Question

The sum of 2 natural numbers is 9 and the sum of their reciptocal is 1\2. Find the numbers

Answer 1

Hey !!!

here is ur ans

Let the Two natural no. be ..'x' and '9 - x'

Whose sum of reciprocal is 1/2 ....


A/Q
$\frac{1}{x} + \frac{1}{9 - x} = \frac{1}{2} \\ \\ \frac{9 - x + x}{x(9 - x)} = \frac{1}{2} \\ \\ 9 \times 2 = 9x - {x}^{2} \\ \\ {x}^{2} - 9x + 18 = 0 \\ \\ {x}^{2} - 6x - 3x + 18 = 0 \\ \\ x(x - 6) - 3(x - 6) = 0 \\ \\ (x - 3)(x - 6) = 0 \\ \\ x - 3 = 0......or...x - 6 = 0 \\ \\ x = 3....or...x = 6 \\ \\ if \: the \: first \: natural \: is \: 3 \: then \: other \: be \: 9 - 3 = 6 \\ \\ and \: if \: first \: 6 \: then \: other \: be \: 9 - 6 = 3$


So The Two natural no. be 3 , 6 ..and 6, 3


=====================

I HOPE IT WILL HELP YOU

Thank you

☺️

Answer 2

let X and Y be two natural numbers
x+Y= 9...1
1/x+1/Y=2....2
i.e. x*Y/ X+Y = 2
x*Y/9=2
x*Y=18
therefore Y= 18/x....3
x+ 18/x= 9
x^2 +18= 9x
x^2-9x+18=0
on solving qudratic equations we get
X^2-6x-3x+18=0
x(x-6)-3(x-6)=0
(x-6)(x-3)=0
therefore x=6orx=3
if x=3 Y=18/3=6 (from 3)
so two natural numbers are 3and 6