Every algebraic fractions question from 8 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.3. 12 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/23/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 Algebraic Fractions past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q24
Write as a single fraction in its simplest form: x − 2⁄x + 3 + x + 3⁄x − 2
Show solution
- Use the common denominator (x + 3)(x − 2).
- Multiply each top by the other bottom: (x − 2)(x − 2) + (x + 3)(x + 3)⁄(x + 3)(x − 2)
- Expand: (x − 2)² = x² − 4x + 4 and (x + 3)² = x² + 6x + 9.
- Add: x² − 4x + 4 + x² + 6x + 9 = 2x² + 2x + 13.
- 2x² + 2x + 13⁄(x + 3)(x − 2)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q15
Write 8⁄x + 2 + 3⁄x − 4 as a single fraction in its simplest form.
Show solution
- Use the common denominator (x + 2)(x − 4).
- Multiply each top by the other bottom: 8(x − 4) + 3(x + 2)⁄(x + 2)(x − 4)
- Expand the top: 8x − 32 + 3x + 6.
- Collect like terms: 11x − 26.
- 11x − 26⁄(x + 2)(x − 4)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q22
Write 8x² + 24x⁄2x² − 18 as a single fraction in its simplest form.
Show solution
- Factorise the top: the common factor is 8x, so 8x² + 24x = 8x(x + 3).
- Factorise the bottom: take out 2, then use the difference of two squares. 2x² − 18 = 2(x² − 9) = 2(x − 3)(x + 3).
- The fraction is 8x(x + 3)⁄2(x − 3)(x + 3).
- Cancel the (x + 3) and divide 8 by 2: 4x⁄x − 3
Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q21
Simplify 2x² + 3x − 20⁄ax + 4a − 3bx − 12b.
Show solution
- Factorise the top: find two numbers that multiply to 2 × (−20) = −40 and add to 3: 8 and −5.
- 2x² + 8x − 5x − 20 = 2x(x + 4) − 5(x + 4) = (2x − 5)(x + 4).
- Factorise the bottom by grouping: ax + 4a = a(x + 4) and −3bx − 12b = −3b(x + 4).
- So the bottom = (a − 3b)(x + 4).
- Cancel the common factor (x + 4).
- 2x − 5⁄a − 3b
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q7
Simplify (a) p⁄t ÷ 2⁄t (b) 3x⁄4 − x − 1⁄2
Show solution
- To divide by a fraction, multiply by its reciprocal: p⁄t × t⁄2.
- The t on the top cancels the t on the bottom.
- p⁄2
- The common denominator is 4. Change x − 1⁄2 to 2(x − 1)⁄4.
- 3x⁄4 − 2(x − 1)⁄4 = 3x − 2(x − 1)⁄4
- Expand the bracket carefully: 3x − 2x + 2 = x + 2.
- x + 2⁄4
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q25
Simplify 10ax + 6bx − 25a − 15b⁄4x² − 25.
Show solution
- Factorise the top by grouping the terms in pairs.
- 10ax + 6bx = 2x(5a + 3b) and −25a − 15b = −5(5a + 3b).
- So the top = 2x(5a + 3b) − 5(5a + 3b) = (5a + 3b)(2x − 5).
- Factorise the bottom as a difference of two squares: 4x² − 25 = (2x + 5)(2x − 5).
- Cancel the common factor (2x − 5).
- 5a + 3b⁄2x + 5
Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q26
Write as a single fraction in its simplest form. (a) mp⁄25y × 15⁄m (b) 3⁄2x − 5 + 4⁄x − 3
Show solution
- Multiply the tops and the bottoms: mp × 15⁄25y × m = 15mp⁄25my.
- Cancel the m, and divide 15 and 25 by 5: 3p⁄5y
- Use the common denominator (2x − 5)(x − 3).
- Top: 3(x − 3) + 4(2x − 5) = 3x − 9 + 8x − 20 = 11x − 29.
- 11x − 29⁄(2x − 5)(x − 3)
Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q28
Simplify 2x² − 11x − 21⁄x² − 49.
Show solution
- Factorise the top: find two numbers that multiply to 2 × (−21) = −42 and add to −11: −14 and 3.
- 2x² − 14x + 3x − 21 = 2x(x − 7) + 3(x − 7) = (2x + 3)(x − 7).
- Factorise the bottom as a difference of two squares: x² − 49 = (x + 7)(x − 7).
- Cancel the common factor (x − 7).
- 2x + 3⁄x + 7
Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q13
Simplify 7⁄2m + 3⁄8m.
Show solution
- The lowest common denominator of 2m and 8m is 8m.
- Multiply the top and bottom of 7⁄2m by 4: 28⁄8m.
- Add: 28⁄8m + 3⁄8m = 31⁄8m
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q18
One day, Anya runs 12 km at a speed of x km/h. The next day she walks 10 km at a speed of (x − 4) km/h.
(a) Write down an expression, in terms of x, for the time she spends running.
(b) Write down an expression, in terms of x, for the time she spends walking.
(c) The time Anya spends walking is 1 hour more than the time she spends running. Write an equation in terms of x and show that it simplifies to x² − 2x − 48 = 0.
(d) Use factorisation to solve the equation x² − 2x − 48 = 0.
(e) Find the time Anya spends running.
Show solution
- Time = distance ÷ speed.
- Running time = 12⁄x hours
- Walking time = 10⁄x − 4 hours
- Walking takes 1 hour more than running: 10⁄x − 4 − 12⁄x = 1.
- Multiply every term by x(x − 4) to clear the fractions: 10x − 12(x − 4) = x(x − 4).
- Expand: 10x − 12x + 48 = x² − 4x.
- Simplify the left side: −2x + 48 = x² − 4x.
- Move everything to the right side: 0 = x² − 4x + 2x − 48.
- x² − 2x − 48 = 0
- Find two numbers that multiply to −48 and add to −2: −8 and +6.
- (x − 8)(x + 6) = 0, so x − 8 = 0 or x + 6 = 0.
- x = 8 or x = −6
- A speed cannot be negative, so x = 8 km/h.
- Running time = 12⁄8 = 1.5 hours
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q23
Simplify. 2x² + 10x⁄x² − 25
Show solution
- Factorise the top: 2x² + 10x = 2x(x + 5).
- Factorise the bottom using the difference of two squares: x² − 25 = (x − 5)(x + 5).
- Cancel the common factor (x + 5).
- 2x⁄x − 5
May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q19
Write as a single fraction in its simplest form. (a) 5a⁄6 × 3b⁄a (b) p⁄2 + 3t⁄4 (c) 2⁄x − 2 − 3⁄x + 1
Show solution
- Multiply the tops and the bottoms: 5a × 3b⁄6 × a = 15ab⁄6a.
- Cancel the a, and divide 15 and 6 by 3: 5b⁄2
- The common denominator is 4: p⁄2 = 2p⁄4.
- 2p⁄4 + 3t⁄4 = 2p + 3t⁄4
- Use the common denominator (x − 2)(x + 1).
- 2(x + 1)⁄(x − 2)(x + 1) − 3(x − 2)⁄(x − 2)(x + 1)
- Expand the top carefully: 2x + 2 − 3x + 6 = 8 − x.
- 8 − x⁄(x − 2)(x + 1)
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