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

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

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New post 17 May 2016, 07:51
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.

Hi Bunuel,

Can you please help me to understand when to change sign ( + / - )...as you did for the above four conditions.
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

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New post 17 May 2016, 07:58
msk0657 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.

Hi Bunuel,

Can you please help me to understand when to change sign ( + / - )...as you did for the above four conditions.


Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\)


Theory on Absolute Values: math-absolute-value-modulus-86462.html
The E-GMAT Question Series on ABSOLUTE VALUE: the-e-gmat-question-series-on-absolute-value-198503.html
Properties of Absolute Values on the GMAT: properties-of-absolute-values-on-the-gmat-191317.html
Absolute Value: Tips and hints: absolute-value-tips-and-hints-175002.html

DS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Absolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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

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New post 11 Feb 2017, 06:03
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: http://gmatclub.com/forum/math-absolute ... 86462.html

Hope it helps.


Bunuel how did you get 0<=2 for the second range?? x gets crossed off so we get 2<=2 which becomes 0<=0 Could you please explain how you calculated that and how the second range is valid? Thank you.
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2 [#permalink]

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New post 11 Feb 2017, 10:35
OreoShake 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 \(x+1-(x-1)\leq{2}\) --> \(2\leq{2}\) 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: http://gmatclub.com/forum/math-absolute ... 86462.html

Hope it helps.


Bunuel how did you get 0<=2 for the second range?? x gets crossed off so we get 2<=2 which becomes 0<=0 Could you please explain how you calculated that and how the second range is valid? Thank you.


You'll get \(2\leq{2}\). Since this inequality is true, then we can say that for the range we consider (\(-1\leq{x}\leq{1}\)) given inequality holds true.
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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]

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New post 23 Feb 2017, 22:56
VeritasPrepKarishma 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


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





thanks a ton! that's pretty much clear. It would be great if you explain the range of values that satisfy |x+3| - |4-x| = |8+x| in the same method....


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

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New post 24 Feb 2017, 02:26
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umabharatigudipalli wrote:
VeritasPrepKarishma 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


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





thanks a ton! that's pretty much clear. It would be great if you explain the range of values that satisfy |x+3| - |4-x| = |8+x| in the same method....


Thanks,
Uma


This concept with 3 terms has been discussed here:
https://www.veritasprep.com/blog/2016/1 ... es-part-v/

Review the post and then try this question. Ask for help if needed. Will provide the solution.
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2   [#permalink] 24 Feb 2017, 02:26

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