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

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.

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.

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.

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: \(|5|=5\)

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"?

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

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

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.

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.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.

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"?

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?

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.

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

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.

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.

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?