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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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03 May 2014, 00:05
1
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

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Re: PS - Number system  [#permalink]

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03 May 2014, 04:22
PathFinder007 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 $$(|x+y|)^2=(x+y)^2$$ --> $$(|x|-|y|)^2=(x+y)^2$$ --> $$x^2-2|xy|+y^2=x^2+2xy+y^2$$ --> $$|xy|=-xy$$ --> $$xy\leq{0}$$, but as given that $$xy\neq{0}$$, then $$xy<0$$.

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 --> $$xy<0$$.

Hope it helps.

HI Bunnel,

3. --x------0--y----: x<0<y --> |x|-|y|=-x-y and |x+y|=-x-y --> -x-y={-x-y}

could you please clarify above statement. here y is positive so |x+y| should be -x+y

Thanks.

y is positive but x is negative and x is further from zero than y, so x+y is negative hence |x+y|=-(x+y). For example, x=-5 and y=1 --> x+y=-4=negative.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of  [#permalink]

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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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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 $$(|x+y|)^2=(x+y)^2$$ --> $$(|x|-|y|)^2=(x+y)^2$$ --> $$x^2-2|xy|+y^2=x^2+2xy+y^2$$ --> $$|xy|=-xy$$ --> $$xy\leq{0}$$

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 --> $$xy<0$$.

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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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 $$(|x+y|)^2=(x+y)^2$$ --> $$(|x|-|y|)^2=(x+y)^2$$ --> $$x^2-2|xy|+y^2=x^2+2xy+y^2$$ --> $$|xy|=-xy$$ --> $$xy\leq{0}$$

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 --> $$xy<0$$.

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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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 $$(|x+y|)^2=(x+y)^2$$ --> $$(|x|-|y|)^2=(x+y)^2$$ --> $$x^2-2|xy|+y^2=x^2+2xy+y^2$$ --> $$|xy|=-xy$$ --> $$xy\leq{0}$$

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 --> $$xy<0$$.

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

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

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