Q.Solve $|x + 2| < 3$ for real $x$.

[!TLDR] Use $|x-c|<a \iff c-a<x<c+a$ with $c=-2, a=3$. [!ANSWER] $x \in (-5, 1)$.

Write $|x+2| = |x-(-2)|$, so $c=-2$ and $a=3$. By the rule $|x-c|<a \iff c-a<x<c+a$: $-2-3 < x < -2+3$, i.e. $-5 < x < 1$. (Equivalently: $|x+2|<3 \iff -3<x+2<3$; subtracting $2$ throughout gives $-5<x<1$.) [!ANSWER] $x \in (-5, 1)$.

Rewrite $x+2$ as $x-(-2)$ to identify $c=-2$, then apply $c-a<x<c+a$; or expand directly as $-3<x+2<3$ and subtract 2.

A frequent error is writing the centre as $c=2$ instead of $c=-2$, since $x+2$ is easy to misread as $x-(-2)$ backwards, giving the wrong interval $(-1,5)$.