Question

How do you solve `abs(2x)<2+ abs3`?

Answer 1

First of all, simplify the inequality.

Since `|3|=3`, we can rewrite this inequality as
`|2x| < 5`

Both sides of an inequality can be divided by a positive constant without changing the sign of inequality, and the new inequality will be equivalent to an old one.
Let's divide out equation by `2`. The result is:
`|x| < 2.5`

The solution to this inequality is, obviously,
`-2.5 < x < 2.5`

The above solution can be obtained using the following logic:
By definition, `|x|` is defined as
`|x| = x` for all `x>=0` and
`|x| = -x` for all `x<0`.
Therefore, we can look for solutions of our inequality for `x>=0` and, separately, for `x<0`.

Case 1. Assume `x>=0`.
Then `|x|=x` and our inequality looks like `x < 2.5`.
Combining this with a requirement `x>=0`, we have the following solutions:
`0<=x<2.5`

Case 2. Assume `x < 0`.
Then `|x|=-x` and our inequality looks like `-x < 2.5` or `x > -2.5`..
Combining this with a requirement `x < 0`, we have the following solutions:
`-2.5 < x < 0`

Combining two areas that represent solutions,
`0 <= x < 2.5` and `-2.5 < x < 0`,
we obtain one interval for `x` - the solution to our inequality:
`-2.5 < x < 2.5`.