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Reciprocals (Posted on 2009-07-12) Difficulty: 2 of 5
Find all pairs (x,y) of real numbers for which
          1     1      1  
    xy + ——— + ——— = ———— + x + y.
          x     y     xy

  Submitted by Bractals    
Rating: 5.0000 (1 votes)
Solution: (Hide)

Multiply the equation by xy and factor:

   x2y2 - x2y - xy2 + x + y - 1 = 0;

   x2y(y - 1) - x(y2 - 1) + y - 1 = 0;

   (y - 1)[x2y - x(y + 1) + 1] = 0;

   (y - 1)[xy(x - 1) - x + 1] = 0;

   (y - 1)(x - 1)(xy - 1) = 0.

Therefore, the pairs that solve the equation are

   (1,t), (t,1), and (t,1/t) for any nonzero real number t.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Puzzle Thoughts K Sengupta2023-07-13 09:33:48
analytical solutionDaniel2009-07-12 13:32:09
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