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### 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 V
Joined: 02 Sep 2009
Posts: 56277
Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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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.
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Manager  Joined: 02 Jun 2011
Posts: 121
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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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 V
Joined: 02 Sep 2009
Posts: 56277
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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1
1
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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Manager  Joined: 02 Jun 2011
Posts: 121
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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Thankyou Bunuel,
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: 92
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 56277
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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1
1
kashishh wrote:
Thankyou Bunuel,
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.
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Manager  Joined: 26 Dec 2011
Posts: 92
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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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.
Intern  Joined: 08 Oct 2007
Posts: 29
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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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 <=?

Math Expert V
Joined: 02 Sep 2009
Posts: 56277
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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4
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 <=?

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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Intern  Joined: 30 Mar 2012
Posts: 31
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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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]

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Hi Bunnel,

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

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

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5
3
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
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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

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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)
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Manager  Joined: 01 Jan 2015
Posts: 62
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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

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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 <=?

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
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Joined: 09 Aug 2015
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Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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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
Intern  Joined: 03 Jan 2016
Posts: 4
Re: Which values of x are solutions |x + 1| + |x - 1| <= 2  [#permalink]

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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

We need to find the value of x such that the sum of the distance of x from -1 and 1 is less than or equal to 2. If the value of x is taken less than -1, then the distance of x from 1 will be more than 2. Similarly, if value of x is greater than 1, then the distance of x from -1 will be greater than 2. So, the only possible region where the sum can be <=2 is -1<=x<=1  Re: Which values of x are solutions |x + 1| + |x - 1| <= 2   [#permalink] 15 May 2016, 03:15

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