Higher June 2022 Paper 1 Q20
20 Rearrange \(\quad y = \dfrac{5x + 9}{x} \quad\) to make \(x\) the subject. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(xy = 5x + 9\) | M1 | |
| \(xy - 5x = 9\) or \(5x - xy = -9\) | M1dep | oe collection of terms |
| \(x(y - 5) = 9\) or \(x(5 - y) = -9\) or \(\dfrac{9}{y - 5}\) or \(\dfrac{-9}{5 - y}\) | M1dep | |
| \(x = \dfrac{9}{y - 5}\) or \(x = \dfrac{-9}{5 - y}\) | A1 | |
| Alternative method 2 | ||
| \(y = 5 + \dfrac{9}{x}\) or \(y - \dfrac{9}{x} = 5\) | M1 | allow \(\dfrac{5x}{x}\) for 5 |
| \(y - 5 = \dfrac{9}{x}\) or \(5 - y = -\dfrac{9}{x}\) | M1dep | |
| \(\dfrac{1}{y - 5} = \dfrac{x}{9}\) or \(x(y - 5) = 9\) or \(x(5 - y) = -9\) or \(\dfrac{1}{5 - y} = -\dfrac{x}{9}\) or \(\dfrac{9}{y - 5}\) or \(\dfrac{-9}{5 - y}\) | M1dep | |
| \(x = \dfrac{9}{y - 5}\) or \(x = \dfrac{-9}{5 - y}\) | A1 | |
Additional guidance
| \(\dfrac{9}{y - 5}\) on answer line with \(x = \dfrac{9}{y - 5}\) in working | M1M1M1A1 |
| Allow the equation with \(x\) on the right, eg \(\dfrac{9}{y - 5} = x\) | M1M1M1A1 |
| Allow appropriate \(\times\) or \(\div\) signs throughout |