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

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Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1 [#permalink]

15 Aug 2009, 14:48

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Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

guys.. try this one ...it's really good??!!

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Re: Good absolute value ds [#permalink]

16 Aug 2009, 09:08

IMO D.

Statement 1: x = 2
For (x+3) >= 0
x >= -3 [Not a solution as it doesn't satisfies the condition]
x+3=4x-3
x = 2

For (x+3) < 0
x < -3 [Not a solution as it doesn't satisfies the condition]
-(x+3) = 4x-3
x = 0 [Not a solution as it doesn't satisfies the condition]

Statement 2: x = 2
For (x+1) >= 0
x >= -1 [Not a solution as it doesn't satisfies the condition]
x+1 = 2x-1
x = 2

For (x+1) < 0
x < -1 [Not a solution as it doesn't satisfies the condition]
-(x+1) = 2x - 1
x = 0 [Not a solution as it doesn't satisfies the condition]

To save time, we should ignore all -ve cases of x as LHS is +ve and equal to RHS. So, for RHS to be positive x must be +ve.
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Re: Good absolute value ds [#permalink]

16 Aug 2009, 10:45

stmt1

|x+3| = 4x - 3
when |x+3|>0
x+3= 4x -3
3x-6=0
x=2

when |x+3|<0
-(x+3) = 4x-3
5x=0,x=0

not sufficient

stmt2
|x+1| = 2x - 1
when |x+1|>0
x+1=2x-1
x=2

when |x+1|<0
-(x+1)=2x-1
3x=0, x=0
not sufficient

is the answer E?

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Re: Good absolute value ds [#permalink]

16 Aug 2009, 11:09

crejoc wrote:

stmt1
...
when |x+3|<0
-(x+3) = 4x-3
5x=0,x=0
....

Here is a typical mistake with moduli. You forgot to check your answer.
It is possible to use -(x+3) only if |x+3|<0 ! But at x=0, |0+3| > 0.
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Re: Good absolute value ds [#permalink]

16 Aug 2009, 11:46

walker wrote:

crejoc wrote:

stmt1
...
when |x+3|<0
-(x+3) = 4x-3
5x=0,x=0
....

Here is a typical mistake with moduli. You forgot to check your answer.
It is possible to use -(x+3) only if |x+3|<0 ! But at x=0, |0+3| > 0.

thanks walker for pointing out the error,.. it was a stupid mistake..

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Re: Good absolute value ds [#permalink]

23 Oct 2009, 18:28

The question stem asks is x>0??

Stmt 1:|x + 3| = 4x – 3
Two cases:
a. x+3=4x-3 ,x=0
b. -(x+3)=4x-3 ,x=0.

Both cases gives us x=0 hence SUFF.

Stmt 2:|x + 1| = 2x – 1
Again two cases:
a.x+1=2x-1
x-2x=-1-1
x=2

b. -(x+1)=2x-1
x=0
Both case gives us diff. values hence insuff.

IMO: A

Please point out my flaw.
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Re: Good absolute value ds [#permalink]

23 Oct 2009, 19:38

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tejal777 wrote:

IMO: A

Please point out my flaw.

First you made a mistake in calculation: see the red part, solution there is x=2.
Second, which is more important the way of solving, from my point of view, is incorrect.

Let me share my way of thinking, step by step:

Is x > 0?
(1) |x + 3| = 4x – 3
As absolute value NEVER NEGATIVE (in our case |x + 3|), thus RHS in our case 4x-3 must be >=0.
4x-3>=0 --> x>=3/4 (you may think at this point that if the stem asks is x>0) that it's already sufficient, but it's not.

We should check if the equation |x + 3| = 4x – 3, with the condition that x>=3/4 has real roots.

x>=3/4 means that |x + 3|>0 so we'll have x+3=4x-3 --> x=2. Only ONE solution: x=2>0.
SUFFICIENT.

(2) |x + 1| = 2x – 1
The same here: 2x-1>=0 --> x>=1/2 --> only one possibility x+1=2x-1, x=2>0.
SUFFICIENT.

Answer: D.
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Re: Good absolute value ds [#permalink]

23 Oct 2009, 19:42

Nice!Thank you so much!Kudos!:)
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Re: Good absolute value ds [#permalink]

23 Oct 2009, 20:03

I used substituting values as below and got D as answer:

S1)

|x+3| = 4x-3
when x=
-2 eq becomes 1 = -7 -->absurd
-1 eq becomes 2 = -7 -->absurd
0 eq becomes 3 = -3 -->absurd
1 eq becomes 5 = 5
2 eq becomes 6 = 9 -->absurd

so we can tell only x=2 satisifies the eq and hence S1 is sufficient.

similarly for S2)

|x+1| = 2x -1

when x =
-2 eq becomes 1 = -5 -->absurd
-1 eq becomes 0 = -3 -->absurd
0 eq becomes 1 = -1 -->absurd
1 eq becomes 2= 0 -->absurd
2 eq becomes 3 = 3
3 eq becomes 4 = 5 -->absurd

again we can tell x = 2 alone satisifes eq, hence S2 is sufficient.
Answer D.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1 [#permalink]

09 Jan 2013, 11:03

One more easier exp.

Stmt (1)
First we squire both sides of the equality:
|x + 3| = 4x – 3
x^2+6x+9=16x^2-24x+9
15x^2-30x=0 divide both sides to 15
and we have:
x^2-2x=0
x(x-2)=0
x=0
but x=0 is not a root of the equation since |0 + 3| is not equal to 4*0 – 3= -3 because absolute value cannot equal to digit below zero
x-2=0
x=2 is a root of the equality, since |2 + 3| = 4*2 – 3=> 5 both sides
hence (1) is sufficient

stmt (2)
|x + 1| = 2x – 1
squire both sides again and we have:
x^2+2x+1=4x^2-4x+1
3x^2-6x=0, divide both sides to 3 and have:
X^2-2x=0
x(x-2)=0
x=0
but x=0 is not a root of the equation since |0 + 1| is not equal to 2*0 – 1= -1 because absolute value cannot equal to digit below zero
x-2=0
x=2 is a root of the equality, since |2 + 1| = 2*2 – 1=> 3 both sides

hence stmt (2) is sufficient

D

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1 [#permalink] 09 Jan 2013, 11:03

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

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

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