4MA0 Higher June 2016 Paper 4R Q14

EdexcelHigherOld spec8 marksExpanding Brackets Solving Simple Equations

(a) Solve \(2y + \dfrac{2 - 3y}{4} = \dfrac{1}{4}\)
Show clear algebraic working. (3)

(b) Factorise \(3x^2 - 8x - 3\) (2)

Mark scheme (a)

Scheme	Marks
\(4 \times 2y + 4 \times \dfrac{2 - 3y}{4} = 4 \times \dfrac{1}{4}\) or \(\dfrac{8y}{4} + \dfrac{2 - 3y}{4} = \dfrac{1}{4}\) or \(2y = \dfrac{1}{4} - \dfrac{2 - 3y}{4}\) oe	M1
\(8y + (2 - 3y) = 1\) or \(8y = -1 + 3y\) Or \(5y = -1\) oe	M1
−0.2	A1
	(3)

Notes

M1: For multiplying each term by 4 or writing all terms with 4 as a denominator or isolating terms with denominator 4 on one side of equation and \(2y\) or \(-2y\) the other side

M1: A correct equation with no fractions

A1: dep on at least M1 earned

Scheme	Marks
	M1
\((3x + 1)(x - 3)\)	A1
	(2)

Mark scheme (c)

Scheme	Marks
\(4x^2 + 12x\) or \(4x^2 - 12x + 9\) or \(-4x^2 + 12x - 9\) oe	M1
\(4x^2 + 12x - 4x^2 + 12x - 9\)	M1
\(24x - 9\) or \(3(8x - 3)\)	A1
	(3)

Notes

M1: For expansion of \(4x(x + 3)\) or \((2x - 3)^2\) or \(-(2x - 3)^2\)

M1: Fully correct expansions with correct removal of bracket (ie all signs correct)