Equations — Paper 1 · IGCSE Mathematics 0580

5 Solve the equation 5x – 7 = 8. Answer x = [2]

2 marks

Mark scheme: 5 5x = 8 + 7 or better seen. M1 (Correct first step) (x = ) 3 A1

This question in 0580/11 Oct/Nov 2005

13 Solve the simultaneous equations 3x − y = 18, 2x + y = 7. Answer x = y = [3]

3 marks

Mark scheme: 13 (x=) 5, (y=)–3 3 M1 correct method to eliminate y or x. (add equations or correct multiply and subtract) A1, A1 ww allow SC1 for 1 correct answer. ww both correct, full marks.

This question in 0580/11 May/June 2006

11 Solve the equation 5x − 2 = 10x − 8. For Examiner's Use Answer x = [2]

2 marks

Mark scheme: 11 6 2 M1 for -2 + 8 = 10x – 5x oe or better. (x = ) oe isw 5

This question in 0580/11 Oct/Nov 2006

3 Solve the equation 1 – 2x = x + 4. Answer x = [2]

2 marks

Mark scheme: 3 (x = ) − 1 2 M1 for 1 − 4 = x + 2x oe Not embedded unless x = –1 seen.

This question in 0580/11 Oct/Nov 2007

9 Solve the equation 5x + 2 = 53. Answer x = [2]

2 marks

Mark scheme: 9 1 2 M1 for (53 − 2) ÷ 5 soi (x=) 10.2 or 10 isw 5

This question in 0580/11 Oct/Nov 2008

10 Solve the simultaneous equations For Examiner's 5x − y = 15, Use 7x − 5y = 3. Answer x = y = [3]

3 marks

Mark scheme: 10 (x =) 4 and (y =) 5 www 3 M1 for complete correct method for one value A1 for 1 correct answer. ww both correct W3 ww one correct W0 Reversed answer, look in working to be convinced of transcription error.

This question in 0580/11 Oct/Nov 2009

10 Solve the simultaneous equations ForFor Examiner'sExaminer's 5x − 3y = 3, UseUse 6x − y = 14. Answer x = y = [3]

3 marks

Mark scheme: 10 (x =) 3 and (y =) 4 www 3 M1 for complete correct method for one value A1 for 1 correct answer. ww both correct W3 ww one correct W0 Reversed answer, look in working to be convinced of transcription error.

This question in 0580/12 Oct/Nov 2009

md 15 J = 3 (a) Find the value of d when J = 32 and m = 8. Answer(a) d = [2] (b) Make d the subject of the formula. Answer(b) d = [2]

4 marks

Mark scheme: 8d 15 (a) 12 2 M1 for 32 = or better. 3 3 J J d (b) (d =) 2 M1 for 3J = md or = m m 3 3

This question in 0580/12 Oct/Nov 2009

8 Examiner's Use 2x – 7 NOT TO SCALE x + 3 x The lengths, in centimetres, of the sides of a triangle are x , x + 3 and 2x − 7. The perimeter of the triangle is 52 cm. (a) Use this information to write down an equation in x. Answer(a) [1] (b) Find the value of x. Answer(b) x = [2]

3 marks

Mark scheme: 8 (a) x + x + 3 + 2x – 7 = 52 or better 1 (b) 14 2ft W1 for 4x or 56 seen Follow through their (a) if linear and equal to 52 for 1 or 2 marks.

This question in 0580/11 May/June 2010

13 Solve the simultaneous equations. 2x − y = 9 7x + 2y = 26 Answer x = y = [3]

3 marks

Mark scheme: 13 (x =) 4 3 M1 for multiplying and subtracting or adding as (y =) − 1 appropriate. (allow errors in arithmetic operations) or any other correct methods. A1 for one correct variable

This question in 0580/11 May/June 2010

8 Make p the subject of the formula m = p2 − 2. Answer p = [2]

2 marks

Mark scheme: 8 (±)√(m + 2) final answer 2 W1 for p2 = m + 2 or ft square root after incorrect first step(s). SC1 answer of (±)√m + 2

This question in 0580/12 May/June 2010

14 Solve the simultaneous equations. 3x − 2y = 15 2x + y = 17 Answer x = y = [3]

3 marks

Mark scheme: 14 (x) = 7, (y) = 3, www 3 M1 for multiplying and subtracting or adding as appropriate. (allow errors in arithmetic operations) or any other correct methods A1 for one correct variable.

This question in 0580/12 May/June 2010

17 Solve the simultaneous equations. 3x + y = 19 5x – y = 13 Answer x = y = [3]

3 marks

Mark scheme: 17 (x =) 4 (y =) 7 www 3 M1 for adding or multiplying and subtracting (allow errors in arithmetic operations) or any other correct methods A1 for one correct variable.

This question in 0580/13 May/June 2010

12 Solve the simultaneous equations. 3 x + y = 18 4 x − 2 y = 34 Answer x = y = [3]

3 marks

Mark scheme: 12 (x = ) 7 3 M1 for multiplying/dividing and adding/ (y =) −3 subtracting or other complete correct method A1 for one correct variable 4 

This question in 0580/11 Oct/Nov 2010

8 (a) Factorise completely. For 8pq + 12pr Examiner's Use Answer(a) [2] (b) Use your answer to part (a) to make p the subject of the formula below. s = 8pq + 12pr Answer(b) p = [1]

3 marks

Mark scheme: 8 (a) 4p(2q + 3r) 2 B1 for p(8q + 12r) or 2p(4q + 6r) or 4p(aq + br) a, b integers or 4(2pq + 3pr) s (b) (p =) oe 1ft ft if p is a common factor in (a) or in working in (b) 4( 2 q + 3r )

This question in 0580/12 Oct/Nov 2010

6 Solve the equation. 2 x + 1 = 4 3 Answer x = [2]

2 marks

Mark scheme: 6 5.5 2 M1 for 2x + 1 = 3 × 4 or better 2x 1 or = 4 − 3 3

This question in 0580/13 Oct/Nov 2010

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

2 marks

Mark scheme: 13 (x =) 2, (y =) −1 2 M1 for correct method for eliminating one variable. Subtract or multiply by 3 and 5, then subtract IGCSE – October/November 2010 0580 13

This question in 0580/13 Oct/Nov 2010

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

3 marks

Mark scheme: 10 x = 13 3 M1 for consistent multiplication and y = −9 addition/subtraction. A1 for x = 13 or A1 for y = –9 26 7 5 13 7 2 7 1 7 2

This question in 0580/11 May/June 2011

19 Piet, Rob and Sam collect model aeroplanes. For Piet has x aeroplanes. Examiner's Rob has 7 more aeroplanes than Piet. Use Sam has three times as many aeroplanes as Piet. (a) Write down an expression, in terms of x, for (i) the number of aeroplanes Rob has, Answer(a)(i) [1] (ii) the number of aeroplanes Sam has. Answer(a)(ii) [1] (b) The total number of aeroplanes is 32. (i) Use the information in part (a) to write down an equation in x. Answer(b)(i) [1] (ii) Solve your equation. Answer(b)(ii) x = [2] (c) Write down the number of aeroplanes Rob has. Answer(c) [1]

6 marks

Mark scheme: 19 (a) (i) x + 7 1 (ii) 3x 1 (b) (i) x+their (a)(i)+their (a)(ii)=32 1ft ft dependent on 2 algebraic expressions in (a) or better (ii) (x =) 5 2ft M1 for 5x = 32 − 7 oe ft their (b)(i) with M1 for ax = b and A1 if answer is an integer. (c) 12 1ft ft their (b)(ii) substituted into their (a)(i) or their (b)(ii) + 7 evaluated correctly

This question in 0580/11 May/June 2011

16 Solve the simultaneous equations. For x + 2y = 3 Examiner's 2x – 3y = 13 Use Answer x = y = [3]

3 marks

Mark scheme: 16 (x =) 5 (y =) −1 3 M1 for consistent multiplication and add/subtract as appropriate. A1 for 1 correct answer.

This question in 0580/13 May/June 2011

12 y 9 8 7 6 5 4 3 2 1 x – 4 – 3 – 2 – 1 0 1 2 The diagram shows the graph of y = (x + 1)2 for −4 Y x Y 2. (a) On the same grid, draw the line y = 3. [1] (b) Use your graph to find the solutions of (x + 1)2 = 3. Give each solution correct to 1 decimal place. Answer(b) x = or x = [2]

3 marks

Mark scheme: 12 (a) Correct ruled line 1 (b) −2.7, 0.7 1, 1ft B2ft their ruled line through (0, 3) for two intersections given to 1 decimal place or B1 for –2.70 to –2.75 and 0.70 to 0.75 or B1ft their ruled line through (0, 3) for two intersections not given to 1 decimal place

This question in 0580/11 Oct/Nov 2011

11 NOT TO SCALE x° 75° 2x° 2x° (a) For the diagram above, write down an equation in x. Answer(a) [1] (b) Solve your equation. Answer(b) x = [2]

3 marks

Mark scheme: 11 (a) x + 2x + 2x + 75 = 360 1 Allow 4x + x + 75 = 360 or 5x + 75 = 360 or 5x = 285 (b) (x =) 57 cao 2 M1 correct first step after 5x + 75 = 360 ie 5x = 360 − 75 or x + 15 = 72 If zero SC1 for correct solution to their linear equation seen in part (a) or in part (b) if (a) is blank IGCSE – October/November 2011 0580 12 1 6 4 3 13 25

This question in 0580/12 Oct/Nov 2011

13 Solve these simultaneous equations. 5x – 2y = 17 2x + y = 5 Answer x = y = [3]

3 marks

Mark scheme: 13 (x =) 3 (y =) –1 www 3 M1 for consistent multiply and consistent add/ subtract as appropriate Allow computational but not method errors Likely 5x + 4x = 17 + 10 Other methods allowed A1 for correct x or y

This question in 0580/12 Oct/Nov 2011

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

2 marks

Mark scheme: 10 (x =) −3 (y =) 5 2 M1 for correctly eliminating one variable

This question in 0580/13 Oct/Nov 2011

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

2 marks

Mark scheme: 11 (x =) 3.5 2 M1 for 2x − 3 = 2 × 2 or better 2x 3 = 2 + 2 2 5

This question in 0580/13 Oct/Nov 2011

8 Solve the equation 4x – 2 = 7 . Answer x = [2]

2 marks

Mark scheme: 2 7 8 2.25 oe 2 M1 4x = 7 + 2 OR x – = or better 4 4

This question in 0580/11 May/June 2012

16 Solve the simultaneous equations. For 3x + 5y = 24 Examiner's x + 7y = 56 Use Answer x = y = [3]

3 marks

Mark scheme: 16 x = −7 3 M1 for consistent multiplication and addition/ y = 9 subtraction as appropriate. Allow computational errors A1 for x= –7 or y = 9

This question in 0580/11 May/June 2012

12 Solve the simultaneous equations. 4x + y = 18 5x + 3y = 19 Answer x = y = [3]

3 marks

Mark scheme: 12 [x =] 5, [y =] –2 3 M1 for consistent multiply and add/subtract as appropriate. Allow computational errors. Other methods allowed. A1 for correct x or y. 4

This question in 0580/13 May/June 2012

16 (a) Solve the equation 5(x − 3) = 21 . Answer(a) x = [2] (b) Make x the subject of the equation y = 3x − 2 . Answer(b) x = [2]

4 marks

Mark scheme: 16 (a) 7.2 oe 2 M1 for 5x – 15 = 21 or x – 3 = 215 y + 2 (b) [x =] 3 2 M1 for 3x = y + 2 or – 3x = –2 – y

This question in 0580/13 May/June 2012

5 Solve the following equations. (a) x + 9 = 16 Answer(a) x = [1] (b) 6y = 27 Answer(b) y = [1]

2 marks

Mark scheme: 5 (a) 7 1 (b) 4.5 or 4½ 1

This question in 0580/11 Oct/Nov 2012

9 Rearrange this equation to make b the subject. For Examiner's Use b a = – 9 5 Answer b = … [2] _____________________________________________________________________________________

2 marks

Mark scheme: 9 [b = ] 5(a + 9) oe final answer 2 M1 for one correct step

This question in 0580/11 May/June 2013

19 (a) Multiply out the brackets. 5 ( x + 3 ) Answer(a) … [1] (b) Factorise completely. 12xy – 3x2 Answer(b) … [2] (c) Solve. 5x – 24 = 51 Answer(c) x = … [2] _____________________________________________________________________________________ Question 20 is printed on the next page.

5 marks

Mark scheme: 19 (a) 5x + 15 final answer 1 (b) 3x( 4y – x ) final answer 2 B1 for 3( 4xy – x2 ) or x( 12y – 3x ) (c) 15 2 M1 for a correct first step

This question in 0580/11 May/June 2013

22 Solve the equation. 5(2y – 17) = 60 Answer y = … [3] _____________________________________________________________________________________

3 marks

Mark scheme: 22 14.5 oe 3 M2 for complete correct method or M1 for one correct step

This question in 0580/12 May/June 2013

23 (a) Simplify y 0. Answer(a) … [1] 1 (b) Make v the subject of E = 2 mv 2. Answer(b) v = … [3] _____________________________________________________________________________________

4 marks

Mark scheme: 23 (a) 1 1 2 E E E 2 E or or (b) [v =] 1 3 M2 for v2 = m 5.0 m m m 2 E or M1 for mv2 = 2E or ½ v2 = m IGCSE – May/June 2013 0580 12

This question in 0580/12 May/June 2013

5 Solve the equation 3x – 5 = 16 . Answer x = … [2] _____________________________________________________________________________________

2 marks

Mark scheme: 5 [x =] 7 2 M1 for correct first step 5 16 3x = 16 + 5 or x − = 3 3

This question in 0580/13 May/June 2013

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

2 marks

Mark scheme: 12 4.8 oe 2 M1 for 5 + 19 = 3x + 2x oe or better or B1 24 – 2x = 3x oe or 5 = 5x – 19 oe

This question in 0580/11 Oct/Nov 2013

22 (a) Factorise completely. 6ab – 24bc Answer(a) … [2] (b) Rearrange the following formula to make m the subject. m j = n – k Answer(b) m = … [2] _____________________________________________________________________________________

4 marks

Mark scheme: 22 (a) 6b( a – 4c) 2 B1 for answer 6( ab – 4bc ) or 3b( 2a – 8c ) Final answer or 2b( 3a – 12c ) or b( 6a – 24c ) (b) n (j + k) or nj + nk oe 2 M1 for one correct step of a two-step method Final answer or SC1 for [m] = k + jn or [m] = j+ kn

This question in 0580/11 Oct/Nov 2013

17 Solve the simultaneous equations. For Examiner′s 2x + 5y = 26 Use 4x + 3y = 24 Answer x = … y = … [3] _____________________________________________________________________________________

3 marks

Mark scheme: 17 [x =] 3, [y =] 4 3 M1 for correctly eliminating one variable A1 for [x =] 3 A1 for [y =] 4 If zero scored, SC1 for correct substitution and evaluation to find the other variable. 7

This question in 0580/12 Oct/Nov 2013

18 Solve the simultaneous equations. For Examiner′s 5x + 6y = 3 Use 4x – 3y = 18 Answer x = … y = … [3] _____________________________________________________________________________________

3 marks

Mark scheme: 18 (x =) 3, (y =) −2 3 M1 for correctly eliminating one variable A1 for [x = ]3 A1 for [y =] −2 If zero scored, SC1 for correct substitution and evaluation to find the other variable 2 6

This question in 0580/13 Oct/Nov 2013

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

2 marks

Mark scheme: n 6 30 2 M1 for n – 8 = 22 or = 15 2  6 

This question in 0580/11 May/June 2014

9 Solve the simultaneous equations. 2x – y = 7 3x + y = 3 Answer x = … y = … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 9 [x =] 2, [y =] – 3 2 B1 B1 or SC1 for reversed answers [ ] 1 f 28 b

This question in 0580/12 May/June 2014

24 Solve the simultaneous equations. 2x + 3y = 29 5x + y = 27 Answer x = … y = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 24 x = 4 3 M1 for correct method to eliminate y = 7 one variable or (substitution) correct rearrangement of one equation seen substituted into the second equation. A1 for one correct answer. If M0 SC1 for both answers satisfying one of the original equations

This question in 0580/13 May/June 2014

26 (a) Factorise completely. 15a3 – 5ab Answer(a) … [2] (b) Simplify. 3x2y3 × x4y Answer(b) … [2] (c) Multiply out the brackets and simplify. 3(x – 2) – 4(2x – 3) Answer(c) … [2] (d) Solve the equation. 8x + 9 = 3(x + 8) Answer(d) x = … [3] __________________________________________________________________________________________

9 marks

Mark scheme: 26 (a) 5a(3a2−b) 2 B1 for a(15a2 − 5b) or 5(3a3− ab) (b) 3x6 y4 2 B1 for x6 or y4 in a product on answer line (c) 6 – 5x as final answer nfww 2 B1 for 3x – 6 or – 8x + 12 seen or SC1 for 6 or – 5x seen in final answer nfww (d) 3 nfww 3 M2 for 5x = 15 or B1 for 3x +24 seen or M1 for 8x – 3x = 3 × 8 − 9 or better. If zero, SC1 for answer [x =] − 1 5

This question in 0580/13 May/June 2014

20 Solve the simultaneous equations. 5x + 2y = 16 3x – 4y = 7 Answer x = … y = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 20 [x =] 3, [y =] 0.5 3 M1 for correct method to eliminate one variable A1 for [x =] 3 A1 for [y =] 0.5 If zero scored, SC1 for correct substitution and evaluation to find the other variable

This question in 0580/11 Oct/Nov 2014

21 (a) Find the value of 5x2 when x = –4. Answer(a) … [2] (b) Make x the subject of the formula y = 5x2. Answer(b) x = … [2] __________________________________________________________________________________________ Question 22 is printed on the next page.

4 marks

Mark scheme: 21 (a) 2 M1 for 5 × (− 4 )2 or 5× 4 2 or better 80 y y (b) [± ] or 2 M1 for correct first step 5 5 y 2 i.e. = x or y = 5 x Final answer 5 or correct 2nd step after incorrect 1st step seen

This question in 0580/11 Oct/Nov 2014

15 Solve the simultaneous equations. You must show all your working. 9x + 2y = 8 5x + 6y = –20 Answer x = … y = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 15 [x =] 2 [y =] −5 3 M1 for correct method to eliminate one variable A1 for x A1 for y If zero scored SC1 for correct substitution and evaluation to find the other variable. 136

This question in 0580/12 Oct/Nov 2014

10 Make x the subject of the formula y = 6x – 1. Answer x = … [2] __________________________________________________________________________________________

2 marks

Mark scheme: y + 1 y 1 10 oe 2 B1 for y + 1= 6 x or = x − 6 6 6 y − 1 y If B0 SC1 for or + 1 6 6

This question in 0580/13 Oct/Nov 2014

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

3 marks

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

This question in 0580/13 Oct/Nov 2014

18 (a) Solve the simultaneous equations. You must show all your working. 4x + 2y = 31 6x – 2y = 34 Answer(a) x = … y = … [2] (b) Factorise 14p2 + 21pq. Answer(b) … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 18 (a) [x =] 6.5 [y =] 2.5 2 B1 for x = 6.5 B1 for y = 2.5 If zero scored, SC1 for correct substitution and evaluation to find other variable or SC1 no working, 2 correct answers given. (b) 7p(2p + 3q) 2 B1 for 7(2p2 + 3pq) or p(14p + 21q)

This question in 0580/12 Feb/March 2015

18 Solve the equation. 5(3y – 2) = 35 Answer y = … [3] __________________________________________________________________________________________

3 marks

Mark scheme: 18 3 3 B1 for 15y − 10 seen or M1 for 3y – 2 = 35 ÷ 5 and M1 for 15y = 35 + their (5 × 2) or 3y = their (35 ÷ 5) + 2

This question in 0580/11 May/June 2015

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

3 marks

Mark scheme: 15 2.8 oe 3 M2 for 12 + 2 = 8x – 3x or better or M1 for 3x + 12 or 8x – 2

This question in 0580/13 May/June 2015

19 (a) s = 4t + 3u Calculate s when t = 2.6 and u = –0.4 . Answer(a) s = … [2] (b) Solve 5x – 7 = 10. Answer(b) x = … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 19 (a) 9.2 2 M1 for 4 × 2.6 + 3 × (–0.4) or better (b) 3.4 2 M1 for one correct step in a 2-step method

This question in 0580/11 Oct/Nov 2015

22 Solve the simultaneous equations. You must show all your working. 5x + 2y = 8 2x – 3y = 26 Answer x = … y = … [4] __________________________________________________________________________________________

4 marks

Mark scheme: 22 Correctly equating one set of M1 eg 10x + 4y = 16 and 10x − 15y = 130 coefficients or 15x + 6y = 24 and 4x – 6y = 52 Correct method to eliminate one M1 eg 19y = k or hx = 114 or 19x = m or ny = 76 variable [x =] 4 A1 [y =] −6 A1 If zero scored SC1 for 2 values satisfying one of the original equations. SC1 if no working shown, but 2 correct answers given

This question in 0580/13 Oct/Nov 2015

20 Solve the simultaneous equations. You must show all your working. 2x + 3y = 15 5x + 4y = 13 x = … y = … [4]

4 marks

Mark scheme: 20 Correctly equating one set of coefficients M1 Correct method to eliminate one variable M1 Dependent on first M1 scored [x =] −3 [y =] 7 A1 A1 If zero scored, SC1 for 2 values satisfying one of the original equations or 2 correct answers given but no working shown

This question in 0580/11 May/June 2016

19 Solve the simultaneous equations. Show all your working. 3x + 4y = 14 5x + 2y = 21 x = … y = … [3]

3 marks

Mark scheme: 19 Correctly eliminating one variable M1 If zero scored SC1 for x = 4 A1 2 values satisfying one of the original equations y = 0.5 oe A1 or if no working shown, but 2 correct answers given

This question in 0580/12 May/June 2016

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

2 marks

Mark scheme: 1 7 0.5 or 2 M1 for correct first step e.g. 6y + 6 = 9 2 9 or y + 1 = 6  −6 

This question in 0580/13 May/June 2016

16 Solve the simultaneous equations. You must show all your working. 3x + 7y = –21 6x + 4y = 3 x = … y = … [3]

3 marks

Mark scheme: 16 For correctly eliminating one M1 Or correctly rearranging one equation and variable substituting into the other [x =] 3.5 A1 [y =] −4.5 A1 If zero scored SC1 for 2 values satisfying one of the original equations or if no working shown but 2 correct answers given 24

This question in 0580/13 May/June 2016

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

3 marks

Mark scheme: 15 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 Page 4 Mark Sccheme Syllabuss Paper Cambridge IGCSE – Occtober/November 2016 0580 11

This question in 0580/11 Oct/Nov 2016

21 (a) Solve the equation. 4x + 3 = 11 x = … [2] (b) Make x the subject of the formula y = 4x2 – 2. x = … [3]

5 marks

Mark scheme: 21 (a) 2 2 M1 for one correct step 3 11 e.g. 4x = 11 – 3 or x + = 4 4 or better y + 2 (b) [x = ] or ( y + 2 ) /4 3 M1 for one correct step 4 y 2 y + 2 e.g. y + 2 = 4x2 or = x2 − or oe final answer 4 4 2 M1 for a further correct step y + 2 2 y 2 e.g. = x or + = x2 4 4 4

This question in 0580/11 Oct/Nov 2016

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

2 marks

Mark scheme: 11 21 2 M1 for k – 8 = 13 or 6k – 48 = 78 or better (13 + 16 ) × 4

This question in 0580/12 Oct/Nov 2016

21 (a) Simplify. 3p + 5p + p … [1] (b) Multiply out the brackets. 4(q – 3) … [1] (c) Factorise completely. 10t + 15t2 … [2] (d) Solve the simultaneous equations. You must show all your working. 3x + y = 7 5x – y = 17 x = … y = … [2]

6 marks

Mark scheme: 21 (a) 9p final answer 1 (b) 4q – 12 final answer 1 (c) 5t(2 + 3t) final answer 2 M1 for t(10 + 15t) or 5(2t + 3t2) (d) [x = ] 3, [y = ] –2 2 B1 for one correct with working with supporting working If zero scored, SC1 for 2 values satisfying one of the original equations or SC1 if no working shown, but 2 correct answers given

This question in 0580/12 Oct/Nov 2016

23 Solve the simultaneous equations. You must show all your working. 5x + 4y = 17 2x – 3y = 16 x = … y = … [4]

4 marks

Mark scheme: 23 [x =] 5 4 M1 for correctly equating one set of coefficients [y =] –2 M1 for correct method to eliminate one variable A1 for x = 5 A1 for y = –2 If zero scored, SC1 for 2 values satisfying one of the original equations. or SC1 if no working shown, but 2 correct answers given

This question in 0580/13 Oct/Nov 2016

22 Solve. (a) 4x = 10 x = … [1] (b) 5(x + 8) = 75 x = … [2] (c) 37 ÷ 3x = 9 x = … [1]

4 marks

Mark scheme: 1 22 (a) 2.5 or 2 1 2 (b) 7 2 M1 for 5 x + 40 = [ 75] or x + 8 = 75 ÷ 5 or better (c) 5 1

This question in 0580/12 Feb/March 2017

20 Solve the simultaneous equations. You must show all your working. 5x - 2y = 24 7x + 4y = -14 x = … y = … [3]

3 marks

Mark scheme: 20 Correctly eliminating one variable M1 [x =] 2 A1 [y =] −7 A1 If zero scored, SC1 for 2 values satisfying one of the original equations SC1 for both correct but no working

This question in 0580/11 May/June 2017

13 Solve. 2 - x = 5 x + 1 x = … [2]

2 marks

Mark scheme: 13 1 2 M1 for 2 – 1 = 5x + x oe oe 6

This question in 0580/12 May/June 2017

16 (a) 5x ÷ 25 = 56 Find the value of x. x = … [1] (b) Solve the simultaneous equations. You must show all your working. 2x + 3y = 16 - 2x + 5y = 24 x = … y = … [2]

3 marks

Mark scheme: 16(a) 8 1 16(b) [x = ] 0.5 1 [y = ] 5 1 If zero scored, SC1 for correct substitution and evaluation to find the other variable

This question in 0580/13 May/June 2017

18 Solve the simultaneous equations. You must show all your working. 3x + 4y = 6 6x - y =- 15 x = … y = … [3]

3 marks

Mark scheme: 18 Correct method to eliminate one M1 variable [x =] –2 A1 [y =] 3 A1 If zero scored, SC1 for both correct but no or wrong working or SC1 for 2 values satisfying one of the original equations

This question in 0580/11 Oct/Nov 2017

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

4 marks

Mark scheme: 23(a) 5 2 M1 for 7 – 2 = 11n + 3n oe or better or 0.357 or 0.357… 14 23(b) 18 2 p 3 M1 for p – 3 = 3 × 5 or = 3 + 5 5

This question in 0580/12 Oct/Nov 2017

9 Solve the equation. 5x + 4 = 19 + 2x x = … [2]

2 marks

Mark scheme: 9 [x = ] 5 2 M1 for 5x – 2x = 19 – 4 or better

This question in 0580/13 Oct/Nov 2017

8 Solve the equation 8x - 5 = 7. x = … [2]

2 marks

Mark scheme: 8 1.5 oe 2 5 7 M1 for 8x = 7 + 5 or x − = oe 8 8

This question in 0580/11 May/June 2018

24 Solve the simultaneous equations. You must show all your working. 3x + 10y = 106 5x - 4y = 1 x = … y = … [4]

4 marks

Mark scheme: 24 for correctly equating one set of M1 coefficients for correct method to eliminate one M1 variable [x =] 7 A1 [y =] 8.5 A1 If zero scored, SC1 for 2 values satisfying one of the original equations or SC1 for both answers correct but no working

This question in 0580/11 May/June 2018

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

2 marks

Mark scheme: 15(a) [w =] 7 1 15(b) [12x =] 36 1

This question in 0580/12 May/June 2018

14 Solve. 1 - p = 4 3 p = … [2]

2 marks

Mark scheme: 14 – 11 2 M1 for 1 − p = 3 × 4 or better p 1 or − = 4 − or better 3 3

This question in 0580/13 May/June 2018

23 Solve the simultaneous equations. You must show all your working. 3x - 2y = 23 2x + 5y = 9 x = … y = … [4] Question 24 is printed on the next page.

4 marks

Mark scheme: 23 correctly multiplying both equations to M1 reach the same coefficient for one variable correctly adding or subtracting the M1 equations [x =] 7 A1 [y =] −1 A1 If zero scored then SC1 for both answers correct and no supporting working or for two answers that satisfy one of the original equations

This question in 0580/13 May/June 2018

23 (a) Expand the brackets and simplify fully. 5 (x - 3) + 2 (3x + 1) … [2] (b) Solve the simultaneous equations. You must show all your working. 4x - y = 14 3x + 2y = 5 x = … y = … [3]

5 marks

Mark scheme: 23(a) 11x – 13 final answer 2 B1 for 5x – 15 or 6x + 2 or for answer of 11x + j or kx – 13 23(b) Correctly eliminating one variable M1 [x = ] 3 A1 [y = ] –2 A1 If zero scored, SC1 for 2 values satisfying one of the original equations or for 2 correct answers with no working

This question in 0580/11 Oct/Nov 2018

19 Solve the simultaneous equations. You must show all your working. 2x + 5y = 60 3x - 2y = 14 x = … y = … [4]

4 marks

Mark scheme: 19 multiplying both equations to get a M1 common coefficient correctly adding or subtracting their M1 equations [x = ] 10 A1 [y = ] 8 A1 If zero scored then SC1 for two answers which satisfy one of the original equations or for 2 correct answers with no working

This question in 0580/12 Oct/Nov 2018

26 Solve. (a) 3w - 7 = 32 w = … [2] (b) 4(5x + 7) = 42 x = … [3]

5 marks

Mark scheme: 26(a) 13 2 7 32 M1 for 3w = 32 + 7 or w − = or 3 3 better 26(b) 7 3 M1 for first correct step 0.7 or M1 for second correct step, FT their 10 first step

This question in 0580/13 Oct/Nov 2018

20 Solve the simultaneous equations. You must show all your working. 6x - 3y = 12 2x + 3y = 16 x = … y = … [2]

2 marks

Mark scheme: 20 [ x = ] 3.5, [ y = ] 3 2 M1 for a correct equation in terms of x or y. with supporting working If 0 scored SC1 for both answers correct.

This question in 0580/12 Feb/March 2019

23 Rearrange this formula to make x the subject. 5x 2 - 3y = 4y + 8 x = … [3]

3 marks

Mark scheme: 23 7 y + 8 3 M1 for 5 x 2 = 4 y + 3 y + 8 or better x = oe final answer 5 2 their (4 y + 3 y ) + 8 M1 for x = or better 5 their (7 y ) + 8 M1 for x = 5 An incorrect final answer scores a maximum of 2

This question in 0580/12 May/June 2019

16 Solve. 7x - 5 = 16 x = … [2]

2 marks

Mark scheme: 16 3 2 5 16 M1 for 7x = 16 + 5 or x − = or better 7 7

This question in 0580/13 May/June 2019

20 Solve the simultaneous equations. You must show all your working. 5x - 2y = 26 7x + 6y = 10 x = … y = … [3]

3 marks

Mark scheme: 20 For correct method to equate M1 coefficients and eliminate one variable [x =] 4 A2 A1 for each [y =] −3 If 0 scored, SC1 for 2 values satisfying one of the original equations SC1 if no working shown, but 2 correct answers given

This question in 0580/13 May/June 2019

22 Solve. (a) 8 (w + 11 ) = 120 w = … [2] x - 2 (b) = 3 3 x = … [2]

4 marks

Mark scheme: 22(a) 4 2 M1 for 8w + 8 × 11 = 120 or w + 11 = 120 ÷ 8 22(b) 11 2 x 2 M1 for x – 2 = 3 × 3 oe or = 3 + 3 3 oe or better

This question in 0580/11 Oct/Nov 2019

23 Solve the simultaneous equations. You must show all your working. 5x + 4y = 10 7x - 6y = 43 x = … y = … [4]

4 marks

Mark scheme: 23 Correctly equating one set of coefficients M1 Correct method to eliminate one variable M1 [x =] 4 A1 [y = ] –2.5 oe A1 If 0 scored, SC1 for 2 values satisfying one of the original equations or for 2 correct values

This question in 0580/11 Oct/Nov 2019

14 Rovers, United and City are football teams. Rovers scored x goals. United scored 8 goals more than Rovers. City scored 3 goals less than twice the number of goals scored by Rovers. The three teams scored a total of 117 goals. Write down and solve an equation to find the value of x. x = … [4]

4 marks

Mark scheme: 14 x + x + 8 + 2x – 3 = 117 or better M2 or B1 for x + 8 or 2x – 3 4x + 5 = 117 oe or better M1 28 A1 If 0 scored, SC1 for the correct answer with no algebra

This question in 0580/12 May/June 2020

20 Des thinks of two numbers. The sum of his two numbers is -6. The difference between his two numbers is 62. Find the two numbers. … and … [4]

4 marks

Mark scheme: 20 28, − 34 4 Trial and improvement OR B1 for x + y = –6 oe B1 for x – y = 62 oe B1 for 28 or − 34

This question in 0580/13 May/June 2020

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

2 marks

Mark scheme: 16 2 y + 7 2 y 7 2 2 y 7 [x =] oe or [x =] + oe M1 for 2y + 7 = 5x oe or = x − oe 5 5 5 5 5 final answer

This question in 0580/11 Oct/Nov 2020

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

2 marks

Mark scheme: 7 1 6 2 M1 for 6 = 2x + 3x or better 1.2 or 1 or 5 5

This question in 0580/12 Oct/Nov 2020

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

3 marks

Mark scheme: 23 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 0580/13 Oct/Nov 2020

20 Solve the simultaneous equations. You must show all your working. 5x + 6y = 14 2x + 8y = 7 x = … y = … [4]

4 marks

Mark scheme: 20 Correctly equating one set of M1 or making x or y the subject of one equation coefficients correctly Correct method to eliminate one M1 or substitution for x or y for their rearranged variable formula x = 2.5 A2 A1 for one correct value If M0 scored, SC1 for 2 values satisfying one y = 0.25 of the original equations

This question in 0580/12 Feb/March 2021

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

3 marks

Mark scheme: 21 Correctly eliminating one variable M1 [x =] 5 A1 [y =] – 7 A1 If M0 scored, SC1 for two values satisfying one of the original equations

This question in 0580/11 May/June 2021

22 4x° NOT TO SCALE ( 3x + 75 )° ( 87- )x° The diagram shows a quadrilateral. Work out the value of x. x = … [4]

4 marks

Mark scheme: 22 18 4 B3 for 6x = 108 or M2 for 4x + 3x + 75 + 87 – x + 90 =360 oe or better or M1 for 4x + 3x + 75 + 87 – x [+ 90] or better or M1 for 4x + 3x + 75 + 87 – x [+ 90] = k If zero scored, SC1 for ax = 108 (a ≠ 6) or for 6x = b (b ≠ 108)

This question in 0580/12 May/June 2021

8 P = 2n - 3t Find the value of n when P = 9 and t = 8 . n = … [3]

3 marks

Mark scheme: 8 16.5 oe 3 B2 for 2n = 33 9 + 3 × 8 or M2 for n = 2 or M1 for 2n – 3 × 8 = 9 or better

This question in 0580/11 Oct/Nov 2021

16 (a) Solve. 7x + 18 = 4 x = … [2] y 6 18 (b) 7 # 7 = 7 Find the value of y. y = … [1]

3 marks

Mark scheme: 16(a) −2 2 18 4 M1 for 7x = 4 − 18 or x + = 7 7 16(b) 12 cao 1

This question in 0580/12 Oct/Nov 2021

7 r = 2 t + 3u Work out the value of t when r = 18 and u = 4. t = … [2]

2 marks

Mark scheme: 7 3 2 M1 for 18 = 2t + (3 × 4) or better

This question in 0580/13 Oct/Nov 2021

11 Joe thinks of a positive number, n. He squares n, then adds it to - 24 . The answer is 25. Work out n. n = … [2]

2 marks

Mark scheme: 11 7 2 M1 for n 2 − 24 = 25 oe or B1 for 49

This question in 0580/12 Feb/March 2022

19 Joe thinks of a number, n, trebles it, and subtracts 5. The result is 22. Write this as an equation in terms of n, and solve the equation. n = … [3]

3 marks

Mark scheme: 19 3n – 5 = 22 1 9 final answer 2 B2FT their equation providing their equation is in the form an+b=22 where a≠0 or 1 and b≠0 5 22 or M1FT for 3n = 22 + 5 or n − = 3 3

This question in 0580/13 May/June 2022

23 Solve the simultaneous equations. You must show all your working. 4x - 3y = 26 5x + 6y = 13 x = … y = … [3]

3 marks

Mark scheme: 23 For correct method to eliminate one M1 variable [x =] 5 A1 [y =] −2 A1 If zero scored, SC1 for 2 values satisfying one of the original equations or SC1 if no working shown, but 2 correct answers

This question in 0580/13 May/June 2022

23 Natalie buys 4 tomato plants and 3 pepper plants for $9.35 . Samir buys 2 tomato plants and 11 pepper plants for $16.55 . Write down a pair of simultaneous equations and solve them to find the cost of one tomato plant and the cost of one pepper plant. You must show all your working. Tomato plant $ … Pepper plant $ … [5]

5 marks

Mark scheme: 23 4t +3p = 9.35 oe B2 B1 for each 2t + 11p = 16.55 oe Correctly eliminating one variable M1 Follow through their equations [t = ] 1.4[0] A1 [p = ] 1.25 A1 If M0 or M0 FT scored, SC1 for 2 values which satisfy one correct equation or one of their equations

This question in 0580/11 Oct/Nov 2022

8 (a) Simplify. 6a + 3b - 2a - 5b … [2] 1 2 (b) s = 5t + at 2 Find the value of s when t = 6 and a = 3 . s = … [2]

4 marks

Mark scheme: 8(a) 4a – 2b or 2(2a – b) final answer 2 B1 for 4a or – 2b in final answer or 4a + – 2b as final answer or for correct answer seen and spoilt 8(b) 84 2 B1 for 30 or 54 1 or M1 for 5 × 6 + × 3 × 62 oe 2

This question in 0580/12 Oct/Nov 2022

16 (a) Expand. x ( x + 8) … [2] (b) Factorise completely. 6a - 3ab … [2] (c) Solve. 5x - 6 = x + 3 x = … [2]

6 marks

Mark scheme: 16(a) x2 + 8x final answer 2 B1 for x2 or +8x in final answer or correct answer spoilt 16(b) 3a(2 − b) final answer 2 B1 for 3(2a −ab) or a(6 − 3b) final answer or correct answer spoilt 16(c) 1 2 M1 for 5x – x = 3 + 6 or better 2.25 or 2 4

This question in 0580/12 Oct/Nov 2022

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

2 marks

Mark scheme: 20 [x =] 9 2 B1 for each answer [y =] 3

This question in 0580/13 Oct/Nov 2022

20 Solve. 25 - 2 u = 2 3 u = … [2]

2 marks

Mark scheme: 20 9.5 2 M1 for 25 – 2u = 3 × 2 oe 25 2u or for − 2 = 3 3

This question in 0580/12 Feb/March 2023

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

3 marks

Mark scheme: 23 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 0580/12 Feb/March 2023

25 At a cinema, an adult ticket costs $a and a child ticket costs $c. (a) Farah buys 3 adult tickets and 4 child tickets for $38.50 . Complete the equation. 3a + 4c = … [1] (b) Hana buys 6 adult tickets and 5 child tickets for $65.00 . Write down another equation in terms of a and c. … [1] (c) Solve the two simultaneous equations to find the value of a and the value of c. You must show all your working. a = … c = … [3]

5 marks

Mark scheme: 25(a) 38.5 1 25(b) 6a + 5c = 65 1 If 0 scored in part(a) and part(b), SC1 for 3850 in part(a) and 6a + 5c = 6500 in part(b) 25(c) Correctly eliminating one variable M1 Follow through their linear equations [a =] 7.5 A1 [c =] 4 A1 If M0 or M0 FT scored, SC1 for 2 values which satisfy one correct equation or one of their equations

This question in 0580/11 May/June 2023

16 Solve the equation. 5x + 7 = 9x - 3 x = … [2]

2 marks

Mark scheme: 16 2.5 2 M1 for 7 + 3 = 9x – 5x or better

This question in 0580/13 Oct/Nov 2023

23 Solve the simultaneous equations. You must show all your working. 3x + 5y = 23 6x - 4y = 11 x = … y = … [3]

3 marks

Mark scheme: 23 Correctly eliminates one variable M1 [x =] 3.5 A1 [y =] 2.5 A1 If M0 scored, SC1 for 2 values satisfying one of the original equations

This question in 0580/13 Oct/Nov 2023

22 Solve the simultaneous equations. You must show all your working. 6x + 2y = 29 3x - 4y = 17 x = … y = … [3]

3 marks

Mark scheme: 22 Correctly eliminating one variable M1  x  5 A1  y  0.5 A1 If M0 scored SC1 for 2 values satisfying one of the original equations

This question in 0580/11 May/June 2024

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

2 marks

Mark scheme: 19 [t = ] 3 2 B1 for each [w = ] –2

This question in 0580/13 May/June 2024

13 (a) Find the value of 6c + 7d when c = 3 and d =- 4 . … [2] (b) Solve. 6x + 8 = 11x + 4 x = … [2]

4 marks

Mark scheme: 13(a) –10 2 B1 for 18 or –28 or M1 for 6 × 3 + 7 × −4 13(b) 4 2 M1 for 8 – 4 = 11x – 6x or better or 0.8 5

This question in 0580/11 Oct/Nov 2024

21 (a) Factorise. 28x - 35 … [1] r (b) Make r the subject of the formula T = - p . 4 r = … [2]

3 marks

Mark scheme: 21(a) 7(4x – 5) final answer 1 21(b) 4T + 4p or 4(T + p) 2 r M1 for T + p = or 4T = r – 4p final answer 4

This question in 0580/11 Oct/Nov 2024

22 Solve the simultaneous equations. You must show all your working. 5x + 6y = 9 3x - 2y = - 17 x = … y = … [3] Questions 23 and 24 are printed on the next page.

3 marks

Mark scheme: 22 Correctly eliminating one M1 variable

This question in 0580/11 Oct/Nov 2024

3 Solve. (a) 5x = 14 x = … [1] (b) x + 6 = 25 x = … [1]

2 marks

Mark scheme: 3(a) 2.8 1 3(b) 19 1

This question in 0580/12 Oct/Nov 2024

22 Solve the simultaneous equations. You must show all your working. 2x + 7y = 34 3x + 5y = 18 x = … y = … [4]

4 marks

Mark scheme: 22 correctly equating one set of coefficients M1 correct method to eliminate one variable M1 [x =] −4 A1 [y =] 6 A1 If A0 scored SC1 for 2 values satisfying one of the original equations.

This question in 0580/12 Oct/Nov 2024

17 Ed has n books. Sam has 3 times as many books as Ed Jane has 2 books fewer than Sam. The total number of books is 54. Use this information to write down an equation and solve it to find the value of n. n = … [4]

4 marks

Mark scheme: 17 7n = 56 M3 OR M2 for n + 3n + 3n – 2 = 54 or any correct equation which would lead to 7n – 2 = 54 OR M1 for rearranging their equation to give an = b OR B1 for 3n or 3n – 2 8 A1 Dep on at least M2 If 0 scored SC1 for answer 8 without algebra

This question in 0580/13 Oct/Nov 2024

18 (a) Complete the table of values for y = . x x -6 -4 -3 -2 -1 1 2 3 4 6 y -3 -6 6 3 [3] 12 (b) On the grid, draw the graph of y = for - 6 G x G -1 and 1 G x G 6 . x y 12 10 8 6 4 2 x -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -2 -4 -6 -8 -10 -12 [4] (c) On the grid, draw the line y =- 9 . [1] 12 (d) Use your graph to solve =- 9 . x x = … [1]

9 marks

Mark scheme: 18(a) −2 − 4 − 12 12 4 2 3 B2 for 4 correct or B1 for 1 correct 18(b) Correct curve 4 B3FT for 9 points correctly plotted or B2FT for 7 points correctly plotted or B1FT for 5 points correctly plotted 18(c) y = −9 correctly drawn, ruled 1 18(d) −1.4 to − 1.2 1 FT their curve and y = −9

This question in 0580/12 Feb/March 2025

26 Solve the simultaneous equations. 4t - 3w = 11 6t + 2w =- 3 t = … w = … [4]

4 marks

Mark scheme: 26  t =  0.5 4  w =  − 3 M1 for correctly equating one set of coefficients M1 for correct method to eliminate one variable OR M1 for making t or w the subject of one equation M1 for substitution of t or w from their rearranged equation A1 for t = 0.5 A1 for w = −3 If A0 scored SC1 for 2 values satisfying one of the original equations

This question in 0580/12 Feb/March 2025

14 (a) Complete the table of values for y = ( x + 3 )( x - 2 ) . x -4 -3 -2 -1 0 1 2 3 y 6 -4 -4 [3] (b) On the grid, draw the graph of y = ( x + 3 )( x - 2 ) for - 4 G x G 3 . y 6 5 4 3 2 1 - 4 -3 -2 -1 0 1 2 3 x -1 -2 -3 - 4 -5 -6 -7 [4] (c) Write down the coordinates of the lowest point of the graph. ( … , … ) [1] (d) Write down the equation of the line of symmetry of the graph. … [1] (e) Use your graph to solve the equation ( x + 3)( x - 2) = 3 . x = … or x = … [2]

11 marks

Mark scheme: 14(a) [6], 0, [−4], -6, -6, [−4], 0, 6 3 B2 for 3 correct B1 for 1 correct 14(b) Correct curve 4 B3FT for 7 points correctly plotted or B2FT for 5 points correctly plotted or B1FT for 3 points correctly plotted 14(c) ( −0.5 , k) where −6.6 k −6 1 14(d) x = −0.5 oe 1 14(e) −3.6 to −3.4 2.4 to 2.6 2 FT their curve B1 for each or B1 for y = 3 drawn

This question in 0580/12 May/June 2025

15 Beth thinks of a positive number, n. She squares n then subtracts 55. The answer is 9. Work out the value of n. n = … [2]

2 marks

Mark scheme: 15 8 2 B1 for 64 seen or M1 for n 2 − 55 = 9 or better.

This question in 0580/12 May/June 2025

24 Solve the simultaneous equations. 5x + 2y = 3 3x + 4y = 27 x = … y = … [3]

3 marks

Mark scheme: 24  x =  − 3 3 M1 for correctly eliminating one variable  y = 9 A1 for x = −3 nfww A1 for y = 9 If A0 scored SC1 for 2 values satisfying one of the original equations

This question in 0580/12 May/June 2025

21 In this question, all lengths are in centimetres. 2x − 1 2x + 1 9 NOT TO x SCALE x + 3 The perimeter of the triangle is equal to the perimeter of the rectangle. Form an equation and solve it to find the value of x. x = … [4]

4 marks

Mark scheme: 21 3x = 15 M3 M2 for 2x + 1 + x + 3 + 9 = 2(x + 2x – 1) or better or M1 for rearranging their equation to give ax = b or B1 for 3x + 13 or 6x – 2 5 A1 dep on at least M2 If 0 scored, SC1 for answer 5 with no algebra

This question in 0580/13 May/June 2025

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

3 marks

Mark scheme: 25 [x = ] 3 final answer nfww 3 M1 for correctly eliminating one variable [y = ] –4 OR M1 for correct substitution of x or y into the other equation A1 for x = 3 A1 for y = –4 If A0 scored, SC1 for 2 values satisfying one of the original equations.

This question in 0580/13 May/June 2025

8 p = 3 q - 2r Work out the value of q when p = 50 and r = 5 . q = … [2]

2 marks

Mark scheme: 8 20 2 M1 for 50 = 3q – 2 × 5 or better

This question in 0580/11 Oct/Nov 2025

29 Jim buys 15 apples and 10 pears for $42.50 . Li buys 4 apples and 5 pears for $16. Write down a pair of simultaneous equations and solve them to find the cost of one apple and the cost of one pear. Apple $ … Pear $ … [5]

5 marks

Mark scheme: 29 15a + 10p = 42.5 oe 2 B1 for each 4a + 5p = 16 oe [apple = ] 1.5[0] cao 3 M1FT for correctly eliminating one [pear = ] 2[.00] cao variable from their equations. A1 for 1.5 A1 for 2 If M1 A0 or M0 scored, SC1 for two positive costs which satisfy one of original conditions

This question in 0580/11 Oct/Nov 2025

23 There are two charges for crossing a bridge. Cars $a Other vehicles $b On Monday 4 cars and 20 other vehicles pay a total of $130. On Tuesday 6 cars and 15 other vehicles pay a total of $105. Write down two equations and solve them to find the value of a and the value of b. a = … b = … [6]

6 marks

Mark scheme: 23 4a + 20b = 130 oe 2 B1 for each 6a + 15b = 105 oe [a =] 2.5 4 M1FT for correctly equating one set of [b =] 6 coefficients M1FT for correct method to eliminate one variable A1 for a = 2.5 A1 for b = 6 If A0 scored, SC1 for two positive costs which satisfy one of the original conditions

This question in 0580/12 Oct/Nov 2025

12 (a) Simplify. 3y - 4y + 2y … [1] (b) Solve. (i) x + 5 = 19 x = … [1] (ii) 6x - 5 = 7 x = … [2]

4 marks

Mark scheme: 12(a) y 1 12(b)(i) 14 1 12(b)(ii) 2 2 5 7 M1 for 6x = 7 + 5 or x − = or better 6 6

This question in 0580/13 Oct/Nov 2025