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The equation provided is:

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

To solve for \( x \), let’s go through each step.

### Step 1: Eliminate the Denominator
Multiply both sides by \( x - 5 \) (assuming \( x \neq 5 \) since that would make the denominator zero):
\[
x^2 + x - 30 = 1 \cdot (x - 5)
\]

### Step 2: Expand and Simplify
Rewrite the equation after expanding the right side:
\[
x^2 + x - 30 = x - 5
\]

### Step 3: Move All Terms to One Side
Subtract \( x \) and add \( 5 \) to both sides to bring all terms to the left:
\[
x^2 + x - 30 - x + 5 = 0
\]

Simplify the terms:
\[
x^2 - 25 = 0
\]

### Step 4: Solve for \( x \)
This is a difference of squares, which can be factored as:
\[
(x + 5)(x - 5) = 0
\]

Thus, the solutions are:
\[
x = -5 \quad \text{and} \quad x = 5
\]

### Step 5: Check for Extraneous Solutions
Since the original equation has a denominator of \( x - 5 \), \( x = 5 \) is not a valid solution because it would make the denominator zero. Therefore, we discard \( x = 5 \).

### Final Answer
The only valid solution is:
\[
x = -5
\]

Would you like further clarification on any of these steps?

---

Here are some related questions for further exploration:

1. What happens if we add constraints to the values of \( x \) based on different domains?
2. How do we solve other types of rational equations with multiple factors in the denominator?
3. What methods can we use if the quadratic equation isn’t factorable?
4. How can we check solutions for extraneous results in more complex equations?
5. How can we generalize the difference of squares for polynomial expressions?

**Tip:** Always check for extraneous solutions when solving equations with denominators.

Solve the equation (x^2 + x - 30) / (x - 5) = 1

Learn to solve the rational equation (x^2 + x - 30) / (x - 5) = 1 with a detailed, step-by-step approach. This solution covers concepts in quadratic equations, difference of squares, and handling extraneous solutions, ideal for students in Grades 8-10.

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