# Absolute value of a real number

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.

• Note that if $a \le 0$, there is no $x$ such that $|x-4|\lt a$, because $|w|$ is always $\ge 0$. That will com out if you use the general machinery described by robjohn, but you might as well handle it separately. – André Nicolas Jun 2 '12 at 19:07
• Probably the hint should be: depending on whether $x-4$ is negative, positive, or zero. – GEdgar Jun 3 '12 at 23:29

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 $-(x-4)<a$.
• There is no modulus here; there is an absolute value. Try this: draw a number line from $0$ to $10$, then color the part of the number line that is closer than $2$ units to $4$. Try the same thing for the part closer than $3$ units to $4$. – robjohn Jun 2 '12 at 14:33