Question

The sum of two natural numbers is 9 and the sum of their reciprocal is $\frac{1}{2}$. Find the numbers.

Answer 1

Let one number be $x$

and other number be $9-x$

sum of their reciprocal $=\frac{1}{2}$

$\frac{1}{x}+\frac{1}{9-x}=\frac{1}{2}$

$\Rightarrow \frac{9-x+x}{x(9-x)}=\frac{1}{2}$

$\Rightarrow \frac{9}{x(9-x)}=\frac{1}{2}$

$\Rightarrow x(9-x)=18$

$\Rightarrow x^2-9x+18=0$

$\Rightarrow x^2-6x-3x+18=0$

$\Rightarrow x(x-6)-3(x-6)=0$

$\Rightarrow (x-6)(x-3)=0$
$\therefore x=6,3$

Hence, the numbers are 6 and 3.