## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$t\lt-3 \text{ or } t\gt3$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|4t|\gt12 ,$ use the definition of greater than (greater than or equal to) absolute value inequality and solve each resulting inequality. $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} 4t\gt12 \\\\\text{OR}\\\\ 4t\lt-12 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 4t\gt12 \\\\ t\gt\dfrac{12}{4} \\\\ t\gt3 \\\\\text{OR}\\\\ 4t\lt-12 \\\\ t\lt-\dfrac{12}{4} \\\\ t\lt-3 .\end{array} Hence, the solution set is $t\lt-3 \text{ or } t\gt3 .$