Answer
$\displaystyle \frac{7x+1}{(x-2)(x+3)}$
Work Step by Step
Step 1: Find the LCD.
List of factors of the first denominator: $\qquad (x-2)$
List of factors of the second denominator: $\qquad (x+3)$
Build the LCD:
- write all factors of the 1st denominator:$\qquad $
List$= (x-2),...\quad$ (for now)
- add to the list factors of the second denominator that are not already on the list
($(x+3)$ is added to the list)
List = $(x -2),(x+3), $
$LCD=(x-2)(x+3)$
Step 2. Rewrite each expression with the LCD:
$=\displaystyle \frac{3}{x-2}\cdot\frac{(x+3)}{(x+3)}+\frac{4}{(x+3)}\cdot\frac{(x-2)}{(x-2)}= \displaystyle \frac{3(x+3)}{(x-2)(x+3)}+\frac{4(x-2)}{(x-2)(x+3)}=...$
Step 3. Combine numerators over the LCD
$=$ $\displaystyle \frac{3(x+3)+4(x-2)}{(x-2)(x+3)}$
Step 4. Simplify, if possible.
$= \displaystyle \frac{3x+9+4x-8}{(x-2)(x+3)} $
= $\displaystyle \frac{7x+1}{(x-2)(x+3)}$
