1. ## Quick absolute value inequality explanation, please

Hey everyone, I'm struggling with some of the concepts behind absolute value inequalities and found a certain problem I need some assistance with. I'll explain what I tried, please point out my mistake and explain why it is a mistake. Much appreciation.

Basically $\displaystyle (|x|+1)/(|x|-1) < 1$

if x >= 0

$\displaystyle (x+1-x+1) / x-1 < 0$

which gives x < 1

and if x < 0

$\displaystyle (-x+1)/(-x-1) < 1$

which gives x >= 0

$\displaystyle [0;1)U(1;infinity)$

2. Originally Posted by Squared
Basically $\displaystyle (|x|+1)/(|x|-1) < 1$ If x >= 0 $\displaystyle (x+1-x+1) / x-1 < 0$ which gives x < 1

Right, if $\displaystyle x\geq 0$ then:

$\displaystyle \dfrac{|x|+1}{|x|-1}<1 \Leftrightarrow \dfrac{x+1}{x-1}<1\Leftrightarrow 1+\dfrac{2}{x-1}<1\Leftrightarrow \dfrac{2}{x-1}<0\Leftrightarrow x\in[0,1)$

Fernando Revilla

3. write the inequality as $\displaystyle \dfrac2{|x|-1}<0$ so $\displaystyle |x|<1\implies -1<x<1.$

perhaps is not what the OP intented to ask, but it's the faster way on solving it.

4. @fernando

Gracias Um, my question might be a bit confusing, forgive me, what I actually want to know is I did everything like you and got the same answer but when x < 0 should I get (1;infinity), should I make a union between the two and is my final answer correct?

[0;1)U(1,infinity)

5. If $\displaystyle x<0$ :

$\displaystyle \dfrac{|x|+1}{|x|-1}<1 \Leftrightarrow \dfrac{-x+1}{-x-1}<1\Leftrightarrow 1-\dfrac{2}{x+1}<1\Leftrightarrow \dfrac{2}{x+1}>0\Leftrightarrow x\in(-1,0)$

Fernando Revilla

6. Yay, forgot to divide by the -1, works fine now, gracias Fernando, also thanks for the quick method Krizalid.

7. Originally Posted by Squared
Hey everyone, I'm struggling with some of the concepts behind absolute value inequalities and found a certain problem I need some assistance with. I'll explain what I tried, please point out my mistake and explain why it is a mistake. Much appreciation.

Basically $\displaystyle (|x|+1)/(|x|-1) < 1$

if x >= 0

$\displaystyle (x+1-x+1) / x-1 < 0$

which gives x < 1

Good!

and if x < 0

$\displaystyle (-x+1)/(-x-1) < 1$

which gives

$\displaystyle [0;1)U(1;infinity)$
Setting it up as you did for $\displaystyle x\ge0$

to get for $\displaystyle x\le0$

$\displaystyle \displaystyle\frac{-x+1}{-x-1}<1\Rightarrow\frac{-x+1}{-x-1}-\frac{-x-1}{-x-1}<0$

$\displaystyle \displaystyle\frac{-x+1-(-x-1)}{-x-1}<0\Rightarrow\frac{-x+x+1+1}{-x-1}<0$

$\displaystyle \displaystyle\frac{-2}{x+1}<0\Rightarrow\ x+1>0\Rightarrow\ x>-1$

8. Hey again everyone, I just wanted to ask, do any of you perhaps have a link to some advanced absolute value inequalities to solve. I googled but kept getting relatively VERY easy ones compared to the ones we are doing. I then tried searching for compound absolute value inequalities and found some who were harder but still too easy?

9. Originally Posted by Squared
Hey everyone, I'm struggling with some of the concepts behind absolute value inequalities and found a certain problem I need some assistance with. I'll explain what I tried, please point out my mistake and explain why it is a mistake. Much appreciation.
Basically $\displaystyle (|x|+1)/(|x|-1) < 1$
I have a different take on this altogether.
Consider the case $\displaystyle |x|>1$ then $\displaystyle |x|-1>0$.
This leads to $\displaystyle |x|+1<|x|-1$ or $\displaystyle 1<-1$. Impossible.

On the other hand, $\displaystyle |x|<1$ gives $\displaystyle |x|+1>|x|-1$ which is true.
So the solution set is $\displaystyle \{x:-1<x<1\}$

10. Originally Posted by Squared
Hey again everyone, I just wanted to ask, do any of you perhaps have a link to some advanced absolute value inequalities to solve. I googled but kept getting relatively VERY easy ones compared to the ones we are doing. I then tried searching for compound absolute value inequalities and found some who were harder but still too easy?
try this http://www.mathhelpforum.com/math-he...es-132202.html

11. Thanks, I'll work through it this weekend, last 3 examples are basically what I'm looking for, but still require some that are harder.