The denominator of a positive fraction is one more than the twice the numerator. If the sum of fraction and its reciprocal is 2.9; find the fraction.

Let numerator be x, denominator = 2x + 1

According to question,

$\Rightarrow \dfrac{x}{2x + 1} + \dfrac{2x + 1}{x} = 2.9 \\[1em] \Rightarrow \dfrac{x^2 + (2x + 1)^2}{x(2x + 1)} = \dfrac{29}{10} \\[1em] \Rightarrow \dfrac{x^2 + 4x^2 + 1 + 4x}{2x^2 + x} = \dfrac{29}{10} \\[1em] \Rightarrow \dfrac{5x^2 + 4x + 1}{2x^2 + x} = \dfrac{29}{10} \\[1em] \Rightarrow 10(5x^2 + 4x + 1) = 29(2x^2 + x) \\[1em] \Rightarrow 50x^2 + 40x + 10 = 58x^2 + 29x \\[1em] \Rightarrow 58x^2 - 50x^2 + 29x - 40x - 10 = 0 \\[1em] \Rightarrow 8x^2 - 11x - 10 = 0 \\[1em] \Rightarrow 8x^2 - 16x + 5x - 10 = 0 \\[1em] \Rightarrow 8x(x - 2) + 5(x - 2) = 0 \\[1em] \Rightarrow (8x + 5)(x - 2) = 0 \\[1em] \Rightarrow (8x + 5) = 0 \text{ or } x - 2 = 0 \\[1em] \Rightarrow x = -\dfrac{5}{8} \text{ or } x = 2.$

Since, fraction is positive,

∴ x ≠ $-\dfrac{5}{8}$.

Fraction = $\dfrac{x}{2x + 1} = \dfrac{2}{2(2) + 1} = \dfrac{2}{5}.$

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