<PRE><B>Question 1130446</B>
Solve and verify your answer. 
The sum of the reciprocals of two consecutive even integers is 9 / 40.
{{{1/n}}} + {{{1/((n+2))}}} = {{{9/40}}}
Find the integers (smaller and larger integers).
multiply equation by 40n(n+2), cancel the denominators
40(n+2) + 40n = 9n(n+2)
40n + 80 + 40n = 9n^2 + 18n 
80n + 80 = n^2 + 18n
Form a quadratic equation on the right
0 = 9n^2 + 18n - 80n - 80
n^2 - 62 - 80 = 0
Use the quadratic formula a=9; b=-62; c=-80
the positive integer solution
n = 8, and therefore 10 are the integers
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see if that works
{{{1/8}}} + {{{1/10}}} = 
common denominator is 40
{{{5/40}}} + {{{4/40}}} + {{{9/40}}}
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