## College Algebra (10th Edition)

$\color{blue}{x=\left\{1, 3\right\}}$
RECALL: $|x| =a \longrightarrow x = a \text{ or } x = -a$ Thus, using the rule above gives: $|x-2| = 1 \longrightarrow x-2=1 \text{ or } x-2=-1$ Solve each equation to obtain: $x-2=1 \\x=1+2 \\x=3$ or $x-2=-1 \\x=-1+2 \\x=1$ Therefore, the solutions are: $\color{blue}{x=\left\{1, 3\right\}}$