Cybermusings wrote:If x and y are integers and xy does not equal 0, is xy < 0?

(1) y = x^4 – x^3

(2) -12y^2 – y^2x + x^2y^2 > 0

Ok ... for the first statement ... muliplt by x u get xy = x^5-x^4 .. now if x is a positive integer then xy will be >0 .. but if x is a negative intger then x^5 will be negative so xy < 0 .. so no defnite answer so insufficient ...

now consider the second eqn ... for the second eqn we can divide both the sides by y^2 ( remenber the sign wont change bcoz y^2 is positive)... we have -12-x+x^2>0 .... x^2-x-12>0 .... solving the quadratic we have .. (x-4)*(x+3)>0 ..... so the posssible values of x are x>4 or x<-3 ... but in this case we dont know the value of y so insufficient ...

now combine both the statements we have if x>4 ( that is +ve) then y is also +ve in which case xy>0 ... but if x is -ve y is still +ve in which case xy <0 ... so no defnite answer so the answer is E ( I hope ) ..

What's the OA?