Simultaneous Equations

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More questions on this topic: GCSE Algebra: Simultaneous Equations (105) IGCSE Algebra: Simultaneous Equations (145)

June 2025 Paper 1 Q22

AQACurrent spec6 marksCircles in Coordinate Geometry Simultaneous Equations

22 A circle has centre \((2, 3)\) and radius 5

A straight line passes through \((3, 7)\) and has a gradient of 2

Work out the coordinates of the two points where the circle and the straight line intersect. [6 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(y = 2x + 1\)	B1	oe equation of straight line eg \(y - 7 = 2(x - 3)\)
\((x - 2)^2 + (y - 3)^2 = 5^2\)	B1	oe eg \(= 25\) or brackets expanded correctly
Substitutes their \((2x + 1)\) into their equation of a circle in the form \((x - a)^2 + (y - b)^2 =\) their \(5^2\) where \(a\) and \(b \ne 0\)	M1	eg \((x - 2)^2 + ((2x + 1) - 3)^2 = 5^2\) or \(x^2 - 4x + 4 + (2x + 1)^2 - 6(2x + 1) + 9 = 25\)
\(5x^2 - 12x - 17\ (= 0)\)	A1	condone \(5x^2 - 12x = 17\) if completing the square
\((5x - 17)(x + 1)\ (= 0)\) or \(\dfrac{12 \pm \sqrt{144 + 340}}{10}\) or \(5\left[\left(x - \dfrac{6}{5}\right)^2 - \dfrac{121}{25}\right]\)	M1dep	dep on B2M1A1
\((-1, -1)\) and \((3.4, 7.8)\)	A1	oe
Alternative method 2
\(x = \dfrac{y - 1}{2}\)	B1	oe equation of straight line eg \(y - 7 = 2(x - 3)\)
\((x - 2)^2 + (y - 3)^2 = 5^2\)	B1	oe eg \(= 25\) or brackets expanded correctly
Substitutes their \(\dfrac{y - 1}{2}\) into their equation of a circle in the form \((x - a)^2 + (y - b)^2 =\) their \(5^2\) where \(a\) and \(b \ne 0\)	M1	eg \(\left(\left(\dfrac{y - 1}{2}\right) - 2\right)^2 + (y - 3)^2 = 5^2\) or \(\left(\dfrac{y - 1}{2}\right)^2 - 4\left(\dfrac{y - 1}{2}\right) + 4 + y^2 - 6y + 9 = 25\)
\(5y^2 - 34y - 39\ (= 0)\)	A1	condone \(5y^2 - 34y = 39\) if completing the square
\((5y - 39)(y + 1)\ (= 0)\) or \(\dfrac{34 \pm \sqrt{1156 + 780}}{10}\) or \(5\left[\left(y - \dfrac{17}{5}\right)^2 - \dfrac{484}{25}\right]\)	M1dep	dep on B2M1A1 would imply B2M1A1
\((-1, -1)\) and \((3.4, 7.8)\) or \(\left(\dfrac{17}{5}, \dfrac{39}{5}\right)\)	A1

Additional guidance

First M mark can imply the B marks if correct
Expanding the circle equation and then substituting in can gain marks as an oe but won’t if the circle equation has been expanded incorrectly
Condone \((5x - 17)(5x + 5)\ (= 0)\) and \((5y - 39)(5y + 5)\ (= 0)\) for the fifth mark
Correct answer from Trial and Improvement	B2M2A2

June 2025 Paper 2 Q17

AQACurrent spec5 marksSimultaneous Equations

17 Solve

\[\begin{aligned} 2a + 4b + c &= 12 \\ 3a - b + 2c &= -2 \\ a - 5b - 3c &= 6 \end{aligned}\]

Do not use trial and improvement.

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
One correct equation in two variables with terms collected and simplified	M1	eg \(14b + 7c = 0\) or \(14a + 13c = -16\) or \(a + 9b = 26\)
Two correct equations each in two variables with terms collected and simplified	M1dep	eg \(14b + 7c = 0\) and \(14b + 11c = -20\) or \(14a + 13c = -16\) and \(a + 9b = 26\)
Correct equation in one variable with terms collected and simplified	M1dep	eg \(4c = -20\)
Correct equation in one variable with terms collected and simplified and any two of \(a = 3.5 \quad b = 2.5 \quad c = -5\)	A1	implied by final A1 oe values
Correct equation in one variable with terms collected and simplified and \(a = 3.5 \quad b = 2.5 \quad c = -5\)	A1	oe values

Additional guidance

Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
3rd M1 Do not award if the only correct equation is one of the values eg only correct equation in one variable is \(c = -5\)	3rd M0
Some correct equations in two variables are \(14b - c = 40 \qquad 14a + 9c = 4 \qquad 6a - 2b = 16\) \(14b + 3c = 20 \qquad 14a - 7c = 84 \qquad 7a + 7b = 42\) \(14b + 7c = 0 \qquad 14a + 13c = -16 \qquad a + 9b = 26\) \(14b + 11c = -20 \qquad 14a + 5c = 24 \qquad 11a - 13b = 6\) All equations have equivalents
M3 can be implied if two variables are eliminated in one step eg \(4c = -20\) (from equation 2 – equation 1 – equation 3)	M3
Correct values only	Zero
Escalate any responses using a matrix method

June 2024 Paper 1 Q23

AQACurrent spec5 marksSequences Simultaneous Equations

23 The \(n\)th term of a quadratic sequence is \(\quad an^2 - 5n + c \quad\) where \(a\) and \(c\) are integers.

The first four terms of the sequence are

\(\qquad 2 \qquad x \qquad 16 \qquad y\)

Work out the values of \(x\) and \(y\). [5 marks]

Mark scheme

Answer	Mark	Comments
\(x = 6\) and \(y = 32\)	B5	B4 \(x = 6\) or \(y = 32\) and \(a = 3\) or \(c = 4\) (could be embedded in \(3n^2 - 5n + 4\)) B3 \(a = 3\) or \(c = 4\) (could be embedded in \(3n^2 - 5n + 4\)) B2 any two different correct equations set up for a combination of \(a\), \(c\), \(x\) and \(y\) (may include \(b\). Do not include \(b = -5\)) B1 any correct equation set up for a combination of \(a\), \(c\), \(x\) and \(y\) (may include \(b\). Do not include \(b = -5\)) or (second difference) \(= 18 - 2x\) or \(x + y - 32\) (must be correctly simplified)

Additional guidance

Do not accept any equations with \(n\)
\(x = 6\) on answer line with no working and \(y \ne 32\)	B0
\(x = 6\) and \(y = 32\) on answer line with no working	B5

June 2024 Paper 1 Q10

AQACurrent spec5 marksFunctions Simultaneous Equations

10 A function f is given by

\[\begin{aligned} \mathrm{f}(x) &= -\tfrac{1}{2}x + 21 & \qquad 0 &\leqslant x \leqslant 2 \\ &= ax^2 + bx & \qquad 2 &\lt x \leqslant 6 \end{aligned}\]

A sketch of \(\quad y = \mathrm{f}(x) \quad\) is shown.

Sketch of y = f(x): a line segment from the y-axis to P, then a curve from P rising to a maximum and falling to Q(6, 12)

Work out the values of \(a\) and \(b\). [5 marks]

Mark scheme

Answer	Mark	Comments
\(P\) is (2, 20) or \(y\) value of \(P\) is 20	M1	implied by \(4a + 2b = 20\) could be seen on diagram
\(4a + 2b = k\)	M1	oe \(4a + 2b = 20\) \(k\) could be any number (including 0) but must be 20 if first M mark awarded
\(36a + 6b = 12\)	M1	oe
\(a = -2\) or \(b = 14\)	A1
\(a = -2\) and \(b = 14\)	A1

Additional guidance

\(a = -2\) and \(b = 14\) with no working	M3A2
\(a = -2\) or \(b = 14\) with no working	M0A0

June 2023 Paper 2 Q19

AQACurrent spec3 marksSimultaneous Equations

19 In these simultaneous equations, \(k\) is a positive constant.

\[\begin{aligned} 3x + 4y &= k \\ y &= 2kx \end{aligned}\]

Solve the simultaneous equations.

Give the answers in their simplest form in terms of \(k\). [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Eliminates \(y\) then uses substitution
Eliminates \(y\)	M1	eg \(3x + 4(2kx) = k\) or \(3x = k - 8kx\) or \(\dfrac{k - 3x}{4} = 2kx\) or \(x = \dfrac{k}{3 + 8k}\)
Substitutes their \(x\) into one of the equations	M1dep	eg \(y = 2k \times \text{their } \dfrac{k}{3 + 8k}\) or \(3 \times \text{their } \dfrac{k}{3 + 8k} + 4y = k\)
\(x = \dfrac{k}{3 + 8k} \qquad y = \dfrac{2k^2}{3 + 8k}\)	A1
Alternative method 2 Eliminates \(x\) then uses substitution
Eliminates \(x\)	M1	eg \(y = 2k \times \dfrac{k - 4y}{3}\) or \(y = \dfrac{2k^2 - 8ky}{3}\) or \(3 \times \dfrac{y}{2k} + 4y = k\) or \(y = \dfrac{2k^2}{3 + 8k}\)
Substitutes their \(y\) into one of the equations	M1dep	eg their \(\dfrac{2k^2}{3 + 8k} = 2kx\) or \(3x + 4 \times \text{their } \dfrac{2k^2}{3 + 8k} = k\)
\(x = \dfrac{k}{3 + 8k} \qquad y = \dfrac{2k^2}{3 + 8k}\)	A1
Alternative method 3 Eliminates \(y\) and eliminates \(x\)
Eliminates \(y\)	M1	eg \(3x + 4(2kx) = k\) or \(3x = k - 8kx\) or \(\dfrac{k - 3x}{4} = 2kx\) or \(x = \dfrac{k}{3 + 8k}\)
Eliminates \(x\)	M1	eg \(y = 2k \times \dfrac{k - 4y}{3}\) or \(y = \dfrac{2k^2 - 8ky}{3}\) or \(3 \times \dfrac{y}{2k} + 4y = k\) or \(y = \dfrac{2k^2}{3 + 8k}\)
\(x = \dfrac{k}{3 + 8k} \qquad y = \dfrac{2k^2}{3 + 8k}\)	A1

Additional guidance

Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
Correct answers followed by further incorrect work	M2A0
Missing brackets must be recovered
Alt 1 M1 scored then rearrangement errors before substituting can score the second M mark if their \(x\) is in terms of \(k\) (their expression for \(x\) must be seen) eg \(3x + 8kx = k \qquad x = \dfrac{k}{3 - 8k}\) \(y = \dfrac{2k^2}{3 - 8k}\)	M1 M1depA0
Alt 2 M1 scored then rearrangement errors before substituting can score the second M mark if their \(y\) is in terms of \(k\) (their expression for \(y\) must be seen) eg \(y = 2k \times \dfrac{k - 4y}{3} \qquad y = \dfrac{2k}{3 + 8k}\) \(x = \dfrac{2k}{3 + 8k} \div 2k\)	M1 M1depA0
A correct equation is needed for elimination marks to be awarded eg Alt 1 \(\begin{array}{r} 3x + 4y = k \\ \underline{4y = 8kx} \end{array}\; - \qquad\) is M0 unless \(3x = k - 8kx\) also seen

June 2023 Paper 1 Q18

AQACurrent spec5 marksSimultaneous Equations

18 Solve the simultaneous equations

\[\begin{aligned} 2x + y &= 13 \\ x + 3z &= 2 \\ z - 2y &= -7 \end{aligned}\]

Do not use trial and improvement.

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative Method 1 – Elimination of \(x\) from the first 2 equations
Correctly setting up two equations to eliminate \(x\) with an attempt to add or subtract from two equations	M1	eg \(2(x + 3z) - (2x + y)\)
Correct method and terms to eliminate a second variable	M1dep	eg \((6z - y) - 6(z - 2y)\)
Correctly eliminating a variable from their two correct equations to form an equation in one variable	M1dep	eg \(y - 12y = 9 - 42\) or \(11y = 33\) or \(y = 3\)
Two correct values with working seen	A1	eg \(y = 3\) and \(x = 5\)
\(x = 5\) and \(y = 3\) and \(z = -1\) with correct working seen	A1	oe

Additional guidance

For first two M marks LHS manipulation only required	M2
All three M marks can be combined in one equation: ie \(2(x + 3z) - (2x + y) - 6(z - 2y) = 11y\) and \(11y = 33\) and \(y = 3\)	M3
May be multiple attempts. M marks can be awarded for best attempt but A marks would have to be from answers chosen
Stop marking after first error on LHS (but check for restarts)
Stop marking as soon as T&I/inspection methods begin (but check for restarts)

Answer	Mark	Comments
Alternative Method 2 – Elimination of \(y\) from 2 equations
Correctly setting up two equations to eliminate \(y\) with an attempt to add or subtract from two equations	M1	eg \(2(2x + y) + (z - 2y)\)
Correct method and terms to eliminate a second variable	M1dep	eg \((4x + z) - 4(x + 3z)\)
Correctly eliminating a variable from their two correct equations to form an equation in one variable	M1dep	eg \(z - 12z = -7 + 18\) or \(-11z = 11\) or \(z = -1\)
Two correct values with working seen	A1	eg \(z = -1\) and \(x = 5\)
\(x = 5\) and \(y = 3\) and \(z = -1\) with correct working seen	A1	oe

Additional guidance

For first two M marks LHS manipulation only required	M2
All three M marks can be combined in one equation: ie \(2(x + 3z) - (2x + y) - 6(z - 2y) = 11y\) and \(11y = 33\) and \(y = 3\)	M3
May be multiple attempts. M marks can be awarded for best attempt but A marks would have to be from answers chosen
Stop marking after first error on LHS (but check for restarts)
Stop marking as soon as T&I/inspection methods begin (but check for restarts)

Answer	Mark	Comments
Alternative Method 3 – Elimination of \(z\) from 2 equations
Correctly setting up two equations to eliminate \(z\) with an attempt to add or subtract from two equations	M1	eg \((x + 3z) - 3(z - 2y)\)
Correct method and terms to eliminate a second variable	M1dep	eg \(2(x + 6y) - (2x + y)\)
Correctly eliminating a variable from their two correct equations to form an equation in one variable	M1dep	eg \(y - 12y = 9 - 42\) or \(11y = 33\) or \(y = 3\)
Two correct values with working seen	A1	eg \(y = 3\) and \(x = 5\)
\(x = 5\) and \(y = 3\) and \(z = -1\) with correct working seen	A1	oe

Additional guidance

For first two M marks LHS manipulation only required	M2
All three M marks can be combined in one equation: ie \(2(x + 3z) - (2x + y) - 6(z - 2y) = 11y\) and \(11y = 33\) and \(y = 3\)	M3
May be multiple attempts. M marks can be awarded for best attempt but A marks would have to be from answers chosen
Stop marking after first error on LHS (but check for restarts)
Stop marking as soon as T&I/inspection methods begin (but check for restarts)

Answer	Mark	Comments
Alternative Method 4 – Substitution
One equation rearranged and correctly substituted into a second equation to eliminate an unknown	M1	eg \(x = 2 - 3z\) and \(2(2 - 3z) + y\)
Two correct equations formed with the same two unknowns in each equation (must be simplified correctly) and one equation rearranged so it can be substituted into the second equation	M1dep	eg \(-6z + y = 9\) and \(z = 2y - 7\)
Correctly eliminating a variable from their two correct equations to form an equation in one variable	M1dep	eg \(-6(2y - 7) + y = 9\) or \(-11y = -33\) or \(y = 3\)
Two correct values with working seen	A1	eg \(z = -1\) and \(y = 3\)
\(x = 5\) and \(y = 3\) and \(z = -1\) with correct working seen	A1	oe

Additional guidance

Could use a combination of elimination and substitution at any point. Stop marking after first error
May be multiple attempts. M marks can be awarded for best attempt but A marks would have to be from answers chosen
Stop marking as soon as T&I/inspection methods begin (but check for restarts)

June 2023 Paper 1 Q17

AQACurrent spec5 marksDifferentiation Simultaneous Equations

17 The curve \(\quad y = ax^3 + bx^2 + 7 \quad\) has a stationary point at \((-2, 11)\)

Work out the values of \(a\) and \(b\). [5 marks]

Mark scheme

Answer	Mark	Comments
\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 3ax^2 + 2bx\)	M1	either term correct ignore LHS
\(3a(-2)^2 + 2b(-2) = 0\) or \(12a - 4b = 0\) or \(3a - b = 0\) or \(3a(-2)^2 + 2(1 + 2a)(-2) = 0\) (this would come from doing the third M mark before the differentiation)	M1dep	when \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) oe can follow through from an error in first M mark (as long as there are at least two terms in \(x^n\))
\(11 = a(-2)^3 + b(-2)^2 + 7\) or \(-8a + 4b = 4\) or \(-2a + b = 1\) or \(11 = a(-2)^3 + 3a(4) + 7\)	M1	oe
\(a = 1\) or \(b = 3\)	A1	dependent on M3 awarded
\(a = 1\) and \(b = 3\)	A1

Additional guidance

Correct answer could come from incorrect working (follow usual guidance)
\(a = 1\) and \(b = 3\) not from incorrect working will be full marks	M3 A2
Only one correct answer without working is 0 marks	M0 A0

June 2023 Paper 1 Q8

AQACurrent spec6 marksCircles in Coordinate Geometry Simultaneous Equations

8 A circle has centre \((0, 0)\) and radius 5

A straight line has equation \(\quad 2y = x + 5\)

Work out the coordinates of the two points where the circle and straight line intersect.

Do not use trial and improvement.

You must show your working. [6 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(x^2 + y^2 = 25\)	B1	oe equation of circle eg \(x^2 + y^2 = 5^2\) eg \((x - 0)^2 + (y - 0)^2 = 25\)
\((2y - 5)^2 + y^2\)	M1	must be \(2y - 5\) substituted into \(x^2 + y^2\)
\((4y^2 - 20y + 25) + y^2\)	M1	oe ft their \(2y - 5\) correctly expanded
\(5y^2 - 20y = 0\) or \(y^2 - 4y = 0\) or \(5y(y - 4) = 0\) or \(y(y - 4) = 0\)	M1dep	oe (dependent on B1 M2) must be fully correct (no ft)
\(y = 0\) and \(y = 4\) or \((-5, 0)\) or \((3, 4)\)	A1
\((-5, 0)\) and \((3, 4)\)	A1
Alternative method 2
\(x^2 + y^2 = 25\)	B1	oe equation of circle eg \(x^2 + y^2 = 5^2\) eg \((x - 0)^2 + (y - 0)^2 = 25\)
\(x^2 + \left(\dfrac{x + 5}{2}\right)^2\)	M1	must be \(\left(\dfrac{x + 5}{2}\right)\) substituted into \(x^2 + y^2\)
\(x^2 + \left(\dfrac{x^2 + 10x + 25}{4}\right)\) or \(4x^2 + x^2 + 10x + 25 \;(= 100)\)	M1	oe ft their \(\left(\dfrac{x + 5}{2}\right)\) correctly expanded
\(5x^2 + 10x - 75 = 0\) or \(x^2 + 2x - 15 = 0\) or \((x - 3)(x + 5) = 0\)	M1dep	oe (dependent on B1 M2) must be fully correct (no ft)
\(x = 3\) and \(x = -5\) or \((-5, 0)\) or \((3, 4)\)	A1
\((-5, 0)\) and \((3, 4)\)	A1

Additional guidance

Answer could be written as \(x = -5\), \(y = 0\) and \(x = 3\), \(y = 4\)	B1 M3 A2
Answer with no working (for M3) or from trying to draw an accurate graph or T&I	B0 M0 A0
Incorrect circle equation can still score 2 marks eg. \(x^2 + y^2 = 10\) \(x^2 + \left(\dfrac{x + 5}{2}\right)^2\) \(4x^2 + x^2 + 10x + 25 = 40\) \(5x^2 + 10x - 15 = 0\) \((x + 3)(x - 1) = 0\) \(x = -3\) and \(x = 1\) \(y = 1\) and \(y = 3\)	B0 M2 M0 A0 A0 B0 M1 M1 M0 A0 A0
Correct answer can be gained by using the incorrect substitution \(x = 5 - 2y\)	B1 M0 M1 M0 A0
Candidates that rearrange \(x^2 + y^2 = 25\) correctly and substitute into \(2y = x + 5\) will gain first M mark. They should then follow main scheme for other marks	B1 M1 …..

June 2022 Paper 2 Q17

AQACurrent spec5 marksSimultaneous Equations

17 Solve the simultaneous equations

\[\begin{aligned} a + 3b - 2c &= 4 \\ 4a - 3b + 5c &= -5 \\ 2a + b + 3c &= 9 \end{aligned}\]

Do not use trial and improvement.

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
Elimination of one variable making an equation with at least two terms correct	M1	eg1 (elimination of \(b\) by adding 1st and 2nd equations) \(5a + 3c = -1\) with at least two terms correct eg2 (elimination of \(a\) by doubling 1st equation and subtracting 3rd equation) \(5b - 7c = -1\) with at least two terms correct
Elimination of one variable making an equation with at least two terms correct and elimination of the same variable making a different equation with at least two terms correct	M1dep	eg1 (elimination of \(b\) by adding 1st and 2nd equations and elimination of \(b\) by trebling 3rd equation and subtracting 1st equation) \(5a + 3c = -1\) with at least two terms correct and \(5a + 11c = 23\) with at least two terms correct eg2 (elimination of \(a\) by doubling 1st equation and subtracting 3rd equation and elimination of \(a\) by doubling 3rd equation and subtracting 2nd equation) \(5b - 7c = -1\) with at least two terms correct and \(5b + c = 23\) with at least two terms correct
Correct equation in one variable with two correct equations in the same two variables	M1dep	eg \(3c - 11c = -1 - 23\) or \(-8c = -24\) or \(c = 3\) with \(5a + 3c = -1\) and \(5a + 11c = 23\)
Two correct values with two correct equations in the same two variables	A1	eg \(c = 3\) and \(a = -2\) with \(5a + 3c = -1\) and \(5a + 11c = 23\)
\(a = -2 \quad b = 4 \quad c = 3\) with two correct equations in the same two variables	A1	eg \(a = -2 \quad b = 4 \quad c = 3\) with \(5a + 3c = -1\) and \(5a + 11c = 23\)

Additional guidance

The two correct equations in the same two variables referred to in the scheme are a pair from one of these columns

\(15b - 13c = 21\)	\(5a + 3c = -1\)	\(13a + 9b = 10\)
\(5b - 7c = -1\)	\(5a + 11c = 23\)	\(7a + 11b = 30\)
\(5b + c = 23\)	\(10a + 14c = 22\)	\(2a - 14b = -60\)

All equations have equivalents
eg equivalents for \(5a + 3c = -1\) include \(-5a - 3c = 1\) and \(5a = -1 - 3c\)

All equations in two variables must have terms collected
eg \(a + 4a - 2c + 5c = 4 - 5\) requires simplification to \(5a + 3c = -1\)

\(0a + 15b - 13c = 21\) is equivalent to \(15b - 13c = 21\) etc

Equations with two terms correct include
eg1 (For \(5b + c = 23\)) \(\quad 5b + c = 10\) and \(-5b - c = 2\) and \(5b - 3c = 23\)
eg2 (For \(5a + 3c = -1\)) \(\quad 5a + 6c = -1\) and \(-5a - 3c = 4\) and \(5a = 2 - 3c\)

For equations with two terms correct the signs can be ignored if the modulus of the numbers in the correct equation are unchanged
eg For the correct equation \(5b - 7c = -1\) (so modulus 5, 7 and 1)
equations with two terms correct include
\(5b + 7c = 1\) and \(5b - 7c = 1\) and \(-5b - 7c = 1\) and \(-5b - 7c + 1 = 0\)

Up to M3 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts

Elimination of variables may be seen from other approaches
eg rearranges 1st equation to \(a = 4 - 3b + 2c\) and substitutes into the 2nd and 3rd equations

Correct values with no working

Zero

Matrix method involving row reduction is equivalent to the methods in the mark scheme

Correct inverse matrix seen with three correct solutions

M3A2

June 2022 Paper 1 Q16

AQACurrent spec6 marksSimultaneous Equations Solving Quadratics

16 Solve the simultaneous equations

\[\begin{aligned} x - y &= \frac{19}{4} \\ xy &= -3 \end{aligned}\]

Do not use trial and improvement.

You must show your working. [6 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4x^2 - 19x + 12 \;(= 0)\)	M1dep	oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square
\((4x + a)(x + b)\) or \((4x - 3)(4x - 16)\)	M1dep	where \(ab = 12\) or \(a + 4b = -19\)
\((4x - 3)(x - 4)\)	A1
\(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\)	A1
\(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\)	A1	all 4 values must be correct to gain this mark
Alternative method 2
Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4x^2 - 19x + 12 \;(= 0)\)	M1dep	oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square
\(\dfrac{19 \pm \sqrt{19^2 - 4 \times 4 \times 12}}{2 \times 4}\)	M1dep
\(\dfrac{19 \pm \sqrt{169}}{8}\)	A1
\(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\)	A1
\(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\)	A1	all 4 values must be correct to gain this mark
Alternative method 3
Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4x^2 - 19x + 12 \;(= 0)\) or \(x^2 - \dfrac{19}{4}x + 3 \;(= 0)\)	M1dep	oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square
\(4\left[\left(x - \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\) or \(\left[\left(x - \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\)	M1	oe
\(4\left(x - \dfrac{19}{8}\right)^2 - \dfrac{169}{16} = 0\) or \(\left[\left(x - \dfrac{19}{8}\right)^2\right] - \dfrac{169}{64} = 0\)	M1dep
\(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\)	A1
\(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\)	A1	all 4 values must be correct to gain this mark
Alternative method 4
Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4y^2 + 19y + 12 \;(= 0)\)	M1dep	oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square
\((4y + a)(y + b)\) or \((4y + 3)(4y + 16)\)	M1dep	where \(ab = 12\) or \(a + 4b = 19\)
\((4y + 3)(y + 4)\)	A1
\(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1
\(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1	all 4 values must be correct to gain this mark
Alternative method 5
Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4y^2 + 19y + 12 \;(= 0)\)	M1dep	oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square
\(\dfrac{-19 \pm \sqrt{19^2 - 4 \times 4 \times 12}}{2 \times 4}\)	M1dep
\(-\dfrac{19 \pm \sqrt{169}}{8}\)	A1
\(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1
\(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1	all 4 values must be correct to gain this mark
Alternative method 6
Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\)	M1	penalise no brackets unless recovered
\(4y^2 + 19y + 12 \;(= 0)\) or \(y^2 + \dfrac{19}{4}y + 3 \;(= 0)\)	M1dep	oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square
\(4\left[\left(y + \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\) or \(\left[\left(y + \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\)	M1	oe
\(4\left(y + \dfrac{19}{8}\right)^2 - \dfrac{169}{16} = 0\) or \(\left[\left(y + \dfrac{19}{8}\right)^2\right] - \dfrac{169}{64} = 0\)	M1dep
\(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1
\(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\)	A1	all 4 values must be correct to gain this mark

Additional guidance

Correct A marks must come from correct algebra in M marks

November 2021 Paper 1 Q14

AQACurrent spec5 marksSimultaneous Equations

14 Solve the simultaneous equations.

\[\begin{aligned} 4a - b + 3c &= 27 \\ 3a + 2b - c &= 5 \\ 2a - 5c &= -7 \end{aligned}\]

Do not use trial and improvement.

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Eliminates \(b\) from first two equations before eliminating a second variable
Correct attempt to eliminate \(b\) from LHS of first two equations	M1	eg \(2(4a - b + 3c) + 3a + 2b - c\) or \(11a + 5c\) adding or subtracting the two equations can be implied from two terms correct
Correct attempt to eliminate \(a\) or \(c\) from LHS of third equation and their equation in \(a\) and \(c\)	M1dep	eg \(11a + 5c + 2a - 5c\) or \(2(11a + 5c) - 11(2a - 5c)\)
Correct equation in \(a\) or \(c\)	M1dep	eg \(13a = 52\) or \(65c = 195\) implied by \(a = 4\) or \(c = 3\) with M2
Two correct values with M3	A1	eg \(a = 4\) and \(c = 3\) with M3
\(a = 4\) and \(b = -2\) and \(c = 3\) with M3	A1
Alternative method 2 Eliminates \(a\) or \(c\) before eliminating a second variable
Two correct attempts to eliminate the same variable (\(a\) or \(c\)) from LHS	M1	eg (eliminating \(a\)) \(4a - b + 3c - 2(2a - 5c)\) and \(2(3a + 2b - c) - 3(2a - 5c)\) or \(-b + 13c\) and \(4b + 13c\)
Correct attempt to eliminate a second variable from LHS of their two equations	M1dep	eg \(-b + 13c - (4b + 13c)\)
Correct equation in one variable	M1dep	eg \(-5b = 10\) implied by \(b = -2\) with M2
Two correct values with M3	A1	eg \(b = -2\) and \(a = 4\) with M3 or \(b = -2\) and \(c = 3\) with M3
\(a = 4\) and \(b = -2\) and \(c = 3\) with M3	A1

Additional guidance

For the first two marks ignore the RHS of the equations
First two method marks may be seen in one attempt eg Alt1 \(2(4a - b + 3c) + 3a + 2b - c + 2a - 5c\)	M1M1
Elimination may be seen from other approaches eg1 Alt 1 (equates expressions for \(2b\) from first two equations) \(2(4a + 3c - 27) = 5 - 3a + c\) eg2 Alt 2 (rearranges third equation to \(a = 2.5c - 3.5\) and substitutes into first two equations) \(4(2.5c - 3.5) - b + 3c\) and \(3(2.5c - 3.5) + 2b - c\)	M1 M1
Correct values with no working	M0A0

Specimen 2020 Paper 1 Q17

AQACurrent spec5 marksSimultaneous Equations

17 Solve the simultaneous equations

\[\begin{aligned} 2a + b - c &= 8 \\ 4a - 3b - 2c &= -9 \\ 6a + 3b + c &= 0 \end{aligned}\]

[5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
Correct attempt to eliminate two variables from left hand side	M1	eg \(2(2a + b - c) - (4a - 3b - 2c)\)
Correct attempt to eliminate two variables	M1dep	eg \(2(2a + b - c) - (4a - 3b - 2c)\) \(= 2 \times 8 - (-9)\) or \(5b = 25\)
Solves their equation	M1dep	eg \(b = 25 \div 5\) or \(b = 5\)
Substitutes their value into two equations and correct method to eliminate a variable	M1	eg \(2a + 5 - c = 8\) and \(6a + 15 + c = 0\) and \(8a + 20 = 8\)
\(a = -\dfrac{3}{2}\) and \(b = 5\) and \(c = -6\)	A1	oe
Alternative method 2
Two correct attempts to eliminate same variable from left hand side	M1	eg \(3(2a + b - c) + (4a - 3b - 2c)\) and \(4a - 3b - 2c + 6a + 3b + c\)
Two correct attempts to eliminate same variable	M1dep	eg \(3(2a + b - c) + (4a - 3b - 2c)\) \(= 24 - 9\) and \(4a - 3b - 2c + 6a + 3b + c = 0 - 9\) or \(10a - 5c = 15\) and \(10a - c = -9\)
Correct attempt to eliminate a variable from their two equations	M1dep	eg \(10a - 5c - (10a - c) = 15 - 9\) or \(-4c = 24\) or \(c = -6\)
Substitutes their value into two equations and correct method to eliminate a variable	M1	eg \(2a + b + 6 = 8\) and \(4a - 3b + 12 = -9\) and \(2(2a + b + 6) - (4a - 3b + 12)\) \(= 2 \times 8 - (-9)\)
\(a = -\dfrac{3}{2}\) and \(b = 5\) and \(c = -6\)	A1	oe

Specimen 2020 Paper 1 Q15

AQACurrent spec9 marksDifferentiation Simultaneous Equations

15 Point \(A\) lies on the curve

\[y = x^2 + 5x + 8\]

The \(x\)-coordinate of \(A\) is \(-4\)

(a) Show that the equation of the normal to the curve at \(A\) is \(\quad 3y = x + 16\) [5 marks]

(b) The normal at \(A\) also intersects the curve at \(B\).

Work out the \(x\)-coordinate of \(B\). [4 marks]

Mark scheme (a)

Answer	Mark	Comments
\((-4)^2 + 5 \times -4 + 8\) or 4	M1	oe
\(2x + 5\)	M1	\(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
\(2 \times -4 + 5\) or \(-3\)	M1dep	gradient of tangent
\(-\dfrac{1}{\text{their } -3}\) or \(\dfrac{1}{3}\)	M1dep	dep on 2nd and 3rd M1
\(y - 4 = \dfrac{1}{3}(x + 4)\) and \(3y = x + 16\)	A1	must see correct working leading to \(3y = x + 16\)

Mark scheme (b)

Answer	Mark	Comments
\(x + 16 = 3(x^2 + 5x + 8)\)	M1	oe
\(3x^2 + 14x + 8\ (= 0)\)	A1
\((3x + 2)(x + 4)\ (= 0)\) or \(\dfrac{-14 \pm \sqrt{14^2 - 4 \times 3 \times 8}}{2 \times 3}\) or \(-\dfrac{7}{3} \pm \sqrt{\dfrac{25}{9}}\)	M1	oe correct attempt to solve their 3-term quadratic
\(-\dfrac{2}{3}\)	A1

Specimen 2020 Paper 2 Q15

AQACurrent spec4 marksIndices Simultaneous Equations

15 Here is a sketch of the curve \(\quad y = ab^{-x} \quad\) where \(a\) and \(b\) are positive constants.

(0, 3) and (2, 0.48) lie on the curve.

Decreasing curve crossing the y-axis at (0, 3) and passing through (2, 0.48)

Not drawn accurately

Work out the values of \(a\) and \(b\). [4 marks]

Mark scheme

Answer	Mark	Comments
\(a = 3\)	B1
\(0.48 =\) their \(3 \times b^{-2}\)	M1	oe
\(b^2 = \dfrac{\text{their } 3}{0.48}\) or \(b^2 = 6.25\) \(\sqrt{\dfrac{\text{their } 3}{0.48}}\) or \(\sqrt{6.25}\)	M1dep	oe
\(b = 2.5\)	A1ft	ft B0M2

June 2018 Paper 1 Q13

AQAOld spec6 marksSimultaneous Equations

13 Solve the simultaneous equations

\[xy = 2 \quad \text{and} \quad y = 3x + 5\]

Do not use trial and improvement.
You must show your working. [6 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(3x + 5 = \dfrac{2}{x}\) or \(x(3x + 5) = 2\)	M1	oe
\(3x^2 + 5x - 2\ (= 0)\) or \(3x^2 + 5x = 2\)	M1dep
\((3x + a)(x + b)\ (= 0)\)	M1dep	\(ab = -2\) or \(a + 3b = 5\)
\((3x - 1)(x + 2)\ (= 0)\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) or \(x = \tfrac{1}{3} \quad y = 6\) or \(x = -2 \quad y = -1\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\)	A1	either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs
Alternative method 2
\(3x + 5 = \dfrac{2}{x}\) or \(x(3x + 5) = 2\)	M1	oe
\(3x^2 + 5x - 2\ (= 0)\) or \(3x^2 + 5x = 2\)	M1dep
\(x = \dfrac{-5 \pm \sqrt{[(5)^2 - 4(3)(-2)]}}{2(3)}\)	M1dep	allow one sign error … but the \(2 \times 3\) term must be beneath the full numerator
\(x = \dfrac{-5 \pm 7}{6}\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) or \(x = \tfrac{1}{3} \quad y = 6\) or \(x = -2 \quad y = -1\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\)	A1	either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs
Alternative method 3
\(3x + 5 = \dfrac{2}{x}\) or \(x(3x + 5) = 2\)	M1	oe
\(3x^2 + 5x - 2\ (= 0)\) or \(3x^2 + 5x = 2\)	M1dep
\((3 \times)\ (x + {}^{5}\!/_{6})^2 \ldots\ldots\)	M1dep
\(x + {}^{5}\!/_{6} = \pm\,{}^{7}\!/_{6}\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) or \(x = \tfrac{1}{3} \quad y = 6\) or \(x = -2 \quad y = -1\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\)	A1	either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs
Alternative method 4
\(y = 3\left(\dfrac{2}{y}\right) + 5\) or \(\dfrac{y(y - 5)}{3} = 2\)	M1	oe
\(y^2 - 5y - 6 = 0\) or \(y^2 - 5y = 6\)	M1dep
\((y + a)(y + b)\ (= 0)\)	M1dep	\(ab = -6\) or \(a + b = -5\)
\((y - 6)(y + 1)\ (= 0)\)	A1
\(y = 6 \quad y = -1\) or \(y = 6 \quad x = \tfrac{1}{3}\) or \(y = -1 \quad x = -2\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\)	A1	either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs
Alternative method 5
\(y = 3\left(\dfrac{2}{y}\right) + 5\) or \(\dfrac{y(y - 5)}{3} = 2\)	M1	oe
\(y^2 - 5y - 6 = 0\) or \(y^2 - 5y = 6\)	M1dep
\(y = \dfrac{5 \pm \sqrt{[(-5)^2 - 4(1)(-6)]}}{2(1)}\)	M1dep	allow one sign error … but the \(2 \times 1\) term must be beneath the full numerator
\(y = \dfrac{5 \pm 7}{2}\)	A1
\(y = 6 \quad y = -1\) or \(y = 6 \quad x = \tfrac{1}{3}\) or \(y = -1 \quad x = -2\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\)	A1	either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs
Alternative method 6
\(y = 3\left(\dfrac{2}{y}\right) + 5\) or \(\dfrac{y(y - 5)}{3} = 2\)	M1	oe
\(y^2 - 5y - 6 = 0\) or \(y^2 - 5y = 6\)	M1dep
\((y - {}^{5}\!/_{2})^2 \ldots\ldots\)	M1dep
\(y - {}^{5}\!/_{2} = \pm\,{}^{7}\!/_{2}\)	A1
\(y = 6 \quad y = -1\) or \(y = 6 \quad x = \tfrac{1}{3}\) or \(y = -1 \quad x = -2\)	A1
\(x = \tfrac{1}{3} \quad x = -2\) and \(y = 6 \quad y = -1\) or \(x = \tfrac{1}{3} \quad y = 6\) and \(x = -2 \quad y = -1\)	A1	either correct \(x\)’s and correct \(y\)’s or correct coordinate pairs

Additional guidance

Trial and improvement … 0 marks No working shown ….. 0 marks
The instructions were clearly stated in the question.

June 2017 Paper 2 Q22

AQAOld spec5 marksCircles in Coordinate Geometry Simultaneous Equations

22 The equation of a circle is \(\quad (x - 2)^2 + (y - 1)^2 = 16\)

The equation of a line is \(\quad y = 2x + 1\)

The circle and the line intersect at two points.

Work out the coordinates of the two points.

You must show your working.

Do not use trial and improvement. [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\((x - 2)^2 + (2x + 1 - 1)^2 = 16\)	M1	oe Eliminates \(y\)
\(x^2 - 2x - 2x + 4 + 4x^2 = 16\) or \(5x^2 - 4x - 12\ (= 0)\)	M1dep	oe Expands both brackets correctly
\((5x + 6)(x - 2) \quad (= 0)\) or \(\dfrac{--4 \pm \sqrt{(-4)^2 - 4 \times 5 \times -12}}{2 \times 5}\)	M1	oe eg \(\dfrac{2}{5} \pm \sqrt{\dfrac{64}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(4^2\) for \((-4)^2\) Implied by correct solutions to their 3-term quadratic seen
(\(x\) =) \(-1.2\) and (\(x\) =) 2 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5x^2 - 4x - 12\ (= 0)\) seen	A1	oe eg (\(x\) =) \(-\dfrac{6}{5}\) and (\(x\) =) 2 with \(5x^2 - 4x - 12\ (= 0)\) seen
(\(-1.2\), \(-1.4\)) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen	A1	oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen
Alternative method 2
\(x^2 - 2x - 2x + 4 + y^2 - y - y + 1 = 16\)	M1	oe Expands both brackets correctly
\(x^2 - 2x - 2x + 4 + (2x + 1)^2\) \(- (2x + 1) - (2x + 1) + 1 = 16\) or \(5x^2 - 4x - 12\ (= 0)\)	M1dep	oe Eliminates \(y\)
\((5x + 6)(x - 2) \quad (= 0)\) or \(\dfrac{--4 \pm \sqrt{(-4)^2 - 4 \times 5 \times -12}}{2 \times 5}\)	M1	oe eg \(\dfrac{2}{5} \pm \sqrt{\dfrac{64}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(4^2\) for \((-4)^2\) Implied by correct solutions to their 3-term quadratic seen
(\(x\) =) \(-1.2\) and (\(x\) =) 2 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5x^2 - 4x - 12\ (= 0)\) seen	A1	oe eg (\(x\) =) \(-\dfrac{6}{5}\) and (\(x\) =) 2 with \(5x^2 - 4x - 12\ (= 0)\) seen
(\(-1.2\), \(-1.4\)) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen	A1	oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5x^2 - 4x - 12\ (= 0)\) seen
Alternative method 3
\(\left(\left(\dfrac{y - 1}{2}\right) - 2\right)^2 + (y - 1)^2 = 16\)	M1	oe Eliminates \(x\)
\(\left(\dfrac{y - 1}{2}\right)^2 - 2\left(\dfrac{y - 1}{2}\right) - 2\left(\dfrac{y - 1}{2}\right) + 4\) \(+ y^2 - y - y + 1 = 16\) or \(5y^2 - 18y - 35\ (= 0)\)	M1dep	oe Expands \(\left(\left(\dfrac{y - 1}{2}\right) - 2\right)^2\) and \((y - 1)^2\) correctly
\((5y + 7)(y - 5) \quad (= 0)\) or \(\dfrac{--18 \pm \sqrt{(-18)^2 - 4 \times 5 \times -35}}{2 \times 5}\)	M1	oe eg \(\dfrac{9}{5} \pm \sqrt{\dfrac{256}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(18^2\) for \((-18)^2\) Implied by correct solutions to their 3-term quadratic seen
(\(y\) =) \(-1.4\) and (\(y\) =) 5 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen	A1	oe eg (\(y\) =) \(-\dfrac{7}{5}\) and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen
(\(-1.2\), \(-1.4\)) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen	A1	oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen
Alternative method 4
\(x^2 - 2x - 2x + 4 + y^2 - y - y + 1 = 16\)	M1	oe Expands both brackets correctly
\(\left(\dfrac{y - 1}{2}\right)^2 - 2\left(\dfrac{y - 1}{2}\right) - 2\left(\dfrac{y - 1}{2}\right) +\) \(4 + y^2 - y - y + 1 = 16\) or \(5y^2 - 18y - 35\ (= 0)\)	M1dep	oe Eliminates \(x\)
\((5y + 7)(y - 5) \quad (= 0)\) or \(\dfrac{--18 \pm \sqrt{(-18)^2 - 4 \times 5 \times -35}}{2 \times 5}\)	M1	oe eg \(\dfrac{9}{5} \pm \sqrt{\dfrac{256}{25}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow \(18^2\) for \((-18)^2\) Implied by correct solutions to their 3-term quadratic seen
(\(y\) =) \(-1.4\) and (\(y\) =) 5 or (\(x\) =) \(-1.2\) and (\(y\) =) \(-1.4\) or (\(x\) =) 2 and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen	A1	oe eg (\(y\) =) \(-\dfrac{7}{5}\) and (\(y\) =) 5 with \(5y^2 - 18y - 35\ (= 0)\) seen
(\(-1.2\), \(-1.4\)) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen	A1	oe eg \(\left(-\dfrac{6}{5}, -\dfrac{7}{5}\right)\) and (2, 5) with \(5y^2 - 18y - 35\ (= 0)\) seen

Additional guidance

Answers only (no valid working)	Zero
Both solutions from scale drawing	5 marks
(2, 5) is often seen without seeing any correct method	Zero
Allow one miscopy for up to M3A0A0

June 2017 Paper 2 Q14

AQAOld spec5 marksLinear Graphs Simultaneous Equations

14 The line \(\quad y = ax + b \quad\) is perpendicular to the line \(\quad x + 4y = 74\)

The lines intersect at the point where \(x = 2\)

Sketch of the lines y = ax + b and x + 4y = 74 meeting at right angles at the point where x = 2

Not drawn accurately

Work out the values of \(a\) and \(b\). [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(-\dfrac{1}{4}\) or \(-0.25\)	B1	gradient of \(x + 4y = 74\) Do not allow embedded May be implied
(gradient =) \(\dfrac{-1}{\text{their } -\frac{1}{4}}\) or 4	M1	ft their \(-\dfrac{1}{4}\) Only ft a non-zero numerical value Implied by \(y = 4x + b\) or \(a = 4\) (B1M1)
(\(y\) =) \(\dfrac{74 - 2}{4}\) or \(\dfrac{72}{4}\) or 18	M1	oe May be seen on diagram
their \(18 =\) their \(4 \times 2 + b\) or \(y -\) their \(18 =\) their \(4(x - 2)\)	M1dep	oe dep on M2
\(b = 10\)	A1ft	ft \(18 -\) their \(4 \times 2\) if B0M3
Alternative method 2
\(-\dfrac{1}{4}\) or \(-0.25\)	B1	gradient of \(x + 4y = 74\) Do not allow embedded May be implied
(gradient =) \(\dfrac{-1}{\text{their } -\frac{1}{4}}\) or 4	M1	ft their \(-\dfrac{1}{4}\) Only ft a non-zero numerical value Implied by \(y = 4x + b\) or \(a = 4\) (B1M1)
Correct method for elimination of \(y\) from \(x + 4y = 74\) and \(y =\) their \(4x + b\)	M1dep	eg \(x + 4(4x + b) = 74\) or \(17x + 4b = 74\)
Substitutes \(x = 2\) into their equation	M1dep	eg \(34 + 4b = 74\)
\(b = 10\)	A1ft	ft \(18 -\) their \(4 \times 2\) if B0M3
Alternative method 3
\(-\dfrac{1}{4}\) or \(-0.25\)	B1	gradient of \(x + 4y = 74\) Do not allow embedded May be implied
(gradient =) \(\dfrac{-1}{\text{their } -\frac{1}{4}}\) or 4	M1	ft their \(-\dfrac{1}{4}\) Only ft a non-zero numerical value Implied by \(y = 4x + b\) or \(a = 4\) (B1M1)
(\(y\) =) \(\dfrac{74 - 2}{4}\) or \(\dfrac{72}{4}\) or 18	M1	oe May be seen on diagram
Correct method for elimination of \(x\) from \(x + 4y = 74\) and \(y =\) their \(4x + b\) and substitutes \(y =\) their 18	M1dep	eg \(y = 4(74 - 4y) + b\) or \(17y = 296 + b\) and \(306 = 296 + b\) dep on M2
\(b = 10\)	A1ft	ft \(18 -\) their \(4 \times 2\) if B0M3

Additional guidance

\(y = 4x + 10\) will gain full marks unless contradicted

If an error is made in the constant term when rearranging \(x + 4y = 74\)
the B1 can still be awarded for gradient \(= -\dfrac{1}{4}\)
eg \(y = -\dfrac{1}{4}x + 19\) and gradient \(= -\dfrac{1}{4}\) is B1
(all other marks are possible)

In alt 1 and alt 3 the mark for \(y = 18\) will sometimes be the only mark awarded

June 2016 Paper 2 Q4

AQAOld spec4 marksSimultaneous Equations

4 Work out the point of intersection of the lines

\[2x + 3y = 11 \qquad \text{and} \qquad 2y = 13 - 3x\]

[4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Equates coefficients and eliminates a variable
\(4x + 6y = 22\) and \(9x + 6y = 39\) or \(6x + 9y = 33\) and \(6x + 4y = 26\)	M1	oe Equates coefficients of one variable Allow one term error
\(4x - 9x =\) \(22 - 39\) or \(-5x = -17\) or \(9y - 4y =\) \(33 - 26\) or \(5y = 7\)	M1dep	oe Eliminates a variable Must be correct method for their equations Unless correctly eliminated, intention to subtract (or add if appropriate) must be seen with the result of their subtraction
(\(x\) =) 3.4 or (\(y\) =) 1.4	A1	oe eg \(x = \dfrac{17}{5}\)
(3.4, 1.4)	A1	oe eg \(\left(\dfrac{17}{5}, \dfrac{7}{5}\right)\) SC3 (1.4, 3.4) oe
Alternative method 2 Makes a variable the subject in first equation
\(y = \dfrac{11}{3} - \dfrac{2}{3}x\) or \(x = 5.5 - 1.5y\)	M1	oe eg \(x = \dfrac{11 - 3y}{2}\) Makes \(y\) or \(x\) the subject Allow one term error
\(2\left(\dfrac{11}{3} - \dfrac{2}{3}x\right) = 13 - 3x\) or \(2y = 13 - 3(5.5 - 1.5y)\)	M1dep	oe Eliminates a variable Must be correct method for their equations
(\(x\) =) 3.4 or (\(y\) =) 1.4	A1	oe eg \(x = \dfrac{17}{5}\)
(3.4, 1.4)	A1	oe eg \(\left(\dfrac{17}{5}, \dfrac{7}{5}\right)\) SC3 (1.4, 3.4) oe
Alternative method 3 Makes a variable the subject in second equation
\(y = 6.5 - 1.5x\) or \(x = \dfrac{13}{3} - \dfrac{2}{3}y\)	M1	oe eg \(y = \dfrac{13 - 3x}{2}\) Makes \(y\) or \(x\) the subject Allow one term error
\(2x + 3(6.5 - 1.5x) = 11\) or \(2\left(\dfrac{13}{3} - \dfrac{2}{3}y\right) + 3y = 11\)	M1dep	oe Eliminates a variable Must be correct method for their equations
(\(x\) =) 3.4 or (\(y\) =) 1.4	A1	oe eg \(x = \dfrac{17}{5}\)
(3.4, 1.4)	A1	oe eg \(\left(\dfrac{17}{5}, \dfrac{7}{5}\right)\) SC3 (1.4, 3.4) oe
Alternative method 4 Makes same variable the subject in both equations
\(y = \dfrac{11}{3} - \dfrac{2}{3}x\) or \(y = 6.5 - 1.5x\) or \(x = 5.5 - 1.5y\) or \(x = \dfrac{13}{3} - \dfrac{2}{3}y\)	M1	oe eg \(y = \dfrac{13 - 3x}{2}\) Makes \(y\) or \(x\) the subject Allow one term error
\(\dfrac{11}{3} - \dfrac{2}{3}x = 6.5 - 1.5x\) or \(5.5 - 1.5y = \dfrac{13}{3} - \dfrac{2}{3}y\)	M1dep	oe Eliminates a variable Must be correct method for their equations
(\(x\) =) 3.4 or (\(y\) =) 1.4	A1	oe eg \(x = \dfrac{17}{5}\)
(3.4, 1.4)	A1	oe eg \(\left(\dfrac{17}{5}, \dfrac{7}{5}\right)\) SC3 (1.4, 3.4) oe

Additional guidance

Answer only (3.4, 1.4)	M2 A2
One value correct (possibly by drawing) with no incorrect working seen for that variable	M2 A1 A0
If the same method is used for both \(x\) and \(y\) (eg equates coefficients and eliminates a variable), mark the attempt that favours the student
Alt 1 \(\quad \begin{aligned} 6x + 9y &= 33 \\ \underline{6x + 4y} &\underline{{}= 26} \\ 4y &= 6 \end{aligned}\) is M1 M0 unless intention to subtract is seen (eg a subtraction symbol is seen or the word subtract is seen) which would then get M1 M1
Alts 2, 3 and 4 Allow rounding or truncating to 1dp or better for up to M1 M1 eg (Alt 2) \(\ y = 3.6 - 0.6x\) \(2(3.6 - 0.6x) = 13 - 3x\)	M1 M1
(1.4, 3.4) is SC3 or M2 A2 if \(x = 3.4\) and \(y = 1.4\) seen in working

Simultaneous Equations

AQA

AQA · Old spec

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