Foundation June 2025 Paper 2 Q11
11 Here are two expressions, A and B.
| A | B |
|---|---|
| \(7(x - 2) + 4x + 6\) | \(5(3x + 2) - 4x - 18\) |
Show that A and B are equivalent. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: expands brackets and simplifies | ||
| \(7x - 14\) or \(15x + 10\) | M1 | |
| \(7x - 14\ (+ 4x + 6)\) and \(11x - 8\) or \(15x + 10\ (- 4x - 18)\) and \(11x - 8\) | M1dep | A or B fully correct |
| \(7x - 14\ (+ 4x + 6)\) and \(11x - 8\) and \(15x + 10\ (- 4x - 18)\) and \(11x - 8\) | A1 | A and B fully correct SC1 \(11x - 8\) for one or both of A and B with no working seen |
| Alternative method 2: shows the \(x\) terms and constant terms are the same | ||
| Any one of \(7x + 4x = 11x\) or \(15x - 4x = 11x\) or \(-14 + 6 = -8\) or \(10 - 18 = -8\) | M1 | |
| Any two of \(7x + 4x = 11x\) or \(15x - 4x = 11x\) or \(-14 + 6 = -8\) or \(10 - 18 = -8\) | M1dep | |
| \(7x + 4x = 11x\) and \(15x - 4x = 11x\) and \(-14 + 6 = -8\) and \(10 - 18 = -8\) | A1 | SC1 \(11x - 8\) for one or both of A and B with no working seen |
Additional guidance
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Use the scheme that favours the student | |
| Alt 2: \(7x + 4x = 11x\) and \(-14 + 6 = -8\) | M1M1 |
| Alt 2: \(7x + 4x = 11x\) and \(15x - 4x = 11x\) | M1M1 |
| Alt 2: \(11x\) only | M0 |
| Alt 2: \(-8\) only | M0 |
| Do not ignore subsequent incorrect work such as \(11x - 8 = 3\), which may be awarded up to M2 if working shown | |
| Substituting values into the two expressions | M0 |