Question

The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $2\frac{16}{21}$, find the fraction.

Answer 1

Let the numerator of fraction be $x$ Then,

Denominator $=2x+1$

$\therefore$ fraction $=\frac{x}{2x+1}\Rightarrow$ Reciprocal of fraction $=\frac{2x+1}{x}$

$\therefore \frac{x}{2x+1}+\frac{2x+1}{x}=2\frac{16}{21}=\frac{58}{21}$

$\frac{5x^2+4x+1}{2x^2+x}=\frac{58}{21}$

$105x^2+84x+21=116x^2+58x$

$11x^2-26x-21=0$

$11x^2-33x+7x-21=0$

$(11x+7)(x-3)=0$

$x=3,\frac{-7}{11}$

Hence fraction $=\frac{x}{2x+1}=\frac{3}{7}$