So I was once again bored when I decided to to try and see if I could do maybe some simple Algebra and came up with this problem:$$\text{Simplify: }\dfrac{x+2}{x^2-2x-3}+\dfrac{x}{x^2-2x}$$Which I thought that I would be able to do. Here is my attempt at simplifying the aforementioned equation:$$\dfrac{x+2}{x^2-2x-3}+\dfrac{x}{x^2-2x}$$$$=\dfrac{x+2}{x^2-2x-3}+\dfrac{1}{x-2}$$$$=\dfrac{x+2}{(x+1)(x-3)}+\dfrac{1}{x-2}$$$$=\dfrac{x^2-4}{(x+1)(x-3)(x-2)}+\dfrac{(x+1)(x-3)}{(x+1)(x-3)(x-2)}$$$$=\dfrac{2x^2-2x-7}{x^3-4x^2+4x+6}$$$$\text{Or }\dfrac{(x-2)(x+2)+(x+1)(x-3)}{(x+1)(x-2)(x-3)}$$
$$\mathbf{\text{My question}}$$
Is the solution that I have achieved correct, or what could I do to attain the correct simplification/attain it more easily?
Your derivation is fine, but there is an important issue to note: while the original expression is not defined for $x=0$ the second one is a valid expression also when $x=0$, in this sense they are not completely equivalent.