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The given equation is:

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

### Steps to solve:

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

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

3. **Factor the quadratic equation**:
   \[
   (x - 5)(x + 5) = 0
   \]

4. **Solve for \(x\)**:
   \[
   x - 5 = 0 \quad \text{or} \quad x + 5 = 0
   \]
   \[
   x = 5 \quad \text{or} \quad x = -5
   \]

### Check for extraneous solutions:
Since \(x = 5\) makes the denominator zero in the original equation, it is not a valid solution. Therefore, the only valid solution is:

\[
x = -5
\]

Would you like more details on the factoring process or any other part of this solution?

### 5 Related Questions:
1. How can we solve quadratic equations using the quadratic formula?
2. What is an extraneous solution, and how does it arise in rational equations?
3. How do you check if a solution is valid in rational equations?
4. Can you explain how to factor quadratic expressions?
5. How do different types of factoring methods apply to polynomials?

**Tip:** Always verify your solutions by substituting them back into the original equation to check for restrictions like division by zero.

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

Step-by-step guide on how to solve the rational equation (x^2 + x - 30) / (x - 5) = 1 by factoring quadratics and eliminating extraneous solutions. Suitable for students in Grades 8-10.

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