Algebraic expression in its most simplified form I am trying to simplify the algebraic expression:
$$\bigg(x-\dfrac{4}{(x-3)}\bigg)\div \bigg(x+\dfrac{2+6x}{(x-3)}\bigg)$$
I am having trouble though. My current thoughts are:
$$=\bigg(\dfrac{x}{1}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x}{1}+\dfrac{2+6x}{(x-3)}\bigg)$$
$$=\bigg(\dfrac{x(x-3)}{1(x-3)}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)}{1(x-3)}+\dfrac{2+6x}{(x-3)}\bigg)$$
$$=\bigg(\dfrac{x(x-3)+(-4)}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)+2+6x}{(x-3)}\bigg)$$
$$=\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x}$$
$$=\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} $$
$$\boxed{=\dfrac{-4(x-3)}{2(1+3x)} }$$ Which does not appear is not the answer. Am I close? Where exactly did I go wrong? I have tried this question multiple times.
Edit: Figured it out!
$\dfrac{x(x-3) - 4}{x(x - 3) + 2(1 + 3x)}\implies\dfrac{x^2-3x-4}{x^2+3x+2}\implies \dfrac{(x-4)(x+1)}{(x+2)(x+1)}$ 
$(x+1)$'s cancel leaving us with: $\boxed{\dfrac{x-4}{x+2}}$

Comment: After the calculation very well described by AWertheim and amWhy, we end up wit $\dfrac{x^2-3x-4}{x^2+3x+2}$.
Note that the calculation is not over. The top and bottom each factor nicely, and there will be some cancellation.   
