SOLUTION: The sum of the reciprocals of two consecutive integers is equal to 11 times the reciprocal of the product of those integers.what are the two integers?

Algebra ->  Inverses -> SOLUTION: The sum of the reciprocals of two consecutive integers is equal to 11 times the reciprocal of the product of those integers.what are the two integers?      Log On


   

Question 1044351: The sum of the reciprocals of two consecutive integers is equal to 11 times the reciprocal of the product of those integers.what are the two integers?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say that
n= one of the integers involved, with n%3C%3E0 , so that the reciprocal of n exists.
1%2Fn= the reciprocal of the integer above.
For consecutive integers, let's say that
n%2B1= the other integer, also with n%2B1%3C%3E0 , so
1%2F%28n%2B1%29= the reciprocal of the "other integer" above.
So, n%28n%2B1%29= the product of those integers.
Since neither n , nor n%2B1 is zero,
neither is their product: n%28n%2B1%29%3C%3E0 ,so the reciprocal exists.
1%2F%28n%28n%2B1%29%29= the reciprocal of the product of those two consecutive integers.
What the problem says is that
1%2Fn%2B1%2F%28n%2B1%29=11%281%2F%28n%28n%2B1%29%29%29
That is the equation to simplify and solve.
1%2Fn%2B1%2F%28n%2B1%29=11%281%2F%28n%28n%2B1%29%29%29
%28n%2B1%29%2F%28n%28n%2B1%29%29%2Bn%2F%28n%28n%2B1%29%29=11%2F%28n%28n%2B1%29%29
%28n%2Bn%2B1%29%2F%28n%28n%2B1%29%29=11%2F%28n%28n%2B1%29%29
%282n%2B1%29%2F%28n%28n%2B1%29%29=11%2F%28n%28n%2B1%29%29
Since n%28n%2B1%29%3E0 , we can multiply both sides of the equation times n%28n%2B1%29 to get a simpler, equivalent equation.
2n%2B1=11
2n=11-1
2n=10
n=10%2F2
highlight%28n=5%29
n%2B1=5%2B1
highlight%28n%2B1=6%29
The two integers are highlight%285%29 and highlight%286%29 .