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

Step 1: Assume the fraction

Let the numerator of the fraction be $x$.

The denominator of the fraction will be $2x+1$.

Thus, the fraction will be $\frac{x}{2x+1}$.

Step 2: Apply the given condition.

The sum of the fraction and its reciprocal is equal to $2\frac{16}{21}$.

$\begin{array}{rcl}\frac{x}{2x+1}+\frac{2x+1}{x}& =& 2\frac{16}{21}\\ \frac{x^2+{\left(2x+1\right)}^2}{x\left(2x+1\right)}& =& 2\frac{16}{21}\\ \frac{x^2+4x^2+1+4x}{2x^2+x}& =& \frac{58}{21}\\ 21(5x^2+1+4x)& =& 58\left(2x^2+x\right)\\ 105x^2+21+84x& =& 116x^2+58x\\ 11x^2-26x-21& =& 0\end{array}$

Step 3: Apply the quadratic formula to obtain the result.

$\begin{array}{rcl}11x^2-26x-21& =& 0\\ x& =& \frac{-\left(-26\right)\pm \sqrt{{(-26)}^2-4\times \left(11\right)\left(-21\right)}}{2\times 11}\\ x& =& \frac{26\pm \sqrt{676+924}}{22}\\ x& =& \frac{26\pm \sqrt{1600}}{22}\\ x& =& \frac{26\pm 40}{22}\\ x& =& \frac{66}{22}\text{or}-\frac{14}{22}\\ x& =& 3\text{or}-\frac{7}{11}\end{array}$

Neglect the fractional value because the numerator will be an integer.

Step 4: Obtain the fraction

Substitute obtained values:

$\begin{array}{rcl}\frac{x}{2x+1}& =& \frac{3}{2\times 3+1}\\ & =& \frac{3}{7}\end{array}$

Therefore, the required fraction is $\frac{3}{7}$.