Equations — Paper 2 · IGCSE Mathematics 0580

5 Solve the equation For 3 x − 2 Examiner's = 8 . Use 5 Answer x = [2]

2 marks

Mark scheme: 5 14 2* M1 correct movement of 2 terms

This question in 0580/21 Oct/Nov 2004

12 Make c the subject of the formula 3c −5 = b . Answer c = [3]

3 marks

Mark scheme: 12 3* M1 for a correct operation b2 + 5 c = M1 for a second correct operation 3 IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 2 * indicates that it is necessary to look in the working following a wrong answer

This question in 0580/21 Oct/Nov 2004

17 Solve the equation For x 2 + 4 x − 22 = 0 . Examiner's Use Give your answers correct to 2 decimal places. Show all your working. Answer x = or x = [4]

4 marks

Mark scheme: 17 3.10 or -7.10 4 M1 for 42 -4 x 1 x -22 or better −4 ± * * M1 for 2 A1 A1 SCA1 3.09 and -7.09 or 3.1 and -7.1

This question in 0580/21 Oct/Nov 2004

9 Rearrange the formula to make y the subject. y x + = 1 9 Answer y = [3]

3 marks

Mark scheme: 9 (9(1 – x))2 oe 3 M1 1 move completed correctly M1 1 more move completed correctly Mark 3rd move in answer space

This question in 0580/21 May/June 2009

12 Solve the simultaneous equations 2y + 3x = 6, x = 4y + 16. Answer x = y = [3]

3 marks

Mark scheme: 12 x = 4 y = –3 3 M1 consistent multiplication and subtraction of their rearranged eqns. Any other answers must first score M1 to gain an A mark Substitution, matrix and equating methods also permitted 2

This question in 0580/21 May/June 2009

14 (a) There are 109 nanoseconds in 1 second. For Find the number of nanoseconds in 5 minutes, giving your answer in standard form. Examiner's Use Answer(a) [2] (b) Solve the equation 5 ( x + 3 × 106 ) = 4 × 107. Answer(b) x = [2]

4 marks

Mark scheme: 14 (a) 3 × 1011 2 M1 60 × 5 × 109 or better (b) 5 000 000 or 5 × 106 or 5 million 2 M1 0.8 × 107 – 3 × 106 oe or M1 5x = 4 × 107 – 15 × 106 oe If m is used for a million it must be used consistently

This question in 0580/21 May/June 2009

9 Solve the simultaneous equations 6x + 18y = 57, 2x – 3y = −8. Answer x = y = [3]

3 marks

Mark scheme: 9 x = 0.5 y = 3 www 3 M1 consistent × and – for y or consistent × and + for x A1 one correct provided M1 scored

This question in 0580/21 Oct/Nov 2009

13 Solve the simultaneous equations. 2 x + y = 7 2 2 x − y = 17 2 Answer x = y = [3]

3 marks

Mark scheme: 13 x = 12 y = –10 3 M1 consistent addition (& mult) for x or consistent subtraction (& mult) for y A1 only earned if method correct 21 k 1 A1 k 96

This question in 0580/21 May/June 2010

15 Examiner's Use y NOT TO SCALE 4 A y = 4 B x 0 2x + y = 8 3x + y = 18 (a) The line y = 4 meets the line 2x + y = 8 at the point A. Find the co-ordinates of A. Answer(a) A ( , ) [1] (b) The line 3x + y = 18 meets the x axis at the point B. Find the co-ordinates of B. Answer(b) B ( , ) [1] (c) (i) Find the co-ordinates of the mid-point M of the line joining A to B. Answer(c)(i) M ( , ) [1] (ii) Find the equation of the line through M parallel to 3x + y = 18. Answer(c)(ii) [2]

5 marks

Mark scheme: 15 (a) (2, 4) 1 (b) (6, 0) 1 (c) (i) (4, 2) ft 1ft From (a) and (b) (ii) y = – 3x + 14 oe 2 M1 sub their (c)(i) into y = –3x + c oe 1

This question in 0580/22 May/June 2010

19 The braking distance, d metres, for Alex’s car travelling at v km/h is given by the formula Examiner's Use 200d = v(v + 40). (a) Calculate the missing values in the table. v 0 20 40 60 80 100 120 (km/h) d 0 16 48 96 (metres) [2] (b) On the grid below, draw the graph of 200d = v(v + 40) for 0 Y v Y 120. d 100 90 80 70 60 Distance 50 (metres) 40 30 20 10 v 0 10 20 30 40 50 60 70 80 90 100 110 120 Speed (km / h) [3] (c) Find the braking distance when the car is travelling at 110 km/h. Answer(c) m [1] (d) Find the speed of the car when the braking distance is 80 m. Answer(d) km/h [1]

7 marks

Mark scheme: 19 (a) 6, 30, 70 2 B1 for 2 correct (b) graph 3 P2 7 plots correct from table P1 5 or 6 plots correct from table C1 smooth curve through the points in the given range within one small square of the plots or the correct position (c) 82.5 or ft ±1 1ft (d) 108 or ft ±1 1ft

This question in 0580/22 May/June 2010

y Examiner's 14 Solve the equation 3( y − 4 ) + = 9 . Use 2 Answer y = [3]

3 marks

Mark scheme: 14 6 3 M1 for one correct first step which leads towards simplifying y 3 y − 12 + = 9 2 or 6(y – 4) + y = 18 y or y – 4 + = 3 6 M1 correctly collecting their terms to py = q IGCSE – May/June 2010 0580 23

This question in 0580/23 May/June 2010

20 Examiner's y Use 4 3 2 1 x 0 1 2 3 4 Find the three inequalities which define the shaded region on the grid. Answer [5]

5 marks

Mark scheme: 20 x [ 0 1 L1 x R 0 1 1 y [ x oe 2 L1 y R x 2 2 x + y Y 4 oe 2 L1 x + y R 4 where R is any one of = < > Y [ B2 all inequalities correct or B1 2 correct sin 140

This question in 0580/23 May/June 2010

20 Solve the equation. x2 – 8x + 6 = 0 Show all your working and give your answers correct to 2 decimal places. Answer x = or x = [4]

4 marks

Mark scheme: 8 ± k 20 x = 0.84 or 7.16 4 B1 B1 √(82 – 4 × 1 × 6) or better 2 A1 A1

This question in 0580/21 Oct/Nov 2010

g h 16 = 2 i Find i in terms of g and h. Answer i = [3]

3 marks

Mark scheme: 4 h  2 16 3 M1 squaring correctly 2 or h   g  g  M1 clearing denominator correctly M1 dividing by coefficient of i or SC2 for correct unsimplified expression

This question in 0580/23 Oct/Nov 2010

17 Solve the simultaneous equations. 5x – y = – 10 x + 2y = 9 Answer x = y = [3]

3 marks

Mark scheme: 17 x = –1, y = 5 3 M1 consistent multiplication and either add or subtract A1 for one correct after M1 x

This question in 0580/23 Oct/Nov 2010

10 Solve the simultaneous equations. 3x + y = 30 2x – 3y = 53 Answer x = y = [3]

3 marks

Mark scheme: g 10 x = 13 3 M1 for consistent multiplication and y = –9 addition/subtraction A1 for x = 13 or A1 for y = -9

This question in 0580/21 May/June 2011

14 Solve the equation 2x2 + 3x – 6 = 0. Show all your working and give your answers correct to 2 decimal places. Answer x = or x = [4]

4 marks

Mark scheme: 14 –2.64, 1.14 cao with working 4 B1 for 32 − 4(2 )(− 6 ) or better seen anywhere B1 for p = –3 and r = 2 × 2 or better as long as in p + q p − q the form or r r After B0B0, SC1 for –2.6 or –2.637(45…) and 1.1 or 1.137(45…)

This question in 0580/21 May/June 2011

20 f(x) = x3 g(x) = 2x − 3 ForFor Examiner'sExaminer's UseUse (a) Find (i) g(6), Answer(a)(i) [1] (ii) f(2x). Answer(a)(ii) [1] (b) Solve fg(x) = 125. Answer(b) x = [3] (c) Find the inverse function g−1(x) . Answer(c) g –1(x) = [2]

7 marks

Mark scheme: 20 (a) (i) 9 1 (ii) 8x3 cao 1 (b) 4 www 3 M1 for (2x – 3)3 = 125 M1 2x – 3 = 5 (c) x + 3 2 M1 for x ± 3 = 2y or x = y ± 3 2 2

This question in 0580/21 May/June 2011

8 Solve the simultaneous equations. x + 2y = 3 2x – 3y = 13 Answer x = y = [3]

3 marks

Mark scheme: 8 (x =) 5 (y =) –1 3 M1 for consistent multiplication and add/subtract as appropriate A1 for 1 correct answer

This question in 0580/23 May/June 2011

10 The cost of a cup of tea is t cents. For Examiner's The cost of a cup of coffee is (t + 5) cents. Use The total cost of 7 cups of tea and 11 cups of coffee is 2215 cents. Find the cost of one cup of tea. Answer cents [3]

3 marks

Mark scheme: 10 120 3 M1 7t + 11(t + 5) = 2215 A1 18t + 55 = 2215

This question in 0580/23 May/June 2011

l 14 T = 2π g (a) Find T when g = 9.8 and ℓ = 2. Answer(a) T = [2] (b) Make g the subject of the formula. Answer(b) g = [3]

5 marks

Mark scheme: 14 (a) 2.84 2 M1 correct substitution of g and l seen 4 π 2 l (b) oe 3 M1 each correct move but third move marked on T 2 answer line

This question in 0580/21 Oct/Nov 2011

1 For 17 f(x) = (x ≠ O4) Examiner's x + 4 Use g(x) = x2 – 3x h(x) = x3 + 1 (a) Work out fg(1). Answer(a) [2] (b) Find hO1(x). Answer(b) h O1(x) = [2] (c) Solve the equation g(x) = O2. Answer(c) x = or x = [3] Question 18 is printed on the next page.

7 marks

Mark scheme: 1 17 (a) 2 B1 f(–2) seen 2 (b) 3√(x – 1) or 3 x − 1 2 M1 x –1 = y3 or 3√(y – 1) (c) 1 2 3 M2 (x – 1)(x – 2) = 0 or M1 (x + a)(x + b) = 0 where ab = 2 or a + b = –3 If 0 scored give M1 for x2 – 3x + 2 = 0 1

This question in 0580/21 Oct/Nov 2011

3 Solve the simultaneous equations. x + 5y = 22 x + 3y = 12 Answer x = y = [2]

2 marks

Mark scheme: 3 (x =) –3 (y =) 5 2 M1 for correctly eliminating one variable 16 81 k 16 4

This question in 0580/23 Oct/Nov 2011

11 Solve the simultaneous equations. 3x + 5y = 24 x + 7y = 56 Answer x = y = [3]

3 marks

Mark scheme: 11 x = −7 3 M1 for consistent multiplication and addition/ y = 9 subtraction as appropriate. Allow computational errors A1 for x= –7 or y = 9 55 27 25 27

This question in 0580/21 May/June 2012

15 Solve the equation 2x2 + 6x – 3 = 0 . Show your working and give your answers correct to 2 decimal places. Answer x = or x = [4]

4 marks

Mark scheme: 15 –3.44, 0.44 4 B1 for ( 6) 2 − 4( 2)( −3) or better seen p + ( or − ) q B1 if in form , for p = –6 and r = 2×2 oe correct working must be shown r B1, B1 (SC1 –3.4 or –3.436… and 0.4 or 0.436...)

This question in 0580/23 May/June 2012

16 Rearrange the formula y = to make x the subject. x − 4 Answer x = [4]

4 marks

Mark scheme: 16 4 + 2 4 M1 xy – 4y = x + 2 y oe − 1 M1 collecting terms in x on one side y M1 factorising M1 dividing by coeff of x

This question in 0580/21 Oct/Nov 2012

x 3 For 20 f(x) = 4(x + 1) g(x) = O 1 Examiner's 2 Use (a) Write down the value of x when f O1(x) = 2. Answer(a) x = [1] (b) Find fg(x). Give your answer in its simplest form. Answer(b) fg(x)= [2] (c) Find gO1(x). Answer(c) g O1(x) = [3] Question 21 is printed on the next page.

6 marks

Mark scheme: 20 (a) 12 1 (b) 2x3 cao 2 M1 clear evidence of adding 1 then multiplying by 4 to g(x) (c) 3 2 ( x+ )1 oe 3 M1 each correct move

This question in 0580/21 Oct/Nov 2012

16 Make y the subject of the formula. A = πx2 O πy2 Answer y = [3]

3 marks

Mark scheme: πx A 16 oe 3 M1 for second correct move π M1 for third correct move 2 θ 4r

This question in 0580/23 Oct/Nov 2012

23 f(x) = 3x + 5 g(x) = 4x O 1 For Examiner's Use (a) Find the value of gg(3). Answer(a) [2] (b) Find fg(x), giving your answer in its simplest form. Answer(b)fg(x) = [2] (c) Solve the equation. f –1(x) = 11 Answer(c) x = [1] Question 24 is printed on the next page.

5 marks

Mark scheme: 23 (a) 43 2 M1 for g(11) or 4[4(3) – 1] –1 (b) 12x + 2 2 M1 for 3(4x – 1) + 5 (c) 38 1 2 2 2

This question in 0580/23 Oct/Nov 2012

1 1 x Examiner′s , x ¸ 0 h(x) =21 f(x) = 5x + 4 g(x) = 2x c 2 m Use Find (a) fg(5) , Answer(a) … [2] (b) gg(x) in its simplest form, Answer(b) gg(x) = … [2] (c) f –1(x) , Answer(c) f –1(x) = … [2] (d) the value of x when h(x) = 8. Answer(d) x = … [2]

8 marks

Mark scheme: 21 (a) 4.5 oe 2 B1 for [g(5)=] 0.1 oe 1 seen oe M1 for (b) x 2 2 1 ( 2 x ) x − 4 (c) oe 2 M1 for a correct first step 5 4 y e.g. y − 4 = 5x or = x + or 5 5 x = 5y + 4 (d) or − 3 2 M1 for 8 x 1 −x 3 or 2 = oe or 2 = 2 8

This question in 0580/22 May/June 2013

10 Find the value of 2x + y for the simultaneous equations. 3x + 5y = 48 2x – y = 19 Answer 2x + y = … [4] _____________________________________________________________________________________

4 marks

Mark scheme: 10 25 4 M1 for correct method to eliminate one variable A1 for x = 11 A1 for y = 3 B1 FT for 2 × their x + their y correctly evaluated

This question in 0580/23 May/June 2013

14 (a) Solve 3n + 23 < n + 41. Answer(a) … [2] (b) Factorise completely ab + bc + ad + cd. Answer(b) … [2] _____________________________________________________________________________________

4 marks

Mark scheme: 14 (a) n < 9 2 M1 for 2n < 18 or 2n – 18 < 0 oe If 0 scored SC1 for 9 with incorrect inequality. (b) (b + d)(a + c) 2 B1 for b(a + c) + d(a + c) or a(b + d) + c (b + d)

This question in 0580/23 May/June 2013

5 Solve the equation. For Examiner′s 5 – 2x = 3x – 19 Use Answer x = … [2] _____________________________________________________________________________________

2 marks

Mark scheme: 5 4.8 oe 2 M1 for 5 + 19 = 3x + 2x oe or better or B1 for 24 – 2x = 3x oe or 5 = 5x – 19 oe 2

This question in 0580/21 Oct/Nov 2013

15 Find the co-ordinates of the point of intersection of the two lines. For Examiner′s Use 2x – 7y = 2 4x + 5y = 42 Answer ( … , … ) [3] _____________________________________________________________________________________

3 marks

Mark scheme: 15 (8, 2) 3 M1 for correctly eliminating one variable A1 for x = 8 A1 for y = 2 If 0 scored, SC2 for correct substitution and correct evaluation to find the other value.

This question in 0580/22 Oct/Nov 2013

19 f(x) = 2x + 3 g(x) = x2 For Examiner′s Use (a) Find fg(6). Answer(a) … [2] (b) Solve the equation gf(x) = 100. Answer(b) x = … or x = … [3] (c) Find f –1(x). Answer(c) f –1(x) = … [2] (d) Find ff –1(5). Answer(d) … [1]

8 marks

Mark scheme: 19 (a) 75 2 B1 for [g(6) =] 36 (b) 3.5 –6.5 3 M1 for (2x + 3)2 = 100 M1 for 2x + 3 = [±]10 If 0 scored, SC1 for one correct value as answer x − 3 3 y (c) oe final answer 2 M1 for x = 2y + 3 or y – 3 = 2x or = x + 2 2 2 or better (d) 5 1

This question in 0580/23 Oct/Nov 2013

3 Solve the equation. n - 8 = 11 2 Answer n = … [2] __________________________________________________________________________________________

2 marks

Mark scheme: n 3 30 2 M1 for n – 8 = 22 or = 15 2

This question in 0580/21 May/June 2014

7 Make x the subject of the formula. y = (x – 4)2 + 6 Answer x = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 7 4 ± y − 6 3 M1 for their 6 moved correctly M1 for their √ taken correctly M1 for their 4 moved correctly

This question in 0580/21 May/June 2014

10 Solve the equation. x + 5 7 = x 3 Answer x = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 10 3.75 oe 3 M2 for 3 × 5 = 7 x − 3 x oe or M1 for 3( x + 5 ) = 7 x or x + 5 = 73 x or 1 + 5x = 73 or better

This question in 0580/21 Oct/Nov 2014

12 Solve the simultaneous equations. 0.4x – 5y = 27 2x + 0.2y = 9 Answer x = … y = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 5 12 nfww 3 M1 for correctly eliminating one variable − 5 A1 for x = 5 A1 for y = −5 If zero scored SC1 for correct substitution and evaluation to find the other variable

This question in 0580/21 Oct/Nov 2014

220 f(x) = 3x – 2 g(x) = , x ≠ –1 x + 1 (a) Find gf(2). Answer(a) … [2] (b) Solve g(x) = 10. Answer(b) x = … [2] (c) Simplify. f(2x) – f(x + 2) Answer(c) … [3]

7 marks

Mark scheme: 20 (a) 0.4 or 52 2 B1 for [f(2) =] 4 2 or M1 for or better (3 x − 2 ) + 1 (b) –0.8 or − 54 2 M1 for 2 = 10( x + )1 or better (c) 3 x − 6 or 3( x − 2 ) nfww 3 M2 for 3(2 x ) − 2 − (3( x + 2 ) − 2 ) or M1 for [f (2 x ) = ]3(2 x ) − 2 or [f ( x + 2 )] = 3( x + 2 ) − 2

This question in 0580/21 Oct/Nov 2014

6 Solve the equation. 2x + 5 = 8 3 Answer x = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 19 6 9.5 or 3 M2 for 2x = (8 × 3) – 5 or better oe 2 or M1 for 2x + 5 = 8 × 3 or better 360 180× (18 2 )

This question in 0580/23 Oct/Nov 2014

8 Make x the subject of the formula. y = 2 + x - 8 Answer x = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 8 8 + (y – 2)2 oe final answer 3 M1 for y – 2 = √(x – 8) M1 for squaring both sides completed correctly M1 for adding their 8 completed correctly on answer line

This question in 0580/23 Oct/Nov 2014

x - 116 f(x) = (x – 3)2 g(x) = h(x) = x3 4 Find (a) hf(1), Answer(a) … [2] (b) g–1(x), Answer(b) g–1(x) = … [2] (c) gh(x), Answer(c) gh(x) = … [1] (d) the solution to the equation f(x) = 0. Answer(d) x = … [1] __________________________________________________________________________________________

6 marks

Mark scheme: 16 (a) 64 2 B1 for [f(1) =] 4 or M1 for ((x – 3)2)3 or better y − 1 (b) 4x + 1 oe 2 M1 for x = or 4y = x – 1 4 x 3 − 1 (c) oe final answer 1 4 (d) 3 nfww 1

This question in 0580/23 Oct/Nov 2014

23 f(x) = 5 – 3x (a) Find f(6). Answer(a) … [1] (b) Find f(x + 2). Answer(b) … [1] (c) Find ff(x), in its simplest form. Answer(c) … [2] (d) Find f –1(x), the inverse of f(x). Answer(d) f –1(x) = … [2]

6 marks

Mark scheme: 23 (a) −13 1 (b) −3x − 1 or 5 − 3( x + 2 ) 1 (c) 9x − 10 cao 2 M1 for 5 − 3( 5 − 3x) 5 − x (d) final answer oe 2 M1 for correct first step e.g. 3 y 5 y + 3 x = 5 or = − x or y − 5 = − 3 x or 3 3 better or for interchanging x and y, e.g. x = 5 − 3 y , this does not need to be the first step

This question in 0580/21 May/June 2015

21 f(x) = x2 + 4x − 6 (a) f(x) can be written in the form (x + m)2 + n. Find the value of m and the value of n. Answer(a) m = … n = … [2] (b) Use your answer to part (a) to find the positive solution to x2 + 4x – 6 = 0. Answer(b) x = … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 21 (a) m = 2 2 B1 for m = 2 n = –10 B1 for n = –10 If 0 scored SC1 for (x + 2)2 in working or x2 + 2mx + m2 + n and equating coefficients 2m[x] = 4[x] or m2 + n = –6 (b) 1.16 or 1.16[2…] from completing 2FT FT dep on negative n square B1 for (x + their m)2 = –their n or SC1 for correct answer from using formula or for both answers 1.16 and –5.16 whatever method used

This question in 0580/22 May/June 2015

9 Solve the equation. 3(x + 4) = 2(4x – 1) Answer x = … [3]

3 marks

Mark scheme: 9 2.8 oe 3 M2 for 12 + 2 = 8 x − 3 x or better or M1 for 3 x + 12 or 8 x − 2 85 67 5  67 5 

This question in 0580/23 May/June 2015

14 Solve the equation. 2x2 + x – 2 = 0 Show your working and give your answers correct to 2 decimal places. Answer x = … or x = … [4] __________________________________________________________________________________________

4 marks

Mark scheme: 14 12 − 4 ( 2 )( −2 ) B1 If completing the square B1 for + x 1  oe  4  p + q p − q 1  1  2 If in form or B1 B1 for x = − + 1 +   r r 4  4  p = – 1, r = 2(2) or 4 2 1  1  or x = − − 1 +   4  4  – 1.28 B1 If 0 scored for the last two B marks then 0.78 B1 SC1 for – 1.3 and 0.8 or – 1.281 to – 1.280 and 0.781 or 0.7807 to 0.7808 or 1.28 and – 0.78 or – 1.28 and 0.78 seen in the working

This question in 0580/23 May/June 2015

19 Solve the equation 5x 2 - 6x - 3 = 0 . Show all your working and give your answers correct to 2 decimal places. Answer x = … or x = … [4]

4 marks

Mark scheme: 19 (− 6 ) 2 − 4 ( 5)( −3) or better seen B1 If completing the square 2  B1 for − x 3 oe  5  p + q p − q if or seen then B1 r r 3 3  3  2 3 3  3  2 B1 for + +   or − +   oe p = − (− 6) and r = 2 × 5 5 5  5  5 5  5  B1 If B0, SC1 for B1 −0.38 − 0.4 and 1.6 1.58 cao final answers or − 0.379[795..] and 1.579[795..] or − 1.58 and 0.38 as final answers or − 0.38 and 1.58 seen in working

This question in 0580/22 Oct/Nov 2015

21 Solve the equation 3x2 + 4x – 5 = 0. Show all your working and give your answers correct to 2 decimal places. Answer x = … or x = … [4] __________________________________________________________________________________________

4 marks

Mark scheme: 21 ( 4) 2 − 4(3)( −5) or better seen B1 If completing the square p + q p − q  2  2 if or seen then B1 for  x +  oe r r  3  2 5 2 2 2 5 2 2 − + 2 or − 3 3 p = – 4 and r = 2(3) B1 B1 for −3 + 3 + 3 3 2 – 2.12 B1 0.79 final answers B1 If B0, SC1 for 0.786[299] and −2.119[632] – 2.1 and 0.8 or – 2.120 or – 2.119 and 0.786 or 2.12 and –0.79 final answers –2.12 and 0.79 seen not as final answers 1

This question in 0580/23 Oct/Nov 2015

1 Solve (x – 7)(x + 4) = 0. x = … or x = … [1]

1 marks

Mark scheme: Qu. Answers Mark Part Marks 1 7, − 4 1

This question in 0580/22 Feb/March 2016

17 Solve the equation 3x 2 - 11x + 4 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

4 marks

Mark scheme: ( ) ( ) ( )( ) 2 17 2 B1 for ( −11) − 4(3)(4) or better 2 × 3 p + q p − q and, if in form or , r r B1 for p = −(−11) and r = 2(3) 0.41 and 3.26 final ans cao B1B1 SC1 for 0.4 and 3.3 or 0.409... and 3.257... or − 0.41 and −3.26 or 0.41 and 3.26 seen in working

This question in 0580/22 Feb/March 2016

20 The nth term of a sequence is an 2 + bn . (a) Write down an expression, in terms of a and b, for the 3rd term. … [1] (b) The 3rd term of this sequence is 21 and the 6th term is 96. Find the value of a and the value of b. You must show all your working. a = … b = … [4] Question 21 is printed on the next page.

5 marks

Mark scheme: 20 (a) 9a + 3b 1 (b) 36a + 6b = 96 or 9a + 3b = 21 B1 for correct method to eliminate M1 one variable a = 3 A1 If M0 A0 A0 scored SC1 for b = −2 A1 2 values satisfying 36a + 6b = 96 or 9a + 3b = 21 or if no working shown, but 2 correct answers given

This question in 0580/22 Feb/March 2016

7 y = mx + c Find the value of y when m = −2, x = −7 and c = −3. y = … [2]

2 marks

Mark scheme: 7 11 2 M1 for –2 × –7 – 3 soi py

This question in 0580/21 May/June 2016

4 Solve the equation. 6(y + 1) = 9 y = … [2]

2 marks

Mark scheme: 1 4 0.5 or 2 M1 for correct first step e.g. 6y + 6 = 9 2 9 or y + 1 = 6 1 6

This question in 0580/23 May/June 2016

11 Solve the simultaneous equations. You must show all your working. 2x + 3y = 13 x + 2y = 9 x = … y = … [3]

3 marks

Mark scheme: 11 Correctly eliminating one variable M1 [x =] −1 and A1 If zero scored, SC1 for 2 values that satisfy one of the original [y = ] 5 A1 equations or SC1 if no working shown, but 2 correct answers given 1

This question in 0580/21 Oct/Nov 2016

23 Solve the equation 2x 2 + 3 x - 3 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4] Question 24 is printed on the next page.

4 marks

Mark scheme: 2  3 23 (3) − 4(2)( −3) oe or better B1 If completing the square, B1 for  x +  oe  4  −+3 k −−3 k 3 3  3  2 3 3  3  2 or oe B1 B1 for − + +   or − − +   oe 2(2) 2(2) 4 2  4  4 2  4  –2.19, 0.69 B1B1 SC1 for –2.2 or –2.186… and 0.7 or 0.686.. or –2.19 and 0.69 seen but not final answer or 2.19 and –0.69 Maximum score without working is 2 2 2 2

This question in 0580/21 Oct/Nov 2016

3 Solve the equation. 6(k – 8) = 78 k = … [2]

2 marks

Mark scheme: 3 21 2 M1 for k – 8 = 13 or 6k – 48 = 78 or better (13 + 16 ) × 4

This question in 0580/22 Oct/Nov 2016

8 Solve the simultaneous equations. You must show all your working. 1 2 x + y = 8 x - 2y = 2 x = … y = … [3]

3 marks

Mark scheme: 8 correctly eliminating one variable M1 [x = ] 9 A1 [y = ] 3.5 A1 If zero scored, SC1 for 2 values satisfying one of the original equations SC1 if no working shown but 2 correct answers given 300

This question in 0580/22 Oct/Nov 2016

x21 f (x) = - 3 g ( x) = 6x - 7 h (x) = 2x 4 (a) Work out the value of x when f(x) = -0.5 . x = … [2] (b) Find g−1(x). g−1(x) = … [2] (c) Work out the value of x when h(x) = f(13). x = … [2]

6 marks

Mark scheme: x 21 (a) 10 2 M1 for − 3 = − 0.5 4 x + 7 y 7 (b) final answer 2 M1 for y + 7 = 6 x or = x − or 6 6 6 x = 6 y − 7 1 (c) –2 2 M1 for [f(13) =] 4

This question in 0580/22 Feb/March 2017

10 Solve. 2 - x = 5 x + 1 x = … [2]

2 marks

Mark scheme: 10 1 2 M1 for 2 – 1 = 5x + x oe oe 6

This question in 0580/22 May/June 2017

15 Make q the subject of the formula p = 2 q 2 . q = … [2]

2 marks

Mark scheme: 15 p 2 p 2 [ ± ] oe M1 for = q or p = 2 q 2 2 p p or [q =] their or [q =] 2 their 2

This question in 0580/22 May/June 2017

11 y 14 12 10 8 6 4 2 x 0 –2 2 4 6 8 10 12 14 16 18 20 22 –2 By shading the unwanted regions of the grid above, find and label the region R that satisfies the following four inequalities. x H 0 x + y H 7 y H x x + 2y G 20 [3]

3 marks

Mark scheme: 11 Correct region 3 1 2 2 3 1 1 2 2 1 1 1 0 1 2 SC1 for R not marked and reverse shading

This question in 0580/23 May/June 2017

21 Solve the equation 5x 2 + 10x + 2 = 0 . You must show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

4 marks

Mark scheme: 21 2 B1 If completing the square: 10 − 4 × 5 × 2 oe or better B1 for ( x + 1) 2 oe 2 2 B1 for −+1 1 − or −−1 1 − oe 5 5 −10 + q −10 − q B1 or oe 2(5) 2(5) − 0.23, −1.77 final ans cao B1B1 SC1 for − 0.2 or − 0.225... and −1.8 or −1.774... or −1.775 or 0.23 and 1.77 as answer or − 0.23 and −1.77 seen in working Maximum score without working is 2

This question in 0580/23 May/June 2017

18 Solve the simultaneous equations. You must show all your working. x y = 2 2x - y = 1 x = … y = … [3]

3 marks

Mark scheme: 18 Correctly eliminating one M1 variable 2 A1 [x =] or 0.667 or 0.6666… 3 1 A1 If zero scored, SC1 for [y =] or 0.333 or 0.333… 2 values satisfying one of the original equations 3 or if no working shown but 2 correct answers given

This question in 0580/22 Oct/Nov 2017

19 Make x the subject of the formula. y = x 2 + 1 x = … [3]

3 marks

Mark scheme: 19 2 3 M1 for correct squaring [ ± ] y − 1 final answer M1 for correct rearranging for x or x2 term M1 for correct square root

This question in 0580/22 Oct/Nov 2017

24 Solve the equations. (a) 7 - 3n = 11 n + 2 n = … [2] p - 3 (b) = 3 5 p = … [2]

4 marks

Mark scheme: 24(a) 5 2 M1 for 7 – 2 = 11n + 3n oe or better or 0.357 or 0.357… 14 24(b) 18 2 p 3 M1 for p – 3 = 3 × 5 or = 3 + 5 5

This question in 0580/22 Oct/Nov 2017

4 Complete these statements. (a) When w = … , 10w = 70. [1] (b) When 5x = 15, 12x = … [1]

2 marks

Mark scheme: 4(a) [w =] 7 1 4(b) [12x =] 36 1

This question in 0580/22 May/June 2018

9 Solve. 1 - p = 4 3 p = … [2]

2 marks

Mark scheme: 9 – 11 2 M1 for 1 − p = 3 × 4 or better p 1 or − = 4 − or better 3 3

This question in 0580/23 May/June 2018

11 A = 2 r + y x 2 ^ h Rearrange the formula to make x the subject. x = … [2]

2 marks

Mark scheme: 11 A 2 A [ ± ] final answer M1 for = x2 2π + y 2π + y M1 for correctly square rooting their expression in x2 [ ± ] A If zero scored SC1 for 2π + y

This question in 0580/23 May/June 2018

18 Solve the simultaneous equations. You must show all your working. 2x + 3y =- 12 5x + 2y = 14 x = … y = … [4]

4 marks

Mark scheme: 18 for correctly equating one set of M1 coefficients for correct method to eliminate one M1 variable [x =] 6 A2 A1 for each [y =] −8 If M0 scored, SC1 for 2 values satisfying one of the original equations or if no working shown, but 2 correct answers given

This question in 0580/21 Oct/Nov 2018

19 Use the quadratic formula to solve the equation 3x 2 + 7x - 11 = 0 . You must show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

4 marks

Mark scheme: 19 2 B2 2 −±7 ( 7 ) − 4 ( 3 )( −11) B1 for ( 7 ) − 4 ( 3 ( −11) ) or better 2 × 3 −+7 q −−7 q and B1 for or 2(3) 2(3) −3.41 and 1.08 cao B2 B1 for each If B0, SC1 for −3.4 and 1.1 or –3.409 and 1.076 or −3.4089... and 1.0756 … or 3.41 and −1.08 or −3.41 and 1.08 seen in working

This question in 0580/21 Oct/Nov 2018

23 f (x) = 7 + 3x g (x) = x4 h (x) = 3x (a) h (3x) = k x Find the value of k. k = … [2] (b) Find the value of x when f (x) = g (2) . x = … [2] (c) Find f -1 (x) . f -1 (x) = … [2]

6 marks

Mark scheme: 23(a) 27 2 M1 for 33x seen 23(b) 3 2 M1 for 7 + 3x = 24 23(c) x − 7 2 M1 for x = 7 + 3y oe final answer y 7 3 or y – 7 = 3x or –3x = 7 – y or = + x 3 3

This question in 0580/21 Oct/Nov 2018

20 Make m the subject of the formula. 3 m x = 2 - m m = … [4]

4 marks

Mark scheme: 20 2 x 4 M1 for correctly clearing the denominator and oe final answer 3+ x expanding bracket M1 for correctly collecting terms in m on one side and terms not in m on the other M1 for correct factorising M1 for correct division dependent on m appearing only once in a factorised expression

This question in 0580/22 Oct/Nov 2018

10 Solve. 3w - 7 = 32 w = … [2]

2 marks

Mark scheme: 10 13 2 7 32 M1 for 3 w = 32 + 7 or w − = or better 3 3

This question in 0580/23 Oct/Nov 2018

20 Solve the equation 3x 2 - 2x - 2 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

4 marks

Mark scheme: 20 2 B2 2 −−( 2) ± ( − 2) − 4(3)( − 2) B1 for ( − 2 ) – 4 ( 3 )( − 2 ) or better oe 2(3) or −−( 2 ) + q −−( 2 ) − q B1 for or 2 ( 3 ) 2 ( 3 ) –0.55, 1.22 B2 B1 for each If zero scored, SC1 for – 0.6 and 1.2 or –0.549 or –0.548… and 1.215… or 0.55 and −1.22 or –0.55 and 1.22 seen in working

This question in 0580/23 Oct/Nov 2018

2 - 1 1 6 20 (a) Work out e4 3eo- 5 4o. [2] f p 3 - 1 (b) Find the value of x when the determinant of is 5. e- 7 xo x = … [2]

4 marks

Mark scheme: 20(a)  7 8  2 B1 for 2 correct elements    −11 36  20(b) 4 2 M1 for 3 x −−( 1) × ( −7 ) = 5 or better

This question in 0580/22 Feb/March 2019

17 (a) Factorise p 2 - q 2 . … [1] (b) p 2 - q 2 = 7 and p - q = 2 . Find the value of p + q. … [2]

3 marks

Mark scheme: 17(a) ( p − q )( p + q ) final answer 1 17(b) 7 2 M1 for 2 × (p + q) = 7 oe 2 2 2 2 2 or for ( 2 + q ) − q = 7 or p − ( p − 2 ) = 7

This question in 0580/21 May/June 2019

6 Solve the equation. 9f + 11 = 3f + 23 f = … [2]

2 marks

Mark scheme: 6 2 2 M1 for 9f – 3f oe or 23 − 11 oe

This question in 0580/22 May/June 2019

14 Solve the simultaneous equations. You must show all your working. 5x + 8y = 4 1 2 x + 3y = 7 x = … y = … [3]

3 marks

Mark scheme: 14 Correctly eliminating one variable M1 [x =] − 4 A2 A1 for one correct [y =] 3 If M0 scored, SC1 for 2 values satisfying one of the original equations

This question in 0580/22 May/June 2019

20 Solve the equation 3x 2 - 2x - 10 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

4 marks

Mark scheme: 20 2 B2 2 −−( 2 ) ± ( −2 ) − 4 ( 3 )( −10 ) B1 for ( −2 ) − 4 ( 3 )( − 10 ) or better 2 × 3 p + q p − q and if in form or then r r B1 for p = −(− 2) and r = 2(3) −1.52 and 2.19 final ans cao B1B1 If B0B0, SC1 for −1.5 and 2.2 or −1.523 to −1.522... and 2.189 … or 1.52 and −2.19 or −1.52 and 2.19 seen in working

This question in 0580/22 May/June 2019

10 Rearrange 2 (w + h ) = P to make w the subject. w = … [2]

2 marks

Mark scheme: 10 P P − 2 h 2 P [w =] – h or final M1 for w + h = or 2w + 2h = P 2 2 2 answer

This question in 0580/23 May/June 2019

14 One solution of the equation ax 2 + a = 150 is x = 7 . (a) Find the value of a. a = … [2] (b) Find the other solution. x = … [1]

3 marks

Mark scheme: 14(a) 3 2 M1 for a × 72 + a = 150 oe 14(b) –7 1

This question in 0580/23 May/June 2019

24 y 8 7 66 5 4 3 2 1 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 8 x – 1 – 2 By shading the unwanted regions of the grid, draw and label the region R which satisfies the following three inequalities. y G 2 x 1 3 y G x + 4 [5]

5 marks

Mark scheme: 24 Correct lines and region indicated 5 B1 for y = 2 solid line B1 for x = 3 dashed line B1 for y = x + 4 solid line B2, B1 or B0 for region

This question in 0580/23 May/June 2019

6 Solve. x - 2 = 3 3 x = … [2]

2 marks

Mark scheme: 6 11 2 x 2 M1 for x − 2 = 3 × 3 oe or = 3 + oe or 3 3 better

This question in 0580/21 Oct/Nov 2019

21 (a) 3 -2 # 3 x = 81 Find the value of x. x = … [2] 1 (b) x - 3 = 32 x -2 Find the value of x. x = … [3]

5 marks

Mark scheme: 21(a) 6 2 B1 for 34 or 3x–2 or M1 for 3x = 81 × 32 or better 21(b) 8 3 5 M2 for x 3 = 32 or better 1 32 or M1 for = or better 1 x 2 x 3 1 − 2 or x 3 −−= 32 or better

This question in 0580/22 Oct/Nov 2019

- 3 10 21 (a) Work out the inverse of the matrix e 1 - 5 o. [2] f p (b) Work out the value of x and the value of y in this matrix calculation. 1 5 - 4 1 x 46 = e2 yo e 2 9o e6 65 o x = … y = … [3]

5 marks

Mark scheme: 21(a) 1  − 5 − 10  2  − 5 − 10    oe isw M1 for k   or det = 5 soi 5  − 1 −3   − 1 − 3  21(b) [x = ] 6 3 B1 for x = 6 [y = ] 7 B2 for y = 7 or M1 for 2 × 1 + 9y = 65 or 2 × −4 + 2y = 6

This question in 0580/23 Oct/Nov 2019

16 Solve the simultaneous equations. You must show all your working. x = 7 - 3y x 2 - y 2 = 39 x = … y = … x = … y = … [6]

6 marks

Mark scheme: 16 8 y 2 − 42 y + 10[ = 0] or M3 M1 for ( 7 − 3 y ) 2 − y 2 = 39 oe 8 x 2 + 14 x − 400[ = 0] 2 2  7 − x  or x −   = 39 oe  3  M1 for 49 – 21y – 21y + 9y2 or better or 49 –7x –7x + x2 or better or for correct expansion of their quadratic binomial ( 8 y − 2 )( y − 5 ) [ = 0] oe M1 M1 for correct method to solve their quadratic equation e.g. factors, quadratic formula, ( 8 x − 50 )( x + 8 ) [ = 0] oe completing the square x = 6.25 oe y = 0.25 oe B2 B1 for x = 6.25, x = −8 or for y = 0.25, y = 5 x = − 8 y = 5 or for a correct pair of x and y values

This question in 0580/22 Feb/March 2020

20 The curve y = x 2 - 2x + 1 is drawn on a grid. A line is drawn on the same grid. The points of intersection of the line and the curve are used to solve the equation x 2 - 7 x + 5 = 0 . Find the equation of the line in the form y = mx + c. y = … [1]

1 marks

Mark scheme: 20 [y = ] 5x – 4 1

This question in 0580/21 May/June 2020

14 Solve the equation. 1 - x = 5 3 x = … [2]

2 marks

Mark scheme: 14 –14 2 M1 for 1 – x = 3 × 5 or better x 1 or = 5 − or better 3 3

This question in 0580/22 May/June 2020

18 (a) Write x 2 - 18 x - 27 in the form ( x + k) 2 + h . … [2] (b) Use your answer to part (a) to solve the equation x 2 - 18x - 27 = 0 . x = … or x = … [2]

4 marks

Mark scheme: 18(a) ( x − 9 )2 − 108 2 B1 for ( x + h ) 2 − 108 or ( x − 9 ) 2 + h or k = − 9 18(b) 19.4 or 19.39… 2 M1FT x − their 9 = ± their108 −1.39 or −1.392… A1 for 9 ± 108 or 9 ± 6 3

This question in 0580/23 May/June 2020

7 Make x the subject of this formula. 2y = 5x - 7 x = … [2]

2 marks

Mark scheme: 7 2 y + 7 2 y 7 2 2 y 7 [ x = ] oe or [ x = ] + oe M1 for 2 y + 7 = 5 x oe or = x – oe 5 5 5 5 5 final answer

This question in 0580/21 Oct/Nov 2020

3 Solve the equation. 6 - 2x = 3x x = … [2]

2 marks

Mark scheme: 3 1 6 2 M1 for 6 = 2x + 3x or better 1.2 or 15 or 5

This question in 0580/22 Oct/Nov 2020

9 Solve the simultaneous equations. 2x + y = 7 3x - y = 8 x = … y = … [2]

2 marks

Mark scheme: 9 [x =] 3 2 B1 for each [y =] 1

This question in 0580/22 Oct/Nov 2020

10 Solve the simultaneous equations. You must show all your working. 3x - 8y = 22 x + 4y = 4 x = … y = … [3]

3 marks

Mark scheme: 10 Correctly eliminates one variable M1 [x =] 6 A2 A1 for either correct [y =] –0.5 oe If M0 scored, SC1 for 2 values satisfying one of the original equations

This question in 0580/23 Oct/Nov 2020

19 Solve the simultaneous equations. You must show all your working. x - y = 7 x 2 + y = 149 x = … y = … x = … y = … [5]

5 marks

Mark scheme: 19 x2 + x – 156 [=0] M2 2 M1 for x + x = 7 + 149 or y2 + 15y – 100 [=0] or correct substitution ( x − 12 )( x + 13 ) [=0] M1 or for correct factors for their quadratic or ( y − 5 )( y + 20 ) [=0] equation or for correct use of quadratic formula or completing the square for their equation [x =] 12 [y =] 5 B2 B1 for x = 12, x = −13 or for y = 5, y = –20 [x =] −13 [y =] −20 or for a correct pair of x and y values If B0 scored and at least 2 method marks scored SC1 for correct substitution of both of their x values or their y values into x – y = 7 or x2 + y = 149

This question in 0580/22 Feb/March 2021

7 Solve the simultaneous equations. You must show all your working. 2x + y = 3 x - 5y = 40 x = … y = … [3]

3 marks

Mark scheme: 7 correctly eliminating 1 variable M1 x = 5 A1 y = − 7 A1 If M0 scored SC1 for two values satisfying one of the original equations

This question in 0580/21 May/June 2021

11 (a) Simplify fully. ( 4ab 5 ) 4 … [2] 3 = 6 (b) 2p 1 Find the value of p. p = … [1] (c) 812 ' 3 t = 9 Find the value of t. t = … [2]

5 marks

Mark scheme: 11(a) 256a 4 b 20 final answer 2 B1 for two correct elements in final answer 11(b) 27 1 11(c) 6 2 M1 for 3k ÷ 3t = 32 or 38 ÷ 3t = 3 k oe or better or 3t = 729 oe

This question in 0580/21 May/June 2021

19 (a) Sketch the graph of y = tan x for 0 ° G x G 360° . y 0 90° 180° 270° 360° x [2] (b) Solve the equation 5 tanx = 1 for 0° G x G 360 ° . x = … or x = … [2]

4 marks

Mark scheme: 19(a) Correct sketch 2 1 for one correct branch or correct sketch but with branches joined 19(b) 11.3 or 11.30 to 11.31 2 B1 for each and If 0 scored SC1 for two answers with a difference of 180° 191.3 or 191.30 to 191.31

This question in 0580/21 May/June 2021

b 28 a = 5c Find b when a = 5.625 and c = 2 . b = … [2]

2 marks

Mark scheme: 8 [±] 7.5 oe 2 b 2 M1 for 5.625 = or better 2 × 5

This question in 0580/22 May/June 2021

18 f ( x) = x 2 - 25 g ( x) = x + 4 Solve fg (x + 1 ) = gf (x) . x = … [4]

4 marks

Mark scheme: 18 [x =] –2.1 oe 4 M3 for x2 + 10x = x2 – 21 or better OR M1 for (x + 1 + 4)2 – 25 or better M1 for x2 – 25 + 4 or better 11 If 0 scored SC1 for answer − oe 6

This question in 0580/22 May/June 2021

24 Solve. 1 9 + = 1 x + 1 x + 9 x = … or x = … [5]

5 marks

Mark scheme: 24 x = 3, x = –3 5 M2 for x + 9 + 9(x + 1) = (x + 1)(x + 9) oe nfww or better or M1 for x + 9 + 9(x + 1) or (x + 1)(x + 9) oe or better B1 for x 2 + x + 9 x + 9 seen M1 dep for [ 0 = ] x 2 − 9 oe simplified or better

This question in 0580/22 May/June 2021

8 Solve the simultaneous equations. You must show all your working. 4x - 2y =- 13 - 3x + 4y = 11 x = … y = … [3]

3 marks

Mark scheme: 8 Correctly eliminates one M1 variable [x =] – 3 , [y =] 0.5 oe A2 A1 for either correct If M0 scored, SC1 for 2 values satisfying one of the original equations If 0 scored, SC1 for correct answers from no working

This question in 0580/21 Oct/Nov 2021

3x - 217 y = 1 - x Make x the subject of the formula. x = … [4]

4 marks

Mark scheme: 17 y + 2 4 M1 y (1 – x) = 3x – 2 or better [x = ] oe final answer y + 3 M1 for correctly isolating x terms on one side FT their first step/bracket expansion M1dep for correctly removing factor of x FT their previous step M1dep for correct division to isolate x Max 3 marks for an incorrect answer

This question in 0580/21 Oct/Nov 2021

17 Solve. ( 5x - 3)( 2x + 7) = 0 x = … or x = … [1]

1 marks

Mark scheme: 17 3 7 1 oe and – oe 5 2

This question in 0580/22 Oct/Nov 2021

18 Solve the simultaneous equations. You must show all your working. y = x 2 - 9x + 21 y = 2 x - 3 x = … y = … x = … y = … [5]

5 marks

Mark scheme: 18 x² – 11x + 24 [= 0] M2 M1 for x² – 9x + 21 = 2x – 3 oe or 2  y + 3   y + 3  y² – 16y + 39 [= 0] or y =   − 9   + 21 oe  2   2  (x – 8)(x – 3) [= 0] M1 or for correct factors for their quadratic or equation (y – 13)(y – 3) [= 0] or for correct use of quadratic formula for their equation [x =] 3 [y =] 3 B2 B1 for one correct pair or two correct [x =] 8 [y =] 13 x values or two correct y values. If B0 scored and at least 2 method marks scored SC1 for correct substitution of both of their x values or their y values into y = x2 – 9x + 21 or y = 2x – 3

This question in 0580/22 Oct/Nov 2021

x - 3 520 f ( )x = 2 g ( )x = 2 x - 1 h ( )x = x - 4 (a) Find ff(6). … [2] (b) Find g -1 g ( x + 21) . … [1] (c) Find x when f ( x) = h ( 84) . x = … [2]

5 marks

Mark scheme: 20(a) 32 2 M1 for f(6) = 8 ( 2 x – 3 ) – 3 or ff( x ) = 2 oe 20(b) x + 21 1 20(c) –1 2 1 M1 for oe or 2–4 oe 16

This question in 0580/22 Oct/Nov 2021

22 Solve the simultaneous equations. You must show all your working. y = x 2 - 3x - 13 y = x - 1 x = … , y = … x = … , y = … [5]

5 marks

Mark scheme: 22 x2 – 4x – 12 [= 0] M2 M1 for x2 – 3x – 13 = x – 1 or or for y = (y + 1)2 – 3(y + 1) – 13 y2 – 2y – 15 [= 0] (x – 6)(x + 2) [= 0] M1 or for correct factors for their quadratic or equation (y – 5)(y + 3) [= 0] or for correct use of quadratic formula or completing the square for their equation [x =] 6, [y =] 5 B2 B1 for one correct pair or two correct [x =] –2, [y =] –3 x values or two correct y values If B0 scored and at least 2 method marks scored SC1 for correct substitution of both of their x values or their y values into y = x2 – 3x – 13 or y = x – 1

This question in 0580/23 Oct/Nov 2021

18 Mrs Kohli buys a jacket, 2 shirts and a hat. The jacket costs $x. The shirts each cost $24 less than the jacket and the hat costs $16 less than the jacket. Mrs Kohli spends exactly $100. Write down an equation in terms of x. Solve this equation to find the cost of the jacket. $ … [3]

3 marks

Mark scheme: 18 A correct equation leading to 3 M2 for 4 x = 164 41 or M1 for x + 2 ( x − 24 ) + x − 16 = 100 oe or M1 for correctly simplifying their equation to the form kx = c provided at least one part correct from [ 2 ]( x − 24 ) oe or x − 16 or B1 for answer 41 without an equation in x shown

This question in 0580/22 Feb/March 2022

20 Solve the simultaneous equations. You must show all your working. 3x + y = 11 x 2 - 2 y = 18 x = … y = … x = … y = … [5]

5 marks

Mark scheme: 20 x 2 + 6 x − 40 [=0] M2 M1 for correct method to eliminate one 2 variable e.g. or y − 40 y − 41 [=0] x 2 − 2 (11 − 3 x ) = 18 (11 − y ) 2 or − 2 y = 18 32 ( x − 4 )( x + 10 ) [=0] M1 or for correct factors for their quadratic equation or ( y − 41)( y + 1) [=0] or for correct use of quadratic formula for their quadratic equation or for correctly completing the square for their quadratic equation x = 4, y = − 1 B2 B1 for x = 4, x = − 10 x = −10, y = 41 or for y = − 1, y = 41 or for a correct pair of x and y values If B0 scored and at least 1 method mark scored SC1 for correct substitution shown of both of their x values or their y values into 3x + y = 11 or x2 – 2y = 18

This question in 0580/22 Feb/March 2022

23 x 2 + 8x + 10 = ( x + p) 2 + q (a) Find the value of p and the value of q. p = … q = … [2] (b) Solve. x 2 + 8x + 10 = 30 x = … or x = … [2]

4 marks

Mark scheme: 23(a) [p = ] 4 2 B1 for one correct [q = ] –6 2 or  x  4   6 or x2 + px + px + p2 [+ q] 23(b) –10 and 2 2 2 M1 for  x  4   36 or  x  their 4  2  30  their  6  or for correct method to solve quadratic e.g.  x  10  x  2 

This question in 0580/21 May/June 2022

27 The line y = x + 1 intersects the graph of y = x 2 - 3x - 11 at the points A and B. Find the coordinates of A and the coordinates of B. You must show all your working. A ( … , … ) B ( … , … ) [4]

4 marks

Mark scheme: 27 (–2, –1) and (6, 7) 4 B3 for x = – 2 and 6 OR M1 for x 2  3 x  11  x  1 or better M1 for correct method to solve their quadratic e.g.  x  2  x  6  If 0 scored, SC1 for one correct pair of coordinates

This question in 0580/21 May/June 2022

1 7x - 2 3 - 10x19 f( )x = kx2 g ( x) = h ( x) = j( )x = x 5 14 (a) f(- 5)k = 675 Find the value of k. k = … [2] (b) Find gh ( x) . … [1] (c) Find h -1 ( x) + j( x) . Give your answer in its simplest form. … [4]

7 marks

Mark scheme: 19(a) 3 2 M1 for k(–5k)2 = 675 or better 19(b) 5 1 final answer 7 x  2 19(c) 1 4 7 or 0.5 B3 for answer 2 14 OR 5 x  2 B2 for 7 or M1 for correct first step for h –1(x) 7 y  2 e.g. x = 5 y  7 x  2 5 2 7 x y +  5 5 2  5 x  2  3  10 x M1FT for  oe with 14 14 common denominator

This question in 0580/23 May/June 2022

11 Solve the simultaneous equations. x - 3y = 7 2x - 3y = 11 x = … y = … [2]

2 marks

Mark scheme: 11 [x =] 4 2 B1 for each [y =] –1

This question in 0580/21 Oct/Nov 2022

22 y 1 0 x 360° – 1 (a) On the diagram, sketch the graph of y = cos x for 0° G x G 360° . [2] 1 (b) Solve the equation cosx =- for 0° G x G 360° . 2 x = … or x = … [2]

4 marks

Mark scheme: 22(a) Correct sketch 2 To go through (0, 1) and close to (360, 1) and reasonably close to (180, –1) 1111 B1 for correct cosine curve shape through 0.50.50.50.5 (0, 1) 0000 0000 50505050 100100100100 150150150150 200200200200 250250250250 300300300300 350350350350 -0.5-0.5-0.5-0.5 -1-1-1-1 Correct sketch to go through (0, 1), (360, 1) and (180, –1) 22(b) 120, 240 2 B1 for each or for two values with sum of 360

This question in 0580/21 Oct/Nov 2022

11 The graph of y = ( x - 3)( x + b)( x + 2) intersects the y-axis at - 30 . (a) Find the value of b. b = … [2] (b) When x 2 0 the graph crosses the x-axis once. Write down the coordinates of this point. ( … , … ) [1]

3 marks

Mark scheme: 11(a) 5 2 M1 for (0 – 3)(0 + b)(0 + 2) = –30 oe or better 11(b) (3, 0) 1

This question in 0580/22 Oct/Nov 2022

x + 517 f ( x) = x2 g ( x) = h ( x) = 7 x - 3 2 (a) Find f ( - 3) . … [1] (b) Find g -1 ( x) . g -1 ( x) = … [2] (c) Solve gf ( x) = hh -1 ( 63) where x 2 0 . x = … [3]

6 marks

Mark scheme: 17(a) 9 1 17(b) 2x – 5 final answer 2 M1 for correct first step e.g. y + 5 5 x x = or 2y = x + 5 or y – = or 2 2 2 better 17(c) 11 3 x 2 + 5 M1 for 2 M1 for hh–1(63) = 63 soi

This question in 0580/22 Oct/Nov 2022

4 223 Solve + = 3 . x + 1 2x - 5 You must show all your working. x = … or x = … [7]

7 marks

Mark scheme: 23 [0 =] 6x2 – 19x + 3 B5 B4 for 8x – 20 + 2x + 2 = 6x2 + 6x –15x – 15 or better OR M2 for 4(2x – 5) + 2(x + 1) = 3(x + 1)(2x – 5) oe or M1 for 4(2x – 5) + 2(x + 1) or better or common denominator (x + 1)(2x – 5) or better B1 for 2x2 + 2x – 5x – 5 or better seen M1 for correctly simplifying their quadratic to the form [0 =] ax2 + bx + c Correct method to solve their three term M1 e.g. (6x – 1)(x – 3) quadratic 2 −−( 19 )  ( −19 ) −4 6 3 2  6 1 B1 x = 3, x = oe 6

This question in 0580/22 Oct/Nov 2022

9 Solve the simultaneous equations. 3x - 2y = 21 5x + 2y = 51 x = … y = … [2]

2 marks

Mark scheme: 9 [x =] 9 2 B1 for each answer [y =] 3

This question in 0580/23 Oct/Nov 2022

20 (a) y 1 0 x 360° – 1 Sketch the graph of y = sin x for 0° G x G 360° . [2] 13 (b) Solve 3 - 2 sinx = for 0° G x G 360° . 4 x = … or x = … [3]

5 marks

Mark scheme: 20(a) 2 B1 for correct sine curve shape through the origin Correct sketch to go through (0, 0), (180, 0) and (360, 0) 20(b) 187.2 3 B2 for one correct value, if more than two and 352.8 answers given award B2 if any of the correct answers found and may be in the working 1 or M1 for sin x = − oe soi 8 If 0 scored, SC1 for two reflex angles with a sum of 540 or two non-reflex angles with a sum of 180

This question in 0580/23 Oct/Nov 2022

7 Solve. (a) 15t + 8 = 4 - t t = … [2] 25 - 2 u (b) = 2 3 u = … [2]

4 marks

Mark scheme: 7(a) 1 2 M1 for 15t + t = 4 – 8 oe − oe 4 7(b) 9.5 oe 2 M1 for 25 – 2u = 3 × 2 oe 25 2u or for − 2 = 3 3

This question in 0580/22 Feb/March 2023

9 Solve the simultaneous equations. You must show all your working. 3x - 2y = 19 x + y = 3 x = … y = … [3]

3 marks

Mark scheme: 9 Correctly eliminating one variable M1 [x =] 5 A1 [y = ] –2 A1 If M0 scored SC1 for 2 values satisfying one of the original equations.

This question in 0580/22 Feb/March 2023

19 Find the values of x when 6x + y = 10 and y = x 2 - 3x + 10 . x = … or x = … [3]

3 marks

Mark scheme: 19 0 and –3 3 B2 for x2 + 3x [= 0] or better or M1 for 10 – 6x = x2 – 3x + 10 oe or for correct simplification of their quadratic to the form ax2 + bx + c [= 0] or better or finding y = 28 and y = 10

This question in 0580/22 Feb/March 2023

19 (a) On the diagram, sketch the graph of y = cos x for 0° G x G 360° . y 1 0 180° 360° x – 1 [2] (b) Solve the equation 5 cosx + 3 = 0 for 0° G x G 360° . x = … or x = … [3]

5 marks

Mark scheme: 19(a) correct sketch 2 B1 for correct cosine curve shape through (0, 1) Correct sketch to go through (0, 1), (360, 1) and (180, –1) 19(b) 126.9 or 126.86 to 126.87 3 B2 for 1 correct angle 233.1 or 233.13 to 233.14 3 or M1 for cos x =  oe 5 If M1 or 0 scored SC1 for two angles with a sum of 360

This question in 0580/21 May/June 2023

20 The table shows some values for y = 3x 2 - 2x - 1. x -1 -0.5 0 0.5 1 1.5 y 4 -1 0 2.75 (a) Complete the table. [1] (b) On the grid, draw the graph of y = 3x 2 - 2x - 1 for - 1 G x G 1.5 . y 4 3 2 1 – 1 – 0.5 0 0.5 1 1.5 x – 1 – 2 [3] (c) By drawing a suitable straight line, solve the equation 3x 2 - 4x - 2 = 0 for - 1 G x G 1.5 . x = … [3] Question 21 is printed on the next page.

7 marks

Mark scheme: 20(a) 0.75 and –1.25 1 20(b) Correct curve 3 B2 FT for 6 or 5 correct plots or B1 FT for 4 or 3 correct plots 20(c) ruled line y  2 x  1 B2 B1 for correct equation  y  2 x  1 soi or y  2 x  k or y  kx  1 drawn −0.35 to −0.45 B1

This question in 0580/21 May/June 2023

8 Solve. 30 (a) = 6 x x = … [1] (b) 11x - 3 H 2 ( 2x + 9) … [3]

4 marks

Mark scheme: 8(a) 5 1 8(b) x ⩾ 3 final answer 3 M1 for correct first step 11x – 3 ≥ 4x + 18 or 5.5x – 1.5 ⩾ 2x + 9 or better M1 for correctly collecting their x terms on one side and their number terms on the other side e.g. 11x – 4x ⩾ 18 + 3 or better

This question in 0580/22 May/June 2023

12 One solution of the equation ax 2 + b = 181 is x = 8 . a and b are both positive integers greater than 1. (a) Find the value of b. b = … [2] (b) Write down the other solution of the equation ax 2 + b = 181. x = … [1]

3 marks

Mark scheme: 12(a) 53 2 M1 for a × 82 + b = 181 oe seen 12(b) –8 1

This question in 0580/22 May/June 2023

18 Make x the subject of the formula. 3x c = 2x - 5 x = … [4]

4 marks

Mark scheme: 18 5c 4 M1 for correctly clearing the denominator and oe final answer expanding bracket 2c  3 or correctly clearing the denominator and dividing by c M1 for correctly collecting terms in x on one side and terms not in x on the other M1 for correct factorising M1 for correct division dependent on x appearing only once in a factorised expression Maximum 3 marks for an incorrect answer

This question in 0580/22 May/June 2023

4 v = u - 9.8t Find the value of v when u = 4 and t =- 7 . v = … [2]

2 marks

Mark scheme: 4 72.6 2 M1 for 4  9.8 7 or better

This question in 0580/23 May/June 2023

17 Rearrange the formula to make m the subject. 2 ( m - k) R = m m = … [4]

4 marks

Mark scheme: 17 2 k 2 k 4 m  or m   2  R   R  2  M1 for clearing fractions final answer M1 for expanding brackets (or ÷ 2) M1 for collecting terms in m on one side and terms not in m on the other M1 for dividing by a bracket maximum of 3 if final answer incorrect

This question in 0580/23 May/June 2023

19 Solve the equation x 2 + 5x - 7 = 0 . You must show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

4 marks

Mark scheme: 19 2 B2 2 5 5 4 1 7 B1 for 5  4 1 7 2  1 p  q p  q and if in form or r r B1 for p = 5 and r = 2×1 −6.14 and 1.14 cao B2 B1 for 1 correct answer for −6.1 and 1.1 or −6.140... and 1.140... or 6.14 and −1.14 or correct answers seen in working

This question in 0580/23 May/June 2023

20 f ( x) = 6 x - 7 g ( x) = x -3 (a) Find f ( x + 2) . Give your answer in its simplest form. … [2] (b) Find f -1 ( x) . f - 1 ( x) = … [2] (c) Find x when g(x) = f(22) . x = … [2]

6 marks

Mark scheme: 20(a) 6 x  5 cao final answer 2 M1 for 6  x  2   7 oe 20(b) x  7 x 7 2 y 7 or  final answer M1 for x  6 y  7 or y  7  6 x or  x  6 6 6 6 6 20(c) 1 2 M1 for x −3 = 6  22  7 or better or 0.2 5

This question in 0580/23 May/June 2023

3 Complete these statements. (a) When x = … , x + 3 = 8 . [1] (b) When 7y = 63, 10 y = … [1]

2 marks

Mark scheme: 3(a) 5 1 3(b) 90 1

This question in 0580/21 Oct/Nov 2023

11 Solve. 4 ( 2x - 3) H 43 + 3 x … [3]

3 marks

Mark scheme: 11 x ⩾ 11 final answer 3 M1 for 8x – 12 ⩾ 43 + 3x or better M1 for e.g. 8x – 3x ⩾ 43 + 12 oe OR 43 3 x M1 for 2x – 3 ⩾ + 4 4 3 x 43 M1 for 2x – ⩾ +3 4 4

This question in 0580/21 Oct/Nov 2023

14 y = 2w 2 - x Rearrange the formula to make w the subject. w = … [3]

3 marks

Mark scheme: 14 y + x 3 M1 for isolating term in w [±] oe final answer M1 for division by 2 2 M1 for square root Max 2 marks if answer incorrect

This question in 0580/22 Oct/Nov 2023

21 The line y = x + 1 intersects the curve y = x 2 + x - 3 at two points. Find the coordinates of the two points. ( … , … ) ( … , … ) [4]

4 marks

Mark scheme: 21 (2, 3) and (–2, –1) 4 B3 for x = 2 and x = –2 or B2 for x 2 − 4  = 0 or better or for (2, 3) or (–2, –1) or M1 for x + 1 = x 2 + x − 3 oe

This question in 0580/22 Oct/Nov 2023

2wy 2 9 P = 3 Find the positive value of y when P = 108 and w = 8 . y = … [3]

3 marks

Mark scheme: 9 1 9 3 3 P 3 108 4.5, 4 or M2 for y2 = or y2 = or better 2 2 2 w 2  8 2 y8 2 or M1 for 108 = or better 3

This question in 0580/23 Oct/Nov 2023

15 T = 3d - e Rearrange the formula to make d the subject. d = … [3]

3 marks

Mark scheme: 15 T 2 + e 3 M1 for T 2 = 3d – e [d =] oe final answer M1 for isolating term in d 3 M1 for dividing by 3 Max 2 marks if answer incorrect

This question in 0580/23 Oct/Nov 2023

22 Find the coordinates of the point where the line 4x + y = 9 intersects the curve y + x 2 = 5 . You must show all your working. ( … , … ) [5]

5 marks

Mark scheme: 22 x2 – 4x + 4 [= 0] M2 M1 for 9 – 4x = 5 – x2 oe (x – 2)(x – 2) M1 Accept alt methods e.g. use of formula, complete the square for their 3 – term quadratic equation (2, 1) B2 B1 for x = 2

This question in 0580/23 Oct/Nov 2023

23 (a) On the axes, sketch the graph of y = cos x , for 0° G x G 360° . y 1 0 x 180° 360° – 1 [2] (b) Solve the equation cosx = 0 .294 for 0° G x G 360° . x = … or x = … [2]

4 marks

Mark scheme: 23(a) 2 M1 for correct cosine curve shape through (0, 1) Correct sketch to go through (0, 1), close to (360, 1) and reasonably close to (180, –1) 23(b) 72.9 and 287.1 2 B1 for one correct If 0 scored, SC1 for two angles with a sum of 360

This question in 0580/22 Feb/March 2024

12 Solve the simultaneous equations. You must show all your working. 3 x + 5y = 5 2 4x - 3y = 46 x = … y = … [4]

4 marks

Mark scheme: 12 Correctly equating one set of M1 coefficients Correct method to eliminate one M1 variable x = 10, y = –2 A2 A1 for x = 10 A1 for y = –2 If M0 scored SC1 for 2 values satisfying one of the original equations.

This question in 0580/21 May/June 2024

18 y 8 7 6 5 4 3 2 1 x – 3 – 2.5 – 2 – 1.5 – 1 – 0.5 0 0.5 1 1.5 – 1 – 2 – 3 The diagram shows the graph of y = x 3 + 4x 2 - 2 for - 3 G x G 1.5 . By drawing a suitable straight line, solve the equation x 3 + 4x 2 - 2 = 2x for - 3 G x G 1.5 . x = … or x = … [3]

3 marks

Mark scheme: 18 y = 2x ruled B1 x = –0.5 to –0.55 B2 B1 for –0.5 to –0.55 x = 0.85 to 0.9 B1 for 0.85 to 0.9

This question in 0580/21 May/June 2024

21 Solve the simultaneous equations. You must show all your working. 4y + 3x = 13 y = x 2 - 18 x = … y = … or x = … y = … [5]

5 marks

Mark scheme: 21 4 x 2  3 x  85   0  M2 2 2 2 13  3 x x  18  3 x  13 or x  18  or 16 y  113 y  7   0  M1 for 4   4 oe simplified 2  13  4 y  or y     18 oe or better  3  correct method to solve their M1 2 3 3 4 4 85 quadratic equation e.g. factors, oe, (4x – 17)(x + 5) 2  4 quadratic formula, completing the square  113    113  2 4 16  7 oe, 2 16 (16y – 1)(y – 7) x = −5 y = 7 B2 B1 for one correct pair or two correct x values or 17 1 two correct y values x = oe y = oe If B0 scored and at least 2 method marks scored, 4 16 SC1 for correct substitution of both of their x values or their y values into 4y + 3x = 13 or y = x2 – 18

This question in 0580/22 May/June 2024

11 Solve the simultaneous equations. 5t - 2w = 19 3t + 2w = 5 t = … w = … [2]

2 marks

Mark scheme: 11 [t = ] 3 2 B1 for each [w = ] –2

This question in 0580/23 May/June 2024

20 f( )x = 3 x + 2 (a) Find x when f ( )x = 245 . x = … [2] (b) Find x when f - 1 ( )x = 7 . x = … [2]

4 marks

Mark scheme: 20(a) 5 2 M1 for 3x + 2 = 245 20(b) 2189 2 M1 for x = f(7) or 37 + 2

This question in 0580/23 May/June 2024

12 Solve the simultaneous equations. You must show all your working. 5x + 6y = 9 3x - 2y = - 17 x = … y = … [3]

3 marks

Mark scheme: 12 Correctly eliminating one variable M1 x = –3 A1 If A0 scored SC1 for 2 values satisfying one of the original equations. y = 4 A1

This question in 0580/21 Oct/Nov 2024

17 Solve. 3x 2 - 7x - 16 = 0 You must show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

4 marks

Mark scheme: 17  −−  7  ( − 7 ) 2 − 4 ( 3 )( −16 ) B2 B1 for ( − 7 ) 2 − 4 ( 3) ( −16 ) ) or better oe 2  3 p + q p − q and if in the form or then r r B1 for p = – (–7) and r = 2(3) 3.75 and –1.42 B2 B1 for each or SC1 for answers 3.8 or 3.754… and –1.4 or –1.42… or –1.421 or 3.75 and –1.42 seen in working or –3.75 and 1.42 as final answers

This question in 0580/21 Oct/Nov 2024

18 g ( x) = 4x + 3 (a) Find x when g ( x) = 1. … [1] -1 1 (b) Find g e o . 16 … [2]

3 marks

Mark scheme: 18(a) –3 1 18(b) –5 2 1 M1 for or 4–2 4 2

This question in 0580/21 Oct/Nov 2024

22 The graph of y = ( x + 2)( x - 1) 2 is shown on the grid. y 4 3 2 1 x -2 -1 0 1 2 -1 (a) Show that y = ( x + 2 )( x - 1 ) 2 can be written as y = x 3 - 3x + 2 . [2] (b) By drawing a suitable straight line, solve the equation 2x 3 - 5x = 0 . x = … or x = … or x = … [4] Question 23 is printed on the next page.

6 marks

Mark scheme: 22(a) x2 – x – x + 1 M1 or x2 + 2x– x – 2 A correct unsimplified expansion A1 e.g. x3 + 2x2 –x2 –2x – x2 –2x + x + 2 oe leading to [y = ] x3 – 3x + 2 22(b) y = 2 – 0.5x ruled B2 B1 for [y =] 2 – 0.5x soi or for y = 2 – kx drawn or for y = k – 0.5x drawn –1.5 to –1.6 B2 B1 for two correct values 0 1.5 to 1.6

This question in 0580/21 Oct/Nov 2024

22 A curve has equation y = x 3 + x 2 - x . 1 5 The curve has a stationary point at e ,3 - 27 o. (a) Find the coordinates of the other stationary point. ( … , … ) [5] (b) By sketching the graph of y = x 3 + x 2 - x , determine whether the stationary point 1 5 e ,3 - 27 o is a maximum or a minimum. y x O 1 5 e ,3 - 27 o is a … [2] (c) The equation x 3 + x 2 - x = k has fewer than 3 solutions. Find the range of possible values for k. … [2] Question 23 is printed on the next page.

9 marks

Mark scheme: 22(a) (–1, 1) nfww 5 B4 for x = – 1 nfww or answer (–1, k) nfww OR B2 for 3x2 + 2x – 1 or B1 for two terms correct dy M1 for setting their = 0 or dx dy stating = 0 dx M1 for correct method to solve their 3-term quadratic e.g. (3x – 1)(x + 1) 22(b) Correct sketch of positive cubic 2 B1 for correct shape of positive with minimum in correct quadrant cubic and minimum 22(c) 2 B1 strict FT for each If their y coordinate Strict FT: from (a) is: or SC1FT for non-inclusive versions of both correct strict FT 5 k their yin ( a ) − inequalities 27 5 k − 27 5 k their yin ( a ) − 27 5 k − 27

This question in 0580/22 Feb/March 2025

18 One day, Anya runs 12 km at a speed of x km/h. The next day she walks 10 km at a speed of ( x - 4 ) km/h. (a) Write down an expression, in terms of x, for the time she spends running. … h [1] (b) Write down an expression, in terms of x, for the time she spends walking. … h [1] (c) The time Anya spends walking is 1 hour more than the time she spends running. Write an equation in terms of x and show that it simplifies to x 2 - 2 x - 48 = 0 . [4] (d) Use factorisation to solve the equation x 2 - 2 x - 48 = 0 . x = … or x = … [3] (e) Find the time Anya spends running. … h [1]

10 marks

Mark scheme: 18(a) 12 1 x 18(b) 10 1 x − 4 18(c) 10 12 M1 their – their = 1 oe x − 4 x 10x – 12x + 48 = x2 – 4x M2 Correctly multiplying their brackets and clearing algebraic fractions  12  e.g. ( x − 4 )  + 1  = 10  x  48 leading to 12 − + x − 4 = 10 and then x 12 x − 48 + x 2 − 4 x = 10 x or M1 for correctly clearing, or correctly collecting into a single fraction, two fractions both with different algebraic denominators e.g. 10x – 12(x – 4) = x(x – 4) or 10 x − 12 ( x − 4 ) [= 1] x ( x − 4 ) Leading to 0 = x2 – 2x – 48 A1 With no errors or omissions seen, dep on M3 18(d) (x + 6)(x – 8) M2 M1 for x(x – 8) + 6(x – 8) or x(x + 6) – 8(x + 6) or (x + a)(x + b) where ab = –48 or a + b = –2 –6, 8 B1 18(e) 1 1 12 1.5 or 1 FT 2 their 8

This question in 0580/21 May/June 2025

6 Solve. (a) 8x + 7 = 39 x = … [2] (b) 2 ( 5y - 1 ) = 24 y = … [3]

5 marks

Mark scheme: 6(a) 4 2 M1 for 8x = 39 – 7 or better 6(b) 13 3 M1 for correct first step e.g. 2.6 or oe 5 24 5y – 1 = or 10y – 2 = 24 or better 2 M1 for correctly isolating terms in y FT their first step e.g. 5y = 12 + 1 or 10y = 24 + 2

This question in 0580/22 May/June 2025

15 y 5 4 3 2 1 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 8 x – 1 – 2 – 3 – 4 – 5 2 The diagram shows the graph of y = - 1. x (a) Write down the coordinates of the point where the graph crosses the x-axis. ( … , … ) [1] (b) Write down the equation of each asymptote. … … [2] 2 (c) By drawing a suitable straight line on the grid, solve - x - 1 = 0 . x x = … or x = … [3]

6 marks

Mark scheme: 15(a) (2, 0) 1 15(b) x = 0, y = –1 2 B1 for each 15(c) y = x ruled B1 x = –2 and x = 1 B2 B1 for one correct or for two correct answers FT from their line

This question in 0580/22 May/June 2025

24 The line y = 7x + 3 intersects the curve y = x 2 + 5x - 12 at the points A and B. Find the coordinates of A and B. A ( … , … ) B ( … , … ) [5]

5 marks

Mark scheme: 24 (5, 38) and (–3, –18) 5 B4 for one correct coordinate or for x = 5 and x = –3 OR M2 for x2 – 2x – 15 [= 0] or y2 – 20y – 684 [= 0] or M1 for 7x + 3 = x2 + 5x – 12 oe  y − 3  2  y − 3  or y = + 5 − 12      7   7  M1 for correct method to solve their three- term quadratic (x – 5)(x + 3) −−( 2 )  ( −2 ) 2 −−4 1 15 oe 2  1 If B0 scored and at least 2 method marks scored, SC1 for correct substitution of both of their x values or their y values into y = 7x + 3 or y = x2 + 5x – 12

This question in 0580/22 May/June 2025

9 (a) Solve. 5x 2 = 12 - 17 x x = … or x = … [4] (b) ax 2 + a = b where a and b are integers. One solution of this equation is x = 6 . Write down the other solution. x = … [1]

5 marks

Mark scheme: 9(a) 3 4 M1 for 5x2 + 17x – 12 [= 0] or 0.6 5 M2 for correct method to solve their and –4 ax2 + bx – c [= 0] e.g. factorising (5x – 3)(x + 4) [= 0], completing the square or using the formula or M1 for (5x + a)(x + b) where ab = –12 or 5b + a = 17 or correct partial factorisation e.g. 5x(x + 4) – 3(x + 4) or for 17 2 − 4 ( 5 )( −12 ) or better −17 + d −17 − d or or 2 ( 5 ) 2 ( 5 ) 9(b) –6 1

This question in 0580/23 May/June 2025

10 Solve the simultaneous equations. 4x - 5y = 13 3x - 2y = 8 x = … y = … [4]

4 marks

Mark scheme: 10 [x =] 2 4 M1 for correctly equating one set of [y =] –1 coefficients or for making x or y the subject of one equation M1 for correct method to eliminate one variable A1 for x = 2 A1 for y = –1 If M0 scored, SC1 for 2 values satisfying one of the original equations

This question in 0580/23 May/June 2025

18 Make t the subject of the formula. m ( 1 - t) 2 = pt t = … [4]

4 marks

Mark scheme: 18 m 4 M1 for correctly clearing their fraction  t = final answer M1 for correct expansion 2 p + m M1 for correctly collecting their terms in t on one side and other terms on the other side, not dividing by 1 – t M1 for correct factorisation and division of their two-term expression in t To a maximum of 3 marks for an incorrect answer

This question in 0580/23 May/June 2025

8 The cost of one orange is t cents. The cost of one apple is w cents. The total cost of 3 oranges and 1 apple is 51 cents. The total cost of 6 oranges and 5 apples is 129 cents. Use simultaneous equations to find the value of t and the value of w. You must show all your working. t = … w = … [5]

5 marks

Mark scheme: 8 3t + w = 51 2 B1 for each 6t + 5w = 129 Correctly eliminating one variable from M1 e.g. 6t + 2w = 102 and 6t + 5w = 129 their equations leading to 3w = 27 or w = 51 − 3t and 6t + 5(51 − 3t ) = 129 [t =] 14 A2 A1 for [t =] 14 [w =] 9 A1 for [w =] 9 If M1A0A0 scored, M1 SC1 for two values satisfying one of their original equations or if M0 scored, SC1 for 2 correct answers

This question in 0580/21 Oct/Nov 2025

19 Solve. x 1 x + 4 e o = 9 3 x = … [3]

3 marks

Mark scheme: 19 8 2 3 M2 for –x = 2(x + 4) oe − or −2 oe 2 ( x + 4 ) 3 3 or M1 for (3–1)x, 3−x , (32)x+4 , 3 or −2( x + 4 ) 1  oe 3

This question in 0580/21 Oct/Nov 2025

22 Solve. 2 x = x - 1 x + 2 x = … or x = … [5]

5 marks

Mark scheme: 22 –1 and 4 5 B2 for x2 – 3x – 4 [= 0] or M1 for 2(x + 2) = x(x – 1) or better M2 for a correct method to solve their three-term quadratic in the numerator or M1 for (x + a)(x + b) where ab = –4 or a + b = –3 or x ( x − 4) + [1]( x − 4) or x ( x + 1) − 4( x + 1) or M1 for ( −3)2 −4 [1] −4 or better p + q p − q or if in the form or then r r M1 for p = −(−3) and r = 2(1) or better

This question in 0580/21 Oct/Nov 2025

27 Solve the simultaneous equations. y = x 2 - 8x + 22 y + 2 = 3x x = … , y = … x = … , y = … [6]

6 marks

Mark scheme: 27 [x =] 3, [y =] 7 6 M2 for x² – 11x + 24 [= 0] oe simplified [x =] 8, [y =] 22 or M1 for x² – 8x + 22 = 3x – 2 oe or better M2 for correct method to solve their three- term quadratic e.g. (x – 8)(x – 3) [= 0] or M1 for x(x – 3) – 8(x – 3) or x(x – 8) – 3(x – 8) or (x + a)(x + b) where ab = 24 or a + b = –11 B1 for x = 3 and x = 8 or y = 7 and y = 22 or one correct pair If B0 scored and at least two method marks scored SC1 for correct substitution of both of their x-values into y + 2 = 3x or y = x2 – 8x + 22

This question in 0580/23 Oct/Nov 2025