Question

The sum of two natural numbers is 9 and the sum of their reciprocals is $\frac{1}{2}$. Find the numbers. [CBSE 2012]

Answer 1

Let the required natural numbers be x and (9 − x).

According to the given condition,
$$\begin{gathered} \frac{1}{x}+\frac{1}{9-x}=\frac{1}{2} \\ \Rightarrow \frac{9-x+x}{x\left(9-x\right)}=\frac{1}{2} \\ \Rightarrow \frac{9}{9x-x^2}=\frac{1}{2} \\ \Rightarrow 9x-x^2=18 \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2-9x+18=0 \\ \Rightarrow x^2-6x-3x+18=0 \\ \Rightarrow x\left(x-6\right)-3\left(x-6\right)=0 \\ \Rightarrow \left(x-3\right)\left(x-6\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x-3=0\mathrm{or}x-6=0 \\ \Rightarrow x=3\mathrm{or}x=6 \end{gathered}$$

When x = 3,
9 − x = 9 − 3 = 6

When x = 6,
9 − x = 9 − 6 = 3

Hence, the required natural numbers are 3 and 6.