The absolute-value inequality |x - c| < r is usually taught as a mechanical rule: "drop the bars, write -r < x - c < r, add c everywhere." That works, but it hides the one idea that actually makes absolute-value inequalities easy — the idea that turns every problem in this article into a sentence you can read at a glance. The idea is this: |x - c| is the distance from x to c on the number line, and an inequality on that distance is a condition about how far x can wander from the centre c. Less-than is "stay within r." Greater-than is "get at least r away." You do not solve. You just read off the interval.
The widget below is a practice surface for that reading. You set the centre c, the radius r, and pick one of the four comparison signs (<, \le, >, \ge). The number line shades the solution set, puts open (hollow) or closed (filled) markers at the boundaries depending on strictness, and prints the interval in standard notation. Move the sliders around until the picture and the notation match your mental model without effort. That is the whole lesson.
The widget
Reading the inequality as a distance
Every absolute-value inequality in one variable can be rewritten in the canonical form |x - c|\ \square\ r, where c is a real number called the centre, r \ge 0 is a real number called the radius, and \square is one of <, \le, >, \ge. Once the inequality is in this form, the whole thing is a single sentence.
The key translation is:
You can see why: if x \ge c then x - c \ge 0, so |x - c| = x - c is exactly how far x sits to the right of c. If x < c then x - c < 0, so |x - c| = -(x - c) = c - x, which is how far x sits to the left of c. Either way, |x - c| is the unsigned gap between the two numbers. The absolute value throws away which side x is on and keeps only how far.
Now the four inequalities read themselves:
- |x - c| < r says "the distance from c is less than r" — stay strictly within r of c. Solution: the open ball (c - r, c + r).
- |x - c| \le r says "at most r from c" — stay within r, endpoints included. Solution: the closed ball [c - r, c + r].
- |x - c| > r says "strictly more than r from c" — be outside. Solution: (-\infty, c - r) \cup (c + r, \infty).
- |x - c| \ge r says "at least r from c" — outside, endpoints included. Solution: (-\infty, c - r] \cup [c + r, \infty).
The word "ball" is standard in higher mathematics. A ball of radius r around c is the set of points within distance r of c. In the real line this is just an interval; in the plane it is a disk; in space it is a solid sphere. The one-dimensional case hides behind interval notation, but the idea is the same — and calling it a ball is a good habit because it keeps the geometry in front of you.
Two worked examples
Example 1. |x - 3| < 2.
Centre c = 3, radius r = 2, sign <. The solution is "within 2 of 3" — the open ball (3 - 2, 3 + 2) = (1, 5). Both endpoints are excluded because the inequality is strict: at x = 1 or x = 5, the distance is exactly 2, not less than 2. Set the widget to c = 3, r = 2, sign <, and you will see the green segment between hollow circles at 1 and 5.
Example 2. |x + 2| > 4.
First rewrite in canonical form. Since x + 2 = x - (-2), the centre is c = -2 (not +2 — pay attention to the sign inside the bars). The radius is r = 4. The sign is strict greater-than, so you need to be more than 4 away from -2. The boundary points are -2 - 4 = -6 and -2 + 4 = 2, both excluded. The solution is the union of two rays: (-\infty, -6) \cup (2, \infty). Set the widget to c = -2, r = 4, sign >, and you will see two red rays shooting outward with hollow circles at -6 and 2.
Notice what changes between the two examples. Strict vs non-strict controls whether the endpoints are filled or hollow. Less-than vs greater-than controls whether the shaded set is the inside (one interval) or the outside (two rays). The centre c slides the whole picture left and right without changing its shape. The radius r stretches or shrinks the gap. Four independent knobs, and every absolute-value inequality is somewhere in that four-dimensional space.
Edge cases the widget handles
Set r = 0 with the < sign. The inequality becomes |x - c| < 0, which asks for a negative distance. Distances are never negative, so the solution is empty — no x works. The widget prints \varnothing. With \le and r = 0, the only x satisfying |x - c| \le 0 is x = c itself; the solution collapses to the single point \{c\}. With > and r = 0, every x \ne c works, giving (-\infty, c) \cup (c, \infty). With \ge and r = 0, every x satisfies |x - c| \ge 0, so the solution is all of \mathbb{R}. These four cases are worth pushing the sliders to — they show that the "ball" picture still works when the ball has zero radius, you just have to track whether the centre itself belongs.
For more on the algebra that accompanies this geometric reading — including the two-case split, the case |x - c| < r with r < 0 (always empty), and how to combine absolute-value inequalities with other constraints — see the main article on absolute-value inequalities.