The difference of two natural numbers is 3 and the difference of their reciprocal is $\frac{3}{28}$. Find the numbers.

Given:

The difference of two natural numbers is 3 and the difference of their reciprocal is $\frac{3}{28}$.

 To do:

We have to find the numbers.

Solution:

Let the two natural numbers be $x$ and $x+3$.

According to the question,

$\frac{1}{x}-\frac{1}{x+3}=\frac{3}{28}$

$\frac{1(x+3)-1(x)}{x(x+3)}=\frac{3}{28}$

$\frac{x+3-x}{x^2+3x}=\frac{3}{28}$

$\frac{3}{x^2+3x}=\frac{3}{28}$

$3(28)=3(x^2+3x)$   (On cross multiplication)

$28=x^2+3x$

$x^2+3x-28=0$

Solving for $x$ by factorization method, we get,

$x^2+7x-4x-28=0$

$x(x+7)-4(x+7)=0$

$(x+7)(x-4)=0$

$x+7=0$ or $x-4=0$

$x=-7$ or $x=4$

We need only natural numbers. Therefore, the value of $x$ is $4$.

$x+3=4+3=7$

The required natural numbers are $4$ and $7$. 

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Updated on: 10-Oct-2022

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