4400 Higher June 2005 Paper 3 Q21
21 Solve the simultaneous equations
\(y = 3x^2\)
\(y = 2x + 5\)
(6)
| Scheme | Marks |
|---|---|
| \(3x^2 = 2x + 5\) | M1 |
| \((3x - 5)(x + 1) = 0\) | M1 |
| \(x = \dfrac{5}{3}\) and \(x = -1\) | A1 A1 |
| e.g. \(2 \times \text{“}\frac{5}{3}\text{”} + 5\) \(2 \times \text{“}{-1}\text{”} + 5\) | M1 |
| \(x = \dfrac{5}{3},\ y = \dfrac{25}{3}\) \(x = -1,\ y = 3\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for correct factorisation
A1: dep on both method marks
M1: for substituting both their \(x\) values into one of the original equations
A1: for both pairs; dep on first M1
Alternative
| \(y = 3\left(\dfrac{y - 5}{2}\right)^2\) | M1 |
| \((3y - 25)(y - 3) = 0\) | M1 |
| \(y = \dfrac{25}{3}\) and \(y = 3\) | A1 A1 |
| e.g. \(\dfrac{25}{3} = 2x + 5\) \(3 = 2x + 5\) | M1 |
| \(x = \dfrac{5}{3},\ y = \dfrac{25}{3}\) \(x = -1,\ y = 3\) | A1 |
Notes as for the main method (for substituting both their values into one of the original equations).