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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 00:18

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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

Answer is E
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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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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]

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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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PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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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]

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22 Oct 2012, 04:02

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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]

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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.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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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]

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14 Jan 2015, 10:21

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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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19 May 2016, 04:56

Hello from the GMAT Club BumpBot!

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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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08 Jun 2016, 20:54

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 \(\,\,xy\leq{0}\).

I think xy=0 wouldn't be the answer because when xy = 0 , either only x or only y can also be zero. Then if x=0 but y is not, then it won't hold true for xy=0. Let me know if I am wrong

Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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18 Jun 2017, 01:07

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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

If lxl - lyl = lx+yl and 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 --> \(xy<0\).

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