Mathematics/Lessons/Algebraic Methods/Algebraic Fractions
Pure · Algebraic Methods · Year 2

Algebraic Fractions

The same rules as numeric fractions — factorise first, then cancel, multiply, divide or combine. Get comfortable here, and partial fractions in the next lesson is just this in reverse.

9 min read 4 worked examples

Factorise before you cancel

An algebraic fraction is just a fraction with algebraic expressions on the top and/or bottom, like $\dfrac{x^2-9}{x^2+x-6}$. It behaves exactly like a numeric fraction — the only extra step is that anything that needs cancelling is usually hidden inside unfactorised expressions. So the first move, almost always, is to factorise fully, top and bottom, before touching anything else.

Once both parts are factorised, any factor common to numerator and denominator cancels — exactly as $\dfrac{6}{15}$ simplifies to $\dfrac{2}{5}$ by cancelling a common factor of $3$.

Common mistake

Cancelling terms that are added, not multiplied. In $\dfrac{x+3}{x+5}$, the $x$'s cannot be cancelled — there is no common factor here, only a common term inside a sum. Cancelling is only ever valid across a common factor that multiplies the whole of both top and bottom.

Multiplying and dividing

To multiply two algebraic fractions, factorise everything first, then cancel any factor that appears in a numerator and a denominator anywhere in the product — even diagonally across the two fractions — before multiplying what's left. To divide by a fraction, flip it (take its reciprocal) and multiply instead, exactly as with numbers.

Adding and subtracting

Adding or subtracting algebraic fractions needs a common denominator, found the same way as for numbers: multiply the two denominators together (or use their lowest common multiple, if they share a factor). Write each fraction over that common denominator, then combine the numerators into one.

Looking ahead

Combining several small fractions into one tangled fraction, as in the worked examples below, is exactly the process that partial fractions — the next lesson — reverses: starting from one combined fraction and splitting it back apart into simple pieces.

Worked example 1

Simplify by factorising and cancelling
Simplify $\dfrac{x^2+5x+6}{x^2-4}$.
MethodFactorise both parts, then cancel the shared factor
1
Factorise the numerator: $x^2+5x+6 = (x+2)(x+3)$.
Two numbers multiplying to 6, summing to 5: 2 and 3.
2
Factorise the denominator: $x^2-4 = (x-2)(x+2)$.
Difference of two squares.
3
So the fraction is $\dfrac{(x+2)(x+3)}{(x-2)(x+2)}$.
Both parts now fully factorised.
4
Cancel the common factor $(x+2)$: $\dfrac{x+3}{x-2}$.
A genuine shared factor, safe to cancel.
▪ solved

Worked example 2

Multiplying two algebraic fractions
Simplify $\dfrac{x}{x-3} \times \dfrac{x^2-9}{x^2+2x}$.
MethodFactorise everything, cancel across the product, then multiply
1
Factorise: $x^2-9=(x-3)(x+3)$ and $x^2+2x=x(x+2)$.
Difference of squares; common factor of x.
2
The product is $\dfrac{x}{x-3} \times \dfrac{(x-3)(x+3)}{x(x+2)} = \dfrac{x(x-3)(x+3)}{(x-3)\,x(x+2)}$.
Multiply numerators together and denominators together.
3
Cancel the common factors $x$ and $(x-3)$: $\dfrac{x+3}{x+2}$.
Each appears once in the top and once in the bottom.
▪ solved
Examiner's eye

Cancelling is allowed across the whole product, not just within each individual fraction — a factor in one numerator can cancel with a factor in the other denominator. Factorise everything first and it becomes obvious what cancels where.

Worked example 3

Dividing algebraic fractions
Simplify $\dfrac{x^2-1}{x+4} \div \dfrac{x-1}{2x+8}$.
MethodFlip the second fraction, then multiply
1
Rewrite as a multiplication: $\dfrac{x^2-1}{x+4} \times \dfrac{2x+8}{x-1}$.
Dividing by a fraction means multiplying by its reciprocal.
2
Factorise: $x^2-1=(x-1)(x+1)$ and $2x+8=2(x+4)$.
Difference of squares; common factor of 2.
3
The product is $\dfrac{(x-1)(x+1)}{x+4} \times \dfrac{2(x+4)}{x-1}$.
Substitute the factorised forms.
4
Cancel $(x-1)$ and $(x+4)$: $2(x+1)$.
Each cancels between a numerator and a denominator.
▪ solved

Worked example 4

Combining over a common denominator
Express $\dfrac{3}{x+1} - \dfrac{2}{x-3}$ as a single fraction.
MethodWrite both fractions over the common denominator, then combine
1
Common denominator: $(x+1)(x-3)$.
The two denominators share no common factor, so multiply them.
2
$\dfrac{3}{x+1} = \dfrac{3(x-3)}{(x+1)(x-3)}$ and $\dfrac{2}{x-3} = \dfrac{2(x+1)}{(x+1)(x-3)}$.
Multiply top and bottom of each by the missing factor.
3
Combine: $\dfrac{3(x-3) - 2(x+1)}{(x+1)(x-3)}$.
Same denominator now, so subtract the numerators.
4
Expand the numerator: $3x-9-2x-2 = x-11$.
Careful with the minus sign across the second bracket.
5
Result: $\dfrac{x-11}{(x+1)(x-3)}$.
A single combined fraction.
▪ solved
Look deeper — running this backwards

This is precisely a partial fractions problem in reverse: given $\dfrac{x-11}{(x+1)(x-3)}$, the next lesson finds constants $A=3$ and $B=-2$ that split it back into $\dfrac{3}{x+1}-\dfrac{2}{x-3}$ — undoing exactly the combination performed here.

Check yourself

Simplify $\dfrac{x^2-2x-15}{x^2-25}$.
(Hint: factorise both the numerator and the denominator first.)

Method
1
Factorise: $x^2-2x-15=(x-5)(x+3)$ and $x^2-25=(x-5)(x+5)$.
2
Cancel the common factor $(x-5)$.
Genuine shared factor.
3
Result: $\dfrac{x+3}{x+5}$.
▪ solved
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Proof by Contradiction