Higher June 2019 Paper 1 Q12

EdexcelHigherCurrent spec8 marksFactorising Solving Simple Equations

(a) Factorise \(2x^2 - 7x + 6\) (2)

(b) Solve \(\dfrac{4m + 9}{3} = 7 - 2m\)
Show clear algebraic working. (4)

Mark scheme (a)

Scheme	Marks
\((2x - 3)(x - 2)\)	B2
	(2)

Notes

B2: or \((3 - 2x)(2 - x)\)
(B1 for \((2x + a)(x + b)\) where \(ab = 6\) or \(2b + a = -7\) eg \((2x + 3)(x + 2)\), \((2x - 5)(x - 1)\)), etc or for

Mark scheme (b)

Scheme	Marks
\(4m + 9 = 3(7 - 2m)\)	M1
\(4m + 9 = 21 - 6m\)	M1
\(4m + 6m = 21 - 9\) or \(10m = 12\) or \(-21 + 9 = -6m - 4m\) or \(-10m = -12\)	M1
Working required Answer: \(\dfrac{12}{10}\) oe	A1
	(4)

Notes

M1: for removing fraction

M1: for correct expansion of bracket in a correct equation

M1: for a correct equation with \(m\) terms isolated on one side
ft their equation if first M1 awarded

A1: dep on at least M2
[SC: B2 for an answer of \(m = 2\) with working shown (from \(4m + 9 = 21 - 2m\) oe) or \(m = -0.2\) oe with working shown (from \(4m + 9 = 7 - 6m\) oe)]

Mark scheme (b) ALT

Scheme	Marks
\(\dfrac{4}{3}m + 3 = 7 - 2m\)	M1
\(\dfrac{4}{3}m + 2m = 7 - 3\) oe	M1
\(10m = 3 \times 4\) oe	M1
Working required Answer: \(\dfrac{12}{10}\) oe	A1

Notes

M1: Division of each term on LHS by 3

M1: for a correct equation with \(m\) terms isolated on one side
ft their equation if first M1 awarded

M1: For removing fraction in a fully correct equation

A1: dep on at least M2

Scheme	Marks
\(\dfrac{y^{\frac{1}{4}}}{y}\) or \(\sqrt[4]{y} = y^{\frac{1}{4}}\) or \(y^{\frac{1}{4} - 1}\)	M1
\(y^{-\frac{3}{4}}\)	A1
	(2)
	(8 marks)