An endpoint marks the boundary separating allowed and disallowed values. Solving the related equality often finds that boundary. Whether the endpoint is included depends on the original inequality: less than and greater than exclude equality, while less than or equal to and greater than or equal to include it.

For a number-line graph, an open circle conventionally excludes the endpoint and a filled circle includes it. These are mathematical statements, not decorative styles. A student who shades the right direction but fills the wrong circle has described a different solution set.

Subtract one from both sides to get 2x < 8, then divide by positive two to obtain x < 4. The boundary is four. Test x = 4: 2 × 4 + 1 = 9, and 9 < 9 is false. Therefore four must be excluded.

Test a value on each side to check direction. At x = 3, the left side is seven, so the inequality holds. At x = 5 it is eleven, so it fails. The graph has an open circle at four and extends left. Endpoint inclusion and shading direction have now been checked separately.

For 2x + 1 ≤ 9, the algebra gives x ≤ 4. At x = 4 the original statement becomes 9 ≤ 9, which is true. The endpoint is included, so use a filled circle at four and extend left. The direction is unchanged, but the boundary membership differs.

Do not decide inclusion from whether the algebra produced a whole number. A fractional boundary can be included or excluded too. For 2x ≤ 5, the boundary 2.5 belongs; for 2x < 5, it does not. The inequality symbol, not the neatness of the number, controls this choice.

The statement −2 < x ≤ 3 requires both conditions at once. Values must be greater than negative two and no greater than three. Draw an open endpoint at −2, a filled endpoint at 3 and the segment between them. The two ends need not use the same convention.

Test both endpoints independently. Negative two fails because it is not strictly greater than itself. Three succeeds because equality is allowed at the upper end. A value such as zero satisfies both conditions. A value such as four fails the upper condition even though it satisfies the lower one.

The real-number interval −2 < x ≤ 3 contains infinitely many real values. If the question specifies integer x, the permitted values are −1, 0, 1, 2 and 3. The interval condition stays the same, but the domain restricts which points are considered.

For x > 3 over the real numbers, there is no smallest solution: any proposed solution has another allowed value between it and three. Over the integers, the smallest solution is four. Read the domain before answering a question about smallest or largest values. Do not silently replace a real interval with a whole-number list.

The direction reverses, but the distinction between strict and non-strict remains. From −2x ≥ 6, divide by −2 to obtain x ≤ −3. Equality is still allowed. At x = −3 the original left side equals six, confirming that the endpoint belongs.

For −2x > 6, the solution is x < −3 and the endpoint is excluded. A sign-reversal correction should not accidentally erase or introduce the equality part. Keep these as two separate checks: relation direction after a negative operation, then boundary membership in the original inequality.

Question 1: solve 3x + 2 < 14 and describe the endpoint. Question 2: solve 5 − x ≤ 2. Question 3: list integer solutions of −1 ≤ x < 3. Question 4: does x < 5 have a largest real solution? Does it have a largest integer solution?

Answers: 1) x < 4; open at four, extend left. 2) x ≥ 3; filled at three, extend right. 3) −1, 0, 1, 2. 4) No largest real solution; the largest integer solution is four. Ask for one substitution check, not only the final symbol.

Bring the original inequality, algebra and number-line sketch. A precise note such as ‘solves x < 4 but fills the circle at four’ distinguishes a representation issue from an equation-solving issue. A small-group 3-pax lesson can compare two almost identical graphs and ask which statement each actually represents.

For parents considering Secondary 2 Mathematics tuition in Punggol, the next-question routine is to solve, check the boundary in the original inequality, check one interior value and read the domain. This keeps a small circle from becoming an unexamined habit and connects the graph to the actual allowed values.