Intermediate June 2023 Unit 1 Q3
3.
(a) Solve the following equations.
(i) \(\dfrac{x}{3} = 8\) [1]
(ii) \(3x - 10 = 17\) [2]
(b) Simplify \(6f - 4g + 2f - 9g\). [2]
| Scheme | Marks | Comments |
|---|---|---|
| (i) \((x =)\ 24\) | B1 | Mark final answer. Allow B1 for a correct embedded answer BUT B0 if contradicted by \(x \neq 24\). |
| (ii) \(3x = 27\) | B1 | Mark final answer. |
| \((x =)\ 9\) | B1 | FT from \(3x = k\). Unsupported answer of 9 is awarded B1B1. \(x = \dfrac{27}{3}\) is awarded B1B0. If FT leads to a whole number answer, it must be shown as a whole number. Otherwise accept a fraction (e.g. if \(3x = 7\), then \(x = \dfrac{7}{3}\) is awarded B0B1, but \(x = 7 \div 3\) is awarded B0B0). Allow B1B1 for a correct embedded answer BUT B1B0 if contradicted by \(x \neq 9\). |
| Scheme | Marks | Comments |
|---|---|---|
| \(8f - 13g\) | B2 | Mark final answer. Must be an expression for B2. Award B1 for one of the following:
|