Algebraic Fractions

Simplifying algebraic fractions by cancelling common factors. Multiplying and dividing algebraic fractions.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel IGCSE Maths 4MA1 · 99 real past-paper questions below

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What an Algebraic Fraction Is

An algebraic fraction is a fraction with letters in it, like $\frac{x + 3}{x - 2}$ .

Good news: it follows exactly the same rules as the number fractions you already know.

Example

With numbers you simplify $\frac{6}{10}$ like this:

\frac{6}{10} = \frac{2 \times 3}{2 \times 5} = \frac{3}{5}

You cancelled the 2 because it multiplies the whole top and the whole bottom.

Rule

You can only cancel something that multiplies the whole top and the whole bottom. That something is called a factor. You can never cancel part of a sum.

Cannot cancel

\frac{x + 3}{x + 5}

The x is added, not multiplied. There is nothing to cancel.

Can cancel

\frac{( x + 3 ) ( x - 1 )}{x + 3}

(x + 3) multiplies the whole top and the whole bottom, so it cancels to x − 1.

Key idea

This one rule is the whole topic: factorise first, so you can see what is multiplying, then cancel.

Key idea

Adding fractions is at least 3,500 years old. The Rhind Papyrus, copied by the Egyptian scribe Ahmes around 1550 BC, is full of fraction sums. Egyptians wrote almost every fraction as a sum of different fractions with 1 on top, like $\frac{3}{4} = \frac{1}{2} + \frac{1}{4}$ .

Simplify Fully

When the question says Simplify fully, it wants you to factorise the top and the bottom, then cancel.

Steps

1
Factorise the top.
2
Factorise the bottom.
3
Cancel any bracket that appears in both.
4
Write down what is left.

Worked example

1
$\frac{2 x ^{2} + 7 x + 3}{4 x ^{2} - 1}$ — The question. The top is a quadratic and the bottom is a difference of two squares, so both will factorise.
2
$= \frac{( 2 x + 1 ) ( x + 3 )}{4 x ^{2} - 1}$ — Factorise the top. Check by expanding: $(2 x + 1) (x + 3) = 2 x^{2} + 6 x + x + 3 = 2 x^{2} + 7 x + 3$ .
3
$= \frac{( 2 x + 1 ) ( x + 3 )}{( 2 x + 1 ) ( 2 x - 1 )}$ — Factorise the bottom. $4 x^{2} - 1 = (2 x)^{2} - 1^{2}$ , so it is $(2 x + 1) (2 x - 1)$ . Now (2x + 1) is on the top and the bottom.
4
$= \frac{( 2 x + 1 ) ( x + 3 )}{( 2 x + 1 ) ( 2 x - 1 )}$ — (2x + 1) multiplies the whole top and the whole bottom, so it is a factor. Dividing both by it is cancelling.
5
$= \frac{x + 3}{2 x - 1}$ — What is left is the answer. The top and bottom share nothing else, so it is fully simplified.

Common mistake

Cancelling the $x^{2}$ in $\frac{x ^{2} + 5}{x ^{2} + 3}$ . The x² terms are added, not multiplied — this fraction does not simplify.

Brackets Back to Front

Sometimes the top and bottom share a bracket written back to front, like (3 − x) and (x − 3). They look different, but one is minus the other.

Formula

a - b = - (b - a)

$a$ and $b$ are any two terms. Swapping their order changes the sign, so a minus goes in front. Check with numbers: $7 - 2 = 5$ and $- (2 - 7) = - (- 5) = 5$ .

Worked example

1
$\frac{3 x - x ^{2}}{x ^{2} - 9}$ — The question. Both the top and the bottom will factorise.
2
$= \frac{x ( 3 - x )}{x ^{2} - 9}$ — Factorise the top: x is in both terms, so take it out. $3 x \div x = 3$ and $x^{2} \div x = x$ .
3
$= \frac{x ( 3 - x )}{( x + 3 ) ( x - 3 )}$ — Factorise the bottom: $x^{2} - 9 = x^{2} - 3^{2}$ , a difference of two squares.
4
$= \frac{- x ( x - 3 )}{( x + 3 ) ( x - 3 )}$ — (3 − x) is (x − 3) back to front. Use the formula: $3 - x = - (x - 3)$ . The minus moves to the front.
5
$= \frac{- x ( x - 3 )}{( x + 3 ) ( x - 3 )}$ — Now (x − 3) multiplies the whole top and the whole bottom, so cancel it.
6
$= \frac{- x}{x + 3}$ — What is left. Keep the minus: it is part of the answer.

Key idea

Check with x = 1: $\frac{3 - 1}{1 - 9} = \frac{2}{- 8} = - \frac{1}{4}$ and $\frac{- 1}{1 + 3} = - \frac{1}{4}$ . They match.

Common mistake

Cancelling (3 − x) with (x − 3) as if they were the same. They cancel to −1, not 1, so the answer loses its minus sign.

Four Terms? Group First

If the top has four terms and nothing divides all four, factorise it by grouping. Then cancel as usual.

Worked example

1
$\frac{x y + 4 y - 2 x - 8}{x ^{2} - 16}$ — The question. The top has four terms and no common factor, so split it into two pairs.
2
$= \frac{y ( x + 4 ) - 2 x - 8}{x ^{2} - 16}$ — Take y out of the first pair: $x y \div y = x$ and $4 y \div y = 4$ .
3
$= \frac{y ( x + 4 ) - 2 ( x + 4 )}{x ^{2} - 16}$ — Take −2 out of the second pair: $- 2 x \div (- 2) = x$ and $- 8 \div (- 2) = + 4$ . Taking out −2, not 2, gives the same bracket again.
4
$= \frac{( x + 4 ) ( y - 2 )}{x ^{2} - 16}$ — (x + 4) is in both parts, so take it out. What is left from each part is y and −2.
5
$= \frac{( x + 4 ) ( y - 2 )}{( x + 4 ) ( x - 4 )}$ — Factorise the bottom: $x^{2} - 16 = x^{2} - 4^{2}$ , a difference of two squares.
6
$= \frac{( x + 4 ) ( y - 2 )}{( x + 4 ) ( x - 4 )}$ — (x + 4) multiplies the whole top and the whole bottom, so cancel it.
7
$= \frac{y - 2}{x - 4}$ — What is left is the answer.

Common mistake

Taking out +2 from −2x − 8 gives 2(−x − 4). That bracket does not match (x + 4), so the grouping stops. Take out −2 instead.

Key idea

Need a refresher on pairing terms? See .

Adding and Subtracting

This is the most common algebraic fractions question in the exam. The clue is the wording write as a single fraction.

Rule

You can only add fractions when their bottoms are the same. That is true for $\frac{1}{2} + \frac{1}{3}$ , and it is true with letters.

Steps

1
Find a common bottom. Multiplying the two bottoms always works, but if they share a factor use a smaller one.
2
Multiply each top by whatever its bottom was missing.
3
Put both tops over the common bottom, each top in brackets.
4
Expand the brackets and collect like terms on the top.

Worked example

1
$\frac{x + 1}{3} + \frac{2 x - 1}{4}$ — The question. The bottoms are 3 and 4. They must match before you can add.
2
$= \frac{( x + 1 ) \times 4}{3 \times 4} + \frac{( 2 x - 1 ) \times 3}{4 \times 3}$ — 12 is the smallest number 3 and 4 both go into. Multiply top and bottom by what each bottom is missing: × 4 and × 3. The value does not change.
3
$= \frac{4 ( x + 1 )}{12} + \frac{3 ( 2 x - 1 )}{12}$ — Keep each top in a bracket with its number in front, so nothing gets lost. Both are now twelfths.
4
$= \frac{4 ( x + 1 ) + 3 ( 2 x - 1 )}{12}$ — The bottoms match, so add the tops and keep the 12.
5
$= \frac{4 x + 4 + 6 x - 3}{12}$ — Expand each bracket: $4 \times x + 4 \times 1 = 4 x + 4$ and $3 \times 2 x - 3 \times 1 = 6 x - 3$ .
6
$= \frac{10 x + 1}{12}$ — Collect like terms: 4x + 6x = 10x and 4 − 3 = 1.

Worked example

1
$\frac{3}{x + 2} - \frac{2}{x - 1}$ — The question. The bottoms (x + 2) and (x − 1) share no factor, so the common bottom is their product.
2
$= \frac{3 ( x - 1 )}{( x + 2 ) ( x - 1 )} - \frac{2 ( x + 2 )}{( x + 2 ) ( x - 1 )}$ — Each bottom is missing the other bracket. Multiply 3’s top and bottom by (x − 1), and 2’s by (x + 2).
3
$= \frac{3 ( x - 1 ) - 2 ( x + 2 )}{( x + 2 ) ( x - 1 )}$ — Same bottom now, so subtract the tops. Keep the second top in a bracket so the minus reaches all of it.
4
$= \frac{3 x - 3 - 2 x - 4}{( x + 2 ) ( x - 1 )}$ — Expand. The minus multiplies all of 2(x + 2): −2 × x = −2x and −2 × 2 = −4.
5
$= \frac{x - 7}{( x + 2 ) ( x - 1 )}$ — Collect: 3x − 2x = x and −3 − 4 = −7. Leave the bottom in brackets; it is already factorised.

Common mistake

A minus in front of a fraction applies to its whole top: $- 2 (x + 2) = - 2 x - 4$ , not $- 2 x + 4$ .

The Smallest Common Bottom

When the bottoms share a factor, multiplying them makes a bottom that is too big, and you must cancel later. Use the smallest bottom that both go into.

Worked example

1
$\frac{5}{6} + \frac{x - 2}{4 x}$ — The question. 6 × 4x = 24x would work, but 6 and 4 share a 2, so a smaller bottom exists.
2
$common bottom = 12 x$ — 12 is the smallest number 6 and 4 both go into. The x must be there too, because 4x has one.
3
$= \frac{5 \times 2 x}{6 \times 2 x} + \frac{( x - 2 ) \times 3}{4 x \times 3}$ — 6 needs × 2x to make 12x, and 4x needs × 3. Multiply each top and bottom by that.
4
$= \frac{10 x + 3 ( x - 2 )}{12 x}$ — Same bottom, so add the tops. 5 × 2x = 10x; keep 3(x − 2) in its bracket.
5
$= \frac{10 x + 3 x - 6}{12 x}$ — Expand: 3 × x = 3x and 3 × (−2) = −6.
6
$= \frac{13 x - 6}{12 x}$ — Collect: 10x + 3x = 13x. Top and bottom share no factor, so this is the simplest form.

Bottoms that are quadratics. Factorise the bottoms first. Then you can see which bracket they share.

Worked example

1
$\frac{6}{x ^{2} - 9} + \frac{1}{x + 3}$ — The question. The first bottom is a quadratic, so factorise it.
2
$= \frac{6}{( x + 3 ) ( x - 3 )} + \frac{1}{x + 3}$ — $x^{2} - 9 = (x + 3) (x - 3)$ , a difference of two squares. Both bottoms contain (x + 3).
3
$= \frac{6}{( x + 3 ) ( x - 3 )} + \frac{1 \times ( x - 3 )}{( x + 3 ) ( x - 3 )}$ — The common bottom is (x + 3)(x − 3). The second fraction is missing (x − 3), so multiply its top and bottom by it.
4
$= \frac{6 + x - 3}{( x + 3 ) ( x - 3 )}$ — Same bottom, so add the tops: 1 × (x − 3) = x − 3.
5
$= \frac{x + 3}{( x + 3 ) ( x - 3 )}$ — Collect: 6 − 3 = 3, so the top is x + 3. Look: it matches a bracket on the bottom.
6
$= \frac{1}{x - 3}$ — Cancel (x + 3). When the whole top cancels, 1 is left on top.

Key idea

After adding, always try to factorise the top. “Simplest form” questions are often built so something cancels at the end.

Common mistake

Using (x² − 9)(x + 3) as the bottom. It works, but it hides an extra (x + 3) that you must still cancel, and most students forget.

Multiplying and Dividing

When fractions are multiplied or divided you do not need a common bottom. Factorise everything and cancel instead.

Steps

1
Dividing? Flip the second fraction and change ÷ to ×.
2
Factorise every top and every bottom.
3
Cancel any bracket that is on a top and on a bottom.
4
Multiply what is left.

Worked example

1
$\frac{x ^{2} - 9}{2 x + 4} \times \frac{x + 2}{x + 3}$ — The question. It is ×, so no common bottom is needed. Factorise, then cancel.
2
$= \frac{( x - 3 ) ( x + 3 )}{2 x + 4} \times \frac{x + 2}{x + 3}$ — Factorise the first top: $x^{2} - 9 = (x - 3) (x + 3)$ , a difference of two squares.
3
$= \frac{( x - 3 ) ( x + 3 )}{2 ( x + 2 )} \times \frac{( x + 2 )}{( x + 3 )}$ — Factorise the first bottom: 2 goes into 2x and 4, so $2 x + 4 = 2 (x + 2)$ . Matching brackets get the same colour.
4
$= \frac{( x - 3 ) ( x + 3 )}{2 ( x + 2 )} \times \frac{( x + 2 )}{( x + 3 )}$ — In a product, any top can cancel with any bottom. (x + 3) and (x + 2) are each on a top and a bottom, so both cancel.
5
$= \frac{x - 3}{2}$ — Multiply what is left: x − 3 on the top, 2 on the bottom.

Example

Why flip when you divide? Dividing by $\frac{1}{2}$ asks how many halves fit in. There are 6 halves in 3, so $3 \div \frac{1}{2} = 3 \times 2$ .

Worked example

1
$\frac{x ^{2} - 1}{3 x} \div \frac{x + 1}{6 x ^{2}}$ — The question. It is ÷, so flip the second fraction first.
2
$= \frac{x ^{2} - 1}{3 x} \times \frac{6 x ^{2}}{x + 1}$ — Turn the second fraction upside down and change ÷ to ×. Dividing by a fraction is multiplying by it flipped.
3
$= \frac{( x - 1 ) ( x + 1 )}{3 x} \times \frac{6 x ^{2}}{x + 1}$ — Factorise the top: $x^{2} - 1 = (x - 1) (x + 1)$ , a difference of two squares.
4
$= \frac{( x - 1 ) ( x + 1 )}{3 x} \times \frac{6 x ^{2}}{( x + 1 )}$ — (x + 1) is on a top and a bottom, so cancel it.
5
$= (x - 1) \times \frac{6 x ^{2}}{3 x}$ — What is left: (x − 1) times $\frac{6 x ^{2}}{3 x}$ .
6
$= 2 x (x - 1)$ — Divide: 6 ÷ 3 = 2 and $x^{2} \div x = x$ , so 6x² ÷ 3x = 2x.

Single terms, no brackets. Short questions often have no brackets at all. Flip if dividing, then cancel numbers and letters.

Worked example

1
$\frac{4 p}{9 q} \div \frac{2 p}{3}$ — The question. There is nothing to factorise, so go straight to flipping.
2
$= \frac{4 p}{9 q} \times \frac{3}{2 p}$ — Flip $\frac{2 p}{3}$ to $\frac{3}{2 p}$ and change ÷ to ×.
3
$= \frac{4}{9 q} \times \frac{3}{2}$ — p is on a top and a bottom, so it cancels: p ÷ p = 1.
4
$= \frac{4 \times 3}{9 q \times 2} = \frac{12}{18 q}$ — Multiply the tops together and the bottoms together.
5
$= \frac{2}{3 q}$ — 12 and 18 share the factor 6: 12 ÷ 6 = 2 and 18 ÷ 6 = 3.

Key idea

Cancel before you multiply. The numbers stay small and your answer is already simplified.

The Big Exam Question

The hardest questions, worth 4 or 5 marks, mix a multiplication with an addition or subtraction.

Rule

Same order as with numbers: in $2 + 3 \times 4$ you multiply first. Here too — do the × part first, add last.

Steps

1
Do the × or ÷ part first: factorise and cancel.
2
Turn a whole term like (4 − x) into a fraction over 1.
3
Give everything the same bottom and add or subtract the tops.
4
Expand and collect like terms.

Worked example

1
$\frac{x ^{2} - 1}{x - 4} \times \frac{x ^{2} - 6 x + 8}{x ^{2} + 2 x - 3} + (4 - x)$ — The question: write it as a single fraction. It has × and +, so do the × part first, just like 2 + 3 × 4.
2
$\frac{( x - 1 ) ( x + 1 )}{x - 4} \times \frac{x ^{2} - 6 x + 8}{x ^{2} + 2 x - 3}$ — Factorise the first top: $x^{2} - 1 = (x - 1) (x + 1)$ , a difference of two squares.
3
$\frac{( x - 1 ) ( x + 1 )}{( x - 4 )} \times \frac{( x - 2 ) ( x - 4 )}{x ^{2} + 2 x - 3}$ — Factorise the second top: −2 and −4 multiply to 8 and add to −6, so it is (x − 2)(x − 4).
4
$\frac{( x - 1 ) ( x + 1 )}{( x - 4 )} \times \frac{( x - 2 ) ( x - 4 )}{( x + 3 ) ( x - 1 )}$ — Factorise the second bottom: 3 and −1 multiply to −3 and add to 2, so it is (x + 3)(x − 1).
5
$= \frac{( x + 1 ) ( x - 2 )}{x + 3}$ — (x − 1) and (x − 4) are each on a top and a bottom, so both cancel. Multiply what is left.
6
$\frac{( x + 1 ) ( x - 2 )}{x + 3} + \frac{( 4 - x ) ( x + 3 )}{x + 3}$ — Now the + part. Give (4 − x) the bottom x + 3 by multiplying its top and bottom by (x + 3).
7
$= \frac{( x + 1 ) ( x - 2 ) + ( 4 - x ) ( x + 3 )}{x + 3}$ — Same bottom, so add the tops and keep x + 3.
8
$= \frac{x ^{2} - x - 2 + ( 4 - x ) ( x + 3 )}{x + 3}$ — Expand the first top: $x^{2} - 2 x + x - 2 = x^{2} - x - 2$ .
9
$= \frac{x ^{2} - x - 2 - x ^{2} + x + 12}{x + 3}$ — Expand the second top: $4 x + 12 - x^{2} - 3 x$ , which is −x² + x + 12.
10
$= \frac{10}{x + 3}$ — Collect: x² − x² = 0, −x + x = 0 and −2 + 12 = 10. Only 10 is left.

Key idea

When the x² and x terms disappear like this, it is a good sign you are right. Exam questions are often built that way.

Key idea

Show every line. In a 4 or 5 mark question, each correct factorising and the common-bottom line earn marks on their own, so a slip at the end still leaves you most of the marks.

A Fraction Inside a Fraction

Some questions give two formulas, like $p = \frac{4}{x + 2}$ and $x = \frac{2}{y - 3}$ , and ask for p in terms of y as a single fraction. Put one inside the other, then clear the small fraction.

Worked example

1
$p = \frac{4}{x + 2}$ — The question. To get p in terms of y, replace the x with its formula.
2
$p = \frac{4}{\frac{2}{y - 3} + 2}$ — Put $\frac{2}{y - 3}$ in place of x. Now there is a small fraction inside the big one.
3
$p = \frac{4 \times ( y - 3 )}{( \frac{2}{y - 3} + 2 ) \times ( y - 3 )}$ — Multiply the whole top and the whole bottom of the big fraction by (y − 3). Same on top and bottom, so the value does not change.
4
$p = \frac{4 ( y - 3 )}{2 + 2 ( y - 3 )}$ — On the bottom, $\frac{2}{y - 3} \times (y - 3) = 2$ and $2 \times (y - 3) = 2 (y - 3)$ . The small fraction has gone.
5
$p = \frac{4 ( y - 3 )}{2 + 2 y - 6}$ — Expand the bottom: 2 × y = 2y and 2 × (−3) = −6.
6
$p = \frac{4 ( y - 3 )}{2 y - 4}$ — Collect: 2 − 6 = −4.
7
$p = \frac{4 ( y - 3 )}{2 ( y - 2 )}$ — Take the common factor 2 out of the bottom, to see if anything cancels.
8
$p = \frac{2 ( y - 3 )}{y - 2}$ — Cancel the 2: 4 ÷ 2 = 2. This is a single fraction in its simplest form.

Key idea

Check with y = 5: $x = \frac{2}{2} = 1$ , so $p = \frac{4}{3}$ . The answer gives $\frac{2 \times 2}{3} = \frac{4}{3}$ . They match.

Common mistake

Multiplying only the small fraction by (y − 3). You must multiply the whole top and the whole bottom of the big fraction, or its value changes.

Which Method? Read the Question

Before you write anything, read the words of the question and look at the sign between the fractions.

The question has	You do
“Simplify fully”	Factorise top and bottom, then cancel
A bracket back to front, like (3 − x) and (x − 3)	Write 3 − x = −(x − 3), then cancel
Four terms on the top	Group in pairs, then cancel
Adding or subtracting, “single fraction”	Common bottom, then add or subtract the tops
× or ÷	Factorise everything, flip if ÷, cancel
× and + together	× part first, then a common bottom
One formula put inside another	Substitute, then multiply top and bottom by the small bottom

Key idea

Every method starts the same way: factorise. If factorising still feels slow, revisit first.

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