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

Given,

⇒ x > y

$\Rightarrow \dfrac{1}{x} \lt \dfrac{1}{y}$

Difference between two positive numbers x and y (x > y) is 4.

∴ x - y = 4

⇒ x = y + 4 ……..(1)

Difference between their reciprocal is $\dfrac{4}{21}$.

$\Rightarrow \dfrac{1}{y} - \dfrac{1}{x} = \dfrac{4}{21} \\[1em] \Rightarrow \dfrac{x - y}{xy} = \dfrac{4}{21}\\[1em] \Rightarrow 21(x - y) = 4xy \\[1em] \Rightarrow 21x - 21y - 4xy = 0 \text{ ……….(2)}$

Substituting value of x from equation (1) in (2), we get :

⇒ 21(y + 4) - 21y - 4y(y + 4) = 0

⇒ 21y + 84 - 21y - 4y² - 16y = 0

⇒ 4y² + 16y - 84 = 0

⇒ 4(y² + 4y - 21) = 0

⇒ y² + 7y - 3y - 21 = 0

⇒ y(y + 7) - 3(y + 7) = 0

⇒ (y - 3)(y + 7) = 0

⇒ y - 3 = 0 or y + 7 = 0

⇒ y = 3 or y = -7.

Since, numbers are positive.

∴ y = 3.

⇒ x = y + 4 = 3 + 4 = 7.

Hence, numbers are 3 and 7.