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

$\left\{−4,4\right\}$
Add 13 to both sides: $4|x|−13+13=3+13 \\4|x|=16$ Divide 4 to both sides: $\dfrac{4|x|}{4}=\dfrac{16}{4} \\|x|=4$ RECALL: $|x|=a \longrightarrow x=a \text{ or } x=−a$ Thus, $|x|=4 \longrightarrow x=4 \text{ or } x=−4$ Therefore, the solution set is: $\left\{−4,4\right\}$