Inequalities Involving Absolute Value

Two master rules handle every absolute-value inequality:

Rule 1. $|x|<r\iff-r<x<r$ (and this forces $r>0$, since $|x|\ge0$). Proof idea: if $x\ge0$, $|x|=x$, so $|x|<r$ means $x<r$; if $x<0$, $|x|=-x$, so $|x|<r$ means $-x<r$, i.e. $x>-r$. Combining both cases gives $-r<x<r$.

Rule 2. $|x|>r\iff x<-r$ or $x>r$. (If $r<0$, every real $x$ satisfies this automatically, since $|x|\ge0>r$.) Proof idea: same two cases as Rule 1, but with the inequality reversed.

Shifted versions (replacing $x$ by $x-a$): for any $a\in R$,

Worked pattern. $|x-9|<2\iff-2<x-9<2\iff7<x<11$. And $\left|\dfrac2{x-4}\right|>1$ (with $x\ne4$) rearranges to $2>|x-4|$, i.e. $-2<x-4<2$, i.e. $2<x<6$ (excluding $x=4$), giving the solution set $(2,4)\cup(4,6)$. …