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

$$\begin{gathered} \text{Let the numerator be}x. \\ \therefore \text{Denominator =}x+3 \\ \therefore \text{Original number =}\frac{x}{x+3} \\ \text{According to the question:} \\ \frac{x}{x+3}+\frac{1}{\left(\frac{x}{x+3}\right)}=2\frac{9}{10} \\ \Rightarrow \frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10} \\ \Rightarrow \frac{x^2+{(x+3)}^2}{x(x+3)}=\frac{29}{10} \\ \Rightarrow \frac{x^2+x^2+6x+9}{x^2+3x}=\frac{29}{10} \\ \Rightarrow \frac{2x^2+6x+9}{x^2+3x}=\frac{29}{10} \\ \Rightarrow 29x^2+87x=20x^2+60x+90 \\ \Rightarrow 9x^2+27x-90=0 \\ \Rightarrow 9\left(x^2+3x-10\right)=0 \\ \Rightarrow x^2+3x-10=0 \\ \Rightarrow x^2+5x-2x-10=0 \\ \Rightarrow x\left(x+5\right)-2\left(x+5\right)=0 \\ \Rightarrow \left(x-2\right)\left(x+5\right)=0 \\ \Rightarrow x-2=0\mathrm{or}x+5=0 \\ \Rightarrow x=2\mathrm{or}x=-5\left(\mathrm{rejected}\right) \\ \mathrm{So},\mathrm{numerator}=x=2 \\ \mathrm{denominator}=x+3=2+3=5 \\ \mathrm{So},\mathrm{required}\mathrm{fraction}=\frac{2}{5} \end{gathered}$$