The absolute value bars on your page are a disguise. Underneath every |x - c| inequality is an interval waiting to be written down. Once you see the pattern, you stop "solving" these problems — you translate them, and the interval appears in one line.

This is the recognition skill: the moment you read |x - c| < r, your pen should already be writing (c - r,\, c + r). No double inequality, no case-split algebra — just the rewrite, because you know what the symbols mean geometrically.

The instant rewrite, in one table

Let r > 0 throughout (negative r is a separate and much easier conversation — more on that below).

You see Read it as Write Because
\lvert x - c \rvert < r "distance from c is less than r" (c - r,\ c + r) open ball of radius r around c
\lvert x - c \rvert \le r "distance from c is at most r" [c - r,\ c + r] closed ball
\lvert x - c \rvert > r "distance from c is more than r" (-\infty,\ c - r) \cup (c + r,\ \infty) outside the ball — two open rays
\lvert x - c \rvert \ge r "distance from c is at least r" (-\infty,\ c - r] \cup [c + r,\ \infty) outside, endpoints included

That is the whole skill. The rest of this article is the why (so the table sticks), the two traps that bite most, and a worked example of each kind so you can see the translation happening live.

Why this works — one picture

On the number line, |x - c| is the distance between x and the point c. The inequality sets a budget on that distance.

The < / \le distinction controls the bracket type, exactly as it does for ordinary inequalities: strict means round (open, excluded), non-strict means square (closed, included).

Flowchart from an absolute-value inequality to its interval formA decision flowchart. A starting rectangle reads "|x − c| ? r with r > 0". Two arrows split to left and right. The left arrow, labelled "less than", leads to a rectangle that says "inside the ball", which points to the output "(c − r, c + r) — use square brackets if ≤". The right arrow, labelled "greater than", leads to a rectangle that says "outside the ball — two rays", which points to the output "(−∞, c − r) ∪ (c + r, ∞) — use square brackets if ≥". |x − c| ? r, r > 0 < or ≤ > or ≥ inside the ball outside — two rays (c − r, c + r) square brackets if ≤ (−∞, c − r) ∪ (c + r, ∞) square brackets if ≥
The absolute-value inequality is a distance condition. Less-than is inside the ball (one interval). Greater-than is outside the ball (union of two rays). The bracket type is decided by whether the inequality is strict.

Worked example: less-than

Rewrite |x + 2| < 7 as an interval.

The trick is to read |x + 2| as |x - (-2)| — so the "centre" is c = -2, not +2. The radius is r = 7.

Apply the rule:

|x - (-2)| < 7 \quad\Longleftrightarrow\quad x \in (-2 - 7,\ -2 + 7) = (-9,\ 5).

Done. No double-inequality bookkeeping, no sign case-splits. One line.

If you want a sanity check: the interval should be centred at c = -2 with radius 7. Midpoint of (-9, 5) is \frac{-9 + 5}{2} = -2. Half-width is \frac{5 - (-9)}{2} = 7. Both match. The rewrite is correct.

Worked example: greater-than

Rewrite |2x - 1| > 5 as a union of intervals.

Here the expression inside is 2x - 1, not x - c. But the pattern still applies — first to the linear expression, then divided out. One clean route is to rewrite 2x - 1 as 2\!\left(x - \tfrac{1}{2}\right) and then use |2 E| = 2|E|:

|2x - 1| > 5 \iff 2\!\left|x - \tfrac{1}{2}\right| > 5 \iff \left|x - \tfrac{1}{2}\right| > \tfrac{5}{2}.

Now apply the "greater than" rule with c = \tfrac{1}{2} and r = \tfrac{5}{2}: the solution is

\left(-\infty,\ \tfrac{1}{2} - \tfrac{5}{2}\right) \cup \left(\tfrac{1}{2} + \tfrac{5}{2},\ \infty\right) = (-\infty,\ -2) \cup (3,\ \infty).

Same answer as you would get by case-splitting 2x - 1 < -5 or 2x - 1 > 5, but reached by recognition instead of algebra. Once you have done this rewrite a few times, the intermediate step of factoring out the 2 becomes automatic — your eye sees |2x - 1| > 5 and immediately computes the two endpoints x = -2 and x = 3 from 2x - 1 = \pm 5.

Two traps that punish the rewrite

Trap 1. Negative r. The rule |x - c| < r \Leftrightarrow x \in (c - r, c + r) assumes r > 0. If you see |x - 3| < -4, stop — the left side is always \ge 0, so it can never be less than -4. The solution is the empty set \varnothing. Conversely, |x - 3| > -4 is satisfied by every real number, because any non-negative number is greater than -4; the solution is \mathbb{R} = (-\infty, \infty). These look like exotic edge cases in textbooks and surprisingly often show up on exam papers exactly to catch students who forgot to check the sign of the right-hand side.

Trap 2. The sign of the thing inside the bars. The pattern is |x - c| with a minus. If you read |x + 5| \le 2 and write x \in (5 - 2, 5 + 2) = (3, 7), you have the wrong centre. The correct reading is |x - (-5)| \le 2, giving x \in [-7, -3]. A good habit: always rewrite the inside explicitly as |x - c| before you pull out c. The plus sign in |x + 5| hides a centre at -5, not +5.

When the inside is not x - c — a quick route

If you see |a x + b| < r for some constants a \neq 0 and b, factor a out of the inside:

|a x + b| = |a|\cdot\left|x + \tfrac{b}{a}\right| = |a| \cdot \left|x - \left(-\tfrac{b}{a}\right)\right|.

So |a x + b| < r becomes |x - c| < \tfrac{r}{|a|} with c = -\tfrac{b}{a}. The centre is the root of ax + b = 0 and the radius is \tfrac{r}{|a|}. You can usually spot the centre without writing this out: it is wherever the inside of the bars equals zero.

Example: for |3x - 6| \le 9, the inside is zero at x = 2, so c = 2. The effective radius is 9 / |3| = 3. Answer: [2 - 3,\ 2 + 3] = [-1, 5]. Try it with the long method — you will get the same answer, just several steps later.

Why recognition beats algebra here

Case-splitting |x - c| < r into -r < x - c < r and solving is never wrong — it is just slow. On a JEE paper with a time budget of two minutes per question, the student who reads |x - 4| \le 3 and writes [1, 7] in six seconds is thirty seconds ahead of the student who sets up the double inequality, adds 4 to all three parts, and then writes [1, 7]. Thirty seconds, five questions deep, is more than an extra problem attempted.

Recognition is not a shortcut. It is a reading. You have done the case-split so many times, in your head, that the intermediate steps disappear and the interval is just what the symbols mean. That is the goal.

For a visual anchor — especially if the "ball around c" picture has not clicked yet — play with |x − c| < r as a Ball of Radius r Around c. Once the geometry is in your bones, the rewrite becomes automatic.