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



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waterineyes
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If we subtract them then it will work or not ??

mukushla
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it will work

waterineyes
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\[(yx)(((x+y)+ 1) = xy(yx)\]

waterineyes
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\[\implies 1  x  y = xy\]

Neemo
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S={(a;a)/ a in B}.....B={1;0;(1+sqrt(5))/2;(1sqrt(5))/2}

mukushla
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@waterineyes
u forgot the case \(y=x\)

Ishaan94
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x^2  y^2  (xy) = xy(xy)
(xy)(x+y 1+xy) = 0
=> x+y1+xy = 0 => y = (1x)/(1+x)
1 + x =/= 0

waterineyes
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Yeah in that case we cannot divide both sides..

Ishaan94
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x=1, y=0
y=1. x=0

Ishaan94
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i can't get other points

mukushla
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3 solutions from \(x=y\)
and any from \(x+y+xy=1\)

waterineyes
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For x = y there are three solutions we will get @mukushla ??
Asking..

mukushla
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yes...