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Manager  Joined: 25 Jun 2009
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If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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

(1) x = 2y+1

(2) y <= -1
Math Expert V
Joined: 02 Sep 2009
Posts: 58320
Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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17
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.
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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2
The way I usually approach these problems is with plugging in examples, like -1, 0, and 1 to see when the equations hold true.

1)
x = 2y +1

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

2) y<=-1

Alone, this obviously tells us nothing about x! Insufficient. 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:

x=2y+1
x=2(-1)+1 = -1
x=2(-2)+1 = -3
x=2(-3)+1 = -5
Obviously, this will continue as a series.

Therefore, we clearly know that x will never be >0 and therefore, C is the answer.
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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.

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?
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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.

Hi!

I did a bit of a unnecessarily long way but got to the wrong answer and was hoping someone could tell me what my logical error is.

i squared both sides of statement 1 and got to $$x^2 = 4y^2 + 4y +1$$

then I replaced $$x^2$$ with $$y^2$$ and got the same as everyone else that y = -1 or y = -(1/3) NOT sufficient

Using statement B i eliminated y=-(1/3) but where I got it wrong is that I thought that since y = -1 then x could be equal to +/- 1 so I chose E

Am i missing something?

Thank you in advance or any responses!
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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Hello Alexpavlos,

You are right in assuming that since y=-1 and x^2=y^2, x=+/-1. However, remember that equation in statement 1 also gives you a relationship between statement x and y. The correct answer should satisfy all the data. Now the question you should ask yourself is whether it is possible to use the information in statement 2 and 1 together to get a single value for x.

Statement 1 mentions that
x=2y+1
Substituting y=-1 in the this equation, we get

x=-1 and hence, x<0. Hence, together the two statements suffice.

Hope this helps! Let me know if you need any further clarification.

alexpavlos 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!

I did a bit of a unnecessarily long way but got to the wrong answer and was hoping someone could tell me what my logical error is.

i squared both sides of statement 1 and got to $$x^2 = 4y^2 + 4y +1$$

then I replaced $$x^2$$ with $$y^2$$ and got the same as everyone else that y = -1 or y = -(1/3) NOT sufficient

Using statement B i eliminated y=-(1/3) but where I got it wrong is that I thought that since y = -1 then x could be equal to +/- 1 so I chose E

Am i missing something?

Thank you in advance or any responses!
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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1
burnttwinky wrote:
If x^2 = y^2, is true that x>0?

(1) x=2y+1

(2) y<= -1

Nothing new to add. Maybe another approach:

From F.S 1, we know that x = 2y+1.
Also $$x^2 = y^2$$
Upon adding,$$x^2+x = (y+1)^2$$.Thus, x(x+1)> 0--> x>0 OR x<-1.Insufficient.

From F.S 2, we know that y<-1. Clearly Insufficient.

On combining both, we know that y = (x-1)/2 --> (x-1)/2<-1-->x<-1. Sufficient.

C.
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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

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

from 1

x^2 = 4y^2 + 4y+1 so now 4y^2 + 4y+1 = y^2 therefore 3y^2 + 4y+1 = 0 ..... (3y + 1) (y+1) = 0 and thus y is either -1/3 or y = -1

subst in 1

x = -2/3 + 1 > 0 or x = -2+1<0 ......insuff

from 2 alone obviously not suff

both ........ y = -1 ........x = -1 too ....suff.......C
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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!
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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1
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
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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.

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?
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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

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...
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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

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.
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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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sinhap07 wrote:
Bunuel wrote:
sinhap07 wrote:

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.

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 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1  [#permalink]

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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.

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? Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1   [#permalink] 20 Jan 2017, 03:42

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