Find all $(x,y)$ such that $1/x+1/y=1/7$. While working in a workbook, I got stuck on the following problem:
Find all the ordered pairs $(x,y)$ such that they satisfy the equation $1/x + 1/y = 1/7$.
I made a bit of progress, but got stuck. 
So far, I have multiplied the entire equation by $7xy$ to get $7y+7x=xy$, then subtracted $7y$ and $7x$ from both sides to get $xy-7x-7y=0$.
From here, I'm not sure how to progress, maybe with SFFT?
Could someone guide me in what to do next?
Thanks!
 A: Assuming that $x$ and $y$ are supposed to be integers...  Abiessu's solution in the comments is very good and a great approach.  This answer provides a more elementary solution.
There are two cases to consider, either $x$ and $y$ are both positive, or one is positive and the other is negative.  Let's deal with both cases separately.


*

*If $x$ and $y$ are both positive, then let's assume that $x\leq y$ (the other case is symmetric).  Therefore, $\frac{1}{x}\geq\frac{1}{y}$.  Since $\frac{1}{x}+\frac{1}{y}=\frac{1}{7}$, it must be that $\frac{1}{7}>\frac{1}{x}\geq\frac{1}{14}$.  If $\frac{1}{x}$ were larger than $\frac{1}{7}$, then $\frac{1}{y}$ would have to be negative.  If $\frac{1}{x}$ were smaller than $\frac{1}{14}$, then $\frac{1}{y}$ would be larger.  This gives $7$ possible values for $x$, $x=8,\cdots,14$.  You can check each one for a possible $y$-value.

*If one of $x$ and $y$ is negative, assume that $y$ is negative and $x$ is positive (the other case is symmetric).  Then, $1\geq\frac{1}{x}>\frac{1}{7}$.  The LHS inequality occurs because $x$ is positive, while the RHS inequality occurs because $\frac{1}{x}$ must be larger than $\frac{1}{7}$ so that we can subtract $\frac{1}{|y|}$ from it.  This leads to $6$ cases for $x$, $x=1,\cdots,6$.  You can check each one for a possible $y$-value.
A: A simple way is to take  $7y+7x=xy$, subtract $7y$ from both sides, and then factor out the $y$ on the right, then divide both sides by $x-7$ to obtain the equation $y=7x/(x-7)$. $x=7$ isn't a solution to the original equation so you don't have to worry about the domain restriction. 
If you are indeed looking for the integer values, (-42,6) is the lowest value of x that can give an integer value for y and (56,8) is the highest. Checking the others finds (0,0), (6,-42), (8,56), and (14,14), but checking in the original equation shows that (0,0) is not a solution. This happened because multiplying by 7xy is zero when x or y are zero. 
