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Intern  Joined: 16 May 2009
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If |x| - |y| = |x+y| and xy not equal zero , which of the following  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58452
Re: PS - Number system  [#permalink]

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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 $$xy<0$$.

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 --> $$xy<0$$.

Hope it helps.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of  [#permalink]

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I get the mechanics of flipping the signs when y is negative, but I guess I don't understand the logic.

If I take an absolute value of a number (always positive) then subtract from it an absolute value of a smaller number, which is negative how does that end up being X+Y? Are we looking just for the values of x and y, as opposed to the values of |x|-|y|?
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Intern  Joined: 03 Sep 2009
Posts: 28
Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following  [#permalink]

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6
|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

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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Posts: 7984
Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following  [#permalink]

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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]

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5
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

Originally posted by divakarbio7 on 13 Jul 2010, 23:00.
Last edited by Bunuel on 04 Dec 2012, 02:11, edited 1 time in total.
Renamed the topic and edited the question.
Intern  Joined: 09 Oct 2009
Posts: 42
Re: PS - Number system  [#permalink]

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2
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.
Manager  Joined: 06 Apr 2010
Posts: 64
Re: PS - Number system  [#permalink]

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1
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]

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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

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).
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following  [#permalink]

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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]

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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]

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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]

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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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GMAT 1: 710 Q47 V40 Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following  [#permalink]

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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]

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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.

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Re: I need a strategy for this one.  [#permalink]

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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]

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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]

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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 $$xy<0$$.

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 --> $$xy<0$$.

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]

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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]

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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, 05:44

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