The problem asks to write the following algebraic expression as a single fraction:

$\frac{x^2 + 7x - 18}{2x^2 - x - 6}$

Step 1: Factor the numerator and denominator.

Numerator: $x^2 + 7x - 18$

We need to factor this quadratic. We are looking for two numbers that multiply to $-18$ and add up to $7$.

The factors are $9$ and $-2$, so: $x^2 + 7x - 18 = (x + 9)(x - 2)$

Denominator: $2x^2 - x - 6$

We factor this quadratic expression. We are looking for two numbers that multiply to $2 \times -6 = -12$ and add to $-1$ (the coefficient of $x$).

The factors are $3$ and $-4$. We split the middle term: $2x^2 - x - 6 = 2x^2 - 4x + 3x - 6$ Now factor by grouping: $= 2x(x - 2) + 3(x - 2)$ Factor out $(x - 2)$: $= (x - 2)(2x + 3)$

Step 2: Write the expression as a single fraction.

Now that both numerator and denominator are factored, substitute these factored forms back into the original expression:

$\frac{(x + 9)(x - 2)}{(x - 2)(2x + 3)}$

Step 3: Simplify the fraction.

Cancel out the common factor $(x - 2)$ from both the numerator and denominator:

$= \frac{x + 9}{2x + 3}$

Thus, the simplified single fraction is:

$\boxed{\frac{x + 9}{2x + 3}}$

Would you like more details on any step or have any questions? Here are a few related questions to further explore this topic:

1. How do you factor quadratics in general?
2. What if the common factor in the numerator and denominator is a more complex expression?
3. Can you explain why it’s important to check for excluded values (like $x = 2$ in this case)?
4. How do you apply the distributive property to check factored forms?
5. What other methods exist for simplifying complex rational expressions?

Tip: Always verify factored forms by expanding them back to the original expression to ensure accuracy.