Question

The sum of two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers.

Answer 1

Let one of the number be x then other number is (18 − x).

Then according to question,

$$\begin{gathered} \frac{1}{x}+\frac{1}{18-x}=\frac{1}{4} \\ \Rightarrow \frac{18-x+x}{x(18-x)}=\frac{1}{4} \\ \Rightarrow 18\times 4=18x-x^2 \\ \Rightarrow 72=18x-x^2 \\ \Rightarrow x^2-18x+72=0 \\ \Rightarrow x^2-12x-6x+72=0 \\ \Rightarrow x(x-12)-6(x-12)=0 \\ \Rightarrow (x-6)(x-12)=0 \\ \Rightarrow x-6=0\mathrm{or}x-12=0 \\ \Rightarrow x=6\mathrm{or}x=12 \end{gathered}$$

Since, x being a number,

Therefore,

When $x=12$ then another number will be

$18-x=18-12=6$

And when $x=6$ then another number will be

$18-x=18-6=12$

Thus, the two numbers are 6 and 12.