## Algebra 1

$\frac{2x^2 + 17x - 45}{(x-5)(x+4)}$
To solve: $\frac{2x}{x-5} + \frac{9}{x+4}$, you will first need to make sure both fractions have common denominators. To do this, we must find the LCD, or the Least Common Multiple. Examining both denominators, you can see that they have no common factors between them, therefore the LCD will be $(x-5)(x+4)$. Now we can rewrite the fractions using the new LCD we found. $\frac{2x}{x-5} + \frac{9}{x+4}$ = $\frac{2x \times (x+4)}{(x-5) \times (x+4)} + \frac{9 \times (x-5)}{(x+4) \times (x-5)}$ =$\frac{2x^2 - 8x}{(x-5)(x+4)} + \frac{9x - 45}{(x-5)(x+4)}$ Now that we have common denominators, we can add the numerators: $\frac{2x^2 + 8x}{(x-5)(x+4)} + \frac{9x - 45}{(x-5)(x+4)}$ = $\frac{(2x^2 + 8x) + (9x - 45)}{(x-5)(x+4)}$ = $\frac{2x^2 + 8x + 9x - 45}{(x-5)(x+4)}$ = $\frac{2x^2 + 17x - 45}{(x-5)(x+4)}$ The numerator cannot be factored, therefore, this is our final answer.