1. **State the problem:** Solve for $x$ in the equation $$\frac{\sqrt{9x + 27}}{2x + 10} + \frac{\sqrt{9x + 27}}{x + 5} = 1.$$\n\n2. **Simplify the equation:** Notice that $2x + 10 = 2(x + 5)$. Rewrite the equation as $$\frac{\sqrt{9x + 27}}{2(x + 5)} + \frac{\sqrt{9x + 27}}{x + 5} = 1.$$\n\n3. **Find a common denominator:** The common denominator is $2(x + 5)$. Rewrite the second term: $$\frac{\sqrt{9x + 27}}{x + 5} = \frac{2\sqrt{9x + 27}}{2(x + 5)}.$$\n\n4. **Combine the fractions:** $$\frac{\sqrt{9x + 27}}{2(x + 5)} + \frac{2\sqrt{9x + 27}}{2(x + 5)} = \frac{3\sqrt{9x + 27}}{2(x + 5)} = 1.$$\n\n5. **Isolate the square root term:** Multiply both sides by $2(x + 5)$ to get $$3\sqrt{9x + 27} = 2(x + 5).$$\n\n6. **Divide both sides by 3:** $$\sqrt{9x + 27} = \frac{2}{3}(x + 5).$$\n\n7. **Square both sides to eliminate the square root:** $$9x + 27 = \left(\frac{2}{3}(x + 5)\right)^2 = \frac{4}{9}(x + 5)^2.$$\n\n8. **Multiply both sides by 9 to clear the denominator:** $$9(9x + 27) = 4(x + 5)^2,$$ which simplifies to $$81x + 243 = 4(x^2 + 10x + 25).$$\n\n9. **Expand the right side:** $$81x + 243 = 4x^2 + 40x + 100.$$\n\n10. **Bring all terms to one side:** $$0 = 4x^2 + 40x + 100 - 81x - 243,$$ which simplifies to $$0 = 4x^2 - 41x - 143.$$\n\n11. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $a=4$, $b=-41$, $c=-143$.\n\nCalculate the discriminant: $$\Delta = (-41)^2 - 4 \cdot 4 \cdot (-143) = 1681 + 2288 = 3969.$$\n\n12. **Find the roots:** $$x = \frac{41 \pm \sqrt{3969}}{8} = \frac{41 \pm 63}{8}.$$\n\nSo, $$x_1 = \frac{41 + 63}{8} = \frac{104}{8} = 13,$$ and $$x_2 = \frac{41 - 63}{8} = \frac{-22}{8} = -\frac{11}{4} = -2.75.$$\n\n13. **Check for extraneous solutions:** The original equation has square roots and denominators. The radicand $9x + 27$ must be non-negative: $$9x + 27 \geq 0 \Rightarrow x \geq -3.$$\n\nDenominators $2x + 10 = 2(x + 5)$ and $x + 5$ cannot be zero, so $$x \neq -5.$$ Both $13$ and $-2.75$ satisfy $x \geq -3$ and are not $-5$.\n\n14. **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:** $$x = 13 \text{ or } x = -\frac{11}{4}.$$