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If |x| - |y| = |x+y| and xy not equal zero , which of the following

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If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 09 Sep 2009, 09:11
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If |x| - |y| = |x+y| and xy not equal zero , which of the following must be true ?

A. x-y> 0
B. x-y< 0
C. x+y> 0
D. xy>0
E. xy<0
[Reveal] Spoiler: OA
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 09 Sep 2009, 09:44
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|x| - |y| = |x+y|
=> (|x| - |y|)^2 = |x+y|^2 = (x+y)^2
=> x^2 - 2|x||y| + y^2 = x^2 + 2xy + y^2
=> |xy| = -xy
because xy != 0 so xy <0
Ans: E


Pedros wrote:
if |x| - |y| = |x+y| and xy not equal zero , which of the following must be true ?

a) x-y> 0
b) x-y< 0
c) x+y> 0
d) xy>0
e) xy<0


this one is from manhattan, the answer is
[Reveal] Spoiler:
E


I want to know wether there is a rule involved that i am missing, or an effective strategy to tackle that kind of questions

Thank you
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 09 Sep 2009, 10:35
there can be 4 instances:-
i)x +,y -....|x| - |y| = |x+y| => x-y=x-y
ii) both +... x-y=x+y.. not possible as xy not equal zero
iii) both -....x-y=-(x+y) .. not possible as xy not equal zero
iv) x-,y+...x-y=y-x......x=y..
therefore x and y are of different signs, when multiplied ,it should be -ive
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If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 13 Jul 2010, 22:00
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If |x| - |y| = |x+y| and xy does not equal to 0, which of the following must be true?

A. x-y > 0
B. x-y < 0
C. x+y > 0
D. xy > 0
E. xy < 0

Last edited by Bunuel on 04 Dec 2012, 01:11, edited 1 time in total.
Renamed the topic and edited the question.
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Re: PS - Number system [#permalink] New post 14 Jul 2010, 04:56
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divakarbio7 wrote:
if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?

A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0


\(|x|-|y|=|x+y|\) --> square both sides --> \((|x|-|y|)^2=(|x+y|)^2\) --> note that \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\).

Answer: E.

Another way:

Right hand side, \(|x+y|\), is an absolute value, which is always non-negative, but as \(xy\neq{0}\), then in this case it's positive --> \(RHS=|x+y|>0\). So LHS must also be more than zero \(|x|-|y|>0\), or \(|x|>|y|\).

So we can have following 4 scenarios:
1. ------0--y----x--: \(0<y<x\) --> \(|x|-|y|=x-y\) and \(|x+y|=x+y\) --> \(x-y\neq{x+y}\). Not correct.
2. ----y--0------x--: \(y<0<x\) --> \(|x|-|y|=x+y\) and \(|x+y|=x+y\) --> \(x+y={x+y}\). Correct.
3. --x------0--y----: \(x<0<y\) --> \(|x|-|y|=-x-y\) and \(|x+y|=-x-y\) --> \(-x-y={-x-y}\). Correct.
4. --x----y--0------: \(x<y<0\) --> \(|x|-|y|=-x+y\) and \(|x+y|=-x-y\) --> \(-x+y\neq{-x-y}\). Not correct.

So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) --> \(x\) and \(y\) have opposite signs --> \(xy<0\).

Answer: E.

Hope it helps.
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Re: PS - Number system [#permalink] New post 15 Jul 2010, 04:12
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divakarbio7 wrote:
if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?

A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0


Also, if you're short on time, and since you know that xy is not 0, then you know it MUST be greater than or less than 0, so you can narrow your choices down to D or E.
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Re: PS - Number system [#permalink] New post 01 Sep 2010, 04:03
SnehaC wrote:
divakarbio7 wrote:
if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?

A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0


Also, if you're short on time, and since you know that xy is not 0, then you know it MUST be greater than or less than 0, so you can narrow your choices down to D or E.


the only way lxl - lyl can equal lx+yl is when one number is positive and one number is negative or both are zero. So then xy must be negative. I think I read if somewhere in Bunuel's post
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 14 Jun 2011, 02:39
solution exists only for x,y<0 which is x+y = 0
thus xy < 0
E
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 14 Jun 2011, 20:05
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Pedros wrote:
if |x| - |y| = |x+y| and xy not equal zero , which of the following must be true ?

a) x-y> 0
b) x-y< 0
c) x+y> 0
d) xy>0
e) xy<0


this one is from manhattan, the answer is
[Reveal] Spoiler:
E


I want to know wether there is a rule involved that i am missing, or an effective strategy to tackle that kind of questions

Thank you


mnpqxyzt has given a great solution above. I would like to add here that it is possible that it doesn't occur to you that you should square both sides. If you do get stuck with such a question, notice that it says 'which of the following MUST be true'. So as a back up, you can rely on plugging in numbers. If you get even one set of values for which the condition does not hold, the condition is not your answer.

|x| - |y| = |x+y|
First set of non-zero values that come to mind is x = 1, y = -1
This set satisfies only options (A) and (E).
Now, the set x = -1, y = 1 will also satisfy the given equation.
But this set will not satisfy option (A).
Hence answer (E).
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 18 Jun 2011, 20:49
Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 19 Jun 2011, 03:57
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AnkitK wrote:
Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.


Ok.
Given: |x| - |y| = |x+y|
There are infinite set of values for x and y that satisfy this equation. Let us try one of them. Say x = 1, y = -1
a) x-y> 0 .......... 1 - (-1) > 0; 2 > 0; True
b) x-y< 0........... 1 - (-1) < 0; 2 < 0; Eliminate
c) x+y> 0........... 1 + (-1) > 0; 0 > 0; Eliminate
d) xy>0.............. 1 *(-1) > 0; -1 > 0; Eliminate
e) xy<0.............. 1 *(-1) < 0; -1 < 0; True

So I have two options that satisfy the assumed values of x and y.
We need to eliminate one of them.
They are
a) x-y> 0
e) xy<0

We see that x = 1, y = -1 satisfies both these inequalities. But option (a) is not symmetric i.e. if you interchange the values of x and y, it will not hold. That is, if x = -1 and y = 1, our original equation |x| - |y| = |x+y| is still satisfied but
a) x-y> 0 .............. (-1) - (1) > 0; -2>0; False. Eliminate
e) xy<0................. (-1)(1) < 0; True
Since option (e) still holds, it is the answer.

xy<0 is certainly true when one of them is negative and the other is positive.

Takeaways:

|x| + |y| = |x+y|
when x and y have the same signs - either both are positive or both are negative (or one or both of them are 0)

|x| - |y| = |x+y|
when x and y have opposite signs - one is positive, the other negative (or y is 0 or both are 0)
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 21 Jun 2011, 09:25
VeritasPrepKarishma wrote:
AnkitK wrote:
Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.


Ok.
Given: |x| - |y| = |x+y|
There are infinite set of values for x and y that satisfy this equation. Let us try one of them. Say x = 1, y = -1
a) x-y> 0 .......... 1 - (-1) > 0; 2 > 0; True
b) x-y< 0........... 1 - (-1) < 0; 2 < 0; Eliminate
c) x+y> 0........... 1 + (-1) > 0; 0 > 0; Eliminate
d) xy>0.............. 1 *(-1) > 0; -1 > 0; Eliminate
e) xy<0.............. 1 *(-1) < 0; -1 < 0; True

So I have two options that satisfy the assumed values of x and y.
We need to eliminate one of them.
They are
a) x-y> 0
e) xy<0

We see that x = 1, y = -1 satisfies both these inequalities. But option (a) is not symmetric i.e. if you interchange the values of x and y, it will not hold. That is, if x = -1 and y = 1, our original equation |x| - |y| = |x+y| is still satisfied but
a) x-y> 0 .............. (-1) - (1) > 0; -2>0; False. Eliminate
e) xy<0................. (-1)(1) < 0; True
Since option (e) still holds, it is the answer.

xy<0 is certainly true when one of them is negative and the other is positive.

Takeaways:

|x| + |y| = |x+y|
when x and y have the same signs - either both are positive or both are negative (or one or both of them are 0)

|x| - |y| = |x+y|
when x and y have opposite signs - one is positive, the other negative (or y is 0 or both are 0)


Karishma, I think the 2nd takeaway has some extra condition missed out :

|x|-|y|=|x+y|
if and only if
1) Both have opposite signs and
2) |x| >= |y|

because
x, y, |x|-|y| , |x+y|
5 ,-6, -1,1
5 ,-5 , 0 , 0
5, -4, 1 , 1

so, |5| cannot be greater than |-6| for the condition to occur
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 21 Jun 2011, 10:42
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krishp84 wrote:

Karishma, I think the 2nd takeaway has some extra condition missed out :

|x|-|y|=|x+y|
if and only if
1) Both have opposite signs and
2) |x| >= |y|

because
x, y, |x|-|y| , |x+y|
5 ,-6, -1,1
5 ,-5 , 0 , 0
5, -4, 1 , 1

so, |5| cannot be greater than |-6| for the condition to occur


The takeaway is that if x and y satisfy the condition |x|-|y|=|x+y|, then they must have opposite signs (or y is 0 or both are 0).

But, x and y having opposite signs is not sufficient to satisfy the condition |x|-|y|=|x+y|. As you said, in that case we will also need to check for their absolute values. (Good thinking, btw)
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink] New post 08 Jul 2011, 20:23
I plugged in numbers:

X = -6, Y =3 and X = 6, Y = -3. The answer is E.
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If |x| - |y| = |x+y| and xy does not equal to 0, which of the fo [#permalink] New post 03 Dec 2012, 19:27
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When you see |x| or absolute values being tested in the GMAT, this means testing
(1) positive and negative signs
(2) zeroes
(3) nature of |x| and |y| against each other

Looking at the equation we know |x| - |y| = |x + y| where xy is not 0.
(1) We figure that |x| and |y| are non-zeroes
(2) We figure that |x| - |y| > 0. This means |x| > |y|

Now, we figure the signs of the two variables by lining up possibilities
(1) both negative ==> x=-5,y=-4 ==> |-5| - |-4| = |-9| FALSE
(2) both positive ==> x=5,y=4 ==> |5| - |4| = |5 + 4| FALSE
(3) x is positive, y is negative ==> |5| - |-4| = |5-4| TRUE
(4) y is positive, x is negative ==> |-5| - |4| = |-5+4| TRUE

Now, let's find the answer
A) x-y > 0 ==> -5-(-4) = -1 FALSE
E) xy <0 ==> This is exactly what we need. x and y have both different signs.

Answer: E
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Re: I need a strategy for this one. [#permalink] New post 03 Dec 2012, 23:46
I think the best strategy is to square up the equation.

|x| - |y| = |x+y|

> (|x| - |y|)^2 = (|x+y|)^2

> x^2 + y^2 - 2|x|.|y| = x^2 + y^2 + 2xy

> |x|.|y| = - xy

'Cause xy#0 and |x|.|y|>=0 --> xy<0
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 06 Jan 2013, 18:22
this is specific to Eden,

For your rule. what if X = -1 and Y = 2 then it would be l - 1 l - l 2 l = l -1 + 2 l ... 1 - 2 = 1 incorrect. is this correct?
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Re: PS - Number system [#permalink] New post 10 Feb 2013, 09:31
Bunuel wrote:
divakarbio7 wrote:
if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?

A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0


\(|x|-|y|=|x+y|\) --> square both sides --> \((|x|-|y|)^2=(|x+y|)^2\) --> note that \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\).

Answer: E.

Another way:

Right hand side, \(|x+y|\), is an absolute value, which is always non-negative, but as \(xy\neq{0}\), then in this case it's positive --> \(RHS=|x+y|>0\). So LHS must also be more than zero \(|x|-|y|>0\), or \(|x|>|y|\).

So we can have following 4 scenarios:
1. ------0--y----x--: \(0<y<x\) --> \(|x|-|y|=x-y\) and \(|x+y|=x+y\) --> \(x-y\neq{x+y}\). Not correct.
2. ----y--0------x--: \(y<0<x\) --> \(|x|-|y|=x+y\) and \(|x+y|=x+y\) --> \(x+y={x+y}\). Correct.
3. --x------0--y----: \(x<0<y\) --> \(|x|-|y|=-x-y\) and \(|x+y|=-x-y\) --> \(-x-y={-x-y}\). Correct.
4. --x----y--0------: \(x<y<0\) --> \(|x|-|y|=-x+y\) and \(|x+y|=-x-y\) --> \(-x+y\neq{-x-y}\). Not correct.

So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) --> \(x\) and \(y\) have opposite signs --> \(xy<0\).

Answer: E.

Hope it helps.


Hi Bunuel,
I thought an absolute value of a product can never be a negative. Could you please explain in your equation how did you progress after getting this negative?
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 10 Feb 2013, 17:03
Basically, remember in these exercises to do square root in both sides to simplify: (|x+y|)^2 = (x+y)^2 or (|x|)^2 = x^2
And also remember that: |x||y|=|xy|

Therefore, with: -|x||y| = xy I would say: xy is always equal to something negative. Solution: E.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 12 Mar 2013, 04:44
Let us separate LHS and RHS :
1] |x| - |y| =0 It follows the curve x=y in 1st and 2nd quadrant

2] |x+y|=0. It follows he curve x=-y in 4rth and 2nd quadrant
The common set is 2nd quadrant which implies E with the exception of origin as xy<>0

Am I wrong in my interpretation ? But this is the way I visualize the problem.
Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of   [#permalink] 12 Mar 2013, 04:44

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1 Experts publish their posts in the topic If xy + z = x(y + z), which of the following must be true? above720 3 08 Feb 2007, 20:24
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If |x| - |y| = |x+y| and xy not equal zero , which of the following

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