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

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.

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?

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.

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

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.

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: \(|5|=5\)

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.

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.

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.

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

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?