1. ## LCD help

Could someone please explain to me how they get this LCD?

Thanks

2. Hello, silencecloak!

That is the dumbest way to write the problem ... and the answer.

Could someone please explain to me how they get this LCD?

$-\frac{2^2}{(-x+3)(x+3)} - \frac{2x+1}{x(x-3)} \;=\;\frac{x(-2^2) + [-(x+3)][-(2x+1)]}{(-x+3)(x+3)x}$

I'd write it as: . $-\frac{4}{-(x-3)(x+3)} - \frac{2x+1}{x(x-3)} \;=\; \frac{4}{(x-3)(x+3)} - \frac{2x+1}{x(x+3)}$

The LCD is $x(x-3)(x+3)$: . ${\color{blue}\frac{x}{x}}\cdot\frac{4}{(x-3)(x+3)} \:- \:{\color{blue}\frac{x+3}{x+3}}\cdot\frac{2x+1}{x( x-3)}$

. . $= \;\;\frac{4x}{x(x-3)(x+3)} - \frac{(x+3)(2x+1)}{x(x-3)(x+3)} \;\;=\;\;\frac{4x - (x+3)(2x+1)}{x(x-3)(x+3)}\;\;\hdots\;\;\text{etc.}$