# When? does the inequality solve

• Jan 1st 2007, 12:15 PM
Freaky-Person
When? does the inequality solve
In what cases would you have to change the inequality sign?

I mean, if you have an equation like,

|x| + |2x-5| < 5

You would have 3 cases right?

So what would the sign (< and >) look like for those 3 and why does it change some times?
• Jan 1st 2007, 12:54 PM
ThePerfectHacker
Quote:

Originally Posted by Freaky-Person
In what cases would you have to change the inequality sign?

I mean, if you have an equation like,

|x| + |2x-5| < 5

You would have 3 cases right?

So what would the sign (< and >) look like for those 3 and why does it change some times?

The traditional way you are taught in high-schools, that means look at the plus minus signs DOES NOT WORK in general. It only works in the problems they give you.
This happens to be a more advanced problem.
We need to use the following definition,
$\displaystyle |n|=\left\{ \begin{array}{c}n \mbox{ for }n\geq 0 \\ -n\mbox{ for }n<0 \end{array} \right\}$
Thus, we the the following cases,
$\displaystyle |x|=x , -x$
And it depends on, $\displaystyle x\geq 0, x<0$
Also we have,
$\displaystyle |2x-5|=2x-5,-2x+5$
And it depends on, $\displaystyle 2x-5\geq 0,2x-5<0$
Solve the inequality,
$\displaystyle x\geq 2.5$ and $\displaystyle x<2.5$.
In combination with,
$\displaystyle x\geq 0$ and $\displaystyle x<0$.
We have the following cases,

1)$\displaystyle x<0\to x<2.5$
$\displaystyle |x|=-x$ and $\displaystyle |2x-5|=-2x+5$

2)$\displaystyle x\geq 0 \mbox{ and }x<2.5$
$\displaystyle |x|=x$ and $\displaystyle |2x-5|=-2x+5$

3)$\displaystyle x\geq 2.5 \to x\geq 0$
$\displaystyle |x|=x$ and $\displaystyle |2x-5|=2x-5$.

Thus, consider the inequality
$\displaystyle |x|+|2x-5|<5$
If, the first case,
$\displaystyle -x-2x+5<5$
$\displaystyle -3x+5<5$
$\displaystyle -3x<0$
$\displaystyle x>0$
But that is a contradiction because $\displaystyle x<0$.
If, the second case,
$\displaystyle x-2x+5<5$
$\displaystyle -x+5<5$
$\displaystyle -x<0$
$\displaystyle x>0$
This will be a contradiction unless $\displaystyle 0<x<2.5$
If, the third case,
$\displaystyle x+2x-5<5$
$\displaystyle 3x-5<5$
$\displaystyle 3x<10$
$\displaystyle x<10/3$
This will be a contradiction unless $\displaystyle 2.5\leq x<10/3$
Thus,
$\displaystyle x=(0,2.5)\cup [2.5,10/3)=(0,10/3)$