Maths · 2 Algebra

2.3 Algebraic fractions

Before you start

Adding, subtracting, multiplying and dividing number fractions, such as $\frac{2}{3} + \frac{1}{4} = \frac{11}{12}$ .
Expanding a bracket with a negative in front: $- 3 (x - 5) = - 3 x + 15$ .
Expanding two brackets: $(x + 1) (x - 2) = x^{2} - x - 2$ .
Factorising by a common factor, by the difference of two squares, and quadratics such as $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ .
The index rule for dividing powers: $\frac{a^{5}}{a^{2}} = a^{3}$ .

Syllabus objectives covered (2)

Manipulate algebraic fractions.
Factorise and simplify rational expressions.

Part 1 of 6

An algebraic fraction is a number fraction in disguise

An algebraic fraction is a fraction with letters in it, such as $\frac{3 a}{4}$ or $\frac{x + 1}{x - 3}$ . Every letter stands for a number, so every rule you already use for number fractions still works. There are no new rules in this subtopic, only the old ones applied to letters.
The syllabus also says rational expression. It means one expression in powers of $x$ divided by another, such as $\frac{x^{2} - 2 x}{x^{2} - 5 x + 6}$ . For this subtopic you can treat the two names as the same thing.
Keep one picture in your head: the fraction line acts as a bracket. $\frac{x + 1}{x - 3}$ means $(x + 1) \div (x - 3)$ . The whole top is divided by the whole bottom. Each trap in this lesson is a way of forgetting that.
Here are the four number rules you will copy. Multiply tops and bottoms. To divide, turn the second fraction upside down and multiply. To add, first make the denominators the same.
$\frac{2}{3} \times \frac{5}{7} = \frac{10}{21} \frac{2}{3} \div \frac{5}{7} = \frac{2}{3} \times \frac{7}{5} = \frac{14}{15} \frac{2}{3} + \frac{5}{7} = \frac{14}{21} + \frac{15}{21} = \frac{29}{21}$
The fourth rule is simplifying: divide the top and the bottom by the same number. $\frac{12}{18} = \frac{2}{3}$ , because you divide both by 6. This is what 'cancelling' means, and the next section is about doing it with letters.
There is a check that works on every answer in this subtopic. Choose a value for the letter, such as $x = 5$ . Work out the original expression and your answer with that value. If the two numbers are different, your answer is wrong. If they match, your answer is very probably right.
Choose the test value with care. Do not use a value that makes a denominator zero. Avoid $0$ , $1$ and $2$ as well: with $x = 1$ you get $x^{2} = x$ , and with $x = 2$ you get $x + 2 = 2 x$ , so some errors give the correct number by accident.

Part 2 of 6

Simplifying: cancel factors, never terms

When the top and the bottom are single terms, cancel numbers and powers separately. For $\frac{6 a^{2} b}{9 a b^{3}}$ , divide the top and the bottom by 3, then by $a$ , then by $b$ .
$\frac{6 a^{2} b}{9 a b^{3}} = \frac{2 a}{3 b^{2}}$
You may only cancel a factor: something that multiplies the whole top and the whole bottom. A term is a part that is joined on by $+$ or $-$ . You can never cancel a term.
The trap looks like this: $\frac{x + 3}{x}$ . It is tempting to cross out the two $x$ s and leave 3. Test it with $x = 1$ : $\frac{1 + 3}{1} = 4$ , not 3. The $x$ on top is a term, joined to the 3 by a plus. It does not multiply the whole top, so it is not a factor. $\frac{x + 3}{x}$ is already in its simplest form.
So when you see a $+$ or a $-$ on the top or the bottom, your first job is to factorise both of them completely. Factorising turns a sum into a product, and a product has factors that you can cancel.
Take $\frac{x^{2} - 2 x}{x^{2} - 5 x + 6}$ . The top has a common factor of $x$ . The bottom is a quadratic: you need two numbers that multiply to $6$ and add to $- 5$ , which are $- 2$ and $- 3$ .
$\frac{x^{2} - 2 x}{x^{2} - 5 x + 6} = \frac{x (x - 2)}{(x - 2) (x - 3)}$
Now the bracket $(x - 2)$ multiplies the whole top and the whole bottom, so it is a common factor and it cancels.
$\frac{x (x - 2)}{(x - 2) (x - 3)} = \frac{x}{x - 3}$
Stop there. The $x$ on top and the $x$ inside $(x - 3)$ do not cancel, because the $x$ in $(x - 3)$ is a term. A bracket cancels only as a whole bracket.
Look for these factorisations, in this order. First a common factor: $x^{2} - 2 x = x (x - 2)$ . Then the difference of two squares: $x^{2} - 9 = (x + 3) (x - 3)$ . Then a quadratic with three terms: $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ , or $2 x^{2} + 5 x - 3 = (2 x - 1) (x + 3)$ .
Watch for two brackets that differ only in sign. $3 - x = - (x - 3)$ , so $\frac{3 - x}{x - 3} = - 1$ . Take out the $- 1$ , and then the brackets cancel and leave a minus sign.
$\frac{x (3 - x)}{(x - 3) (x + 1)} = \frac{- x (x - 3)}{(x - 3) (x + 1)} = \frac{- x}{x + 1}$

Part 3 of 6

Multiplying and dividing

Multiplying is the easiest operation. Multiply the tops together and the bottoms together, exactly as with numbers.
$\frac{3 a}{4} \times \frac{9 a}{10} = \frac{27 a^{2}}{40}$
Before you multiply, look across the whole product for factors to cancel. Any top can cancel with any bottom, because in a product everything on the top multiplies everything on the bottom. This keeps the numbers small.
$\frac{4 x}{9} \times \frac{15}{8 x^{2}} = \frac{1}{3} \times \frac{5}{2 x} = \frac{5}{6 x}$
In that example, 4 cancels with 8 to leave 1 and 2. Then 15 cancels with 9 to leave 5 and 3. Then $x$ cancels with $x^{2}$ to leave 1 and $x$ .
To divide, turn the second fraction upside down and multiply. The upside-down fraction is the reciprocal. Dividing by $\frac{9 a}{10}$ is the same as multiplying by $\frac{10}{9 a}$ , because $\frac{9 a}{10} \times \frac{10}{9 a} = 1$ .
$\frac{3 a}{4} \div \frac{9 a}{10} = \frac{3 a}{4} \times \frac{10}{9 a} = \frac{30 a}{36 a} = \frac{5}{6}$
The $a$ cancels completely here, so the answer is a number with no letter. That is a correct answer: it means the value is $\frac{5}{6}$ for every value of $a$ except $0$ .
Only the second fraction turns upside down. The first fraction is the thing that you divide, so it does not change. Write the line with the $\times$ and the reciprocal before you do anything else.
When there are brackets or quadratics, factorise every top and every bottom first. Then cancel. Then multiply what is left. If you multiply out first, you get a longer expression that you must factorise again.
$\frac{x^{2} - 9}{4 x} \times \frac{8}{x + 3} = \frac{(x + 3) (x - 3)}{4 x} \times \frac{8}{x + 3} = \frac{2 (x - 3)}{x}$

Part 4 of 6

Adding and subtracting: numbers in the denominators

You can only add fractions that have the same denominator, because only then are the pieces the same size. So the first move is to find a common denominator. The lowest common multiple of the denominators gives the simplest one.
For $\frac{x}{3} + \frac{x - 4}{2}$ , the lowest common denominator is 6. Multiply the top and the bottom of each fraction by the number that makes its denominator 6.
$\frac{x}{3} + \frac{x - 4}{2} = \frac{2 x}{6} + \frac{3 (x - 4)}{6}$
Put a bracket round $x - 4$ when you multiply it by 3. The 3 multiplies the whole numerator, because the fraction line acts as a bracket.
Now there is one denominator, so add the numerators and collect like terms.
$\frac{2 x + 3 x - 12}{6} = \frac{5 x - 12}{6}$
Subtraction has a sign trap. For $\frac{2 x}{3} - \frac{3 (x - 5)}{2}$ , the common denominator is again 6. The minus sign subtracts the whole of the second numerator, so keep it in a bracket.
$\frac{2 x}{3} - \frac{3 (x - 5)}{2} = \frac{4 x - 9 (x - 5)}{6}$
Now expand with care. $- 9 \times x = - 9 x$ and $- 9 \times - 5 = + 45$ .
$\frac{4 x - 9 x + 45}{6} = \frac{45 - 5 x}{6}$
Last, check whether the numerator has a factor in common with the denominator. $45 - 5 x = 5 (9 - x)$ , but 5 is not a factor of 6. So $\frac{45 - 5 x}{6}$ is in its simplest form. $\frac{5 (9 - x)}{6}$ is the same answer written differently.
This is an expression, not an equation. There is no equals sign with another side, so you cannot multiply by 6 and remove the denominator. That move is for solving equations, where both sides change together. Here the 6 stays in your answer.

Part 5 of 6

Adding and subtracting: letters in the denominators

The method is the same. The common denominator is made from the denominators you have. When they have no factor in common, multiply them together.
For $\frac{1}{x - 2} + \frac{x + 1}{x - 3}$ , the common denominator is $(x - 2) (x - 3)$ . Multiply the top and the bottom of each fraction by the bracket that it does not have.
$\frac{1}{x - 2} + \frac{x + 1}{x - 3} = \frac{1 (x - 3)}{(x - 2) (x - 3)} + \frac{(x + 1) (x - 2)}{(x - 2) (x - 3)}$
Expand the numerators only. $(x + 1) (x - 2) = x^{2} - x - 2$ , so the numerator is $x - 3 + x^{2} - x - 2$ .
$= \frac{x^{2} - 5}{(x - 2) (x - 3)}$
Leave the denominator factorised. It is a correct final answer, and it shows at once which brackets could cancel. Expand it only when the question gives you a form to reach, as a Show (that) question can.
Then look for a numerator factor that matches a bracket in the denominator. $x^{2} - 5$ does not factorise into brackets with whole numbers, so nothing cancels and you have finished.
When the denominators already share a factor, do not multiply them together. Factorise them first and use each different bracket once. In $\frac{3}{x^{2} - 9} - \frac{1}{x + 3}$ , the first denominator is $(x + 3) (x - 3)$ . It already contains $(x + 3)$ , so $(x + 3) (x - 3)$ is the common denominator.
$\frac{3}{(x + 3) (x - 3)} - \frac{x - 3}{(x + 3) (x - 3)} = \frac{3 - (x - 3)}{(x + 3) (x - 3)} = \frac{6 - x}{(x + 3) (x - 3)}$
The second fraction needed only $(x - 3)$ on its top and bottom. Notice the bracket round $x - 3$ again: the minus subtracts all of it, so $- (x - 3) = - x + 3$ .
If you multiply the two denominators together anyway, your working is still correct, but you get a longer fraction. You then have to factorise the numerator and cancel to reach the simplest form.

Part 6 of 6

Choosing the method and knowing when to stop

Read the sign between the fractions before you write anything. For $\times$ or $\div$ : factorise, cancel, multiply. For $+$ or $-$ : common denominator, combine the numerators, then factorise the numerator to look for a cancel.
Simplify means simplify fully. You have finished when the top and the bottom are fully factorised and they have no factor in common. The answer can be written factorised or expanded; both are the same expression.
Write the factorised form on the page before you cancel. The method mark is for a valid method, and the examiner can see that you cancelled correctly only if the brackets are written down.
In a Show (that) question the answer is printed for you, so the marks are for the steps. Write the common denominator, then the numerator before you expand it, then the numerator after you expand it, each on its own line.
Finish with the number check from the first section. It takes less than a minute and finds sign errors that you cannot see when you read your own working.

Worked examples

Example 1. Simplify $\frac{2 x^{2} + 5 x - 3}{x^{2} - 9}$ .

Show the working

There are plus and minus signs on the top and the bottom, so nothing can cancel yet. Factorise both first.
The bottom is a difference of two squares, because $9 = 3^{2}$ .
$x^{2} - 9 = (x + 3) (x - 3)$
The top has $2 x^{2}$ , so use $a c = 2 \times (- 3) = - 6$ . Find two numbers that multiply to $- 6$ and add to $5$ : they are $6$ and $- 1$ . Split the middle term with them.
$2 x^{2} + 6 x - x - 3 = 2 x (x + 3) - 1 (x + 3) = (2 x - 1) (x + 3)$
Write the fraction with both factorisations. $(x + 3)$ multiplies the whole top and the whole bottom, so it cancels.
$\frac{(2 x - 1) (x + 3)}{(x + 3) (x - 3)} = \frac{2 x - 1}{x - 3}$
Nothing else cancels: $(2 x - 1)$ and $(x - 3)$ are different brackets.
Check with $x = 5$ . Original: $\frac{50 + 25 - 3}{25 - 9} = \frac{72}{16} = 4.5$ . Answer: $\frac{10 - 1}{5 - 3} = \frac{9}{2} = 4.5$ . They match.

Answer: $\frac{2 x - 1}{x - 3}$

Example 2. Simplify $\frac{x^{2} - 4}{3 x} \div \frac{x + 2}{6 x^{2}}$ .

Show the working

It is a division, so turn the second fraction upside down and change $\div$ to $\times$ . Write this line first.
$\frac{x^{2} - 4}{3 x} \times \frac{6 x^{2}}{x + 2}$
Factorise everything that can be factorised. $x^{2} - 4$ is a difference of two squares.
$\frac{(x + 2) (x - 2)}{3 x} \times \frac{6 x^{2}}{x + 2}$
Cancel across the product. $(x + 2)$ on top cancels with $(x + 2)$ underneath. $6 x^{2}$ on top and $3 x$ underneath leave $2 x$ on top.
Multiply what is left.
$(x - 2) \times 2 x = 2 x (x - 2)$
Check with $x = 3$ . Original: $\frac{5}{9} \div \frac{5}{54} = \frac{5}{9} \times \frac{54}{5} = 6$ . Answer: $2 \times 3 \times 1 = 6$ . They match.

Answer: $2 x (x - 2)$ , which you can also write as $2 x^{2} - 4 x$ .

Example 3. Write $\frac{x + 2}{4} - \frac{2 x - 3}{6}$ as a single fraction in its simplest form.

Show the working

The lowest common multiple of 4 and 6 is 12, so 12 is the common denominator.
The first fraction needs $\times 3$ on top and bottom. The second needs $\times 2$ . Keep each numerator in a bracket.
$\frac{3 (x + 2)}{12} - \frac{2 (2 x - 3)}{12} = \frac{3 (x + 2) - 2 (2 x - 3)}{12}$
Expand. The minus sign multiplies both terms in the second bracket: $- 2 \times 2 x = - 4 x$ and $- 2 \times - 3 = + 6$ .
$\frac{3 x + 6 - 4 x + 6}{12}$
Collect like terms in the numerator.
$\frac{12 - x}{12}$
$12 - x$ has no factor in common with 12, so this is the simplest form. Do not cancel the 12s: the 12 on top is a term, not a factor.
Check with $x = 6$ . Original: $\frac{8}{4} - \frac{9}{6} = 2 - 1.5 = 0.5$ . Answer: $\frac{12 - 6}{12} = 0.5$ . They match.

Answer: $\frac{12 - x}{12}$

Example 4. Simplify $\frac{3}{x - 1} - \frac{6}{x^{2} - 1}$ .

Show the working

Factorise the denominators first. $x^{2} - 1 = (x + 1) (x - 1)$ , which already contains $(x - 1)$ .
So the common denominator is $(x + 1) (x - 1)$ . The first fraction needs $(x + 1)$ on its top and bottom. The second fraction already has this denominator.
$\frac{3 (x + 1)}{(x + 1) (x - 1)} - \frac{6}{(x + 1) (x - 1)}$
Combine the numerators.
$\frac{3 x + 3 - 6}{(x + 1) (x - 1)} = \frac{3 x - 3}{(x + 1) (x - 1)}$
Do not stop yet. Factorise the numerator to look for a cancel: $3 x - 3 = 3 (x - 1)$ .
$\frac{3 (x - 1)}{(x + 1) (x - 1)} = \frac{3}{x + 1}$
Check with $x = 3$ . Original: $\frac{3}{2} - \frac{6}{8} = 1.5 - 0.75 = 0.75$ . Answer: $\frac{3}{4} = 0.75$ . They match.

Answer: $\frac{3}{x + 1}$

Where people go wrong

Mistake: Cancelling a term that is joined on by $+$ or $-$ , e.g. turning $\frac{x + 3}{x}$ into $3$ , or $\frac{x^{2} - 3 x - 10}{x^{2} - 25}$ into $\frac{- 3 x - 10}{- 25}$ .

Instead: Factorise the top and the bottom fully, then cancel only factors or brackets that multiply the whole top and the whole bottom. $\frac{x + 3}{x}$ does not simplify.

Cancelling is dividing the top and the bottom by the same thing. $(x + 3) \div x$ is $1 + \frac{3}{x}$ , not 3, so crossing out the $x$ s changes the value. With $x = 1$ , $\frac{x + 3}{x} = 4$ .

Mistake: Adding the tops and adding the bottoms: $\frac{1}{x} + \frac{1}{y} = \frac{2}{x + y}$ .

Instead: Use a common denominator: $\frac{1}{x} + \frac{1}{y} = \frac{y}{x y} + \frac{x}{x y} = \frac{x + y}{x y}$ .

The denominator tells you the size of the pieces, and you can only add pieces of the same size. With $x = y = 1$ the wrong version gives $1$ , but $\frac{1}{1} + \frac{1}{1} = 2$ .

Mistake: Losing a sign when subtracting, e.g. writing $\frac{4 x - 9 (x - 5)}{6}$ as $\frac{4 x - 9 x - 45}{6}$ .

Instead: Keep the second numerator in a bracket and multiply every term in it by the minus: $- 9 (x - 5) = - 9 x + 45$ .

The minus sign subtracts the whole of the second fraction, not only its first term.

Mistake: Treating an expression as an equation: multiplying $\frac{x}{3} + \frac{x - 4}{2}$ by 6 and giving $5 x - 12$ as the answer.

Instead: The answer is $\frac{5 x - 12}{6}$ . The common denominator stays in the answer.

You may multiply by 6 in an equation, because both sides change together. An expression has no other side, so multiplying it by 6 gives a different expression, six times as big.

Mistake: Turning the wrong fraction upside down in a division, or turning both.

Instead: Keep the first fraction, change $\div$ to $\times$ , and turn only the second fraction upside down.

Dividing by a fraction is multiplying by its reciprocal. The first fraction is the thing that you divide, so it does not change.

Check yourself

Simplify $\frac{12 a^{3} b}{8 a b^{2}}$ .
Answer
Divide top and bottom by $4 a b$ : $\frac{3 a^{2}}{2 b}$ .
Simplify $\frac{x^{2} - 16}{x^{2} + 4 x}$ .
Answer
$\frac{(x + 4) (x - 4)}{x (x + 4)} = \frac{x - 4}{x}$ .
Simplify $\frac{x^{2} + x - 12}{x^{2} - 6 x + 9}$ .
Answer
$\frac{(x + 4) (x - 3)}{(x - 3) (x - 3)} = \frac{x + 4}{x - 3}$ .
Does $\frac{x + 6}{x + 3}$ simplify to $2$ ? Give a reason.
Answer
No. The 6 and the 3 are terms, not factors of the whole top and the whole bottom, so nothing cancels. Test with $x = 1$ : $\frac{7}{4}$ , not $2$ .
Write $\frac{2 x}{5} + \frac{x + 1}{3}$ as a single fraction.
Answer
Common denominator 15: $\frac{6 x + 5 (x + 1)}{15} = \frac{11 x + 5}{15}$ .
Simplify $\frac{5 a}{6} \times \frac{4}{15 a^{2}}$ .
Answer
$\frac{20 a}{90 a^{2}} = \frac{2}{9 a}$ .
Simplify $\frac{2 y}{9} \div \frac{8 y^{2}}{3}$ .
Answer
$\frac{2 y}{9} \times \frac{3}{8 y^{2}} = \frac{6 y}{72 y^{2}} = \frac{1}{12 y}$ .
Write $\frac{3}{x + 2} - \frac{2}{x - 1}$ as a single fraction.
Answer
$\frac{3 (x - 1) - 2 (x + 2)}{(x + 2) (x - 1)} = \frac{3 x - 3 - 2 x - 4}{(x + 2) (x - 1)} = \frac{x - 7}{(x + 2) (x - 1)}$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Factorise $x^{2} - 3 x - 10$ . 2 marks No calculator

Mark scheme

B2 (x − 5)(x + 2). B1 for (x + a)(x + b) where ab = −10 or a + b = −3

(b) Hence simplify $\frac{x^{2} - 3 x - 10}{x^{2} - 25}$ . 2 marks No calculator

Mark scheme

M1 x² − 25 = (x − 5)(x + 5) seen
A1 (x + 2)/(x + 5) oe as final answer

(c) Another student writes $\frac{x^{2} - 3 x - 10}{x^{2} - 25} = \frac{- 3 x - 10}{- 25}$ . Explain why this is wrong. 1 mark No calculator

Mark scheme

B1 x² is a term, not a factor of the whole numerator and the whole denominator, so it cannot be cancelled oe; or a correct value that gives different results, e.g. x = 1 gives 1/2 on the left and 13/25 on the right

Question 2.

(a) Write $\frac{3 x}{4} - \frac{2 (x - 1)}{3}$ as a single fraction in its simplest form. 3 marks No calculator

Mark scheme

M1 common denominator 12 with at least one numerator correct, e.g. 9x/12 or 8(x − 1)/12
M1 9x − 8x + 8 seen, i.e. the minus applied to both terms of 8(x − 1)
A1 (x + 8)/12 cao

(b) Simplify $\frac{x^{2} + 2 x}{x^{2} - 1} \div \frac{x + 2}{x + 1}$ . 3 marks No calculator

Mark scheme

M1 second fraction inverted and ÷ changed to ×
M1 x(x + 2) and (x − 1)(x + 1) both seen
A1 x/(x − 1) oe as final answer

Question 3. A rectangle has length $\frac{x + 3}{2}$ cm and width $\frac{x - 1}{3}$ cm, where $x > 1$ .

(a) Show that the perimeter of the rectangle is $\frac{5 x + 7}{3}$ cm. 3 marks No calculator

Mark scheme

M1 2((x + 3)/2 + (x − 1)/3) oe, or the four sides added
M1 common denominator with correct numerators, e.g. (3(x + 3) + 2(x − 1))/6 or (3x + 9 + 2x − 2)/6
A1 reaches (5x + 7)/3 with no errors; nfww

(b) Find an expression for the area of the rectangle. Give your answer as a single fraction. 2 marks No calculator

Mark scheme

M1 (x + 3)/2 × (x − 1)/3 oe
A1 (x + 3)(x − 1)/6 or (x² + 2x − 3)/6

Question 4.

(a) Show that $\frac{4}{x + 1} + \frac{3}{x - 2}$ can be written as $\frac{7 x - 5}{x^{2} - x - 2}$ . 3 marks No calculator

Mark scheme

M1 common denominator (x + 1)(x − 2) soi
M1 4(x − 2) + 3(x + 1) over the common denominator
A1 4x − 8 + 3x + 3 and (x + 1)(x − 2) = x² − x − 2 both seen, reaching the given result; nfww

How this is examined

Factorising the top and the bottom of a rational expression, then cancelling the common bracket: A 3- or 4-mark Simplify, or a Factorise part followed by a 'Hence simplify' part that uses it Both the numerator and the denominator written in fully factorised form, e.g. $\frac{(x - 5) (x + 2)}{(x - 5) (x + 5)}$ , then the common bracket cancelled to give $\frac{x + 2}{x + 5}$ . The method mark is for the factorisation, so write it on the page before you cancel.
Adding or subtracting two fractions with brackets in the denominators: A 3-mark Write as a single fraction, or a Show (that) with the final fraction printed in the question The common denominator as the product of the different brackets, e.g. $(x + 1) (x - 2)$ ; each numerator multiplied by the bracket it did not have, e.g. $4 (x - 2) + 3 (x + 1)$ ; then the expanded numerator, $7 x - 5$ . In a Show (that), also write the expanded denominator if the printed answer has it expanded.
Subtracting fractions with number denominators without losing a sign: A 3-mark Write as a single fraction in its simplest form The common denominator with both numerators over it and the second in a bracket, e.g. $\frac{9 x - 8 (x - 1)}{12}$ ; then the expanded numerator $9 x - 8 x + 8$ ; then $\frac{x + 8}{12}$ , with the 12 kept in the answer.
Dividing one algebraic fraction by another: A 3-mark Simplify with a quadratic in at least one of the fractions The second fraction turned upside down and $\div$ changed to $\times$ , written down; every top and bottom factorised; then the answer after cancelling, e.g. $\frac{x}{x - 1}$ .
Saying why a cancellation is wrong: A 1-mark Explain about another person's working A statement that only a factor of the whole numerator and the whole denominator can be cancelled, and that the part crossed out (such as $x^{2}$ ) is a term, not a factor. A value of $x$ that gives two different results also shows the error.

Words to know

Algebraic fraction: A fraction with letters in the numerator, the denominator or both, such as $\frac{x + 1}{x - 3}$ .
Rational expression: One expression in powers of $x$ divided by another, such as $\frac{x^{2} - 2 x}{x^{2} - 5 x + 6}$ . In this subtopic it means the same as algebraic fraction.
Factor: A part that multiplies the rest of an expression. In $x (x - 2)$ the factors are $x$ and $(x - 2)$ . Only a factor of the whole top and the whole bottom can be cancelled.
Term: A part of an expression joined to the rest by $+$ or $-$ . In $x^{2} - 2 x$ the terms are $x^{2}$ and $- 2 x$ . A term cannot be cancelled.
Common denominator: A denominator that every fraction in a sum can be written over. The lowest common denominator is the simplest one.
Reciprocal: A fraction turned upside down. The reciprocal of $\frac{9 a}{10}$ is $\frac{10}{9 a}$ , and a fraction times its reciprocal is $1$ .
Simplest form: A fraction whose numerator and denominator, when fully factorised, have no factor in common.

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