#### Question

Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 7/10.

#### Solution

Let the numbers be x and x + 3

From the given information

`1/x + 1/(x + 3) = 7/10`

`(x + 3 + x)/(x(x + 3)) = 7/10`

`(2x + 30)/(x^2 + 3x) = 7/10`

`20x + 30 = 7x^2 + 21x`

`7x^2 + x - 30 = 0`

`7x^2 - 14x + 15x - 30 = 0`

7x(x - 2) + 15(x - 2) = 0

(x - 2)(7x + 15) = 0

`x = 2, (-15)/7`

Since x is a natural number so x= 2

Thus the number are 2 and 5

Is there an error in this question or solution?

Advertisement

Advertisement

Two Natural Numbers Differ by 3. Find the Numbers, If the Sum of Their Reciprocals is 7/10. Concept: Quadratic Equations.

Advertisement