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absolute [#permalink] New post 04 Dec 2006, 05:18
SOLVE FOR X

1) l x+1 l / x-2 l <3

2) l n+2 / n l >=4

i will highly appreciate detailed conceptual answers
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Re: absolute [#permalink] New post 04 Dec 2006, 08:35
2) l n+2 / n l >=4

It's more solving for n than for x, here ;)

| n+2 / n | >=4
<=> |n+2| / |n| >=4
<=> |n+2| >= 4*|n| as |n| >= 0
<=> |n+2| - 4*|n| >= 0

o If n > 0 then the inequation becomes:
n+2 - 4*n >= 0
<=> n <= 2/3
=> 0 < n <= 2/3

o If -2 < n < 0 then the inequation becomes:
n+2 - 4*(-n) >= 0
<=> n >= -2/5
=> -2/5 <= n < 0

o If n < -2 then the inequation becomes:
-(n+2) - 4*(-n) >= 0
<=> n >= 2/3
Impossible.... no values for n here.

Finally, we have:
o 0 < n <= 2/3
or
o -2/5 <= n < 0
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 [#permalink] New post 04 Dec 2006, 08:47
we don't need to even solve for n because question stem is asking for value of x. so 2 is automatically insuff. we only look at statement 1
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 [#permalink] New post 04 Dec 2006, 09:31
Hermione wrote:
we don't need to even solve for n because question stem is asking for value of x. so 2 is automatically insuff. we only look at statement 1


Following previous post of Yezz, I understand it more as 2 independent excercices :).... and not a DS :).... A bit like in this thread : http://www.gmatclub.com/phpbb/viewtopic.php?p=274350

Maybe, I'm wrong.... Let Yezz clarifies it :)
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 [#permalink] New post 04 Dec 2006, 09:33
Okie :-D
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 [#permalink] New post 05 Dec 2006, 01:46
FIG is on target here, sorry folks for confusion , i was sending this in a hurry during mu lunch break, thanks FIG and Hermoine, a million :wink:
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 [#permalink] New post 08 Dec 2006, 23:43
Fig, i know i'm going back to basics here...

In this problem, how did you know to test for:

(1) n>o
(2) -2 < n < 0
(3) n < -2

Normally for a problem like this, I would just test for n>0 or n<0 but apparently this is not the case.
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 [#permalink] New post 09 Dec 2006, 00:40
Hermione wrote:
Fig, i know i'm going back to basics here...

In this problem, how did you know to test for:

(1) n>o
(2) -2 < n < 0
(3) n < -2

Normally for a problem like this, I would just test for n>0 or n<0 but apparently this is not the case.


Any question has always to be asked :), especially before going to an exam such as the GMAT ;)... and u are welcome :)

Concerning the reason, the principle is to look for the crucial "point" or value s of n,x... when those values make flip the sign of the expression in the absolute value. This is done for every absolute values of an inequation.

Concretely, we decompose the whole inquation in module to study:
o |n+2|:
# n+2 = 0 <=> n = -2

Also,
# n+2 > 0 <=> n > -2
# n+2 < 0 <=> n < -2

> The original inequation has to be studied on those 2 domains n < -2 and n > -2. Indeed, for those 2 domains, the simplified expression of |n+2| differs. That is |n+2| = n+2 when n > -2 and |n+2| = -(n+2) when n < -2.

o |n|:
# n = 0

Thus,
# n > 0
# n < 0

Finally, we have to study all combined domains, from |n+2| and from |n|, in the original inequation.

That gives : n < -2, -2 < n < 0, and n > 0.
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 [#permalink] New post 09 Dec 2006, 04:25
Thanks so much for explaining Fig!!! That was a bit complicated... But I think I got it... Can you suggest any sites where I can brush up on absolute value fundamentals? :)
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  [#permalink] 09 Dec 2006, 04:25
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