How can I graph an inequality solution on a number line?
Quick answer
For a solution such as x > 3, mark 3 with an open circle because 3 is not included, then shade to the right for values greater than 3. For x ≥ 3, use a closed circle because 3 is included, and shade right as before. Shade left for values less than the boundary. Check that the endpoint and shading together match the inequality.
The question, the details, the next step.
I solved an inequality and got x > 3. How do I turn that into a number-line graph?
First mark the boundary value, 3, on the number line. Since the sign is > and does not include equality, draw an open circle at 3. Then shade to the right: numbers farther right are greater than 3. The graph represents every number greater than 3, but not 3 itself.
A quick way to decide the shading direction is to read the inequality aloud: “x is greater than 3” points right; “x is less than 3” points left. The endpoint tells whether the boundary itself belongs in the solution. An open endpoint excludes it; a closed endpoint includes it. [S1]
What changes for x ≥ 3? I’m not sure when to make the circle closed.
The shading still goes right because the solutions are greater than 3. But ≥ means “greater than or equal to,” so 3 is also a solution. Draw a closed circle at 3 to show that it is included. The same rule works on the less-than side: x ≤ 3 gets a closed circle at 3 and shading to the left.
Here is the symbol-to-graph guide:
- > or <: open circle; the boundary is excluded.
- ≥ or ≤: closed circle; the boundary is included.
- > or ≥: shade right.
- < or ≤: shade left.
The endpoint choice distinguishes an excluded boundary from an included one. [S1]
Could you show a full example where I solve the inequality first and then graph it?
Suppose the inequality is 2x + 1 < 7. Subtract 1 from both sides to get 2x < 6, then divide by 2 to get x < 3. On the number line, put an open circle at 3 and shade left.
For a second example, suppose you get −2x ≥ 6. Dividing both sides by −2 reverses the inequality sign, so the result is x ≤ −3. Graph it with a closed circle at −3 and shade left. When solving an inequality, multiplying or dividing by a negative reverses its sign. [S1]
A check can catch graphing mistakes: choose one value in the shaded region and see whether it satisfies the final inequality, then check whether the boundary value should be included. For x < 3, 2 is in the shaded region and satisfies the inequality; 3 does not.
How can I check that my finished graph matches the inequality?
Check the endpoint and the shading separately. For x < 3, the open circle shows that 3 is excluded, and shading left shows values less than 3. Try a value on the shaded side, such as 2: it satisfies x < 3. Then try the boundary, 3: it does not. For x ≥ 3, the circle should be closed and the shading should go right; the boundary is included.
If either the endpoint or the direction disagrees with the inequality, revise the graph. The graph should represent exactly the values that satisfy the final inequality.
A practical checklist
Solve the inequality and simplify the result before graphing.
Mark the boundary value on the number line.
Use an open circle for < or >, and a closed circle for ≤ or ≥.
Shade right for greater-than solutions and left for less-than solutions.
Test a shaded value and check whether the boundary is included.
A few more useful answers
Does an open circle mean the boundary value is not a solution?
Yes. An open circle marks a value excluded by a strict inequality such as < or >. A closed circle marks a boundary included by ≤ or ≥.
Why does the inequality sign reverse when dividing by a negative?
Multiplying or dividing an inequality by a negative reverses its direction, so the sign must be switched before graphing. For example, −2x ≥ 6 becomes x ≤ −3. [S1]
How can I tell whether to shade left or right?
Greater-than solutions extend to the right of the boundary; less-than solutions extend to the left. Read the final inequality aloud to confirm the direction. [S1]
Sources
References used for this guide.
- [S1] OpenStax Intermediate Algebra 2e: Solve Linear Inequalities openstax.org · Checked