Question

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

Answer 1

Let numerator of the fraction be $x.$

Then, Denominator $=2x+1$
$\therefore$ Fraction $=\frac{x}{2x+1}$ and its reciprocal $=\frac{2x+1}{x}$

According to the given condition, we have

$\frac{x}{2x+1}+\frac{2x+1}{x}=\frac{58}{21}$ [$\because 2\frac{16}{21}=\frac{58}{21}$]

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

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

$\Rightarrow 21x^2+84x^2+84x+21-116x^2-58x=0$

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

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

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

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

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

$\Rightarrow x=3or\frac{-7}{11}\left(Neglect\right)$ [On taking positive value of $x$]

So, numerator $=x=3$

and, denominator $=2x+1=2\times 3+1=7$

$\therefore$ Required fraction is $\frac{3}{7}$