Foundation June 2017 Paper 2 Q17
17
(a) Rearrange the equation to make \(x\) the subject.\[y = 7x - 3\]
[2]
(b) Factorise.
(i) \(x^2 - xy\) [1]
(ii) \(x^2 + 8x + 12\) [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{y + 3}{7}\) or \(\dfrac{-y - 3}{-7}\) final answer | 2 | M1 for \(y + 3 = 7x\) or \(\dfrac{y}{7} = x - \dfrac{3}{7}\) Or for correct FT completion to answer after incorrect first step has been shown | For M1, accept the ‘negative terms’ versions |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(x(x - y)\) final answer | 1 | Condone omission of final bracket Condone \([1]x([1]x - [1]y)\) | |
| (ii) \((x + 6)(x + 2)\) final answer | 2 | M1 for \((x + a)(x + b)\) where \(ab = \pm 12\) or \(a + b = \pm 8\) or for \(x(x + 6) + 2(x + 6)\) seen or \(x(x + 2) + 6(x + 2)\) seen | \(a\), \(b\) integers For 2 marks, condone solutions after correct factors For 2 marks or M1, condone omission of final bracket |