TheJason Administrator Staff member Administrator Moderator Jan 25, 2018 2,507 545 113 jasonyankee.info Jul 21, 2018 #2 \(\displaystyle \dfrac{1}{a} + \dfrac{1}{b}\) \(\displaystyle \dfrac{(1)(b) + (a)(1)}{(a)(b)} = \dfrac{b + a}{ab}\)

\(\displaystyle \dfrac{1}{a} + \dfrac{1}{b}\) \(\displaystyle \dfrac{(1)(b) + (a)(1)}{(a)(b)} = \dfrac{b + a}{ab}\)

harpazo Pure Mathematics Banned Mar 20, 2018 5,788 361 83 NYC Jul 22, 2018 #3 TheJason said: \(\displaystyle \dfrac{1}{a} + \dfrac{1}{b}\) \(\displaystyle \dfrac{(1)(b) + (a)(1)}{(a)(b)} = \dfrac{b + a}{ab}\) Click to expand... The answer can also be expressed as (a + b)/(ab).

TheJason said: \(\displaystyle \dfrac{1}{a} + \dfrac{1}{b}\) \(\displaystyle \dfrac{(1)(b) + (a)(1)}{(a)(b)} = \dfrac{b + a}{ab}\) Click to expand... The answer can also be expressed as (a + b)/(ab).