Question

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

Answer 1

$$\begin{gathered} \text{Let the numbers be}x\text{and}(x+4). \\ \text{According to the question:} \\ \frac{1}{x}-\frac{1}{x+4}=\frac{4}{21} \\ \Rightarrow \frac{\cancel{x}+4-\cancel{x}}{x(x+4)}=\frac{4}{21} \\ \Rightarrow \frac{{\cancel{4}}^1}{x^2+4x}=\frac{{\cancel{4}}^1}{21} \\ \Rightarrow \frac{1}{x^2+4x}=\frac{1}{21} \\ \Rightarrow x^2+4x=21 \\ \Rightarrow x^2+4x-21=0 \\ \Rightarrow x^2+(7-3)x-21=0 \\ \Rightarrow x^2+7x-3x-21=0 \\ \Rightarrow x(x+7)-3(x+7)=0 \\ \Rightarrow (x+7)(x-3)=0 \\ \Rightarrow x+7=0\text{or}x-3=0 \\ \Rightarrow x=-7\text{or}x\text{= 3} \\ \mathrm{If}x=-7,\text{the numbers are}-\text{7 and -3. (∵}-\text{7 + 4 =}-\text{3)} \\ \text{If}x=3,\text{the numbers are 3 and 7. (∵ 3 + 4}=7) \\ \text{Hence, the numbers are}(-\text{7,}-\text{3})\text{or}\left(\text{3, 7}\right)\text{.} \end{gathered}$$