Question

How do you solve `abs(2x+9)-2x>=0`?

Answer 1

`abs(2x+9)-2x>=0` is true for all `x in RR`

Explanation:

`abs(2x+9)-2x>=0 -> abs(2x+9)-(2x+9)+9 ge 0`

then, for `x ne -9/2`

`1-(2x+9)/abs(2x+9)+9/abs(2x+9) ge 0`

Calling `9/abs(2x+9) = epsilon ge 0` we have

`1 pm 1 + epsilon ge 0`. This is always true so

`abs(2x+9)-2x>=0` is true for all `x in RR`