Foundation November 2019 Paper 1 Q16
16. Make \(x\) the subject of the following formula. [2]
\[y = \dfrac{x + 3}{4}\]| Scheme | Marks | Comments |
|---|---|---|
| \(x = 4y - 3\) | B2 | B1 for \(4y = x + 3\) or \(y - \dfrac{3}{4} = \dfrac{x}{4}\) If no marks then SC1 for a final answer of \(x = 4y + 3\) |
| (2) |