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

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If x^2 = y^2, is true that x>0? [#permalink] New post 20 May 2012, 18:02
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If x^2 = y^2, is true that x>0?

(1) x=2y+1

(2) y<= -1
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 20 May 2012, 23:38
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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.
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 23 May 2012, 11:02
picked C. if x^2 = y^2 means that |x| = |y| with statement 1 considering when y is negative only if y =-1 mod x and mod y will be equal (really both will be -1). (if y is positive x will not be equal to y) statement B says y less than or equal to -1 i.e. x =-1 hence we have a unique ans.
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 26 May 2012, 04:09
for those of you who hate absolute value:
Given x^2=y^2
Is x>0?
1. y=(x-1)/2, x^=(x-1)^2/4. Solving for x [1/3, -1]. Not sufficient
2. y<=-1 Nothing about x. Not Sufficient.
Together: (x-1)/2 <= -1, x<= -1
Answer is NO. C.
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If x^2 = y^2, is it true that x > 0? [#permalink] New post 18 Jun 2012, 15:44
If x^2 = y^2, is it true that x > 0?

(1) x = 2y + 1

(2) y<= -1


I got the correct answer, but I got stuck. Below is my solution.

When x ^ 2 = Y ^2 then x = y and x = -y

Considering statement 1

x = 2y + 1------------when x = y then y = -1 and x = -1 i.e negative.

When y =1 then x is positive. Therefore, insufficient.

Considering statement 2

if x = y then x < = -1 i.e. negative

if x = -y then -x <=-1 or x >=1 i.e. positive . Therefore, insufficient.

Now I am strugling to combine the two and get the answer. can someone please help?
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Re: Is x > 0? [#permalink] New post 18 Jun 2012, 16:16
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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: Is x > 0? [#permalink] New post 19 Jun 2012, 00:10
Here's my take on the problem

Statement (1) is clearly insufficient. Matter of fact, it tells us that x is odd. LOL wtf?

Statement (2) is clearly insufficient since there's no mention of x.

Now let's see what we have:

X = 2y + 1

and

y < - 1

I then make y < - 1 into y = Lessthan(-1) <--- I am just translating it. Don't panic.

Now:

plug in y into x = 2y + 1

we have

x = 2(lessthan -1) + 1

x = (less than -2) + 1 <--- by simple arithmetic

x = less than -1

which means that x < -1

Now we can conclude that x is indeed less than 0 :) Sufficient.

*by the way, vandygrad11's way is also superb. need to familiarize ourselves in both ways.

P.S. haven't replied to vandygrad11's reply yet in re: Big4 reputation. will do so when I get home.

Cheers!
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 28 Aug 2012, 09:13
Sorry to open an old thread...
if we solve like this...

X^2 =Y^2

from 1st
x=2y+1
if we plug in the value in the above equation for x
we get
(2y+1)^2=y^2
4y^2 +4y+1 =y^2
3y^2 +4y +1=0
3y^2 +3y+y+1=0
3y(y+1)+(y+1)=0
(3y+1)(y+1)=0
we get 2 values of y for which x=4y+1 and x^2=y^2
i.e y=-1/3 and -1
and x = 1/3 and -1 respectively

the second condition tells us that
y<= -1
hence y can be both -1/3 and -1
and corresponding x values would be -1 and 1/3
a positive and negative
Hence, answer should be E

am i wrong somewhere...?
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 29 Aug 2012, 00:48
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venom2330 wrote:
Sorry to open an old thread...
if we solve like this...

X^2 =Y^2

from 1st
x=2y+1
if we plug in the value in the above equation for x
we get
(2y+1)^2=y^2
4y^2 +4y+1 =y^2
3y^2 +4y +1=0
3y^2 +3y+y+1=0
3y(y+1)+(y+1)=0
(3y+1)(y+1)=0
we get 2 values of y for which x=4y+1 and x^2=y^2
i.e y=-1/3 and -1
and x = 1/3 and -1 respectively

the second condition tells us that
y<= -1
hence y can be both -1/3 and -1
and corresponding x values would be -1 and 1/3
a positive and negative
Hence, answer should be E

am i wrong somewhere...?


-1/3 is greater than -1, so y cannot be -1/3.

Hope it helps.
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 30 Aug 2012, 19:17
i am struggling with dsinequalities

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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 31 Aug 2012, 06:48
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 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? [#permalink] New post 31 Aug 2012, 07:17
Expert's post
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.
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 31 Aug 2012, 07:41
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.


Hi Bunnel

I made a blunder by not considering -1/3, anyway thanks for the quick response..........
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 20 Jan 2013, 20:14
great solution, thanks bunuel.
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 02 Mar 2013, 09:38
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!

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? [#permalink] New post 02 Mar 2013, 13:10
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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.

Answer-C

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.

Answer: C.

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? [#permalink] New post 10 May 2013, 17:02
burnttwinky wrote:
If x^2 = y^2, is true that x>0?

(1) x=2y+1

(2) y<= -1


My simple approach to solve the question in minimum time.

Pick 2 first as it looks straightforward condition. Clearly it's not sufficient.
Now look at 1, doesn't help in determining the sign of x.

Combine 1 and 2:

(x-1)/2 <=-1
x <=-1.
So x can't be positive. Hence C.
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Re: If x^2 = y^2, is true that x>0? [#permalink] New post 10 May 2013, 21:50
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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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If x^2 = y^2, is it true that x > 0? [#permalink] New post 19 May 2013, 08:47
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.
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Re: If x^2 = y^2, is it true that x > 0? [#permalink] New post 19 May 2013, 10:08
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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
Re: If x^2 = y^2, is it true that x > 0?   [#permalink] 19 May 2013, 10:08
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