Simplify and calculate algebraic fractions
Factor before cancelling multiplicative factors · Add or subtract over a common denominator · Multiply or divide fractions while retaining exclusions
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Factorisation
- Fraction arithmetic
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Factor before cancelling multiplicative factors
- Add or subtract over a common denominator
- Multiply or divide fractions while retaining exclusions
Factors cancel; separate terms do not
List zero-denominator exclusions before simplifying. Factor complete numerators and denominators, then cancel common non-zero factors. (x²−1)/(x−1) = x+1 only for x ≠ 1; cancellation does not repair the original hole.
Three operations, three conditions
For addition/subtraction, use a common denominator: 1/(x−1)−1/(x+1) = 2/(x²−1), x ≠ ±1. For multiplication, factor and cancel across the product. For division, multiply by the second fraction's reciprocal; that divisor must also be non-zero, so its numerator adds an exclusion.
For example, (x²−1)/(x+2) × 1/(x−1) = (x+1)/(x+2), with x ≠ −2, 1. Keep the original restrictions even if a final denominator looks simpler.
Sources for this lesson
Pause and explain
After cancelling x−1 from (x²−1)/(x−1), is x = 1 allowed?
Worked example
Simplify (x²−9)/(x²+2x−3).
Show the worked solution
- Factor to (x−3)(x+3)/[(x+3)(x−1)].
- Exclude x = −3 and 1; cancel the shared factor elsewhere.
Answer (x−3)/(x−1), x ≠ −3, 1.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Simplify (x²−9)/(x²+2x−3), stating exclusions.
Give me a hint
Factor both complete expressions.
Compare my reasoning
- (x−3)(x+3)/[(x+3)(x−1)].
- Cancel x+3 where it is non-zero; exclude −3 and 1.
(x−3)/(x−1), x ≠ −3, 1.
Look for these in your work
- Factor before cancelling
- Retain original exclusions
Add 1/(x+1) + 2/(x−1).
Give me a hint
Use denominator (x+1)(x−1).
Compare my reasoning
- Numerator (x−1)+2(x+1) = 3x+1.
- Denominator x²−1; exclude x = ±1.
(3x+1)/(x²−1), x ≠ ±1.
Look for these in your work
- Use the whole numerator
- Check both denominator roots
Simplify (2x)/(3y) ÷ (4x)/(9y), including all original restrictions.
Give me a hint
The second fraction cannot be zero.
Compare my reasoning
- Require y ≠ 0 and x ≠ 0, then multiply 2x/(3y) by 9y/(4x).
- The product reduces to 18/12 = 3/2.
3/2, with x ≠ 0 and y ≠ 0.
Look for these in your work
- Invert the divisor
- Retain its non-zero numerator condition
Common mistakes
- Cancel factors, not terms within a sum.
- Division needs a non-zero divisor as well as non-zero denominators.
What can you explain now?
After cancelling x−1 from (x²−1)/(x−1), is x = 1 allowed?
Compare with the explanation
No; the original denominator is zero there.
The simplified rule is equivalent only on the original domain.
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Make it stick
After cancelling x−1 from (x²−1)/(x−1), is x = 1 allowed?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.