Two positive numbers differ by 4, and the difference of their reciprocals is $\frac{1}{24}$. What is the sum of the two numbers?

Explanation

1. 01

   Let the numbers be $y$ and $y + 4$: $\frac{1}{y} - \frac{1}{y + 4} = \frac{4}{y(y + 4)} = \frac{1}{24}$.

2. 02

   So $y(y + 4) = 96$, i.e. $y^2 + 4y - 96 = 0$.

3. 03

   $(y + 12)(y - 8) = 0$; positive root $y = 8$.

4. 04

   The numbers are 8 and 12, so the sum is 20.