inequalities

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fruti_yum: Master | Next Rank: 500 Posts; Posts: 128; Joined: Thu Jul 30, 2009 1:46 pm; Thanked: 1 times

by fruti_yum » Mon Aug 03, 2009 12:23 pm

Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

Some body hellllllpppppppppppppp... My exam is in less than a week and I can't solve this question.. I'm freaking out!!

Please dwell on the approach.. not just the answer!! appreciate it

OA is C

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Source: Beat The GMAT — Data Sufficiency | Mon Aug 03, 2009 12:23 pm

ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

by ssmiles08 » Mon Aug 03, 2009 1:35 pm

1) |x + 1| = 2|x – 1|

x + 1 = 2(x-1)

x + 1 = 2x - 2

x = 3

x + 1 = -[2(x - 1)]

x + 1 = -2x + 2

3x = 1

x = 1/3

Insufficient as there are two solutions.

2) |x – 3| > 0

x - 3 > 0

x > 3

-(x -3) > 0

x < 3

Insufficient as the answer can be yes or no.

Together:

x cannot be = 3 according to the second statement, which leaves us with only 1 solution x = 1/3.

is |1/3| < 1? ---> YES.

Sufficient. (C)

You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.

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mehravikas: Legendary Member; Posts: 1161; Joined: Mon May 12, 2008 2:52 amFollowed by: 1 member; Location: Sydney; Thanked: 23 times

by mehravikas » Mon Aug 03, 2009 10:02 pm

Can I ask you, when considering negative values for x, why have you not done:

-(x + 1) = -[2(x - 1)]

-x - 1 = -2x + 2
x = 1

ssmiles08 wrote:1) |x + 1| = 2|x – 1|

x + 1 = 2(x-1)

x + 1 = 2x - 2

x = 3

x + 1 = -[2(x - 1)]

x + 1 = -2x + 2

3x = 1

x = 1/3

Insufficient as there are two solutions.

2) |x – 3| > 0

x - 3 > 0

x > 3

-(x -3) > 0

x < 3

Insufficient as the answer can be yes or no.

Together:

x cannot be = 3 according to the second statement, which leaves us with only 1 solution x = 1/3.

is |1/3| < 1? ---> YES.

Sufficient. (C)

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ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

by ssmiles08 » Tue Aug 04, 2009 4:34 am

mehravikas wrote:Can I ask you, when considering negative values for x, why have you not done:

-(x + 1) = -[2(x - 1)]

-x - 1 = -2x + 2
x = 1

Making both sides negative will give you the same solution as the first one.

IMO you made a calculation error. -x + 2x = 2+ 1
x = 3

-hope that helps.

You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.

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mehravikas: Legendary Member; Posts: 1161; Joined: Mon May 12, 2008 2:52 amFollowed by: 1 member; Location: Sydney; Thanked: 23 times

by mehravikas » Tue Aug 04, 2009 12:05 pm

I guess that's not a choice, that's what we have to do.

When x > 0

x + 1 = 2x - 2

When x < 0

- (x + 1) = -2(x-2)

Please correct if I am wrong.

ssmiles08 wrote:
mehravikas wrote:Can I ask you, when considering negative values for x, why have you not done:

-(x + 1) = -[2(x - 1)]

-x - 1 = -2x + 2
x = 1

Making both sides negative will give you the same solution as the first one.

IMO you made a calculation error. -x + 2x = 2+ 1
x = 3

-hope that helps.

Like this post

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