Where's my extra xy?

[tontonio]

Hi. I was trying to solve the quant challenge for september 16, 2002 called 'Complex Simplification' (http://www.manhattangmat.com/ChallengeP ... ChallID=66) and apparently I'm doing something wrong with the simplification, I just don't know what. Can you help me?

Stem is:
8xy^3 + 8x^3y = {2x^2y^2}/2^-3

So I simplified:

8xy(y^2 + x^2) = 2^3(2x^2y^2)
8xy(y^2 + x^2) = 8(2)(x^2)(y^2)
8xy(y^2 + x^2) = 8(2)(x)(x)(y)(y), therefore 8xy(2xy) <===== I'm guessing this is where I'm wrong. Is this a valid operation?
y^2 + x^2 = 2xy
y^2 + -2xy + x^2 = 0
(y-x)(y-x)=0
y=x

I followed your explanation and it makes sense but I can't explain what's wrong with my simplification.

Btw, MGMAT's answer is:
xy(y^2 + -2xy + x^2) = 0
xy (y-x)(y-x) = 0
xy = 0
-or-
y=x

Thanks in advance!

[Guest]

Interesting.

You canceled the "missing" xy along with 8 (in step 3)

I believe the rule is not to cancel variables because they could give us a potential solution but I would like a confirmation from the experts.

[RA]

The rule is not to divide both sides of the equation by a variable unless you are sure that the varibale is not zero. This is becaise division by 0 is undefined.

This rule is given in the EIV strategy guide.

[RonPurewal]

the posters above already nailed this one.

in general, DON'T DIVIDE BY VARIABLES, unless you are absolutely sure that those variables are nonzero.

instead, just MOVE EVERYTHING OVER TO ONE SIDE, AND FACTOR OUT the common variables. (in fact, this is still a good protocol to follow even if you do know that the variables are nonzero.) this is what's done in the mgmat problem.

if you are moved by the muses to divide out by the variable(s), then you MUST consider, as a separate case, whether those variables can equal zero. if you don't consider such cases, then you are almost certain to miss out on certain solutions to the problem (as you did here).