The difference of two number is 4. If the difference of their reciprocals is $\frac{4}{21}$, find the numbers.

Given: 

The difference of two number is 4. The difference of their reciprocals is $\frac{4}{21}$.

To do: 

We have to find the numbers.

Solution: 

Let the two numbers be $x$ and $x-4$.

According to the question,

Difference of the reciprocals $=\frac{4}{21}$.

This implies,

$\frac{1}{x-4} -\frac{1}{x} =\frac{4}{21}$

$\frac{1(x)-1(x-4)}{(x-4)x} =\frac{4}{21}$

$\frac{x-x+4}{x^2-4x} =\frac{4}{21}$

$\frac{4}{x^2-4x}=\frac{4}{21}$

$4(21)=4(x^2-4x)$    (By cross multiplication)

$21=x^2-4x$

$x^2-4x-21=0$

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

$x^2-7x+3x-21=0$

$x(x-7)+3(x-7)=0$

$(x-7)(x+3)=0$

$x-7=0$ or $x+3=0$

$x=7$ or $x=-3$

If $x=7$, $x-4=7-4=3$

If $x=-3$, $x-4=-3-4=-7$

The required numbers are $3$ and $7$ or $-3$ and $-7$.

Updated on: 10-Oct-2022

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