Representation on the Number Line

Once the solution set of a one-variable linear inequality has been found algebraically, it can be pictured directly on the number line — a single horizontal line marked with the real numbers in increasing order from left to right. This picture makes the size and boundedness of a solution set visible in a way a written interval alone does not.

Circles mark the endpoints. At any boundary point $p$ of the solution set:

• An open (unfilled) circle at $p$ means $p$ itself is not included — used for a strict inequality ($<$ or $>$) at that end.
• A closed (filled) circle at $p$ means $p$ is included — used for a slack inequality ($\le$ or $\ge$) at that end.

One-sided (unbounded) solution sets. If the solution has the form $x > k$ or $x \ge k$, the picture is a circle at $k$ (open or closed, per the rule above) together with an arrow extending indefinitely to the right, showing every real number beyond $k$ in that direction belongs to the solution set, without end. Similarly, $x < k$ or $x \le k$ is a circle at $k$ with an arrow extending indefinitely to the left.

Two-sided (bounded) solution sets. If the solution is a compound inequality such as $-2 \le x < 5$, the picture is a solid shaded segment joining the two endpoints, with the circle at each endpoint chosen independently according to whether that particular end is strict or slack — the two ends need not match. No arrow is drawn beyond either endpoint, since the set is bounded on both sides.

Reading the modulus solutions of Section 3 the same way. Since $|x-c| < a$ solves to the bounded interval $c-a < x < c+a$, it is drawn as an open circle at $c-a$, an open circle at $c+a$, and a shaded segment joining them. Since $|x-c| > a$ solves to two separate unbounded rays ($x < c-a$ or $x > c+a$), it is drawn as two disconnected pieces — an open circle at $c-a$ with an arrow going left, and a separate open circle at $c+a$ with an arrow going right — with nothing shaded in between, because the region strictly between the two points is explicitly excluded by a "greater than" modulus inequality. …