Question

The difference of two numbers is $4$, if the difference of their reciprocals is $\frac{4}{21}$, find the numbers.

Answer 1

Step-1: Formation of equation:

Assume that the numbers are $a\mathrm{and}b$.$(a>b)$

Their difference is $4$. Determine the equation of the difference:

$\begin{array}{rcl}a-b& =& 4\\ & \Rightarrow & a=4+b\end{array}$

The difference of their reciprocals is $\frac{4}{21}$.

Determine the equation for the difference of reciprocals:

$\begin{array}{rcl}\frac{1}{b}-\frac{1}{a}& =& \frac{4}{21}\\ & \Rightarrow & \frac{1}{b}-\frac{1}{b+4}=\frac{4}{21}\\ & \Rightarrow & \frac{b+4-b}{b^2+4b}=\frac{4}{21}\\ & \Rightarrow & \frac{4}{b^2+4b}=\frac{4}{21}\end{array}$

This implies that $b^2+4b=21\mathrm{or}{b}^2+4b-21=0$.

Step-2 : Determine the velues:

Solve the equation:

$$\begin{gathered} b^2+4b-21=0 \\ \Rightarrow \left(b+7\right)\left(b-3\right)=0 \\ \Rightarrow b=3\mathrm{or}-7 \end{gathered}$$

Determine the second number:

$\begin{array}{rcl}a& =& b+4\\ & \Rightarrow & a=3+4\mathrm{or}a=-7+4\\ & \Rightarrow & a=7\mathrm{or}a=-3\end{array}$

Hence, the numbers are $3\mathrm{and}7\mathrm{or}-7\mathrm{and}-3$.