# Thread: When? does the inequality solve

1. ## 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?

2. 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,
$|n|=\left\{ \begin{array}{c}n \mbox{ for }n\geq 0 \\ -n\mbox{ for }n<0 \end{array} \right\}$
Thus, we the the following cases,
$|x|=x , -x$
And it depends on, $x\geq 0, x<0$
Also we have,
$|2x-5|=2x-5,-2x+5$
And it depends on, $2x-5\geq 0,2x-5<0$
Solve the inequality,
$x\geq 2.5$ and $x<2.5$.
In combination with,
$x\geq 0$ and $x<0$.
We have the following cases,

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

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

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

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