Determine all distinct positive integers $x$ and $y$ such that

$$
\frac{1}{x}+\frac{1}{y}=\frac{2}{7}
$$
The original equation assumes $x \neq 0, y \neq 0$ and asks for positive, distinct integer solutions, so we can consider the equivalent equation

$$
2 x y=7(x+y), \quad \text { with } \quad x>0, y>0 \text { and } x \neq y
$$

Since 2 and 7 are prime numbers, it follows that 7 divides $x$ or $y$. Since the equation is symmetric in $x$ and $y$, we can assume that 7 divides $x$. Then, $x=7 k$, for some integer $k>0$, and it follows that $2 \times 7 k y=7(7 k+y)$, or, simplifying, $2 k y=7 k+y$ or, still,

$$
(2 k-1) y=7 k=x
$$

If 7 divided $y$, we would have $y=7 m$, for some integer $m>0$. In this case, we would have $49 \times 2 k m=2 x y=7(x+y)=49(k+m)$, leading to $2 k m=k+m$. But then

$$
2=\frac{k+m}{k m}=\frac{1}{k}+\frac{1}{m} \leq 1+1=2
$$

which means that $k=m=1$ and, therefore, $x=7=y$. Since we want $x \neq y$, we conclude that 7 does not divide $y$, so 7 divides $2 k-1$. Taking $k=4$, we get $x=28$ and

$$
y=\frac{7 k}{2 k-1}=\frac{28}{7}=4
$$

yielding the solution

$$
\frac{1}{28}+\frac{1}{4}=\frac{2}{7}
$$

Note: The solution obtained is unique. Indeed, since $2 k-1$ is always odd and 7 divides $2 k-1$, the multiple of 7 that equals $2 k-1$ must be odd. Thus, there is some integer $n>0$ such that

$$
7(2 n-1)=2 k-1
$$

This implies that $k=7 n-3$ and, therefore,

$$
y=\frac{7 k}{2 k-1}=\frac{7(7 n-3)}{7(2 n-1)}=\frac{7 n-3}{2 n-1}=\frac{3(2 n-1)+n}{2 n-1}=3+\frac{n}{2 n-1}
$$

Since $y$ must be an integer, we conclude that $2 n-1$ divides $n$, so $2 n-1 \leq n$. However, $n \geq 1$ and, therefore, $2 n-1 \geq n$. The only possibility is $2 n-1=n$ and, therefore, $n=1$. It follows that $k=4$ and $a=28$ give the unique solution.