Answer
$\displaystyle \frac{7x^{2}-4x+12}{(x+4)(x-2)(x-1)}$
Work Step by Step
1. Find the LCD .
1st denominator = $(x+4)(x-2)$
2nd denominator = $(x-2)(x-1)$
List factors of the 1st denominator.
From each next denominator, add only those factors that do not yet appear in the list.
LCD = $(x+4)(x-2)(x-1)$
$ \displaystyle \frac{7x}{(x+4)(x-2)}+\frac{3}{(x-2)(x-1)}=$
2. Rewrite each rational expression with the the LCDas the denominator
= $\displaystyle \frac{7x(x-1)}{(x+4)(x-2)(x-1)}+\frac{3(x+4)}{(x-2)(x-1)(x+4)}$
3. Add or subtract numerators, placing the resulting expression over the LCD.
= $\displaystyle \frac{7x(x-1)+3(x+4)}{(x+4)(x-2)(x-1)}$
4. If possible, simplify.
= $\displaystyle \frac{7x^{2}-7x+3x+12}{(x+4)(x-2)(x-1)} $
= $\displaystyle \frac{7x^{2}-4x+12}{(x+4)(x-2)(x-1)}$
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.
... we can't find factors of $84$ whose sum is $-4$
... we can't factor the numerator, so we leave it as it is.
