Higher June 2019 Paper 2 Q15
15 Make \(x\) the subject of the formula \(y = \sqrt{\dfrac{3x - 2}{x + 1}}\)
(4)
| Scheme | Marks |
|---|---|
| \(y^2 = \dfrac{3x - 2}{x + 1}\) | M1 |
| \(xy^2 + y^2 = 3x - 2\) oe | M1 |
| \(y^2 + 2 = x(3 - y^2)\) oe | M1 |
| \(x = \dfrac{2 + y^2}{3 - y^2}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: squaring both sides to get a correct equation
M1: for multiplying by the denominator and expanding the bracket
M1: for isolating terms in \(x\) and factorising the correct expression of the equation
A1: accept \(x = \dfrac{-2 - y^2}{y^2 - 3}\) oe