Algebraic fractions and excluded values
Treat algebraic fractions as quotients of complete expressions and keep their original restrictions.
Extended only.
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
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.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources