You know that $|x - a|$ is the distance between $x$ and $a$, from absolute value equations, and how “and” and “or” work from compound inequalities. Put the two together and absolute value inequalities read like sentences about distance.
Why it works
$|x - 3| < 5$ says “$x$ is less than $5$ units from $3$.” Those numbers are all within $5$ of $3$: between $-2$ and $8$. So it becomes the “and” statement $-5 < x - 3 < 5$. But $|x - 3| > 5$ says “$x$ is more than $5$ units from $3$,” which happens on either side: $x - 3 < -5$ or $x - 3 > 5$. Less than means one segment; greater than means two rays.
Key idea
Isolate the absolute value first. Then, for $k > 0$: $|u| < k$ means $-k < u < k$ (one segment), and $|u| > k$ means $u < -k$ or $u > k$ (two rays). The same holds with $\le$ and $\ge$.
Worked examples
Worked example: Less than: one segment
Solve $|2x - 1| \le 7$.
- Distance at most $7$ from $0$: $-7 \le 2x - 1 \le 7$.
- Add $1$ to all three parts: $-6 \le 2x \le 8$.
- Divide by $2$: $-3 \le x \le 4$.
Answer: $-3 \le x \le 4$
Worked example: Greater than: two rays
Solve $|x + 4| - 3 > 2$.
- Isolate first: add $3$ to get $|x + 4| > 5$.
- More than $5$ from $0$: $x + 4 < -5$ or $x + 4 > 5$.
- Solve each: $x < -9$ or $x > 1$.
Answer: $x < -9$ or $x > 1$.
Worked example: When the number is negative
Solve $|x - 2| < -1$, then $|x - 2| \ge -1$.
- An absolute value is never negative, so it can never be less than $-1$: no solution.
- For the same reason it is always at least $-1$: every real number works.
Answer: No solution; all real numbers.
Common mistake
Writing $|x| > 3$ as $-3 > x > 3$. That chain says $-3 > 3$, which is false, so it describes no numbers at all. The two pieces point away from each other, so they must be joined with “or”: $x < -3$ or $x > 3$. Check with a number like $5$: $|5| > 3$ is true, so $5$ must be in your answer.