Question

the sum of two natural number is 9 and sum of their reciprocal is 9 by 20 find the numbers​

Answer 1

Answer:

Step-by-step explanation:

Let the numbers be x and y

Given,  

(i) Sum of the two natural numbers is 9

x + y = 9 …(1)

(ii) Sum of their reciprocals is 9/20

1/x + 1/y = 9/20

(x + y)/xy = 9/20 … (2)

Using (1) in (2)

9/xy = 9/20

xy = 20 …(3)

Using (1) in (3)

x(9 - x) = 20

x^2  -9x + 20 = 0

(x - 5)(x - 4) = 0  

x = 5 or x = 4 …(4)

Using (4) and (1),  

If x = 5, then y = 4.

If x = 4, then y = 5.

The two natural numbers are 4 and 5.

Answer 2

Answer:

Step-by-step explanation:

Let the numbers are x and 9-x

Then given that:

$\frac1x+\frac{1}{9-x} =\frac{9}{20} \\\\20(9-x)+20x=9x(9-x)\\\\180-20x+20x=81x-9x^2\\\\9x^2-81x+180=0\\\\x^2-9x+20=0\\\\x^2-5x-4x+20=0\\\\x(x-5)-4(x-5)=0\\\\(x-5)(x-4)=0\\\\\bf x=5\ or \ 4\\\\Thus \ the \ numbers \ are : \ 4 \ and \ 5$