@MarkFL @harpazo this seems like this might go on forever.

\(\displaystyle \dfrac{3}{5b} - \dfrac{2}{4b} + \dfrac{1}{7c} + \dfrac{7}{21c}\)

\(\displaystyle \dfrac{-2}{4b} + \dfrac{1}{7c}\)

\(\displaystyle \dfrac{(-2)(7c) + (4b)(1)}{(4b)(7c)} = \dfrac{(-2)(7c)}{(4b)(7c)} + \dfrac{(4b)(1)}{(4b)(7c)}\)

Cancel out stuff.

\(\displaystyle \dfrac{-2}{4b} + \dfrac{1}{7c}\)

\(\displaystyle \dfrac{3}{5b} -\dfrac{2}{4b} + \dfrac{1}{7c}+ \dfrac{7}{21c}\)