In a certain positive fraction , the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator , the fraction is decreased by $\dfrac{1}{14}$. Find the fraction.

Let the numerator of the fraction be x

Given, the denominator is greater than numerator by 3 hence, denominator = x + 3

Fraction = $\dfrac{x}{x + 3}$

Given, if 1 is subtracted from both the numerator and denominator , the fraction is decreased by $\dfrac{1}{14}$

$\Rightarrow \dfrac{x}{x + 3} - \dfrac{x - 1}{x + 3 - 1} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x}{x + 3} - \dfrac{x - 1}{x + 2} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x(x + 2) - (x - 1)(x + 3)}{(x + 3)(x + 2)} = \dfrac{1}{14} \text{ (On taking L.C.M) } \\[1em] \Rightarrow \dfrac{x^2 + 2x - (x^2 + 3x - x - 3)}{(x^2 + 2x + 3x + 6)} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x^2 - x^2 + 2x - 2x + 3}{x^2 + 5x + 6} = \dfrac{1}{14} \\[1em] \Rightarrow 3 \times 14 = x^2 + 5x + 6 \text{ (Cross multiplying) } \\[1em] \Rightarrow x^2 + 5x + 6 = 42 \\[1em] \Rightarrow x^2 + 5x + 6 - 42 = 0 \\[1em] \Rightarrow x^2 + 5x - 36 = 0 \\[1em] \Rightarrow x^2 + 9x - 4x - 36 = 0 \\[1em] \Rightarrow x(x + 9) - 4(x + 9) = 0 \\[1em] \Rightarrow (x - 4)(x + 9) = 0 \\[1em] x = 4 \text{ or } x = -9 \\[1em]$

If x = -9 , Fraction = $\dfrac{x}{x + 3} = \dfrac{9}{6}$ In this case numerator > denominator which is incorrect according to the question hence, x ≠ -9.

∴ x = 4 , Fraction = $\dfrac{x}{x + 3} = \dfrac{4}{7}$

Hence, the fraction is $\dfrac{4}{7}$.