Higher November 2018 Paper 1 Q22
22 Solve \(\quad \dfrac{x}{x + 4} + \dfrac{7}{x - 2} = 1\)
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Any two of \(x(x - 2)\) and \(7(x + 4)\) and \((x - 2)(x + 4)\) | M1 | oe \(x(x - 2)\) and \(7(x + 4)\) cannot be denominators |
| correct equation including \(x(x - 2)\) and \(7(x + 4)\) and \((x - 2)(x + 4)\) | M1dep | |
| \(x^2 - 2x + 7x + 28 = x^2 + 4x - 2x - 8\) | M1dep | oe all brackets must be expanded |
| \(-12\) | A1 | |
| Alternative method 2 | ||
| \(\dfrac{x(x - 2)}{x + 4} + 7 = x - 2\) | M1 | |
| \(\dfrac{x(x - 2)}{x + 4} = x - 9\) or \(x(x - 2) = (x - 9)(x + 4)\) | M1dep | |
| \(x^2 - 2x = x^2 - 9x + 4x - 36\) | M1dep | oe all brackets must be expanded |
| \(-12\) | A1 | |
| Alternative method 3 | ||
| \(x + \dfrac{7(x + 4)}{x - 2} = x + 4\) | M1 | |
| \(\dfrac{7(x + 4)}{x - 2} = 4\) or \(7(x + 4) = 4(x - 2)\) | M1dep | |
| \(7x + 28 = 4x - 8\) | M1dep | oe all brackets must be expanded |
| \(-12\) | A1 | |
Additional guidance
In Alt 1, do not allow \(x \times x - 2\) or \(7 \times x + 4\) unless recovered