# How do you solve abs(x+1)+ abs(x-1)<=2?

May 30, 2015

With moduli it is useful to split into cases at the point that the sign of the enclosed expression changes. $\left(x + 1\right)$ changes sign at $x = - 1$ and $\left(x - 1\right)$ changes sign at $x = 1$....

Case (a) : $x < - 1$

$x + 1 < 0$ so $\left\mid x + 1 \right\mid = - \left(x + 1\right)$
$x - 1 < 0$ so $\left\mid x - 1 \right\mid = - \left(x - 1\right)$

$2 \ge \left\mid x + 1 \right\mid + \left\mid x - 1 \right\mid = - \left(x + 1\right) + - \left(x - 1\right)$
$= - x - 1 - x + 1$
$= - 2 x$

Dividing both ends by $- 2$ and reversing the inequality we get:
$x \ge - 1$

Notice that we have to reverse the inequality, because we are dividing by a negative number.

So in Case (a) we have $x < - 1$ and $x \ge - 1$
These conditions cannot be satisfied at the same time, so Case (a) yields no solutions.

Case (b) : $- 1 \le x \le 1$

$x + 1 \ge 0$ so $\left\mid x + 1 \right\mid = \left(x + 1\right)$
$x - 1 \le 0$ so $\left\mid x - 1 \right\mid = - \left(x - 1\right)$

$\left\mid x + 1 \right\mid + \left\mid x - 1 \right\mid = \left(x + 1\right) + - \left(x - 1\right) = 1 + 1 = 2 \le 2$

So the target inequality is satisfied for all $x$ in Case (b).
That is $- 1 \le x \le 1$.

Case (c) : $x > 1$

$x + 1 > 0$ so $\left\mid x + 1 \right\mid = \left(x + 1\right)$
$x - 1 > 0$ so $\left\mid x - 1 \right\mid = \left(x - 1\right)$

$\left\mid x + 1 \right\mid + \left\mid x - 1 \right\mid = \left(x + 1\right) + \left(x - 1\right) = 2 x > 2$
So in Case (c) the inequality is never satisfied.

So the solution is $- 1 \le x \le 1$