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

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Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 21 May 2012, 08:14
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Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 21 May 2012, 08:29
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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.
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 22 May 2012, 00:28
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"?
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 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}.

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.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 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"?
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 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?
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 23 May 2012, 05:29
Expert's post
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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 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.
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 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
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 23 May 2012, 23:51
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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.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 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.
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink] New post 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
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2   [#permalink] 23 Nov 2012, 23:18
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