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: $\boxed{\dfrac{x-4}{x+2}}$
 A: You did not distribute the term $(x - 3)$ in the denominator when you wrote:
$$\begin{align} & =\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x} \\ \\ & =\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} \end{align}$$
What would be correct is the following denominator:
$$\begin{align} & \quad\color{blue}{(x-3)}[x(x-3)+2+6x] \\ \\ & = \color{blue}{(x-3)}x(x-3)+\color{blue}{(x-3)}(2+6x)\end{align} $$

But note
$$\dfrac{x(x-3)+(-4)}{\color{blue}{\bf (x-3)}}\times \dfrac{\color{blue}{\bf(x-3)}}{x(x-3)+2+6x}$$
The highlighted terms cancel, leaving you:
$$\begin{align} & =\dfrac{x(x-3)+(-4)}{1}\times \dfrac{1}{x(x-3)+2+6x} \\ \\
& = \frac{x(x-3) - 4}{x(x - 3) + 2 + 6x} \\ \\
& = \dfrac{x^2-3x-4}{x^2+3x+2} \tag{$\diamondsuit$}
\end{align} $$
Now, all both the numerator and denominator of $\diamondsuit$ factor very nicely, and in fact, share a common factor, and hence, can be further simplified.

Recall: $$\frac{[b + c]a}{a[d+e]} = \frac{a[b+c]}{a[d+e]} = \frac{b+c}{d+e}$$
Or: $$\frac{a[b+c]}{a[d+e]} = \frac{ab+ac}{ad+ae}= \frac{b+c}{d+e}$$
A: 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.   
