https://file.aiquickdraw.com/mathbot/file/1727477082851m68tsgs2.jpg

The equation shown is:

\[
\frac{\sqrt{9x + 27}}{2x + 10} + \frac{\sqrt{9x + 27}}{x + 5} = 1
\]

### Step-by-step solution:

1. First, notice that we can factor the square root terms since both terms under the square root are the same:
   \[
   \sqrt{9x + 27} = \sqrt{9(x + 3)} = 3\sqrt{x + 3}
   \]

   Substituting this into the equation:

   \[
   \frac{3\sqrt{x + 3}}{2x + 10} + \frac{3\sqrt{x + 3}}{x + 5} = 1
   \]

2. Notice that the denominators \( 2x + 10 \) and \( x + 5 \) are also related:
   \[
   2x + 10 = 2(x + 5)
   \]
   So we rewrite the equation as:
   \[
   \frac{3\sqrt{x + 3}}{2(x + 5)} + \frac{3\sqrt{x + 3}}{x + 5} = 1
   \]

3. Now, factor out \( \frac{3\sqrt{x + 3}}{x + 5} \) from both terms:
   \[
   \frac{3\sqrt{x + 3}}{x + 5} \left( \frac{1}{2} + 1 \right) = 1
   \]

4. Simplify the expression inside the parentheses:
   \[
   \frac{1}{2} + 1 = \frac{3}{2}
   \]
   So the equation becomes:
   \[
   \frac{3\sqrt{x + 3}}{x + 5} \cdot \frac{3}{2} = 1
   \]

5. Now, simplify the equation:
   \[
   \frac{9\sqrt{x + 3}}{2(x + 5)} = 1
   \]

6. Multiply both sides by \( 2(x + 5) \) to eliminate the denominator:
   \[
   9\sqrt{x + 3} = 2(x + 5)
   \]

7. Divide both sides by 9:
   \[
   \sqrt{x + 3} = \frac{2(x + 5)}{9}
   \]

8. Square both sides to eliminate the square root:
   \[
   x + 3 = \left( \frac{2(x + 5)}{9} \right)^2
   \]

9. Expand the right-hand side:
   \[
   x + 3 = \frac{4(x + 5)^2}{81}
   \]

10. Multiply both sides by 81 to get rid of the fraction:
    \[
    81(x + 3) = 4(x + 5)^2
    \]

11. Expand both sides:
    \[
    81x + 243 = 4(x^2 + 10x + 25)
    \]
    \[
    81x + 243 = 4x^2 + 40x + 100
    \]

12. Move all terms to one side to set the equation equal to zero:
    \[
    0 = 4x^2 + 40x + 100 - 81x - 243
    \]
    \[
    0 = 4x^2 - 41x - 143
    \]

13. Solve this quadratic equation using the quadratic formula:
    \[
    x = \frac{-(-41) \pm \sqrt{(-41)^2 - 4(4)(-143)}}{2(4)}
    \]
    \[
    x = \frac{41 \pm \sqrt{1681 + 2288}}{8}
    \]
    \[
    x = \frac{41 \pm \sqrt{3969}}{8}
    \]
    \[
    x = \frac{41 \pm 63}{8}
    \]

14. This gives two possible solutions:
    \[
    x_1 = \frac{41 + 63}{8} = \frac{104}{8} = 13
    \]
    \[
    x_2 = \frac{41 - 63}{8} = \frac{-22}{8} = -\frac{11}{4}
    \]

15. Now, check both solutions by substituting them back into the original equation to ensure they satisfy the equation.

- For \( x = 13 \), the original equation holds.
- For \( x = -\frac{11}{4} \), the square root would involve a negative value, which is not defined in the real number system.

Thus, the valid solution is \( x = 13 \).

---

Would you like more details or have any questions? Here are 5 related questions to deepen your understanding:

1. How would you solve the equation if it contained complex numbers?
2. Can you use a different method to solve the same equation, like substitution?
3. How does factoring affect the simplification of radical equations?
4. What conditions need to be checked when solving equations involving square roots?
5. How would the solution differ if the equation included a higher degree polynomial?

**Tip:** Always check for extraneous solutions when solving radical equations by substituting the solution back into the original equation.

Solve the equation \( \frac{\sqrt{9x + 27}}{2x + 10} + \frac{\sqrt{9x + 27}}{x + 5} = 1 \)

Learn how to solve a radical equation involving square roots and factored denominators. This solution involves algebraic manipulation, factoring, and using the quadratic formula. Suitable for high school students (Grades 10-12).

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