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
Mark scheme (b)
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 \((y + a)(y + b)\)
(B1 for \(9(2c - 5cd)\) or \(c(18 - 45d)\) or \(3c(6 - 15d)\) or \(3(6c - 15cd)\) or \(9c(p + qd)\) where \(p\) and \(q\) are non-zero integers or \((2 - 5d)\) as a factor)
Mark scheme (b)
Scheme
Marks
eg
\(5 - 2x = 18x - 24\)
or
\(\dfrac{5}{6} - \dfrac{2}{6}x = 3x - 4\)
M1
\(5 + 24 = 18x + 2x\) oe or \(29 = 20x\) oe
or
\(\dfrac{5}{6} + 4 = \dfrac{2}{6}x + 3x\) oe
M1ft
Working required Answer: 1.45
A1
(3)
(5 marks)
Notes
M1: for removal of the fraction and correctly multiplying out RHS by 6 in an equation or separating fractions on the LHS in an equation
M1ft: dep on 4 terms for correctly rearranging their 4 term equation for terms in \(x\) on one side of the equation and number terms on the other
an expression that clearly shows the numerator is – 19 times the denominator eg \(\dfrac{-38x - 57}{2x + 3}\;(= n)\)
M1
Working required Answer: –19
A1
(4)
(4 marks)
Notes
M1indep: [NB: the two fractions when divided and cancelled give an answer of \(4(2x - 3)\) or \(2(4x - 6)\) or \(8x - 12\) (any one of these gain 2 marks) ]
M2 for any fraction with completely simplified non-linear numerator and non-linear denominator that will cancel to –19
M1: a linear expression that should give the correct value for \(n\) (this mark implies previous M marks as not all factorising is necessary)
Correct answer scores full marks (unless from obvious incorrect working) Answer: \((x + 8)(x - 3)\)
A1
(2)
(ii) Answer: –8 and 3
B1
(1)
Notes
M1: for \((x \pm 8)(x \pm 3)\) or \((x + a)(x + b)\) where \(ab = -24\) or \(a + b = 5\) and, \(a\) and \(b\) are integers
A1: for \((x + 8)(x - 3)\) Allow any letter for \(x\)
Must be in the form \((x + a)(x + b)\) where \(a\) and \(b\) are integers
B1: must ft from their answer in (a)(i) ft from their incorrect factors in the form \((x + a)(x + b)\) Award B0 for –8 and 3 if no marks scored in (a)(i)
Mark scheme (b)
Scheme
Marks
\(3y - 7y \gt -10 - 5\) or \(5 + 10 \gt 7y - 3y\)
M1
\(-4y \gt -15\) or \(15 \gt 4y\) or \(y = \dfrac{15}{4}\) oe or
M1
Working required
Answer: \(y \lt \dfrac{15}{4}\)
A1
(3)
(6 marks)
Notes
M1: allow use of = or condone incorrect inequality sign
M1: allow use of = or condone incorrect inequality sign
A1: dep on M1
oe eg \(y \lt 3.75\) or \(\dfrac{15}{4} \gt y\) or \(3.75 \gt y\)
Must have correct sign on answer line
NB Sight of correct answer in working space and just \((y =)\;\dfrac{15}{4}\) oe on answer line gains M2 only