Which values of x are solutions |x + 1| + |x - 1| <= 2 : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 21:05

### 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 June
Open Detailed Calendar

# Which values of x are solutions |x + 1| + |x - 1| <= 2

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 28 Feb 2012
Posts: 29
Followers: 1

Kudos [?]: 37 [1] , given: 1

Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

21 May 2012, 08:14
1
KUDOS
18
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

81% (02:35) correct 19% (00:40) wrong based on 134 sessions

### HideShow timer Statistics

Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93289 [11] , given: 10555

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

21 May 2012, 08:29
11
KUDOS
Expert's post
21
This post was
BOOKMARKED
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this

$$|x + 1|+|x - 1|\leq{2}$$ --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If $$x<-1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$-(x+1)-(x-1)\leq{2}$$ --> $$x\geq{-1}$$, not a valid range since we are considering $$x<-1$$;

If $$-1\leq{x}\leq{1}$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1-(x-1)\leq{2}$$ --> $$0\leq{2}$$ which is true, so for $$-1\leq{x}\leq{1}$$ given inequality holds true;

If $$x>1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1+x-1\leq{2}$$ --> $$x\leq{1}$$, not a valid range since we are considering $$x>1$$.

So, finally we have that $$|x + 1|+|x - 1|\leq{2}$$ holds true for $$-1\leq{x}\leq{1}$$.

For more check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

Hope it helps.
_________________
Manager
Joined: 02 Jun 2011
Posts: 159
Followers: 1

Kudos [?]: 75 [0], given: 11

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

22 May 2012, 00:28
1
This post was
BOOKMARKED
Bunuel wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this

$$|x + 1|+|x - 1|\leq{2}$$ --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If $$x<-1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$-(x+1)-(x-1)\leq{2}$$ --> $$x\geq{-1}$$, not a valid range since we are considering $$x<-1$$;

If $$-1\leq{x}\leq{1}$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1-(x-1)\leq{2}$$ --> $$0\leq{2}$$ which is true, so for $$-1\leq{x}\leq{1}$$ given inequality holds true;

If $$x>1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1+x-1\leq{2}$$ --> $$x\leq{1}$$, not a valid range since we are considering $$x>1$$.

So, finally we have that $$|x + 1|+|x - 1|\leq{2}$$ holds true for $$-1\leq{x}\leq{1}$$.

For more check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

Hope it helps.

Dear Bunuel,

what these kind of inequality questions are asking for?
is it the values of X, for which the solution of equation is <=2?
i had been through the link provided and have the same doubt regarding the example 1 -
|x+3|-|4-x|=|8+x| -How many solutions does the equation have?
How you have taken "four conditions given"?
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93289 [1] , given: 10555

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

22 May 2012, 09:25
1
KUDOS
Expert's post
kashishh wrote:
Bunuel wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this

$$|x + 1|+|x - 1|\leq{2}$$ --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If $$x<-1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$-(x+1)-(x-1)\leq{2}$$ --> $$x\geq{-1}$$, not a valid range since we are considering $$x<-1$$;

If $$-1\leq{x}\leq{1}$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1-(x-1)\leq{2}$$ --> $$0\leq{2}$$ which is true, so for $$-1\leq{x}\leq{1}$$ given inequality holds true;

If $$x>1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1+x-1\leq{2}$$ --> $$x\leq{1}$$, not a valid range since we are considering $$x>1$$.

So, finally we have that $$|x + 1|+|x - 1|\leq{2}$$ holds true for $$-1\leq{x}\leq{1}$$.

For more check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

Hope it helps.

Dear Bunuel,

what these kind of inequality questions are asking for?
is it the values of X, for which the solution of equation is <=2?
i had been through the link provided and have the same doubt regarding the example 1 -
|x+3|-|4-x|=|8+x| -How many solutions does the equation have?
How you have taken "four conditions given"?

The question asks about the range(s) of x for which the given inequality holds true. We have that it holds for $$-1\leq{x}\leq{1}$$, for example if x=0 then $$|x + 1|+|x - 1|\leq{2}$$ is true but if x is out of this range, say x=3, then $$|x + 1|+|x - 1|\leq{2}$$ is NOT true.

Hope it's clear.
_________________
Manager
Joined: 02 Jun 2011
Posts: 159
Followers: 1

Kudos [?]: 75 [0], given: 11

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

23 May 2012, 00:45
Thankyou Bunuel,
It is asking for range.
i understand,so is in this question "|x+3|-|4-x|=|8+x| -How many solutions does the equation have?

Please can you elaborate on "four conditions given"?
Manager
Joined: 26 Dec 2011
Posts: 117
Followers: 1

Kudos [?]: 32 [0], given: 17

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

23 May 2012, 05:05
If for the same question, |x+3|-|4-x|=|8+x|, we had to identify for what what value of x will the equation be valid. How would we do that?
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93289 [1] , given: 10555

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

23 May 2012, 05:29
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
kashishh wrote:
Thankyou Bunuel,
It is asking for range.
i understand,so is in this question "|x+3|-|4-x|=|8+x| -How many solutions does the equation have?

Please can you elaborate on "four conditions given"?

pavanpuneet wrote:
If for the same question, |x+3|-|4-x|=|8+x|, we had to identify for what what value of x will the equation be valid. How would we do that?

How many solutions does $$|x+3| - |4-x| = |8+x|$$ have?

Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.

If $$x < -8$$ then $$|x+3| - |4-x| = |8+x|$$ expands as $$-(x+3)-(4-x)=-(8+x)$$ --> $$x = -1$$, which is not a valid solution since we are considering $$x < -8$$ range and -1 is out of it;

If $$-8\leq{x}\leq{-3}$$ then $$|x+3| - |4-x| = |8+x|$$ expands as $$-(x+3)-(4-x)=(8+x)$$ --> $$x=-15$$, which is also not a valid solution since we are considering $$-8\leq{x}\leq{-3}$$ range;

If $$-3<x<4$$ then $$|x+3| - |4-x| = |8+x|$$ expands as $$(x+3)-(4-x)=(8+x)$$ --> $$x = 9$$, which is also not a valid solution since we are considering $$-3<x<4$$ range;

If $$x\geq{4}$$ then $$|x+3|-|4-x|=|8+x|$$ expands as $$(x+3)+(4-x)=(8+x)$$ --> $$x = -1$$, which is also not a valid solution since we are considering $$x>4$$ range.

So we have that no solution exist for $$|x+3|-|4-x|=|8+x|$$ (no x can satisfy this equation).

Hope it's clear.
_________________
Manager
Joined: 26 Dec 2011
Posts: 117
Followers: 1

Kudos [?]: 32 [0], given: 17

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

23 May 2012, 06:29
Inequalities have been quite a nightmare for me.. I understood the overall solution but, based on what do we decide to x<-8 and not x<=-8....?? I know this may be a very basic question, but can you please explain this concept for these ranges. Will be a great help! Thanks in advance.
Intern
Joined: 08 Oct 2007
Posts: 33
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

23 May 2012, 20:03
Hi, Bunuel!
How many solutions does have?

In your solution when you define intervals, how do you define what inequality sign to put? I mean < or> or<=
For example, why the second interval is -8<=x<=-3, why both inequality signs are <=?

Thank you in advance
Math Expert
Joined: 02 Sep 2009
Posts: 36583
Followers: 7087

Kudos [?]: 93289 [3] , given: 10555

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

23 May 2012, 23:51
3
KUDOS
Expert's post
pavanpuneet wrote:
Inequalities have been quite a nightmare for me.. I understood the overall solution but, based on what do we decide to x<-8 and not x<=-8....?? I know this may be a very basic question, but can you please explain this concept for these ranges. Will be a great help! Thanks in advance.

Rice wrote:
Hi, Bunuel!
How many solutions does have?

In your solution when you define intervals, how do you define what inequality sign to put? I mean < or> or<=
For example, why the second interval is -8<=x<=-3, why both inequality signs are <=?

Thank you in advance

We should include "check points" (-8, -3, and 4) in either of the ranges but it really doesn't matter in which one. So, for example we could have written:
If $$x\leq{-8}$$ ...
If $$-8<x<-3$$ ...
If $$-3\leq{x}<\leq{4}$$ ...
If $$x>4$$ ...

Hope it's clear.
_________________
Intern
Joined: 30 Mar 2012
Posts: 36
Followers: 0

Kudos [?]: 2 [0], given: 11

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

24 May 2012, 01:52
Bunuel wrote:
How many solutions does $$|x+3| - |4-x| = |8+x|$$ have?

Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.

If $$x < -8$$ then $$|x+3| - |4-x| = |8+x|$$ expands as $$-(x+3)-(4-x)=-(8+x)$$ --> $$x = -1$$, which is not a valid solution since we are considering $$x < -8$$ range and -1 is out of it;

If $$-8\leq{x}\leq{-3}$$ then $$|x+3| - |4-x| = |8+x|$$ expands as $$-(x+3)-(4-x)=(8+x)$$ --> $$x=-15$$, which is also not a valid solution since we are considering $$-8\leq{x}\leq{-3}$$ range;

If $$-3<x<4$$ then $$|x+3| - |4-x| = |8+x|$$ expands as $$(x+3)-(4-x)=(8+x)$$ --> $$x = 9$$, which is also not a valid solution since we are considering $$-3<x<4$$ range;

If $$x\geq{4}$$ then $$|x+3|-|4-x|=|8+x|$$ expands as $$(x+3)+(4-x)=(8+x)$$ --> $$x = -1$$, which is also not a valid solution since we are considering $$x>4$$ range.

So we have that no solution exist for $$|x+3|-|4-x|=|8+x|$$ (no x can satisfy this equation).

Hope it's clear.

The implicit understanding is that when a equation for a range is simplified and they contradict the basic premises of the considered range then the range is not considered as the solution.
_________________

This time its personal..

Intern
Joined: 27 Aug 2012
Posts: 20
Followers: 0

Kudos [?]: 4 [0], given: 55

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

23 Nov 2012, 23:18
Hi Bunnel,

I follow all your solutions. Really great!!!. Thanks for taking time to answer us.
I read the anwer explanations for the ques above. But still I am little unclear. Please help me.

I know |x-1| = 4 can be written as (x-1) = 4 and -(x-1) = 4. Fine.

But for the Ques |x+3| - |4-x| = |x+8|
I understand the key points are -8, -3, 4

For x<= - 8, First how did you take the condition -(x+3)-(4-x)=-(8+x) ?
Why not (x+3)-(4-x) = (x+8) or (x+3)+(4-x) = (8+x)?

And why x <= -8 is tested as condition? why not x >= -8.

Regards,
Sree
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13468
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

29 Dec 2014, 14:35
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Current Student
Joined: 23 Nov 2014
Posts: 60
Location: India
GMAT 1: 730 Q49 V40
GPA: 3.14
WE: Sales (Consumer Products)
Followers: 2

Kudos [?]: 27 [0], given: 64

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

14 May 2015, 11:00
Bunuel wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this

$$|x + 1|+|x - 1|\leq{2}$$ --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If $$x<-1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$-(x+1)-(x-1)\leq{2}$$ --> $$x\geq{-1}$$, not a valid range since we are considering $$x<-1$$;

If $$-1\leq{x}\leq{1}$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1-(x-1)\leq{2}$$ --> $$0\leq{2}$$ which is true, so for $$-1\leq{x}\leq{1}$$ given inequality holds true;

If $$x>1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1+x-1\leq{2}$$ --> $$x\leq{1}$$, not a valid range since we are considering $$x>1$$.

So, finally we have that $$|x + 1|+|x - 1|\leq{2}$$ holds true for $$-1\leq{x}\leq{1}$$.

For more check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

Hope it helps.

Hi Bunuel,

Should the ranges for checking be:

a) x < -1
b) -1 =< x < 1
c) x >= 1
?

Have you consciously altered the position of the greater than or equal and less than or equal signs for the three ranges because it would make no difference?

Hope my question is clear.

TIA.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2137

Kudos [?]: 13673 [4] , given: 222

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

14 May 2015, 22:38
4
KUDOS
Expert's post
3
This post was
BOOKMARKED
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this

You can solve this question using the number line approach too.

"Distance of x from -1" + "Distance of x from 1" <= 2

----------(-2)-----(-1)-----0-----(1)-----(2)--------

Note that distance between -1 and 1 is 2 so whenever x is between these two values, the sum of distance from -1 and 1 will be 2

----------(-2)-----(-1)---x--0-----(1)-----(2)--------

Hence all values between -1 and 1 (inclusive) will satisfy this condition.

When you go to the right of 1 or left of -1, the sum of distances from -1 and 1 will exceed 2. So the only range that satisfies the inequality is
-1 <= x <= 1
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Joined: 08 Sep 2012
Posts: 66
Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)
Followers: 0

Kudos [?]: 9 [0], given: 251

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

11 Sep 2015, 22:34
Bunuel wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this

$$|x + 1|+|x - 1|\leq{2}$$ --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If $$x<-1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$-(x+1)-(x-1)\leq{2}$$ --> $$x\geq{-1}$$, not a valid range since we are considering $$x<-1$$;

If $$-1\leq{x}\leq{1}$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1-(x-1)\leq{2}$$ --> $$0\leq{2}$$ which is true, so for $$-1\leq{x}\leq{1}$$ given inequality holds true;

If $$x>1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1+x-1\leq{2}$$ --> $$x\leq{1}$$, not a valid range since we are considering $$x>1$$.

So, finally we have that $$|x + 1|+|x - 1|\leq{2}$$ holds true for $$-1\leq{x}\leq{1}$$.

For more check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

Hope it helps.

Dear Bunuel
I didnt get one point in highlighted part:
x + 1 - (x - 1) <= 2
=> x + 1 - x + 1 <=2
=> 2 <= 2
But you are getting 0 <=2
Did i miss anything?
_________________

+1 Kudos if you liked my post! Thank you!

Director
Joined: 10 Mar 2013
Posts: 608
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Followers: 15

Kudos [?]: 266 [0], given: 200

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

19 Oct 2015, 02:09
Bunuel wrote:
sandal85 wrote:
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this

$$|x + 1|+|x - 1|\leq{2}$$ --> we have two check points -1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check.

If $$x<-1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$-(x+1)-(x-1)\leq{2}$$ --> $$x\geq{-1}$$, not a valid range since we are considering $$x<-1$$;

If $$-1\leq{x}\leq{1}$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1-(x-1)\leq{2}$$ --> $$0\leq{2}$$ which is true, so for $$-1\leq{x}\leq{1}$$ given inequality holds true;

If $$x>1$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1+x-1\leq{2}$$ --> $$x\leq{1}$$, not a valid range since we are considering $$x>1$$.

So, finally we have that $$|x + 1|+|x - 1|\leq{2}$$ holds true for $$-1\leq{x}\leq{1}$$.

For more check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

Hope it helps.

Hi Bunuel,
I have one questions regarding absolute value equations: why can't we use the same logis (as stated below) to solve this kind of equations --> |x + 1| + |x - 1| <= 2. We check here ranges, but in the example below we don't need doing so. Thanks in advance.
|2x - 1| = |4x + 9|
Solution: x =-4/3 or-5
2x-1=4x+9 or 2x-1=-(4x+9)
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Manager
Joined: 01 Jan 2015
Posts: 63
Followers: 0

Kudos [?]: 60 [0], given: 14

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

19 Oct 2015, 18:04
Bunuel wrote:
If $$-1\leq{x}\leq{1}$$ then $$|x + 1|+|x - 1|\leq{2}$$ expands as $$x+1-(x-1)\leq{2}$$ --> $$0\leq{2}$$ which is true, so for $$-1\leq{x}\leq{1}$$ given inequality holds true;

Bunuel, there is a slight typo. It should read $$2\leq{2}$$ instead of $$0\leq{2}$$
Intern
Joined: 26 Oct 2014
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 7

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

12 Nov 2015, 20:06
Bunuel wrote:
pavanpuneet wrote:
Inequalities have been quite a nightmare for me.. I understood the overall solution but, based on what do we decide to x<-8 and not x<=-8....?? I know this may be a very basic question, but can you please explain this concept for these ranges. Will be a great help! Thanks in advance.

Rice wrote:
Hi, Bunuel!
How many solutions does have?

In your solution when you define intervals, how do you define what inequality sign to put? I mean < or> or<=
For example, why the second interval is -8<=x<=-3, why both inequality signs are <=?

Thank you in advance

We should include "check points" (-8, -3, and 4) in either of the ranges but it really doesn't matter in which one. So, for example we could have written:
If $$x\leq{-8}$$ ...
If $$-8<x<-3$$ ...
If $$-3\leq{x}<\leq{4}$$ ...
If $$x>4$$ ...

Hope it's clear.

Hi Bunuel,

For the question posted, if the ranges where
x<= -1, then -(x+1)-(x-1) <=2, gives x>=-1 (Not Satisfied)
-1<x<1, then (x+1) -(x-1) <=2 , gives 2 <=2 (Satisfied)
x>=1, then (x+1)+(x-1)<=2, gives x<=1(Not Satisfied)

So I will choose the range as -1<x<1. But the correct answer is -1<=x<=1.
What am i missing?

Regards,
Deepthi
Intern
Joined: 09 Aug 2015
Posts: 4
Followers: 0

Kudos [?]: 0 [0], given: 220

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

### Show Tags

12 May 2016, 23:12
Hi,

is it correct if i follow below approach to solve this

|x+1| + |x-1| <=2

|2x| <=2 --> this expands to

2x <=2 or -2x <=2

x<=1 or x>=-1 i.e. -1<=x<=1
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2   [#permalink] 12 May 2016, 23:12

Go to page    1   2    Next  [ 23 posts ]

Similar topics Replies Last post
Similar
Topics:
Given that 1/x < -2/3, which of the following cannot be the value of x 1 18 Nov 2016, 02:58
If 1/(x(x+1))+1/((x+1)(x+2))=1/(x+1), what is the value of x? 3 20 Aug 2016, 23:04
3 If (x – 1)^2 = 144, which of the following could be the value of 2x? 5 12 Aug 2016, 08:09
If 3x + 1 < 2x < 9 - x, which of the following must be true? 5 11 Feb 2014, 06:08
68 If x is an integer and |1-x|<2 then which of the following 22 03 Sep 2012, 22:59
Display posts from previous: Sort by

# Which values of x are solutions |x + 1| + |x - 1| <= 2

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.