Sum of reciprocals of integers is given - find integers

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The problem involves finding two consecutive odd integers whose reciprocals sum to 28/195. The equation derived from the problem is 1/x + 1/(x-2) = 28/195. By multiplying through by 195x(x-2), a quadratic equation is formed, leading to two integer solutions. The initial attempt at solving yielded a non-integer result of 3.44, indicating a miscalculation in the application of the quadratic formula.

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Homework Statement


The sum of the reciprocals of two consecutive odd integers is 28/195. Find the two integers.


Homework Equations


?


The Attempt at a Solution


28/195= 1/x + 1/(x-2)
Solved using quadratic formula and got 3.44, which does not seem to be right.. Any help?
 
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jackleyt said:

Homework Statement


The sum of the reciprocals of two consecutive odd integers is 28/195. Find the two integers.


Homework Equations


?


The Attempt at a Solution


28/195= 1/x + 1/(x-2)
Solved using quadratic formula and got 3.44, which does not seem to be right.. Any help?

"Two consecutive odd integers...", you have then two numbers to start with:
2n+1, and 2n+3. The 'n' is a variable and the factor of 2 ensures that the entire expression will be ODD. Note that 'n' will be an integer.
 


Using "x" and "x-2" is just as valid as "2n+3" and "2n+1" so the equation you have, 1/x+ 1/(x-2)= 28/195 is perfectly good. Unfortunately, you have left out the important part: you don't show how you got 3.44. 1/x+ 1/(x-2)= 28/195, after you multiply the entire equation by 195x(x-2) becomes a quadratic formula and I get two integer solutions.
 

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