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Question Number 217064 by ArshadS last updated on 28/Feb/25

$$\mathrm{Two}\mathrm{numbers}\mathrm{differ}\mathrm{by}6.\mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{their}\mathrm{reciprocals}\mathrm{is}\frac{2}{15}.$$$$\mathcal{D}eterminethenumbers.$$

Answered by A5T last updated on 28/Feb/25

$$\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}=\frac{2}{15}\Rightarrow 15(\mathrm{a}+\mathrm{b})=2\mathrm{ab}$$$$\mathrm{a}-\mathrm{b}=\underset{-}{+}6\Rightarrow \mathrm{a}=\mathrm{b}\underset{-}{+}6$$$$\mathrm{a}=\mathrm{b}+6\Rightarrow 15(\mathrm{b}+3)=\mathrm{b}(\mathrm{b}+6)$$$$\Rightarrow {\mathrm{b}}^2-9\mathrm{b}-45=0\Rightarrow \mathrm{b}=\frac{9\underset{-}{+}\sqrt{261}}{2}$$$$\Rightarrow \mathrm{a}=\frac{21\underset{-}{+}\sqrt{261}}{2}\Rightarrow (\mathrm{a},\mathrm{b})=(\frac{21\underset{-}{+}\sqrt{261}}{2},\frac{9\underset{-}{+}\sqrt{261}}{2})$$$$\mathrm{a}=\mathrm{b}-6\Rightarrow 15(\mathrm{b}-3)=\mathrm{b}(\mathrm{b}-6)$$$$\Rightarrow {\mathrm{b}}^2-21\mathrm{b}+45\mathrm{b}=0\Rightarrow \mathrm{b}=\frac{21\underset{-}{+}\sqrt{261}}{2}\Rightarrow \mathrm{a}=\frac{9\underset{-}{+}\sqrt{261}}{2}$$

Commented by ArshadS last updated on 28/Feb/25

$$\mathrm{Thanks}\mathrm{sir}!$$

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