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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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Re: If x^2 = y^2, is true that x>0? [#permalink]
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
Re: If x^2 = y^2, is true that x>0? [#permalink]
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.
Answer: C.
Hope it's clear.
Hi Bunuel
Why cant answer be B
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?
Re: If x^2 = y^2, is true that x>0? [#permalink]
12 Nov 2014, 05:29
Expert's post
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.
Answer: C.
Hope it's clear.
Hi Bunuel
Why cant answer be B
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... _________________
Re: If x^2 = y^2, is true that x>0? [#permalink]
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.
Answer: C.
Hope it's clear.
Hi Bunuel
Why cant answer be B
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.
gmatclubot
Re: If x^2 = y^2, is true that x>0?
[#permalink]
12 Nov 2014, 09:39
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