The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is $2\dfrac{9}{10}$. Find the fraction.

Let the fraction be $\dfrac{a}{b}$.

It is given that the denominator of a fraction is 3 more than its numerator.

⇒ b = a + 3

And, the sum of the fraction and its reciprocal is $2\dfrac{9}{10}$.

$\Rightarrow \dfrac{a}{b} + \dfrac{b}{a} = 2\dfrac{9}{10}\\[1em] \Rightarrow \dfrac{a}{a + 3} + \dfrac{a + 3}{a} = \dfrac{29}{10}\\[1em] \Rightarrow \dfrac{(a \times a) + [(a + 3)(a + 3)] }{(a + 3)a} = \dfrac{29}{10}\\[1em] \Rightarrow \dfrac{a^2 + (a + 3)^2}{(a + 3)a} = \dfrac{29}{10}\\[1em] \Rightarrow \dfrac{a^2 + (9 + 6a + a^2)}{(a + 3)a} = \dfrac{29}{10}\\[1em] \Rightarrow \dfrac{a^2 + 9 + 6a + a^2}{3a + a^2} = \dfrac{29}{10}\\[1em] \Rightarrow 10(a^2 + 9 + 6a + a^2) = 29(3a + a^2)\\[1em] \Rightarrow 10(2a^2 + 9 + 6a) = 29(3a + a^2)\\[1em] \Rightarrow 20a^2 + 90 + 60a = 87a + 29a^2\\[1em] \Rightarrow 20a^2 + 90 + 60a - 87a - 29a^2 = 0\\[1em] \Rightarrow -9a^2 - 27a + 90 = 0\\[1em] \Rightarrow a^2 + 3a - 10 = 0\\[1em] \Rightarrow a^2 + 5a - 2a - 10 = 0\\[1em] \Rightarrow a(a + 5) - 2(a + 5) = 0\\[1em] \Rightarrow (a + 5)(a - 2) = 0\\[1em] \Rightarrow (a + 5) = 0 \text{ or }(a - 2) = 0\\[1em] \Rightarrow a = -5 \text{ or } a = 2\\[1em]$

If a = -5, then b = 3 + (-5) = -2

And, if a = 2, then b = 3 + 2 = 5

So, the fraction = $\dfrac{-5}{-2} = \dfrac{5}{2} \text{ or }\dfrac{2}{5}$

Since, denominator is 3 more than the numerator,

∴ The fraction is $\dfrac{2}{5}$

Hence, the fraction = $\dfrac{2}{5}$.