June 2023 Paper 1 Q13
13 Write \(\quad \dfrac{2x - 3}{x} - \dfrac{1}{3x} + 1 \quad\) as a single fraction.
Give your answer in its simplest form. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Common denominator found and used for all terms at some point during the method and at least one new numerator correct for that denominator | M1 | eg \(\dfrac{6x - 9}{3x} - \dfrac{1}{3x} + \dfrac{3x}{3x}\) or \(\dfrac{3x(2x - 3)}{3x^2} - \dfrac{x}{3x^2} + \dfrac{3x^2}{3x^2}\) or \(\dfrac{6x - 10}{3x} + \dfrac{3x}{3x}\) |
| Numerator of \(6x - 9 - 1 + 3x\) for denominator of \(3x\) or numerator of \(6x^2 - 9x - x + 3x^2\) for denominator of \(3x^2\) | M1dep | oe – will depend on the denominator but all numerator terms must be correct for the denominator used. could be a single fraction or separate fractions with the same denominator |
| \(\dfrac{9x - 10}{3x}\) | A1 | \(3 - \dfrac{10}{3x}\) is not a single fraction so is A0 |
Additional guidance
| Penalise incorrect further working | M2 A0 |