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mukushla

  • 3 years ago

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

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  1. waterineyes
    • 3 years ago
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    If we subtract them then it will work or not ??

  2. mukushla
    • 3 years ago
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    it will work

  3. waterineyes
    • 3 years ago
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    \[(y-x)(-((x+y)+ 1) = xy(y-x)\]

  4. waterineyes
    • 3 years ago
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    \[\implies 1 - x - y = xy\]

  5. Neemo
    • 3 years ago
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    S={(a;a)/ a in B}.....B={-1;0;(1+sqrt(5))/2;(1-sqrt(5))/2}

  6. mukushla
    • 3 years ago
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    @waterineyes u forgot the case \(y=x\)

  7. Ishaan94
    • 3 years ago
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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

  8. waterineyes
    • 3 years ago
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    Yeah in that case we cannot divide both sides..

  9. Ishaan94
    • 3 years ago
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    x=1, y=0 y=1. x=0

  10. Ishaan94
    • 3 years ago
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    i can't get other points

  11. mukushla
    • 3 years ago
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    3 solutions from \(x=y\) and any from \(x+y+xy=1\)

  12. waterineyes
    • 3 years ago
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    For x = y there are three solutions we will get @mukushla ?? Asking..

  13. mukushla
    • 3 years ago
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    yes...

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