Pre-calculus $\frac 27 = \frac 1a + \frac 1b$ I had this questions from a previous exam that I couldn't answer, I am apologizing for any English mistakes or for any stupid questions, I tried to solve them and I searched the internet and I couldn't find answers. 
4-if $\frac 27$ could be written in a unique way in the form of $$\frac 1a +\frac 1b$$ $$a,b \in \mathbb Z^+$$ $$a\neq b$$What is a+b ?
*It seemed so easy maybe I am missing something isn't there infinite values that satisfy this equation?
$7(a+b) = 2ab$
Thanks for taking the time to read the question !

To write ${2\over7}={1\over a}+{1\over b}$ with, say $a\lt b$, it's clear that $1\le a\le6$, since otherwise we would have ${1\over a}+{1\over b}\le{1\over7}+{1\over8}\lt{2\over7}$.  It's also easy to see that we must have $a\ge4$, since ${1\over3}\gt{2\over7}$.  So this leaves just three possibilities to check: $a=4$, $5$, and $6$.  It's easy to see that only $a=4$ works:  ${2\over7}-{1\over4}={1\over28}$, so $a+b=4+28=32$.
