June 2017 Paper 2 Q15
15 Rearrange \(\qquad w = \dfrac{8x - y}{y} \qquad\) to make \(y\) the subject. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(wy = 8x - y\) or \(w = \dfrac{8x}{y} - 1\) or \(w + 1 = \dfrac{8x}{y}\) | M1 | |
| \(wy + y = 8x\) or \(y(w + 1) = 8x\) or \(\dfrac{w + 1}{8x} = \dfrac{1}{y}\) | M1dep | oe \(y\) term(s) collected eg \(-wy - y = -8x\) M2 \(\dfrac{8x}{w + 1}\) or \(\dfrac{-8x}{-w - 1}\) or \(\dfrac{-8x}{-(w + 1)}\) |
| \(y = \dfrac{8x}{w + 1}\) or \(y = \dfrac{-8x}{-w - 1}\) or \(y = \dfrac{-8x}{-(w + 1)}\) | A1 | oe eg \(y = \dfrac{4x}{0.5w + 0.5}\) Must have \(y =\) |
| Alternative method 2 | ||
| \(y = \dfrac{8x}{w} - \dfrac{y}{w}\) | M1 | |
| \(y + \dfrac{y}{w} = \dfrac{8x}{w}\) or \(y\left(1 + \dfrac{1}{w}\right) = \dfrac{8x}{w}\) | M1dep | oe \(y\) term(s) collected M2 \(\dfrac{\frac{8x}{w}}{1 + \frac{1}{w}}\) |
| \(y = \dfrac{\frac{8x}{w}}{1 + \frac{1}{w}}\) | A1 | oe Must have \(y =\) |
Additional guidance
| \(y = \dfrac{8x}{w + 1}\) in working with \(\dfrac{8x}{w + 1}\) on answer line etc | M2A1 |
| Allow multiplications signs and 1s throughout | |
| \(w = \dfrac{8x}{y} - \dfrac{y}{y}\) with no further simplification | M0 |
| Correct answer followed by incorrect further work | M2A0 |