Question

The sum of two numbers is 18 and the sum of their reciprocals is $\frac{1}{4}$. Find the numbers.

Answer 1

$$\begin{gathered} \text{Let the numbers be}x\text{and}(18-x). \\ \text{According to the question:} \\ \frac{1}{x}+\frac{1}{18-x}=\frac{1}{4} \\ \Rightarrow \frac{18-x+x}{x(18-x)}=\frac{1}{4} \\ \Rightarrow \frac{18}{18x-x^2}=\frac{1}{4} \\ \Rightarrow 18x-x^2=72 \\ \Rightarrow x^2-18x+72=0 \\ \Rightarrow x^2-(12+6)x+72=0 \\ \Rightarrow x^2-12x-6x+72=0 \\ \Rightarrow x(x-12)-6(x-12)=0 \\ \Rightarrow (x-12)(x-6)=0 \\ \Rightarrow x-12=0\text{or}x-6=0 \\ \Rightarrow x=12\text{or}x\text{= 6} \\ \mathrm{If}x=12,\mathrm{t}\text{he numbers are 12 and 6. (∵ 18}-\text{12 = 6)} \\ \mathrm{If}x=6,\text{the numbers are 6 and12 (∵ 18}-\text{6}=12) \\ \text{Hence, the numbers are 12 and 6.} \end{gathered}$$