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

May 26, 2015

First of all, simplify the inequality.

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

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 $x \ge 0$ and
$| x | = - x$ for all $x < 0$.
Therefore, we can look for solutions of our inequality for $x \ge 0$ and, separately, for $x < 0$.

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

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,
$0 \le x < 2.5$ and $- 2.5 < x < 0$,
we obtain one interval for $x$ - the solution to our inequality:
$- 2.5 < x < 2.5$.