Question

How do you solve `abs(3 + 2x)>abs(4 - x)`?

Answer 1

The solution is `x in (-oo, -7) uu(1/3, +oo)`

Explanation:

This is an inequality with absolute values

`|3+2x| >|4-x|`

`|3+2x| -|4-x| >0`

Let `f(x)=|3+2x| -|4-x|`

`{(3+2x>=0),(4-x>=0):}`, `<=>`, `{(x>=-3/2),(x<=4):}`

The sign chart is as follows :

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaaaa)` `-3/2` `color(white)(aaaaaaaaaaa)` `4` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `3+2x` `color(white)(aaaaa)` `-` `color(white)(aaaaaaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaaaaaaa)` `+`

`color(white)(aaaa)` `4-x` `color(white)(aaaaaa)` `+` `color(white)(aaaaaaa)` #color(white)(aaaaa)+# `color(white)(aaaaa)` `0` `color(white)(aaa)` `+`

`color(white)(aaaa)` `|3+2x|` `color(white)(aaaaa)` `-3-2x` `color(white)(aa)` `0` `color(white)(aaa)` `3+2x` `color(white)(aaaaaa)` `3+2x`

`color(white)(aaaa)` `|4-x|` `color(white)(aaaaaaaa)` `4-x` `color(white)(aaa)` #color(white)(aaaa)4-x# `color(white)(aaa)` `0` `color(white)(aa)` `-4+x`

In the interval `(-oo, -3/2)`,

`f(x)=-3-2x-4+x=-7-x`

`f(x)>0`, `=>`, `-7-x>0`, `=>`, `x<-7`

In the interval `[-3/2, 4]`,

`f(x)=3+2x-4+x=-1+3x`

`f(x)>0`, `=>`, `-1+3x>0`, `x>1/3`

In the interval `( 4, +oo)`,

`f(x)=3+2x+4-x=7+x`

`f(x)>0`, `=>`, `7+x>0`, `=>`, `x<-7`

This result is not valid since `x !in`(4, +oo)

The solution is

`x in (-oo, -7) uu(1/3, +oo)`