Higher June 2023 Unit 2 Q1
1.
(a) Solve the equation \(7 + 5(x - 2) = 3x + 8\). [3]
(b) Make \(f\) the subject of the formula \(h = 13 - 2f\). [2]
(c) Factorise \(15x - 35y\). [1]
| Scheme | Marks | Comments | ||||||
|---|---|---|---|---|---|---|---|---|
| \(7 + 5x - 10 = 3x + 8\) or equivalent. | B1 | FT until 2nd error. Bracket must be expanded or correct division by 5 e.g \(x - 2 = \dfrac{3x}{5} + \dfrac{1}{5}\) (but not \(x - 2 = \dfrac{3x + 1}{5}\)) | ||||||
| \(2x = 11\) OR \(-11 = -2x\) | B1 | Or equivalent Correctly simplifying the equation to a single \(x\) term and number term (e.g. \(2x - 11 = 0\)). | ||||||
| \(x = \dfrac{11}{2}\) or 5·5 or equivalent. | B1 | Mark final answer. Correct answer implies B1B1B1. Do not allow \(-x = \dfrac{-11}{2}\) or \(x = \dfrac{-11}{-2}\) A final answer of ‘11 ÷ 2’ is B1B1B0. If FT leads to a whole number answer, it must be shown as a whole number. Otherwise, accept a fraction. Allow any decimal answer to be rounded or truncated to 1 or more decimal place. Allow B1B1B1 for a correct embedded answer BUT only B1B1B0 if contradicted by \(x \ne \dfrac{11}{2}\) or equivalent. Note:
|
| Scheme | Marks | Comments |
|---|---|---|
| \(2f = 13 - h\) or \(h - 13 = -2f\) | B1 | Or equivalent. |
| \(f = \dfrac{13 - h}{2}\) or \(\dfrac{h - 13}{-2} = f\) or equivalent | B1 | Or equivalent. Must not come from incorrect working. Mark final answer. FT only from \(\pm 2f = \pm 13 \pm h\). Unsupported \(f = \dfrac{\pm 13 \pm h}{\pm 2}\) implies B0B1 unless B2. Award B1B0 for \(-f = \dfrac{h - 13}{2}\) or equivalent. If no marks, award SC1 for a final answer of either:
|
| Scheme | Marks | Comments |
|---|---|---|
| \(5(3x - 7y)\) | B1 | Mark final answer. Allow \(-5(-3x + 7y)\) or \(5(3x + -7y)\). |