 3\sqrt{2}} = \frac{(6 + 5\sqrt{2})(4 - 3\sqrt{2})}{(3\sqrt{2} + 4)(4 - 3\sqrt{2})}$$ 4. **Expand numerator:** $$6 \times 4 = 24$$ $$6 \times (-3\sqrt{2}) = -18\sqrt{2}$$ $$5\sqrt{2} \times 4 = 20\sqrt{2}$$ $$5\sqrt{2} \times (-3\sqrt{2}) = -15 \times 2 = -30$$ Sum numerator terms: $$24 - 18\sqrt{2} + 20\sqrt{2} - 30 = (24 - 30) + (-18\sqrt{2} + 20\sqrt{2}) = -6 + 2\sqrt{2}$$ 5. **Expand denominator using difference of squares:** $$(3\sqrt{2})^2 - 4^2 = 9 \times 2 - 16 = 18 - 16 = 2$$ 6. **Final expression:** $$\frac{-6 + 2\sqrt{2}}{2} = \frac{-6}{2} + \frac{2\sqrt{2}}{2} = -3 + \sqrt{2}$$ **Answer:** $$-3 + \sqrt{2}$$