Question

Sum of two natural numbers is $8$ and the difference of their reciprocals is $\frac{2}{15}$. Find the numbers.

Answer 1

Step 1: Formation of quadratic equation

Let the two natural numbers be $\mathit{x}$ and $\mathit{y}$.

According to the question, their sum, i.e. $x+y=8$.

Also, the difference of their reciprocals, i.e. $\frac{1}{x}-\frac{1}{y}=\frac{2}{15}$.

Put $y=8-x$ in $\frac{1}{x}-\frac{1}{y}=\frac{2}{15}$, we get

$$\begin{gathered} \Rightarrow \frac{1}{x}-\frac{1}{8-x}=\frac{2}{15} \\ \Rightarrow \frac{8-x-x}{x(8-x)}=\frac{2}{15} \\ \Rightarrow [8-2x]\times 15=2\times \left[x\right(8-x\left)\right] \\ \Rightarrow [4-x]\times 15=\left[x\right(8-x\left)\right] \\ \Rightarrow 60-15x=-x^2+8x \\ \Rightarrow x^2-23x+60=0 \\ \Rightarrow x^2-3x-20x+60=0 \\ \Rightarrow x(x-3)-20(x-3)=0 \\ \Rightarrow (x-20)(x-3)=0 \\ \Rightarrow x=20,3 \end{gathered}$$

As the total sum of numbers is $8$, so $x\ne 20$

Step 2: Calculation for other number

Replace $\mathit{x}\mathbf{=}\mathbf{3}$ in $\mathit{x}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{8}$

$$\begin{gathered} x+y=8 \\ \Rightarrow y=8-x \\ \Rightarrow y=8-3 \\ \Rightarrow y=5 \end{gathered}$$

Thus, the two natural numbers are $3$ and $5$.