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# If |x|-|y|=|x+y|, then which of the following must be true?

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If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  21 Oct 2012, 23:42
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Question Stats:

66% (02:19) correct 33% (02:15) wrong based on 3 sessions
If |x|-|y|=|x+y|, then 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

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree
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Kudos [?]: 277 [0], given: 45

Re: if |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  22 Oct 2012, 00:18
If x & y are both equal to 0. Then none of the options are true. So if want to find which MUST be true then answer is none. Question should be missing some part i guess.

If the question states that x and y are non zero. Then we can see that x and y should be off opposite polarity to satisfy the equation.
Illustration :

x = 5, y = -1

1)true 2)false 3)true 4) false 5) true

x= -5, y = 1

1)false 2)true 3)false 4) false 5)true

So, answer should be

xy<0

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Kudos [?]: 328 [1] , given: 53

Re: if |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  22 Oct 2012, 02:34
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Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.
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Re: if |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  22 Oct 2012, 03:24
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Marcab wrote:
if |x|-|y|=|x+y|, then which of the following must be true?
1) x-y>0
2) x-y<0
3) x+y>0
4) xy>0
5) xy<0

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree

The correct answer is E, but should be xy\leq{0} and not xy<0. Otherwise, none of the answers is correct.
The given equality holds for x=y=0, for which none of the given answers is correct.

The given equality can be rewritten as |x| = |y| + |x + y|.
If y=0, the equality becomes |x|=|x|, obviously true.
From the given answers, D cannot hold, and A,B or C holds, depending on the value of x. Corrected E holds.
If y>0, then necessarily x must be negative, because if x>0, then |x+y|>|x| (x+y>x), and the given equality cannot hold.
If y<0, then necessarily x must be positive, because if x<0, then again |x+y|>|x| (-x-y>-x) and the given equality cannot hold.
It follows that x and y must have opposite signs or y=0.

Answer corrected version of E \,\,xy\leq{0}.
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Kudos [?]: 277 [2] , given: 45

Re: if |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  22 Oct 2012, 03:28
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Marcab wrote:
Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.

Squaring both sides we get :

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

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

So.,

|x||y| = -xy

So -xy is positive (since modulus cannot be negative) and hence xy should be negative.
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Re: if |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  22 Oct 2012, 03:57
Macfauz , the solution was perfect. Thanks.
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Kudos [?]: 328 [0], given: 53

Re: if |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  22 Oct 2012, 04:02
EvaJager wrote:
Marcab wrote:
if |x|-|y|=|x+y|, then which of the following must be true?
1) x-y>0
2) x-y<0
3) x+y>0
4) xy>0
5) xy<0

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree

The correct answer is E, but should be xy\leq{0} and not xy<0. Otherwise, none of the answers is correct.
The given equality holds for x=y=0, for which none of the given answers is correct.

The given equality can be rewritten as |x| = |y| + |x + y|.
If y=0, the equality becomes |x|=|x|, obviously true.
From the given answers, D cannot hold, and A,B or C holds, depending on the value of x. Corrected E holds.
If y>0, then necessarily x must be negative, because if x>0, then |x+y|>|x| (x+y>x), and the given equality cannot hold.
If y<0, then necessarily x must be positive, because if x<0, then again |x+y|>|x| (-x-y>-x) and the given equality cannot hold.
It follows that x and y must have opposite signs or y=0.

Answer corrected version of E \,\,xy\leq{0}.

Many thanks for the explanation.
It will be great if you elaborate on how to solve split modulus questions such as given above.
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Re: if |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  22 Oct 2012, 04:26
MacFauz wrote:
Marcab wrote:
Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.

Squaring both sides we get :

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

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

So.,

|x||y| = -xy

So -xy is positive (since modulus cannot be negative) and hence xy should be negative.

OR 0, that's why the correct answer should be xy\leq{0}.
Otherwise, very nice solution.
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Kudos [?]: 277 [0], given: 45

Re: if |x|-|y|=|x+y|, then which of the following must be true? [#permalink]  22 Oct 2012, 05:01
EvaJager wrote:
MacFauz wrote:
Marcab wrote:
Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.

Squaring both sides we get :

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

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

So.,

|x||y| = -xy

So -xy is positive (since modulus cannot be negative) and hence xy should be negative.

OR 0, that's why the correct answer should be xy\leq{0}.
Otherwise, very nice solution.

I was solving on the basis of my previous comment where I had just added the phrase "where x and y are non zero" to the question.

But seeing as how it is much more probable to leave out a <= sign than an entire sentence, I guess the question frame is right and the answer should be xy <= 0
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Re: if |x|-|y|=|x+y|, then which of the following must be true?   [#permalink] 22 Oct 2012, 05:01
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