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# If x^2 = y^2, is true that x>0?

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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28 Nov 2014, 13:04
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Bunuel wrote:
If x^2 = y^2, is true that x>0?

$$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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.

Hope it's clear.

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

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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29 Nov 2014, 05:57
VeritasPrepKarishma wrote:
WholeLottaLove wrote:
Question time:

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

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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30 Nov 2014, 21:24
sinhap07 wrote:

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

|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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18130 [0], given: 236 Manager Joined: 13 Dec 2013 Posts: 53 Kudos [?]: 7 [0], given: 21 GPA: 2.71 Re: If x^2 = y^2, is true that x>0? [#permalink] ### Show Tags 03 Jan 2015, 03:54 Bunuel wrote: If x^2 = y^2, is true that x>0? $$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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. Kudos [?]: 7 [0], given: 21 Senior Manager Affiliations: SPG Joined: 15 Nov 2006 Posts: 320 Kudos [?]: 911 [0], given: 28 If x^2 = y^2, is true that x>0? [#permalink] ### Show Tags 20 Jan 2017, 02:42 Top Contributor Bunuel wrote: If x^2 = y^2, is true that x>0? $$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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. Download the Ultimate SC Flashcards Kudos [?]: 911 [0], given: 28 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7796 Kudos [?]: 18130 [1], given: 236 Location: Pune, India Re: If x^2 = y^2, is true that x>0? [#permalink] ### Show Tags 22 Jan 2017, 23:55 1 This post received KUDOS Expert's post dimitri92 wrote: Bunuel wrote: If x^2 = y^2, is true that x>0? $$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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). _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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27 Jun 2017, 07:49
Bunuel wrote:
If x^2 = y^2, is true that x>0?

$$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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.

Hope it's clear.

Hello Bunuel,

Why did u take only two possibilities?

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
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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27 Jun 2017, 08:48
ydmuley wrote:
Bunuel wrote:
If x^2 = y^2, is true that x>0?

$$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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.

Hope it's clear.

Hello Bunuel,

Why did u take only two possibilities?

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.
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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27 Jun 2017, 09:51
Bunuel wrote:
ydmuley wrote:
Bunuel wrote:
If x^2 = y^2, is true that x>0?

$$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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.

Hope it's clear.

Hello Bunuel,

Why did u take only two possibilities?

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.
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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27 Jun 2017, 22:55
burnttwinky wrote:
If x^2 = y^2, is true that x>0?

(1) x=2y+1

(2) y<= -1

Given : x^2 = y^2 => x=+/-y
DS : x>0

Option 1 : x = 2y+1
x^2 = [(x-1)/2]^2
=> 4x^2 = x^2 +1 -2x
=> 3x^2 +2x -1 = 0
=> (x+1) (3x-1) = 0
=> x= -1, 1/3

NOT SUFFICIENT

Option 2 :
y <= -1
So there can be many values of x . x<=-1 or x>=1
NOT SUFFICIENT

Combined :
x= -1,1/3
so y = 1, -1, 1/3, -1/3
But y <=-1
So x = -1 is possible

SUFFICIENT

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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01 Jul 2017, 04:03
Bunuel wrote:
If x^2 = y^2, is true that x>0?

$$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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.

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
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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01 Jul 2017, 04:25
rekhabishop wrote:
Bunuel wrote:
If x^2 = y^2, is true that x>0?

$$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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.

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.
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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01 Jul 2017, 05:04
Bunuel wrote:
rekhabishop wrote:
Bunuel wrote:
If x^2 = y^2, is true that x>0?

$$x^2 = y^2$$ --> $$|x|=|y|$$ --> 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.

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
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If x^2 = y^2, is true that x>0? [#permalink]

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13 Nov 2017, 18:13
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.

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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13 Nov 2017, 20:03
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alekkx wrote:
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.
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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13 Nov 2017, 22:49
Test for -1, 0, 1 condition

x = 2y +1

Here, we actually don't need to do anything. we can't know if x>0 if we don't know the value of y.

2) y<=-1

Answer is either C or E.

1+2)

Let's plug a few values of y that are <= -1 into the equation from situation #1 to see how they affect x:
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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19 Nov 2017, 09:54
Bunuel wrote:
alekkx wrote:
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?

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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19 Nov 2017, 10:01
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Expert's post
alekkx wrote:
Bunuel wrote:
alekkx wrote:
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.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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19 Nov 2017, 10:31
Bunuel wrote:
|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.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----

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.

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Re: If x^2 = y^2, is true that x>0? [#permalink]

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19 Nov 2017, 10:35
alekkx wrote:
Bunuel wrote:
|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.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----

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.

10. Absolute Value

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Kudos [?]: 135660 [0], given: 12705

Re: If x^2 = y^2, is true that x>0?   [#permalink] 19 Nov 2017, 10:35

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