Higher June 2019 Paper 6 Q9
9 Martha’s solution to the inequality \(8x + 5 \leqslant 3x - 10\) is shown on the number line.
Is her solution correct?
Explain your reasoning. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct solution is \(x \leqslant -3\) from algebraic working | M3 | M2 for \(8x - 3x \leqslant -10 - 5\) or better, or for \(5 + 10 \leqslant 3x - 8x\) or better or M1 for \(8x - 3x\), or \(3x - 8x\), or [±]\(5x\), or –10 – 5, or 5 + 10, or [±]15 seen | For M2 and M1 condone incorrect inequality sign or “equals”. Alternative method 3 trials for values of \(x\) where \(x \lt -3\), \(x = -3\) and \(x \gt -3\) and correct conclusion can score full marks. Without the correct conclusion, maximum for this approach is SC1 for only the 3 correct trials (as described below) See table of values below |
| No and number line shows \(x \geqslant -3\) oe or No and draws the correct inequality on number line or No and “the arrow points the wrong way” oe | A1dep | A1 dep on M3 After 0 scored, allow SC1 for number line shows \(x \geqslant -3\) or “the arrow points the wrong way” oe but only if no incorrect working shown or correct substitution of a value ≠ –3 and conclusion that inequality is false oe | |
Table of values for Q9 trials
| \(x\) | –6 | –5 | –4 | –3 | –2 | –1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(8x + 5\) | –43 | –35 | –27 | –19 | –11 | –3 | 5 | 13 | 21 | 29 | 37 |
| < | < | < | = | > | > | > | > | > | > | > | |
| \(3x - 10\) | –28 | –25 | –22 | –19 | –16 | –13 | –10 | –7 | –4 | –1 | 2 |