# Quick absolute value inequality explanation, please

• Feb 9th 2011, 10:17 AM
Squared
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 $(|x|+1)/(|x|-1) < 1$

if x >= 0

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

which gives x < 1

and if x < 0

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

which gives x >= 0

$[0;1)U(1;infinity)$
• Feb 9th 2011, 10:39 AM
FernandoRevilla
Quote:

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

Right, if $x\geq 0$ then:

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

Fernando Revilla
• Feb 9th 2011, 10:51 AM
Krizalid
write the inequality as $\dfrac2{|x|-1}<0$ so $|x|<1\implies -1

perhaps is not what the OP intented to ask, but it's the faster way on solving it. :D
• Feb 9th 2011, 10:58 AM
Squared
@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)
• Feb 9th 2011, 11:11 AM
FernandoRevilla
If $x<0$ :

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

Fernando Revilla
• Feb 9th 2011, 11:21 AM
Squared
Yay, forgot to divide by the -1, works fine now, gracias Fernando, also thanks for the quick method Krizalid.
• Feb 9th 2011, 12:07 PM
Quote:

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 $(|x|+1)/(|x|-1) < 1$

if x >= 0

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

which gives x < 1

Good!

and if x < 0

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

which gives

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

Setting it up as you did for $x\ge0$

to get for $x\le0$

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

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

$\displaystyle\frac{-2}{x+1}<0\Rightarrow\ x+1>0\Rightarrow\ x>-1$
• Feb 10th 2011, 06:08 AM
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?
• Feb 10th 2011, 06:43 AM
Plato
Quote:

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 $(|x|+1)/(|x|-1) < 1$

I have a different take on this altogether.
Consider the case $|x|>1$ then $|x|-1>0$.
This leads to $|x|+1<|x|-1$ or $1<-1$. Impossible.

On the other hand, $|x|<1$ gives $|x|+1>|x|-1$ which is true.
So the solution set is $\{x:-1
• Feb 10th 2011, 07:53 AM
Krizalid
Quote:

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
• Feb 10th 2011, 09:00 AM
Squared
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.