Equations — Paper 2 · IGCSE Mathematics - International 0607

9 Solve the simultaneous equations. 4x + 3y = 0 2x - y = 5 x = … y = … [3] J N

3 marks

Mark scheme: 9 [ x =]1.5 3 M1 for correct method to eliminate one variable [ y =] − 2 B1 for x = 1.5 B1 for y = −2 If 0 scored SC1 for correct substitution into one of original equations and correct evaluation to find other variable

This question in 0607/21 May/June 2017

12 Solve. 2x 2 - 5x - 7 = 0 x = … or x = … [3]

3 marks

Mark scheme: 12 7 3 M2 for (2 x − 7)( x + 1) x = , − 1 oe 2 or M1 for (2 x + a )( x + b ) where a + 2b = − 5 or ab = − 7 or M1 for 2 x ( x + 1) − 7( x + 1) or x (2 x − 7) + 1(2 x − 7)

This question in 0607/21 May/June 2017

10 y = x + 1 and y = 2 - x Find the value of x. x = … [2]

2 marks

Mark scheme: 10 0.5 oe 2 M1 for x + 1 = 2 – x or for correctly eliminating x

This question in 0607/23 May/June 2017

5 Solve the simultaneous equations. x – 3y = 4 5x – 6y = –7 x = … y = … [3]

3 marks

Mark scheme: 5 x = –5 3 M1 for correctly eliminating one variable y = –3 B1 for each answer If zero scored SC1 for correct substitution and evaluation to find the other variable

This question in 0607/21 Oct/Nov 2017

8 Solve the equation. 90 45 - = 15 x x = … [3]

3 marks

Mark scheme: 8 3 3 90 M2 for 45 − 15 or M1 for correct first step

This question in 0607/23 Oct/Nov 2017

9 The mean of two numbers is 46. The difference between the two numbers is 12. Find the two numbers. … and … [2]

2 marks

Mark scheme: 9 40 and 52 2 B1 for 92

This question in 0607/23 Oct/Nov 2017

10 Solve the equation. x 2 - 5x - 24 = 0 x = … or x = … [3]

3 marks

Mark scheme: 10 8, –3 3 M2 for ( x − 8)( x + 3) oe or M1 for ( x + a )( x + b ) where ab = −24 or a + b = −5

This question in 0607/23 Oct/Nov 2017

14 Rearrange the formula to make x the subject. 3x A = 2x - 5 x = … [3] Questions 15 and 16 are printed on the next page.

3 marks

Mark scheme: 14 5 A − 5 A 3 M1 for correctly eliminating fractions x = or x = M1 for correctly collecting their x terms 2 A − 3 3 − 2 A M1 for correct final division of their terms final answer

This question in 0607/23 Oct/Nov 2017

3 Solve these simultaneous equations. x - 3y = 7 x - 2y = 5 x = … y = … [2]

2 marks

Mark scheme: 3 [x = ] 1 2 B1 for each [y = ] – 2 If 0 scored SC1 for correct substitution and evaluation to find the other variable

This question in 0607/21 May/June 2018

5 Solve. 7x + 9 = 5x + 17 x = … [2]

2 marks

Mark scheme: 5 4 2 M1 for correctly moving at least one term

This question in 0607/22 May/June 2018

14 Solve the simultaneous equations. 3x + 2y = 4 2x - 3y = 7 x = … y = … [4]

4 marks

Mark scheme: 14 [ x = ] 2 4 M1 for correctly equating one set of [ y = ] − 1 coefficients or M1 for equation x = or y = from one equation M1 for correct substitution into other equation A1 for one correct value If 0 scored SC1 for correct substitution into one of original equations and evaluation

This question in 0607/21 Oct/Nov 2018

7 Solve the simultaneous equations. 3t - u = -5 3t + 2u = 1 t = … u = … [2]

2 marks

Mark scheme: 7 [t =] –1 2 B1 for each [u =] 2 If 0 scored, SC1 for their answers satisfying one equation.

This question in 0607/23 Oct/Nov 2018

8 Solve the simultaneous equations. a + b = 16 2a - b = 17 a = … b = … [2]

2 marks

Mark scheme: 8 [a = ] 11 2 B1 for each [b = ] 5 If 0 scored, SC1 for their values of a and b satisfying one equation

This question in 0607/22 May/June 2019

13 Make a the subject of s = ut + at 2 . 2 a = … [3]

3 marks

Mark scheme: Question Answer Marks Partial Marks 7 9 3 27 5 M2 for × oe 5 3 27 5 or B1 for or 5 3 8(a) 1.002 × 10[1] 2 B1 for 10.02 8(b) 5 × 10 2 2 M1 for correct answer not in standard form 9 0.09 oe 2 M1 for 0.3 × 0.3 oe 10(a)  0  2 0  k    B1 for  or    32  k  32  10(b) 10 2 M1 for 6 2 + 8 2 soi by 100 11(a) 250 1 11(b) 96 2 120 M1 for 100 × oe 125 12 (7, − 1) 2 B1 for (7, k) or (k, –1) If 0 scored SC1 for (2.5, 2) 13 2 s − 2ut M1 for correctly multiplying by 2 [ a = ] oe 2 3 M1 for correctly collecting terms t M1 for correctly dividing by 2t

This question in 0607/23 May/June 2019

10 Solve. 4w 2 - 8w - 5 = 0 w = … or w = … [3]

3 marks

Mark scheme: 10 1 5 3 B2 for (2 w + 1)(2 w − 5) − oe , oe 2 2 B1 for ( aw + b )( cw + d ) with two correct from ac = 4, bd = –5, ad + bc = –8

This question in 0607/21 Oct/Nov 2019

14 Make l the subject of the formula T = 2 r . g l = … [3] 1

3 marks

Mark scheme: 14  T  2 gT 2 3 M1 for correct division by 2π or 4π 2 g   or 2 final answer 2  2 π  4 π M1 for correct multiplication by g or g M1 for correct squaring

This question in 0607/21 Oct/Nov 2019

5 Dippi buys 5 burgers and 4 bags of chips for a total cost of $8.10 . Burgers cost $1.10 each. Find the cost of one bag of chips. $ … [3]

3 marks

Mark scheme: 5 0.65 3 8.10 −×5 1.1 M2 for oe 4 or M1 for 5 × 1.1

This question in 0607/22 May/June 2020

8 Solve the simultaneous equations. 3x + 2y =- 1 7x - y = 26 x = … y = … [3]

3 marks

Mark scheme: 8 [x =] 3 3 M1 for correct method to eliminate [y =] –5 one variable B1 for each If 0 scored, SC1 for their answers satisfying one equation

This question in 0607/22 May/June 2020

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

3 marks

Mark scheme: 10 x = –2, y = 3 3 M1 for correct equation in x or equalising coefficients of x or y and adding / subtracting appropriately. B1 for one correct solution If 0 scored, SC1 for a correct substitution and evaluation of the other variable

This question in 0607/23 May/June 2020

5 Solve the simultaneous equations. 2p - 3q = 7 p + 3q = 2 p = … q = … [2]

2 marks

Mark scheme: 5 [p =] 3 2 B1 each 1 If 0 scored SC1 for correct substitution and [q =] − oe evaluation of the other variable 3

This question in 0607/23 Oct/Nov 2020

4 Solve. x + 2 = 7 … [1]

1 marks

Mark scheme: 4 5 and –5 1

This question in 0607/22 Feb/March 2021

5 Solve. 2 ( 4x - 1) = 3 ( 2x + 1) x = … [3]

3 marks

Mark scheme: 5 1 3 M1 for 8x – 2 = 6x + 3 2 oe M1FT for 8x – 6x = 3 + 2 2

This question in 0607/21 Oct/Nov 2021

8 Solve the simultaneous equations. 3x - 2y = 12 5x + y = 7 x = … y = … [3]

3 marks

Mark scheme: 8 Correctly eliminating one variable M1 x = 2 A2 A1 for each y = –3 If 0 scored, SC1 for correct substitution and evaluation to find other variable.

This question in 0607/21 Oct/Nov 2021

15 Solve. 8 - x x + 1 = 3 2 x = … [3]

3 marks

Mark scheme: 15 2.6 oe 3 B2 for 16 – 3 = 3x + 2x or better or M1 for 2(8 − x ) = 3( x + 1)

This question in 0607/22 Oct/Nov 2021

5 There are 640 students in a school. The table shows the favourite colour of each of the students. Favourite colour Blue Green Red Yellow Number of students 120 2x 280 x (a) Find the value of x. x = … [2] (b) Find the relative frequency of students whose favourite colour is red. Give your answer as a fraction in its lowest terms. … [2]

4 marks

Mark scheme: 5(a) 80 2 M1 for (640 − 120 − 280) ÷ (2 + 1) 5(b) 7 2 280 cao M1 for oe 16 640

This question in 0607/23 Oct/Nov 2021

10 Solve the simultaneous equations. You must show all your working. 4x + 3y =- 10 3x - 4y = 5 x = … y = … [4]

4 marks

Mark scheme: 10 Correctly equating one set of coefficients M1 Correct method to eliminate one variable M1 [ x = ] − 1 A1 [ y = ] − 2 A1 If 0 scored, SC1 for answers that satisfy one equation

This question in 0607/23 Oct/Nov 2021

9 Solve the simultaneous equations. 5x + 2y =- 12 3x - y =- 5 x = … y = … [3]

3 marks

Mark scheme: 9 Correct method to eliminate one M1 If 0 scored, SC1 for answers that satisfy one variable equation [ x ] 2 A1 [ y ] 1 A1

This question in 0607/22 May/June 2022

11 Solve. 6x 2 - 5x - 6 = 0 x = … or x = … [3]

3 marks

Mark scheme: 11 2 3 3 M2 for (3 x  2)(2 x  3)  , 3 2 or M1 for ( ax  b )( cx  d ) with two of ac  6, bd 6, ad  bc 5 OR M2 for correct use of formula or M1 for one error in substituting into formula

This question in 0607/22 May/June 2022

11 Solve. (a) 4x 2 - 5x - 6 = 0 x = … or x = … [3] (b) 2x + 1 = 3 … [2]

5 marks

Mark scheme: 11(a) 3 3 M2 for (4 x + 3)( x − 2) 2 , − oe 4 or M1 for ( ax + b)(cx + d ) where ac = 4 and bd = −6 OR 5  11 M2 for 8 −−( 5)  ( −5) 2 −−4 4 ( 6) or M1 for condone 1 slip 2  4 11(b) –2, 1 2 B1 for each

This question in 0607/21 Oct/Nov 2022

6 Solve. 5x - 10 = 3 x - 6 x = … [2]

2 marks

Mark scheme: 6 2 2 M1 for 5x − 3x = −+6 10 oe

This question in 0607/22 Feb/March 2023

11 Solve the simultaneous equations. 5x - 2y = 12 3x + 4y = 2 x = … y = … [3]

3 marks

Mark scheme: 11 [ x =] 2 3 M1 for correct method to eliminate [ y =] − 1 one variable A1 for each If 0 scored SC1 for answers that satisfy one equation

This question in 0607/22 Feb/March 2023

11 Solve the simultaneous equations. 1 1 x - y = 7 2 3 3x + y = 6 x = … y = … [3]

3 marks

Mark scheme: 11 x = 6, y = –12 3 M1 for correct method to eliminate one variable B1 for x = 6 B1 for y = –12 If 0 scored, SC1 for a pair of solutions that satisfy one equation.

This question in 0607/22 May/June 2023

15 Rearrange the equation to make x the subject. A + 4y = A ( 2 - 3x) x = … [3] Question 16 is printed on the next page.

3 marks

Mark scheme: 15 A  4 y 3 B1 for 2 A  3 Ax [ x  ] oe 3 A M1 for correctly isolating their x term M1 for correct division to find their x in a 3-term equation Maximum of 2 if answer not fully correct

This question in 0607/23 May/June 2023

12 Solve x 2 - 2x - 6 = 0. Give your answer in the form a ! b where a and b are integers. … [3]

3 marks

Mark scheme: 12 1  7 3 2  28 M2 for 2 −−( 2)  ( −2) 2 −−4 1 ( 6) or M1 for 2  1

This question in 0607/22 Oct/Nov 2023

8 Solve the simultaneous equations. 1 2 x + y = 8 2 3 3x - y = 18 x = … y = … [3]

3 marks

Mark scheme: 8 Correctly eliminating x or y M1 [x =] 8 A2 A1 for each [y =] 6 If zero scored SC1 for correct substitution and evaluation to find the other variable

This question in 0607/22 Feb/March 2024

9 These are the equations of two lines. 4y = x + 7 y + 4x = 6 (a) Find the coordinates of the point where these two lines intersect. ( … , … ) [3] (b) Are the two lines perpendicular? Give a reason for your answer. … because … … [2]

5 marks

Mark scheme: 9(a) (1, 2) final answer 3 M1 for correctly equating one set of coefficients A1 for x = 1 A1 for y = 2 Correct answers spoiled scores SC2 If 0 scored SC1 for their solutions satisfying one equation 9(b) 1 B1 both gradients seen –4 and 4 1 B1 FT their gradients and correct conclusion Yes and −4 = −1 4

This question in 0607/22 Feb/March 2025

15 Solve. 2x 2 - 5 x - 3 = 0 x = … or x = … [3]

3 marks

Mark scheme: 15 1 3 B2 for (2x + 1)(x – 3) – , 3 or B1 for (2x + a)(x + b) where ab = –3 2 or a + 2b = –5 or for 2x(x – 3) + (x – 3) or x(2x + 1) – 3(2x + 1). OR −−( 5)  ( −5) 2 −−4 2 ( 3) B2 for oe 2  2 −−( 5) + q −−( 5) − q or B1 for or 2  2 2  2 or ( −5) 2 −−4 2 ( 3) OR 5  5  2 3 B2 for  + oe   4  4  2  5  2 or B1 for  x −   4 

This question in 0607/21 May/June 2025

5 Ganpreet and Rahul share $240 in the ratio 7 : 5. (a) Show that the value of Rahul’s share is $100. [1] (b) Ganpreet spends $x of her share. Rahul spends $x of his share. The ratio of their remaining money is Ganpreet : Rahul = 2 : 1. Find the value of x. x = … [3]

4 marks

Mark scheme: 5(a) 240 240 1  5 or  5 7 + 5 12 5(b) 60 3 M2 for 140 – x = 2(100 – x) oe or M1 for 140 – x or 100 – x

This question in 0607/23 May/June 2025

6 The cost of a bar of chocolate is $c. The cost of a packet of sweets is $p. The total cost of 5 bars of chocolate and 2 packets of sweets is $13. The total cost of 7 bars of chocolate and 4 packets of sweets is $20. By forming a pair of simultaneous equations, find the cost of one bar of chocolate and the cost of one packet of sweets. cost of one bar of chocolate = $ … cost of one packet of sweets = $ … [5]

5 marks

Mark scheme: 6 5c + 2p = 13 2 B1 for each 7c + 4p = 20 [c = ] 2 3 M1 for correctly eliminating one variable [p = ] 1.5[0] A1 for 2 A1 for 1.5 If 0 scored, SC1 for answers that satisfy one of their equations

This question in 0607/23 May/June 2025

15 f ( x) = 3x - 1 g ( x) = 3 - 2x h ( x) = x 2 - 2x + 3 (a) Find f ( -3) . … [1] (b) Find f -1 ( 10 ) . … [2] (c) Find and simplify gh(x). … [2] (d) Find g -1 ( x) . g -1 ( x) = … [2] (e) Solve h ( x) = f ( x) . x = … or x = … [4]

11 marks

Mark scheme: 15(a) –10 1 15(b) 11 2 x + 1 oe M1 for 3x – 1 = 10 or 3 3 15(c) –2x2 + 4x – 3 final answer 2 M1 for 3 – 2 (x2 – 2x + 3) 15(d) 3 − x 2 M1 for x = 3 – 2y or y – 3 = –2x or oe final answer y 3 2 = − x 2 2 15(e) 1, 4 4  9 B3 for (x – 1)(x – 4) or  x = 5 2 or B2 for x2 – 5x + 4 [=0] or B1 for x 2 − 2 x + 3 = 3 x − 1 If 0 scored SC1 for one correct solution

This question in 0607/21 Oct/Nov 2025