Solve simultaneous equations in two — Paper 2 · IGCSE Mathematics - Additional 0606

11 y y = x3 + 4x2 – 5x + 5 A B C y = 5 E O D x The diagram shows part of the curve y = x 3 + 4 x 2 - 5 x + 5 and the line y = 5. The curve and the line intersect at the points A, B and C. The points D and E are on the x-axis and the lines AE and CD are parallel to the y-axis. (i) Find y ( x 3 + 4x 2 - 5x + 5 )d x . [2] (ii) Find the area of each of the rectangles OEAB and OBCD. [4] (iii) Hence calculate the total area of the shaded regions enclosed between the line and the curve. You must show all your working. [4] Question 12 is printed on the next page.

10 marks

Mark scheme: 11(i) x 4 4 x 3 5 x 2 B2 B1 for any 3 correct terms + − + 5 x [ + c ] isw 4 3 2 11(ii) x 3 + 4 x 2 − 5 x + 5 = 5 and rearrange to B1 y x ( x 2 + 4 x − 5 ) = 0 oe soi A(−5, 5) B C(1, 5) E (−5, 0) O D(1, 0) x Solves their x 2 + 4 x − 5[ = 0] soi M1 x = –5, x = 1 soi A1 OEAB = 25, OBCD = 5 A1 11(iii) Correct or correct FT substitution of 0, their −5 M1 dependent on at least B1 in (i)  x 0 4 4 x 3 5 x 2  seen in  + − + 5 x   4 3 2  their − 5 Correct or correct FT substitution of their 1, 0 seen M1 dependent on at least B1 in (i)  x their 1 4 4 x 3 5 x 2  in  + − + 5 x   4 3 2  0 1175 49 M1 for the strategy needed to combine the their − theirOEAB + theirOBCD − their oe areas; may be in steps; 12 12 97.916& − 25 + 5 − 4.083& 886 5 A1 all method steps must be seen; not from oe or 73 oe or 73.83& rot to 3 or more sig wrong working 12 6 figs If M0 then allow SC3 for 0 3 2 1 3 2 ∫−5 ( x + 4 x − 5 x ) dx − ∫0 ( x + 4 x − 5 x ) d x oe  x 4 4 x 3 5 x 2  0  x 4 4 x 3 5 x 2  1 =  + −  −  + −   4 3 2  − 5  4 3 2  0   625 500 125     1 4 5   =  0 −  − −   −   + −  − 0    4 3 2     4 3 2   443 = oe 6 or SC2 for 0 3 2 their 1 3 2 ∫their ( − 5) ( x + 4 x − 5 x ) d x − ∫0 ( x + 4 x − 5 x ) dx oe  x 4 4 x 3 5 x 2  0  x 4 4 x 3 5 x 2  their 1 =  + −  −  + −   4 3 2  their ( −5)  4 3 2  0 = [ F (0) − F (their ( −5)) ] − [ F (their1) − F (0) ]

This question in 0606/22 May/June 2017

8 The matrix A is . 4 3 (i) Find 2A -1 . [3] ^ h (ii) Hence solve the simultaneous equations 2y + 4x + 5 = 0 , 6y + 8x + 9 = 0 . [4]

7 marks

Mark scheme: 8(i)  4 2  B1 2A=    8 6  −1 1  6 − 2  B2  6 − 2  ( 2 A ) =   B1 for   8  − 8 4   − 8 4  1 B1 for 8 8(ii) 4 x + 2 y = −5 B1 8 x + 6 y = − 9  − 5  M1 Allow recovery Pre multiply   by a 2 × 2 matrix.  − 9   x  1  6 − 2  − 5  M1  − 5    =    Pre multiply their   by their  y  8  − 8 4  − 9   − 9  answer to (i)  x  1  − 12   − 1.5  A2 A1 for x value   =   =   A1 for y value oe  y  8  4   0.5  Allow both unsimplified

This question in 0606/22 Oct/Nov 2017

2 F C There are 105 boys in a year group at a school. Some boys play football (F) and some play cricket (C). • x boys play both football and cricket. • The number of boys that play neither game is the same as the number of boys that play both. • 40 boys play cricket. • The number of boys that only play football is twice the number of boys that only play cricket. Complete the Venn diagram and find the value of x. [5]

5 marks

Mark scheme: 2 ′ B1 n ( F ∩ C ) = n ( F ∪ C ) = x n ( C ∩ F ′ ) = 40 − x B1 n ( F ∩ C ′ ) = 80 − 2 x or 2 ( 40 −x ) B1 x + x + 40 − x + 80 − 2 x = 105 M1 x = 15 A1 cao

This question in 0606/22 Oct/Nov 2018

5 Solve the simultaneous equations 8 p + 1 11 q = 2 , 4 3 2p + 5 3q = 9 . [5] 1 27 3

5 marks

Mark scheme: 5 3( p +1) 2 p + 5 M1 2 11 3 2 ( 3 q ) = 2 or = 3 2 2 q 3 1  3  3  x a a − b a b a + b M1 Use = x or x × x = x x b 3 p + 3 − 2 q = 11 and 2 p + 5 −=1 6 q A1 Allow unsimplified M1 solve p = 4 and q = 2 A1

This question in 0606/22 Oct/Nov 2018

4 Do not use a calculator in this question. Solve the following simultaneous equations, giving your answers for both x and y in the form a + b 2 , where a and b are integers. 2x + y = 5 3x - 2 y = 7 [5]

5 marks

Mark scheme: 4 Eliminate x or y M1 7 + 5 2 1 A1 x = or y = 3 + 2 2 3 + 2 2 Multiply numerator and denominator by M1 3 − 2 2 x = 1 + 2 A1 y = 3 − 2 2 A1

This question in 0606/21 Oct/Nov 2019

6 Do not use a calculator in this question. The curve xy = 11x + 5 cuts the line y = x + 10 at the points A and B. The mid-point of AB is the point C. Show that the point C lies on the line x + y = 11. [7]

7 marks

Mark scheme: 6 Eliminate y M1 x 2 − x − 5 = 0 A1 Use formula M1 1 ± 21 A1 x = 2 21 ± 21 A1 y = 2 Find mid-point M1 ( 0.5 ,1 0.5 ) Show that mid-point lies on x + y = 11 A1

This question in 0606/21 Oct/Nov 2019

7 (a) Solve the simultaneous equations 10 x + 2 y = 5, 10 3 x + 4 y = 50 , giving x and y in exact simplified form. [4] 3 - x 3 - 10 = 0. [3] (b) Solve 2x 2 1

7 marks

Mark scheme: 7(a) x + 2y = lg5 or B1 3x + 4y = lg50 Solves their linear simultaneous M1 equations x = lg2 or equivalent simplified form A1 1 5 A1 If A0 A0 then SC1 for a correct pair of y = lg or equivalent simplified unsimplified values or a correct pair of 2 2 decimal values correct to at least 3sf form 7(b)  1  1  M1  x 3 + 2  2 x 3 − 5  oe       1 5 M1 x 3 = −,2 2 125 A1 x = –8, 8

This question in 0606/21 May/June 2020

5 Solutions to this question by accurate drawing will not be accepted. The points A and B are (4, 3) and (12, −7) respectively. (a) Find the equation of the line L, the perpendicular bisector of the line AB. [4] (b) The line parallel to AB which passes through the point (5, 12) intersects L at the point C. Find the coordinates of C. [4]

8 marks

Mark scheme: 5(a) Finds coordinates of mid-point B1 (8, –2) 3 + 7  5  B1 m AB =  = −  oe soi 4 − 12  4  −1 M1 mL = oe − 54 4 A1 y + 2 = ( x − 8) oe isw 5 5(b) 5 B1 y − 12 = − ( x − 5) 4 Attempts to solve their equations M1 (13, 2) A2 A1 for x = 13 or y = 2

This question in 0606/22 May/June 2020

2 Find the coordinates of the points of intersection of the curve x 2 + xy = 9 and the line y = x - 2 . 3 [5]

5 marks

Mark scheme: 2 2  2  M1 Eliminate y x + x  x − 2  = 9  3  5 x 2 − 6 x − 27 = 0 A1 ( x − 3 )( 5 x + 9 ) = 0 M1 Factorise or formula (3, 0) A1 Or both x values  9 16  A1  − , −   5 5 

This question in 0606/21 Oct/Nov 2020

2 Solve the simultaneous equations. x 2 + 3xy = 4 2x + 5y = 4 [5]

5 marks

Mark scheme: 2 2  4 − 2 x  M1 eliminate x or y x + 3 x   = 4  5  x 2 − 12 x + 20 ( = 0 ) A1 3 terms on one side if eliminating y 5 y 2 + 16 y ( = 0 ) if eliminating x ( x − 2 )( x − 10 ) ( = 0 ) M1 or y ( 5 y + 16 ) ( = 0 ) x = 2 or x = 10 nfww A1 or correct pair 16 A1 y = 0 or y = − nfww 5

This question in 0606/23 Oct/Nov 2020

5 Solve the following simultaneous equations. 3 x # 9 y - 1 = 243 1 2 x + 1 y - 2 2 8 # 2 = [5] 4 2

5 marks

Mark scheme: 5 express an equation correctly in powers M1 of 3 or powers of 2 x + 2 y − 2 = 5 oe (x + 2y = 7) A1 accept unsimpified 2 x + 1 − 2.5 = 3 + y − 0.5 oe (2x – y = 4) A1 accept unsimplified solve correct equations for x or y M1 x = 3 and y = 2 A1

This question in 0606/23 Oct/Nov 2020

9 Solve the following simultaneous equations. 4 x 2 + 3 xy + y 2 = 8 xy + 4 = 0 [6]

6 marks

Mark scheme: 9 Correctly eliminates x or y e.g. M1 2  4   4  2 4 x + 3 x  −  +  −  = 8 oe  x   x   4  2  4  2 or 4  −  + 3  −  y + y = 8 oe  y   y  Rearranges to a 3-term quadratic in x2 A1 or y2 soi e.g. 4 x 4 − 20 x 2 + 16 = 0 or y 4 − 20 y 2 + 64 = 0 Factorises or solves their 3-term M1 quadratic in x2 or y2 soi : (x2 – 1)(x2 − 4) or (y2 – 16)(y2 − 4) x 2 = 1 , x 2 = 4 oe, nfww A1 or y 2 = 16 , y 2 = 4 oe, nfww x = ±1 x = ± 2 A2 A1 for all 4 x values or all 4 y values y = ∓4 y = ∓2 oe, nfww

This question in 0606/22 May/June 2021

2 Solve the following simultaneous equations. xy + x 2 = 15 y + 3 x = 11 [5]

5 marks

Mark scheme: 2 Eliminate one unknown M1 x(11− 3x) + x2 =15 2 A1 2x −11x + 15[ = 0] Factorises or solves their 3-term quadratic M1 5 7 A2 5 x = , y = A1 for x = , x = 3 nfww 2 2 2 x = 3, y = 2 7 or y = , y = 2 nfww 2

This question in 0606/21 Oct/Nov 2021

5 (a) Solve the following simultaneous equations. e x + e y = 5 2 e x - 3e y = 8 [5] (b) Solve the equation e ( 2 t - 1 ) = 5e ( 5 t - 3 ) . [4]

9 marks

Mark scheme: 5(a) Solves 3e x + 3e y = 15 and 2e x − 3e y = 8 oe M1 by elimination as far as 3 e x + 2e x = 23 or substitutes e y = 5 − e x into 2e x − 3e y = 8 oe OR Solves 2e x + 2e y = 10 and 2e x − 3e y = 8 oe by elimination as far as 2e y + 3e y = 2 or substitutes e x = 5 − e y into 2e x − 3e y = 8 oe x 23 y 2 A1 e = or e = oe 5 5 x = ln4.6 [ = 1.53] oe A1 If M0 scored SC1 for using their expression of the form cex = d or y = ln0.4 [ = −0.916 ] oe d d to give x = ln provided > 0 c c Finds the other value, e y or e x , by substituting their ex M1 FT their ex or ey or ey y = ln0.4 [ = −0.916 ] oe A1 or x = ln4.6 [ = 1.53] oe 5(b) 2 t −−1 ( 5 t − 3 ) 5 t −−3 ( 2 t 1) 1 M1 e = 5 or e −= oe 5 2 − 3 t 3 2 1 A1 e = 5 or e t−= 5 1 M1 FT their e a − bt = 5 or 2 − 3t = ln5 or 3t − 2 = ln 5 ct d 1 their e −= where a, b, c and 5 d are positive integers 2 − ln5 2 + ln0.2 A1 t = or t = or 0.13[0] oe 3 3 Alternative method ln e 2 t −1 = ln5 + lne 5 t − 3 oe (M1) (2t − 1)[lne] = ln5 + (5t − 3) [ lne ] oe (A1) 5t – 2t = 3 – 1 – ln5 oe (M1) Dep on one correct log law applied with at most one sign error 2 − ln5 2 + ln0.2 (A1) t = or t = or 0.13[0] oe 3 3

This question in 0606/22 Oct/Nov 2021

2 Solve the following simultaneous equations. Give your answers in the form a + b 3 , where a and b are rational. x + y = 3 2x - 3 y = 5 [5]

5 marks

Mark scheme: 2 Solves 2 x + 2 y = 6 and 2 x − 3 y = 5 oe M1 by elimination as far as 2 y + 3 y = 1 or substitutes x = 3 – y into 2 x − 3 y = 5 oe OR solves 3 x + 3 y = 3 3 and 2 x − 3 y = 5 oe by elimination as far as 2 x + 3 x = 3 3 + 5 or substitutes y = 3 – x into 2 x − 3 y = 5 oe 1 3 3 + 5 A1 y = or x = 2 + 3 2 + 3 1 2 − 3 3 3 + 5 2 − 3 M1 FT their value of x or y providing y = × oe or x = × oe of equivalent difficulty 2 + 3 2 − 3 2 + 3 2 − 3 y = 2 − 3 and x = 1 + 3 A2 A1 for either and no extra values

This question in 0606/23 Oct/Nov 2021

1 Solve the following simultaneous equations, giving your answers in the form a + b 7 where a and b are integers. x + 3y = 11 x - 7 y = 7 [5]

5 marks

Mark scheme: Question Answer Marks Guidance 1 Finds by elimination 3 y + 7 y = 4 oe M1 or substitutes x = 11 − 3 y into x − 7 y = 7 oe OR Finds by elimination 3 y + 7 y = 21 + 11 7 oe 11 − x or substitutes y = into x − 7 y = 7 3 oe 4 A1 y = 3 + 7 21 + 11 7 or x = 3 + 7 4 3 − 7 M1 FT their value of x or y providing of y =  oe equivalent difficulty 3 + 7 3 − 7 21 + 11 7 3 − 7 or x =  oe 3 + 7 3 − 7 y = 6 − 2 7 and x = 6 7 − 7 A2 A1 for either and no extra values

This question in 0606/21 Oct/Nov 2022

1 Solve the following simultaneous equations. x + 5y =- 4 3y - xy = 6 [5]

5 marks

Mark scheme: Question Answer Marks Guidance 1 3 y −−( 5 y − 4) y = 6 oe M1  −−4 x   −−4 x  or 3   − x   = 6  5   5   6  or x + 5   = − 4 oe  3 − x  5 y 2 + 7 y − 6 = 0 A1 or x 2 + x − 42 = 0 Factorises their 3-term quadratic M1 expression or solves their 3-term quadratic equation, e.g. (5y – 3)(y + 2) [= 0] or (x – 6)(x + 7) [= 0] x = 6, y = −2 A2 A1 for either x = 6, x = −7 x = −7, y = 0.6 or y = −2, y = 0.6 or for an x, y pair from a correct factorisation or correct solving of a correct equation. The method of solution must be seen in this case.

This question in 0606/22 Oct/Nov 2022

4 The line y = kx + 6 intersects the curve y = x 3 - 4x 2 + 3kx + 2 at the point where x = 2 . (a) Find the value of k. [2] (b) Show that, for this value of k, the line cuts the curve only once. [4]

6 marks

Mark scheme: 4(a) 2k + 6 = 8 − 16 + 6k + 2 oe M1 For equating line to curve and substituting x = 2, or vice versa k = 3 A1 4(b) x3 − 4 x 2 + ( 2  theirk ) x − 4  = 0 M1 FT their k in correct cubic or x3 − 4 x 2 + 6 x − 4  = 0 x 2 − 2 x + 2 A2 Correct quadratic factor from correct cubic A1 for a quadratic factor with two terms correct, from correct cubic ( −2) 2 − 4 (1)( 2 ) < 0 oe A1 Uses discriminant correctly on the correct quadratic factor or 4 – 8 < 0 oe [and so x = 2 is the only solution]

This question in 0606/23 Oct/Nov 2022

3 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Show that x + 3 is a factor of - 12 + 23x + 3x 2 - 2x 3 . [1] (b) The curve y =- 5 + 33 x + 3x 2 - 2x 3 and the line y = 10x + 7 intersect at three points, A, B and C. These points are such that the x-coordinate of A has the least value and the x-coordinate of C has the greatest value. Show that B is the mid-point of AC. [7]

8 marks

Mark scheme: 3(a) 12 – 69 + 27 + 54 = 0 B1 3(b) 10 x  7 2 x 3  3 x 2  33 x  5 oe, soi M1 Uses the correct factor x + 3 to find a M1 quadratic factor of the polynomial from part (a) oe with at least 2 terms correct 2 x 2  9 x  4 A1 or 2 x 2  9 x  4 Factorises or solves their 3-term quadratic DM1 dep on previous M1 factor = 0: (2 x  1)( x 4) or ( 2 x  1)( x  4) or (2 x  1)( x  4) oe x = 3, x = 0.5, x = 4 nfww A1 dep on at least M0 M1 A1 DM1 awarded A(3, 23), B(0.5, 12), C(4, 47) oe B2 dep on x = 3, x = 0.5, x = 4 nfww and correct method to show mid-point e.g.:  3 4 23  47   1  B1 dep on x = 3, x = 0.5, x = 4 nfww for ,  ,12 oe      2 2   2  A(3, 23), B(0.5, 12), C(4, 47) oe   0.5   3   3.5  AB   or            and  12   23   35  or   4   0.5   3.5  [x-coordinate of the mid-point ]  BC     47    12    35  oe 3 4 1  oe 3 4 1 2 2 OR [x-coordinate mid-point]  oe 2 2 and valid comment e.g. The points are collinear [so B is the mid-point of AC].

This question in 0606/22 May/June 2023

2 Solve the following simultaneous equations. 5x - 3 ln y = 2 [4] x + ln y = 1

4 marks

Mark scheme: 2 Correct method to eliminate y: M1 5 x − 3 (1 − x ) = 2 oe or adds 3x + 3lny = 3 5x – 3lny = 2 to obtain 3x + 5x = 3 + 2 or better 5 A1 x = oe 8 3 M1 ln y = oe 8 3 A1 y = e 8 oe or 1.45 2 Alternative method Correct method to eliminate x: (M1) 5(1 – ln y) – 3ln y = 2 oe or subtracts 5x – 3lny = 2 from 5x + 5lny = 5 to obtain 5lny – (–3lny) = 5 – 2 or better 3 (M1) ln y = oe 8 3 (A1) y = e 8 oe or 1.45 5 (A1) x = oe 8

This question in 0606/21 Oct/Nov 2023

1 (a) A straight line passes through the points (4, 23) and (-8, 29). Find the point of intersection, P, of this line with the line y = 2x + 5 . [5] (b) Find the distance of P from the origin. [2]

7 marks

Mark scheme: Question Answer Marks Guidance 1(a) 1 3 29 − 23 1 y = − x + 25 isw M1 for m = oe or − 2 −−8 4 2 and y − 23  1  M1 FT for = their  −  oe x − 4  2  or 1 1 y = their − x + c and 23 = −  4 + c oe ( 2 ) 2 OR M1 for solving 23 = 4m + c 29 = –8m + c 1 for m = − or c = 25 2 and M1 FT for correctly using their m or their c to find c or m Solves their linear equation simultaneously M1 1 FT their y = − x + 25 oe with y = 2x + 5 to find x or y 2 (8, 21) A1 1(b) 2 2 M1 FT their (8, 21) 8 + 21 oe 505 isw or 22.5 A1 or 22.47[22…] rot to 2 or more dp

This question in 0606/22 Oct/Nov 2023

3 (a) Solve the following simultaneous equations. 3 log 2 x + 2 log 2 y = 24 5 log 2 x - 3 log 2 y = 2 [5] 2 t + 4 (b) Solve the equation = 512 . [4] l - 2 t 2

9 marks

Mark scheme: 3(a) Correctly eliminates log 2 x or log 2 y M1 A correct equation in log 2 x only or log 2 y only x = 16 or y = 64 A2 A1 for log 2 x = 4 or log 2 y = 6 y = 64 or x = 16 A2 A1 for log 2 y = 6 or log 2 x = 4 Alternative method 3 2 24 x 5 2 (M1) x y = 2 and = 2 oe y 3 y = 64 or x = 16 (A2) A1 for y19 = 2114 oe or x19 = 276 oe x = 16 or y = 64 (A2) A1 for x3 642 = 224 oe x 5 2 or 3 = 2 oe 64 OR 163  y2 = 224 oe 16 5 2 or 3 = 2 oe y 3(b) t + 4 −−(1 2 t ) 9 B2 t + 4 −−(1 2 t ) 2 = 2 B1 for 2 = 512 or 2t + 4 = 29 +1− 2 t oe, soi 2 t + 4 9 or 1− 2 t = 2 soi 2 OR OR t + 4 − (1 − 2t ) = log 2 512 oe, soi ( t + 4 ) log a 2 − (1 − 2t )log a 2 = log a 512 log a 512 or t + 4 − (1 − 2t ) = oe log a 2 3t + 3 = 9 or better M1 t = 2 A1

This question in 0606/23 Oct/Nov 2023

7 (a) The curves 4x 2 - 3y 2 + xy = 24 and y = intersect at the points P and Q. Find the coordinates x of P and Q. [5] (b) Find the length of PQ. Give your answer in the form a b, where a is rational and b is the smallest possible integer. [2]

7 marks

Mark scheme: 7(a) Correctly eliminates x or y e.g. M1 2  2  2  2  4 x − 3   + x   = 24 oe or  x   x   2  2 2  2  4 − 3 y + y = 24 oe      y   y  Rearranges to a 3-term quadratic in x2 or y2 A1 soi e.g. 4 x 4 − 22 x 2 − 12  = 0  or 2 x 4 − 11x 2 − 6  = 0  or 3 y 4 + 22 y 2 − 16  = 0  oe Factorises or solves their 3-term quadratic in M1 x2 or y2 soi e.g. (2x2 + 1)(x2 − 6) or (3y2 – 2)(y2 + 8) 2 2 2 A1 x = 6 oe, nfww or y = nfww 3  2   2  A1 and no other values;  6,    or   6,   oe, nfww dep on at least the first M1 A1  6   3  7(b) 2 2 M1 FT providing their xP, xQ and their yP, ( xP − xQ ) + ( y P − yQ ) oe, soi yQ are non-zero 4 A1 15 3

This question in 0606/22 Feb/March 2024

4 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Find the exact distance between the two points where the curve 9 ( x - 1) 2 + 4 ( y - 3) 2 = 36 cuts the y-axis. [4] (b) Find the coordinates of the points where the curve with equation 2x 2 + 83 xy = x 3 y - 20 x 1 intersects the curve with equation y = . Give each of your answers in the form a + b c , where x a and b are rational and c is the smallest integer possible. [6]

10 marks

Mark scheme: 4(a) 4( y  3) 2  36  9 or 4 y 2  24 y  9[  0] M1 27 A1 y  3  or exact equivalent, soi 4 27  27  M1 FT their a  b providing b is not a 3    3   oe 4  4  square number 27 or 3 3 or exact equivalent, nfww A1 4(b) Eliminates one unknown and simplifies M1 terms: 2 x 2  83  x 2  20 x oe, soi x 2  20 x  83  0 A1 Applies quadratic formula or completes the M1 FT their 3-term quadratic 20  20 2  4[1](83) square: x  2 x 10  17 A1 1 10  17 M1 FT their x = a  b providing y   10  17 10  17 previous M1 awarded 10 17 A1 dep on all marks previously awarded x 10  17 , y   83 83

This question in 0606/21 May/June 2024