# How do you solve absx> x - 1 ?

May 24, 2015

When dealing with moduli it's often useful to split into cases where the enclosed value changes sign. For this example, that means examine the $x < 0$ and $x \ge 0$ cases separately.

Case (a) : $x < 0$

If $x < 0$ then $| x | = - x$ and the inequality takes the form

$- x > x - 1$

Add $x + 1$ to both sides to get:

$2 x < 1$

Divide both sides by $2$ to get:

$x < \frac{1}{2}$

Since we're dealing with the case $x < 0$ this condition is already satisfied.

Case (b) : $x \ge 0$

If $x \ge 0$ then $| x | = x$ and the inequality takes the form

$x > x - 1$

Subtract $x$ from both sides to get:

$0 > - 1$

...which is always true.

Putting these two cases together, we find that the original inequality is satisfied for all $x \in \mathbb{R}$.