IGCSE Maths · Algebra and Graphs · Extended only

Algebraic Fractions: Simplifying, Combining and Cancelling Extended only

Algebraic fractions follow exactly the same rules as ordinary numeric fractions — the only new skill is that factorising is usually the key that unlocks simplifying and combining them. This entire note is Extended-only; Core has no equivalent section at all.

Why this matters Every rule in this note — common denominators, cancelling common factors, “keep-flip-multiply” for division — is identical to what you already do with numeric fractions like 3/4 or 5/6. If a step here feels unfamiliar, it’s worth checking the equivalent numeric fraction skill from the Number unit first; the algebra is rarely the hard part, the fraction mechanics are.

1. Simplifying Algebraic Fractions

To simplify an algebraic fraction, factorise the numerator and denominator fully, then cancel any factor that appears in both.

Worked example Simplify (x² − 2x) / (x² − 5x + 6).
Factorise the numerator: x² − 2x = x(x − 2)
Factorise the denominator: x² − 5x + 6 = (x − 2)(x − 3)
So the fraction is x(x − 2) / (x − 2)(x − 3) — cancel the shared factor (x − 2):
Answer: x / (x − 3)
Worked example — difference of two squares Simplify (x² − 9) / (x² + 5x + 6).
Factorise the numerator: x² − 9 = (x − 3)(x + 3)
Factorise the denominator: x² + 5x + 6 = (x + 2)(x + 3)
Cancel the shared factor (x + 3):
Answer: (x − 3) / (x + 2)
Try it — Simplifying
  1. Simplify (x² + 3x) / (x² + 5x + 6).
  2. Simplify (x² − 16) / (x² + 7x + 12).
  3. Simplify (2x² − 6x) / (x² − x − 6).
Show answers
  1. x(x+3) / [(x+2)(x+3)] = x / (x + 2)
  2. (x−4)(x+4) / [(x+3)(x+4)] = (x − 4) / (x + 3)
  3. 2x(x−3) / [(x−3)(x+2)] = 2x / (x + 2)

2. Multiplying Algebraic Fractions

Multiply numerators together and denominators together, then simplify — factorising first often makes the cancelling much easier.

Worked example — numeric coefficients Simplify (3a/4) × (9a/10).
Multiply straight across: (3a × 9a) / (4 × 10) = 27a² / 40
Worked example — with cancelling brackets Simplify [(x + 2)/(x − 1)] × [(x − 1)/(x + 5)].
The (x − 1) in the first denominator cancels with the (x − 1) in the second numerator:
Answer: (x + 2) / (x + 5)
Try it — Multiplying
  1. Simplify (2a/5) × (10a/3).
  2. Simplify [(x + 3)/(x + 4)] × [(x + 4)/(x − 2)].
  3. Simplify [(x − 5)/(2x)] × [(4x)/(x − 5)].
Show answers
  1. 20a² / 15 = 4a²/3
  2. (x + 3) / (x − 2)
  3. 4x / 2x = 2

3. Dividing Algebraic Fractions

Dividing by a fraction means multiplying by its reciprocal — flip the second fraction, then multiply as usual.

Worked example Simplify (3a/4) ÷ (9a/10).
Flip and multiply: (3a/4) × (10/9a) = 30a / 36a
Simplify: 5/6 (the a’s cancel, and 30/36 simplifies to 5/6)
Worked example — with brackets Simplify [(x + 3)/(2x)] ÷ [(x + 3)/(4x)].
Flip and multiply: [(x + 3)/(2x)] × [4x/(x + 3)]
The (x + 3) terms cancel, leaving 4x/2x = 2
Try it — Dividing
  1. Simplify (5a/6) ÷ (10a/9).
  2. Simplify [(x − 2)/(3x)] ÷ [(x − 2)/(6x)].
  3. Simplify [(x + 1)/(x − 4)] ÷ [(x + 1)/(2x − 8)] (hint: factorise 2x − 8 first).
Show answers
  1. (5a/6) × (9/10a) = 45a/60a = 3/4
  2. [(x−2)/3x] × [6x/(x−2)] = 6x/3x = 2
  3. 2x − 8 = 2(x − 4), so: [(x+1)/(x−4)] × [2(x−4)/(x+1)] = 2

4. Adding and Subtracting — Same Denominator

When the denominators already match, combine the numerators directly.

Worked examples Simplify 3/(x + 1) + 5/(x + 1).
Same denominator, so add the numerators: 8/(x + 1)

Simplify (2x + 1)/(x − 2) − (x − 3)/(x − 2).
Same denominator, so subtract the numerators (careful with brackets!):
[(2x + 1) − (x − 3)] / (x − 2) = (2x + 1 − x + 3) / (x − 2) = (x + 4)/(x − 2)
Try it — Same Denominator
  1. Simplify 4/(x + 3) + 2/(x + 3).
  2. Simplify 7/(2x) − 3/(2x).
  3. Simplify (3x + 2)/(x − 1) − (x − 5)/(x − 1).
Show answers
  1. 6/(x + 3)
  2. 4/(2x) = 2/x
  3. (3x+2−x+5)/(x−1) = (2x + 7)/(x − 1)

5. Adding and Subtracting — Different Denominators

Just like numeric fractions, you need a common denominator first — multiply each fraction so both share the same bottom, then combine the numerators.

Worked example — numeric denominators Simplify x/3 + (x − 4)/2.
Common denominator is 6: 2x/6 + 3(x − 4)/6
Combine: [2x + 3(x − 4)] / 6 = [2x + 3x − 12] / 6 = (5x − 12)/6
Worked example — subtracting with numeric denominators Simplify 2x/3 − 3(x − 5)/2.
Common denominator is 6: 4x/6 − 9(x − 5)/6
Combine: [4x − 9(x − 5)] / 6 = [4x − 9x + 45] / 6 = (45 − 5x)/6
Worked example — algebraic denominators Simplify 1/(x − 2) + (x + 1)/(x − 3).
The common denominator is the product of both: (x − 2)(x − 3)
Rewrite each fraction: (x − 3)/[(x − 2)(x − 3)] + (x + 1)(x − 2)/[(x − 2)(x − 3)]
Combine the numerators: (x − 3) + (x + 1)(x − 2)
Expand (x + 1)(x − 2) = x² − x − 2
Add: (x − 3) + (x² − x − 2) = x² − 5
Answer: (x² − 5) / [(x − 2)(x − 3)]
Try it — Different Denominators
  1. Simplify x/4 + (x + 2)/3.
  2. Simplify 3x/5 − (x − 1)/2.
  3. Simplify 2/(x + 1) + 3/(x + 4).
  4. Simplify 1/x − 1/(x + 3) (a good one to double-check by expanding fully).
Show answers
  1. Common denominator 12: 3x/12 + 4(x+2)/12 = [3x+4x+8]/12 = (7x + 8)/12
  2. Common denominator 10: 6x/10 − 5(x−1)/10 = [6x−5x+5]/10 = (x + 5)/10
  3. Common denominator (x+1)(x+4): [2(x+4) + 3(x+1)] / [(x+1)(x+4)] = [2x+8+3x+3] / [(x+1)(x+4)] = (5x + 11) / [(x+1)(x+4)]
  4. Common denominator x(x+3): [(x+3) − x] / [x(x+3)] = 3 / [x(x+3)]
Common mistakes to watch for
  • Cancelling terms instead of factors — you can only cancel something that’s multiplied across the whole numerator and whole denominator, never a term that’s just added or subtracted somewhere inside.
  • Forgetting brackets when subtracting a numerator with more than one term — (2x+1) − (x−3) is not 2x+1−x−3; the second bracket’s signs must flip: 2x+1−x+3.
  • Multiplying straight across when dividing, instead of flipping the second fraction first.
  • Simplifying a fraction before fully factorising — a factor hiding inside an unfactorised expression is easy to miss.
  • Using a common denominator that isn’t actually the product of both original denominators when they share no common factor.

6. Quick-Fire Challenge Round

True or False?
  1. (x + 3)/(x + 3) simplifies to 1, provided x ≠ −3.
  2. You can cancel the x’s in (x + 2)/x to get just 2.
  3. Dividing by a fraction is the same as multiplying by its reciprocal.
  4. To add two algebraic fractions with different denominators, you must first find a common denominator.
Show answers
  1. True.
  2. False — x is a term added to 2 in the numerator, not a factor of the whole numerator, so it cannot be cancelled with the x in the denominator.
  3. True.
  4. True.
Spot the Error
  1. “(x² + 2x)/(2x) simplifies to x² + 1.”
  2. “3/(x+1) + 2/(x+1) = 5/(2x+2).”
  3. “(2/3) ÷ (4/5) = (2/3) × (4/5).”
Show answers
  1. x² + 2x = x(x + 2), so (x²+2x)/2x = x(x+2)/2x = (x+2)/2, not x² + 1.
  2. The denominators already match, so only the numerators add: 3/(x+1) + 2/(x+1) = 5/(x+1), not 5/(2x+2) — that would double-count the denominator.
  3. Dividing means multiplying by the reciprocal of the second fraction: (2/3) ÷ (4/5) = (2/3) × (5/4), not (2/3) × (4/5).

7. Mixed Practice — Bringing It All Together

  1. Simplify (x² − 4x)/(x² − x − 12).
  2. Simplify [(x + 5)/(3x)] × [(6x)/(x + 5)].
  3. Simplify (7a/8) ÷ (14a/9).
  4. Simplify 5/(x − 3) + 2/(x − 3).
  5. Simplify x/5 + (x − 2)/4.
  6. Simplify 3/(x + 2) − 1/(x − 1).
Show answers
  1. x(x−4)/[(x−4)(x+3)] = x/(x + 3)
  2. 2
  3. (7a/8) × (9/14a) = 63a/112a = 9/16
  4. 7/(x − 3)
  5. Common denominator 20: 4x/20 + 5(x−2)/20 = (9x − 10)/20
  6. Common denominator (x+2)(x−1): [3(x−1) − (x+2)] / [(x+2)(x−1)] = (3x−3−x−2)/[(x+2)(x−1)] = (2x − 5)/[(x+2)(x−1)]

Practice at Home

A large set again — take it steadily across a few sittings.

Part A — Simplifying
  1. Simplify (x² + 4x)/(x² + 6x + 8).
  2. Simplify (x² − 25)/(x² + 2x − 15).
  3. Simplify (3x² − 9x)/(x² − 9).
Part B — Multiplying and dividing
  1. Simplify (4a/9) × (3a/8).
  2. Simplify [(x − 6)/(x + 1)] × [(x + 1)/(x + 7)].
  3. Simplify (5x/12) ÷ (10x/3).
  4. Simplify [(x + 4)/(5x)] ÷ [(x + 4)/(10x)].
Part C — Adding and subtracting, same denominator
  1. Simplify 6/(x + 5) + 3/(x + 5).
  2. Simplify (4x + 1)/(x − 3) − (x − 2)/(x − 3).
Part D — Adding and subtracting, different denominators
  1. Simplify x/6 + (x + 3)/4.
  2. Simplify 2x/3 − (x − 1)/5.
  3. Simplify 4/(x + 2) + 1/(x − 3).
  4. Simplify 2/x − 3/(x + 5).
Part E — Applied / challenge
  1. Simplify (x² − 1)/(x² + 3x − 4) fully — factorise both the numerator and denominator first.
  2. Simplify [(x + 2)/(x − 5)] ÷ [(x² − 4)/(2x − 10)] fully (hint: factorise everything before flipping and multiplying).
  3. Show that 1/(x − 1) − 1/(x + 1) simplifies to 2/(x² − 1).

Key Vocabulary Recap

Rational expression — a fraction where the numerator and/or denominator are algebraic expressions
Cancel — remove a factor common to both the numerator and denominator
Common denominator — a shared bottom used to combine fractions by adding or subtracting
Reciprocal — a fraction flipped upside down; used when dividing

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section E2.3 (Extended subject content) — Algebraic fractions: manipulating them (adding, subtracting, multiplying, dividing) and factorising to simplify rational expressions. There is no Core equivalent for this section.

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