|x|<1? i |x+1|=2|x-1| ii |x-3| is not equal to zero Pls

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|x|<1? i |x+1|=2|x-1| ii |x-3| is not equal to zero Pls [#permalink]

19 Mar 2006, 13:50

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3.|x|<1?
i |x+1|=2|x-1|
ii |x-3| is not equal to zero

Pls explain.. I would like to know the various possibilities when there is absolute value on both the sides.. Like if I take st1.. how do i find all the possible values of X??

Thanks
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Re: DS + Absolute values [#permalink]

19 Mar 2006, 15:34

bewakoof wrote:

ii) not sufficient.

i) |x+1|=|2x-2|
|x+1| = |-x-1|
|2x-2| = |-2x+2|

we would have 4 options

1) -x-1 = 2x-2
2) -x-1 = -2x+2
3) x+1 = 2x-2
4) x+1 = -2x+2

so there are 4 solutions for x:

X = 1/3 , 3, 3, 1/3

so is insufficient.

taking both statements

|x|<1

then for |x|<1 is a C

Last edited by conocieur on 19 Mar 2006, 15:41, edited 1 time in total.

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Re: DS + Absolute values [#permalink]

19 Mar 2006, 15:39

bewakoof wrote:

3.|x|<1?
i |x+1|=2|x-1|
ii |x-3| is not equal to zero

lets take statement i. there are four possibles.

a. both modulas are +ve; x = 3.
b. both modulas are -ve; x = 3.
c. |x+1| is +ve and (2|x-1|) is -ve; x = 1/3.
d. |x+1| is -ve and (2|x-1|) is +ve; x = 1/3

so x is either 1/3 or 3. so statement i is not suff.

statement ii says |x-3| = 0. this alone is also not enough cuz we do not know the value of x and only know x is not equal to 3.

from i and ii, both, we know that x = 1/3 which is less tha 1.

So C is answer.

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[#permalink]

19 Mar 2006, 15:41

I see.. so you guys are considering all 4 possibilities.. I thought there was a shortcut... thanks.. appreciate it!!
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[#permalink]

19 Mar 2006, 15:49

bewakoof wrote:

I see.. so you guys are considering all 4 possibilities.. I thought there was a shortcut... thanks.. appreciate it!!

I wonder, is always the case for absolute values that the solution for both negatives is the same as for both positives, and also the solution of one negative and one positive is the same no matter what is the order?

in that case you would have your shortcut, I have no clue if it works all the times but it wouldn't hurt trying or even finding a proof for it somewhere.

[#permalink] 19 Mar 2006, 15:49

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|x|<1? i |x+1|=2|x-1| ii |x-3| is not equal to zero Pls

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|x|<1? i |x+1|=2|x-1| ii |x-3| is not equal to zero Pls

|x|<1? i |x+1|=2|x-1| ii |x-3| is not equal to zero Pls

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