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# y|>|y+1| A. y<-1/3 B. -1<=y<-1/2 C. y<-1/2 D.

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VP
Joined: 09 Jul 2007
Posts: 1103
Location: London
Followers: 6

Kudos [?]: 105 [0], given: 0

y|>|y+1| A. y<-1/3 B. -1<=y<-1/2 C. y<-1/2 D. [#permalink]

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03 Nov 2007, 14:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

|y|>|y+1|

A. y<-1/3
B. -1<=y<-1/2
C. y<-1/2
D. y<-1/4
E. -1<y<-1/2
SVP
Joined: 01 May 2006
Posts: 1797
Followers: 9

Kudos [?]: 152 [0], given: 0

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03 Nov 2007, 15:14
(C) for me

|y|>|y+1|

The crucial point are -1 and 0...

o If y < -1 then
|y|>|y+1|
<=> -y > -(y+1)
<=> 0 > -1 .... all y such that y < -1 are solutions

o If 0 > y >= -1 then
|y|>|y+1|
<=> -y > (y+1)
<=> y < -1/2 .... all y such that -1/2 > y >= -1 are solutions

o If y >=0 then
|y|>|y+1|
<=> y > (y+1)
<=> 0 > 1 .... Impossible... No y here.

Finally,
y < -1/2
VP
Joined: 09 Jul 2007
Posts: 1103
Location: London
Followers: 6

Kudos [?]: 105 [0], given: 0

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03 Nov 2007, 16:45
can we take squares of each side of the equation and solve?
03 Nov 2007, 16:45
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