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

$\bf{\text{Solution Outline:}}$ To solve the given inqeuality, $|x+2|\gt x ,$ use the definition of a greater than (greater than or equal to) absolute value inequality and solve each resulting inequality. Finally, graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ 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} x+2\gt x \\\\\text{OR}\\\\ x+2\gt -x .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} x+2\gt x \\\\ x-x\gt -2 \\\\ 0\gt -2 \text{ (TRUE)} \\\\\text{OR}\\\\ x+2\gt -x \\\\ x+x\gt -2 \\\\ 2x\gt -2 \\\\ x\gt -\dfrac{2}{2} \\\\ x\gt -1 .\end{array} Hence, the solution is the set of all real numbers.