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 the fraction be $x$, then its denominator is $2x+1$.

So the fraction is $\frac{x}{2x+1}$

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

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

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

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

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

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

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

$\Rightarrow 11x=-7$ or $x=3$

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

$\Rightarrow x=3$ since $x$ is a natural number $\therefore x\ge 0$

$\therefore$ Required fraction is $\frac{x}{2x+1}=\frac{3}{2\times 3+1}=\frac{3}{7}$