My question is:
Solve: $|x-4|< a$, where $a$ belongs to the real numbers. Solve this by considering various cases depending upon whether $a$ is negative, positive or zero.
What I have tried so far: If $a>0$ then: $x < a+4$ and $x>4-a$, if $a=0$ then there is no solution.
My doubt is: Should I consider the case $a<0$ as again $|x-4|<a$ which is not possible as absolute value cannot be negative.
Hint: $|x-4|<a$ means that $x$ is closer than $a$ units to $4$.
Another Hint: $|x-4|<a$ means that $(x-4)<a$ and $-(x-4)<a$.