1MA0 Higher June 2015 Paper 1 Q20
20 Make \(a\) the subject of the formula \(p = \dfrac{3a + 5}{4 - a}\) (4)
| Answer | Mark | Notes |
|---|---|---|
| \(a = \dfrac{4p - 5}{3 + p}\) or | 4 | M1 for multiplying both sides by \(4 - a\) as the first step M1 (dep) for isolating their two terms in \(a\) M1 (dep) for factorising their two term expression in \(a\) A1 for \(a = \dfrac{4p - 5}{3 + p}\) or \(a = \dfrac{5 - 4p}{-p - 3}\) |