The numerator of a fraction is 3 less than its denominator. If one is added to the denominator, the fraction is decreased by $\dfrac{1}{15}$. Find the fraction.

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

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

⇒ a = b - 3

And, one is added to the denominator, the fraction is decreased by $\dfrac{1}{15}$.

$\Rightarrow \dfrac{a}{b + 1} = \dfrac{a}{b} - \dfrac{1}{15}\\[1em] \Rightarrow \dfrac{b - 3}{b + 1} = \dfrac{b - 3}{b} - \dfrac{1}{15}\\[1em] \Rightarrow \dfrac{b - 3}{b + 1} - \dfrac{b - 3}{b} = - \dfrac{1}{15}\\[1em] \Rightarrow \dfrac{b(b - 3) - (b + 1)(b - 3)}{b(b + 1)} = - \dfrac{1}{15}\\[1em] \Rightarrow \dfrac{(b^2 - 3b) - (b^2 - 3b + b - 3)}{b(b + 1)} = - \dfrac{1}{15}\\[1em] \Rightarrow \dfrac{b^2 - 3b - b^2 + 3b - b + 3}{b(b + 1)} = - \dfrac{1}{15}\\[1em] \Rightarrow \dfrac{ - b + 3}{b^2 + b} = - \dfrac{1}{15}\\[1em] \Rightarrow 15(-b + 3) = - 1(b^2 + b)\\[1em] \Rightarrow - 15b + 45 = - b^2 - b\\[1em] \Rightarrow - 15b + 45 + b^2 + b = 0\\[1em] \Rightarrow b^2 - 14b + 45 = 0\\[1em] \Rightarrow b^2 - 9b - 5b + 45 = 0\\[1em] \Rightarrow b(b - 9) - 5(b - 9) = 0\\[1em] \Rightarrow (b - 9)(b - 5) = 0\\[1em] \Rightarrow (b - 9) = 0 \text{ or }(b - 5) = 0\\[1em] \Rightarrow b = 9 \text{ or }b = 5\\[1em]$

If b = 9, then a = 9 - 3 = 6

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

So, the fraction = $\dfrac{6}{9} = \dfrac{2}{3} \text{ or }\dfrac{2}{5}$

Hence, the fraction = $\dfrac{2}{3} \text{ or } \dfrac{2}{5}$.