absolute value and inequality

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absolute value and inequality [#permalink]

07 Jul 2009, 17:06

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|y|>|y+1|

what is the value of y?
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Re: absolute value and inequality [#permalink]

07 Jul 2009, 19:11

Y is negative for sure from the given info. But it seems the question is incomplete, bcoz we cannot get a unique value of Y

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Re: absolute value and inequality [#permalink]

09 Jul 2009, 10:28

Y<-0.5

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Re: absolute value and inequality [#permalink]

09 Jul 2009, 15:10

y < = -1
that is the value when 1>0 or absolute y > absolute (y-1)

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Re: absolute value and inequality [#permalink]

13 Jul 2009, 11:44

square it on both sides... we get

y^2 > (y+1)^2

and hence y < -1/2

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Re: absolute value and inequality [#permalink]

15 Jul 2009, 18:56

Thanks for the replies..
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Re: absolute value and inequality [#permalink]

19 Jul 2009, 10:28

y^2>(y+1)^2

how do you get y<-1/2 ??

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Re: absolute value and inequality [#permalink]

19 Jul 2009, 12:56

Quote:

y^2>(y+1)^2

how do you get y<-1/2 ??

expand the equation.

y^2 > y^2 + 1+2y

0 > 1+2y

1+2y < 0

y<-1/2

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Re: absolute value and inequality [#permalink]

19 Jul 2009, 19:00

vannu wrote:

|y|>|y+1|

what is the value of y?

2 scenerios:

1. When y >= 0:
The inequality, |y| > |y+1|, is never possible because y cannot be > y+1 when y is +ve..

2. When y<0:
-y > y+1
-y - y > 1
-2y > 1
y < -1/2

Thats the solution.
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Re: absolute value and inequality [#permalink]

20 Jul 2009, 03:15

Oh! I must have been blind

Ok but we can solve it like this:

since |x| = √x^2

the term can be expressed as

√y^2 > √(y+1)^2 | than we square
y^2 > (y+1)^2 | binomial
y^2 > y^2+2y+1
0 > 2y+1
y < -1/2

that's also the right way to think isn't it?
Thanks a lot folks!

Re: absolute value and inequality [#permalink] 20 Jul 2009, 03:15

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absolute value and inequality

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