Answer
$\displaystyle \frac{3x-4}{(x+2)(x-3)}$
Work Step by Step
1. Find the LCD .
1st denominator = $(x+2)(x-1)$
2nd denominator = $(x-3)(x-1)$
List factors of the 1st denominator.
From each next denominator, add only those factors that do not yet appear in the list.
LCD = $(x+2)(x-1)(x-3)$
2. Rewrite each rational expression with the the LCDas the denominator
$\displaystyle \frac{3x(x-3)}{(x+2)(x-1)(x-3)}+\frac{2(x+2)}{(x-3)(x-1)(x+2)}$
3. Add or subtract numerators, placing the resulting expression over the LCD.
$= \displaystyle \frac{3x(x-3)+2(x+2)}{(x+2)(x-1)(x-3)}$
4. If possible, simplify.
$= \displaystyle \frac{3x^{2}-9x+2x+4}{(x+2)(x-1)(x-3)} $
$= \displaystyle \frac{3x^{2}-7x+4}{(x+2)(x-1)(x-3)}$
The numerator is of the form $ax^{2}+bx+c.$
To factor, find factors of $ac$ whose sum is $b.$
If successful, rewrite $bx$ and factor in pairs.
$3x^{2}-7x+4$=$3x^{2}-3x-4x+4$
=$3x(x-1)-4(x-1)=(x-1)(3x-4)$
= $\displaystyle \frac{(x-1)(3x-4)}{(x+2)(x-1)(x-3)}\qquad$... cancel the common factor
= $\displaystyle \frac{3x-4}{(x+2)(x-3)}$
