Algebra Session 10 Homework, Q29

GRE Quant Problem Solving
1

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If I choose numbers, I am getting the answer as 1, but Greg said the answer can not always be 1. Can someone explain this?

Also, by taking the algebra route, my answer is coming out as 1/y, which is not in the answer choices (I can square root 1/y and arrive at the correct answer choice E, but I am not sure if my method is correct).

2

What numbers did you choose?

How?

Can you simplify \frac{x \sqrt{y}}{y \sqrt{x^2}} noting that \sqrt{x^2} = x but since x > 0 then |x| = x.

3

I chose x= 2 and y =1. Here is how I solved it using both the ways

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4

And this is how Greg solved it:

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5

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You have to “undo” the fact that you squared both sides.

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Yeah that’s for (x,y) = (2,1) and you can’t just generalize this just because one particular example yielded this. For example, now consider (x,y) = (2,2). Clearly the answer isn’t 1 now.

6

Okay so, if I am squaring the term and arrive at an answer, I have to “undo” the square for my final answer choice?
And we do this because the square was not initially a part of the problem?

7

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You have that \left( \frac xy \sqrt{\frac{y}{x^2}}\right)^2 = \frac 1y, but the question asks you what \left( \frac xy \sqrt{\frac{y}{x^2}}\right) simplifies to. Thus, it only makes sense to consider square root-ing both sides of your expression to then have \left( \frac xy \sqrt{\frac{y}{x^2}}\right)=\sqrt{\frac 1y} = \frac{1}{\sqrt{y}}.

Anyway, as a side note, since x,y>0 then squaring (and other shenanigans) is fine to do. However, since squaring and square rooting are not always inverse operations, you’d generally want to be more careful in other cases

8

Thanks a lot!