Question

How to solve complex absolute value inequalities like `absx+ abs(x-2)< 5`?

Answer 1

The solution is `x in (-3/2, 7/2)`

Explanation:

This is an inequality with absolute values.

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

`|x|+|x-2|-5<0`

Let `f(x)=|x|+|x-2|-5`

`{(x>=0),(x-2>=0):}`, `<=>`, `{(x>=0),(x>=2):}`

Let's build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaaaa)` `0` `color(white)(aaaaaaaaa)` `2` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaaaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `x-2` `color(white)(aaaaaaaa)` `-` `color(white)(aaaaaa)` `-` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `|x|` `color(white)(aaaaaaaaaa)` `-x` `color(white)(aaaaaaa)` `x` `color(white)(aaaaaaaa)` `x`

`color(white)(aaaa)` `|x-2|` `color(white)(aaaaaa)` `-x+2` `color(white)(aa)` `-x+2` `color(white)(aaaa)` `x-2`

In the interval `(-oo,0)`

`-x-x+2-5<0`, `=>`, `2x > -3`, `=>`, `x> -3/2`

`-3/2 in (-oo, 0)`

In the interval `(0,2)`

`x-x+2-5<0`, `=>`, `0-3<0`, `=>`, solution

In the interval `(2,+oo)`

`x+x-2-5<0`, `=>`, `2x<7`, `=>`, `x<7/2`

`7/2 in (2, +oo)`

The solution is `x in (-3/2, 7/2)`

graph{|x|+|x-2|-5 [-10, 10, -5, 5]}