Algebraic fractions and excluded values
Treat algebraic fractions as quotients of complete expressions and keep their original restrictions.
Section Paper 2 · Extended · Non-calculator
Extended only. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Use common denominators with numerical fractions.
- Factorise quadratics and common factors.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 1/3 + 1/4?
What you will learn
- Add, subtract, multiply and divide algebraic fractions.
- Factorise before cancelling.
- State values excluded by the original denominators.
Cancel factors rather than terms
The quotient (x² − 9)/(x² + 3x) becomes [(x − 3)(x + 3)]/[x(x + 3)]. Cancel the complete factor x + 3 to obtain (x − 3)/x. The original expression is undefined at x = 0 and x = −3, and both restrictions remain even though the simplified form appears defined at −3.
You cannot cancel the x in (x + 2)/x because x + 2 is a sum, not a product containing x as a factor. Division distributes to give 1 + 2/x when x ≠ 0, which shows why simply leaving two would be wrong. Write factored numerators and denominators before drawing any cancellation marks.
Use a common denominator for sums
For 2/x + 3/(x + 1), use x(x + 1) as the common denominator. The numerator is 2(x + 1) + 3x = 5x + 2. The result is (5x + 2)/[x(x + 1)], with x ≠ 0, −1. Adding the denominators would not give fractions of the same-sized unit.
A subtraction sign acts on the entire second numerator. For 1/(x − 1) − 2/(x + 1), the new numerator is (x + 1) − 2(x − 1) = 3 − x. Brackets prevent the sign error −2x − 2. Factor the resulting numerator if possible and check for cancellation only afterwards.
Multiply factors; division takes a reciprocal
To multiply algebraic fractions, multiply numerators and denominators, using factorisation to cancel common non-zero factors. For (3x/4) × (8/9x), cancellation gives 2/3 for x ≠ 0. Keep the restriction even when the final answer contains no letter.
For (x/3) ÷ [(x + 1)/6], invert the whole divisor: (x/3) × [6/(x + 1)] = 2x/(x + 1). Here x ≠ −1 because the original divisor must not be zero. If a divisor contains its own denominator, that denominator must also be non-zero. A valid numerical substitution is a useful check of signs and scale.
Pause and explain
Why must x = 2 stay excluded after cancelling (x − 2)/(x − 2)?
Worked example
Simplify (x² − 16)/(x² − 2x − 8), giving every excluded value.
Show the worked solution
- Factorise the numerator as (x − 4)(x + 4) and denominator as (x − 4)(x + 2).
- The original denominator excludes x = 4 and x = −2.
- Cancel x − 4 where it is non-zero; the remaining quotient is (x + 4)/(x + 2).
Answer (x + 4)/(x + 2), x ≠ 4, −2
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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 6x²/(9x), with its restriction.
Give me a hint
Separate numerical and variable factors.
Compare my reasoning
- The original denominator is zero at x = 0, so exclude it.
- Reduce 6/9 to 2/3 and x²/x to x.
- The result is 2x/3, for x ≠ 0.
2x/3, x ≠ 0
Look for these in your work
- Cancelled multiplicative factors.
- Kept the original excluded value.
Simplify 2/(x − 1) − 1/(x + 1).
Give me a hint
Use (x − 1)(x + 1) and bracket both numerators.
Compare my reasoning
- Exclude x = 1 and −1.
- The common numerator is 2(x + 1) − (x − 1) = x + 3.
- The quotient is (x + 3)/(x² − 1), with the restrictions unchanged.
(x + 3)/(x² − 1), x ≠ ±1
Look for these in your work
- Applied subtraction to the entire numerator.
- Used both denominator restrictions.
A rectangular panel has area (x² − 4) cm² and width (x + 2) cm, with x > 2. Express its length and find it at x = 5.
Give me a hint
Length is area divided by width; factorise the area.
Compare my reasoning
- Length = (x² − 4)/(x + 2).
- Use the difference of squares to simplify to x − 2, valid under x > 2.
- At x = 5 the length is 3 cm; 3 × 7 = 21 cm² checks the area.
x − 2 cm; 3 cm at x = 5
Look for these in your work
- Formed the correct quotient from units.
- Checked the physical domain and area.
Common mistakes
- Cancelling a term inside an unfactorised sum.
- Adding denominators in a fraction sum.
- Dropping an excluded value after cancellation.
What can you explain now?
Simplify (x/5) ÷ (2x/15), including its restriction.
Compare with the explanation
3/2, x ≠ 0
Multiply x/5 by 15/(2x). The original divisor must be non-zero, excluding x = 0.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Create a rational expression with a factor that cancels. Explain why its simplified form still needs the original restriction.
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.