Graphical solution and number-line representation

Every solution of a linear inequality in one variable is a subset of $\mathbb{R}$ , almost always an interval or a union of intervals. The number line is the picture that shows the solution. This subtopic teaches you to draw and read these pictures fluently.

Drawing solutions on a number line

To represent the solution set of an inequality:

1. Mark the boundary points as dots.
2. Open circle ($\circ$) for strict $<$ or $>$ , point excluded.
3. Closed circle ($\bullet$) for $\le$ or $\ge$ , point included.
4. Shade the segment(s) covering the solution.
5. Arrows at the end indicate unboundedness ($\to \pm \infty$).

Examples:

• $x \ge 2$: closed dot at $2$, shade to the right, arrow to $\infty$.
• $-1 < x < 3$: open dots at $-1$ and $3$, shade in between.
• $x < 0$ or $x > 5$: shade in two separate pieces.

Geometric meaning of the manipulation rules

Adding a constant shifts the entire shaded region. Multiplying by a positive constant scales it. Multiplying by a negative constant flips it across the origin , and that flip is exactly why the inequality reverses.

This picture turns the abstract sign-rule into a visual one.

Compound inequalities as set operations

For two inequalities $A$ and $B$:

• "$A$ and $B$" (both must hold): take the intersection of the two number-line shadings.
• "$A$ or $B$" (either may hold): take the union.

Worked examples

Example 1. Sketch the solution of $x > 3$ on a number line.

Open circle at $3$, shade to the right, arrow to $+\infty$. In interval notation: $(3, \infty)$.

Example 2. Sketch the solution of $-2 \le x \le 4$.

Closed dot at $-2$, closed dot at $4$, shade between. Notation: $[-2, 4]$.

Example 3. Sketch $x \le -1$ or $x \ge 2$.

Two separate shaded rays , one going left from $-1$ (closed), one going right from $2$ (closed). Notation: $(-\infty, -1] \cup [2, \infty)$.

Example 4. Sketch and write the solution of the system $\{x > 0, x \le 5\}$.

Intersection: $0 < x \le 5$, i.e. $(0, 5]$. Open dot at $0$, closed dot at $5$.

Example 5 (harder). Sketch the solution of $\dfrac{x - 1}{x + 2} > 0$.

This is positive when both numerator and denominator have the same sign:

• Both positive: $x > 1$ (so $x > 1$).
• Both negative: $x < -2$ (so $x < -2$).

Union: $(-\infty, -2) \cup (1, \infty)$.

On the number line: open dot at $-2$, shade left to $-\infty$; open dot at $1$, shade right to $\infty$.

Try it yourself

1. Sketch and write in interval notation: $x \ge -3$.
2. Sketch: $x < 0$ or $x > 4$.
3. Sketch: $-5 < x \le 2$.
4. Sketch the system $\{x \le 4, x \ge 1\}$.
5. Sketch the system $\{x > 1\}$ and $\{x < 5\}$ on the same number line and find the intersection.
6. Sketch $\{x > -1\}$ or $\{x \le -3\}$ and find the union.
7. Sketch the integer solutions of $-3 \le x \le 2$.
8. Sketch $\dfrac{x}{x - 1} < 0$.
9. Sketch $|x - 1| < 2$ (preview of next subtopic).
10. Mark all integers in $(-1, 4)$.
11. Sketch $(x - 2)(x + 3) > 0$.
12. Sketch $\dfrac{1}{x} > 1$ for $x > 0$.

Pitfalls / Tricks

• Always indicate clearly which dots are open and which are closed.
• For "or" statements, the shaded region has two pieces , do not accidentally merge them.
• When the inequality has a strict $<$ in $\dfrac{f(x)}{g(x)}$, remember the denominator can never be zero; mark that point as excluded.
• Insight. Whenever a problem asks "how many integers satisfy ...", first sketch the solution on a number line, then count the integers in the shaded region.