Higher June 2018 Paper 1 Q26
26. Solve the equation
\[\frac{4}{2x - 3} + \frac{12}{x + 2} = 7.\][6]
| Scheme | Marks | Comments |
|---|---|---|
| Clears the fractions \(4(x + 2) + 12(2x - 3) = 7(2x - 3)(x + 2)\) | M2 | May be seen or implied later e.g. \(28x - 28 = 7(2x^2 + x - 6)\) M1 for \(\dfrac{4(x + 2) + 12(2x - 3)}{(2x - 3)(x + 2)}\ [= 7]\) For first M1 or M2, allow recovery from omission of brackets e.g. \(4\times x + 2 + 12\times2x - 3 = 7\times2x - 3\times x + 2\) |
| Expands the brackets and collects all terms on one side \([4x + 8 + 24x - 36 = 7(2x^2 + x - 6)]\) \([28x - 28 = 14x^2 + 7x - 42]\) \(14x^2 - 21x - 14 = 0\) or \(2x^2 - 3x - 2 = 0\) | M2 | For M2 and M1 allow one error in expansion or collection M1 for ‘their \(28x - 28\)’ = ‘their \(14x^2 + 7x - 42\)’ FT for M2 or M1 only if of equivalent difficulty, so e.g. if brackets have been omitted, they must ‘work as if they are there’ for these marks |
| Factorises or solves their 3-term quadratic equation \((2x + 1)(x - 2)\ [= 0]\) or equivalent | M1 | May use the quadratic formula; must be correct for their equation; their equation must have real roots. |
| \(x = -\dfrac{1}{2},\ x = 2\) | A1 | CAO |
| (6) |