x^2 = xy => x(x-y) = 0 => x=0, or x=y

x^2-y^2 = xy -5x + 5y - 25

If x=y, 0 = x^2 - 25 => x=+/-5 we do not know whether x=+/-5 but x may rake any value so not suff
If x=0, -y^2 = 5(y-5) => y^2+5y-25=0 => y= (-5 + (25+100)^1/2)/2 and (-5 - (25+100)^1/2)/2
so we get two values of y, but y can take any value so not sff

So A willn ot work

hi all
very unpleasant problem to me. i`ll try to prove the case that x=y is not always valid here,but i failed to prove if x=0
x=y=k. then k^2-k^2=(k+5)(k-5). k^2=25, k=5, k=-5.
the problem here is that
x^2-y^2=(x+5)(y-5) is valid for two different values of x,y, i mean when they are equal and when they are not (opposite)
say x=5,y=5, 5^2-5^2=0 and (5+5)(5-5)=0 also, here x=y and x^2=x*y we proved yes case, but on the other hand
x=-5,y=5, then
(-5)^2-5^2=0 and(-5+5)(5-5)=0 also, but x^2=(-5)^2=25, does not equal xy=(-5)*(5)=-25. 25=/=(-25) (not equal)
thus 1 st insuff

maihuna, that was neat!