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If |x| - |y| = |x+y| and xy does not equal to 0, which of

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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
[Reveal] Spoiler: OA

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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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.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 09 May 2013, 06:15
Interpreting the given

the distance between x and 0 is larger than that between y and 0 and this difference in distance of each of x and y from zero is equal to the distance between x and -y

draw this on a number line

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

this means that x and y has to be on opposite sides from zero , i.e. different signs thus xy<0
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 12 May 2013, 06:56
can we do the following


Generally /x/ - /-y/ <= /x-(-y)/ and the equality hold true only when -xy>o or xy<0 and /x/>/-y/ or /x/>/y/

given xy not = 0 , and given /x/-/y/ = /x+y/ , i.e. /x/ - /-y/ = /x-(-y)/ therefore /x/> /-y/ and -xy>o ,i.e xy<0
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 30 May 2013, 14:25
How does x^2-2|xy|+y^2 become x^2+2xy+y^2? What happened to the -2|xy|? Even if xy were negative |xy| will be positive and pos. * neg (In this case, the -2) = negative, right?

Also, how do we know that xy is negative?

Last edited by WholeLottaLove on 30 May 2013, 14:46, edited 1 time in total.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 30 May 2013, 14:45
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WholeLottaLove wrote:
How does x^2-2|xy|+y^2 become x^2+2xy+y^2? What happened to the -2|xy|? Even if xy were negative |xy| will be positive and pos. * neg (In this case, the -2) = negative, right?


It does not become xy.

|x| - |y| = |x+y| , square both sides (|x| - |y|)^2 = (|x+y|)^2, x^2 - 2|xy|=y^2 = x^2+2xy+y^2, eliminate similar terms and divide by 2
-|xy|=xy from here multiply by -1 just for clarity and obtain |xy|=-xy
remember that |abs|\geq{0} to the equation translates into -xy\geq{0} or xy\leq{0}

The text says that xy\neq{0} so xy\leq{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 30 May 2013, 14:49
but I don't understand how the -2|xy| simplifies to + 2xy.

And why do we multiply by -1?!

Zarrolou wrote:
|x| - |y| = |x+y| , square both sides (|x| - |y|)^2 = (|x+y|)^2, x^2 - 2|xy|=y^2 = x^2+2xy+y^2, eliminate similar terms and divide by 2
-|xy|=xy from here multiply by -1 just for clarity and obtain |xy|=-xy
remember that |abs|\geq{0} to the equation translates into -xy\geq{0} or xy\leq{0}

The text says that xy\neq{0} so xy\leq{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 30 May 2013, 14:54
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WholeLottaLove wrote:
but I don't understand how the -2|xy| simplifies to + 2xy.

And why do we multiply by -1?!



You arrive at -2|xy|=2xy right? Now divide by 2: -|xy|=xy.
From here the -1 multiplication is not necessary to arrive at the result, but IMO it helps.

Keeping in mind that |abs|\geq{0}, we can write -|abs|\leq{0}, right?
The abs expression stands for any value or expression we find inside "| |", xy included.

Now we can merge the equations xy=-|xy| with -|abs|\leq{0} into xy=-|xy|\leq{0} => xy\leq{0}
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 30 May 2013, 15:06
Yes, but why does -2|xy| = 2xy?

and is this example posted by Bunuel...
2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct.

How does |x| - |y| = x+y? wouldn't it be x-y?

I am sorry about the stupid questions. This is an extremely difficult topic for me and I am having a tough time picking it up.

Zarrolou wrote:
WholeLottaLove wrote:
but I don't understand how the -2|xy| simplifies to + 2xy.

And why do we multiply by -1?!



You arrive at -2|xy|=2xy right? Now divide by 2: -|xy|=xy.
From here the -1 multiplication is not necessary to arrive at the result, but IMO it helps.

Keeping in mind that |abs|\geq{0}, we can write -|abs|\leq{0}, right?
The abs expression stands for any value or expression we find inside "| |", xy included.

Now we can merge the equations xy=-|xy| with -|abs|\leq{0} into xy=-|xy|\leq{0} => xy\leq{0}
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 30 May 2013, 15:06
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Another 15 sec approach:

1. The equation isn't sensitive to changing signs of x and y simultaneously. In other words, the equation is the same for (-x,-y). So, A, B, C are out.
2. If x and y had the same sign, |x+y| would be always greater than |x| (and |x| - |y|). So, x and y have different signs and only E remains.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 30 May 2013, 15:12
WholeLottaLove wrote:
Yes, but why does -2|xy| = 2xy?

and is this example posted by Bunuel...
2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct.

How does |x| - |y| = x+y? wouldn't it be x-y?

I am sorry about the stupid questions. This is an extremely difficult topic for me and I am having a tough time picking it up.


The text says so.

You start from here |x| - |y| = |x+y|, square both side ...(refer to the passages above)
Eventually you obtain -|xy|=xy

In the Bunuel's example:
2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct.

x is greater than 0 => |x|=x
y is less than 0 => |y|=-y
So |x|-|y|=x-(-y)=x+y

But those two examples are from two different approaches, do not mix them
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink] New post 30 May 2013, 15:32
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By the way, I would be extremely careful with using the "squaring" approach to solve absolute value problems.

For example, if |x|=-|x|, what is x?

The answer is obvious. x=0

But,

(|x|)^2 = (-|x|)^2 --> x^2 = x^2 --> x can be any numbers (incorrect!).
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of   [#permalink] 30 May 2013, 15:32
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