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The given equation is:

\[
\frac{2}{x^2 - 3x - 10} = \frac{3}{x^2 - x - 6} + \frac{9 - x}{x^2 - 8x + 15}
\]

### Steps to solve:

1. **Factor all quadratic expressions** in the denominators to simplify the equation.

   - \( x^2 - 3x - 10 \) factors as \( (x - 5)(x + 2) \).
   - \( x^2 - x - 6 \) factors as \( (x - 3)(x + 2) \).
   - \( x^2 - 8x + 15 \) factors as \( (x - 5)(x - 3) \).

So the equation becomes:

\[
\frac{2}{(x - 5)(x + 2)} = \frac{3}{(x - 3)(x + 2)} + \frac{9 - x}{(x - 5)(x - 3)}
\]

2. **Find a common denominator** on both sides of the equation. The least common denominator (LCD) is \( (x - 5)(x + 2)(x - 3) \).

3. **Multiply through by the LCD** to eliminate the denominators:

\[
2(x - 3) = 3(x - 5) + (9 - x)(x + 2)
\]

4. **Expand both sides** of the equation:

   - \( 2(x - 3) = 2x - 6 \)
   - \( 3(x - 5) = 3x - 15 \)
   - \( (9 - x)(x + 2) = 9x + 18 - x^2 - 2x = -x^2 + 7x + 18 \)

Thus, the equation becomes:

\[
2x - 6 = 3x - 15 - x^2 + 7x + 18
\]

5. **Simplify** both sides:

   \[
2x - 6 = -x^2 + 10x + 3
\]

6. **Rearrange all terms** to one side to form a quadratic equation:

\[
x^2 - 8x - 9 = 0
\]

7. **Solve the quadratic equation** \( x^2 - 8x - 9 = 0 \).

   Using the quadratic formula:
   \[
   x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-9)}}{2(1)} = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm \sqrt{100}}{2} = \frac{8 \pm 10}{2}
   \]
   
   Therefore:
   \[
   x = \frac{8 + 10}{2} = 9 \quad \text{or} \quad x = \frac{8 - 10}{2} = -1
   \]

8. **Check for extraneous solutions** by ensuring that neither \( x = 9 \) nor \( x = -1 \) make the original denominators zero.

   - For \( x = 9 \), none of the denominators are zero.
   - For \( x = -1 \), none of the denominators are zero either.

Thus, the solutions are \( x = 9 \) and \( x = -1 \).

Would you like further clarification or details on any step?

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Here are 5 related questions to expand your understanding:

1. How do you factor quadratic expressions such as \( x^2 - 3x - 10 \)?
2. What is the significance of finding the least common denominator in rational equations?
3. Why is it important to check for extraneous solutions in equations involving rational expressions?
4. How can the quadratic formula be used for solving any quadratic equation?
5. How do you know when a solution makes a denominator undefined in a rational equation?

**Tip:** Always factor and simplify quadratic expressions early in the problem to make the solution process more efficient.

Solve the equation \( \frac{2}{x^2 - 3x - 10} = \frac{3}{x^2 - x - 6} + \frac{9 - x}{x^2 - 8x + 15} \)

This problem explores solving rational equations with quadratic expressions. It involves factoring, finding common denominators, and using the quadratic formula. Suitable for students in Grades 9-11.

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