June 2016 Paper 2 Q16
16 Rearrange \(\quad y = \dfrac{8(w - x)}{x} \quad\) to make \(x\) the subject. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(yx = 8(w - x)\) or \(y = \dfrac{8w - 8x}{x}\) | M1 | |
| \(yx = 8w - 8x\) | M1dep | oe eg \(yx - 8w + 8x = 0\) Implies M1 M1 |
| \(yx + 8x = 8w\) or \(x(y + 8) = 8w\) or \(\dfrac{8w}{y + 8}\) | M1dep | oe dep on M1 M1 Implies M1 M1 M1 |
| \(x = \dfrac{8w}{y + 8}\) | A1 | oe eg \(\dfrac{-8w}{-y - 8}\) Must have \(x =\) SC2 \(x = \dfrac{8w}{y + 1}\) SC1 \(\dfrac{8w}{y + 1}\) |
| Alternative method 2 | ||
| \(y = \dfrac{8w}{x} - 8\) or \(y = \dfrac{8w}{x} - \dfrac{8x}{x}\) | M1 | |
| \(y + 8 = \dfrac{8w}{x}\) | M1dep | oe eg \(y + 8 - \dfrac{8w}{x} = 0\) Implies M1 M1 |
| \(yx + 8x = 8w\) or \(x(y + 8) = 8w\) or \(\dfrac{1}{y + 8} = \dfrac{x}{8w}\) or \(\dfrac{8w}{y + 8}\) | M1dep | oe dep on M1 M1 Implies M1 M1 M1 |
| \(x = \dfrac{8w}{y + 8}\) | A1 | oe eg \(\dfrac{-8w}{-y - 8}\) Must have \(x =\) SC2 \(x = \dfrac{8w}{y + 1}\) SC1 \(\dfrac{8w}{y + 1}\) |
| Alternative method 3 | ||
| \(yx = 8(w - x)\) | M1 | |
| \(\dfrac{yx}{8} = w - x\) | M1dep | oe eg \(\dfrac{yx}{8} - w + x = 0\) Implies M1 M1 |
| \(\dfrac{yx}{8} + x = w\) or \(x\left(\dfrac{y}{8} + 1\right) = w\) or \(\dfrac{w}{\frac{y}{8} + 1}\) | M1dep | oe dep on M1 M1 Implies M1 M1 M1 |
| \(x = \dfrac{w}{\frac{y}{8} + 1}\) | A1 | oe eg \(x = \dfrac{-w}{-\frac{y}{8} - 1}\) Must have \(x =\) SC2 \(x = \dfrac{8w}{y + 1}\) SC1 \(\dfrac{8w}{y + 1}\) |
Additional guidance
| \(x = \dfrac{8w}{y + 8}\) in working with \(\dfrac{8w}{y + 8}\) on answer line | M3 A1 |
| \(x = \dfrac{8w}{y + 1}\) in working with \(\dfrac{8w}{y + 1}\) on answer line | SC2 |
| 3rd M1 is for collecting terms in \(x\) (or \(x\) in numerator in Alt 2) | |
| Allow multiplications signs and 1s throughout | |
| Correct answer followed by incorrect further work | M3 A0 |