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Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 15:40

1

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00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

60% (02:23) correct
40% (01:27) wrong based on 627 sessions

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|?

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 17:07

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

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
30 May 2013, 22:47

Expert's post

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. _________________

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
31 May 2013, 00:12

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 _________________

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]
31 May 2013, 02:05

Expert's post

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?

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.

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

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

Re: PS - Number system [#permalink]
23 Apr 2014, 04:51

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

Re: PS - Number system [#permalink]
23 Apr 2014, 06:15

Expert's post

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. _________________

Re: PS - Number system [#permalink]
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

Re: PS - Number system [#permalink]
03 May 2014, 03:22

Expert's post

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.

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.

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. _________________

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
06 May 2014, 20:00

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

Re: PS - Number system [#permalink]
18 Jun 2014, 05:57

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},

Re: PS - Number system [#permalink]
18 Jun 2014, 07:13

Expert's post

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.

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]
18 Jun 2014, 11:41

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.

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