It seems you're asking to simplify or solve the expression:

\[
\frac{2x^2 + x - 3}{x^2 - 6x + 5}
\]

### Step 1: Factor the numerator and denominator
#### Numerator: \( 2x^2 + x - 3 \)
We need to factor this quadratic expression.

Looking for two numbers that multiply to \( 2 \times (-3) = -6 \) and add to \( 1 \), the middle term:
- The pair \( 3 \) and \( -2 \) work, because \( 3 \times -2 = -6 \) and \( 3 + (-2) = 1 \).

So we can rewrite the numerator:
\[
2x^2 + x - 3 = 2x^2 - 2x + 3x - 3 = 2x(x - 1) + 3(x - 1)
\]
Now factor out the common term \( (x - 1) \):
\[
2x^2 + x - 3 = (2x + 3)(x - 1)
\]

#### Denominator: \( x^2 - 6x + 5 \)
We factor the denominator:
\[
x^2 - 6x + 5 = (x - 5)(x - 1)
\]

### Step 2: Simplify the expression
Now that both the numerator and denominator are factored, we have:
\[
\frac{(2x + 3)(x - 1)}{(x - 5)(x - 1)}
\]

The \( (x - 1) \) terms cancel out (as long as \( x \neq 1 \)), leaving:
\[
\frac{2x + 3}{x - 5}
\]

### Final Simplified Expression:
\[
\frac{2x + 3}{x - 5}, \quad \text{for} \quad x \neq 1 \quad \text{and} \quad x \neq 5
\]

Would you like more details on any of these steps?

Here are five related questions to explore:
1. How do you factor quadratics?
2. How do you identify terms that cancel in rational expressions?
3. What happens if \(x = 1\) or \(x = 5\) in the original expression?
4. How can you apply the quadratic formula if factoring is difficult?
5. What are restrictions in rational expressions?

**Tip:** Always check for common factors before simplifying a fraction.

Learn how to simplify the rational expression (2x² + x - 3) ÷ (x² - 6x + 5) with a clear, step-by-step solution. This guide focuses on factoring quadratics and simplifying rational expressions, perfect for students in Grades 8-10.

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