Absolute value inequality $3 > |x + 4| \geq 1$

Question

I've just started with absolute value equations and I have a real hard time understanding how to solve this. I got the following question, and I can't make heads or tails out of it.

Assume that $x, y$ are points on the real line. Explain what $|x − y|$ means geometrically. Use this to illustrate the inequalities (as subsets of the real line) below and write the inequality without absolute values.

$3 > |x + 4| \geq 1$

All help will be much appreciated, even as much as small pointers will be more than welcome...

Answer 1

$|x-y|$ is the distance between $x$ and $y$ on a number line.

Answer 2

Building up on what Aleksandar already said. Your inequality can be written as:$$1\le|x-(-4)|<3$$ So geometrically, $x$ are the numbers on the real line such that they are at least a distance 1 from -4 and at most (but not equal to) a distance 3 from -4. So $$-7<x\le-5$$ or $$-3\le x<-1$$ A more systematic way of doing such questions is to consider the 2 possibilities i.e $1\le x+4<3$ and $1\le -(x+4)<3$ and simplying should give the same answer as above.