https://file.aiquickdraw.com/mathbot/file/1729702710942pnyqadxt.jpg

The given equation from the image is:

\[
\frac{x^2 + x - 30}{x - 5} = 1
\]

### Steps to solve this:

1. Multiply both sides of the equation by \( x - 5 \) to eliminate the denominator:
\[
x^2 + x - 30 = 1(x - 5)
\]
This simplifies to:
\[
x^2 + x - 30 = x - 5
\]

2. Move all terms to one side of the equation:
\[
x^2 + x - 30 - x + 5 = 0
\]
Simplifying the terms:
\[
x^2 - 25 = 0
\]

3. Now solve this quadratic equation:
\[
x^2 = 25
\]

4. Take the square root of both sides:
\[
x = \pm 5
\]

### Solution:
The possible solutions are \( x = 5 \) and \( x = -5 \).

However, we need to check for any extraneous solutions. If \( x = 5 \), the denominator of the original equation \( (x - 5) \) becomes zero, which is undefined. Therefore, \( x = 5 \) is not a valid solution.

Thus, the only valid solution is \( x = -5 \).

Would you like more details on this solution? Here are some related questions you can explore:

1. What would happen if the equation had been \( \frac{x^2 + x - 30}{x - 6} = 1 \)?
2. How do you determine if a quadratic equation has real or complex roots?
3. What is the significance of extraneous solutions in rational equations?
4. How can you graphically interpret the solution to this equation?
5. What are some strategies for solving more complex rational equations?

**Tip:** Always check for extraneous solutions when solving equations with variables in the denominator to avoid dividing by zero.

Discover how to solve the rational equation involving a quadratic expression, x^2 + x - 30 / (x - 5) = 1. This step-by-step solution covers multiplying both sides, solving quadratic equations, and checking for extraneous solutions. Ideal for students in Grades 9-11.

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