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Re: GRE Math Challenge #137- (x+y) and 2(x+y) [#permalink]
What's wrong with dividing both sides by (x+y)? If we do so, Quantity B (2) will always be greater than A (1).
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Re: GRE Math Challenge #137- (x+y) and 2(x+y) [#permalink]
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TAYYABGRE wrote:
What's wrong with dividing both sides by (x+y)? If we do so, Quantity B (2) will always be greater than A (1).



Brent explained WHY we cannot beautifully above. Please read Sir

Essentially, we cannot because we DO NOT know the values of x and y
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Re: GRE Math Challenge #137- (x+y) and 2(x+y) [#permalink]
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TAYYABGRE wrote:
What's wrong with dividing both sides by (x+y)? If we do so, Quantity B (2) will always be greater than A (1).


Whenever you divide both sides by an unknown quantity such as x+y, you are risking the possibility of that x+y equals 0 or some negative value.

This video explains this concept in much greater detail: https://www.greenlighttestprep.com/modu ... video/1097
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GRE Math Challenge #137- (x+y) and 2(x+y) [#permalink]
Thank you so much.
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