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Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3

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Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3 [#permalink] New post 30 Mar 2008, 11:46
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Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|
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Re: DS: absolute value [#permalink] New post 30 Mar 2008, 11:58
Expert's post
A

1. |x + 3| = 4x – 3 --> 4x-3>0 --> x>3/4 --> sufficient

2. |x – 3| = |2x – 3|. x=0 and x=2 satisfy the condition. insufficient
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Re: DS: absolute value [#permalink] New post 30 Mar 2008, 13:16
E?
How about Ix+3I> 0
Solves to be X=2

Both S1 and S2 give X=0/2
E
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Re: DS: absolute value [#permalink] New post 30 Mar 2008, 19:31
walker wrote:
A

1. |x + 3| = 4x – 3 --> 4x-3>0 --> x>3/4 --> sufficient

2. |x – 3| = |2x – 3|. x=0 and x=2 satisfy the condition. insufficient



Walker, you always walk ahead! :lol:

1. specific x = 3 so sufficient
2. the same above

A
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Re: DS: absolute value [#permalink] New post 30 Mar 2008, 19:57
I got statement 2 to be insuff by just eye balling the equation and realizing that 2 and 0 were solutions..however I am having a hard time putting statemnt 2 in a equation and finding the critical points...can someone pls expand and show how to get the solutions to the following equation
|x-3|=|2x-3|...
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Re: DS: absolute value [#permalink] New post 30 Mar 2008, 21:07
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fresinha12 wrote:
I got statement 2 to be insuff by just eye balling the equation and realizing that 2 and 0 were solutions..however I am having a hard time putting statemnt 2 in a equation and finding the critical points...can someone pls expand and show how to get the solutions to the following equation
|x-3|=|2x-3|...


1. we have to find key points, in which modules are zero:
1.1. x-3=0 --> x=3
1.2. 2x-3=0 --> x=1.5

2. Now, we carefully eliminate modulas:
2.1. for x e (-∞,1.5)&(3,+∞): x-3=2x-3 --> x=0 (x is in the range)
2.2. for x e [1.5,3]: x-3=-2x+3 --> 3x=6 --> x=2 (x is in the range)

So, we have two solutions: 0 and 2
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Re: DS: absolute value [#permalink] New post 30 Mar 2008, 23:52
chineseburned wrote:
Is x > 0?

(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|


1: |x + 3| = 4x – 3

if |x + 3| is +ve, x + 3 = 4x – 3 and x = 2.
if |x + 3| is -ve, -x -3 = 4x – 3 and x = 0.

x = 2 works for the equality but x = 0 doesnot make the equation equal. so x cannot be 0. then x must be 2. so suff.

2:|x – 3| = |2x – 3|

if |x – 3| and |2x – 3| are either -ves (or +ves), x – 3 = 2x – 3. x = 0
if |x – 3| is +ve (or -ve) and |2x – 3| is -ve (or +ve), x - 3 = -2x + 3. x = 2

x = 0 and 2 both satisfiy the equation. so not suff....

A is it...
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Re: DS: absolute value [#permalink] New post 31 Mar 2008, 04:36
Yes, A it is. I see where I messed up.
I solved the inequality but did not check whether both the conditions are satisfied with the value "x=0".
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Re: DS: absolute value   [#permalink] 31 Mar 2008, 04:36
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