IGCSE Maths 0580 Equations and Formulae Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Algebra and Graphs · 2.5

Every equations and formulae question from 28 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.5. 52 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Algebra and Graphs past paper questions.

0580 Equations and Formulae past paper questions: Core (Papers 1 and 3)

11 markCore

May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q21(a)

Make T the subject of the formula P = Ty.

Show solution
  1. T is multiplied by y, so do the opposite: divide both sides by y.
  2. P⁄y = T, so T = P⁄y
Answer T = P⁄y
24 marksCore

May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q23

Solve the simultaneous equations 3x + 4y = 1 and 2x − 3y = 12.

Show solution
  1. Label the equations: 3x + 4y = 1 (1) and 2x − 3y = 12 (2).
  2. Make the y coefficients match: (1) × 3 gives 9x + 12y = 3 and (2) × 4 gives 8x − 12y = 48.
  3. The y terms have opposite signs, so add the equations: 17x = 51.
  4. x = 51 ÷ 17 = 3.
  5. Substitute x = 3 into (1): 9 + 4y = 1, so 4y = −8 and y = −2.
  6. x = 3, y = −2
  7. Check in (2): 2(3) − 3(−2) = 6 + 6 = 12 ✓
Answer x = 3, y = −2
34 marksCore

May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q10

Phil has 7y − 52 toy cars. Ann has 78 − 6y toy cars. Phil and Ann have the same number of cars.

(a) Write down an equation to show this information. (b) Solve your equation and work out the number of cars they each have.

Show solution
Part (a)
  1. They have the same number of cars, so the two expressions are equal.
  2. 7y − 52 = 78 − 6y
Answer 7y − 52 = 78 − 6y
Part (b)
  1. Add 6y to both sides: 13y − 52 = 78.
  2. Add 52 to both sides: 13y = 130.
  3. Divide by 13: y = 10.
  4. Number of cars = 7 × 10 − 52 = 70 − 52 = 18.
  5. Check: Ann has 78 − 6 × 10 = 18 ✓. They each have 18 cars
Answer 18 cars
42 marksCore

May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q28

Make T the subject of the formula Q = 3 − 4T.

Show solution
  1. Start with Q = 3 − 4T.
  2. Add 4T to both sides: Q + 4T = 3.
  3. Subtract Q from both sides: 4T = 3 − Q.
  4. Divide by 4: T = 3 − Q⁄4
Answer T = 3 − Q⁄4
53 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q22

A rectangle has sides 6 cm and (x − 1) cm. The area, in cm², of the rectangle is 5x + 8. Write an equation to show this information and solve it to find the value of x.

0580/13 May/June 2026 Q22 – rectangle with sides 6 and x − 1x − 16NOT TOSCALE
Show solution
  1. Area of a rectangle = length × width, so the area is 6(x − 1).
  2. 6(x − 1) = 5x + 8
  3. Expand the bracket: 6x − 6 = 5x + 8.
  4. Subtract 5x from both sides: x − 6 = 8.
  5. Add 6: x = 14
  6. Check: 6 × 13 = 78 and 5 × 14 + 8 = 78 ✓
Answer 6(x − 1) = 5x + 8 x = 14
63 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q23

Solve the simultaneous equations x + 5y = 9 and 4x + 6y = 1.

Show solution
  1. Label the equations: x + 5y = 9 (1) and 4x + 6y = 1 (2).
  2. Rearrange (1) to make x the subject: x = 9 − 5y.
  3. Substitute into (2): 4(9 − 5y) + 6y = 1.
  4. Expand: 36 − 20y + 6y = 1, so 36 − 14y = 1.
  5. Subtract 36: −14y = −35, so y = 2.5.
  6. Substitute back: x = 9 − 5 × 2.5 = 9 − 12.5 = −3.5.
  7. x = −3.5, y = 2.5
  8. Check in (2): 4(−3.5) + 6(2.5) = −14 + 15 = 1 ✓
Answer x = −3.5, y = 2.5
73 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q8

Solve. (a) 6 + x = 2 (b) 5x + 2 = 18

Show solution
Part (a)
  1. Subtract 6 from both sides: x = 2 − 6.
  2. x = −4
Answer x = −4
Part (b)
  1. Subtract 2 from both sides: 5x = 16.
  2. Divide by 5: x = 3.2
Answer x = 3.2
83 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q20(a)

P = 5x + 4y. Find the value of x when P = 18 and y = −3.

Show solution
  1. Substitute P = 18 and y = −3: 18 = 5x + 4 × (−3).
  2. Simplify: 18 = 5x − 12.
  3. Add 12 to both sides: 30 = 5x.
  4. Divide by 5: x = 6
Answer x = 6
97 marksCore

May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q14

There are 60 counters in a jar. Kan adds C counters into the jar. Mo adds 2C + 7 counters into the jar.

(a) Write an expression, in terms of C, for the number of counters that are now in the jar. Give your answer in its simplest form.

(b) Kan and Mo share the counters equally. (i) Write down an expression, in terms of C, for the number of counters they each receive. (ii) They each receive 86 counters. Work out the value of C.

Show solution
Part (a)
  1. Start with 60 and add both amounts: 60 + C + 2C + 7.
  2. Collect the C terms: C + 2C = 3C. Collect the numbers: 60 + 7 = 67.
  3. 3C + 67
Answer 3C + 67
Part (b)(i)
  1. Sharing equally between 2 people means dividing by 2.
  2. Each gets 3C + 67⁄2
Answer 3C + 67⁄2
Part (b)(ii)
  1. Set the expression equal to 86: 3C + 67⁄2 = 86.
  2. Multiply both sides by 2: 3C + 67 = 172.
  3. Subtract 67: 3C = 105.
  4. Divide by 3: C = 35
  5. Check: 60 + 35 + 77 = 172, and 172 ÷ 2 = 86 ✓
Answer C = 35
104 marksCore

May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q13

Solve. (a) 32.4⁄a = 6 (b) 3y − 7 = 5y − 10

Show solution
Part (a)
  1. Multiply both sides by a: 32.4 = 6a.
  2. Divide both sides by 6: a = 32.4 ÷ 6 = 5.4
Answer 5.4
Part (b)
  1. Subtract 3y from both sides: −7 = 2y − 10.
  2. Add 10 to both sides: 3 = 2y.
  3. Divide by 2: y = 1.5
  4. Check: 3 × 1.5 − 7 = −2.5 and 5 × 1.5 − 10 = −2.5 ✓
Answer y = 1.5
114 marksCore

Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q20

Solve the simultaneous equations −5x − 7y = 1 and 6x + 8y = −2.

Show solution
  1. Label the equations: −5x − 7y = 1 (1) and 6x + 8y = −2 (2).
  2. Make the x coefficients opposite: (1) × 6 gives −30x − 42y = 6 and (2) × 5 gives 30x + 40y = −10.
  3. Add the equations so x cancels: −2y = −4.
  4. y = −4 ÷ −2 = 2.
  5. Substitute y = 2 into (1): −5x − 14 = 1, so −5x = 15 and x = −3.
  6. x = −3, y = 2
  7. Check in (2): 6(−3) + 8(2) = −18 + 16 = −2 ✓
Answer x = −3, y = 2
122 marksCore

Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q27

(a) A teacher asks a student to make x the subject of the formula y = x⁄5 + 2. The student’s working is: line A: y = x⁄5 + 2; line B: 5y = x + 2; line C: 5y − 2 = x; line D: x = 5y − 2. Complete this statement: line … is not correct because …

(b) Make p the subject of the formula A = pt².

Show solution
Part (a)
  1. Line A to line B multiplies both sides by 5. Every term must be multiplied, but the student did not multiply the 2.
  2. Line B is not correct because the 2 should also be multiplied by 5
  3. The correct working is 5y = x + 10, so x = 5y − 10.
Answer Line B the 2 should also be multiplied by 5
Part (b)
  1. p is multiplied by t², so divide both sides by t².
  2. p = A⁄t²
Answer p = A⁄t²
133 marksCore

Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q14

Solve 5(2x − 7) = 79.

Show solution
  1. Expand the bracket: 5 × 2x − 5 × 7 = 10x − 35, so 10x − 35 = 79.
  2. Add 35 to both sides: 10x = 114.
  3. Divide by 10: x = 11.4
Answer x = 11.4
142 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q8

p = 3q − 2r. Work out the value of q when p = 50 and r = 5.

Show solution
  1. Substitute p = 50 and r = 5 into p = 3q − 2r: 50 = 3q − 2 × 5.
  2. Simplify: 50 = 3q − 10.
  3. Add 10 to both sides: 60 = 3q.
  4. Divide by 3: q = 20
Answer q = 20
155 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q29

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.

Show solution
  1. Let one apple cost $a and one pear cost $p.
  2. Jim: 15a + 10p = 42.5 (1). Li: 4a + 5p = 16 (2).
  3. Double equation (2) so the p terms match: 8a + 10p = 32 (3).
  4. Subtract (3) from (1): 15a − 8a = 42.5 − 32, so 7a = 10.5 and a = 1.5.
  5. Substitute a = 1.5 into (2): 4 × 1.5 + 5p = 16, so 6 + 5p = 16, 5p = 10 and p = 2.
  6. Apple $1.50, pear $2.00
  7. Check in (1): 15 × 1.50 + 10 × 2 = 22.50 + 20 = 42.50 ✓
Answer 15a + 10p = 42.5 4a + 5p = 16 Apple $1.50, pear $2.00
166 marksCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q23

A bridge charges $a for cars and $b for other vehicles. 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.

Show solution
  1. Monday: 4 cars and 20 other vehicles cost $130: 4a + 20b = 130 (1)
  2. Tuesday: 6 cars and 15 other vehicles cost $105: 6a + 15b = 105 (2)
  3. Make the a terms match: (1) × 3 gives 12a + 60b = 390 and (2) × 2 gives 12a + 30b = 210.
  4. Subtract: 30b = 180, so b = 6.
  5. Substitute b = 6 into (1): 4a + 120 = 130, so 4a = 10 and a = 2.5.
  6. a = 2.5, b = 6
  7. Check in (2): 6 × 2.5 + 15 × 6 = 15 + 90 = 105 ✓
Answer 4a + 20b = 130 6a + 15b = 105 a = 2.5, b = 6
174 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q12

(a) Simplify 3y − 4y + 2y. (b) Solve (i) x + 5 = 19 (ii) 6x − 5 = 7.

Show solution
Part (a)
  1. All the terms are y terms, so combine the numbers in front: 3 − 4 + 2 = 1.
  2. 1y is written y, so the answer is y
Answer y
Part (b)(i)
  1. Subtract 5 from both sides: x = 19 − 5.
  2. x = 14
Answer x = 14
Part (b)(ii)
  1. Add 5 to both sides: 6x = 12.
  2. Divide by 6: x = 2
Answer x = 2
182 marksCore

Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q5

Solve. (a) 8x = 5 (b) x + 8 = 5

Show solution
Part (a)
  1. Divide both sides by 8: x = 5 ÷ 8.
  2. x = 5⁄8 (= 0.625)
Answer x = 5⁄8
Part (b)
  1. Subtract 8 from both sides: x = 5 − 8.
  2. x = −3
Answer x = −3
192 marksCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q24

Make p the subject of the formula m = 9p + 32.

Show solution
  1. We want p on its own. First subtract 32 from both sides: m − 32 = 9p.
  2. Then divide both sides by 9.
  3. p = m − 32⁄9
Answer p = m − 32⁄9
202 marksCore

May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q12

Solve. 5x + 8 = 3x − 2

Show solution
  1. Start with 5x + 8 = 3x − 2.
  2. Subtract 3x from both sides to get the x terms together: 2x + 8 = −2
  3. Subtract 8 from both sides: 2x = −10
  4. Divide both sides by 2: x = −5
  5. Check: 5(−5) + 8 = −17 and 3(−5) − 2 = −17 ✓
Answer −5
214 marksCore

May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q26

Solve the simultaneous equations. 2x + 5y = 5 and 3x + 4y = 11

Show solution
  1. Label the equations: 2x + 5y = 5 (1) and 3x + 4y = 11 (2).
  2. Make the x coefficients the same: (1) × 3 gives 6x + 15y = 15 and (2) × 2 gives 6x + 8y = 22.
  3. Subtract the second from the first: 15y − 8y = 15 − 22, so 7y = −7.
  4. y = −1.
  5. Substitute y = −1 into (1): 2x + 5(−1) = 5, so 2x − 5 = 5 and 2x = 10.
  6. x = 5.
  7. x = 5, y = −1
  8. Check in (2): 3(5) + 4(−1) = 15 − 4 = 11 ✓
Answer x = 5, y = −1
222 marksCore

May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q15

Beth thinks of a positive number, n. She squares n then subtracts 55. The answer is 9. Work out the value of n.

Show solution
  1. Write the words as an equation: n² − 55 = 9.
  2. Add 55 to both sides: n² = 64.
  3. Take the square root: n = 8 or n = −8.
  4. n is positive, so n = 8
Answer 8
233 marksCore

May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q24

Solve the simultaneous equations. 5x + 2y = 3 and 3x + 4y = 27

Show solution
  1. Label the equations: 5x + 2y = 3 (1) and 3x + 4y = 27 (2).
  2. Make the y coefficients the same: (1) × 2 gives 10x + 4y = 6 (3).
  3. Subtract (2) from (3): 10x − 3x = 6 − 27, so 7x = −21 and x = −3.
  4. Substitute x = −3 into (1): −15 + 2y = 3, so 2y = 18 and y = 9.
  5. x = −3, y = 9
  6. Check in (2): 3(−3) + 4(9) = −9 + 36 = 27 ✓
Answer x = −3, y = 9
244 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q21

In this question, all lengths are in centimetres. A triangle has sides 2x + 1, x + 3 and 9. A rectangle has sides 2x − 1 and x. 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.

0580/13 May/June 2025 Q21 – triangle with sides 2x + 1, x + 3 and 9 and rectangle with sides 2x − 1 and x2x + 19x + 32x − 1xNOT TOSCALE
Show solution
  1. Perimeter of the triangle = (2x + 1) + (x + 3) + 9 = 3x + 13.
  2. Perimeter of the rectangle = 2 × length + 2 × width = 2(2x − 1) + 2x = 6x − 2.
  3. The perimeters are equal: 3x + 13 = 6x − 2.
  4. Subtract 3x from both sides: 13 = 3x − 2.
  5. Add 2: 15 = 3x, so x = 5
  6. Check: triangle 11 + 8 + 9 = 28 and rectangle 2 × 9 + 2 × 5 = 28 ✓
Answer 5
252 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q22

g = h⁄3 − 8. Rearrange the formula to make h the subject.

Show solution
  1. Start with g = h⁄3 − 8.
  2. Add 8 to both sides: g + 8 = h⁄3.
  3. Multiply both sides by 3: 3(g + 8) = h.
  4. h = 3(g + 8) (or h = 3g + 24)
Answer h = 3(g + 8)
263 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q25

Solve the simultaneous equations. 8x + 5y = 4 and 2x − y = 10

Show solution
  1. Label the equations: 8x + 5y = 4 (1) and 2x − y = 10 (2).
  2. Make the y terms opposite: (2) × 5 gives 10x − 5y = 50 (3).
  3. Add (1) and (3) so y cancels: 18x = 54, so x = 3.
  4. Substitute x = 3 into (2): 6 − y = 10, so y = −4.
  5. x = 3, y = −4
  6. Check in (1): 8(3) + 5(−4) = 24 − 20 = 4 ✓
Answer x = 3, y = −4

0580 Equations and Formulae past paper questions: Extended (Papers 2 and 4)

272 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q2

Solve 7x + 4 = 25.

Show solution
  1. Subtract 4 from both sides: 7x = 25 − 4 = 21.
  2. Divide both sides by 7: x = 3
  3. Check: 7 × 3 + 4 = 25 ✓
Answer x = 3
282 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q6

Solve the simultaneous equations 2x − y = 3 and x + y = 12.

Show solution
  1. Label the equations: 2x − y = 3 (1) and x + y = 12 (2).
  2. The y terms have opposite signs, so add the equations: 3x = 15.
  3. x = 15 ÷ 3 = 5.
  4. Substitute x = 5 into (2): 5 + y = 12, so y = 7.
  5. x = 5, y = 7
  6. Check in (1): 2 × 5 − 7 = 3 ✓
Answer x = 5, y = 7
293 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q6(b)

E = 5c² − d². Rearrange the formula to write c in terms of E and d.

Show solution
  1. Start with E = 5c² − d².
  2. Add d² to both sides: E + d² = 5c².
  3. Divide both sides by 5: c² = E + d²⁄5.
  4. Take the square root: c = ±√(E + d²⁄5)
Answer c = ±√(E + d²⁄5)
305 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q25

The straight line y = 7 − 4x intersects the curve y = 4 + 3x − 2x² at two points. Find the coordinates of these two points.

Show solution
  1. At the intersection points the line and the curve have the same y: 7 − 4x = 4 + 3x − 2x².
  2. Move everything to the left side: 2x² − 4x − 3x + 7 − 4 = 0, so 2x² − 7x + 3 = 0.
  3. Factorise: (2x − 1)(x − 3) = 0, so x = 0.5 or x = 3.
  4. Find y using the line y = 7 − 4x. x = 0.5: y = 7 − 2 = 5. x = 3: y = 7 − 12 = −5.
  5. (0.5, 5) and (3, −5)
Answer (0.5, 5) and (3, −5)
314 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q17(b)

y = 2(x − 3)⁄x − 5. Rearrange the formula to write x in terms of y.

Show solution
  1. Multiply both sides by (x − 5) to remove the fraction: y(x − 5) = 2(x − 3).
  2. Expand both brackets: xy − 5y = 2x − 6.
  3. Get all the x terms on one side: xy − 2x = 5y − 6.
  4. Factorise out x: x(y − 2) = 5y − 6.
  5. Divide by (y − 2): x = 5y − 6⁄y − 2
Answer x = 5y − 6⁄y − 2
326 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q18

The diagram shows a rectangle measuring (x + 6) cm by (x + 2) cm, with an unshaded right-angled triangle with two sides of x cm. The shaded area is 30 cm².

0580/23 May/June 2026 Q18 – rectangle with sides 6 + x and x + 2, and an unshaded triangle with two sides x cm6 cmx cmx cm2 cmNOT TOSCALE

(a) Show that x² + 16x − 36 = 0.

(b) Solve by factorisation x² + 16x − 36 = 0 to find the length x cm.

Show solution
Part (a)
  1. Shaded area = area of rectangle − area of triangle.
  2. Rectangle: (x + 6)(x + 2) = x² + 2x + 6x + 12 = x² + 8x + 12.
  3. Triangle: the two sides of x cm meet at the right angle, so area = 1⁄2 × x × x = 1⁄2x².
  4. Shaded area = x² + 8x + 12 − 1⁄2x² = 1⁄2x² + 8x + 12.
  5. It equals 30: 1⁄2x² + 8x + 12 = 30, so 1⁄2x² + 8x − 18 = 0.
  6. Multiply every term by 2: x² + 16x − 36 = 0 ✓
Answer x² + 16x − 36 = 0
Part (b)
  1. Find two numbers that multiply to −36 and add to 16: 18 and −2.
  2. (x + 18)(x − 2) = 0, so x = −18 or x = 2.
  3. A length cannot be negative, so x = 2
Answer x = 2
333 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q9

Solve the equation w + 8⁄4 = 2w − 3⁄3.

Show solution
  1. The denominators are 4 and 3. Multiply both sides by 12 to clear the fractions.
  2. 3(w + 8) = 4(2w − 3).
  3. Expand: 3w + 24 = 8w − 12.
  4. Subtract 3w from both sides: 24 = 5w − 12.
  5. Add 12: 36 = 5w, so w = 36 ÷ 5 = 7.2
Answer 7.2
348 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q18

In this question, all lengths are in centimetres. Triangle ABC has AB = x, BC = 2x and angle ABC = 30°. Triangle DEF is right-angled at E with DE = (x − 4) and EF = (2x − 6).

0580/42 May/June 2026 Q18 – triangle ABC with sides x and 2x and angle 30°, and right-angled triangle DEF with sides x − 4 and 2x − 62xx30°(x − 4)(2x − 6)BCAEFDNOT TOSCALE

(a) Show that the area of triangle ABC is 1⁄2x² cm².

(b) The area of triangle ABC is equal to the area of triangle DEF. Show that x² − 14x + 24 = 0.

(c) By factorising and solving x² − 14x + 24 = 0, find the area of triangle ABC.

Show solution
Part (a)
  1. Use Area = 1⁄2ab sin C with the two sides and the angle between them.
  2. Area = 1⁄2 × x × 2x × sin 30° = x² × sin 30°.
  3. sin 30° = 1⁄2, so Area = 1⁄2x² ✓
Answer 1⁄2x²
Part (b)
  1. Triangle DEF is right-angled at E, so its area = 1⁄2 × base × height = 1⁄2(2x − 6)(x − 4).
  2. Take out the 2: 1⁄2 × 2(x − 3)(x − 4) = (x − 3)(x − 4) = x² − 7x + 12.
  3. Set the areas equal: 1⁄2x² = x² − 7x + 12.
  4. Multiply both sides by 2: x² = 2x² − 14x + 24.
  5. Subtract x² from both sides: x² − 14x + 24 = 0 ✓
Answer x² − 14x + 24 = 0
Part (c)
  1. Find two numbers that multiply to 24 and add to −14: −2 and −12.
  2. (x − 2)(x − 12) = 0, so x = 2 or x = 12.
  3. If x = 2, then DE = x − 4 = −2, and a length cannot be negative. So x = 12.
  4. Area of ABC = 1⁄2 × 12² = 1⁄2 × 144 = 72 cm²
Answer 72 cm²
354 marksExtended

Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q7

p² = r² − 8t. (a) Find p when r = −15 and t = 7. (b) Rearrange p² = r² − 8t to make t the subject.

Show solution
Part (a)
  1. Substitute r = −15 and t = 7: p² = (−15)² − 8 × 7.
  2. (−15)² = 225 (a negative number squared is positive), and 8 × 7 = 56.
  3. p² = 225 − 56 = 169.
  4. p = √169. Both 13 and −13 square to 169: p = ±13
Answer p = ±13
Part (b)
  1. Add 8t to both sides: p² + 8t = r².
  2. Subtract p² from both sides: 8t = r² − p².
  3. Divide by 8: t = r² − p²⁄8
Answer t = r² − p²⁄8
365 marksExtended

Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q10

The diagram shows a rhombus ABCD. Angle DAB = (−6x + 9y)° and angle ADC = (2x + 3y)°. Angle BCD is 117°. Write two simultaneous equations and solve them to find the value of x and the value of y. You must show all your working.

0580/42 February/March 2026 Q10 – rhombus ABCD with angles (−6x + 9y)° and (2x + 3y)°(−6x + 9y)°(2x + 3y)°ABCDNOT TOSCALE
Show solution
  1. In a rhombus, opposite angles are equal. Angle DAB is opposite angle BCD.
  2. Equation 1: −6x + 9y = 117
  3. Next-door angles in a rhombus add up to 180°, so angle ADC = 180 − 117 = 63°.
  4. Equation 2: 2x + 3y = 63
  5. Multiply equation 2 by 3 so the x terms cancel: 6x + 9y = 189.
  6. Add this to equation 1: (−6x + 6x) + (9y + 9y) = 117 + 189, so 18y = 306 and y = 17.
  7. Substitute y = 17 into equation 2: 2x + 51 = 63, so 2x = 12 and x = 6.
  8. x = 6, y = 17
  9. Check: −6 × 6 + 9 × 17 = −36 + 153 = 117 ✓
Answer −6x + 9y = 117 2x + 3y = 63 x = 6, y = 17
373 marksExtended

Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q12

Solve 4x − 3⁄5 = 7 − 2x⁄3.

Show solution
  1. The denominators are 5 and 3. Multiply both sides by 15 to clear the fractions.
  2. 3(4x − 3) = 5(7 − 2x).
  3. Expand: 12x − 9 = 35 − 10x.
  4. Add 10x to both sides: 22x − 9 = 35.
  5. Add 9: 22x = 44, so x = 2
  6. Check: (8 − 3) ÷ 5 = 1 and (7 − 4) ÷ 3 = 1 ✓
Answer x = 2
385 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q8

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.

Show solution
  1. Write an equation for each sentence.
  2. 3 oranges and 1 apple cost 51 cents: 3t + w = 51 (1)
  3. 6 oranges and 5 apples cost 129 cents: 6t + 5w = 129 (2)
  4. Double (1) so the t terms match: 6t + 2w = 102 (3).
  5. Subtract (3) from (2): 3w = 27, so w = 9.
  6. Substitute w = 9 into (1): 3t + 9 = 51, so 3t = 42 and t = 14.
  7. t = 14, w = 9
  8. Check in (2): 6 × 14 + 5 × 9 = 84 + 45 = 129 ✓
Answer 3t + w = 51 6t + 5w = 129 t = 14, w = 9
395 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q17

I = M(k² + c²). (a) Find the value of I when M = 7, k = 3 and c = 2. (b) Rearrange the formula to make k the subject.

Show solution
Part (a)
  1. Substitute the values: I = 7 × (3² + 2²).
  2. Work out the bracket first: 9 + 4 = 13.
  3. I = 7 × 13 = 91
Answer 91
Part (b)
  1. Divide both sides by M: I⁄M = k² + c².
  2. Subtract c² from both sides: k² = I⁄M − c².
  3. Take the square root: k = √(I⁄M − c²)
Answer k = √(I⁄M − c²)
405 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q22

Solve 2⁄x − 1 = x⁄x + 2.

Show solution
  1. Cross-multiply to remove the fractions: 2(x + 2) = x(x − 1).
  2. Expand both sides: 2x + 4 = x² − x.
  3. Make one side 0: 0 = x² − x − 2x − 4, so x² − 3x − 4 = 0.
  4. Factorise: find two numbers that multiply to −4 and add to −3: −4 and +1. (x − 4)(x + 1) = 0.
  5. x = −1 or x = 4
Answer x = −1 or x = 4
415 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q10

b = dm + 2mk

(a) d = 3.14, m = 7.92 and k = 10.16. By rounding each value correct to 1 significant figure, work out an estimate for b.

(b) Rearrange the formula to make m the subject.

Show solution
Part (a)
  1. Round each value to 1 significant figure: d = 3.14 → 3, m = 7.92 → 8, k = 10.16 → 10.
  2. dm = 3 × 8 = 24. 2mk = 2 × 8 × 10 = 160.
  3. b ≈ 24 + 160 = 184
Answer 184
Part (b)
  1. m is in both terms, so factorise it out: b = m(d + 2k).
  2. Divide both sides by (d + 2k): m = b⁄d + 2k
Answer m = b⁄d + 2k
426 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q27

Solve the simultaneous equations y = x² − 8x + 22 and y + 2 = 3x.

Show solution
  1. Rearrange the second equation to make y the subject: y = 3x − 2.
  2. Substitute into the first equation: 3x − 2 = x² − 8x + 22.
  3. Move everything to one side: 0 = x² − 8x − 3x + 22 + 2, so x² − 11x + 24 = 0.
  4. Factorise: two numbers that multiply to 24 and add to −11 are −3 and −8. (x − 3)(x − 8) = 0.
  5. x = 3 or x = 8. Find y using y = 3x − 2: x = 3 gives y = 7, and x = 8 gives y = 22.
  6. x = 3, y = 7 and x = 8, y = 22
Answer x = 3, y = 7 x = 8, y = 22
432 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q2

Solve 11 − 2x = 4.

Show solution
  1. Start with 11 − 2x = 4.
  2. Add 2x to both sides: 11 = 4 + 2x.
  3. Subtract 4 from both sides: 7 = 2x.
  4. Divide both sides by 2: x = 3.5
  5. Check: 11 − 2 × 3.5 = 11 − 7 = 4 ✓
Answer x = 3.5
4411 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q24

Martha walks a distance of 10 km at a speed of x km/h. She then runs a distance of 5 km at a speed of (x + 4) km/h. The total time taken for the whole journey is 3.5 hours.

(a) Write down an expression in terms of x for the time Martha is walking.

(b) Show that 7x² − 2x − 80 = 0.

(c) Solve 7x² − 2x − 80 = 0, giving your answers correct to 2 decimal places. You must show all your working.

(d) Calculate the difference between the time Martha is walking and the time she is running. Give your answer in hours and minutes, correct to the nearest minute.

Show solution
Part (a)
  1. Time = distance ÷ speed.
  2. Walking time = 10⁄x hours
Answer 10⁄x hours
Part (b)
  1. Running time = 5⁄x + 4 hours.
  2. Total time is 3.5 hours: 10⁄x + 5⁄x + 4 = 3.5
  3. Multiply every term by x(x + 4) to clear the fractions: 10(x + 4) + 5x = 3.5x(x + 4)
  4. Expand: 10x + 40 + 5x = 3.5x² + 14x, so 15x + 40 = 3.5x² + 14x.
  5. Multiply by 2 to remove the decimal: 30x + 80 = 7x² + 28x.
  6. Rearrange so one side is 0: 7x² − 2x − 80 = 0 ✓
Answer 7x² − 2x − 80 = 0
Part (c)
  1. Use the quadratic formula with a = 7, b = −2, c = −80.
  2. x = −b ± √(b² − 4ac)⁄2a = 2 ± √((−2)² − 4 × 7 × (−80))⁄2 × 7
  3. b² − 4ac = 4 + 2240 = 2244, and √2244 = 47.371…
  4. x = 2 + 47.371…⁄14 = 3.526… or x = 2 − 47.371…⁄14 = −3.240…
  5. x = 3.53 or x = −3.24
Answer x = 3.53 or x = −3.24
Part (d)
  1. A speed cannot be negative, so x = 3.526… km/h.
  2. Walking time = 10 ÷ 3.526… = 2.836 h.
  3. Running time = 5 ÷ (3.526… + 4) = 5 ÷ 7.526… = 0.664 h.
  4. Difference = 2.836 − 0.664 = 2.171… h.
  5. 0.171… h × 60 = 10.3 minutes, so the difference is 2 h 10 min
Answer 2 h 10 min
454 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q26

Solve the simultaneous equations y = 2x² − 3x − 7 and y = 2x − 7. You must show all your working.

Show solution
  1. Both equations equal y, so set them equal to each other: 2x² − 3x − 7 = 2x − 7
  2. Subtract 2x and add 7 to both sides: 2x² − 5x = 0
  3. Factorise (x is common): x(2x − 5) = 0
  4. So x = 0 or 2x − 5 = 0, which gives x = 2.5.
  5. Find y using y = 2x − 7. When x = 0: y = −7. When x = 2.5: y = 5 − 7 = −2.
  6. x = 0, y = −7 and x = 2.5, y = −2
Answer x = 0, y = −7 x = 2.5, y = −2
462 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q2

Solve 8x − 17 = 27.

Show solution
  1. Add 17 to both sides: 8x = 27 + 17 = 44.
  2. Divide both sides by 8: x = 44 ÷ 8.
  3. x = 5.5
Answer x = 5.5
476 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q19

Solve the simultaneous equations y = 5 − 2x and y = 3x² − 7x − 6. You must show all your working and give your answers correct to 2 decimal places.

Show solution
  1. Both equations equal y, so set them equal: 3x² − 7x − 6 = 5 − 2x.
  2. Move everything to one side: 3x² − 7x + 2x − 6 − 5 = 0, so 3x² − 5x − 11 = 0.
  3. This does not factorise, so use the quadratic formula with a = 3, b = −5, c = −11.
  4. x = 5 ± √((−5)² − 4 × 3 × (−11))⁄2 × 3 = 5 ± √157⁄6.
  5. √157 = 12.529…, so x = 17.529…⁄6 = 2.921… or x = −7.529…⁄6 = −1.254…
  6. Find y using y = 5 − 2x. x = 2.921… gives y = −0.843…; x = −1.254… gives y = 7.509…
  7. x = 2.92, y = −0.84 and x = −1.25, y = 7.51
Answer x = 2.92, y = −0.84 x = −1.25, y = 7.51
482 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q10

Solve the simultaneous equations 3x + 5y = 5 and 2x − 5y = 45. You must show all your working.

Show solution
  1. The y terms are +5y and −5y, so adding the two equations will remove y.
  2. (3x + 2x) + (5y − 5y) = 5 + 45, so 5x = 50 and x = 10.
  3. Substitute x = 10 into the first equation: 30 + 5y = 5.
  4. Subtract 30: 5y = −25, so y = −5.
  5. x = 10, y = −5
  6. Check: 2 × 10 − 5 × (−5) = 20 + 25 = 45 ✓
Answer x = 10, y = −5
495 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q14

These expressions are all equal in value: 5x − 2⁄3, 10 − x and y + 11. Find the value of y.

Show solution
  1. Use the first two expressions to find x: 5x − 2⁄3 = 10 − x.
  2. Multiply both sides by 3: 5x − 2 = 30 − 3x.
  3. Add 3x to both sides: 8x − 2 = 30. Add 2: 8x = 32, so x = 4.
  4. The value of each expression is 10 − 4 = 6.
  5. So y + 11 = 6, and y = 6 − 11 = −5
Answer −5
503 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q17

Make t the subject of the formula x = 7t⁄5 − t.

Show solution
  1. Multiply both sides by (5 − t) to clear the fraction: x(5 − t) = 7t.
  2. Expand: 5x − xt = 7t.
  3. Get the t terms on one side: add xt to both sides, so 5x = 7t + xt.
  4. Factorise the t out: 5x = t(7 + x).
  5. Divide by (7 + x): t = 5x⁄x + 7
Answer t = 5x⁄x + 7
515 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q6

Solve. (a) 8x + 7 = 39 (b) 2(5y − 1) = 24

Show solution
Part (a)
  1. Subtract 7 from both sides: 8x = 39 − 7 = 32.
  2. Divide by 8: x = 4
Answer 4
Part (b)
  1. Divide both sides by 2: 5y − 1 = 12.
  2. Add 1: 5y = 13.
  3. Divide by 5: y = 2.6
Answer 2.6
525 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q24

The line y = 7x + 3 intersects the curve y = x² + 5x − 12 at the points A and B. Find the coordinates of A and B.

Show solution
  1. At A and B the line and the curve have the same y, so set them equal: x² + 5x − 12 = 7x + 3.
  2. Move everything to one side: x² + 5x − 7x − 12 − 3 = 0, so x² − 2x − 15 = 0.
  3. Factorise: two numbers that multiply to −15 and add to −2 are −5 and 3. (x − 5)(x + 3) = 0.
  4. x = 5 or x = −3.
  5. Find y using the line y = 7x + 3. x = 5: y = 35 + 3 = 38. x = −3: y = −21 + 3 = −18.
  6. (5, 38) and (−3, −18)
Answer (5, 38) and (−3, −18)

Question papers used on this page

All equations and formulae questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/11/M/J/260580_s26_qp_11May/June 2026 · Paper 1 (Core)Q21(a), Q23
0580/12/M/J/260580_s26_qp_12May/June 2026 · Paper 1 (Core)Q10, Q28
0580/13/M/J/260580_s26_qp_13May/June 2026 · Paper 1 (Core)Q22, Q23
0580/21/M/J/260580_s26_qp_21May/June 2026 · Paper 2 (Extended)Q2, Q6
0580/22/M/J/260580_s26_qp_22May/June 2026 · Paper 2 (Extended)Q6(b), Q25
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q17(b), Q18
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q8, Q20(a)
0580/32/M/J/260580_s26_qp_32May/June 2026 · Paper 3 (Core)Q14
0580/33/M/J/260580_s26_qp_33May/June 2026 · Paper 3 (Core)Q13
0580/42/M/J/260580_s26_qp_42May/June 2026 · Paper 4 (Extended)Q9, Q18
0580/12/F/M/260580_m26_qp_12Feb/Mar 2026 · Paper 1 (Core)Q20, Q27
0580/32/F/M/260580_m26_qp_32Feb/Mar 2026 · Paper 3 (Core)Q14
0580/42/F/M/260580_m26_qp_42Feb/Mar 2026 · Paper 4 (Extended)Q7, Q10, Q12
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q8, Q29
0580/12/O/N/250580_w25_qp_12Oct/Nov 2025 · Paper 1 (Core)Q23
0580/13/O/N/250580_w25_qp_13Oct/Nov 2025 · Paper 1 (Core)Q12
0580/21/O/N/250580_w25_qp_21Oct/Nov 2025 · Paper 2 (Extended)Q8, Q17, Q22
0580/22/O/N/250580_w25_qp_22Oct/Nov 2025 · Paper 2 (Extended)Q10
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q27
0580/32/O/N/250580_w25_qp_32Oct/Nov 2025 · Paper 3 (Core)Q5
0580/33/O/N/250580_w25_qp_33Oct/Nov 2025 · Paper 3 (Core)Q24
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q2, Q24, Q26
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q2, Q19
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q10, Q14, Q17
0580/11/M/J/250580_s25_qp_11May/June 2025 · Paper 1 (Core)Q12, Q26
0580/12/M/J/250580_s25_qp_12May/June 2025 · Paper 1 (Core)Q15, Q24
0580/13/M/J/250580_s25_qp_13May/June 2025 · Paper 1 (Core)Q21, Q22, Q25
0580/22/M/J/250580_s25_qp_22May/June 2025 · Paper 2 (Extended)Q6, Q24

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