Question

all solutions to

\(x^2+y=xy^2\)
\(y^2+x=yx^2\)

\(x\) , \(y\) are real numbers

Answer 1

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

Answer 2

it will work

Answer 3

\[(y-x)(-((x+y)+ 1) = xy(y-x)\]

Answer 4

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

Answer 5

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

Answer 6

@waterineyes

u forgot the case \(y=x\)

Answer 7

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

Answer 8

Yeah in that case we cannot divide both sides..

Answer 9

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

Answer 10

i can't get other points

Answer 11

3 solutions from \(x=y\)

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

Answer 12

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

Asking..

Answer 13

yes...