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

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23 Jan 2017, 00:55
1
dimitri92 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 & 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?

Basically there are two distinct strategies - Number plugging (for example a percentages question in which one set will certainly give the answer) and Number testing (such as this question in which you need to test for various values)

I swear by number plugging but number testing is almost never my strategy of choice. I might want to try one set to understand the question/logic better but I will rarely try to deduce my answer from number testing.
So when I see x^2 = y^2, I automatically think
|x| = |y| which implies x = y or x = -y
or x^2 - y^2 = (x + y) * (x - y) which implies x = -y or x = y

This makes me jump to stmnt 2 immediately since it is clearly insufficient. This brings me to (C) or (E).
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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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27 Jun 2017, 08:49
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.

Hello Bunuel,

Why did u take only two possibilities?

x2=y2x2=y2 --> |x|=|y||x|=|y| --> either y=xy=x or y=−xy=−x

we can have -x = - y and -x = y

I'm I missing something?

Posted from my mobile device
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Posts: 46319
Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1 [#permalink]

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27 Jun 2017, 09:48
ydmuley 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.

Hello Bunuel,

Why did u take only two possibilities?

x2=y2x2=y2 --> |x|=|y||x|=|y| --> either y=xy=x or y=−xy=−x

we can have -x = - y and -x = y

I'm I missing something?

Posted from my mobile device

-x = -y is the same as x = y and -x = y is the same as x = -y.
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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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27 Jun 2017, 23:55
burnttwinky wrote:
If x^2 = y^2, is true that x>0?

(1) x=2y+1

(2) y<= -1

Given : x^2 = y^2 => x=+/-y
DS : x>0

Option 1 : x = 2y+1
x^2 = [(x-1)/2]^2
=> 4x^2 = x^2 +1 -2x
=> 3x^2 +2x -1 = 0
=> (x+1) (3x-1) = 0
=> x= -1, 1/3

NOT SUFFICIENT

Option 2 :
y <= -1
So there can be many values of x . x<=-1 or x>=1
NOT SUFFICIENT

Combined :
x= -1,1/3
so y = 1, -1, 1/3, -1/3
But y <=-1
So x = -1 is possible

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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01 Jul 2017, 05:03
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.

In the following statement,

(1)+(2) Since from (2) $$y\leq{-1}$$ then from (1) $$y=x=-1$$, so the answer to the question is NO. Sufficient.

what if, y=-x=-1 ?
Then x=1>0 and hence (1)+(2) = insufficient
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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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01 Jul 2017, 05:25
rekhabishop 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.

In the following statement,

(1)+(2) Since from (2) $$y\leq{-1}$$ then from (1) $$y=x=-1$$, so the answer to the question is NO. Sufficient.

what if, y=-x=-1 ?
Then x=1>0 and hence (1)+(2) = insufficient

From (1) we have only two possibilities:
a. $$y=x=-1$$
b. $$y=-x=-\frac{1}{3}$$

y = -1 and x = 1 does not satisfy the first statement.
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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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13 Nov 2017, 19:13
Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.
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Posts: 46319
Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1 [#permalink]

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13 Nov 2017, 21:03
1
alekkx wrote:
Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.

You could use number plugging to check whether you are right.

x^2 = y^2:

The above could be true for example if x = y = -1 or if x = 1 and y = -1. These examples answer no to both of your question. Generally x^2 = y^2 means that |x| = |y|, so x and y are at the same distance from 0 on the number line.
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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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19 Nov 2017, 10:54
Bunuel wrote:
alekkx wrote:
Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.

You could use number plugging to check whether you are right.

x^2 = y^2:

The above could be true for example if x = y = -1 or if x = 1 and y = -1. These examples answer no to both of your question. Generally x^2 = y^2 means that |x| = |y|, so x and y are at the same distance from 0 on the number line.

Thanks. What are the implications of |y|<|x|, is it "y<x or y>-x" or something different?
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Posts: 46319
Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1 [#permalink]

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19 Nov 2017, 11:01
1
alekkx wrote:
Bunuel wrote:
alekkx wrote:
Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.

You could use number plugging to check whether you are right.

x^2 = y^2:

The above could be true for example if x = y = -1 or if x = 1 and y = -1. These examples answer no to both of your question. Generally x^2 = y^2 means that |x| = |y|, so x and y are at the same distance from 0 on the number line.

Thanks. What are the implications of |y|<|x|, is it "y<x or y>-x" or something different?

|x| is the distance from x to 0 on the number line, so |y| < |x| means that x if further from 0 than y is.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----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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19 Nov 2017, 11:31
Bunuel wrote:
|x| is the distance from x to 0 on the number line, so |y| < |x| means that x if further from 0 than y is.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----

Thanks Bunuel. It looks like I did not capture all cases with my inequalities. For example, when y>0, I have y is always less than x, which is not true based on case 2. When faced with this condition on DS, is the best way to draw out the number line like this? How does this compare to getting a condition that says |x-5|>2 in terms of how you would get the constraints.
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Posts: 46319
Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1 [#permalink]

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19 Nov 2017, 11:35
alekkx wrote:
Bunuel wrote:
|x| is the distance from x to 0 on the number line, so |y| < |x| means that x if further from 0 than y is.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----

Thanks Bunuel. It looks like I did not capture all cases with my inequalities. For example, when y>0, I have y is always less than x, which is not true based on case 2. When faced with this condition on DS, is the best way to draw out the number line like this? How does this compare to getting a condition that says |x-5|>2 in terms of how you would get the constraints.

10. Absolute Value

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Re: If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1   [#permalink] 19 Nov 2017, 11:35

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