https://file.aiquickdraw.com/mathbot/file/1728952759831wyksdom5.jpg

The equation in the image you provided is:

\[
\frac{3}{x-8} + \frac{x}{x-3} = \frac{-20x - 65}{x^2 - 11x + 24}
\]

We need to solve for \(x\). Let's proceed step by step:

### Step 1: Factor the denominator on the right-hand side
The quadratic expression in the denominator, \(x^2 - 11x + 24\), can be factored:
\[
x^2 - 11x + 24 = (x - 8)(x - 3)
\]

Thus, the equation becomes:

\[
\frac{3}{x-8} + \frac{x}{x-3} = \frac{-20x - 65}{(x-8)(x-3)}
\]

### Step 2: Combine the left-hand side
Now, let's combine the fractions on the left-hand side:

\[
\frac{3}{x-8} + \frac{x}{x-3} = \frac{3(x-3) + x(x-8)}{(x-8)(x-3)}
\]

Simplifying the numerator:

\[
3(x-3) = 3x - 9
\]
\[
x(x-8) = x^2 - 8x
\]

Thus, the combined numerator is:

\[
3x - 9 + x^2 - 8x = x^2 - 5x - 9
\]

So the equation becomes:

\[
\frac{x^2 - 5x - 9}{(x-8)(x-3)} = \frac{-20x - 65}{(x-8)(x-3)}
\]

### Step 3: Eliminate the denominators
Since the denominators on both sides are the same, we can cancel them out and solve the remaining equation:

\[
x^2 - 5x - 9 = -20x - 65
\]

### Step 4: Move all terms to one side
Bring all terms to the left-hand side:

\[
x^2 - 5x - 9 + 20x + 65 = 0
\]

Simplify:

\[
x^2 + 15x + 56 = 0
\]

### Step 5: Factor the quadratic
Now, we factor the quadratic equation:

\[
x^2 + 15x + 56 = (x + 7)(x + 8) = 0
\]

### Step 6: Solve for \(x\)
Set each factor equal to 0:

\[
x + 7 = 0 \quad \text{or} \quad x + 8 = 0
\]
\[
x = -7 \quad \text{or} \quad x = -8
\]

### Step 7: Check for restrictions
The original equation has restrictions where the denominators cannot be zero. From the original equation, \(x \neq 8\) and \(x \neq 3\).

Thus, the only solution is:

\[
x = -7
\]

Let me know if you need further details or have any questions!

### Related questions:
1. How do you factor quadratic equations?
2. Why is it important to check for restrictions in a rational equation?
3. Can this equation be solved using a different method?
4. What are other types of equations that involve factoring?
5. How do you deal with complex solutions in quadratic equations?

**Tip:** Always check for restrictions from denominators when solving rational equations to avoid invalid solutions!

Solve for x: \( \frac{3}{x-8} + \frac{x}{x-3} = \frac{-20x - 65}{x^2 - 11x + 24} \)

Learn how to solve the rational equation involving factoring: \( \frac{3}{x-8} + \frac{x}{x-3} = \frac{-20x - 65}{x^2 - 11x + 24} \). This detailed guide covers factoring, solving quadratics, and checking for restrictions. Suitable for students in Grades 9-11.

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