I've come with the following equation, after a lot of simplification, but can't reduce further. Any chances it can be solved by reducing the $b$ and get the value of $a$?
$$a = \frac{1000(1000 - 2b)}{2(1000-b)}$$
Maybe by getting something like $1000 - b$ simplified?
Thanks in advance.
Sometimes "simple" is all in the eye of the beholder. With a quantity like yours, some might prefer to keep it as is. There are a number of things you could do to that quantity but they aren't necessarily better. For example, $$\begin{align}\frac{1000(1000 - 2b)}{2(1000-b)} &= \frac{500(1000-b-b)}{1000-b} \\ &= \frac{500(1000-b)-500b}{1000-b} \\ &= \frac{500(1000-b)}{1000-b} - \frac{500b}{1000-b}\\ &= 500 - \frac{500b}{1000-b}\\ &= 500\left[ 1- \frac{b}{1000-b}\right] \end{align}$$ Not sure if this is what you are looking for, but this equation should be consistent with yours for $b \neq 1000$. Another thing you could do at this point (if $b \neq 0$) and you want to eliminate occurences of $b$ is to multiply through by $\frac{b^{-1}}{b^{-1}}$ to get $$500\left[1 - \frac{1}{1000b^{-1}-1}\right]$$