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How to do this ?

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How to do this ? [#permalink]

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New post 03 Nov 2008, 07:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

how do we solve for x in this expression \(|\frac{2}{(x-4)}| > 1\)

Please note the expression \(\frac{2}{(x-4)}\) is fully in modulus
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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 08:09
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amitdgr wrote:
how do we solve for x in this expression \(|\frac{2}{(x-4)}| > 1\)

Please note the expression \(\frac{2}{(x-4)}\) is fully in modulus



so if x>4 then
2/(x-4)>1

2>(x-4)
6>x..or 4<X<6

if x<4

-2/(x-4)>1
2<-(x-4)
2<-x+4
-2<-x or 2>x

so you have the range 2<x<6

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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 08:23
fresinha12 wrote:
amitdgr wrote:
how do we solve for x in this expression \(|\frac{2}{(x-4)}| > 1\)

Please note the expression \(\frac{2}{(x-4)}\) is fully in modulus



so if x>4 then
2/(x-4)>1

2>(x-4)
6>x..or 4<X<6

if x<4

-2/(x-4)>1
2<-(x-4)
2<-x+4
-2<-x or 2>x

so you have the range 2<x<6


I get it now ... so \(|\frac{2}{(x-4)}|\) means we consider the cases for \(|x-4|\)

Thanks....

So if the equation was something like \(|\frac{(x+2)}{(x-4)}| > 1\)

then we have to consider scenarios for both nr and dr ... right ?
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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 11:32
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amitdgr wrote:
So if the equation was something like \(|\frac{(x+2)}{(x-4)}| > 1\)

then we have to consider scenarios for both nr and dr ... right ?


In this case there are two points of consideration, x = -2 and x = 4.

Hence, the expression should be evaluated for
case 1: x<-2
case 2: -2<=x<4
Case 3: 4<=x.

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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 11:39
scthakur wrote:
amitdgr wrote:
So if the equation was something like \(|\frac{(x+2)}{(x-4)}| > 1\)

then we have to consider scenarios for both nr and dr ... right ?


In this case there are two points of consideration, x = -2 and x = 4.

Hence, the expression should be evaluated for
case 1: x<-2
case 2: -2<=x<4
Case 3: 4<=x.


Thanks scthakur and fresinha :) +1 for you both
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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 14:40
fresinha12 wrote:
amitdgr wrote:
how do we solve for x in this expression \(|\frac{2}{(x-4)}| > 1\)

Please note the expression \(\frac{2}{(x-4)}\) is fully in modulus



so if x>4 then
2/(x-4)>1

2>(x-4)
6>x..or 4<X<6

if x<4

-2/(x-4)>1
2<-(x-4)
2<-x+4
-2<-x or 2>x

so you have the range 2<x<6


How can you come up with the range 2<x<6?
From (1), you have 4<x<6, right?
From (2), you have x<2
If you combine, how can you have the range of 2<x<6?

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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 15:27
I agree, shouldn't it be 4<x<6, if it's 2<x<6 then when we plug in 3 we get -2 which is not greater than 1, thus fails.

Am I missing something?

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New post 03 Nov 2008, 15:38
smarinov wrote:
I agree, shouldn't it be 4<x<6, if it's 2<x<6 then when we plug in 3 we get -2 which is not greater than 1, thus fails.

Am I missing something?


Actually, when you plug 3 in, you will get the result of 2, not -2 becuase of the sign of the absolute value.

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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 15:41
ah that's right, that makes sense now then 2 < x < 6.

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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 15:53
fresinha12 wrote:
amitdgr wrote:
how do we solve for x in this expression \(|\frac{2}{(x-4)}| > 1\)

Please note the expression \(\frac{2}{(x-4)}\) is fully in modulus



so if x>4 then
2/(x-4)>1

2>(x-4)
6>x..or 4<X<6

if x<4

-2/(x-4)>1
2<-(x-4)
2<-x+4
-2<-x or 2>x

so you have the range 2<x<6


Fresinha12,
I went over and over to your explanation and here is what i figured out.

1. I agree with your conclusion of 4<x<6.
2. when x<4, then:
2/(-x+4)>1
2>-x+4
x>2
so from (2), we have 2<x<4

Combine (1) and (2), we have 2<x<6 and x is not equal 4.
Am I right?

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Re: How to do this ? [#permalink]

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New post 03 Nov 2008, 20:21
I agree with nganle08, we ought to have x is not equal to 4 in the solution set, because at x=4 the expression becomes indefinite ...
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Re: How to do this ?   [#permalink] 03 Nov 2008, 20:21
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