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ajmer_sasan
- Newbie | Next Rank: 10 Posts
- Posts: 1
- Joined: Mon Mar 18, 2013 9:05 am
Hi, please help me with below.
The question is:-
8xy^3 + 8x^3y = 2x^2y^2 / 2^-3, What is xy?
(1) y > x (2) x < 0
Here is how I did it and got wrong answer:-
8xy^3 + 8x^3y = 2x^2y^2 / 2^-3
=8xy(x^2+y^2) = 16x^2Y^2
=x^2+y^2 = 2xy (I cancelled xy on left hand side with xy of right hand side)
=(x-y)^2 = 0
=> x-y = 0 or x = y.
However the solution is:-
8xy^3 + 8x^3y = 2x^2y^2 / 2^-3
=> 8xy^3 + 8x^3y - 16x^2y^2 = 0
=> xy(x^2+y^2-2xy) = 0
=> xy(x-y)^2 = 0
=> xy = 0 or x-y = 0
=> xy = 0 or x=y.
So in my solution I missed out xy=0 option.
My doubt is why can't I cancel out xy on two sides of equality? Please help me understand.
Regards,
Ajmer
The question is:-
8xy^3 + 8x^3y = 2x^2y^2 / 2^-3, What is xy?
(1) y > x (2) x < 0
Here is how I did it and got wrong answer:-
8xy^3 + 8x^3y = 2x^2y^2 / 2^-3
=8xy(x^2+y^2) = 16x^2Y^2
=x^2+y^2 = 2xy (I cancelled xy on left hand side with xy of right hand side)
=(x-y)^2 = 0
=> x-y = 0 or x = y.
However the solution is:-
8xy^3 + 8x^3y = 2x^2y^2 / 2^-3
=> 8xy^3 + 8x^3y - 16x^2y^2 = 0
=> xy(x^2+y^2-2xy) = 0
=> xy(x-y)^2 = 0
=> xy = 0 or x-y = 0
=> xy = 0 or x=y.
So in my solution I missed out xy=0 option.
My doubt is why can't I cancel out xy on two sides of equality? Please help me understand.
Regards,
Ajmer
