S3 Maths: Linear Inequalities in One Unknown (Paper 1: basics)
This first paper keeps to one and two step inequalities, focusing on sign changes when moving terms and on dividing by a positive number. Substitute the answer back into the original inequality each time.
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S3 Maths: Linear Inequalities in One Unknown |
| Level | S3 Maths |
| # | Answer | One-line note |
|---|---|---|
| 1 | x > 7 | Subtract 5 from both sides: 12 – 5 = 7, and the inequality sign stays the same. |
| 2 | x <= 7 | Add 4 to both sides: 3 + 4 = 7, and the “less than or equal” sign is unchanged. |
| 3 | x < 5 | Divide both sides by 3; because 3 is positive the sign does not reverse. |
| 4 | x >= 8 | Multiply both sides by 4: 2 x 4 = 8, and multiplying by a positive number keeps the sign. |
| 5 | x <= 4 | Subtract 3 to get 2x <= 8, then divide by 2 to get 4. |
| 6 | x > 5 | Add 7 to get 5x > 25, then divide by 5 to get 5. |
| 7 | x > 8 | Greater than uses > and excludes 8; only “not less than” uses >=. |
| 8 | x < 6 | Subtract 6 to get 4x < 24, then divide by 4 to get 6. |
| 9 | x <= 8 | Divide both sides by 7: 56 / 7 = 8, and the sign is unchanged. |
| 10 | The correct answer is x <= -4 | Moving a term changes its sign, so it is 3 – 7 = -4 rather than 3 + 7. |
Three basics: moving a term changes its sign, multiplying or dividing by a positive number keeps the inequality sign, and the answer is written as x greater than, or x not less than, a value. Write the steps before the solution set.Tutor Registration →
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