The interval property reads: Given $b\in \mathbb{R}$ with $b>0$, we have $|x|<b$ if and only if $-b\lt x \lt b$. Corollary reads: Let $a,b\in \mathbb{R}$ with $b\gt 0$ be given. For all $y\in \mathbb{R}$ we have $|y-a|<b$ if and only if $a-b < y < a+b$.
This clearly works when an equation is something like $|x+2|<1$, but fails when $|x+2|>1$. Why?
i.e. the false solution is $-2-1>x>-2+1\iff-3>x>-1$. This clearly fails as $-3\not\gt-1$.
Because by definition of an absolute value we have: $$|y-a|>b$$ it's $$y-a>b$$ or $$y-a<-b$$ For example $$|y+2|>1$$ it's $$y+2>1$$ or $$y+2<-1,$$ which gives $$(-\infty,-3)\cup(-1,+\infty).$$ Also, $$|y+2|>-1$$ it's $\mathbb R$, which we can get by the previous way: $$y+2>-1$$ or $$y+2<1,$$ which is $$(-3,+\infty)\cup(-\infty,-1),$$ which is $\mathbb R.$