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\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

thank you for this. Strangely enough, GMAT Prep states x=2/3 (when y=(-1/3)!

for #2.) we are given y<=-1. This is not sufficient because of the following:

|x| = |y|

x=y OR x=-y

x=-y OR x=-(-y) x=y

Correct?

I am not really sure what you have done here. The 4 cases will be x = y x = -y -x = y -x = -y which are equivalent to just two cases: x = y or x = -y. Statement 2 is not sufficient because all we know now is that y is negative. IF x = y, x is negative. If x = -y, x is positive. So we still don't know whether x is positive or not.

WholeLottaLove wrote:

Another thing that has bothered me is this. If x=-y, and for example, y=5, then would x=(-5)?

As always, thanks for the help!

Yes. If x = -y and y = 5, then x = -5

Hi Karishma

I did get C as the answer applying the same method that you have indicated here. But if we solve through using variables, are we not getting x in the positive range? Applying the scenario where x=-y or y=-x

I did get C as the answer applying the same method that you have indicated here. But if we solve through using variables, are we not getting x in the positive range? Applying the scenario where x=-y or y=-x

|x| = |y| implies either x = y or x = -y. Which is the actual case, we do not know.

Stmnt 2 gives us y is negative. But do we know whether x = y or x = -y? No. If x = y, x is negative. If x = -y, x is positive.

So stmnt 2 alone is not sufficient to say whether x is positive or negative. It could be either.
_________________

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

would I be making a mistake if I had plugged in values. The first one statement was insufficient because we had two values of X based on whether Y was positive or Negative.

The second statement was insufficient because it had no information about x.

Combining both, we can easily observe for all negative values of Y, X is also Negative.

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel & Karishma, I will appreciate if you can comment:

1- How can we see this question and decide whether we need to pick numbers or not?

2- In other words, combining the two statements here is fairly easy, but how do we get it to a 50-50 between C and E in 5 to 10 seconds?

3 - In other words, how do we ensure that we don't waste time plugging in numbers in statement 1?
_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel & Karishma, I will appreciate if you can comment:

1- How can we see this question and decide whether we need to pick numbers or not?

2- In other words, combining the two statements here is fairly easy, but how do we get it to a 50-50 between C and E in 5 to 10 seconds?

3 - In other words, how do we ensure that we don't waste time plugging in numbers in statement 1?

Basically there are two distinct strategies - Number plugging (for example a percentages question in which one set will certainly give the answer) and Number testing (such as this question in which you need to test for various values)

I swear by number plugging but number testing is almost never my strategy of choice. I might want to try one set to understand the question/logic better but I will rarely try to deduce my answer from number testing. So when I see x^2 = y^2, I automatically think |x| = |y| which implies x = y or x = -y or x^2 - y^2 = (x + y) * (x - y) which implies x = -y or x = y

This makes me jump to stmnt 2 immediately since it is clearly insufficient. This brings me to (C) or (E).
_________________

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

x2=y2x2=y2 --> |x|=|y||x|=|y| --> either y=xy=x or y=−xy=−x

we can have -x = - y and -x = y

I'm I missing something?

Posted from my mobile device _________________

"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Worried About IDIOMS?Here is a Daily Practice List: https://gmatclub.com/forum/idiom-s-ydmuley-s-daily-practice-list-250731.html#p1937393

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

x2=y2x2=y2 --> |x|=|y||x|=|y| --> either y=xy=x or y=−xy=−x

we can have -x = - y and -x = y

I'm I missing something?

Posted from my mobile device

-x = -y is the same as x = y and -x = y is the same as x = -y.

My bad, so dumb of me not to think that, looks like my brain was freezed after seeing this question. Thank you Bunuel.
_________________

"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Worried About IDIOMS?Here is a Daily Practice List: https://gmatclub.com/forum/idiom-s-ydmuley-s-daily-practice-list-250731.html#p1937393

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

In the following statement,

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

what if, y=-x=-1 ? Then x=1>0 and hence (1)+(2) = insufficient _________________

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

In the following statement,

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

what if, y=-x=-1 ? Then x=1>0 and hence (1)+(2) = insufficient

From (1) we have only two possibilities: a. \(y=x=-1\) b. \(y=-x=-\frac{1}{3}\)

y = -1 and x = 1 does not satisfy the first statement.
_________________

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

In the following statement,

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

what if, y=-x=-1 ? Then x=1>0 and hence (1)+(2) = insufficient

From (1) we have only two possibilities: a. \(y=x=-1\) b. \(y=-x=-\frac{1}{3}\)

y = -1 and x = 1 does not satisfy the first statement.

Oh, my bad! Thanks a ton, Bunuel _________________

Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.

Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.

You could use number plugging to check whether you are right.

x^2 = y^2:

The above could be true for example if x = y = -1 or if x = 1 and y = -1. These examples answer no to both of your question. Generally x^2 = y^2 means that |x| = |y|, so x and y are at the same distance from 0 on the number line.
_________________

Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.

You could use number plugging to check whether you are right.

x^2 = y^2:

The above could be true for example if x = y = -1 or if x = 1 and y = -1. These examples answer no to both of your question. Generally x^2 = y^2 means that |x| = |y|, so x and y are at the same distance from 0 on the number line.

Thanks. What are the implications of |y|<|x|, is it "y<x or y>-x" or something different?

Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.

You could use number plugging to check whether you are right.

x^2 = y^2:

The above could be true for example if x = y = -1 or if x = 1 and y = -1. These examples answer no to both of your question. Generally x^2 = y^2 means that |x| = |y|, so x and y are at the same distance from 0 on the number line.

Thanks. What are the implications of |y|<|x|, is it "y<x or y>-x" or something different?

|x| is the distance from x to 0 on the number line, so |y| < |x| means that x if further from 0 than y is.

Thanks Bunuel. It looks like I did not capture all cases with my inequalities. For example, when y>0, I have y is always less than x, which is not true based on case 2. When faced with this condition on DS, is the best way to draw out the number line like this? How does this compare to getting a condition that says |x-5|>2 in terms of how you would get the constraints.

Thanks Bunuel. It looks like I did not capture all cases with my inequalities. For example, when y>0, I have y is always less than x, which is not true based on case 2. When faced with this condition on DS, is the best way to draw out the number line like this? How does this compare to getting a condition that says |x-5|>2 in terms of how you would get the constraints.