Question

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

Answer 1

Let the first number is $x.$

Then second number$=9-x.$

Their reciprocals=$\frac{1}{x},\frac{1}{9-x}$

Then according to the question

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

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

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

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

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

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

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

$\Rightarrow x-3=0,x-6=0$

$\Rightarrow x=3,6$

First number$=3$

$\therefore$ Second number $=9-3=6.$