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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x [#permalink]
tonebeeze wrote:
Thanks Bunuel. I apologize for the bad formatting. Won't happen again!


And just to emphasize that we can divide both sides by y ONLY BECAUSE we are explicitly told that x and y are both different from zero as Bunuel clearly explained above.
Cheers

J :)
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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x [#permalink]
Bunuel - Can you please add the math formula formatting tag to the OP. For the first minute, I was solving for y as a power of x. :l
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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x [#permalink]
jlgdr wrote:
tonebeeze wrote:
Thanks Bunuel. I apologize for the bad formatting. Won't happen again!


And just to emphasize that we can divide both sides by y ONLY BECAUSE we are explicitly told that x and y are both different from zero as Bunuel clearly explained above.
Cheers

J :)



You are absolutely right...
if xy≠0 wasn't mentioned then A would be insufficient as y can be zero too.
and hence E would have been the answer..
Thanks
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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x [#permalink]
A is correct. Here's why:

Manipulate orig eq to the following --> (x^5)(y^4) = ?

(1) (x^3)(y^5) = 6y
= (x^5)(y^4) = 6

SUFFICIENT

(2) x = 2y --> plug into orig eq
=(2y)^5[(y^4)] = 32(y^9)

NOT SUFFICIENT
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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x [#permalink]
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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x [#permalink]
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