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 Denominator$=2x+1$
Fraction $=\frac{x}{2x+1}$
Reciprocal of the fraction $=\frac{2x+1}{x}$
It is given that the sum fo the fraction and its reciprocal is $2\frac{16}{21}$
$\therefore \frac{x}{2x+1}+\frac{2x+1}{x}=2\frac{16}{21}$

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

$\Rightarrow 21(5x^2+4x+1)=58(2x^2+x)\Rightarrow 105x^2+84x+21=116x^2+58x$

$\Rightarrow 11x^2-26x-21=0\Rightarrow 11x^2-33x+7x-21=0\Rightarrow 11x(x-3)+7(x-3)=0$

$\Rightarrow (11x+7)(x-3)=0\Rightarrow x=3,-\frac{7}{11}$

$\because$ $x$ can not be a negative

$\Rightarrow x=3$

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