1. **State the problem:** Solve the equation $$\frac{\sqrt{9x + 27}}{2x + 10} + \frac{\sqrt{9x + 27}}{x + 5} = 1.$$\n\n2. **Simplify the expression inside the square root:** Notice that $$9x + 27 = 9(x + 3).$$ So, $$\sqrt{9x + 27} = \sqrt{9(x + 3)} = 3\sqrt{x + 3}.$$\n\n3. **Rewrite the equation using this simplification:**\n$$\frac{3\sqrt{x + 3}}{2x + 10} + \frac{3\sqrt{x + 3}}{x + 5} = 1.$$\n\n4. **Factor denominators where possible:**\n$$2x + 10 = 2(x + 5).$$\nSo the equation becomes:\n$$\frac{3\sqrt{x + 3}}{2(x + 5)} + \frac{3\sqrt{x + 3}}{x + 5} = 1.$$\n\n5. **Find a common denominator:** The common denominator is $$2(x + 5).$$\nRewrite the second fraction:\n$$\frac{3\sqrt{x + 3}}{x + 5} = \frac{3\sqrt{x + 3} \cdot 2}{2(x + 5)} = \frac{6\sqrt{x + 3}}{2(x + 5)}.$$\n\n6. **Combine the fractions:**\n$$\frac{3\sqrt{x + 3}}{2(x + 5)} + \frac{6\sqrt{x + 3}}{2(x + 5)} = \frac{9\sqrt{x + 3}}{2(x + 5)} = 1.$$\n\n7. **Solve for $$\sqrt{x + 3}$$:**\nMultiply both sides by $$2(x + 5)$$:\n$$9\sqrt{x + 3} = 2(x + 5).$$\n\n8. **Isolate $$\sqrt{x + 3}$$:**\n$$\sqrt{x + 3} = \frac{2(x + 5)}{9}.$$\n\n9. **Square both sides to eliminate the square root:**\n$$x + 3 = \left(\frac{2(x + 5)}{9}\right)^2 = \frac{4(x + 5)^2}{81}.$$\n\n10. **Multiply both sides by 81 to clear the denominator:**\n$$81(x + 3) = 4(x + 5)^2.$$\n\n11. **Expand both sides:**\nLeft side: $$81x + 243.$$\nRight side: $$4(x^2 + 10x + 25) = 4x^2 + 40x + 100.$$\n\n12. **Set the equation to zero:**\n$$81x + 243 = 4x^2 + 40x + 100$$\n$$0 = 4x^2 + 40x + 100 - 81x - 243$$\n$$0 = 4x^2 - 41x - 143.$$\n\n13. **Solve the quadratic equation:**\nUse the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=4$$, $$b=-41$$, $$c=-143$$.\nCalculate the discriminant:\n$$\Delta = (-41)^2 - 4 \cdot 4 \cdot (-143) = 1681 + 2288 = 3969.$$\n\n14. **Calculate the roots:**\n$$x = \frac{41 \pm \sqrt{3969}}{8} = \frac{41 \pm 63}{8}.$$\n\n15. **Find the two solutions:**\n- $$x_1 = \frac{41 + 63}{8} = \frac{104}{8} = 13.$$\n- $$x_2 = \frac{41 - 63}{8} = \frac{-22}{8} = -\frac{11}{4} = -2.75.$$\n\n16. **Check for extraneous solutions:**\nThe original equation has square roots and denominators.\n- The radicand $$9x + 27 = 9(x + 3)$$ must be $$\geq 0$$, so $$x \geq -3.$$ Both solutions satisfy this.\n- Denominators $$2x + 10 = 2(x + 5)$$ and $$x + 5$$ cannot be zero, so $$x \neq -5$$. Both solutions are valid here.\n\n17. **Verify solutions in the original equation:**\n- For $$x=13$$, the equation holds true.\n- For $$x=-2.75$$, the equation also holds true after substitution.\n\n**Final answer:** $$x = 13 \text{ or } x = -\frac{11}{4}.$$