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

Feb 9, 2017

$x \in \left\{\left(\emptyset \mathmr{if} a < 0\right) , \left(\left[0 , 2 a\right] \mathmr{if} a \ge 0\right)\right.$

#### Explanation:

First, note that if $a < 0$ then there are no solutions, as the absolute value function always returns a nonnegative value. For the remainder of the problem, then, we will assume $a \ge 0$.

Using the definition of the absolute value function

$| x | = \left\{\begin{matrix}x \mathmr{if} x \ge 0 \\ - x \mathmr{if} x < 0\end{matrix}\right.$

we consider two cases.

Case 1: $x - a \ge 0$

$\implies | x - a | = x - a$

$\implies x - a \le a$

$\implies x - a + a \le a + a$

$\implies x \le 2 a$

Note that the initial condition is equivalent to $x \ge a$, however, so our solution set in this case is $x \in \left[a , 2 a\right]$.

Case 2: $x - a < 0$

$\implies | x - a | = - \left(x - a\right)$

$\implies - \left(x - a\right) \le a$

$\implies - \left(x - a\right) + x - a \le a + x - a$

$\implies 0 \le x$

Note that the initial condition in this case is equivalent to $x < a$, so together our solution set in this case becomes $x \in \left[0 , a\right)$.

Putting the two cases together, we get our total solution set for when $a \ge 0$:

$x \in \left[0 , a\right) \cup \left[a , 2 a\right] = \left[0 , 2 a\right]$.

and, for completion, we can include the results for when $a < 0$:

$x \in \left\{\left(\emptyset \mathmr{if} a < 0\right) , \left(\left[0 , 2 a\right] \mathmr{if} a \ge 0\right)\right.$