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

May 17, 2015

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

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

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

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

In case (e);
$| 2 x | = 2 x$ and $| x - 3 | = x - 3$
So the original inequality is equivalent to $2 x \le x - 3$
Subtracting $x$ from both sides we get $x \le - 3$.
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:
$- 3 \le x \le 1$