The sum of two numbers is 18.

Solution:

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

Then according to question,

$\frac{1}{x}+\frac{1}{18-x}=\frac{1}{4}$

$\Rightarrow \frac{18-x+x}{x(18-x)}=\frac{1}{4}$

$\Rightarrow 18 \times 4=18 x-x^{2}$

$\Rightarrow 72=18 x-x^{2}$

$\Rightarrow x^{2}-18 x+72=0$

 

$\Rightarrow x^{2}-12 x-6 x+72=0$

$\Rightarrow x(x-12)-6(x-12)=0$

$\Rightarrow(x-6)(x-12)=0$

$\Rightarrow x-6=0$ or $x-12=0$

 

$\Rightarrow x=6$ or $x=12$

Since, x being a number,

Therefore,

When x=12">x=12x=12 then another number will be

$18-x=18-12=6$

And when x=6">x=6x=6 then another number will be

$18-x=18-6=12$

Thus, the two numbers are 6 and 12.