GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Jan 2019, 09:53

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### Key Strategies to Master GMAT SC

January 26, 2019

January 26, 2019

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
• ### Free GMAT Number Properties Webinar

January 27, 2019

January 27, 2019

07:00 AM PST

09:00 AM PST

Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.

# Absolutes QUESTION - MGMAT Article

Author Message
TAGS:

### Hide Tags

Intern
Joined: 31 Aug 2016
Posts: 46
Absolutes QUESTION - MGMAT Article  [#permalink]

### Show Tags

30 Nov 2017, 06:49
Hello guys,

It has this example: |x + 1| + |x - 3| = 6 where the critical points are -1 and 3 and defines the critical points as the numbers that zero | |.

Why in the inequality below the critical point is not -3? It is -6 and 0.
|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)

Thanks
Math Expert
Joined: 02 Aug 2009
Posts: 7213
Re: Absolutes QUESTION - MGMAT Article  [#permalink]

### Show Tags

30 Nov 2017, 07:09
standyonda wrote:
Hello guys,

It has this example: |x + 1| + |x - 3| = 6 where the critical points are -1 and 3 and defines the critical points as the numbers that zero | |.

Why in the inequality below the critical point is not -3? It is -6 and 0.
|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)

Thanks

hi.

the critical point is -3 itself..
at x<-3, x+3 will be NEGATIVE, so -(x+3)>3.....x+3<-3....x<-6...lower limit -Inf
at x>=-3, x+3>3...x>0.. Higher limit +inf
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 31 Aug 2016
Posts: 46
Re: Absolutes QUESTION - MGMAT Article  [#permalink]

### Show Tags

30 Nov 2017, 07:14
chetan2u wrote:
standyonda wrote:
Hello guys,

It has this example: |x + 1| + |x - 3| = 6 where the critical points are -1 and 3 and defines the critical points as the numbers that zero | |.

Why in the inequality below the critical point is not -3? It is -6 and 0.
|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)

Thanks

hi.

the critical point is -3 itself..
at x<-3, x+3 will be NEGATIVE, so -(x+3)>3.....x+3<-3....x<-6...lower limit -Inf
at x>=-3, x+3>3...x>0.. Higher limit +inf

Thanks!! So to find the range you use the critical point that in the end it has nothing to do with the range. The CP is a step.
Re: Absolutes QUESTION - MGMAT Article &nbs [#permalink] 30 Nov 2017, 07:14
Display posts from previous: Sort by