Question: User: $\frac{x+2}{x-3}$ + $\frac{2x-1}{x+4}$
1. **State the problem:** Simplify the expression $$\frac{x+2}{x-3} + \frac{2x-1}{x+4}$$.
2. **Find a common denominator:** The denominators are $x-3$ and $x+4$. The common denominator is their product: $$(x-3)(x+4)$$.
3. **Rewrite each fraction with the common denominator:**
$$\frac{x+2}{x-3} = \frac{(x+2)(x+4)}{(x-3)(x+4)}$$
$$\frac{2x-1}{x+4} = \frac{(2x-1)(x-3)}{(x+4)(x-3)}$$
4. **Add the numerators:**
$$\frac{(x+2)(x+4) + (2x-1)(x-3)}{(x-3)(x+4)}$$
5. **Expand the numerators:**
$$(x+2)(x+4) = x^2 + 4x + 2x + 8 = x^2 + 6x + 8$$
$$(2x-1)(x-3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$$
6. **Sum the expanded numerators:**
$$x^2 + 6x + 8 + 2x^2 - 7x + 3 = (x^2 + 2x^2) + (6x - 7x) + (8 + 3) = 3x^2 - x + 11$$
7. **Write the combined fraction:**
$$\frac{3x^2 - x + 11}{(x-3)(x+4)}$$
8. **Check for factorization:** The numerator $3x^2 - x + 11$ does not factor nicely over the integers, so this is the simplified form.
**Final answer:**
$$\frac{3x^2 - x + 11}{(x-3)(x+4)}$$