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 !
 A: We have to solve $$2ab=7a+7b$$ in the positive integers.
We have $$(2a-7)(2b-7)=4ab-14a-14b+49=2(2ab-7a-7b)+49$$
Because of $2ab=7a+7b$ we get $$(2a-7)(2b-7)=49$$
If $a\ne b$, we can assume $a<b$ WLOG
We have $2a-7=1$ and $2b-7=49$ giving $a=4$ and $b=28$
So, we have $$\frac{2}{7}=\frac{1}{4}+\frac{1}{28}$$ giving $a+b=4+28=32$
A: The given problem boils down to finding lattice points on a hyperbola. The equation
$$ 2ab=7(a+b) \tag{1}$$
is equivalent to
$$ (2a-7)(2b-7) = 49\tag{2}$$
and $49=7^2$ cannot be written as a product of integers in too many ways. From the assumption $(2a-7)=1$ and $(2b-7)=49$ we get the non-trivial solution $\color{red}{(a,b)=(4,28)}.$
A: 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 $a+b=4+28=32$.
A: Express the equation in terms of the desired unknown, $s:=a+b$.
$$\frac1a+\frac1{s-a}=\frac27$$ yields the quadratic equation
$$2a^2-2sa+7s=0$$ with solutions
$$a=\frac{s\pm\sqrt{s^2-14s}}2=\frac{s\pm\sqrt{(s-7)^2-49}}2.$$
Now $s-7$ and $7$ are members of a Pythagorean triple. The only factorization of $7$ is $7\cdot1$, so that the triple must be $7=(4+3)(4-3),24=2\cdot3\cdot4,25=4^2+3^2$.
From this, $s=25+7=\color{green}{32}$ is the only admissible solution (and $a=4$ or $a=28$).

Note that the degenerate Pythagorean triple $49=(7+0)(7-0),0=2\cdot7\cdot0,49=7^2+0^2$ is also possible, giving $s=14$, and $a=b=7$.
A: If $a=b$ we have $a=b=7$.
Let $a>b$.
Thus, $\frac{2}{7}=\frac{1}{a}+\frac{1}{b}<\frac{2}{b}$, which gives $b<7$.
In another hand, since $\frac{1}{b}<\frac{2}{7}$ we get $b>3.5$.
Id est, $4\leq b\leq6$.
Now, for $b=4$ we get $a=28$ and $b=5$ and $b=6$ they impossible.
Thus, we have three solutions $(7,7)$, $(4,28)$ and $(28,4)$, which gives the answer: $\{14,32\}$.
