Answer
$\frac{4x}{(3x+4)(3x-4)}$
Work Step by Step
The given expression is
$=\frac{3x^2}{9x^2-16}-\frac{x}{3x+4}$
Factor $9x^2-16$.
$=(3x)^2-4^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(3x+4)(3x-4)$
$=\frac{3x^2}{(3x+4)(3x-4)}-\frac{x}{3x+4}$
LCM $=$ greatest power of all the prime factors.
LCM $=(3x+4)(3x-4)$
$=\frac{3x^2}{(3x+4)(3x-4)}-\frac{(3x-4)}{(3x-4)}\frac{x}{3x+4}$
Simplify.
$=\frac{3x^2}{(3x+4)(3x-4)}-\frac{x(3x-4)}{(3x+4)(3x-4)}$
Add both numerators because both denominators are equal.
$=\frac{3x^2-x(3x-4)}{(3x+4)(3x-4)}$
Simplify.
$=\frac{3x^2-3x^2+4x}{(3x+4)(3x-4)}$
$=\frac{4x}{(3x+4)(3x-4)}$.
