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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]
21 May 2012, 08:29

5

This post received KUDOS

Expert's post

7

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 \(-1\leq{x}\leq{1}\).

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]
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 \(-1\leq{x}\leq{1}\).

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"?

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]
22 May 2012, 09:25

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 \(-1\leq{x}\leq{1}\).

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.

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]
23 May 2012, 05:29

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).

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

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

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]
23 May 2012, 23:51

1

This post received 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 \(x>4\) ...

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

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

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

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]
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 \(-1\leq{x}\leq{1}\).

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?

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 _________________

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]
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 \(-1\leq{x}\leq{1}\).

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!

gmatclubot

Re: Which values of x are solutions |x + 1| + |x - 1| <= 2
[#permalink]
11 Sep 2015, 22:34

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...