4400 Higher June 2008 Paper 4 Q1
1 Solve
(a) \(6x + 13 = 2x + 7\) (3)
(b) \(\dfrac{y}{5} - 2 = 4\) (2)
| Scheme | Marks |
|---|---|
| \(6x - 2x = 7 - 13\) or \(2x - 6x = 13 - 7\) | M1 |
| \(4x = -6\) or \(-4x = 6\) | M1 |
| \(x = -1\frac{1}{2}\) oe | A1 |
| (3) |
Notes
M1: \(6x - 2x + 13 - 7 = 0\) or \(2x - 6x - 13 + 7 = 0\)
A1: Accept \(-6/4\) or \(-3/2\) (not \(6/{-4}\) or \(3/{-2}\))
| Scheme | Marks |
|---|---|
| \(y - 2 \times 5 = 4 \times 5\) or \(y/5 = 4 + 2\) | M1 |
| \(y = 30\) | A1 |
| (2) | |
| (5 marks) |