$50$ is divided into two parts such that the sum of their reciprocals is $\dfrac{1}{12}$. Find the two parts.
Let $50$ be divided into $x$ and $50 - x$
Reciprocal of the parts: $\;$ $\dfrac{1}{x}$ $\;$ and $\;$ $\dfrac{1}{50 - x}$
Given: Sum of reciprocal of parts $= \dfrac{1}{12}$
i.e. $\;$ $\dfrac{1}{x} + \dfrac{1}{50 - x} = \dfrac{1}{12}$
i.e. $\;$ $\dfrac{50 - x + x}{x \left(50 - x\right)} = \dfrac{1}{12}$
i.e. $\;$ $600 = 50 x - x^2$
i.e. $\;$ $x^2 - 50 x + 600 = 0$
i.e. $\;$ $x^2 - 30x - 20x + 600 = 0$
i.e. $\;$ $x \left(x - 30\right) - 20 \left(x - 30\right) = 0$
i.e. $\;$ $\left(x - 20\right) \left(x - 30\right) = 0$
$\implies$ $x = 20$ $\;$ or $\;$ $x = 30$
$\therefore \;$ The two parts are $\;$ $20$ $\;$ and $\;$ $30$