Foundation June 2018 Paper 2R Q15
15
(a) Solve \(8 - 2p = 15\) (2)
(b) Solve \(\dfrac{7x - 2}{4} = 3x + 1\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
| Scheme | Marks |
|---|---|
| \(-2p = 15 - 8\) or \(8 = 2p + 15\) or \(\dfrac{8}{2} - p = \dfrac{15}{2}\) oe | M1 |
| −3.5 | A1 |
| (2) |
Notes
A1: oe
| Scheme | Marks |
|---|---|
| eg \(7x - 2 = 4(3x + 1)\) oe | M1 |
| \(7x - 12x = 4 + 2\) oe or \(-2 - 4 = 12x - 7x\) oe | M1 |
Working required Answer: \(-\dfrac{6}{5}\) | A1 |
| (3) | |
| (5 marks) |
Notes
M1: correct first step
M1: for rearranging the \(x\) terms on one side and the numerical terms on the other. ft rearranging their expansion \(ax + b = cx + d\) eg \(7x - 2 = 12x + 4\)
A1: oe, dep on M1