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The sum of the reciprocals of 2 consecutive intergers is 5/6. Find the numbers

User Ambran
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1 Answer

30 votes
30 votes

hmmmm let's say our numbers are "a" and "b", but we know they're consecutive integer, therefore "b" is either a-1 or a+1, hmm let's use the latter

a = first integer

a+1 = second integer

we know their reciprocals add up to 5/6, so


\begin{cases} a\\ a+1 \end{cases}\qquad \stackrel{\textit{sum of their reciprocals}}{\cfrac{1}{a}+\cfrac{1}{a+1}}=\cfrac{5}{6}

let's use the LCD of (a)(a+1)(6) and multiply both sides by it, to do away with the denominators


6a(a+1)\left[ \cfrac{1}{a}+\cfrac{1}{a+1} \right]=6a(a+1)\left[ \cfrac{5}{6} \right] \\\\\\ 6(a+1)+6a = 5a(a+1)\implies 6a+6+6a=5a^2+5a \\\\\\ 12a+6=5a^2+5a\implies 0 = 5a^2-7a-6\implies 0 = (a-2)(5a+3) \\\\\\ \begin{cases} 2=a&\textit{\large \checkmark}\\ -(3)/(5)=a&\textit{not an integer} \end{cases}~\hfill \begin{cases} a &= 2\\ a+1 &= 3 \end{cases}

User Nishant Gupta
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