# How to solve 2 absolute value inequalities abs(x+1)+abs(x-1)<=2?

Apr 7, 2015

With respect to the two absolute value expressions
$\left\mid x + 1 \right\mid$ and $\left\mid x - 1 \right\mid$
there are two values of $x$ for which the absolute value operation becomes significant:
$x = - 1$ and $x = + 1$

This gives us three ranges to consider:
(1) $x \le - 1$
(2)$- 1 < x < + 1$ and
(3)$x \ge + 1$

Case 1 : $x \le - 1$
Both $\left(x + 1\right)$ and and $\left(x - 1\right)$ are zero or negative
so the original inequality can be re-written as
$- \left(x + 1\right) - \left(x - 1\right) \le 2$
$- 2 x \le 2$
$x \ge - 1$
The only value that satisfies $\left(x \le - 1\right)$ and $\left(x \ge - 1\right)$
is $\left(x = - 1\right)$

Case 2: $- 1 < x < + 1$
Of the two terms involving $x$ only $\left(x - 1\right)$ will be negative and the inequality can be re-written as
$\left(x + 1\right) - \left(x - 1\right) \le 2$
$2 \le 2$
This is true for all values of $x$ in the "Case 2" range.

Case 3: $x \ge + 1$
Both of the terms involving $x$ are positive and the inequality can be re-written as
$\left(x + 1\right) + \left(x - 1\right) \le 2$
$2 x \le 2$
$x \le 1$

Solution
Combining all three cases we get
$- 1 \le x \le + 1$
as the solution range.