Higher November 2018 Paper 2 Q7
7 Work out the values of \(a\) and \(b\) in the identity
\(5(7x + 8) + 3(2x + b) \equiv ax + 13\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(35x + 6x = ax\) or \(35 + 6 = a\) or \(41x = ax\) | M1 | |
| \(a = 41\) | A1 | |
| \(40 + 3b = 13\) | M1 | oe |
| \(b = {-9}\) | A1 | SC3 \(a = 41\), \(b = {-27}\) or \(a = 41\), \(b = \dfrac{5}{3}\) |
| Alternative method 2 | ||
| \(35x + 40 + 6x + 3b\) or \(41x + 40 + 3b\) | M1 | |
| \(35x + 6x = ax\) or \(35 + 6 = a\) and \(40 + 3b = 13\) | M1dep | oe eg \(41x = ax\) and \(3b = {-27}\) |
| \(a = 41\) | A1 | implies first M1 only |
| \(b = {-9}\) | A1 | SC3 \(a = 41\), \(b = {-27}\) or \(a = 41\), \(b = \dfrac{5}{3}\) |
Additional guidance
| \(a = 41\) and \(b = {-9}\) | M1A1M1A1 |
| \(a = 41\) or \(b = {-9}\) | M1A1 |
| \(35x\), 40, \(6x\) and \(3b\) seen without addition signs shown or implied | M0 |
| \(35x + 40 + 6x + \boldsymbol{b}\) leading to an answer of \(a = 41\) and \(b = {-27}\) | SC3 |
| \(35x + \mathbf{8} + 6x + 3b\) leading to an answer of \(a = 41\) and \(b = \dfrac{5}{3}\) | SC3 |
| \(35x + \mathbf{8} + 6x + \boldsymbol{b}\) leading to an answer of \(a = 41\) and \(b = 5\) | M1A1 |
| \(a = 41x\) | M0 |
| For \(\dfrac{5}{3}\) accept 1.66… or 1.67 | |
| Condone multiplication signs eg \(35 \times x\) for \(35x\) |