Higher June 2017 Paper 2 Q11
11 Solve \(\dfrac{3x - 2}{4} - \dfrac{2x + 5}{3} = \dfrac{1 - x}{6}\) (4)
| Answer | Mark | Notes |
|---|---|---|
| \(9\dfrac{1}{3}\) | M1 | for writing at least 2 fractions with a common denominator eg. \(\dfrac{3(3x - 2)}{12}\), \(\dfrac{4(2x + 5)}{12}\), \(\dfrac{2(1 - x)}{12}\) with at least one correct numerator or for \(\dfrac{3x}{4} - \dfrac{2}{4} - \dfrac{2x}{3} - \dfrac{5}{3} = \dfrac{1}{6} - \dfrac{x}{6}\) (accept \(+\dfrac{5}{3}\) instead of \(-\dfrac{5}{3}\)) |
| M1 | (dep) for a method to eliminate all fractions in an equation, ignore errors in any expanded terms eg. \(3(3x - 2) - 4(2x + 5) = 2(1 - x)\) or \(6 \times [3(3x - 2) - 4(2x + 5)] = 12 \times [1 - x]\) or \(3 \times 3x - 3 \times 2 - 4 \times 2x - 4 \times 5 = 2 \times 1 - 2 \times x\) OR for the correct expansion of brackets leading to \(\dfrac{9x - 6 - 8x - 20}{12} = \dfrac{2 - 2x}{12}\) | |
| M1 | (dep on M2) for correctly isolating terms in \(x\) and number terms of their linear equation e.g. \(9x - 8x + 2x = 2 + 6 + 20\) | |
| A1 | for \(9\dfrac{1}{3}\) oe |