Two natural numbers differ by 5 and their reciprocals differ by 5/14. The larger number is:

Let the numbers be $x$ and $x-5$ with $x$ > 5. Then $1/(x-5)-1/x=5/(x(x-5))=5/14$, so $x^2-5x=14$, i.e. $x^2-5x-14=0$ and $(x-7)(x+2)=0$. The natural-number solution is $x=7$. (2 is the smaller number of the pair; 9 and 4 also differ by 5 but their reciprocals differ by 5/36.) Check: 1/2 − 1/7 = 5/14.