Higher June 2025 Paper 1 Q6
6
(a) Solve \(x - 4 = \dfrac{3 + 2x}{6}\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
(b)
(i) Factorise \(y^2 - 11y + 30\) (2)
(ii) Hence solve \(y^2 - 11y + 30 = 0\) (1)
| Scheme | Marks |
|---|---|
| \(6x - 24 = 3 + 2x\) or \(x - 4 = \dfrac{3}{6} + \dfrac{2}{6}x\) oe | M1 |
\(6x - 2x = 3 + 24\) or \(4x = 27\) or \(-24 - 3 = 2x - 6x\) or \(-27 = -4x\) oe or \(x - \dfrac{2}{6}x = \dfrac{3}{6} + 4\) oe or \(-4 - \dfrac{3}{6} = \dfrac{2}{6}x - x\) oe | M1ft |
Working required Answer: \(\dfrac{27}{4}\) | A1 |
| (3) |
Notes
M1: for correct removal of fraction and expansion of bracket in a correct equation
or
separating fraction (RHS) in an equation
M1ft: (dep on 4 terms) correctly rearranging their 4 term equation for terms in \(x\) on one side of equation and number terms on the other
A1: oe eg 6.75 or \(6\dfrac{3}{4}\), dep on M1
| Scheme | Marks |
|---|---|
| (i) \((y \pm 6)(y \pm 5)\) or \((6 \pm y)(5 \pm y)\) or \(y(y - 6) - 5(y - 6)\) or \(y(y - 5) - 6(y - 5)\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \((y - 6)(y - 5)\) | A1 |
| (2) | |
| (ii) Answer: (\(y\) =) 6, (\(y\) =) 5 | B1 |
| (1) | |
| (6 marks) |
Notes
M1: for \((y \pm 6)(y \pm 5)\) or \((6 \pm y)(5 \pm y)\) or for \((y + a)(y + b)\) where \(ab = 30\) or \(a + b = -11\)
or
\(y(y + a) + b(y + a)\) or
\(y(y + b) + a(y + b)\)
where \(ab = 30\) or \(a + b = -11\)
A1: oe, allow any letter for \(y\)
B1: must ft from their answer in (b)(i) ft from their factors in the form \((y + a)(y + b)\)