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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 it true that x > 0? [#permalink]

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19 May 2013, 12:46
yezz wrote:
... and thus y is either -1/3 or y = -1

I believe -1/3 is discounted in the question where it states x^2 = y^2.

If y = -1/3 then x = -2/3; these values for x and y contradict the question. y = -1/3 is rejected, leaving only y = -1.

Why is my position not correct?

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

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20 May 2013, 03:32
Bunuel wrote:
Please refer to the solutions provided.

That is what I am doing, and the solutions provided sometimes appear to me to be incorrect. This is one of those times.

Here is my working:

stormbind wrote:
If x^2 = y^2, is it true that x > 0?

1. x = 2y + 1
2. y <= -1

I answered "Statement 1 alone is sufficient"; Can you please show why my working is not correct?

x^2 = y^2
x^2 = (2y + 1)(2y + 1)

If x=1 or x=-1 then x^2 = 1.

1 = (2y + 1)(2y + 1)

Using number substitution, only -1 is substitutable for y without breaking the x^2 = y^2 statement in the question:

i.e. (2*-1 + 1)(2*-1 + 1) = (4 - 2 -2 + 1) = 1 = x^2.

Therefore y = -1

x = 2y + 1; so x = -1 also.

Statement 1 is sufficient. Where did I go wrong?

Thanks.

Official Solution states Statement 1 is insufficient because it renders y = -1/3 or y = -1. However, I believe that -1/3 is not a possible value for y because the question states x^2 = y^2.

Explanation: If y = -1/3 then x = -2/3, which contradicts the question. Thus -1/3 is rejected, leaving only y = -1. This would make Statement 1 sufficient.

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

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20 May 2013, 09:50
stormbind wrote:
Bunuel wrote:
Please refer to the solutions provided.

Explanation: If y = -1/3 then x = -2/3, which contradicts the question. Thus -1/3 is rejected, leaving only y = -1. This would make Statement 1 sufficient.

This part is not correct. x = 2y+1. If y = -1/3, x is not equal to -2/3.
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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20 May 2013, 12:20
Thanks vinaymimani!

Not seeing that was driving me round the bend.

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

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20 May 2013, 14:43
(1) X = 2y + 1
$$x^2 = 4y^2 + 1 + 4y$$
$$y^2 = 4y^2 + 1 + 4y$$
$$3y^2 + 4y + 1 = 0$$
$$3y^2 + 3y +y +1 = 0$$
$$(3y+1)(y+1) = 0$$
y = -1/3 or -1
x = 1/3 (Yes) or -1 (No) – Insufficient

(2) Alone is insuff –

Combining – we know Y = -1
Therefore x = -1 and thus NO it is not greater than 0
IMO (C)
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Re: If x^2 = y^2, is it true that x > 0? [#permalink]

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20 May 2013, 23:56
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Expert's post
stormbind wrote:
Bunuel wrote:
Please refer to the solutions provided.

That is what I am doing, and the solutions provided sometimes appear to me to be incorrect. This is one of those times.

Here is my working:

stormbind wrote:
If x^2 = y^2, is it true that x > 0?

1. x = 2y + 1
2. y <= -1

I answered "Statement 1 alone is sufficient"; Can you please show why my working is not correct?

x^2 = y^2
x^2 = (2y + 1)(2y + 1)

If x=1 or x=-1 then x^2 = 1.

1 = (2y + 1)(2y + 1)

Using number substitution, only -1 is substitutable for y without breaking the x^2 = y^2 statement in the question:

i.e. (2*-1 + 1)(2*-1 + 1) = (4 - 2 -2 + 1) = 1 = x^2.

Therefore y = -1

x = 2y + 1; so x = -1 also.

Statement 1 is sufficient. Where did I go wrong?

Thanks.

Official Solution states Statement 1 is insufficient because it renders y = -1/3 or y = -1. However, I believe that -1/3 is not a possible value for y because the question states x^2 = y^2.

Explanation: If y = -1/3 then x = -2/3, which contradicts the question. Thus -1/3 is rejected, leaving only y = -1. This would make Statement 1 sufficient.

Consider y=-x=-1/3 --> x^2=y^2 and x=2y+1.

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

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11 Jun 2013, 15:57
Why do we have to solve for 'y' when we are looking for the value of x? I know how to get x=-1 and x=1/3 but I'm still not sure why the values of y are relevant.

Bunuel wrote:
kotela 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.

Hi Bunnel,

I need a clarification here, when i put statement 1 in X^2 = Y^2 i get the below

3Y^2= -4Y-1,

and since LHS is positive so the RHS needs to be positive so in order to RHS to be positive Y should be "-ve" and when Y is negative, Now

considering statement 1 we can get to know that X is negative.....

So my point is again statement 2 needed?

Please clarify me if i am wrong?

How did you get that if y is negative x must be positive?

For (1) we have:
$$x^2 = y^2$$ and $$x=2y+1$$. Solving gives: $$x=-1$$ and $$y=-1$$ OR $$x=\frac{1}{3}$$ and $$y=-\frac{1}{3}$$, just substitute these values to check that they satisfy both equations.

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

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11 Jun 2013, 17:03
Can someone tell me if my reasoning is sound?

x^2=y^2 and thus, x=y or x=-y

1.)

I.) x=2y+1

y=2y+1
-y=1
y=-1

y=2(-1)+1
x=-1

II.) x=2y+1

-y=2y+1
-3y=1
y=-1/3

x=2(-1/3)+1
x=1/3

Insufficient because we have two values for x. (however, using the same reasoning to arrive at c. cant we say that both answers we arrive at show that x ≠ 0 and thus, a is sufficient?)

2. y≤-1 Not sufficent

1+2 2. says that y≤-1. In #1, the only case where y≤-1 is y=2y+1

-y=1
y=-1

y=2(-1)+1
x=-1

Here we get an answer of x=-1 which is obviously ≠ to 0.

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

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11 Jun 2013, 21:33
WholeLottaLove wrote:
Can someone tell me if my reasoning is sound?

x^2=y^2 and thus, x=y or x=-y

1.)

I.) x=2y+1

y=2y+1
-y=1
y=-1

y=2(-1)+1
x=-1

II.) x=2y+1

-y=2y+1
-3y=1
y=-1/3

x=2(-1/3)+1
x=1/3

Insufficient because we have two values for x. (however, using the same reasoning to arrive at c. cant we say that both answers we arrive at show that x ≠ 0 and thus, a is sufficient?)

2. y≤-1 Not sufficent

1+2 2. says that y≤-1. In #1, the only case where y≤-1 is y=2y+1

-y=1
y=-1

y=2(-1)+1
x=-1

Here we get an answer of x=-1 which is obviously ≠ to 0.

Everything you did is correct except that you misunderstood the question.
The question is:

Is x positive? Is x > 0?
It does not ask you whether x is equal to 0.

Statement I tells you that x could be positive or negative. So not sufficient.
Both statements together tell you that x is negative. Hence it is not positive. It answers the question with 'No'. Sufficient.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18473 [0], given: 237 Manager Joined: 03 Mar 2013 Posts: 85 Kudos [?]: 10 [0], given: 6 Location: India Concentration: General Management, Marketing GPA: 3.49 WE: Web Development (Computer Software) Re: If x^2 = y^2, is true that x>0? [#permalink] ### Show Tags 18 Jun 2013, 09:33 burnttwinky wrote: If x^2 = y^2, is true that x>0? (1) x=2y+1 (2) y<= -1 S1: put x= y once and -y once so we get 1/2 and -1 two values hence NS now s2: NS s1 and s2 : sufficient and answer is C Questions above 700 are either direct math or pure logic Kudos [?]: 10 [0], given: 6 Senior Manager Joined: 13 May 2013 Posts: 456 Kudos [?]: 211 [0], given: 134 Re: If x^2 = y^2, is true that x>0? [#permalink] ### Show Tags 20 Jun 2013, 15:45 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? 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! Kudos [?]: 211 [0], given: 134 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7866 Kudos [?]: 18473 [1], given: 237 Location: Pune, India Re: If x^2 = y^2, is true that x>0? [#permalink] ### Show Tags 20 Jun 2013, 19:55 1 This post received KUDOS Expert's post 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 _________________ 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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21 Jun 2013, 19:49
Correct answer is E not C
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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22 Jun 2013, 02:40
trafficspinners wrote:
Correct answer is E not C

The correct answer is C, check under the spoiler in the original post.
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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27 Jun 2013, 12:05
If x^2 = y^2, is true that x>0?

If x^2 = y^2, then we can get the square root of both sides:
√x^2 = √y^2
|x| = |y|
x=y
OR
x=-y

1) x=2y+1
(2) y<= -1

1) x=2y+1
So, we are looking for the value of x.
x=y or -x=y

I.) x=2y+1
x=2x+1
-x=1
x=-1

II.) x=2y+1
x=-2x+1
3x=1
x=1/3

x could be greater than zero or less than zero.
NOT SUFFICIENT

(2) y<= -1

x could = y or -y. For example.
|x| = |y|
|2| = |-2|
OR
|-2| = |-2|

NOT SUFFICIENT

1 + 2) x=2y+1 and y<= -1

In 1) we have two cases, where x is negative and where x is positive. However, when x is negative y is positive and when x is positive y is negative. 2) tells us that y is negative meaning x must be positive or greater than zero.
SUFFICIENT

(c)

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

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06 Sep 2014, 12:04
If x^2=y^2, is it true that X>0?
|x|=|y| ---> x=y or x=-y

1. x=2y+1, x is also equal to y. --> 2y+1=y or y=-1. If y=-1, x=2(-1)+1=-2+1=-1
x=2y+1, x is also equal to -y. --> 2y+1=-y or y=-(1/3). If y=-(1/3), x=-(2/3)+1=(-2+3)/3=1/3
we see that x is positive in second case and not in the first. NOT SUFFICIENT

2. y<=-1. This one is easy to notice that x can be anything as long as its magnitude is same as y's.

1+2: y can only be -1 from combining these two statement. This is also our case 1 from statement 1. Hence, x=-1<0. SUfficient

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

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12 Nov 2014, 04:48
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.

Hi Bunuel

From x^2=y^2
x and y can take the following values:
1 and -1 or
-1 and 1
as x and y different so not considering same integers like both positive or both negative

from 1st statement none of the values satisfy the given equation
from 2nd statement, y is negative which makes x as positive. Where am I wrong?

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

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12 Nov 2014, 05:29
sinhap07 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.

Hi Bunuel

From x^2=y^2
x and y can take the following values:
1 and -1 or
-1 and 1
as x and y different so not considering same integers like both positive or both negative

from 1st statement none of the values satisfy the given equation
from 2nd statement, y is negative which makes x as positive. Where am I wrong?

x and y can be the same number. Generally, unless it is explicitly stated otherwise, different variables CAN represent the same number.

So, for (2) it's possible that x = y = -1, or x= 1 and y = -1, or x = y = -2, or x = y = -1.5, or x = 100 and y = -100...
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Re: If x^2 = y^2, is true that x>0? [#permalink]

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12 Nov 2014, 09:39
Bunuel wrote:
sinhap07 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.

Hi Bunuel

From x^2=y^2
x and y can take the following values:
1 and -1 or
-1 and 1
as x and y different so not considering same integers like both positive or both negative

from 1st statement none of the values satisfy the given equation
from 2nd statement, y is negative which makes x as positive. Where am I wrong?

x and y can be the same number. Generally, unless it is explicitly stated otherwise, different variables CAN represent the same number.

So, for (2) it's possible that x = y = -1, or x= 1 and y = -1, or x = y = -2, or x = y = -1.5, or x = 100 and y = -100...

Agreed Bunuel. Tried that too. But by that logic, Statement 1 holds true only when both x and y are negative and hence we could get A as the answer as we would know that y is negative along with x.

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

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12 Nov 2014, 09:48
sinhap07 wrote:
Bunuel wrote:
sinhap07 wrote:

Hi Bunuel

From x^2=y^2
x and y can take the following values:
1 and -1 or
-1 and 1
as x and y different so not considering same integers like both positive or both negative

from 1st statement none of the values satisfy the given equation
from 2nd statement, y is negative which makes x as positive. Where am I wrong?

x and y can be the same number. Generally, unless it is explicitly stated otherwise, different variables CAN represent the same number.

So, for (2) it's possible that x = y = -1, or x= 1 and y = -1, or x = y = -2, or x = y = -1.5, or x = 100 and y = -100...

Agreed Bunuel. Tried that too. But by that logic, Statement 1 holds true only when both x and y are negative and hence we could get A as the answer as we would know that y is negative along with x.

That's not correct.

For (1) it's possible that x = -1 and y = -1 or x = 1/3 and y = -1/3.

The correct answer is C, not B or A.
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Re: If x^2 = y^2, is true that x>0?   [#permalink] 12 Nov 2014, 09:48

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