33
The sum of two numbers is $18$ and the sum of their reciprocals is $\frac{1}{4}$. Find the numbers.
Show SolutionHide Solution↓
Let two numbers be $x$ and $$\begin{aligned}& (18 - x) \\ & A.T.Q. \\ & \frac{1}{x} + \frac{1}{18-x} = \frac{1}{4} \\ & \Rightarrow x^2 - 18x + 72 = 0 \\ & \Rightarrow (x - 12)(x - 6) = 0 \\ & \Rightarrow x = 12, x = 6 \\ & \therefore\end{aligned}$$ two numbers are $12$ and $6$.