Algebraic fractions follow exactly the rules you already use for numeric fractions. The only new skill is factorising first, so that you can see what cancels.
Today's hook, You can only cancel factors, never terms. That single rule is behind almost every mark lost in this topic, and factorising is what makes the factors visible.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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A student writes $\dfrac{x^2 + 6}{x} = x + 6$. Test it with $x = 2$. What went wrong, and what would have to be true for the cancellation to be legal?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Factorise first, then cancel factors
Work through the core explanation before applying it.
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Factorise first, then cancel factors
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An algebraic fraction simplifies only when the numerator and denominator share a common factor. Factorising both is what turns a sum into a product so the common factor can be seen and cancelled.
$\dfrac{x^2 - 9}{x + 3}$ looks stuck until you factorise the top: $\dfrac{(x-3)(x+3)}{x+3}$. Now $(x+3)$ is a factor of both, so it cancels, leaving $x - 3$.
In $\frac{x^2+6}{x}$ nothing cancels, because $x^2$ and 6 are terms being added, not factors being multiplied.
Divide means multiply by the reciprocal
Flip the second fraction and multiply. Never try to cancel across a division sign before flipping.
Factorise before finding a denominator
The lowest common denominator is usually obvious once every denominator is in factorised form.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
An algebraic fraction is simplified by cancelling common factors, not common terms.
To multiply, multiply numerators and denominators, cancelling common factors first.
To divide, multiply by the reciprocal of the second fraction.
To add or subtract, rewrite over a common denominator, then combine numerators.
Understand
Concepts
Why only factors can be cancelled, and what goes wrong when a term is cancelled.
Why factorising every numerator and denominator is the first move in every question type.
Why the lowest common denominator is built from the factorised denominators.
Can do
Skills
Simplify an algebraic fraction by factorising and cancelling.
Multiply and divide algebraic fractions, cancelling before multiplying out.
Add and subtract algebraic fractions with different denominators.
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Key terms
Algebraic fractionA fraction whose numerator or denominator contains a variable. Like this: $\frac{x+1}{x-2}$ is an algebraic fraction.
FactorSomething being multiplied. Only factors can be cancelled. Like this: in $\frac{3(x+1)}{5(x+1)}$ the bracket $(x+1)$ is a factor of both, so it cancels to give $\frac{3}{5}$.
TermSomething being added or subtracted. Terms can never be cancelled. Like this: in $\frac{x+1}{x}$ the $x$ on top is a term, so it does not cancel with the $x$ underneath.
ReciprocalA fraction turned upside down, used to turn a division into a multiplication. Like this: dividing by $\frac{x}{3}$ is the same as multiplying by $\frac{3}{x}$.
Lowest common denominatorThe simplest expression every denominator divides into, used to add or subtract fractions. Like this: for $\frac{1}{x+1}$ and $\frac{1}{x-1}$ it is $(x+1)(x-1)$.
RestrictionA value the variable cannot take, because it would make a denominator zero. Like this: $\frac{x+2}{x-3}$ has the restriction $x \neq 3$.
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Simplifying by factorising
Work through the core explanation before applying it.
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Simplifying by factorising
core concept
A fraction such as $\dfrac{x^2 - 9}{x^2 + x - 12}$ cannot be simplified as it stands. Factorise both parts first: $\dfrac{(x-3)(x+3)}{(x-3)(x+4)}$.
Now $(x-3)$ is a common factor of the numerator and the denominator, so it cancels, leaving $\dfrac{x+3}{x+4}$.
Cancelling is only legal for factors. In $\dfrac{x^2 + 6}{x}$ the numerator is a sum, so nothing cancels. Testing with $x = 2$ gives $\frac{10}{2} = 5$, while the illegal cancellation would give 8.
State restrictions when asked. $\dfrac{(x-3)(x+3)}{(x-3)(x+4)}$ requires $x \neq 3$ and $x \neq -4$, because either would make the original denominator zero.
Quick check: simplify $\dfrac{x^2 - 4}{x + 2}$.
Factorise the numerator and the denominator fully, then cancel any bracket appearing in both. Only factors cancel; a term in a sum never cancels. Values making the original denominator zero are excluded.
Pause, copy the factorise-then-cancel method with the worked $\frac{x^2-9}{x^2+x-12} = \frac{x+3}{x+4}$, and the counter-example showing $\frac{x^2+6}{x} \neq x+6$, into your book.
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Multiplying and dividing
core concept
We just saw that cancelling requires a common factor in a single fraction. That raises a question: what happens when two fractions are multiplied or divided, and can you still cancel? This card answers it → with multiplication you may cancel across the two fractions, and division becomes multiplication once you flip the second fraction.
To multiply, factorise everything, then cancel any factor on a top against the same factor on a bottom, even across the two fractions. Only then multiply what is left.
To divide, multiply by the reciprocal of the second fraction: $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$. Flip first, cancel second.
Cancelling before multiplying keeps the numbers small and usually removes the need to expand anything at all.
Flip before you cancel. In a division you must invert the second fraction before any cancelling. Cancelling across a division sign is not valid.
Which step is NOT allowed?
Multiply: factorise, cancel any factor top against bottom across both fractions, then multiply the remainder. Divide: flip the second fraction first, then treat it as a multiplication.
Pause, copy the multiply rule (factorise, cancel across, then multiply), the divide rule (multiply by the reciprocal), and the warning that you must flip before cancelling, into your book.
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Adding and subtracting
core concept
We just saw that multiplying and dividing allow cancelling once everything is factorised. That raises a question: why can you not simply cancel when adding, and what has to happen instead? This card answers it → addition needs a common denominator first, because only fractions with the same denominator can be combined.
To add or subtract, every fraction must be written over the same denominator. Factorise each denominator, then build the lowest common denominator from all the distinct factors.
For $\dfrac{3}{x+1} + \dfrac{2}{x-1}$ the common denominator is $(x+1)(x-1)$. Rewrite each fraction over it: $\dfrac{3(x-1)}{(x+1)(x-1)} + \dfrac{2(x+1)}{(x+1)(x-1)}$.
Now combine the numerators: $\dfrac{3x - 3 + 2x + 2}{(x+1)(x-1)} = \dfrac{5x - 1}{(x+1)(x-1)}$. Leave the denominator factorised.
Subtraction changes every sign. $\dfrac{a}{d} - \dfrac{b + c}{d} = \dfrac{a - b - c}{d}$. Bracket the second numerator before subtracting, or you will lose a sign.
Fill the blank: the lowest common denominator of $\dfrac{1}{x}$ and $\dfrac{1}{x+2}$ is $x(x+$ $)$.
To add or subtract, factorise the denominators, build the lowest common denominator from all distinct factors, rewrite each fraction over it, then combine numerators. Bracket the numerator you are subtracting so every sign changes.
Pause, copy the common-denominator method with the worked $\frac{3}{x+1} + \frac{2}{x-1} = \frac{5x-1}{(x+1)(x-1)}$, and the subtraction sign warning, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · SIMPLIFYING
Simplify $\dfrac{x^2 - 9}{x^2 + x - 12}$ and state any restrictions.
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Numerator: $x^2 - 9 = (x-3)(x+3)$
Difference of two squares.
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Denominator: $x^2 + x - 12 = (x-3)(x+4)$
Two numbers multiplying to $-12$ and adding to 1 are $-3$ and 4.
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$\dfrac{(x-3)(x+3)}{(x-3)(x+4)} = \dfrac{x+3}{x+4}$, with $x \neq 3$ and $x \neq -4$
Cancel the common factor; exclude values making the original denominator zero.
Expand and collect the numerator; leave the denominator factorised.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
Simplify $\dfrac{2x + 6}{x + 3}$.
Simplify $\dfrac{x^2 - 25}{x - 5}$.
Simplify $\dfrac{a}{4} \times \dfrac{8}{a^2}$.
Simplify $\dfrac{1}{x} + \dfrac{1}{x + 1}$.
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Revisit the illegal cancellation
Run the quick drill and copy the summary into your book.
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Revisit the illegal cancellation
You tested $\dfrac{x^2 + 6}{x} = x + 6$ at the start. Write one sentence a student could memorise that would stop them making that mistake again, and give a fraction where the cancellation IS legal.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. Simplify $\dfrac{x^2 + 5x + 6}{x^2 - 4}$ and state any restrictions on $x$. (3 marks)
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ApplyBand 43 marks
Q2. Simplify $\dfrac{2}{x - 3} - \dfrac{1}{x + 2}$, leaving your answer as a single fraction. (3 marks)
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UnderstandBand 32 marks
Q3. Explain why $\dfrac{x + 5}{x}$ cannot be simplified to $5$, using a numerical substitution as part of your answer. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $\frac{2(x+3)}{x+3} = 2$. Practice 2: $x + 5$. Practice 3: $\frac{2}{a}$. Practice 4: $\frac{2x+1}{x(x+1)}$.
Q3 (2 marks): The $x$ on top is a term in a sum, not a factor, so it cannot cancel [1]. Substituting $x = 1$ gives $\frac{6}{1} = 6$, not 5, which shows the cancellation is invalid [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Fraction Fixer
earn bronze · silver · gold
Simplify, multiply, divide and combine algebraic fractions, cancelling only what is legal to cancel. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.