Q.Solve the modulus inequality $|x - 3| < 4$ for real $x$, and represent the solution on the number line.

[!TLDR] Use $|x-c|<a \iff c-a<x<c+a$ with $c=3, a=4$. [!ANSWER] $x \in (-1, 7)$; number line: open circles at $- …

The inequality $|x-3|<4$ has the form $|x-c|<a$ with $c=3$ and $a=4$. By the modulus rule, this is equivalent to $c-a < x < c+a$, i.e. $3-4 < x < 3+4$, i.e. $-1 < x < 7$. (Equivalently, by definition, $|x-3|<4$ means $-4 < x-3 < 4$; adding $3$ throughout gives $-1<x<7$ directly.) On the number line, since both ends are strict, this …

Apply $|x-c|<a \iff c-a<x<c+a$ directly, or equivalently expand to $-4<x-3 …

A common error is writing the solution as $|{-1}| < x < |7|$ or otherwise forgetting to shift the interval by $c=3$ …