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mukushla Group Title

all solutions to \(x^2+y=xy^2\) \(y^2+x=yx^2\) \(x\) , \(y\) are real numbers

  • one year ago
  • one year ago

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  1. waterineyes Group Title
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    If we subtract them then it will work or not ??

    • one year ago
  2. mukushla Group Title
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    it will work

    • one year ago
  3. waterineyes Group Title
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    \[(y-x)(-((x+y)+ 1) = xy(y-x)\]

    • one year ago
  4. waterineyes Group Title
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    \[\implies 1 - x - y = xy\]

    • one year ago
  5. Neemo Group Title
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    S={(a;a)/ a in B}.....B={-1;0;(1+sqrt(5))/2;(1-sqrt(5))/2}

    • one year ago
  6. mukushla Group Title
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    @waterineyes u forgot the case \(y=x\)

    • one year ago
  7. Ishaan94 Group Title
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    x^2 - y^2 - (x-y) = -xy(x-y) (x-y)(x+y -1+xy) = 0 => x+y-1+xy = 0 => y = (1-x)/(1+x) 1 + x =/= 0

    • one year ago
  8. waterineyes Group Title
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    Yeah in that case we cannot divide both sides..

    • one year ago
  9. Ishaan94 Group Title
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    x=1, y=0 y=1. x=0

    • one year ago
  10. Ishaan94 Group Title
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    i can't get other points

    • one year ago
  11. mukushla Group Title
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    3 solutions from \(x=y\) and any from \(x+y+xy=1\)

    • one year ago
  12. waterineyes Group Title
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    For x = y there are three solutions we will get @mukushla ?? Asking..

    • one year ago
  13. mukushla Group Title
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    yes...

    • one year ago
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