Absolute value question

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voodoo_child: Master | Next Rank: 500 Posts; Posts: 405; Joined: Thu Feb 10, 2011 1:44 amFollowed by: 1 member; Thanked: 3 times

by voodoo_child » Sun Mar 06, 2011 3:56 pm

What would be the solution for

|X| - |Y| = |X+Y| ; XY != 0 ; X and Y are integers.

Can anyone please provide algebraic method ?

Here's what I did:-

Squaring on both sides and using Schwarz Inequality,

-2|X||Y| =< 2|XY|

Therefore, 4|XY| > 0

therefore, XY > 0 ??? The answer is wrong.

I am lost. Can anyone please help ?

[spoiler]OS : XY < 0[/spoiler]

Thanks
Voodoo

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Source: Beat The GMAT — Quantitative Reasoning | Sun Mar 06, 2011 3:56 pm

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Mar 06, 2011 5:59 pm

voodoo_child wrote:What would be the solution for

|X| - |Y| = |X+Y| ; XY != 0 ; X and Y are integers.

Can anyone please provide algebraic method ?

Here's what I did:-

Squaring on both sides and using Schwarz Inequality,

-2|X||Y| =< 2|XY|

Therefore, 4|XY| > 0

therefore, XY > 0 ??? The answer is wrong.

I am lost. Can anyone please help ?

[spoiler]OS : XY < 0[/spoiler]

Thanks
Voodoo

Using graph, there are two possible sets (you can deduce this algebraically by considering the cases : x>0, y>0 ; x>0, y<0 and x+y > 0; x>0, y<0 and x+y < 0 ; x<0, y<0; x<0, y>0 and x+y < 0 ; x<0, y> 0 and x+y > 0, in fact check all the equalities also , so for eg. x>=0 instead of x>0 wherever mentioned)

x > 0, y< 0 and x+y > 0
x <0 , y > 0 and x+y < 0

The drawn area is the solution (and you can see the value of xy < 0 in that area; please note that origin is not part of the solution as xy=0 there.)

Thanks
Anshu

(Every mistake is a lesson learned )

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krusta80: Master | Next Rank: 500 Posts; Posts: 143; Joined: Mon Mar 14, 2011 3:13 amFollowed by: 5 members; Thanked: 34 times

by krusta80 » Mon Mar 14, 2011 3:19 am

Algebraically, here is where you went wrong:

1. |X| - |Y| = |X+Y|

Square both sides...

2. (|X| - |Y|)^2 = (|X+Y|)^2

Break out the left side...note that anything squared loses its "absolute valueness"...

3. X^2 - 2*|X|*|Y| + Y^2 = (|X+Y|)^2

NOW WHERE YOU MIS-STEPPED...breaking out the right side. Since the entire quantity of |X+Y| is being squared, we drop the absolute value right away! In other words, (x+y)^2 always equals |x+y|^2

4. X^2 - 2*|X|*|Y| + Y^2 = X^2 + 2*X*Y + Y^2

Now we see the difference...there are no absolute value signs on the right hand side.

5. X*Y = - (|X|*|Y|)

Since the product of two positive integers |X|, |Y| is always positive, XY must be negative.

XY < 0

voodoo_child wrote:What would be the solution for

|X| - |Y| = |X+Y| ; XY != 0 ; X and Y are integers.

Can anyone please provide algebraic method ?

Here's what I did:-

Squaring on both sides and using Schwarz Inequality,

-2|X||Y| =< 2|XY|

Therefore, 4|XY| > 0

therefore, XY > 0 ??? The answer is wrong.

I am lost. Can anyone please help ?

[spoiler]OS : XY < 0[/spoiler]

Thanks
Voodoo

Like this post

shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Mon Mar 14, 2011 6:50 am

voodoo_child wrote:What would be the solution for

|X| - |Y| = |X+Y| ; XY != 0 ; X and Y are integers.

Can anyone please provide algebraic method ?

Here's what I did:-

Squaring on both sides and using Schwarz Inequality,

-2|X||Y| =< 2|XY|

Therefore, 4|XY| > 0

therefore, XY > 0 ??? The answer is wrong.

I am lost. Can anyone please help ?

[spoiler]OS : XY < 0[/spoiler]

Thanks
Voodoo

ok lets start a bit algebrically.....and you will find it very simple

case 1: X>0, Y>0 --> |X|=X, |Y|=Y & |X+Y|= X+Y
so |X| - |Y| = |X+Y| --> X-Y = X+Y which is not true.....considering XY !=0

case 2: X>0, Y<0 --> |X|=X, |Y|=-Y & |X+Y|= |X+Y|
so |X| - |Y| = |X+Y| --> X+Y = |X+Y| which is true only if X+Y>0 --> X>-Y............(i)

case 3: X<0, Y>0 --> |X|=-X, |Y|=Y & |X+Y|= |X+Y|
so |X| - |Y| = |X+Y| --> -X-Y = |X+Y| which is true only if X+Y<0 --> X<-Y............(ii)

case3: X<0, Y<0 --> |X|=-X, |Y|=-Y & |X+Y|= -X-Y
so |X| - |Y| = |X+Y| --> -X+Y = -X-Y which is not true.....considering XY !=0

so from (i) & (ii) required conditions are X>0,Y<0 and X+Y>0 & X<0, Y>0 and X+Y<0

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