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Mathematics

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6th - 8th Grade

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Easy

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Roman Hall

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1. Positive real numbers  aa  and  bb  satisfy  a+1b=7a+\frac{1}{b}=7  and  b+1a=5b+\frac{1}{a}=5  . Compute  ab+1abab+\frac{1}{ab}  .

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Solution to Problem 1

Multiply the two equations together to get  (a+1b)(b+1a)=ab+1+1+1ab=35⟹ab+1ab=33\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right)=ab+1+1+\frac{1}{ab}=35\Longrightarrow ab+\frac{1}{ab}=33  

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