Answer
$$\left| {x - 1} \right| \geqslant 3$$
Work Step by Step
$$\eqalign{
& \left( { - \infty , - 2} \right]{\text{or}}\left[ {4,\infty } \right) \cr
& {\text{Use the interval notations for an Unbounded Intervals}} \cr
& \left( {page{\text{ 1152}}} \right) \cr
& \left( { - \infty ,b} \right] = \left\{ {x:x \leqslant b} \right\},{\text{ }}\left[ {a,\infty } \right) = \left\{ {x:x \geqslant a} \right\} \cr
& \left\{ {x:x \leqslant - 2} \right\}{\text{or}}\left\{ {x:x \geqslant 4} \right\} \cr
& {\text{Substracting 1 from each side on both inequations}} \cr
& \left\{ {x:x - 1 \leqslant - 3} \right\}{\text{or}}\left\{ {x:x - 1 \geqslant 3} \right\} \cr
& {\text{,Then using the definition of absolute value}} \cr
& \left| {x - a} \right| \geqslant b:{\text{ }}x - a \geqslant b{\text{ or }}x - a \leqslant - b \cr
& {\text{Set }}a = 1{\text{ and }}b = 3 \cr
& \left| {x - 1} \right| \geqslant 3:{\text{ }}x - 1 \geqslant 3{\text{ or }}x - 1 \leqslant - 3 \cr
& {\text{The answer is:}} \cr
& \left| {x - 1} \right| \geqslant 3 \cr} $$
