Question

How do you solve `abs(2x)<=abs(x-3)`?

Answer 1

When dealing with moduli, it is often helpful to split into cases at values where the sign of the enclosed value changes.

For our example, `2x` changes sign at `x=0` and `x-3` changes sign at `x=3`. So split into cases:
(a) `x < 0`
(b) `x = 0`
(c) `0 < x < 3`
(d) `x = 3`
(e) `x > 3`

In case (a):
`|2x| = -2x` and `|x - 3| = -x+3`
So the original inequality is equivalent to `-2x <= -x+3`
Adding `2x-3` to both sides we get `-3 <= x`
Since this is case (a), we have `-3 <= x < 0`

In case (b):
`|2x| = |0| = 0` and `|x - 3| = |-3| = 3`
So the inequality `|2x| < |x-3|` is satisfied.
So `x=0` is also a solution.

In case (c):
`|2x| = 2x` and `|x - 3| = -x+3`
So the original inequality is equivalent to `2x <= -x+3`
Add x to both sides and divide both sides by 3 to get: `x <= 1`
Since this is case (c), we also require `0 < x < 3`, so this gives us solutions: `0 < x <= 1`.

In case (d):
`|2x| = 6` and `|x - 3| = 0`, so the original inequality is not satisfied.

In case (e);
`|2x| = 2x` and `|x - 3| = x - 3`
So the original inequality is equivalent to `2x <= x - 3`
Subtracting `x` from both sides we get `x <= -3`.
Since this is case (e), we also require `x > 3`, which cannot be satisfied at the same time.

The union of our solutions from cases (a)-(c) gives us:
`-3 <= x <= 1`