Higher June 2018 Paper 3 Q18
18 Show that, for \(\;x \neq 0\)
\[\frac{x + 4}{3x} - \frac{5}{2x}\]can be written in the form \(\;\dfrac{ax + b}{cx}\;\) where \(a\), \(b\) and \(c\) are integers. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{2(x + 4)}{6x}\) or (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x + 8}{6x}\) or (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x(x + 4)}{6x^2}\) or (–)\(\dfrac{15x}{6x^2}\) or \(\dfrac{2x^2 + 8x}{6x^2}\) or (–)\(\dfrac{15x}{6x^2}\) | M1 | oe A correct fraction using a common denominator for one of the given fractions Accept for this mark only eg \(\;2(3x)\) for \(6x\) 3(5) for 15 \((2x)(3x)\) for \(6x^2\) First fraction can be written as separate fractions eg \(\dfrac{2x}{2(3x)} + \dfrac{8}{2(3x)}\) |
| \(\dfrac{2(x + 4)}{6x}\) and (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x + 8}{6x}\) and (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x(x + 4)}{6x^2}\) and (–)\(\dfrac{15x}{6x^2}\) or \(\dfrac{2x^2 + 8x}{6x^2}\) and (–)\(\dfrac{15x}{6x^2}\) | A1 | oe A correct fraction using a common denominator for both of the given fractions First fraction can be written as separate fractions eg \(\dfrac{2x}{6x} + \dfrac{8}{6x}\) |
| \(\dfrac{2x - 7}{6x}\) or \(\dfrac{2\mathrm{k}x - 7\mathrm{k}}{6\mathrm{k}x}\), where k is a constant value | A1 | Accept eg \(\dfrac{2x + -7}{6x}\) Do not ignore further working |
| Alternative method 2 | ||
| \(\dfrac{2(x + 4)}{6x}\) or (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x + 8}{6x}\) or (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x(x + 4)}{6x^2}\) or (–)\(\dfrac{15x}{6x^2}\) or \(\dfrac{2x^2 + 8x}{6x^2}\) or (–)\(\dfrac{15x}{6x^2}\) | M1 | oe A correct fraction using a common denominator for one of the given fractions Accept for this mark only eg \(\;2(3x)\) for \(6x\) 3(5) for 15 \((2x)(3x)\) for \(6x^2\) First fraction can be written as separate fractions eg \(\dfrac{2x}{2(3x)} + \dfrac{8}{2(3x)}\) |
| \(\dfrac{2x + 8 - 15}{6x}\) or \(\dfrac{2x - 7}{6x}\) or \(\dfrac{2\mathrm{k}x - 7\mathrm{k}}{6\mathrm{k}x}\), where k is a constant value | A1 | Allow one error in numerator Accept eg \(\dfrac{2x + -7}{6x}\) Must be \(6x\) or a multiple of \(6x\) |
| \(\dfrac{2x - 7}{6x}\) or \(\dfrac{2\mathrm{k}x - 7\mathrm{k}}{6\mathrm{k}x}\), where k is a constant value | A1 | Accept eg \(\dfrac{2x + -7}{6x}\) Do not ignore further working |
Additional guidance
| Use the method that gives the greater mark | |
| \(\dfrac{2x^2 - 7x}{6x^2}\) | M1A1 |
| \(\dfrac{2x - 7}{6x} = \dfrac{-5}{6x}\) | M1A1A0 |
| \(\dfrac{15x}{6x^2} - \dfrac{2x^2 + 8x}{6x^2}\) (order of fractions reversed) | M1A0A0 |