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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 12 May 2013, 06:56
can we do the following


Generally /x/ - /-y/ <= /x-(-y)/ and the equality hold true only when -xy>o or xy<0 and /x/>/-y/ or /x/>/y/

given xy not = 0 , and given /x/-/y/ = /x+y/ , i.e. /x/ - /-y/ = /x-(-y)/ therefore /x/> /-y/ and -xy>o ,i.e xy<0

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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 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.

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

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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 \(xy\leq{0}\)=>\(xy<0\)
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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 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 \(xy\leq{0}\)

The text says that \(xy\neq{0}\) so \(xy\leq{0}\)=>\(xy<0\)

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

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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 \(xy=-|xy|\leq{0}\) => \(xy\leq{0}\)
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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 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 \(xy=-|xy|\leq{0}\) => \(xy\leq{0}\)

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

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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.
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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 30 May 2013, 15:12
WholeLottaLove wrote:
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.


The text says so.

You start from here \(|x| - |y| = |x+y|\), square both side ...(refer to the passages above)
Eventually you obtain \(-|xy|=xy\)

In the Bunuel's example:
2. ----y--0------x--: y<0<x --> |x|-|y|=x+y and |x+y|=x+y --> x+y={x+y}. Correct.

x is greater than 0 => \(|x|=x\)
y is less than 0 => \(|y|=-y\)
So \(|x|-|y|=x-(-y)=x+y\)

But those two examples are from two different approaches, do not mix them
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]

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By the way, I would be extremely careful with using the "squaring" approach to solve absolute value problems.

For example, if |x|=-|x|, what is x?

The answer is obvious. x=0

But,

(|x|)^2 = (-|x|)^2 --> x^2 = x^2 --> x can be any numbers (incorrect!).
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]

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I get the mechanics of flipping the signs when y is negative, but I guess I don't understand the logic.

If I take an absolute value of a number (always positive) then subtract from it an absolute value of a smaller number, which is negative how does that end up being X+Y? Are we looking just for the values of x and y, as opposed to the values of |x|-|y|?

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

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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

xy not equal to 0 means neither of them is zero , the given equation can't hold if both are positive because in that case you can remove the mod sign and it will give you y=0 which is not possible

if both x,y are positive ,x - y = x + y

Same goes for x and y both being negative .

However, one of them can be neg and one positive , x=-2 and y = 1 satisfies the equation and hence ans 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 30 May 2013, 22:47
WholeLottaLove wrote:
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?


|x|-|y| will always be (x-y) ONLY for non-negative values of BOTH x and y. If even one of them is negative, then |x|-|y| will never (x-y). It will be either -x-y (In case x is negative and y positive) OR x-(-y)-->x+y(x positive and y negative).

When y is negative, as |y| is always a non-negative entity, thus we can't write |y| = y. Thus, we attach a negative sign to 'y' to make the entire term (-y) as positive.

Also, even though (x+y) does look like addition of two numbers, it actually isn't, as y is negative.
It will be easier to understand this concept if you use valid nos.
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]

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WholeLottaLove wrote:
I get the mechanics of flipping the signs when y is negative, but I guess I don't understand the logic.

If I take an absolute value of a number (always positive) then subtract from it an absolute value of a smaller number, which is negative how does that end up being X+Y? Are we looking just for the values of x and y, as opposed to the values of |x|-|y|?



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

x is greater than 0 => \(|x|=x\)
y is less than 0 => \(|y|=-y\). You take -y from |y| because y<0
So \(|x|-|y|=x-(-y)=x+y\)

This is why \(|x|-|y|\) becomes \(x+y\). If you want you can try with real numbers, example: \(x=5\)>0 and \(y=-3\)<0
\(|x|-|y|=|5|-|-3|=5-3=2\)
\(|x|-|y|=x+y=5-3=2\)
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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]

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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?

Also, how do we know that xy is negative?


Step by step:

Step 1: square \(|x|-|y|=|x+y|\) --> \((|x|-|y|)^2=(|x+y|)^2\) (note that \((|x+y|)^2=(x+y)^2\)) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\)

Step 2: cancel x^2+y^2 in both sides to get \(-2|xy|=2xy\)

Step 3: reduce by -2 to get \(|xy|=-xy\)

Step 4: apply absolute value property, which says that the absolute value is always non-negative --> \(LHS=|xy|\) is always non negative, thus \(|xy|\geq{0}\), therefore RHS is also non-negative: \(|xy|=-xy\geq{0}\) --> \(-xy\geq{0}\).

Step 5: multiply by -1 and flip the sign of the inequality --> \(xy\leq{0}\)

Step 6: apply info given in the stem --> since given that \(xy\neq{0}\), then \(xy<0\).

Hope it's clear.
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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 31 May 2013, 02:37
walker wrote:
By the way, I would be extremely careful with using the "squaring" approach to solve absolute value problems.

For example, if |x|=-|x|, what is x?

The answer is obvious. x=0

But,

(|x|)^2 = (-|x|)^2 --> x^2 = x^2 --> x can be any numbers (incorrect!).



|x|=-|x|

you shall write it as

2 |x| = 0 and then square it

you will get x=0

even without squaring you will get 0.

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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 10 Jul 2013, 09:10
If |x| - |y| = |x+y| and xy does not equal to 0, which of the following must be true?

|x| - |y| = |x+y|
|x|-|y| = an absolute value which means that |x|-|y| is positive. Therefore, we can square both sides.
(|x| - |y|)^2 = (|x+y|)^2
(|x| - |y|)*(|x| - |y|) = (|x+y|)*(|x+y|)
x^2-2(xy|) + y^2 = x^2+2xy+y^2
-2|xy|=2xy
(I take it we cannot add -2|xy| to 2xy?)
|xy|=-xy
-xy must be positive as it is equal to an absolute value. For that to be possible:
|xy|=-xy
|xy|=-(-xy)
|xy|=xy
xy=xy

So, xy < 0

(E)

A. x-y > 0
B. x-y < 0
C. x+y > 0
D. xy > 0
E. xy < 0

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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 22 Jul 2013, 22:44
its better testing with some smart numbers.obviously will take some seconds more but worth it.
As per this question there can be four cases:
1 )+ +
2)+ -
3)- +
4)- -. Check for the validity and you will see that choices 2 and 3 only suffice.So the product is going to be <0 with either case.Hence E

Answer which you 'll derive using smart numbers will be beyond doubt and perfect.Get used to smart numbers in areas like absolute values,percentages,ratios.They help a lot :))

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Re: PS - Number system [#permalink]

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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.


Could you please explain following line as above defined in solution

note that (|x+y|)^2=(x+y)^2

Thanks.

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Re: PS - Number system [#permalink]

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New post 23 Apr 2014, 06:15
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 \(xy<0\).

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.


Could you please explain following line as above defined in solution

note that (|x+y|)^2=(x+y)^2

Thanks.


Generally, \(|x|^2=x^2\). For example, \(|-2|^2=4=2^2\) or \(|3|^2=9=3^2\).
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Re: PS - Number system [#permalink]

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New post 02 May 2014, 22:45
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 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

Please clarify

Thanks.

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Re: PS - Number system   [#permalink] 02 May 2014, 22:45

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