Graphing Linear Inequalities on a Number Line

A one-variable inequality represents a set of possible values. Its number-line graph shows the boundary and every value that satisfies the statement. Most errors come from choosing the wrong endpoint or shading the wrong direction.

Solve before graphing

Use inverse operations to isolate the variable, as with an equation. If you multiply or divide both sides by a negative number, reverse the inequality sign. For example, −2x < 6 becomes x > −3 after dividing by −2. The direction changes because dividing reverses the order of negative and positive quantities.

Choose the endpoint symbol

• x < a or x > a: use an open circle at a because the endpoint is excluded.
• x ≤ a or x ≥ a: use a closed circle at a because the endpoint is included.
• Use the solved boundary value, not an earlier value from the unsimplified inequality.

Shade toward the solutions

For x < a or x ≤ a, shade to the left because those values are smaller than a. For x > a or x ≥ a, shade to the right because those values are larger. You can test a number on the shaded side in the original inequality to confirm the graph.

Example

Solve 3x + 2 ≥ 11: subtract 2 to get 3x ≥ 9, then divide by positive 3 to get x ≥ 3. Graph a closed circle at 3 and shade right. If the last step had divided by a negative number, the sign would reverse.

Check a compound inequality

For a statement such as −2 < x ≤ 4, use an open circle at −2 and a closed circle at 4, then shade only the segment between them. “And” means values satisfying both boundaries; “or” can create two separate rays. Translate the words before drawing.

Translate an interval into a graph

An inequality such as −2 < x ≤ 4 uses an open endpoint at −2 and a closed endpoint at 4, with the segment between them shaded. In interval notation, that is (−2, 4]. Parentheses exclude an endpoint; brackets include it. For x < −1 or x ≥ 3, graph two rays: an open circle at −1 shaded left and a closed circle at 3 shaded right. The word or means either region is a solution.

Use a test value to choose the direction

If a solution is x > 3, try 4 in the original inequality; if it makes the statement true, shade right. A test value is especially useful after dividing by a negative number or when a graph seems ambiguous. Do not test only the boundary value: a strict inequality excludes that value, and the boundary may not satisfy the original statement.

Graph compound inequalities

For an AND statement, shade only values satisfying both conditions. The solution to x > −2 and x ≤ 4 is the overlap from −2 to 4. For an OR statement, include values satisfying either part, which may create separate rays. A compound statement can also have no solution if its conditions cannot both be true, or all real numbers if the union covers the number line.

Distinguish boundary from solution

The number used to place the endpoint is the value after the inequality has been fully solved. For 2x + 3 ≤ 9, the boundary is x = 3, not 9. The open or closed circle comes from the symbol attached to the final variable. If both sides are multiplied or divided by a negative, reverse the sign before deciding which direction to shade.

Check the interval and notation

Read interval notation from left to right and check that endpoints are ordered from smaller to larger. Use negative infinity or infinity with parentheses because infinity is not a number included in a solution set. For x ≥ 3, write [3, ∞); for x < −1, write (−∞, −1). Confirm that the notation, number-line graph, and inequality describe the same values.

Key takeaway

Open means excluded; closed means included. Less than shades left, greater than shades right. Reverse the sign only when multiplying or dividing by a negative.

Graph the boundary and choose open or closed

A one-variable inequality describes a set of values. For x < 4, mark an open circle at 4 because the boundary is excluded; shade left because values less than 4 are solutions. For x ≥ −2, mark a closed circle at −2 because the boundary is included; shade right. The circle type encodes the symbol, and the shading encodes the direction.

The solution can be checked with a test value from the shaded region and one outside it. For x < 4, x = 3 should satisfy the inequality and x = 5 should not. If both do, the boundary or shading is wrong. The number line represents every real value in the interval, not just the labeled integers.

Graph compound inequalities

An “and” statement gives an intersection. For −1 < x ≤ 3, shade values greater than −1 and at most 3; the solution is the interval (−1, 3]. An “or” statement gives a union. For x < −2 or x ≥ 4, shade two separate rays: (−∞, −2) ∪ [4, ∞). Use the correct endpoint bracket in interval notation.

If an inequality is written as 2 < x + 1 ≤ 7, subtract 1 from all three parts to get 1 < x ≤ 6. Adding or subtracting the same number preserves both inequality directions. Multiplying or dividing all parts by a negative reverses both symbols: −6 < −2x ≤ 4 becomes 3 > x ≥ −2 after dividing by −2, which is equivalent to −2 ≤ x < 3 when written in increasing order.

Connect an algebraic solution to the graph

Solve the inequality first, then translate it to a boundary, circle, and direction. For 3x + 2 ≤ 11, subtract 2 and divide by 3 to get x ≤ 3; graph a closed circle at 3 and shade left. Keep the equal sign attached through each operation. If solving creates an impossible statement, there is no shaded solution; if it creates a true identity, shade the entire number line.

Use interval notation when asked for a symbolic description, and use union notation for separate intervals. Infinity is never included, so always use parentheses at −∞ or ∞. A graph, inequality, and interval should all represent the same set; converting among them provides a strong check.

• Use an open circle for strict inequalities and a closed circle for inclusive inequalities.
• Shade left for less than and right for greater than.
• Interpret “and” as overlap and “or” as union.
• Reverse the symbol when multiplying or dividing by a negative.
• Check a sample value and use parentheses at infinity.