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Re: PS - Number system [#permalink]
14 Jul 2010, 04:56

6

This post received KUDOS

Expert's post

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.

Re: PS - Number system [#permalink]
15 Jul 2010, 04:12

1

This post received KUDOS

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.

Re: PS - Number system [#permalink]
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

If |x| - |y| = |x+y| and xy does not equal to 0, which of the fo [#permalink]
03 Dec 2012, 19:27

2

This post received KUDOS

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.

Re: PS - Number system [#permalink]
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?

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

More:"All I wish someone had told me about GMAT beforehand" There are many things you want to know before doing the GMAT exam (how is exam day, what to expect, how to think, to do's...), and you have them in this blog, in a simple way

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

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

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 14:45

2

This post received KUDOS

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 14:54

1

This post received KUDOS

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} _________________

It is beyond a doubt that all our knowledge that begins with experience.

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

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 15:06

1

This post received KUDOS

Expert's post

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. _________________