Question

How do you solve `3abs(x+2)-2>7`?

Answer 1

`x < -5`
or
`x > 1`

Explanation:

Given
`color(white)("XXXX")` `3abs(x+2)-2 >7`
Adding 2 to both sides maintains the inequality, so
`color(white)("XXXX")` `3abs(x+2) > 9`
Dividing both sides by a positive amount also maintains the inequality, so
`color(white)("XXXX")` `abs(x+2) > 3`

There are two possibilities.

Possibility 1: `(x+2) < 0 rarr x<-2`
In this case
`color(white)("XXXX")` `abs(x+2) = -(x+2)`
So
`color(white)("XXXX")` `abs(x+2) > 3`
becomes
`color(white)("XXXX")` `-x-2 > 3`
Adding 2 to both sides (which maintains the inequality)
`color(white)("XXXX")` `-x > 5`
Multiplying by `(-1)` (remembering that multiplication by a negative reverses the inequality)
`color(white)("XXXX")` `x < -5`
The two conditions for Possibility 1: `x < -2` and `x < -5` simplify to
`color(white)("XXXX")` `x<-5`

Possibility 2: `(x+2)>=0 rarr x>=-2`
In this case
`color(white)("XXXX")` `abs(x+2) = x+2`
So
`color(white)("XXXX")` `abs(x+2) >3`
becomes
`color(white)("XXXX")` `x+2 > 3`
Subtracting 2 from both sides (inequality maintained)
`color(white)("XXXX")` `x>1`
The two conditions for Possibility 2: `x>= -2` and `x > 1` simplify to
`color(white)("XXXX")` `x > 1`

Combining
Either Possibility 1 or Possibility 2
so
`color(white)("XXXX")` `x < -5` or `x > 1`