# How do you solve absolute value inequality -3abs(2x-5)<9?

Apr 9, 2015

Normally we would consider the two cases $\left(2 x - 5\right)$ is negative and $\left(2 x - 5\right)$ is positive or zero, separately and evaluate each case for restrictions on the value of $x$.

However, in this case
$\left(- 3\right) \left\mid 2 x - 5 \right\mid < 9$
can be re-written as
$\left\mid 2 x - 5 \right\mid > - 3$
(by dividing both sides by $\left(- 3\right)$ and reversing the inequality)

$\left\mid 2 x - 5 \right\mid \ge 0 \succ 3$
for all values of $x$ (based on the definition of absolute value

so the solution to the given inequality is $x \epsilon \left(- \infty , + \infty\right)$