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

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Manager
Joined: 19 Aug 2007
Posts: 170
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Kudos [?]: 19 [0], given: 0

Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3 [#permalink]  31 Oct 2007, 06:49
00:00

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|
Manager
Joined: 02 Aug 2007
Posts: 146
Followers: 1

Kudos [?]: 16 [0], given: 0

1) INSUFF - could be x=3 or x=0
Positive case:
x+3=4x-3
x = 3
Negative case:
-x-3=4x-3
x = 0

2) SUFFICIENT - x = 0

B.
Manager
Joined: 25 Apr 2007
Posts: 50
Followers: 1

Kudos [?]: 2 [0], given: 0

yuefei wrote:
1) INSUFF - could be x=3 or x=0
Positive case:
x+3=4x-3
x = 3
Negative case:
-x-3=4x-3
x = 0

2) SUFFICIENT - x = 0

B.

yuefei,

help me out here. we get -x=0 and x=0 for both the +ve & -ve case of stmt B, is that the reason we are saying its SUFF?

also did you calculate both, the +ve & -ve cases for B ?

Thanks !!!
Manager
Joined: 02 Aug 2007
Posts: 146
Followers: 1

Kudos [?]: 16 [0], given: 0

Yes. For statement B, x = 0 is the solution for both +ve and -ve cases.

Negative case:
-x = 0 is x = 0/-1 = 0
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

(A) for me

Stat 1
|x + 3| = 4x – 3

o If x >= -3 then
|x + 3| = 4x – 3
<=> x+3 = 4*x - 3
<=> x = 2 >>> Ok, as x >= -3

o If x < -3 then
|x + 3| = 4x – 3
<=> -(x+3) = 4*x - 3
<=> x = 0 >>> Out as x < -3

Thus, x=2 is the only solution and it's positive.

SUFF.

Stat 2
|x – 3| = |2x – 3|

We have to look at 3 intervals : x < 3/2, 3/2 =< x < 3 and x >=3

o If x < 3/2 then
|x – 3| = |2x – 3|
<=> -(x-3) = -(2*x-3)
<=> x-3 = 2*x - 3
<=> x = 0 >>>> OK as x < 3/2

o If 3/2 =< x < 3 then
|x – 3| = |2x – 3|
<=> -(x-3) = 2*x-3
<=> 3*x = 6
<=> x = 2 >>>> OK as 3/2 =< x < 3

o If x >= 3 then
|x – 3| = |2x – 3|
<=> x-3 = 2*x - 3
<=> x = 0 >>>> Out as x >= 3

Finally, we have x = 0 or x = 2

INSUFF.
Director
Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 80 [0], given: 0

Fig wrote:
(A) for me

Stat 1
|x + 3| = 4x – 3

o If x >= -3 then
|x + 3| = 4x – 3
<=> x+3 = 4*x - 3
<x>>> Ok, as x >= -3

o If x < -3 then
|x + 3| = 4x – 3
<=> -(x+3) = 4*x - 3
<x>>> Out as x < -3

Thus, x=2 is the only solution and it's positive.

SUFF.

Stat 2
|x – 3| = |2x – 3|

We have to look at 3 intervals : x < 3/2, 3/2 =< x <3>=3

o If x < 3/2 then
|x – 3| = |2x – 3|
<=> -(x-3) = -(2*x-3)
<=> x-3 = 2*x - 3
<x>>>> OK as x < 3/2

o If 3/2 =< x < 3 then
|x – 3| = |2x – 3|
<=> -(x-3) = 2*x-3
<=> 3*x = 6
<x>>>> OK as 3/2 =< x <3>= 3 then[/b]
|x – 3| = |2x – 3|
<=> x-3 = 2*x - 3
<x>>>> Out as x >= 3

Finally, we have x = 0 or x = 2

INSUFF.

thanks Fig.. i also got A with similar reasoning but didnt feel like retyping my work
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

beckee529 wrote:
Fig wrote:
(A) for me

Stat 1
|x + 3| = 4x – 3

o If x >= -3 then
|x + 3| = 4x – 3
<=> x+3 = 4*x - 3
<x>>> Ok, as x >= -3

o If x < -3 then
|x + 3| = 4x – 3
<=> -(x+3) = 4*x - 3
<x>>> Out as x < -3

Thus, x=2 is the only solution and it's positive.

SUFF.

Stat 2
|x – 3| = |2x – 3|

We have to look at 3 intervals : x < 3/2, 3/2 =< x <3>=3

o If x < 3/2 then
|x – 3| = |2x – 3|
<=> -(x-3) = -(2*x-3)
<=> x-3 = 2*x - 3
<x>>>> OK as x < 3/2

o If 3/2 =< x < 3 then
|x – 3| = |2x – 3|
<=> -(x-3) = 2*x-3
<=> 3*x = 6
<x>>>> OK as 3/2 =< x <3>= 3 then[/b]
|x – 3| = |2x – 3|
<=> x-3 = 2*x - 3
<x>>>> Out as x >= 3

Finally, we have x = 0 or x = 2

INSUFF.

thanks Fig.. i also got A with similar reasoning but didnt feel like retyping my work

Some what long to type I admit too
Manager
Joined: 02 Aug 2007
Posts: 146
Followers: 1

Kudos [?]: 16 [0], given: 0

" |x – 3| = |2x – 3|
We have to look at 3 intervals : x < 3/2, 3/2 =< x <3>=3 "

Fig, how did you quickly get that we need to look at 3 intervals?
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

yuefei wrote:
" |x – 3| = |2x – 3|
We have to look at 3 intervals : x < 3/2, 3/2 =< x <3>=3 "

Fig, how did you quickly get that we need to look at 3 intervals?

Well, the technic consist of looking at values make fliping the sign in each absolute value.... at 3, x-3 changes its sign... also at 3/2, it's 2*x-3 that change its sign

Then, every singularity point, which flip a sign of an absolute, is the limit of 2 intervals

If u want more practices... I suggest u this : http://www.gmatclub.com/forum/t39533
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