Use the definition of absolute value to rewrite each equation or inequality.
[tex]|x| = \begin{cases} x & \text{if } x\ge0 \\ -x & \text{if }x < 0 \end{cases}[/tex]
1. (E) From the definition it follows that
[tex]x-4 \ge 0 \implies |x-4| = x-4 = 8 \implies x = 12[/tex]
[tex]x-4 < 0 \implies |x-4| = -(x-4) = 4-x = 8 \implies x = -4[/tex]
Then the solution set contains exactly 2 elements, and we write it as shown in choice (E), [tex]x\in\{-4,12\}[/tex].
2. (D) We solve the inequality in essentially the same way as in (1), we just need to keep track of the direction of the inequality.
[tex]x-4\ge0 \implies|x-4|=x-4\ge8 \implies x\ge12[/tex]
[tex]x-4<0 \implies|x-4|=4-x\ge8 \implies x\le-4[/tex]
Note the inclusion of [tex]x=12[/tex] and [tex]x=-4[/tex]. We write this as a union of two half-closed intervals, [tex]x\in(-\infty,-4]\cup[12,\infty)[/tex].
3. (C) This follows from the same steps as in (2). This time the inequality is strict, so we exclude the endpoints and with open intervals write [tex]x\in(-\infty,4)\cup(12,\infty)[/tex].
4. (A) Since [tex]|x|[/tex] always returns a non-negative number, any real number [tex]x[/tex] satisfies the inequalty [tex]|x-4|<\infty[/tex]. We write the solution set as the entire real line, [tex]x\in(-\infty,\infty)[/tex].
5. This leaves us with (B) for the last solution set. This inequality is complementary to the one in (3), which means the solution set to [tex]|x-4|\le8[/tex] is the complement of the set we found in (3). That interval removes everything between and including -4 and 12 from the real line. So the solution in this case is what we omit from the solution to (3), and we write [tex]x\in[-4,12][/tex].