The sum of numerator and denominator of a certain positive fraction is 8 . If 2 is added to both the numerator and denominator, the fraction is increased by $\dfrac{4}{35}.$ Find the fraction.

Let the denominator be = x so, numerator = 8 - x.

Fraction = $\dfrac{8 - x}{x}$

Given , if 2 is added to both the numerator and denominator, the fraction is increased by $\dfrac{4}{35}.$

$\Rightarrow \dfrac{8 - x + 2}{x + 2} - \dfrac{8 - x}{x} = \dfrac{4}{35} \\[1em] \Rightarrow \dfrac{x(10 - x) - (x + 2)(8 - x)}{x(x + 2)} = \dfrac{4}{35} \text{ (On taking L.C.M) }\\[1em] \Rightarrow \dfrac{10x - x^2 - (8x - x^2 + 16 - 2x)}{x^2 + 2x} = \dfrac{4}{35} \\[1em] \Rightarrow 35(-x^2 + x^2 - 8x + 10x + 2x - 16) = 4(x^2 + 2x) \\[1em] \Rightarrow 35(4x - 16) = 4x^2 + 8x \\[1em] \Rightarrow 140x - 560 = 4x^2 + 8x \\[1em] \Rightarrow 4x^2 + 8x - 140x + 560 = 0 \\[1em] \Rightarrow 4x^2 - 132x + 560 = 0 \\[1em] \Rightarrow 4(x^2 - 33x + 140) = 0 \\[1em] \Rightarrow x^2 - 33x + 140 = 0 \\[1em] \Rightarrow x^2 - 28x - 5x + 140 = 0 \\[1em] \Rightarrow x(x - 28) - 5(x - 28) = 0 \\[1em] \Rightarrow (x - 28)(x - 5) = 0 \\[1em] \Rightarrow x - 28 = 0 \text{ or } x - 5 = 0 \\[1em] x = 28 \text{ or } x = 5$

If x = 28 , 8 - x = -20 which will make fraction = $-\dfrac{20}{28}$ negative hence, x ≠ 28

∴ x = 5 , 8 - x = 3 ,fraction = $\dfrac{3}{5}$

Hence, the fraction is $\dfrac{3}{5}$.