What are algebraic fractions and how do they differ from numerical fractions?

Algebraic fractions are fractions where the numerator, the denominator, or both contain algebraic expressions, such as variables and constants. Unlike numerical fractions, which only involve numbers, algebraic fractions require understanding of algebraic operations to simplify or solve them. This concept is crucial in the Cambridge IGCSE Mathematics curriculum as it builds foundational skills for more advanced algebra topics.

How do you simplify algebraic fractions?

To simplify algebraic fractions, you need to factor both the numerator and the denominator to their simplest forms and then cancel out any common factors. For example, in the fraction (x^2 - 4)/(x^2 - 2x), factor the numerator as (x + 2)(x - 2) and the denominator as x(x - 2). The common factor (x - 2) can be canceled, simplifying the fraction to (x + 2)/x. This process is a key skill in the Cambridge IGCSE Mathematics syllabus.

How do you add or subtract algebraic fractions?

To add or subtract algebraic fractions, you must first find a common denominator, similar to numerical fractions. Once a common denominator is established, adjust the numerators accordingly and perform the addition or subtraction. For example, to add (1/x) + (2/y), the common denominator is xy, resulting in (y + 2x)/(xy). Mastery of this technique is essential for success in Cambridge IGCSE Algebra.

What is the process for multiplying and dividing algebraic fractions?

Multiplying algebraic fractions involves multiplying the numerators together and the denominators together. For example, (a/b) * (c/d) = (ac)/(bd). For division, multiply by the reciprocal of the divisor: (a/b) ÷ (c/d) = (a/b) * (d/c) = (ad)/(bc). Simplifying the resulting fraction is often necessary. These operations are fundamental in the Cambridge IGCSE Mathematics curriculum, reinforcing algebraic manipulation skills.

How do you solve equations involving algebraic fractions?

To solve equations with algebraic fractions, first eliminate the fractions by multiplying every term by the least common denominator (LCD) of all the fractions involved. This transforms the equation into a polynomial equation, which can be solved using standard algebraic methods. For example, solving (1/x) + (2/y) = 3 involves multiplying through by xy to get y + 2x = 3xy, then solving for the variable. This technique is a critical component of the Cambridge IGCSE Mathematics syllabus.

Why is "Cancelling individual terms instead of factors" a common mistake in Algebraic Fractions?

Why it happens: Students see x + 3/x + 1 and try to cancel the x's, leaving 3/1 = 3. How to avoid it: Cancellation only works for **shared factors** (the whole bracket). Factorise first, then cancel. Source: 0580/42 — examiner reports, recurring

Why is "Adding fractions without a common denominator" a common mistake in Algebraic Fractions?

Why it happens: Adding numerators and denominators directly: 2/x+1 + 3/x-2 ≠ 5/2x - 1. How to avoid it: Always rewrite over a common denominator first.

Why is "Flipping the wrong fraction in a division" a common mistake in Algebraic Fractions?

Why it happens: Students flip the first fraction instead of the second. How to avoid it: Rule: keep the first, change division to multiplication, flip the SECOND.

Why is "Not factorising before simplifying" a common mistake in Algebraic Fractions?

Why it happens: Skipping the factorisation step leaves no shared factors to cancel. How to avoid it: Step 1 is **always** factorise — both numerator and denominator. Then look for shared factors.

Cambridge IGCSE Mathematics (0580)

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Short Notes - Algebraic Fractions

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Algebraic Fractions — Cambridge IGCSE 0580 Maths Extended (2026)

Same arithmetic as ordinary fractions, just with letters. Add, subtract, multiply, divide and simplify rational expressions confidently — Paper 4 will reward fluency.

At a glance

Simplify by FACTORISING top and bottom, then cancelling COMMON FACTORS.
Cannot cancel additive terms: $\dfrac{x + 3}{x}$ is NOT $1 + 3$ .
Add/subtract: find a common denominator (often the product or LCM).
Multiply: top × top, bottom × bottom — cancel before multiplying.
Divide: multiply by the reciprocal (flip the second fraction).
Watch the SIGN when subtracting a fraction — distribute the minus.
Final answers should always be in simplest form.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.6 — Manipulate algebraic fractions including addition, subtraction, multiplication and division.

Simplifying algebraic fractions

Factorise top and bottom; cancel common FACTORS, never common terms inside an addition.

Procedure.

Factorise the numerator.
Factorise the denominator.
Cancel any factor common to both.

Worked. Simplify $\dfrac{6x^{2}}{9x}$ .

Both: factor numbers and letters separately. $\dfrac{6}{9} = \dfrac{2}{3}$ . $\dfrac{x^{2}}{x} = x$ .
Result: $\dfrac{2x}{3}$ .

Worked. Simplify $\dfrac{x^{2} - 9}{x + 3}$ .

Top is difference of two squares: $(x + 3)(x - 3)$ .
$\dfrac{(x + 3)(x - 3)}{x + 3} = x - 3$ .

Worked. Simplify $\dfrac{x^{2} + 5x + 6}{x + 2}$ .

Factorise top: $(x + 2)(x + 3)$ .
Cancel $(x + 2)$ : $x + 3$ .

Factorise the numerator, then cancel the common factor (x + 2).

Crucial caveat. $\dfrac{x + 3}{x}$ does NOT simplify to $1 + 3 = 4$ . The $x$ on the bottom is not a common FACTOR — only common factors cancel.

A factor multiplying the whole line cancels; a term that is only added does not.

Factorise BEFORE cancelling.
Cancel only common FACTORS.
Difference of two squares often appears on top.
Final answers in simplest form.

Adding and subtracting algebraic fractions

Find a common denominator first. Distribute the minus sign carefully when subtracting.

Same denominator → just combine numerators. $\dfrac{2x}{5} + \dfrac{x}{5} = \dfrac{3x}{5}.$

Different denominators → find a common one. The simplest common denominator is usually the PRODUCT, but the LCM is more efficient if there's a shared factor.

Worked. $\dfrac{2}{x} + \dfrac{3}{x + 1}$ .

Common denominator: $x(x + 1)$ .
$\dfrac{2(x + 1)}{x(x + 1)} + \dfrac{3x}{x(x + 1)} = \dfrac{2(x + 1) + 3x}{x(x + 1)} = \dfrac{2x + 2 + 3x}{x(x + 1)} = \dfrac{5x + 2}{x(x + 1)}$ .

Rewrite each fraction over x(x + 1), then add the numerators.

Worked (subtraction). $\dfrac{x + 1}{2} - \dfrac{x - 3}{4}$ .

Common denominator: $4$ .
$\dfrac{2(x + 1)}{4} - \dfrac{x - 3}{4} = \dfrac{2(x + 1) - (x - 3)}{4}$ .
Distribute the minus: $\dfrac{2x + 2 - x + 3}{4} = \dfrac{x + 5}{4}$ .

The sign trap. When subtracting, the minus applies to EVERY term in the numerator of the second fraction. Use brackets to keep yourself honest: $- (x - 3)$ becomes $-x + 3$ .

Common denominator first.
Combine numerators after equalising denominators.
Distribute the minus sign across the entire second numerator.
Simplify the final fraction if possible.

Multiplying and dividing algebraic fractions

Multiply: top × top, bottom × bottom, then simplify. Divide: flip the second fraction and multiply.

Multiplication. $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}$ . Cancel BEFORE multiplying.

Worked. $\dfrac{x + 2}{3} \times \dfrac{6}{x^{2} - 4}$ .

Factorise: $x^{2} - 4 = (x + 2)(x - 2)$ .
$\dfrac{(x + 2)}{3} \times \dfrac{6}{(x + 2)(x - 2)}$ .
Cancel $(x + 2)$ : $\dfrac{1}{3} \times \dfrac{6}{x - 2} = \dfrac{6}{3(x - 2)} = \dfrac{2}{x - 2}$ .

Factorise, cancel the common (x + 2), and reduce 6/3 to 2.

Division. $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ . Flip the second fraction, then multiply.

Dividing by c/d is the same as multiplying by its reciprocal d/c.

Worked. $\dfrac{x + 1}{2x} \div \dfrac{(x + 1)^{2}}{4}$ .

Flip: $\dfrac{x + 1}{2x} \times \dfrac{4}{(x + 1)^{2}}$ .
Cancel $(x + 1)$ once: $\dfrac{1}{2x} \times \dfrac{4}{x + 1} = \dfrac{4}{2x(x + 1)} = \dfrac{2}{x(x + 1)}$ .

Multiply: top × top, bottom × bottom.
Divide: multiply by the reciprocal of the second fraction.
Cancel BEFORE multiplying — saves work.
Always factorise both fractions first.

Quick recap

Simplify: factorise top and bottom, cancel common FACTORS only.
$\dfrac{x + 3}{x}$ does NOT simplify (no factor of $x$ on top).
Add/subtract: common denominator first; distribute the minus sign.
Multiply: top × top, bottom × bottom, cancel first.
Divide: flip the second fraction and multiply.
Final answers should be fully simplified.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Algebraic fraction — a fraction whose numerator or denominator (or both) contains algebraic expressions.
Common denominator — a single denominator shared by two or more fractions, used for adding/subtracting.
Reciprocal — for $\dfrac{a}{b}$ , the reciprocal is $\dfrac{b}{a}$ (for non-zero $a$ ).

How it’s examined

Algebraic fractions show up most years on Paper 4 as 3-4 mark questions: simplify, then sometimes use the result inside an equation. Paper 2 has the simpler form (single fraction simplification, 1-2 marks). Examiner reports flag the sign mistake when subtracting (forgetting to distribute the minus) and cancelling individual additive terms.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.6); 0580/22 May/Jun 2024 — Q14 (simplify fraction); 0580/42 Oct/Nov 2024 — Q11 (combine fractions); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Algebraic Fractions

Step-by-step solutions to past-paper-style questions on algebraic fractions, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Algebraic Fractions

Almost every algebraic fractions exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — simplify or express as a single fraction in simplest form — on one fraction or one $+,-,\times,\div$ between two fractions.

How to approach it

Factorise every numerator and denominator first. For $+$ or $-$ build a common denominator; for $\div$ flip the second fraction and multiply; then cancel shared factors.

Common trap

Cancelling individual terms instead of whole factors — $\dfrac{x+3}{x+1}$ does not become $3$ . Examiner reports flag this every series; only entire brackets cancel.

Multi-step problem

Recognise it by

A fraction equation to solve, or a chained combine then factorise then simplify task — several methods linked together.

How to approach it

For an equation, multiply every term by the common denominator to clear fractions, rearrange to a standard quadratic, then solve (factorise or formula) and check the discriminant.

Common trap

Not stating the discriminant when there are no real roots, or forgetting to clear every term by the common denominator. The mark scheme still credits the clearing step even if no real solutions exist.

1Simplify by factorising and cancelling
ExtendedDirect calculation• simplifying
Question
Simplify $\dfrac{x^{2} - 9}{x^{2} - 4x + 3}$ .
Step-by-step solution
1. Step 1
  Factorise numerator (difference of two squares).
  $x^{2} - 9 = (x + 3)(x - 3)$
2. Step 2
  Factorise denominator.
  $x^{2} - 4x + 3 = (x - 1)(x - 3)$
3. Step 3
  Cancel the common factor $(x-3)$ .
  $\dfrac{(x+3)(x-3)}{(x-1)(x-3)} = \dfrac{x+3}{x-1}$
Answer
$\dfrac{x+3}{x-1}$
Examiner tip
You can only cancel factors, never individual terms. $\dfrac{x+3}{x+1}$ does NOT simplify to $\dfrac{3}{1} = 3$ .
2Add two algebraic fractions
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q14• addition
Question
Express $\dfrac{2}{x + 1} + \dfrac{3}{x - 2}$ as a single fraction in simplest form.
Step-by-step solution
1. Step 1
  Common denominator is $(x+1)(x-2)$ .
  $\dfrac{2(x-2)}{(x+1)(x-2)} + \dfrac{3(x+1)}{(x+1)(x-2)}$
2. Step 2
  Combine numerators.
  $\dfrac{2(x-2) + 3(x+1)}{(x+1)(x-2)}$
3. Step 3
  Expand and simplify the numerator.
  $\dfrac{2x - 4 + 3x + 3}{(x+1)(x-2)} = \dfrac{5x - 1}{(x+1)(x-2)}$
Answer
$\dfrac{5x - 1}{(x + 1)(x - 2)}$
3Multiply algebraic fractions
ExtendedDirect calculation• multiplication
Question
Simplify $\dfrac{x^{2} - 1}{x + 2} \times \dfrac{x + 2}{x - 1}$ .
Step-by-step solution
1. Step 1
  Factorise the difference of two squares.
  $\dfrac{(x+1)(x-1)}{x+2} \times \dfrac{x+2}{x-1}$
2. Step 2
  Cancel $(x+2)$ and $(x-1)$ .
  $= x + 1$
Answer
$x + 1$
4Divide algebraic fractions
ExtendedDirect calculation• division
Question
Simplify $\dfrac{2x + 6}{x^{2}} \div \dfrac{x + 3}{4x}$ .
Step-by-step solution
1. Step 1
  Flip the second fraction and multiply.
  $\dfrac{2x + 6}{x^{2}} \times \dfrac{4x}{x + 3}$
2. Step 2
  Factorise $2x + 6 = 2(x + 3)$ .
  $\dfrac{2(x+3)}{x^{2}} \times \dfrac{4x}{x+3}$
3. Step 3
  Cancel $(x+3)$ and one $x$ .
  $= \dfrac{2 \cdot 4}{x} = \dfrac{8}{x}$
Answer
$\dfrac{8}{x}$
5Add a constant and a fraction with a linear denominator
CoreDirect calculation• addition, common denominator
Question
Express $\dfrac{3}{x} + \dfrac{2}{5}$ as a single fraction.
Step-by-step solution
1. Step 1
  The common denominator is $5x$ .
  $\dfrac{3}{x} = \dfrac{15}{5x},\quad \dfrac{2}{5} = \dfrac{2x}{5x}$
2. Step 2
  Add the numerators over the common denominator.
  $\dfrac{15 + 2x}{5x}$
Answer
$\dfrac{2x + 15}{5x}$
Examiner tip
Examiners reward candidates who write each term over the common denominator before combining. Adding numerators directly without matching denominators is a guaranteed zero.
6Subtract a fraction with a factorisable denominator
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q15• subtraction, factorising
Question
Express $\dfrac{3}{x - 2} - \dfrac{1}{x^{2} - 4}$ as a single fraction in simplest form.
Step-by-step solution
1. Step 1
  Factorise $x^{2} - 4 = (x - 2)(x + 2)$ .
2. Step 2
  The common denominator is $(x - 2)(x + 2)$ . Rewrite the first fraction.
  $\dfrac{3}{x - 2} = \dfrac{3(x + 2)}{(x - 2)(x + 2)}$
3. Step 3
  Subtract the numerators over the common denominator.
  $\dfrac{3(x + 2) - 1}{(x - 2)(x + 2)}$
4. Step 4
  Expand and simplify the numerator.
  $= \dfrac{3x + 6 - 1}{(x - 2)(x + 2)} = \dfrac{3x + 5}{(x - 2)(x + 2)}$
Answer
$\dfrac{3x + 5}{(x - 2)(x + 2)}$
Examiner tip
The 2024 mark scheme awards method marks for spotting that $x^{2} - 4$ already contains $(x - 2)$ as a factor — so the LCM is $(x-2)(x+2)$ , not $(x-2)(x^{2}-4)$ .
7Simplify a fraction by factorising a trinomial
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q15• simplifying, trinomial
Question
Simplify $\dfrac{x^{2} - 9}{x^{2} - x - 6}$ .
Step-by-step solution
1. Step 1
  Factorise the numerator (difference of two squares).
  $x^{2} - 9 = (x + 3)(x - 3)$
2. Step 2
  Factorise the denominator: find numbers whose product is $-6$ and sum is $-1$ . Try $-3$ and $2$ .
  $x^{2} - x - 6 = (x - 3)(x + 2)$
3. Step 3
  Cancel the common factor $(x - 3)$ .
  $\dfrac{(x + 3)(x - 3)}{(x - 3)(x + 2)} = \dfrac{x + 3}{x + 2}$
Answer
$\dfrac{x + 3}{x + 2}$
Examiner tip
The examiner report flags candidates often try to cancel terms ( $x^{2}$ in numerator with $x^{2}$ in denominator) before factorising. Only entire factors cancel — factorise first, then cancel.
8Solve an equation containing algebraic fractions
ExtendedMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q17• solving, equations
Question
Solve $\dfrac{4}{x + 1} - \dfrac{1}{x - 2} = 1$ .
Step-by-step solution
1. Step 1
  Multiply every term by the common denominator $(x + 1)(x - 2)$ .
  $4(x - 2) - 1(x + 1) = (x + 1)(x - 2)$
2. Step 2
  Expand each side.
  $4x - 8 - x - 1 = x^{2} - x - 2$
3. Step 3
  Collect on the left, then move to one side.
  $3x - 9 = x^{2} - x - 2$
4. Step 4
  Rearrange to a standard quadratic.
  $0 = x^{2} - 4x + 7 \Rightarrow x^{2} - 4x + 7 = 0$
5. Step 5
  Compute the discriminant: $\Delta = (-4)^{2} - 4(1)(7) = 16 - 28 = -12 < 0$ . There are no real solutions because the discriminant is negative.
Answer
No real solutions ( $\Delta < 0$ ).
Examiner tip
The 2024 mark scheme still awards method marks for correctly clearing the fractions and reaching a quadratic, even when there are no real roots. State the discriminant value to secure the final mark.
9Multiply algebraic fractions requiring factorising
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q13• multiplication, factorising
Question
Simplify $\dfrac{x^{2} + 3x}{x^{2} - 4} \times \dfrac{x - 2}{2x}$ .
Step-by-step solution
1. Step 1
  Factorise where possible.
  $x^{2} + 3x = x(x + 3),\ x^{2} - 4 = (x - 2)(x + 2)$
2. Step 2
  Rewrite the product.
  $\dfrac{x(x + 3)}{(x - 2)(x + 2)} \times \dfrac{x - 2}{2x}$
3. Step 3
  Cancel $x$ and $(x - 2)$ .
  $= \dfrac{x + 3}{2(x + 2)}$
Answer
$\dfrac{x + 3}{2(x + 2)}$
Examiner tip
The examiner report flags candidates often multiply numerators and denominators first, then try to factor a quartic mess. Factorise everything FIRST, then cancel.
10Combine, factorise and simplify
ChallengeMulti-step problem• Adapted from 0580/42 Feb/Mar 2024 Q20• multi-step, quadratic
Question
Simplify $\dfrac{2x^{2} - 8}{x^{2} + 5x + 6} \div \dfrac{x - 2}{x + 3}$ .
Step-by-step solution
1. Step 1
  Rewrite the division as multiplication by the reciprocal.
  $\dfrac{2x^{2} - 8}{x^{2} + 5x + 6} \times \dfrac{x + 3}{x - 2}$
2. Step 2
  Factorise each polynomial: $2x^{2} - 8 = 2(x - 2)(x + 2)$ and $x^{2} + 5x + 6 = (x + 2)(x + 3)$ .
  $\dfrac{2(x - 2)(x + 2)}{(x + 2)(x + 3)} \times \dfrac{x + 3}{x - 2}$
3. Step 3
  Cancel $(x + 2)$ , $(x + 3)$ and $(x - 2)$ .
  $= 2$
Answer
$2$
Examiner tip
A* candidates secure full marks by factorising every polynomial completely before cancelling. The mark scheme allows partial credit for any correct factorisation step, so always show the factorisations explicitly.

Key Formulae — Algebraic Fractions

The formulae you need to memorise for algebraic fractions on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Cancellation rule
$\dfrac{ac}{bc} = \dfrac{a}{b},\quad c \neq 0$
When to use
Only when the same FACTOR appears in numerator and denominator — never with individual terms.
Adding fractions over a common denominator
$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}$
When to use
Adding or subtracting any two fractions with different denominators.
Dividing fractions
$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$
When to use
Always rewrite a division as multiplication by the reciprocal.

Key Definitions and Keywords — Algebraic Fractions

Definitions to memorise and the exact keywords mark schemes credit for algebraic fractions answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Algebraic fraction
Examiner keyword
A fraction whose numerator and/or denominator contain algebraic expressions (variables).
Example
$\dfrac{x + 1}{x - 2}$
Common denominator
Examiner keyword
A denominator shared by two or more fractions, used to combine them by addition or subtraction.
Reciprocal of a fraction
The reciprocal of $\dfrac{a}{b}$ is $\dfrac{b}{a}$ .
Excluded values
Values of the variable that make any denominator zero — these are excluded from the domain because division by zero is undefined.
Example
In $\dfrac{1}{x - 2}$ , $x = 2$ is excluded.

Common Mistakes and Misconceptions — Algebraic Fractions

The traps other students keep falling into on algebraic fractions questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Cancelling individual terms instead of factors
0580/42 — examiner reports, recurring
Why it happens
Students see $\dfrac{x + 3}{x + 1}$ and try to cancel the $x$ 's, leaving $\dfrac{3}{1} = 3$ .
How to avoid it
Cancellation only works for shared factors (the whole bracket). Factorise first, then cancel.
✕Adding fractions without a common denominator
Why it happens
Adding numerators and denominators directly: $\dfrac{2}{x+1} + \dfrac{3}{x-2} \neq \dfrac{5}{2x - 1}$ .
How to avoid it
Always rewrite over a common denominator first.
✕Flipping the wrong fraction in a division
Why it happens
Students flip the first fraction instead of the second.
How to avoid it
Rule: keep the first, change division to multiplication, flip the SECOND.
✕Not factorising before simplifying
Why it happens
Skipping the factorisation step leaves no shared factors to cancel.
How to avoid it
Step 1 is always factorise — both numerator and denominator. Then look for shared factors.

Algebraic Fractions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Mathematics (0580)

Algebraic Fractions

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Short Notes - Algebraic Fractions

Detailed Study Notes

Detailed notes on Algebra for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Algebraic Fractions — Cambridge IGCSE 0580 Maths Extended (2026)

Same arithmetic as ordinary fractions, just with letters. Add, subtract, multiply, divide and simplify rational expressions confidently — Paper 4 will reward fluency.

At a glance

Simplify by FACTORISING top and bottom, then cancelling COMMON FACTORS.
Cannot cancel additive terms: $\dfrac{x + 3}{x}$ is NOT $1 + 3$ .
Add/subtract: find a common denominator (often the product or LCM).
Multiply: top × top, bottom × bottom — cancel before multiplying.
Divide: multiply by the reciprocal (flip the second fraction).
Watch the SIGN when subtracting a fraction — distribute the minus.
Final answers should always be in simplest form.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.6 — Manipulate algebraic fractions including addition, subtraction, multiplication and division.

Simplifying algebraic fractions

Factorise top and bottom; cancel common FACTORS, never common terms inside an addition.

Procedure.

Factorise the numerator.
Factorise the denominator.
Cancel any factor common to both.

Worked. Simplify $\dfrac{6x^{2}}{9x}$ .

Both: factor numbers and letters separately. $\dfrac{6}{9} = \dfrac{2}{3}$ . $\dfrac{x^{2}}{x} = x$ .
Result: $\dfrac{2x}{3}$ .

Worked. Simplify $\dfrac{x^{2} - 9}{x + 3}$ .

Top is difference of two squares: $(x + 3)(x - 3)$ .
$\dfrac{(x + 3)(x - 3)}{x + 3} = x - 3$ .

Worked. Simplify $\dfrac{x^{2} + 5x + 6}{x + 2}$ .

Factorise top: $(x + 2)(x + 3)$ .
Cancel $(x + 2)$ : $x + 3$ .

Factorise the numerator, then cancel the common factor (x + 2).

Crucial caveat. $\dfrac{x + 3}{x}$ does NOT simplify to $1 + 3 = 4$ . The $x$ on the bottom is not a common FACTOR — only common factors cancel.

A factor multiplying the whole line cancels; a term that is only added does not.

Factorise BEFORE cancelling.
Cancel only common FACTORS.
Difference of two squares often appears on top.
Final answers in simplest form.

Adding and subtracting algebraic fractions

Find a common denominator first. Distribute the minus sign carefully when subtracting.

Same denominator → just combine numerators. $\dfrac{2x}{5} + \dfrac{x}{5} = \dfrac{3x}{5}.$

Different denominators → find a common one. The simplest common denominator is usually the PRODUCT, but the LCM is more efficient if there's a shared factor.

Worked. $\dfrac{2}{x} + \dfrac{3}{x + 1}$ .

Common denominator: $x(x + 1)$ .
$\dfrac{2(x + 1)}{x(x + 1)} + \dfrac{3x}{x(x + 1)} = \dfrac{2(x + 1) + 3x}{x(x + 1)} = \dfrac{2x + 2 + 3x}{x(x + 1)} = \dfrac{5x + 2}{x(x + 1)}$ .

Rewrite each fraction over x(x + 1), then add the numerators.

Worked (subtraction). $\dfrac{x + 1}{2} - \dfrac{x - 3}{4}$ .

Common denominator: $4$ .
$\dfrac{2(x + 1)}{4} - \dfrac{x - 3}{4} = \dfrac{2(x + 1) - (x - 3)}{4}$ .
Distribute the minus: $\dfrac{2x + 2 - x + 3}{4} = \dfrac{x + 5}{4}$ .

The sign trap. When subtracting, the minus applies to EVERY term in the numerator of the second fraction. Use brackets to keep yourself honest: $- (x - 3)$ becomes $-x + 3$ .

Common denominator first.
Combine numerators after equalising denominators.
Distribute the minus sign across the entire second numerator.
Simplify the final fraction if possible.

Multiplying and dividing algebraic fractions

Multiply: top × top, bottom × bottom, then simplify. Divide: flip the second fraction and multiply.

Multiplication. $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}$ . Cancel BEFORE multiplying.

Worked. $\dfrac{x + 2}{3} \times \dfrac{6}{x^{2} - 4}$ .

Factorise: $x^{2} - 4 = (x + 2)(x - 2)$ .
$\dfrac{(x + 2)}{3} \times \dfrac{6}{(x + 2)(x - 2)}$ .
Cancel $(x + 2)$ : $\dfrac{1}{3} \times \dfrac{6}{x - 2} = \dfrac{6}{3(x - 2)} = \dfrac{2}{x - 2}$ .

Factorise, cancel the common (x + 2), and reduce 6/3 to 2.

Division. $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ . Flip the second fraction, then multiply.

Dividing by c/d is the same as multiplying by its reciprocal d/c.

Worked. $\dfrac{x + 1}{2x} \div \dfrac{(x + 1)^{2}}{4}$ .

Flip: $\dfrac{x + 1}{2x} \times \dfrac{4}{(x + 1)^{2}}$ .
Cancel $(x + 1)$ once: $\dfrac{1}{2x} \times \dfrac{4}{x + 1} = \dfrac{4}{2x(x + 1)} = \dfrac{2}{x(x + 1)}$ .

Multiply: top × top, bottom × bottom.
Divide: multiply by the reciprocal of the second fraction.
Cancel BEFORE multiplying — saves work.
Always factorise both fractions first.

Quick recap

Simplify: factorise top and bottom, cancel common FACTORS only.
$\dfrac{x + 3}{x}$ does NOT simplify (no factor of $x$ on top).
Add/subtract: common denominator first; distribute the minus sign.
Multiply: top × top, bottom × bottom, cancel first.
Divide: flip the second fraction and multiply.
Final answers should be fully simplified.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Algebraic fraction — a fraction whose numerator or denominator (or both) contains algebraic expressions.
Common denominator — a single denominator shared by two or more fractions, used for adding/subtracting.
Reciprocal — for $\dfrac{a}{b}$ , the reciprocal is $\dfrac{b}{a}$ (for non-zero $a$ ).

How it’s examined

Take this whole topic with you

Step-by-step worked examples — Algebraic Fractions

Step-by-step solutions to past-paper-style questions on algebraic fractions, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Algebraic Fractions

Almost every algebraic fractions exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — simplify or express as a single fraction in simplest form — on one fraction or one $+,-,\times,\div$ between two fractions.

How to approach it

Factorise every numerator and denominator first. For $+$ or $-$ build a common denominator; for $\div$ flip the second fraction and multiply; then cancel shared factors.

Common trap

Cancelling individual terms instead of whole factors — $\dfrac{x+3}{x+1}$ does not become $3$ . Examiner reports flag this every series; only entire brackets cancel.

Multi-step problem

Recognise it by

A fraction equation to solve, or a chained combine then factorise then simplify task — several methods linked together.

How to approach it

For an equation, multiply every term by the common denominator to clear fractions, rearrange to a standard quadratic, then solve (factorise or formula) and check the discriminant.

Common trap

1Simplify by factorising and cancelling
ExtendedDirect calculation• simplifying
Question
Simplify $\dfrac{x^{2} - 9}{x^{2} - 4x + 3}$ .
Step-by-step solution
1. Step 1
  Factorise numerator (difference of two squares).
  $x^{2} - 9 = (x + 3)(x - 3)$
2. Step 2
  Factorise denominator.
  $x^{2} - 4x + 3 = (x - 1)(x - 3)$
3. Step 3
  Cancel the common factor $(x-3)$ .
  $\dfrac{(x+3)(x-3)}{(x-1)(x-3)} = \dfrac{x+3}{x-1}$
Answer
$\dfrac{x+3}{x-1}$
Examiner tip
You can only cancel factors, never individual terms. $\dfrac{x+3}{x+1}$ does NOT simplify to $\dfrac{3}{1} = 3$ .
2Add two algebraic fractions
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q14• addition
Question
Express $\dfrac{2}{x + 1} + \dfrac{3}{x - 2}$ as a single fraction in simplest form.
Step-by-step solution
1. Step 1
  Common denominator is $(x+1)(x-2)$ .
  $\dfrac{2(x-2)}{(x+1)(x-2)} + \dfrac{3(x+1)}{(x+1)(x-2)}$
2. Step 2
  Combine numerators.
  $\dfrac{2(x-2) + 3(x+1)}{(x+1)(x-2)}$
3. Step 3
  Expand and simplify the numerator.
  $\dfrac{2x - 4 + 3x + 3}{(x+1)(x-2)} = \dfrac{5x - 1}{(x+1)(x-2)}$
Answer
$\dfrac{5x - 1}{(x + 1)(x - 2)}$
3Multiply algebraic fractions
ExtendedDirect calculation• multiplication
Question
Simplify $\dfrac{x^{2} - 1}{x + 2} \times \dfrac{x + 2}{x - 1}$ .
Step-by-step solution
1. Step 1
  Factorise the difference of two squares.
  $\dfrac{(x+1)(x-1)}{x+2} \times \dfrac{x+2}{x-1}$
2. Step 2
  Cancel $(x+2)$ and $(x-1)$ .
  $= x + 1$
Answer
$x + 1$
4Divide algebraic fractions
ExtendedDirect calculation• division
Question
Simplify $\dfrac{2x + 6}{x^{2}} \div \dfrac{x + 3}{4x}$ .
Step-by-step solution
1. Step 1
  Flip the second fraction and multiply.
  $\dfrac{2x + 6}{x^{2}} \times \dfrac{4x}{x + 3}$
2. Step 2
  Factorise $2x + 6 = 2(x + 3)$ .
  $\dfrac{2(x+3)}{x^{2}} \times \dfrac{4x}{x+3}$
3. Step 3
  Cancel $(x+3)$ and one $x$ .
  $= \dfrac{2 \cdot 4}{x} = \dfrac{8}{x}$
Answer
$\dfrac{8}{x}$
5Add a constant and a fraction with a linear denominator
CoreDirect calculation• addition, common denominator
Question
Express $\dfrac{3}{x} + \dfrac{2}{5}$ as a single fraction.
Step-by-step solution
1. Step 1
  The common denominator is $5x$ .
  $\dfrac{3}{x} = \dfrac{15}{5x},\quad \dfrac{2}{5} = \dfrac{2x}{5x}$
2. Step 2
  Add the numerators over the common denominator.
  $\dfrac{15 + 2x}{5x}$
Answer
$\dfrac{2x + 15}{5x}$
Examiner tip
Examiners reward candidates who write each term over the common denominator before combining. Adding numerators directly without matching denominators is a guaranteed zero.
6Subtract a fraction with a factorisable denominator
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q15• subtraction, factorising
Question
Express $\dfrac{3}{x - 2} - \dfrac{1}{x^{2} - 4}$ as a single fraction in simplest form.
Step-by-step solution
1. Step 1
  Factorise $x^{2} - 4 = (x - 2)(x + 2)$ .
2. Step 2
  The common denominator is $(x - 2)(x + 2)$ . Rewrite the first fraction.
  $\dfrac{3}{x - 2} = \dfrac{3(x + 2)}{(x - 2)(x + 2)}$
3. Step 3
  Subtract the numerators over the common denominator.
  $\dfrac{3(x + 2) - 1}{(x - 2)(x + 2)}$
4. Step 4
  Expand and simplify the numerator.
  $= \dfrac{3x + 6 - 1}{(x - 2)(x + 2)} = \dfrac{3x + 5}{(x - 2)(x + 2)}$
Answer
$\dfrac{3x + 5}{(x - 2)(x + 2)}$
Examiner tip
The 2024 mark scheme awards method marks for spotting that $x^{2} - 4$ already contains $(x - 2)$ as a factor — so the LCM is $(x-2)(x+2)$ , not $(x-2)(x^{2}-4)$ .
7Simplify a fraction by factorising a trinomial
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q15• simplifying, trinomial
Question
Simplify $\dfrac{x^{2} - 9}{x^{2} - x - 6}$ .
Step-by-step solution
1. Step 1
  Factorise the numerator (difference of two squares).
  $x^{2} - 9 = (x + 3)(x - 3)$
2. Step 2
  Factorise the denominator: find numbers whose product is $-6$ and sum is $-1$ . Try $-3$ and $2$ .
  $x^{2} - x - 6 = (x - 3)(x + 2)$
3. Step 3
  Cancel the common factor $(x - 3)$ .
  $\dfrac{(x + 3)(x - 3)}{(x - 3)(x + 2)} = \dfrac{x + 3}{x + 2}$
Answer
$\dfrac{x + 3}{x + 2}$
Examiner tip
The examiner report flags candidates often try to cancel terms ( $x^{2}$ in numerator with $x^{2}$ in denominator) before factorising. Only entire factors cancel — factorise first, then cancel.
8Solve an equation containing algebraic fractions
ExtendedMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q17• solving, equations
Question
Solve $\dfrac{4}{x + 1} - \dfrac{1}{x - 2} = 1$ .
Step-by-step solution
1. Step 1
  Multiply every term by the common denominator $(x + 1)(x - 2)$ .
  $4(x - 2) - 1(x + 1) = (x + 1)(x - 2)$
2. Step 2
  Expand each side.
  $4x - 8 - x - 1 = x^{2} - x - 2$
3. Step 3
  Collect on the left, then move to one side.
  $3x - 9 = x^{2} - x - 2$
4. Step 4
  Rearrange to a standard quadratic.
  $0 = x^{2} - 4x + 7 \Rightarrow x^{2} - 4x + 7 = 0$
5. Step 5
  Compute the discriminant: $\Delta = (-4)^{2} - 4(1)(7) = 16 - 28 = -12 < 0$ . There are no real solutions because the discriminant is negative.
Answer
No real solutions ( $\Delta < 0$ ).
Examiner tip
The 2024 mark scheme still awards method marks for correctly clearing the fractions and reaching a quadratic, even when there are no real roots. State the discriminant value to secure the final mark.
9Multiply algebraic fractions requiring factorising
ExtendedDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q13• multiplication, factorising
Question
Simplify $\dfrac{x^{2} + 3x}{x^{2} - 4} \times \dfrac{x - 2}{2x}$ .
Step-by-step solution
1. Step 1
  Factorise where possible.
  $x^{2} + 3x = x(x + 3),\ x^{2} - 4 = (x - 2)(x + 2)$
2. Step 2
  Rewrite the product.
  $\dfrac{x(x + 3)}{(x - 2)(x + 2)} \times \dfrac{x - 2}{2x}$
3. Step 3
  Cancel $x$ and $(x - 2)$ .
  $= \dfrac{x + 3}{2(x + 2)}$
Answer
$\dfrac{x + 3}{2(x + 2)}$
Examiner tip
The examiner report flags candidates often multiply numerators and denominators first, then try to factor a quartic mess. Factorise everything FIRST, then cancel.
10Combine, factorise and simplify
ChallengeMulti-step problem• Adapted from 0580/42 Feb/Mar 2024 Q20• multi-step, quadratic
Question
Simplify $\dfrac{2x^{2} - 8}{x^{2} + 5x + 6} \div \dfrac{x - 2}{x + 3}$ .
Step-by-step solution
1. Step 1
  Rewrite the division as multiplication by the reciprocal.
  $\dfrac{2x^{2} - 8}{x^{2} + 5x + 6} \times \dfrac{x + 3}{x - 2}$
2. Step 2
  Factorise each polynomial: $2x^{2} - 8 = 2(x - 2)(x + 2)$ and $x^{2} + 5x + 6 = (x + 2)(x + 3)$ .
  $\dfrac{2(x - 2)(x + 2)}{(x + 2)(x + 3)} \times \dfrac{x + 3}{x - 2}$
3. Step 3
  Cancel $(x + 2)$ , $(x + 3)$ and $(x - 2)$ .
  $= 2$
Answer
$2$
Examiner tip
A* candidates secure full marks by factorising every polynomial completely before cancelling. The mark scheme allows partial credit for any correct factorisation step, so always show the factorisations explicitly.

Key Formulae — Algebraic Fractions

The formulae you need to memorise for algebraic fractions on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Cancellation rule
$\dfrac{ac}{bc} = \dfrac{a}{b},\quad c \neq 0$
When to use
Only when the same FACTOR appears in numerator and denominator — never with individual terms.
Adding fractions over a common denominator
$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}$
When to use
Adding or subtracting any two fractions with different denominators.
Dividing fractions
$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$
When to use
Always rewrite a division as multiplication by the reciprocal.

Key Definitions and Keywords — Algebraic Fractions

Definitions to memorise and the exact keywords mark schemes credit for algebraic fractions answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Algebraic fraction
Examiner keyword
A fraction whose numerator and/or denominator contain algebraic expressions (variables).
Example
$\dfrac{x + 1}{x - 2}$
Common denominator
Examiner keyword
A denominator shared by two or more fractions, used to combine them by addition or subtraction.
Reciprocal of a fraction
The reciprocal of $\dfrac{a}{b}$ is $\dfrac{b}{a}$ .
Excluded values
Values of the variable that make any denominator zero — these are excluded from the domain because division by zero is undefined.
Example
In $\dfrac{1}{x - 2}$ , $x = 2$ is excluded.

Common Mistakes and Misconceptions — Algebraic Fractions

The traps other students keep falling into on algebraic fractions questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Cancelling individual terms instead of factors
0580/42 — examiner reports, recurring
Why it happens
Students see $\dfrac{x + 3}{x + 1}$ and try to cancel the $x$ 's, leaving $\dfrac{3}{1} = 3$ .
How to avoid it
Cancellation only works for shared factors (the whole bracket). Factorise first, then cancel.
✕Adding fractions without a common denominator
Why it happens
Adding numerators and denominators directly: $\dfrac{2}{x+1} + \dfrac{3}{x-2} \neq \dfrac{5}{2x - 1}$ .
How to avoid it
Always rewrite over a common denominator first.
✕Flipping the wrong fraction in a division
Why it happens
Students flip the first fraction instead of the second.
How to avoid it
Rule: keep the first, change division to multiplication, flip the SECOND.
✕Not factorising before simplifying
Why it happens
Skipping the factorisation step leaves no shared factors to cancel.
How to avoid it
Step 1 is always factorise — both numerator and denominator. Then look for shared factors.

Algebraic Fractions — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Algebraic Fractions

Short Notes - Algebraic Fractions

Detailed Study Notes

At a glance

What you’ll learn

Simplifying algebraic fractions

Adding and subtracting algebraic fractions

Multiplying and dividing algebraic fractions

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Algebraic Fractions

1Simplify by factorising and cancelling

2Add two algebraic fractions

3Multiply algebraic fractions

4Divide algebraic fractions

5Add a constant and a fraction with a linear denominator

6Subtract a fraction with a factorisable denominator

7Simplify a fraction by factorising a trinomial

8Solve an equation containing algebraic fractions

9Multiply algebraic fractions requiring factorising

10Combine, factorise and simplify

Cancellation rule

Adding fractions over a common denominator

Dividing fractions

Algebraic fraction

Common denominator

Reciprocal of a fraction

Excluded values

✕Cancelling individual terms instead of factors

✕Adding fractions without a common denominator

✕Flipping the wrong fraction in a division

✕Not factorising before simplifying

Algebraic Fractions — frequently asked questions

Algebraic Fractions

Short Notes - Algebraic Fractions

Detailed Study Notes

At a glance

What you’ll learn

Simplifying algebraic fractions

Adding and subtracting algebraic fractions

Multiplying and dividing algebraic fractions

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Algebraic Fractions

1Simplify by factorising and cancelling

2Add two algebraic fractions

3Multiply algebraic fractions

4Divide algebraic fractions

5Add a constant and a fraction with a linear denominator

6Subtract a fraction with a factorisable denominator

7Simplify a fraction by factorising a trinomial

8Solve an equation containing algebraic fractions

9Multiply algebraic fractions requiring factorising

10Combine, factorise and simplify

Cancellation rule

Adding fractions over a common denominator

Dividing fractions

Algebraic fraction

Common denominator

Reciprocal of a fraction

Excluded values

✕Cancelling individual terms instead of factors

✕Adding fractions without a common denominator

✕Flipping the wrong fraction in a division

✕Not factorising before simplifying

Algebraic Fractions — frequently asked questions