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If |x| - |y| = |x+y| and xy not equal zero , which of the following

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

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New post 03 May 2014, 00:05
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divakarbio7 wrote:
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


Let us plug in and eliminate:

x = -2 and y = 2 OR x = 2 and y = -2 then |2| - |-2| = |2-2|, let us check the options

A. ELIMINATED (-2 - 2>0)
B. ELIMINATED (2 - (-2) <0)
C. ELIMINATED (2 - 2 > 0)
D. ELIMINATED (-2(2) < 0 )
E. The only one which stays

Hence answer is 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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New post 06 May 2014, 21:00
My Analysis : xy # 0 mean x and y might have either positive or negative Numbers ( Integers)
Concepts : Absolute value , Modules
My Steps : As xy # 0 , So x and y have either positive or negative numbers
By Plugging x and y as positive numbers |x| - |y| = |x+y| the Equation never becomes Equal
So One Number has to be Negative
Ex : x = -6 , y = 3
|-6| - |3| = |-6 + 3|
6 - 3 = |-3 |
3 = 3
So xy = -6*3 = -18

Mean , xy < 0
E

If Any thing is Wrong Pls Reply
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Re: PS - Number system  [#permalink]

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New post 18 Jun 2014, 06:57
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 \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\)

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 ,

Can you please explain how \(|xy|=-xy\) --> \(xy\leq{0}\),
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Re: PS - Number system  [#permalink]

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New post 18 Jun 2014, 08:13
gauravsoni wrote:
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 \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\)

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 ,

Can you please explain how \(|xy|=-xy\) --> \(xy\leq{0}\),


Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html


Hope this 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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New post 18 Jun 2014, 12:41
Bunuel wrote:
gauravsoni 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 \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\)

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 ,

Can you please explain how \(|xy|=-xy\) --> \(xy\leq{0}\),


Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html


Hope this helps.[/quote]

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

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New post 23 Nov 2014, 00:19
Hi,

How did we get this?
Quote:
3. --x------0--y----: x<0<y --> |x|-|y|=-x-y and |x+y|=-x-y --> -x-y={-x-y}. Correct.
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following  [#permalink]

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New post 23 Nov 2014, 05:21
aj0809 wrote:
Hi,

How did we get this?
Quote:
3. --x------0--y----: x<0<y --> |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.


For this case x < 0 < y and x is further from 0 than y is. Now, since x < 0, then |x| = -x and since y > 0, then |y| = y. Also, since x is negative and further from 0 than y, then |x + y|= -(x + y)= - x - y (for example, consider x = -2 and y =1).

Does this make sense?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 25 Jul 2016, 09:37
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

Whatever comes out of |x| will be positive,
whatever comes out of |y| will be positive
since LHS =RHS and RHS is positive
we can say for LHS :- positive - positive = smaller positive
Smaller positive = |x+y| {we are comparing LHS with RHS now as per the equation}
Smaller Positive = ??? WHAT KIND OF POSITIVE ..obviously smaller
SO, The positive on RHS also has to be a smaller positive. Not a big positive
This can be only possible if either x is negative or y is negative because if both x and y have same polarity then subtracting x-y will be smaller positive and x+y will be bigger ; but in that case LHS will not be equal to RHS
so we can infer that both x and y do not have the same polarity either x or y is negative
therefore their product xy will be negative
hence xy<0
ANSWER is E
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Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following  [#permalink]

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