Foundation November 2017 Paper 2 Q27
27 Solve \(\qquad 4(3x - 2) = 2x - 5\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(12x - 8\) | M1 | May be seen in a grid |
| their \(12x - 2x = -5 +\) their 8 or \(10x = 3\) or their \(-8 + 5 = 2x -\) their \(12x\) or \(-3 = -10x\) | M1 | Collecting two terms in \(x\) and two constant terms correctly oe eg \(10x - 3 = 0\) |
| 0.3 or \(\dfrac{3}{10}\) | A1ft | ft M1M0 or M0M1 with exactly one error |
| Alternative method 2 | ||
| \(\dfrac{x}{2} - \dfrac{5}{4}\) | M1 | |
| \(3x -\) their \(\dfrac{x}{2}\) = their \(-\dfrac{5}{4} + 2\) or \(\dfrac{5}{2}x = \dfrac{3}{4}\) or \(-2 +\) their \(\dfrac{5}{4}\) = their \(\dfrac{x}{2} - 3x\) or \(-\dfrac{3}{4} = -\dfrac{5}{2}x\) | M1 | Collecting two terms in \(x\) and two constant terms correctly oe eg \(\dfrac{5}{2}x - \dfrac{3}{4} = 0\) |
| 0.3 or \(\dfrac{3}{10}\) | A1ft | ft M1M0 or M0M1 with exactly one error |
Additional guidance
| \(12x - 2 = 2x - 5\) \(10x = -3\) \(x = -0.3\) | M0 M1 A1ft |
| \(12x - 8 = 2x - 5\) \(10x = -5\) \(x = \dfrac{-5}{10}\) | M1 M0 A1ft |
| \(12x - 8 = 2x - 5\) \(14x = 3\) \(x = \dfrac{3}{14}\) | M1 M0 A1ft |
| \(12x - 8 = 2x - 5\) \(14x = -13\) \(x = -\dfrac{13}{14}\) (two errors) | M1 M0 A0ft |
| \(12x - 8 = 8x - 20\) | M1M0A0 |
| Any ft answer must be exact or rounded or truncated to at least 2 dp | |
| The last two marks can be implied without the collection of terms seen | |
| eg \(12x - 6 = 2x - 5\) and answer 0.1 | M0M1A1ft |
| Collecting terms before the bracket has been expanded | Zero |