https://file.aiquickdraw.com/mathbot/file/1727230463915lwwm8c6t.jpg

The equation from the newly uploaded image is:

\[
\frac{x}{x + 8} + \frac{7}{x - 1} = \frac{18x + 24}{x^2 + 7x - 8}
\]

Let's solve for \( x \).

### Step 1: Factor the denominator on the right-hand side
We need to factor the quadratic expression:
\[
x^2 + 7x - 8
\]
This factors as:
\[
x^2 + 7x - 8 = (x + 8)(x - 1)
\]

Now, the equation becomes:

\[
\frac{x}{x + 8} + \frac{7}{x - 1} = \frac{18x + 24}{(x + 8)(x - 1)}
\]

### Step 2: Combine the fractions on the left-hand side
We need a common denominator for the left-hand side, which will be \((x + 8)(x - 1)\). Rewriting each term:

\[
\frac{x}{x + 8} = \frac{x(x - 1)}{(x + 8)(x - 1)} = \frac{x^2 - x}{(x + 8)(x - 1)}
\]
\[
\frac{7}{x - 1} = \frac{7(x + 8)}{(x + 8)(x - 1)} = \frac{7x + 56}{(x + 8)(x - 1)}
\]

Now, combine the fractions on the left-hand side:

\[
\frac{x^2 - x}{(x + 8)(x - 1)} + \frac{7x + 56}{(x + 8)(x - 1)} = \frac{x^2 - x + 7x + 56}{(x + 8)(x - 1)} = \frac{x^2 + 6x + 56}{(x + 8)(x - 1)}
\]

### Step 3: Set the two sides equal
Now, set the left-hand side equal to the right-hand side:

\[
\frac{x^2 + 6x + 56}{(x + 8)(x - 1)} = \frac{18x + 24}{(x + 8)(x - 1)}
\]

Since the denominators are the same, we can equate the numerators:

\[
x^2 + 6x + 56 = 18x + 24
\]

### Step 4: Solve the quadratic equation
Rearrange the equation to bring all terms to one side:

\[
x^2 + 6x + 56 - 18x - 24 = 0
\]
\[
x^2 - 12x + 32 = 0
\]

### Step 5: Factor or use the quadratic formula
This quadratic can be solved using the quadratic formula:
\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(32)}}{2(1)}
\]
\[
x = \frac{12 \pm \sqrt{144 - 128}}{2}
\]
\[
x = \frac{12 \pm \sqrt{16}}{2}
\]
\[
x = \frac{12 \pm 4}{2}
\]

The solutions are:
\[
x = \frac{12 + 4}{2} = 8 \quad \text{or} \quad x = \frac{12 - 4}{2} = 4
\]

### Step 6: Check for extraneous solutions
We must check if either solution makes the original denominator zero. The original denominators are \(x + 8\) and \(x - 1\). Neither \(x = 8\) nor \(x = 4\) makes these denominators zero, so both solutions are valid.

### Final Answer:
The solutions are:
\[
x = 8 \quad \text{or} \quad x = 4
\]

Would you like further clarification or have any questions?

Here are 5 related questions:
1. How do you handle equations with multiple rational terms?
2. What happens if a solution makes the denominator zero?
3. How do you apply the quadratic formula to different types of equations?
4. What strategies help simplify complex rational equations?
5. Can you solve similar equations graphically?

**Tip:** Always simplify the equation first, especially when factoring can make solving easier!

Solve for x: \( \frac{x}{x + 8} + \frac{7}{x - 1} = \frac{18x + 24}{x^2 + 7x - 8} \).

Learn how to solve the rational equation \( \frac{x}{x + 8} + \frac{7}{x - 1} = \frac{18x + 24}{x^2 + 7x - 8} \). This solution involves factoring quadratic expressions and applying the quadratic formula to find the values of x. Suitable for high school students in Grades 9-11.

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