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

4

This post received KUDOS

Expert's post

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

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

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.

Regards, Sree

gmatclubot

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