(b) Solve \(7 + x \leqslant \dfrac{5x}{2} - 8\) (3)
(c) Solve \(9 \lt 2y + 4 \lt 12\) (2)
Mark scheme (a)
Answer
Mark
Mark scheme
\(6 - 3m\)
B1
for \(6 - 3m\) oe
Additional guidance
Accept \(3(2 - m)\)
Mark scheme (b)
Answer
Mark
Mark scheme
\(x \geqslant 10\)
M1
for a correct first step working with an inequality or an equation, eg \(7 + x + 8 \leqslant \dfrac{5x}{2} - 8 + 8\) or \(15 + x \leqslant \dfrac{5x}{2}\) or \(7 + x - x \leqslant \dfrac{5x}{2} - 8 - x\) or \(7 \leqslant \dfrac{3x}{2} - 8\) or \(7 \times 2 + x \times 2 \leqslant \dfrac{5x}{2} \times 2 - 8 \times 2\) or \(14 + 2x \leqslant 5x - 16\)
M1
(dep M1) for a correct second step, eg subtracts \(x\) from both sides or adds 8 to both sides or subtracts \(2x\) from both sides or multiplies both sides by 2
or gives the critical value of 10
A1
for \(x \geqslant 10\) as final answer
Additional guidance
Can work with an equation or incorrect inequality symbol for both M marks
For M marks step must be carried out not just intention shown. For example, if you see \[\begin{array}{ccc} 7 + x &\leqslant& \dfrac{5x}{2} - 8 \\ +8 && +8 \end{array}\] award M1 for \(k + x \leqslant \dfrac{5x}{2}\) where \(k \gt 7\) or indicating \(-x\) and reaching \(7 \leqslant kx - 8\) where \(k \lt \dfrac{5}{2}\) or indicating multiplying by 2 and obtaining an equation or inequality with no more than one term incorrect and no term unchanged.
The first 2 marks can be awarded for critical value of 10, eg \(x = 10\)
Accept \(10 \leqslant x\)
Mark scheme (c)
Answer
Mark
Mark scheme
\(2.5 \lt y \lt 4\)
M1
for a correct first step, eg \(9 - 4 \lt 2y \lt 12 - 4\) or \(5 \lt 2y \lt 8\) or \(9 \div 2 \lt y + 2 \lt 12 \div 2\) or \(4.5 \lt y + 2 \lt 6\) or showing 2.5 and 4 as the critical values
A1
for \(2.5 \lt y \lt 4\) oe as final answer
Additional guidance
For M mark condone use of “=” and incorrect inequality signs
For M mark step must be carried out not just intention shown. For example, if you see \[\begin{array}{ccccc} 9 &\lt & 2y + 4 &\lt & 12 \\ -4 && -4 && -4 \end{array}\] award M1 for \(a \lt 2y \lt b\) where \(a \lt 9\) and \(b \lt 12\)
or if you see \[\begin{array}{ccccc} 9 &\lt & 2y + 4 &\lt & 12 \\ \div 2 && \div 2 && \div 2 \end{array}\] award M1 for \(a \lt y + 2 \lt b\) where \(a \lt 9\) and \(b \lt 12\)
17 Show that \(\dfrac{6x - y}{10xy} + \dfrac{1}{2x} - \dfrac{2y - 7x}{5xy}\) simplifies to \(\dfrac{k}{y}\) where \(k\) is an integer. (3)
Mark scheme
Answer
Mark
Mark scheme
shown
M1
for use of a common denominator for 2 correct fractions, eg 2 of \(\dfrac{6x - y}{10xy}\) or \(\dfrac{5y}{10xy}\) or \(\dfrac{2(2y - 7x)}{10xy}\) oe
M1
for a correct method to write as single fraction, eg \(\dfrac{6x - y}{10xy} + \dfrac{5y}{10xy} - \dfrac{4y - 14x}{10xy}\) or \(\dfrac{6x - y}{10xy} + \dfrac{5y}{10xy} - \dfrac{(4y - 14x)}{10xy}\) or \(\dfrac{6x - y + 5y - 4y + 14x}{10xy}\) or \(\dfrac{20x}{10xy}\) oe
for factorisation eg \((4 - x^2 =)\ (2 - x)(2 + x)\) and \((x^2 + 3x =)\ x(x + 3)\) or for inversion and multiplication (condone incorrect factorising) eg \(\dfrac{4 - x^2}{x^2 + 3x} \times \dfrac{x + 3}{x + 2}\) oe
M1
for factorisation of both quadratics and inversion and multiplication, eg \(\dfrac{(2 - x)(2 + x)}{x(x + 3)} \times \dfrac{x + 3}{x + 2}\) allow sign errors in one bracket
in the form \(\dfrac{ax + b}{cx + d}\) where \(a\), \(b\), \(c\) and \(d\) are integers. (4)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{6x + 37}{6x - 42}\)
P1
for factorising to get the numerator, eg \((x + 4)(2x - 14)\) or \(2(x + 4)(x - 7)\) or \((2x + 8)(x - 7)\)
P1
For process to work with the division by \(\dfrac{2x^2 - 6x - 56}{2x + 3}\) eg multiplies by \(\dfrac{2x + 3}{2x^2 - 6x - 56}\)
P1
for two correct fractions with a common denominator, eg \(\dfrac{28}{6(x - 7)}\) and \(\dfrac{6x + 9}{6(x - 7)}\) oe or \(\dfrac{28 + 6x + 9}{6(x - 7)}\) oe
for writing a correct equation without fractions eg \((2 - 2x) + 3(x + 4) = (2 - 2x)(x + 4)\) oe
M1
for simplifying to a correct equation in the form \(ax^2 + bx + c\ (= 0)\), eg \(2x^2 + 7x + 6\ (= 0)\)
M1
for method to solve a correct quadratic equation, eg \((2x + 3)(x + 2)\ (= 0)\) or correct substitution into the formula or for correctly completing the square
14 Show that \(\dfrac{x^2 - x - 6}{2x^2 - 5x - 3}\) can be written in the form \(\dfrac{ax + b}{cx + d}\) where \(a\), \(b\), \(c\) and \(d\) are integers. (3)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{x + 2}{2x + 1}\)
M1
for correctly factorising one expression, eg \((x - 3)(x + 2)\) or \((x - 3)(2x + 1)\)
M1
for factorising both expressions, eg \((x - 3)(x + 2)\) and \((x - 3)(2x + 1)\)
Give your answer in the form \(\dfrac{p \pm \sqrt{q}}{2}\) where \(p\) and \(q\) are integers. (4)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{5 \pm \sqrt{15}}{2}\)
M1
for using a common denominator eg \(\dfrac{x - 1}{(2x - 1)(x - 1)} + \dfrac{3(2x - 1)}{(2x - 1)(x - 1)}\ (= 1)\) or \((x - 1) + 3(2x - 1) = (2x - 1)(x - 1)\)
M1
for expanding and rearranging to get \(2x^2 - 10x + 5\ (= 0)\)
M1
(dep M1) ft for a method to solve their 3 term quadratic equation eg \(\dfrac{10 \pm \sqrt{(-10)^2 - 4 \times 2 \times 5}}{2 \times 2}\) or \(\dfrac{10 \pm \sqrt{60}}{4}\) or \(2\left[\left(x - \dfrac{5}{2}\right)^2 - \left(\dfrac{5}{2}\right)^2\right] + 5 = 0\) oe
A1
cao
Additional guidance
Note we don’t need to see “= 0”; just the LHS is sufficient Accept other forms of the 3 term quadratic, eg \(2x^2 - 10x = -5\)
19 Show that \(\dfrac{3x}{x + 2} - \dfrac{2x + 1}{x - 2} - 1\) can be written in the form \(\dfrac{ax + b}{x^2 - 4}\) where \(a\) and \(b\) are integers. (4)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{-11x + 2}{x^2 - 4}\)
M1
for writing at least one of the 3 terms with a denominator of \((x^2 - 4)\) or \((x - 2)(x + 2)\) eg \(\dfrac{3x(x - 2)}{x^2 - 4}\) oe or \(\dfrac{(x + 2)(2x + 1)}{x^2 - 4}\) oe or \(\dfrac{x^2 - 4}{x^2 - 4}\)
M1
for \(\dfrac{3x(x - 2)}{x^2 - 4} - \dfrac{(x + 2)(2x + 1)}{x^2 - 4} - \dfrac{x^2 - 4}{x^2 - 4}\) oe or for \(\dfrac{x^2 - 11x - 2}{x^2 - 4}\ (-\ 1)\) or for \(\dfrac{[x^2 - 11x - 2]}{x^2 - 4} - \dfrac{x^2 - 4}{x^2 - 4}\)
M1
for a numerator of \(3x^2 - 6x - 2x^2 - 5x - 2 - x^2 + 4\)
A1
for \(\dfrac{-11x + 2}{x^2 - 4}\)
Additional guidance
Students may work with a denominator of \((x - 2)(x + 2)\) for the first 3 marks
\([x^2 - 11x - 2]\) denotes their expansion of \(3x(x - 2) - (x + 2)(2x + 1)\)
22 Show that \(\dfrac{6x^3}{(9x^2 - 144)} \div \dfrac{2x^4}{3(x - 4)}\) can be written in the form \(\dfrac{1}{x(x + r)}\) where \(r\) is an integer. (3)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{1}{x(x + 4)}\)
M1
inverting the fraction and multiplying eg \(\dfrac{6x^3}{(9x^2 - 144)} \times \dfrac{3(x - 4)}{2x^4}\)
15 Show that \(\dfrac{4x + 3}{2x} + \dfrac{3}{5}\) can be written in the form \(\dfrac{ax + b}{cx}\) where \(a\), \(b\) and \(c\) are integers. (3)
Mark scheme
Answer
Mark
Mark scheme
\(\dfrac{26x + 15}{10x}\)
M1
for method to write at least one of the fractions with a suitable denominator, eg \(\dfrac{4x + 3}{2x} \times \dfrac{5}{5}\ \left(= \dfrac{20x + 15}{10x}\right)\) or \(\dfrac{3}{5} \times \dfrac{2x}{2x}\ \left(= \dfrac{6x}{10x}\right)\)
M1
for method to combine the fractions, eg \(\dfrac{5(4x + 3)}{5 \times 2x} + \dfrac{3 \times 2x}{5 \times 2x}\) or \(\dfrac{5(4x + 3) + 3 \times 2x}{5 \times 2x}\) or \(\dfrac{20x + 15}{10x} + \dfrac{6x}{10x}\)
A1
for correct algebra leading to \(\dfrac{26x + 15}{10x}\) oe in form \(\dfrac{ax + b}{cx}\)
for factorising the numerator or denominator of the 1st fraction, eg \(\dfrac{7(x - 2)}{(x - 2)(x + 6)}\) or \(\dfrac{7(x - 2)}{x^2 + 4x - 12}\) or \(\dfrac{7x - 14}{(x - 2)(x + 6)}\)
M1
for factorising the denominator of the second fraction, eg \(\dfrac{x - 6}{x(x + 6)(x - 6)}\ \left(= \dfrac{1}{x(x + 6)}\right)\)
(a) Write \(\dfrac{4x^2 - 9}{6x + 9} \times \dfrac{2x}{x^2 - 3x}\) in the form \(\dfrac{ax + b}{cx + d}\) where \(a\), \(b\), \(c\) and \(d\) are integers. (3)
(b) Express \(\dfrac{3}{x + 1} + \dfrac{1}{x - 2} - \dfrac{4}{x}\) as a single fraction in its simplest form. (3)
19 \(2 - \dfrac{x + 2}{x - 3} - \dfrac{x - 6}{x + 3}\) can be written as a single fraction in the form \(\dfrac{ax + b}{x^2 - 9}\) where \(a\) and \(b\) are integers.
Work out the value of \(a\) and the value of \(b\). (4)
Mark scheme
Answer
Mark
Notes
\(a = 4\), \(b = -42\)
M1
for at least two terms from \(2(x - 3)(x + 3)\), \((x + 2)(x + 3)\), \((x - 6)(x - 3)\)
M1
(dep) for the correct expansion of at least two expressions, irrespective of signs, eg. \(2x^2 - 18\), \(x^2 + 2x + 3x + 6\), \(x^2 - 6x - 3x + 18\) oe
for writing at least 2 fractions with a common denominator eg. \(\dfrac{3(3x - 2)}{12}\), \(\dfrac{4(2x + 5)}{12}\), \(\dfrac{2(1 - x)}{12}\) with at least one correct numerator or for \(\dfrac{3x}{4} - \dfrac{2}{4} - \dfrac{2x}{3} - \dfrac{5}{3} = \dfrac{1}{6} - \dfrac{x}{6}\) (accept \(+\dfrac{5}{3}\) instead of \(-\dfrac{5}{3}\))
M1
(dep) for a method to eliminate all fractions in an equation, ignore errors in any expanded terms eg. \(3(3x - 2) - 4(2x + 5) = 2(1 - x)\) or \(6 \times [3(3x - 2) - 4(2x + 5)] = 12 \times [1 - x]\) or \(3 \times 3x - 3 \times 2 - 4 \times 2x - 4 \times 5 = 2 \times 1 - 2 \times x\) OR for the correct expansion of brackets leading to \(\dfrac{9x - 6 - 8x - 20}{12} = \dfrac{2 - 2x}{12}\)
M1
(dep on M2) for correctly isolating terms in \(x\) and number terms of their linear equation e.g. \(9x - 8x + 2x = 2 + 6 + 20\)
M1 for identifying a common denominator in a correct expression or equation eg \(\dfrac{4(x + 1)}{12} + \dfrac{3(2x + 5)}{12}\) (= 2) M1 (dep on M1) for expanding both brackets in an equation eg \(\dfrac{4x + 4}{12} + \dfrac{6x + 15}{12} = 2\) eg \(4x + 4 + 6x + 15 = 12 \times 2\) M1 (dep on M1) for isolating “like terms” on each side of an equation eg \(4x + 6x = 24 - 15 - 4\)
M1 for correct method to expand \((n + 2)^2\) or \((n + 1)^2\) M1 for correct simplification of numerator M1 for factorisation of \(2n^2 + 3n\) or for clearing the fractions on both sides correctly C1 for complete and correct proof
OR M1 for \(\{(n + 2) - (n + 1)\}\{(n + 2) + (n + 1)\}\) M1 for \(1 \times (2n + 3)\) M1 for factorisation of \(2n^2 + 3n\) or for clearing the fractions on both sides correctly C1 for complete and correct proof
OR M1 for \(n\{(n + 2)^2 - (n + 1)^2\} = (2n^2 + 3n) \times 1\) M1 for \(n(n + 2)^2 - n(n + 1)^2\) or for correct expansion of \((n + 2)^2 - (n + 1)^2\) M1 for correct expansion of \(n\{(n + 2)^2 - (n + 1)^2\}\) C1 for complete and correct proof (must include statement recognising the equality of LHS and RHS)
(a) Write as a single fraction in its simplest form \(\dfrac{5}{2 - x} - \dfrac{4}{x}\) (3)
(b) Make \(y\) the subject of the formula \[t = \frac{2 - 3y}{y + 2}\] (4)
Mark scheme (a)
Answer
Mark
Notes
\(\dfrac{9x - 8}{x(2 - x)}\)
3
M1 for method to use a common denominator, e.g. \(\dfrac{5(x) - 4(2 - x)}{x(2 - x)}\) M1 (dep on M1) for correct expansion of brackets and combination of numerators e.g. \(5x - 8 + 4x\) \((= 9x - 8)\) A1 for \(\dfrac{9x - 8}{x(2 - x)}\) or \(\dfrac{9x - 8}{2x - x^2}\)
Mark scheme (b)
Answer
Mark
Notes
\(y = \dfrac{2 - 2t}{t + 3}\)
4
M1 for intention to multiply both sides by \(y + 2\) as a first step e.g. \(t \times y + 2 = 2 - 3y\) M1 for intention to correctly isolate their \(y\) terms on one side and the other terms on the other side, e.g. \(ty + 3y = 2 - 2t\) M1 for intention to factorise, e.g. \(y(t + 3)\) \((= 2 - 2t)\) A1 for \(y = \dfrac{2 - 2t}{t + 3}\) oe
M1 for putting LHS over a common denominator with at least one correct numerator (ignore signs) or for showing an intention to multiply each term on both sides by \(R\) or \(v\) or \(b\) M1 for rearranging correctly to isolate terms in \(m\) M1 for factorising with common factor \(m\) from 2 terms A1 for \(m = \dfrac{tv(R - b)}{b(R - v)}\) oe
(c) Write \(\dfrac{5}{x - 3} - \dfrac{4}{x + 3}\) as a single fraction in its simplest form. (3)
Mark scheme (a)
Answer
Mark
Notes
1
1
B1 cao
Mark scheme (b)
Answer
Mark
Notes
\(\dfrac{5y}{8x^3}\)
2
M1 for correct square root or correct use of reciprocal eg \(\dfrac{8x^3}{5y}\) or \(\dfrac{25y^2}{64x^6}\) A1 for \(\dfrac{5y}{8x^3}\) or \(\dfrac{5}{8}yx^{-3}\) oe
Mark scheme (c)
Answer
Mark
Notes
\(\dfrac{x + 27}{(x - 3)(x + 3)}\)
3
M1 for denominator \((x - 3)(x + 3)\) or \(x^2 - 9\) M1 for \(\dfrac{5(x + 3)}{(x - 3)(x + 3)}\) oe or \(\dfrac{4(x - 3)}{(x - 3)(x + 3)}\) oe (NB The denominator must be \((x - 3)(x + 3)\) or \(x^2 - 9\) or another suitable common denominator) A1 for \(\dfrac{x + 27}{(x - 3)(x + 3)}\) or \(\dfrac{x + 27}{x^2 - 9}\)
(b) Write as a single fraction in its simplest form \[\frac{2}{y + 3} - \frac{1}{y - 6}\] (3)
Mark scheme (a)
Answer
Mark
Notes
4
3
M1 for correct expansion to \(32x - 8\) or multiplying both sides by \(3x\) or dividing both sides by 4 M1 for a compete and correct method to isolate the \(x\) terms and the number terms (condone one arithmetic error in multiplying out the bracket) A1 cao
Mark scheme (b)
Working
Answer
Mark
Notes
\(\dfrac{2(y - 6) - (y + 3)}{(y + 3)(y - 6)}\)
\(\dfrac{y - 15}{(y + 3)(y - 6)}\)
3
M1 for common denominator of \((y + 3)(y - 6)\) M1 for \(\dfrac{2(y - 6)}{(y + 3)(y - 6)} - \dfrac{y + 3}{(y + 3)(y - 6)}\) oe or \(\dfrac{2(y - 6) - (y + 3)}{(y + 3)(y - 6)}\) oe A1 for \(\dfrac{y - 15}{(y + 3)(y - 6)}\) or \(\dfrac{y - 15}{y^2 - 3y - 18}\)
M1 for multiplying throughout by 10 oe or writing LHS as a single fraction e.g \(2(4x - 1) + 5(x + 4) = 3 \times 10\) or \(\dfrac{2(4x - 1) + 5(x + 4)}{10}\) or \(\dfrac{2(4x - 1)}{10} + \dfrac{5(x + 4)}{10}\) M1 (dep) for a complete correct method to obtain linear equation of the form \(ax = b\) (condone one arithmetic error in multiplying out the bracket) A1 for \(\dfrac{12}{13}\) oe (decimal equivalent is 0.923…)
M1 Use of common denominator of 6 (or any other multiple of 6) and at least one numerator correct Eg. \(\dfrac{3(x + 1)}{6}\) or \(\dfrac{2(x + 3)}{6}\) M1 \(\dfrac{3(x + 1)}{6} + \dfrac{2(x + 3)}{6}\) oe A1 cao
M1 for denominator \((x + 2)(x - 2)\) oe or \(x^2 - 4\) M1 for \(\dfrac{4(x - 2)}{(x + 2)(x - 2)}\) oe or \(\dfrac{3(x + 2)}{(x + 2)(x - 2)}\) oe (NB. The denominator must be \((x + 2)(x - 2)\) or \(x^2 - 4\) or another suitable common denominator) A1 for \(\dfrac{7x - 2}{(x + 2)(x - 2)}\) or \(\dfrac{7x - 2}{x^2 - 4}\) SC: If no marks awarded then award B1 for \(\dfrac{4(x - 2)}{x^2 - 2} + \dfrac{3(x + 2)}{x^2 - 2}\) oe
24 Prove that \(\quad \dfrac{60x^4 - 15x^2}{-2x - 1} \times \dfrac{1}{6x - 3} \quad\) can never be positive. [4 marks]
Mark scheme
Answer
Mark
Comments
Partially or fully factorises numerator
M1
eg \(15x(4x^3 - x)\) or \(x^2(60x^2 - 15)\) or \(15x^2(2x - 1)(2x + 1)\)
Factorises at least one denominator or correct multiplication of the denominators
M1
eg \(-(2x + 1)\) and/or \(3(2x - 1)\)
eg \(-12x^2 + (6x - 6x +)\ 3\) may be in a grid
Converts numerators and denominators into terms which can be fully cancelled
M1dep
dep on M1M1 eg \(\dfrac{15x^2(2x - 1)(2x + 1)}{-(2x + 1)3(2x - 1)}\) or \(\dfrac{15x^2(2x - 1)(2x + 1)}{-(2x + 1)} \times \dfrac{1}{3(2x - 1)}\) or \(\dfrac{5x^2(12x^2 - 3)}{-(12x^2 - 3)}\) factorisation and cancelling may be done in stages
\(\dfrac{15x^2}{-3}\) or \(-5x^2\) with M3 awarded and explanation that \(x^2\) cannot be negative
A1
oe with full cancelling of algebraic terms
condone explanation that \(x^2\) must be positive
Additional guidance
\(\dfrac{20x^4 - 5x^2}{(-2x - 1)(2x - 1)}\) or \(\dfrac{20x^4 - 5x^2}{-4x^2 + 1}\) implies M1M1 as 3 has been cancelled from both
M1M1
\(\dfrac{5x^2(2x + 1)}{-(2x + 1)}\) implies complete factorisation of numerator and denominator as 3 and \((2x - 1)\) have been cancelled from both
\(d = k \quad e = 14k^2 \quad f = k\) where \(k\) is a non-zero constant
A1
eg \(d = 1 \quad e = 14 \quad f = 1\) or \(d = 2 \quad e = 56 \quad f = 2\) or \(d = 6 \quad e = 504 \quad f = 6\)
Additional guidance
Take the values on the answer lines as the final answer eg \(\dfrac{2 \pm \sqrt{56}}{2}\) in working with \(d = 2 \quad e = \sqrt{56} \quad f = 2\) on answer lines
M3A0
\(1 \pm \sqrt{14}\) in working with \(d = 1 \quad e = 14 \quad f =\) (blank)
M3A0
For terms seen in a grid accept eg \(3x\) for \(+3x\)
For up to M2 accept algebraic fractions but do not allow 3rd M1 unless recovered eg \(\dfrac{x^2 - 9x + 3x - 27}{x + 3} = \dfrac{12 - 2x^2}{x + 3}\)
condone \(b + a\) for \(a + b\) condone \(30a + 30b\) for \(30(a + b)\)
Additional guidance
Students can gain M1M1 if they incorrectly simplify a correct expression for total time before forming the division eg \(\dfrac{30}{a} + \dfrac{30}{b} = \dfrac{60}{a + b}\) followed by \(60 \div \dfrac{60}{a + b}\)
M1M1A0
Allow correct cancellation of 60 and 30 at any stage of the working
\((x - 2)(x + 2)\) or \(x^2 - 2x + 2x - 4\) must be seen (expansion may be seen in a grid) brackets in any order if the brackets are not shown for the numerators, expansions must be correct may be seen as a single fraction
\(x^2 - 5x + 2x - 10\) or \(x^2 - 3x - 10\) or \(x^2 + 5x - 2x - 10\) or \(x^2 + 3x - 10\)
M1
correct expansion of \((x - 5)(x + 2)\) or \((x + 5)(x - 2)\) ignore denominators may be seen in a grid implied by \(2x^2 - 20\) if no errors seen in expansions
M2 seen with no errors and \(\dfrac{2x^2 - 20}{x^2 - 4}\)
A1
allow M2 seen with no errors and \(a = 2 \quad b = 20\)
Additional guidance
Missing brackets must be recovered but condone missing closing bracket at the end of a numerator or denominator eg \(\dfrac{(x - 5)(x + 2)}{(x - 2)(x + 2)} + \dfrac{(x + 5)(x - 2)}{(x + 2)(x - 2}\)
1st M1
2nd M1 is awarded for four correct terms even if subsequently simplified incorrectly
For terms seen in a grid, signs must be correct (allow eg \(2x\) for \(+\,2x\))
For 1st M1 allow multiplication signs
After M2A1 ignore incorrect values stated eg \(a = 2 \quad b = -20\)
\(\dfrac{2x^2 - 20}{x^2 - 4}\) may come from wrong working or incomplete working eg \(\dfrac{(x - 5)(x + 2)}{(x - 2)(x + 2)} + \dfrac{(x + 5)(x - 2)}{(x + 2)(x - 2)}\) \(\dfrac{x^2 - 10 + x^2 - 10}{x^2 - 4} = \dfrac{2x^2 - 20}{x^2 - 4}\)
\(\dfrac{3}{10}x = \dfrac{20}{5} - \dfrac{29}{5}\) or \(\dfrac{3}{10}x = -\dfrac{9}{5}\)
M1
oe eg \(0.3x = -1.8\) terms must be collected
\(-6\)
A1ft
ft M1A0M1
Additional guidance
Accept decimal answers for follow through correct to 1 dp or better
Apply the principles of alt 1 for any use of other common denominators eg common denominator of 20 (or multiplication through by 20)
\(10(x + 8) + 4(9 - x) = 6x + 116\)
M1A1
\(6x + 116 = 80 \qquad x = -6\)
M1A1
An incorrect simplification of \(5x + 40 + 18 - 2x\) may still gain the third and fourth marks
eg \(5x + 40 + 18 - 2x = 3x + 68\) followed by \(3x + 68 = 40\) and \(x = -\dfrac{28}{3}\)
M1A0M1A1ft
eg \(5x + 40 + 18 - 2x = 2x + 68\) followed by \(2x + 68 = 40\) and \(x = -14\)
M1A0M1A1ft
An incorrect denominator may still gain the third and fourth marks \(\dfrac{5x + 40 + 18 - 2x}{7}\) followed by \(5x + 40 + 18 - 2x = 28\) and \(x = -10\)
M1A0M1A1ft
Denominator not processed \(3x + 58 = 4\) followed by \(3x = -54\) and \(x = -18\)
M1A1M0A0
\((x + 8) + (9 - x) = 40\)
M0A0M1A0
Two errors in the expansion but with brackets seen may go on to get the third and fourth marks \(5(x + 8) + 2(9 - x) = 5x + 8 + 18 - x\)
1st M1A0
Two errors in the expansion and no brackets seen, no follow through allowed \(5x + 8 + 18 - x\) followed by \(4x + 26 = 40\) and \(x = \dfrac{14}{4}\)
\(\left(\dfrac{6}{a} =\right) \dfrac{24}{4a}\) or converts both fractions to a common denominator or correct unsimplified fraction eg \(\dfrac{26}{8a}\) or \(\dfrac{13a}{4a^2}\) or \(\dfrac{3.25}{a}\)
M1
oe eg \(\dfrac{48}{8a}\) and \(\dfrac{22}{8a}\) or \(\dfrac{24a}{4a^2}\) and \(\dfrac{11a}{4a^2}\)
\(\dfrac{13}{4a}\)
A1
Additional guidance
Do not ignore further work eg \(\dfrac{13}{4a}\) followed by answer \(\dfrac{3.25}{a}\)
M1A0
Allow a division sign rather than a fraction line for M1 only eg \(26 \div 8a\) eg \(13 \div 4a\)
\(y(y + 7)\) or \(y^2 + 7y\) with no other working
M1M1M1A1
Answer \(\dfrac{y(y + 7)}{1}\) or \(\dfrac{y^2 + 7y}{1}\)
M1M1M1A0
Ignore the consistent use of a different variable within a factorisation
Award SC1 only if there are no correct factorisations eg correct factorisation to \((y + 7)(y + 3)\) and correct expansion to \(y^4 - 3y^3 + 10y^3 - 30y^2 + 21y^2 - 63y\)
Give your answer as a single fraction in its simplest form. [2 marks]
(b) Sofia is trying to simplify \(\quad \dfrac{6c + 10}{2}\)
Her method is
divide \(6c\) by 2 then add 10
Evaluate her method. [1 mark]
Mark scheme (a)
Answer
Mark
Comments
\(\dfrac{5a^2}{4}\) or \(1\dfrac{1}{4}a^2\)
B2
B1 correct single fraction not in simplest form eg \(\dfrac{50a^2}{40}\) or \(1.25a^2\) or \(\dfrac{5}{4}a\) or \(\dfrac{5a}{4}\) or \(1\dfrac{1}{4}a\)
Additional guidance
Final answer \(1.25a^2\) (even if \(\dfrac{5a^2}{4}\) seen in working)
B1
Mark scheme (b)
Answer
Mark
Comments
Valid evaluation
B1
eg she needs to divide 10 by 2 or the answer should be \(3c + 5\)
Additional guidance
Do not award marks when an incorrect statement or incorrect algebra is seen with a correct statement or correct algebra
She needs to add 5 not 10
B1
She must divide all of the numerator by 2
B1
She must divide everything by 2
B1
She should divide both sides by 2
B0
She needs to work out \(6c + 10\) then divide by 2
A1 \(2(n + 1)\,\dfrac{n + 1}{2}\) or \(\dfrac{2(n + 1)^2}{2}\) allow \((n + 1)(n + 1)\) for \((n + 1)^2\)
Additional guidance
Only substituting in values of \(n\)
M0M0A0A0
Consistently using a different letter to \(n\) can score up to M1M1A1A1
Using two different letters consistently within the two fractions (eg \(n\) replaced by \(x\) in the first equation and \(n\) replaced by \(y\) in the second equation) can score a maximum of M1M1A0A0 unless recovered to the same letter
Multiplying fractions instead of adding can score a maximum of M2A0
Equating to \(n^2\) in working can score a maximum of M1M1A0A0 (equating to eg \(x^2\) can score up to M1M1A1A1)
\(1n\) is allowed for \(n\) throughout
Alts 2 and 3 \(\dfrac{n + 1}{2}(2n + 2)\) with M1 seen scores M2 If they attempt to expand \((n + 1)(2n + 2)\) use Alt 2 If they attempt to expand \(\dfrac{1}{2}(2n + 2)\) use Alt 3
Correct factorisation of numerator \(2(2x - 4x^2)\) or \(4(x - 2x^2)\) or \(x(4 - 8x)\) or \(2x(2 - 4x)\) or \(4x(1 - 2x)\) or correct factorisation of denominator \(2(6x - 3)\) or \(3(4x - 2)\) or \(6(2x - 1)\) or correct cancelling by 2 throughout \(\dfrac{2x - 4x^2}{6x - 3}\)
M1
oe with negative coefficients
Correct fraction with numerator \(4x(1 - 2x)\) or \(-4x(2x - 1)\) and denominator \(6(2x - 1)\) or \(-6(1 - 2x)\) or \(-\dfrac{4x}{6}\) or \(\dfrac{-4x}{6}\) or \(\dfrac{4x}{-6}\) or \(\dfrac{2x(2 - 4x)}{-3(2 - 4x)}\) or \(\dfrac{2x(2 - 4x)}{3(4x - 2)}\)
M1dep
oe with cancelling of 2 throughout eg \(\dfrac{2x(1 - 2x)}{3(2x - 1)}\) or \(\dfrac{2x(1 - 2x)}{-3(1 - 2x)}\)
\(-\dfrac{2x}{3}\) or \(-\dfrac{2}{3}x\)
A1
allow \(\dfrac{-2x}{3}\) or \(\dfrac{2x}{-3}\)
Additional guidance
Allow multiplication signs up to M1M1
Allow \(-0.\dot{6}\) for \(-\dfrac{2}{3}\)
Do not allow \(-0.66\ldots\) for \(-\dfrac{2}{3}\)
For the first M1 only, allow any correct factorisation seen within multiple attempts
can be written in the form \(\;\dfrac{ax + b}{cx}\;\) where \(a\), \(b\) and \(c\) are integers. [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
\(\dfrac{2(x + 4)}{6x}\) or (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x + 8}{6x}\) or (–)\(\dfrac{15}{6x}\)
or \(\dfrac{2x(x + 4)}{6x^2}\) or (–)\(\dfrac{15x}{6x^2}\) or \(\dfrac{2x^2 + 8x}{6x^2}\) or (–)\(\dfrac{15x}{6x^2}\)
M1
oe A correct fraction using a common denominator for one of the given fractions Accept for this mark only eg \(\;2(3x)\) for \(6x\) 3(5) for 15 \((2x)(3x)\) for \(6x^2\) First fraction can be written as separate fractions eg \(\dfrac{2x}{2(3x)} + \dfrac{8}{2(3x)}\)
\(\dfrac{2(x + 4)}{6x}\) and (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x + 8}{6x}\) and (–)\(\dfrac{15}{6x}\)
or \(\dfrac{2x(x + 4)}{6x^2}\) and (–)\(\dfrac{15x}{6x^2}\) or \(\dfrac{2x^2 + 8x}{6x^2}\) and (–)\(\dfrac{15x}{6x^2}\)
A1
oe A correct fraction using a common denominator for both of the given fractions First fraction can be written as separate fractions eg \(\dfrac{2x}{6x} + \dfrac{8}{6x}\)
\(\dfrac{2x - 7}{6x}\) or \(\dfrac{2\mathrm{k}x - 7\mathrm{k}}{6\mathrm{k}x}\), where k is a constant value
A1
Accept eg \(\dfrac{2x + -7}{6x}\) Do not ignore further working
Alternative method 2
\(\dfrac{2(x + 4)}{6x}\) or (–)\(\dfrac{15}{6x}\) or \(\dfrac{2x + 8}{6x}\) or (–)\(\dfrac{15}{6x}\)
or \(\dfrac{2x(x + 4)}{6x^2}\) or (–)\(\dfrac{15x}{6x^2}\) or \(\dfrac{2x^2 + 8x}{6x^2}\) or (–)\(\dfrac{15x}{6x^2}\)
M1
oe A correct fraction using a common denominator for one of the given fractions Accept for this mark only eg \(\;2(3x)\) for \(6x\) 3(5) for 15 \((2x)(3x)\) for \(6x^2\) First fraction can be written as separate fractions eg \(\dfrac{2x}{2(3x)} + \dfrac{8}{2(3x)}\)
\(\dfrac{2x + 8 - 15}{6x}\) or \(\dfrac{2x - 7}{6x}\) or \(\dfrac{2\mathrm{k}x - 7\mathrm{k}}{6\mathrm{k}x}\), where k is a constant value
A1
Allow one error in numerator Accept eg \(\dfrac{2x + -7}{6x}\) Must be \(6x\) or a multiple of \(6x\)
\(\dfrac{2x - 7}{6x}\) or \(\dfrac{2\mathrm{k}x - 7\mathrm{k}}{6\mathrm{k}x}\), where k is a constant value
A1
Accept eg \(\dfrac{2x + -7}{6x}\) Do not ignore further working
Additional guidance
Use the method that gives the greater mark
\(\dfrac{2x^2 - 7x}{6x^2}\)
M1A1
\(\dfrac{2x - 7}{6x} = \dfrac{-5}{6x}\)
M1A1A0
\(\dfrac{15x}{6x^2} - \dfrac{2x^2 + 8x}{6x^2}\) (order of fractions reversed)
\(\dfrac{8x^2 - 8}{4x + 4} \quad\) simplifies to the form \(\quad ax + b \quad\) where \(a\) and \(b\) are integers. [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
Any correct factorisation of the numerator or the denominator
M1
eg \(8(x^2 - 1)\) or \(4(x + 1)\) or \(2(4x^2 - 4)\) or \(2(2x + 2)\) or \(4(2x^2 - 2)\) or \((4x + 4)(2x - 2)\) or \((4x - 4)(2x + 2)\) or \((8x + 8)(x - 1)\) or \((8x - 8)(x + 1)\) or \(-2(-4x^2 + 4)\) does not need to be seen in a fraction may be implied eg \(\dfrac{2x^2 - 2}{x + 1}\) or \(\dfrac{4x^2 - 4}{2x + 2}\)
Correct fraction with a common algebraic factor in the numerator and the denominator
A1
eg \(\dfrac{8(x + 1)(x - 1)}{4(x + 1)}\) or \(\dfrac{2(2x + 2)(2x - 2)}{2(2x + 2)}\) or \(\dfrac{2(x + 1)(x - 1)}{(x + 1)}\) or \(\dfrac{4(x + 1)(2x - 2)}{4(x + 1)}\) or \(\dfrac{(4x + 4)(2x - 2)}{4x + 4}\)
\(2x - 2\) or \(a = 2\) and \(b = -2\) with M1A1 scored
A1
Alternative method 2
\(4ax^2 + 4ax + 4bx + 4b\)
M1
oe expands \((ax + b)(4x + 4)\) to 4 terms with at least 3 terms correct
Any 2 of \(4a = 8 \qquad 4b = -8 \qquad 4a + 4b = 0\)
A1
\(a = 2\) and \(b = -2\) and shows that third equation is satisfied with M1A1 scored
A1
Additional guidance
M1 is implied by the first A1 eg \(\dfrac{8(x + 1)(x - 1)}{4(x + 1)}\)
M1A1
\(1(8x^2 - 8)\) or \(-1(8 - 8x^2)\) etc
M0
\(2x - 2\) without M1A1 scored
M0A0A0
M1A1 scored and \(2x - 2\) followed by attempt to solve \(2x - 2 = 0\)
M1A1A1
M1A1 scored and \(2x - 2\) followed by \(2(x - 1)\)
M1A1A1
M1A1 scored followed by \(2(x - 1)\) but \(2x - 2\) not seen
B2 for \((x + 6)(x - 7)\) isw or M1 for \(x(x + 6) - 7(x + 6)\) seen or \(x(x - 7) + 6(x - 7)\) seen or for \((x + a)(x + b)\) where \(a + b = -1\) or \(ab = -42\) isw
If answer \(x\) – 6/\(x\) – 7 check working and if seen written correctly allow 4 marks
M2 for \((3x - 1)(x - 5)\) [= 0] or M1 for \(3x(x - 5) - 1(x - 5)\) or for \(x(3x - 1) - 5(3x - 1)\) or for \((3x + a)(x + b)\) where \(ab = 5\) or \(3b + a = -16\)
M2 and M1 FT their 3-term quadratic M2 alt method correct substitution into formula or correct complete square M1 condone 1 error in substitution into formula For complete square, correct FT \((x + a)^2\)
If 0 or 1 scored instead award SC2 for correct answers
M2 for \([x](x + 3)(x + 5)\) or M1 for \([x](x(x + 5) + 3(x + 5))\) or \([x](x(x + 3) + 5(x + 3))\) or for \([x](x + a)(x + b)\) where \(a + b = 8\) or \(ab = 15\) and M1 for \([x](x + 3)(x - 3)\) or \(\frac{x(x^2 + 8x + 15)}{x(x^2 - 9)}\) or \(\frac{x^2 + 8x + 15}{x^2 - 9}\) or for \(x^2 + 8x + 15\) and \(x^2 - 9\) or any other partially factorised form of the numerator or denominator
For M2 and M1 marks, if written as a quotient, condone [\(x\)] not being consistently present in both or cancelled out from both
Also award M2 and M1 marks for factorising without first factorising [\(x\)]. eg. \((x^2 + 3x)(x + 5)\) earns M2
They multiplied the 2 by \((x - 1)\) it should have been \((x + 4)\)
1
The brackets have been multiplied by the wrong number it should be \(2(x + 4)\)
1
The \((x - 1)\) and \((x + 4)\) should be the other way round
1
The numerators have been multiplied by the wrong/own denominators
1
(Correct answer is given)
1
They have multiplied the wrong numerator by the wrong denominator
1
It should be this \(\frac{2(x + 4) + 3(x - 1)}{(x - 1)(x + 4)}\) (single fraction only)
1
They multiply the numerator and the denominator rather than the opposite denominator
1(BOD)
They did not cross multiply
0
They have multiplied their denominators by their numerators instead of adding them
0
Mark scheme (b)
Answer
Marks
Part marks and guidance
This question was discounted and all candidates were awarded 1 mark.
1
Notes
Editor’s note: the solution line shown is not actually wrong: \(x = -\frac{7 \pm \sqrt{7^2 - 4 \times 1 \times 5}}{2 \times 1}\) gives the same two values as \(x = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times 5}}{2 \times 1}\), because \(\pm\) covers both signs, which is presumably why the part was discounted.
\(\dfrac{2(x - 5)}{x + 2}\) or \(\dfrac{2x - 10}{x + 2}\) final answer
5
M2 for \(2(x - 5)(x + 5)\) or \((2x - 10)(x + 5)\) or M1 for \(2(x^2 - 25)\) M2 for \((x + 2)(x + 5)\) or M1 for \(x(x + 2) + 5(x + 2)\) or \(x(x + 5) + 2(x + 5)\) or for \((x + a)(x + b)\) where \(a + b = 7\) or \(ab = 10\)
For method marks condone omission of final bracket
M1 for \(x^2 - 4 = (x + 2)(x - 2)\) soi in the denominator
AND M3 for all 3 fractions combined with quadratic common denominator and expanded numerator or M2 for correct products on numerator of at least 2 equivalent fractions that are consistent with their common denominators or M1 for correct product on numerator of 1 fraction that is consistent with an attempted common denominator
AND M1dep for \(\frac{(2x - 1)(x - 2)}{(x + 2)(x - 2)}\) or \(\frac{(2x - 1)(x - 2)}{x^2 - 4}\) (dep on previous M3 earned)
e.g. \(\frac{6x}{(x + 2)(x - 2)}\) seen, or a common denominator of \((x + 2)(x - 2)\) later expanded to \(x^2 - 4\)
e.g. \(\frac{x^2 - 2x + x^2 + x + 2x + 2 - 6x}{(x + 2)(x - 2)}\) soi by \(\frac{2x^2 - 5x + 2}{(x + 2)(x - 2)}\)
e.g. \(\frac{x(x - 2)}{(x + 2)(x - 2)}\) oe and \(\frac{(x + 1)(x + 2)}{(x + 2)(x - 2)}\) oe
e.g. \(\frac{x(x - 2)}{(x + 2)(x - 2)}\) oe
Factorises numerator of combined fraction
Can earn up to M1 + M2 + M0 for common denominator used that is not in its lowest terms.
21 Show that \(\dfrac{5x}{x + 5} + \dfrac{25}{x - 7} - \dfrac{300}{(x + 5)(x - 7)}\) simplifies to an integer. [6]
Mark scheme
Answer
Marks
Part marks and guidance
5 nfww and after \(\dfrac{5(x + 5)(x - 7)}{(x + 5)(x - 7)}\) seen
6
B1 for \((x + 5)(x - 7)\) or \(x^2 + 5x - 7x - 35\) or better seen as a common denominator of the first two fractions AND B3 for numerator \(5x^2 - 10x - 175\) or B2 for numerator \(5x^2 - 10x + 125\) or M1 for \(5x(x - 7)\) and \(25(x + 5)\) AND M1 for \(5(x^2 - 2x - 35)\) or \((5x + 25)(x - 7)\) or \((x + 5)(5x - 35)\) or \(5(x + 5)(x - 7)\)
Condone missing final bracket.
Condone numerators written without any denominators or with an incorrect common denominator
20 Solve this equation, giving your answers correct to 1 decimal place.
\[\frac{5}{x + 2} + \frac{3}{x - 3} = 2\][6]
Mark scheme
Answer
Marks
Part marks and guidance
−0.3 5.3
6
B1 for \((x + 2)(x - 3)\) oe seen
likely seen in \(2(x + 2)(x - 3)\) implied by \(x^2 - x - 6\)
M1 for \(5(x - 3) + 3(x + 2)\) oe or better
implied by \(5x - 15 + 3x + 6\) or \(8x - 9\)
M1FT for \(2x^2 - 10x - 3\) [= 0] or FT their correct attempt to form a quadratic equation with at most two errors
\(2x^2 - 10x - 3\) [=0] seen scores 3 marks, condone e.g. \(2x^2 - 10x = 3\) for 3 marks
M1FT for \(\dfrac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times -3}}{2 \times 2}\) oe condoning at most two errors or better FTtheir ‘quadratic equation’
for completing the square see additional guidance
A1 for each of −0.3 or 5.3 or for both answers correct but to more than 1dp. or A1FT for two answers correct to 1 d.p. FT from their ‘quadratic equation’
For A1 the correct answers are −0.28388… and 5.28388… and can be rounded or truncated. Note: for A1FT they must get M1 first.
(b) \((x + 3)(x - 4)(x + 5)\) is identical to \(x^3 + ax^2 - 17x + b\).
Find the value of \(a\) and the value of \(b\). [2]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(\dfrac{x + 4}{x + 1}\) final answer nfww
4
M1 for \((x + 4)(x - 4)\) AND M2 for \((x - 4)(x + 1)\) Or M1 for \(x(x + 1) - 4(x + 1)\) seen or \(x(x - 4) + 1(x - 4)\) seen or for \((x + a)(x + b)\) where \(a + b = -3\) or \(ab = -4\)
nfww please check working not just answer
Mark scheme (b)
Answer
Marks
Part marks and guidance
4 and −60
2
B1 for each
For 2 marks or B1, accept answers embedded in expression provided no contradiction seen
