Higher November 2020 Paper 1 Q10
10.
(a) Solve \(10(x - 1) - (7x + 9) = x\). [3]
(b) Factorise and hence solve \(x^2 + 3x - 18 = 0\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(10x - 10 - 7x - 9 = x\) | B1 | FT until 2nd error Expands both brackets |
| \(10x - 7x - x = 10 + 9\) oe | B1 | FT; collects terms |
| \((x =)\ \dfrac{19}{2}\) oe | B1 | FT; ignore attempts to convert to a decimal |
| Scheme | Marks | Comments |
|---|---|---|
| \((x - 3)(x + 6)\) oe | B2 | If not B2, award B1 for \((x \ldots 3)(x \ldots 6)\) |
| \(x = 3,\ x = -6\) | B1 | STRICT FT from ‘their \((x \ldots a)(x \ldots b)\)’ where \(a\) and \(b\) are constants; accept ‘\(x = 3\) or \(x = -6\)’, which is correct, allow for e.g. ‘\(x = 3\) and \(x = -6\)’ |
| (6) |