rational expressions #2
J JeffM Elite Member Joined Sep 14, 2012 Messages 7,872 Jul 24, 2013 #2 mathymath said: Click to expand... Start by simplifying each fraction and don't use X for multiplication in algebra if you want to remain sane. \(\displaystyle \dfrac{x - 2}{x + 5} + \dfrac{x^2 - 2x - 3}{x^2 - x - 6} * \dfrac{x^2 + 2x}{x^2 - 4x} =\) \(\displaystyle \dfrac{x - 2}{x + 5} + \dfrac{(x - 3)(x + 1)}{(x - 3)(x + 2)} * \dfrac{x(x + 2)}{x(x - 4)} =\) \(\displaystyle \dfrac{x - 2}{x + 5} + \dfrac{x + 1}{x - 4} = what?\)
mathymath said: Click to expand... Start by simplifying each fraction and don't use X for multiplication in algebra if you want to remain sane. \(\displaystyle \dfrac{x - 2}{x + 5} + \dfrac{x^2 - 2x - 3}{x^2 - x - 6} * \dfrac{x^2 + 2x}{x^2 - 4x} =\) \(\displaystyle \dfrac{x - 2}{x + 5} + \dfrac{(x - 3)(x + 1)}{(x - 3)(x + 2)} * \dfrac{x(x + 2)}{x(x - 4)} =\) \(\displaystyle \dfrac{x - 2}{x + 5} + \dfrac{x + 1}{x - 4} = what?\)
M mathymath New member Joined Jul 24, 2013 Messages 13 Jul 24, 2013 #3 im still getting it wrong, it would just help if someone showed me the steps i could quickly see where i did wrong
im still getting it wrong, it would just help if someone showed me the steps i could quickly see where i did wrong