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# x != -y is (x-y)/(x+y)>1? 1. x>0 2. y <0 is there a

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Manager
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x != -y is (x-y)/(x+y)>1? 1. x>0 2. y <0 is there a [#permalink]

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19 Apr 2006, 16:49
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x != -y is (x-y)/(x+y)>1?

1. x>0
2. y <0

is there a way to do this without picking numbers?
Senior Manager
Joined: 15 Mar 2005
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Re: DS - question from OG [#permalink]

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19 Apr 2006, 17:15
The inequality is
(x-y)/(x+y) > 1

Add 1 to both sides
(x-y)/(x+y) + 1 > 2
(x-y+x+y)/(x+y) > 2
2x/(x+y) > 2
x/(x+y) > 1
1/(1+y/x) > 1

Obviously, if y/x is +ve, this holds true, and if -ve, doesn't hold true.

Thus we need to know both x and y being >/< 0 to know if y/x is >/< 0.
Thus C.

Hope that helps.
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Who says elephants can't dance?

VP
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Re: DS - question from OG [#permalink]

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19 Apr 2006, 19:37
kapslock wrote:
Add 1 to both sides
(x-y)/(x+y) + 1 > 2

i like this one....

but donot you think that x is +ve and y is negative since x!= -y? x! cannot be -ve and if x is +ve, y must be -ve.

Quote:
x!= -y is (x-y)/(x+y)>1?
1. x>0
2. y <0
is there a way to do this without picking numbers?
Director
Joined: 04 Jan 2006
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19 Apr 2006, 21:12
E...

What ever value of x or y.. you cant get it..

try x = 2, ==> y = -2 ==> x-y/x+y = 1
try x = 4, ==> y = -24 ==> 28/24 > 1..
Senior Manager
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20 Apr 2006, 06:37
x! = -y.

option a: x>0 => y has to be -ve and as x!=-y |x| has to be less that |y| and so denominator in below case should be -ve and so we can say that below inequality is false
So, (x-y)/(x+y)>0

drill down to A or D

option b: y<0 => same thing as in A

I go for D

hhmm .. Can we have factorial of -ve nos? (kinda dumb question I know )
VP
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20 Apr 2006, 06:43
I wud go with E

The equation reduces to
y/(x+y)<0

y is always -ve! As a factorial of a number is always +ve!

When x = 0, y = -1!

Hence the numerator and denominator are both -ve, LHS>0

The seconds statement y<0 is useless! as x! = -y, and a factorial is always +ve; y will always be -ve!!

Where was the question obtained from?
Intern
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20 Apr 2006, 10:36
add 0 to the numerator in a form of +2y -2y:

(x-y+2y-2y)/(x+y)
1-2y/(x+y) > 1
2y/(x+y) < 0

numerator and denominator should take opposite sign to satisfy the above statement
i. y<0 => x+y > 0 => x > -y
ii y>0 => x+y < 0 => x < -y

unfortunately the original conditions
x>0, y<0 are insufficient...
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-al

20 Apr 2006, 10:36
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# x != -y is (x-y)/(x+y)>1? 1. x>0 2. y <0 is there a

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