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Which values of x are solutions |x + 1| + |x - 1| <= 2 21 May 2012, 09:14
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Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?


A. \(-2\leq{x}\leq{-1}\).

B. \(0\leq{x}\leq{2}\).

C. \(-7<{x}<{-6}\).

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

E. \(2\leq{x}\leq{3}\).





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Apart from algebra can we think conceptually to solve this
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Which values of x are solutions |x + 1| + |x - 1| <= 2 21 May 2012, 09:29
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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
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\(|x + 1|+|x - 1|\leq{2}\) --> 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: https://gmatclub.com/forum/math-absolute ... 86462.html

Hope it helps.
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Which values of x are solutions |x + 1| + |x - 1| <= 2 14 May 2015, 23:38
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sandal85
Which values of x are solutions to the inequality |x + 1| + |x - 1| <= 2 ?
Apart from algebra can we think conceptually to solve this
Show more

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
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Which values of x are solutions |x + 1| + |x - 1| <= 2 22 May 2012, 01:28
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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.
Show more

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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Which values of x are solutions |x + 1| + |x - 1| <= 2 22 May 2012, 10:25
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Bunuel
sandal85
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"?
Show more

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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Which values of x are solutions |x + 1| + |x - 1| <= 2 23 May 2012, 01:45
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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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Which values of x are solutions |x + 1| + |x - 1| <= 2 23 May 2012, 06:05
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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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Which values of x are solutions |x + 1| + |x - 1| <= 2 23 May 2012, 06:29
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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"?
Show more
pavanpuneet
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?
Show more



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.
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Which values of x are solutions |x + 1| + |x - 1| <= 2 23 May 2012, 07:29
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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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Which values of x are solutions |x + 1| + |x - 1| <= 2 23 May 2012, 21:03
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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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Which values of x are solutions |x + 1| + |x - 1| <= 2 24 May 2012, 00:51
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pavanpuneet
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.
Show more
Rice
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
Show more

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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Which values of x are solutions |x + 1| + |x - 1| <= 2 17 May 2016, 08:51
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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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Which values of x are solutions |x + 1| + |x - 1| <= 2 17 May 2016, 08:58
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msk0657
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.
Show more

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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Which values of x are solutions |x + 1| + |x - 1| <= 2 11 Feb 2017, 07:03
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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: https://gmatclub.com/forum/math-absolute ... 86462.html

Hope it helps.
Show more

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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Which values of x are solutions |x + 1| + |x - 1| <= 2 11 Feb 2017, 11:35
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Bunuel
sandal85
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: https://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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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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Which values of x are solutions |x + 1| + |x - 1| <= 2 23 Feb 2017, 23:56
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VeritasPrepKarishma
sandal85
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
Show more




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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Which values of x are solutions |x + 1| + |x - 1| <= 2 24 Feb 2017, 03:26
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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

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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This concept with 3 terms has been discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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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twobagels
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Which values of x are solutions |x + 1| + |x - 1| <= 2 23 Jan 2021, 16:36
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You asked for a non-algebraic way to deduce the range of values of 'x' that satisfy the inequality. I initially graphed it on paper, but then used calc to graph it out.

since you have |x -1 | and | x+1| on the LHS as one gets smaller the other absolute value gets larger. Since the value of 'x' must must yield output less than equal to to, from looking at the graph, you can see the range is -1 to 1.
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adityasuresh
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Which values of x are solutions |x + 1| + |x - 1| <= 2 28 Mar 2023, 19:49
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First we equate the expression within mod to zero, then we get x to be 1 and -1, we draw a number line which divides the values of x into 3 regions,<-1,between -1&1 &>1. While substituting values of x in the respective region, say in region 1(<-1) so the expression x +1 becomes negative so we multiply by -1, similarly the expression x -1 becomes negative so we multiply the expression by -1. We get -x -1 -x+1<=2, x>= -1.(dividing by -2 flips the sign),similarly we can calculate for other regions in region 2 the values of x are between -1 and 1 simplifying we get 2 <=2 so all values in this region satisfy the expression, finally in region 3 both expressions are positive so writing as is we get 2x <=2, x<=1
So combing the inequalities we get -1<=x<=1
D
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Which values of x are solutions |x + 1| + |x - 1| <= 2 02 Sep 2024, 02:21
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Hello, by solving the equations, I can have −1≤x≤1 and 0≤x≤2.

So I understand D is right, but why B is not right?
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