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# Is x > 0, 1. |x + 3| = 4x - 3 2. |x - 3| = |2x -3|

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Director
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Is x > 0, 1. |x + 3| = 4x - 3 2. |x - 3| = |2x -3| [#permalink]

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18 Aug 2007, 16:13
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Is x > 0,

1. |x + 3| = 4x - 3

2. |x - 3| = |2x -3|

Guys need your help with absolute on both ends; they are tough to fathom

Last edited by asaf on 18 Aug 2007, 17:08, edited 1 time in total.
Director
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18 Aug 2007, 18:53
A is suff

it hold sonly for x=2

I don't know for B

Any inequality masters?
Manager
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18 Aug 2007, 18:56
I find the answer to be A..

Here is how I arrived at it..

Option 1

Substituting values, there is no such value for X , when X is negative that satisfies equation 1 , hence X has to be positive

Option 2

X can be negative or X can be positive to satisfy the equation

Manager
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18 Aug 2007, 18:59

for B x=0 or 2

for A x=2

No A cant be the answer...

form

1. |x + 3| = 4x - 3

x is 2 when we take x+3 = 4x -3
but its mod so other value would be 0 when u take x+3 =-(4x-3)

from 2. 2. |x - 3| = |2x -3

value are 0 and 2..

Manager
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18 Aug 2007, 19:18
chiya wrote:

for B x=0 or 2

for A x=2

No A cant be the answer...

form

1. |x + 3| = 4x - 3

x is 2 when we take x+3 = 4x -3
but its mod so other value would be 0 when u take x+3 =-(4x-3)

from 2. 2. |x - 3| = |2x -3

value are 0 and 2..

when x=0

|x-3| = 3

4x-3 = -3

3 != -3

Thx ! I juss noticed.. A bit weak in inequatily. IS there any good reference which i can follow for it ?
Director
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18 Aug 2007, 19:26
chiya wrote:
chiya wrote:

for B x=0 or 2

for A x=2

No A cant be the answer...

form

1. |x + 3| = 4x - 3

x is 2 when we take x+3 = 4x -3
but its mod so other value would be 0 when u take x+3 =-(4x-3)

from 2. 2. |x - 3| = |2x -3

value are 0 and 2..

when x=0

|x-3| = 3

4x-3 = -3

3 != -3

Thx ! I juss noticed.. A bit weak in inequatily. IS there any good reference which i can follow for it ?

yah I am too lol. I graphed it with Graphmatica to check it out.

Ughmmmmm the way I do the problems is to take conditions, one when the absolute is negative, so in A's case, when x<3>-3

when you have 2 absolutes there are 4 cases, when one is neg, the other pos, both neg, both pos, etc......... inequalities are a biaaaaatch :D
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18 Aug 2007, 20:06
I am no master of absolute values. Giving it a shot anyways.

statement 1: |x+3| = 4x-3

So either, x+3 = 4x-3 i.e., x = 2
or -x-3 = 4x-3 i.e., x=0

Not sufficient.

statement 2: |x-3| = |2x-3|

Either, x-3 = 2x-3 i.e., x=0
or x-3 = -2x+3 i.e., x=2
or -x+3 = 2x-3 i.e., x=2
or -x+3 = -2x+3 i.e., x=0

Not sufficient.

Both combined also not sufficient. Answer E.

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