Question

The denominator of a fraction is $3$ more than its numerator. The sum of the fraction and its reciprocal is $2\frac{9}{10}$. Find the fraction.

Answer 1

Step 1: Assumption of data

Let the numerator of fraction be $x$.

Given that the denominator is $3$ more than its numerator, then denominator will be $x+3$.

Step 2: Determining the quadratic equation

The original fraction becomes $\frac{x}{x+3}$ ……(1)

Also, given that sum of the fraction and its reciprocal is $2\frac{9}{10}=\frac{29}{10}$.

Here,

$$\begin{gathered} \frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10} \\ \Rightarrow \frac{x^2+{\left(x+3\right)}^2}{\left(x+3\right)x}=\frac{29}{10} \\ \Rightarrow \frac{x^2+x^2+9+6x}{\left(x+3\right)x}=\frac{29}{10} \\ \Rightarrow \frac{2x^2+9+6x}{\left(x+3\right)x}=\frac{29}{10} \end{gathered}$$

After cross multiplication,

$$\begin{gathered} \Rightarrow 10\left(2x^2+9+6x\right)=29x\left(x+3\right) \\ \Rightarrow 20x^2+90+60x=29x^2+87x \\ \Rightarrow 29x^2-20x^2+87x-60x-90=0 \\ \Rightarrow 9x^2+27x-90=0 \end{gathered}$$

Step 3: Solving the quadratic equation

Simplifying the equaion in terms of $x$.

$$\begin{gathered} 9\left(x^2+3x-10\right)=0 \\ \Rightarrow x^2+3x-10=0 \\ \Rightarrow x^2+\left(5-2\right)x-10=0 \\ \Rightarrow x^2+5x-2x-10=0 \end{gathered}$$

After simplifying,

$$\begin{gathered} \Rightarrow x\left(x+5\right)-2\left(x+5\right)=0 \\ \Rightarrow \left(x+5\right)\left(x-2\right)=0 \\ \Rightarrow x=-5,2 \end{gathered}$$

Here, $x=-5$ is not considerable.

Substitute $x=2$ in equation (1),

Original fraction $\frac{x}{x+3}=\frac{2}{5}$.

Therefore, the fraction is $\frac{2}{5}$.