Higher November 2020 Paper 2 Q15
15 Make \(x\) the subject of \(y = \dfrac{5 - 2x}{x + 3}\)
(4)
| Scheme | Marks |
|---|---|
| \(xy + 3y = 5 - 2x\) oe | M1 |
| e.g. \(xy + 2x = 5 - 3y\) | M1 |
| eg \(x(y + 2) = 5 - 3y\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(x = \dfrac{5 - 3y}{2 + y}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: multiplying both sides by \((x + 3)\) and expanding the brackets correctly
M1: ft dep on 2 terms on left and \((5 - 2x)\) on right, for collecting all \(x\) terms on one side and non-\(x\) terms on the other side
M1: ft, dep on 2 terms in \(x\), for factorising for \(x\)
A1: oe allow \(\dfrac{5 - 3y}{2 + y}\) as answer so long as previously seen \(x = \dfrac{5 - 3y}{2 + y}\)