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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.

Official Answer: C My Answer: A

Please help me to understand why is my position not correct.

Re: If x^2 = y^2, is it true that x > 0? [#permalink]
20 May 2013, 09:50

Expert's post

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. _________________

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.

Official Answer: C My Answer: A

Please help me to understand why is my position not correct.

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

Answer: C.

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.

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

Re: If x^2 = y^2, is true that x>0? [#permalink]
11 Jun 2013, 21:33

Expert's post

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. _________________

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

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

Re: If x^2 = y^2, is true that x>0? [#permalink]
02 Aug 2014, 10:45

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