Question
The difference between two positive numbers $x$ and $y(x>y$ ) is $4$ and the difference between their reciprocals is $\frac{4}{21}$. Find the numbers.

Answer

Two numbers are $x$ and $y$ respectively such that $x>y$.
Then,
$x-y=4\ldots...(i)$
$\Rightarrow x=4+y$
And,
$\frac{1}{y}-\frac{1}{x}=\frac{4}{21}$
$ \Rightarrow \frac{x-y}{x y}=\frac{4}{21}$
$\Rightarrow \frac{4}{x y}=\frac{4}{21} \dots......$[From $(1)]$
$\Rightarrow x y=21$
$ \Rightarrow(4+y) y=21$
$ \Rightarrow 4 y+y 2=21$
$ \Rightarrow y 2+4 y-21=0$
$ \Rightarrow y 2+7 y-3 y-21=0$
$ \Rightarrow y(y+7)-3(y+7)=0$
$ \Rightarrow(y-3)(y+7)=0$
$ \Rightarrow y=3$ and $y=-7$
We reject $y=-7$ since $y$ is positive.
$\Rightarrow y=3$
$ \Rightarrow x=4+y=4+3=7$
Thus, the two numbers are $7$ and $3$ respectively.

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