Q.Solve $|x - 4| > 2$ for real $x$, and write the solution set as a union of two intervals.

[!TLDR] Use $|x-c|>a \iff x<c-a$ or $x>c+a$ with $c=4, a=2$. [!ANSWER] $x \i …

$|x-4|>2$ has the form $|x-c|>a$ with $c=4$, $a=2$. By the rule, this gives $x < c-a$ or $x > c+a$, i.e. $x < 4-2=2$ or $x>4+2=6$. Check $x=0$: $|0-4|=4>2$ ✓; check $x=4$ (the centre): $|4-4|=0$, not $>2$, correctly exclud …

Identify the centre $c=4$ and radius $a=2$, then apply the "greater than" modulus rule t …

Writing the answer as a single interval $2<x<6$ is the classic mistake here — that is the solution of $|x-4|<2$ …