Question

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

Answer 1

Answer:

No real value of x is possible. so question is incorrect

Step-by-step explanation:

Given,

Sum of two numbers = 9

Sum of their reciprocal = 9 / 20

To find: The numbers

Solution:

Let the one number be x.

Then another number = ( 9 - x)

Then the equation formed will be,

$\dfrac{1}{x} + \dfrac{1}{(9-x)} = \dfrac{9}{20}\\\\\dfrac{(9-x)+x}{x(9-x)} = \dfrac{9}{12}\\\\12 \times 9 = 9 (9x -x^2)\\\\108 = 18x - 9x^2\\\\12 = 2x - x^2\\\\x^2 - 2x + 12 = 0$

No real value of x is possible.

Answer 2

Answer:

The two numbers are 5 and 4

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.