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Is │x│=│y│? (1) x - y = 6 (2) x + y = 0

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Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 02 Jul 2012, 20:25
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Is │x│=│y│?

(1) x - y = 6
(2) x + y = 0
[Reveal] Spoiler: OA
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 02 Jul 2012, 20:38
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Hi,

Using (1),
x-y=6
x=3, y=-3
then, |x|=|y|
but when x=9, y=3
\(|x| \neq |y|\). Insufficient.

Using (2),
x+y=0
or x = -y, when y=1, x=-1
and |x|=|y|, Sufficient.

Thus, Answer (B)

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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 02 Jul 2012, 22:38
Expert's post
Is │x│=│y│?

Is \(|x|=|y|\)? The question basically asks whether \(x=y\) or \(x=-y\)

(1) x - y = 6. If \(x=3\) and \(y=-3\) then the answer is YES but if \(x\) and \(y\) are some other values then the answer is NO.

(2) x + y = 0. Rearrange: \(x=-y\). Sufficient.

Answer: B.
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 16 May 2013, 08:34
catennacio wrote:
Is │x│=│y│?

(1) x - y = 6
(2) x + y = 0




x= y + 6 stmt1

x= -y -> !x! = !y! so stmt1 is the answer B is the choice :)
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 12 Jun 2013, 11:07
My first instinct was to manipulate |x|=|y|

x=y
OR
x=-y

1.) says that x-y=6 which means we can get values for x and y

x=6+y
y=x-6

So, for x=y

6+y=y
6=0 (Invalid)

x=x-6
0=6 (Invalid)

So, for x=-y

6+y=-y
y=-3

x=-x+6
x=3

Why wouldn't we use that methodology on this problem?
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 12 Jun 2013, 11:27
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WholeLottaLove wrote:
My first instinct was to manipulate |x|=|y|

x=y
OR
x=-y

1.) says that x-y=6 which means we can get values for x and y

x=6+y
y=x-6

So, for x=y

6+y=y
6=0 (Invalid)

x=x-6
0=6 (Invalid)

So, for x=-y

6+y=-y
y=-3

x=-x+6
x=3

Why wouldn't we use that methodology on this problem?


Are you saying that A is sufficient?

In the case \(x=3\) and \(y=-3\) => \(|x|=|y|\).

But if \(x=90\) and \(y=84\) then x - y = 6 but \(|x|\neq{|y|}\).

The question asks you if x=y OR x=-y, you cannot assume that it's true in your solution to find the values of x,y for which it holds ture.
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 12 Jun 2013, 11:43
I should have worded that better.

I guess that solving the problem they way I did, I would say A is correct.

I see that for x-y=6 there could be many values for x and y, however, it seems like many similar problems are solved by manipulating statements (i.e. x-y=6) and plugging into x=y or x=-y.

I understand how this problem was solved, but I want to understand WHY it was solved the way it was.

As always, thank you for for help.


Zarrolou wrote:
WholeLottaLove wrote:
My first instinct was to manipulate |x|=|y|

x=y
OR
x=-y

1.) says that x-y=6 which means we can get values for x and y

x=6+y
y=x-6

So, for x=y

6+y=y
6=0 (Invalid)

x=x-6
0=6 (Invalid)

So, for x=-y

6+y=-y
y=-3

x=-x+6
x=3

Why wouldn't we use that methodology on this problem?


Are you saying that A is sufficient?

In the case \(x=3\) and \(y=-3\) => \(|x|=|y|\).

But if \(x=90\) and \(y=84\) then x - y = 6 but \(|x|\neq{|y|}\).

The question asks you if x=y OR x=-y, you cannot assume that it's true in your solution to find the values of x,y for which it holds ture.
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 12 Jun 2013, 11:48
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Expert's post
WholeLottaLove wrote:
My first instinct was to manipulate |x|=|y|

x=y
OR
x=-y

1.) says that x-y=6 which means we can get values for x and y

x=6+y
y=x-6

So, for x=y

6+y=y
6=0 (Invalid)

x=x-6
0=6 (Invalid)

So, for x=-y

6+y=-y
y=-3

x=-x+6
x=3

Why wouldn't we use that methodology on this problem?


When you assume that x=y and solve the equation [x-y = 6], you WILL get invalid solutions as you got because you have anyways assumed that x=y-->
x-y = 0;which contradicts the given fact.

However, when you assume that x=-y-->x+y=0, you have inherently assumed that |x| IS equal to |y| and now you are just solving for the values of x and y. Thus, the equation [x-y=6] would really not make any difference for this method.( x-y) could equal anything and you would still get |x| = |y|.
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 12 Jun 2013, 11:52
Ahhh! That makes sense! I am assuming |x| = |y| when I am trying to verify if it does or does not. Thanks!

vinaymimani wrote:
WholeLottaLove wrote:
My first instinct was to manipulate |x|=|y|

x=y
OR
x=-y

1.) says that x-y=6 which means we can get values for x and y

x=6+y
y=x-6

So, for x=y

6+y=y
6=0 (Invalid)

x=x-6
0=6 (Invalid)

So, for x=-y

6+y=-y
y=-3

x=-x+6
x=3

Why wouldn't we use that methodology on this problem?


When you assume that x=y and solve the equation [x-y = 6], you WILL get invalid solutions as you got because you have anyways assumed that x=y-->
x-y = 0;which contradicts the given fact.

However, when you assume that x=-y-->x+y=0, you have inherently assumed that |x| IS equal to |y| and now you are just solving for the values of x and y. Thus, the equation [x-y=6] would really not make any difference for this method.( x-y) could equal anything and you would still get |x| = |y|.
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 12 Jun 2013, 11:53
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Yes Wotta you should manipolate the statement to understand the question better.

Is \(|x|=|y|\)? it s like saying is x=y or x=-y?

But once you start analyzing the statement you cannot use that info.
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0 [#permalink] New post 01 Mar 2015, 14:44
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Re: Is │x│=│y│? (1) x - y = 6 (2) x + y = 0   [#permalink] 01 Mar 2015, 14:44
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