Algebraic expressions (Grade 10) — Mathematics Grades 8–11

Expand, factorise and simplify algebraic expressions and algebraic fractions.

Gr 10 Term 1 - Algebraic expressionsTerm 1 - 3 weeks (CAPS)

Practise Algebraic expressions in the app →

What gets asked

Expand a binomial times a trinomial and collect like terms.
Factorise a trinomial, including one with a leading coefficient bigger than 1.
Factorise a four-term expression by grouping, or a sum or difference of cubes.
Simplify an algebraic fraction by factorising the top and bottom first.

You must be able to

Multiply every term in one bracket by every term in the other bracket.
Find two numbers with the right product and the right sum to factorise a trinomial.
Group four terms into two pairs that share an identical bracket.
Cancel a common factor in a fraction, never a single term.

Traps that cost marks

Pulling the leading number out the front of a trinomial as if it were a common factor of all three terms. Check first: a common factor must divide every term. If it does not, use the product-and-sum method instead.
Cancelling a letter that appears on the top and bottom, even though it is not a common factor. Only cancel whole factors, never a single term. $\frac{x + 3}{x + 5}$ cannot be simplified at all.
Grouping four terms into two pairs and stopping as soon as each pair has a common factor out. Check the two remaining brackets match exactly. If they do not, reorder the terms and try again.

Worked example

Factorise completely:

2 x^{2} - 5 x - 3

Multiply $a$ and $c$ : $2 \times (- 3) = - 6$ . Find factors of $- 6$ that add to $- 5$ : those are $- 6$ and $1$ .
Rewrite the middle term: $2 x^{2} - 6 x + x - 3$ .
Group and factorise: $2 x (x - 3) + 1 (x - 3)$ , giving $(x - 3) (2 x + 1)$ .

Rational and irrational numbers

Some numbers can be written as a fraction and some never can. Here is how to tell them apart.

real number: Any number with a place on the number line. Fractions and roots count too.
rational number: A number you can write as $\frac{a}{b}$ , where $a$ and $b$ are integers and $b$ is not $0$ .
irrational number: A real number that cannot be written that way. Its decimal never ends and never repeats a block.
recurring decimal: A decimal whose digits repeat forever, like $0, 333 \dots$
surd: A root with no exact fraction answer, like $2$ . A surd stays under the root sign.

Real numbers split into two groups. A rational number can be written as one integer over another. An irrational number cannot.

A decimal that stops, like $0, 25$ , is rational. A recurring decimal is rational too. Only a never-ending decimal with NO repeating block is irrational. $0, 101001000 \dots$ follows a rule yet repeats no block — irrational. So the comma alone tells you nothing.

Roots work the same way. $16 = 4$ is rational, because $16$ is a perfect square. Most numbers are not. When the number under the root is not a perfect square, that root is a surd. Every surd is irrational.

A calculator rounds. It shows $2$ as $1, 414213562$ , which looks as though it stops. It does not. Judge from the number under the root, not the screen.

To place a surd between two integers, find the perfect square just below and just above. $30$ sits between $25$ and $36$ , so it lies between $5$ and $6$ . Give both, smaller one first.

Rules to remember

$n$ is rational only when the whole number $n$ is a perfect square
When $a$ and $b$ are positive, $a^{2} < n < b^{2}$ gives $a < n < b$

Examples

Say whether

49

0, 75

and

7

are rational or irrational.

Worked answer

$49 = 7$ , and $7 = \frac{7}{1}$ , so it is rational.
$0, 75 = \frac{3}{4}$ , so it is rational too.
$7$ is not a perfect square, so $7$ has no exact answer.
So $7$ is a surd, and surds are irrational.

Answer: $49$ and $0, 75$ are rational; $7$ is irrational

A root is irrational only when the number under it is not a perfect square.

Between which two consecutive integers does

50

lie?

Worked answer

Find the perfect squares near $50$ : $7^{2} = 49$ and $8^{2} = 64$ .
$49$ is just below $50$ , and $64$ just above.
So $50$ lies between $49$ and $64$ .
That is between $7$ and $8$ .

Answer: between $7$ and $8$

The perfect squares on either side of $50$ give the two integers.

A square vegetable patch has an area of

30 m^{2}

. How long is one side?

Worked answer

The side of a square is the square root of its area, so each side is $30$ m.
$30$ is not a perfect square, so $30$ is a surd.
$5^{2} = 25$ and $6^{2} = 36$ , so the side lies between $5$ m and $6$ m.
A calculator shows about $5, 48$ , but that is rounded.

Answer: $30$ m, which lies between $5$ m and $6$ m

The exact side stays a surd, and the integers show how big it is.

Traps

Calling $0, 25$ irrational because it is written as a decimal. A decimal that stops or repeats is rational. Here $0, 25 = \frac{1}{4}$ .
Writing $π$ as $3, 14$ and calling $π$ rational. That is rounded. $π$ 's decimal never ends and never repeats a block — irrational. $\frac{1}{3} = 0, 333 \dots$ never ends too, but repeats — rational.
Assuming every square root is irrational, so calling $16$ irrational. Check for a perfect square first. $16 = 4$ , which is rational.
Halving the number under the root, so guessing $50$ is near $25$ . Bracket it between perfect squares. The nearest are $49$ and $64$ , giving $7$ and $8$ .

Learn and practise “Rational and irrational numbers” in the app →

Ordering and rounding

Putting numbers in order, and cutting them down to size without losing the answer.

ascending order: Arranged from smallest to largest.
decimal place: A digit position after the comma. In $3, 472$ the first decimal place holds $4$ .
round: To swap a number for a nearby simpler one with fewer digits.
exact value: The full value of a number, before anything has been rounded off.

To sort a mix of numbers, get them into one form you can compare. A surd has no exact decimal. Bracket it between integers, or use a rounded value just for the comparing. Then list the original numbers in order, not the rounded ones.

Minus signs turn the order around. $10$ is about $3, 16$ , so $- 10$ is about $- 3, 16$ . That sits to the left of $- 3$ on the number line. So $- 10$ is the smaller of the two.

To round to a number of decimal places, count the places after the comma. Then look at the very next digit. If it is $5$ or more, the last digit you keep goes up by one. If it is under $5$ , that digit stays. Drop the rest either way.

Round once, and round from the exact value. Rounding an already rounded number is how $3, 4499$ ends up as $3, 5$ when it should be $3, 4$ .

In a calculation, carry the exact value all the way and round only at the end. Rounding a price early can move a total by rands. Match the accuracy to the job as well: money to cents, a tape measure to millimetres.

Rules to remember

A next digit of $5$ or more sends the last kept digit up by one
Round once, from the exact value, never from an answer you have already rounded

Examples

Write

3

10

\frac{16}{5}

and

3, 4

in ascending order.

Worked answer

$3^{2} = 9$ and $4^{2} = 16$ , so $10$ lies between $3$ and $4$ .
$\frac{16}{5} = 3, 2$ , and $10$ is about $3, 16$ .
Now compare $3$ , $3, 16$ , $3, 2$ and $3, 4$ .
Write the original numbers back in that order.

Answer: $3$ , $10$ , $\frac{16}{5}$ , $3, 4$

The surd needed a decimal value before it could take its place.

Round

7, 3876

to two decimal places.

Worked answer

The second decimal place holds $8$ .
The next digit is $7$ , which is $5$ or more.
So the $8$ goes up to $9$ and the rest falls away.
That makes $7, 3876$ round to $7, 39$ .

Answer: $7, 39$

Only the digit straight after the place decides, not the ones behind it.

Petrol costs R

24, 37

a litre. A taxi driver buys

38

litres. What does that cost, to the nearest rand?

Worked answer

Multiply the exact value first: $24, 37 \times 38 = 926, 06$ .
Now round the answer to the nearest rand, which gives R $926$ .
Had the price been rounded to R $24$ first, you would get $24 \times 38 = 912$ .
That answer is R $14$ out.

Answer: R $926$

Rounding early moved the total, so the exact value had to be carried through.

Traps

Putting $- 10$ after $- 3$ because $10$ is the bigger number. The more negative number comes first. $- 10$ is about $- 3, 16$ , so it sits left of $- 3$ .
Cutting the extra digits off, so $7, 3876$ becomes $7, 38$ . Look at the next digit first. It is $7$ , so the $8$ rounds up and the answer is $7, 39$ .
Rounding twice, so $3, 4499$ becomes $3, 45$ and then $3, 5$ . Round once, from the exact value. To one decimal place $3, 4499$ is $3, 4$ .
Rounding the price before multiplying, so the total comes out rands short. Carry the exact value through the working. Round only the final answer.

Learn and practise “Ordering and rounding” in the app →

Expand and simplify

Multiplying a two-part bracket by a three-part bracket, and tidying up what comes out.

term: A part of an expression that is added or subtracted. In $3 x + 5$ the terms are $3 x$ and $5$ .
like terms: Terms with exactly the same letters, each raised to exactly the same power. $3 x^{2}$ and $2 x^{2}$ are like terms.
binomial: An expression with exactly two terms, like $x + 3$ .
trinomial: An expression with exactly three terms, like $x^{2} + 5 x + 6$ .
expand: To multiply brackets out until no brackets are left.
perfect square trinomial: A trinomial that is the square of a binomial, such as $x^{2} + 6 x + 9 = (x + 3)^{2}$ .

To expand a binomial times a trinomial, every term in the first bracket must meet every term in the second. Two terms times three terms gives six products. Write all six down before you tidy anything. Missing one is the most common mistake here.

Then collect like terms. Only terms with the same letters and the same power can join. So $3 x^{2}$ and $2 x^{2}$ add, but $x^{2}$ and $x$ never do.

Watch a minus sign in front of a bracket. It changes the sign of every term inside, not just the first. Expand the bracket first, then deal with the sign, and you will not lose a term.

A perfect square trinomial comes from squaring a binomial. Square the first term, square the last term, and the middle term is twice the two multiplied. That middle term is the check: two squares alone prove nothing.

The sign of the middle term tells you which bracket it came from. A plus gives $(a + b)^{2}$ and a minus gives $(a - b)^{2}$ . The first and last terms are positive squares either way.

Rules to remember

$a (b + c + d) = ab + a c + a d$
$(a + b)^{2} = a^{2} + 2 ab + b^{2}$
$(a - b)^{2} = a^{2} - 2 ab + b^{2}$

Examples

Expand:

(x + 2) (x^{2} + 3 x + 4)

Worked answer

Multiply each term of $(x + 2)$ by each term of the trinomial.
From the $x$ you get $x^{3} + 3 x^{2} + 4 x$ .
From the $2$ you get $2 x^{2} + 6 x + 8$ .
Collect: $3 x^{2} + 2 x^{2} = 5 x^{2}$ and $4 x + 6 x = 10 x$ .
So the answer is $x^{3} + 5 x^{2} + 10 x + 8$ .

Answer: $x^{3} + 5 x^{2} + 10 x + 8$

Six products came out, and only two pairs of them were like terms.

Simplify:

5 x - (x + 2) (x^{2} + 1)

Worked answer

Expand the bracket in two halves: $x (x^{2} + 1) = x^{3} + x$ .
And $2 (x^{2} + 1) = 2 x^{2} + 2$ .
Together the product is $x^{3} + 2 x^{2} + x + 2$ .
The minus in front flips all four signs.
Now add the $5 x$ : $- x + 5 x = 4 x$ , giving $- x^{3} - 2 x^{2} + 4 x - 2$ .

Answer: $- x^{3} - 2 x^{2} + 4 x - 2$

Every term inside the bracket changed sign, not only the first one.

Find the number that makes

x^{2} + 10 x + c

a perfect square trinomial.

Worked answer

A perfect square trinomial looks like $(x + b)^{2} = x^{2} + 2 b x + b^{2}$ .
Match the middle terms: $2 b = 10$ , so $b = 5$ .
The last term is $b$ squared, and $5^{2} = 25$ .
Check it: $x^{2} + 10 x + 25 = (x + 5)^{2}$ .

Answer: $25$

Halving the number in front of $x$ gives the number in the bracket, and squaring it finishes the job.

Traps

Multiplying only the first terms of each bracket, so three products come out instead of six. Every term in the first bracket meets every term in the second. Two times three is six products.
Adding the powers when collecting, so $x^{2} + x$ becomes $x^{3}$ . Only like terms add, and they keep their power. $x^{2} + x$ cannot be simplified at all.
Changing the sign of only the first term after a minus in front of a bracket. A minus in front flips every term inside. So $- (x^{2} + x + 2)$ becomes $- x^{2} - x - 2$ .
Calling $x^{2} + 7 x + 9$ a perfect square because $x^{2}$ and $9$ are both squares. Check the middle term too. For $(x + 3)^{2}$ it has to be $6 x$ , not $7 x$ .

Learn and practise “Expand and simplify” in the app →

Factorise trinomials

Finding the two brackets that a three-part expression came from.

factor: Something you multiply by. In $3 \times 4 = 12$ , both $3$ and $4$ are factors of $12$ .
factorise: To write an expression as a product, so that it ends up inside brackets.
common factor: A factor that every term shares. In $6 a + 9$ both terms share the factor $3$ .
leading coefficient: The number in front of the term with the highest power. In $3 x^{2} - x + 5$ it is $3$ .
product-and-sum method: Factorising a trinomial by hunting for two numbers with the right product and the right sum.

To factorise a trinomial is to expand it backwards. You are looking for two brackets that multiply back to what you started with. Expanding your answer is always a free check.

Use the product-and-sum method. When the leading coefficient is $1$ , find two numbers whose product is the last number and whose sum is the middle one. Both conditions must hold. A pair that gets the product right and the sum wrong is no good.

The signs follow the last number. If it is positive, both numbers carry the sign of the middle term. If it is negative, one number is positive and one is negative, and the bigger of the two carries the middle sign.

When the leading coefficient is not $1$ , multiply it by the last number. Find factors of that product which add to the middle number. Split the middle term into those two parts. You now have four terms, which you can pair off and factorise.

Fractions do not change the method. Take the fraction out as a common factor first. Factorise the whole-number trinomial that is left. Then write the fraction back in front of your answer.

Rules to remember

$x^{2} + b x + c = (x + p) (x + q)$ when $p + q = b$ and $pq = c$

Examples

Factorise:

x^{2} + 7 x + 12

Worked answer

Look for two numbers with product $12$ and sum $7$ .
Try $2$ and $6$ : $2 \times 6 = 12$ , but $2 + 6 = 8$ .
Try $3$ and $4$ : $3 \times 4 = 12$ and $3 + 4 = 7$ .
That pair works, so $x^{2} + 7 x + 12 = (x + 3) (x + 4)$ .

Answer: $(x + 3) (x + 4)$

The product and the sum both had to work, so the first pair was thrown out.

Factorise:

3 x^{2} + 11 x + 6

Worked answer

Multiply the leading coefficient by the last number: $3 \times 6 = 18$ .
Find factors of $18$ that add to $11$ : $9 \times 2 = 18$ and $9 + 2 = 11$ .
Split the middle term: $3 x^{2} + 9 x + 2 x + 6$ .
Pair them off: $3 x (x + 3) + 2 (x + 3)$ .
Take the common bracket out: $(x + 3) (3 x + 2)$ .

Answer: $(x + 3) (3 x + 2)$

Splitting the middle term turned a trinomial into a four-term pairing job.

Factorise:

\frac{1}{2} x^{2} + \frac{5}{2} x + 3

Worked answer

Every term here can be divided by $\frac{1}{2}$ .
Take it out to get $\frac{1}{2} (x^{2} + 5 x + 6)$ .
Factorise the trinomial: $x^{2} + 5 x + 6 = (x + 2) (x + 3)$ .
Write the $\frac{1}{2}$ back in front.

Answer: $\frac{1}{2} (x + 2) (x + 3)$

Clearing the fraction only helps if you put it back into the answer.

Traps

Pulling the leading number out the front of a trinomial as if it were a common factor of all three terms. Check first: a common factor must divide every term. If it does not, use the product-and-sum method instead.
Choosing $2$ and $6$ for $x^{2} + 7 x + 12$ because they multiply to $12$ . The sum has to work as well. Here $2 + 6 = 8$ , not $7$ , so use $3$ and $4$ .
Writing $x^{2} - 2 x - 8$ as $(x + 4) (x - 2)$ . A negative last number means the signs differ. The bigger number takes the middle sign, giving $(x - 4) (x + 2)$ .
Clearing the fraction and then leaving it out of the final answer. The $\frac{1}{2}$ belongs in the answer: $\frac{1}{2} (x + 2) (x + 3)$ .

Learn and practise “Factorise trinomials” in the app →

Grouping and cubes

Two more ways into brackets: pairing off four terms, and the rules for cubes.

grouping: Splitting an expression into pairs of terms, so that each pair has a common factor of its own.
sum of two cubes: An expression of the form $a^{3} + b^{3}$ : one cube added to another.
difference of two cubes: An expression of the form $a^{3} - b^{3}$ : one cube subtracted from another.
factorise completely: To keep going until no bracket in your answer can be factorised any further.

Grouping is for four terms. Split them into two pairs. Take a common factor out of each pair. If the two brackets that are left match exactly, take that bracket out too. If they do not match, pair the terms up differently and try again.

Sometimes a sign is all that is wrong. Remember that $(b - a) = - (a - b)$ . Taking a negative out of the second pair can make the two brackets agree.

Cubes have their own two rules. A sum of two cubes and a difference of two cubes each factorise into a binomial times a trinomial. The binomial keeps the sign you can see. The middle sign of the trinomial is the opposite one.

To spot a cube, look for $8$ , $27$ , $64$ or $125$ , which are $2^{3}$ , $3^{3}$ , $4^{3}$ and $5^{3}$ . A letter is cubed when its power divides by $3$ .

To factorise completely, take out a common factor first, then use whichever rule fits, then look once more. If a bracket in your answer still factorises, you are not finished.

Rules to remember

$a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$
$a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$

Examples

Factorise by grouping:

3 a x + 3 a y - 2 b x - 2 b y

Worked answer

Four terms, so split them into two pairs.
First pair: $3 a x + 3 a y = 3 a (x + y)$ .
Second pair: $- 2 b x - 2 b y = - 2 b (x + y)$ .
Both pairs left $(x + y)$ behind.
Take it out: $(x + y) (3 a - 2 b)$ .

Answer: $(x + y) (3 a - 2 b)$

The minus had to come out with the $2 b$ , or the brackets would not have matched.

Factorise:

x^{3} - 8

Worked answer

Write $8$ as a cube: $8 = 2^{3}$ .
This is a difference of two cubes with $a = x$ and $b = 2$ .
The rule gives $(a - b) (a^{2} + ab + b^{2})$ .
Put $x$ and $2$ in: $(x - 2) (x^{2} + 2 x + 4)$ .

Answer: $(x - 2) (x^{2} + 2 x + 4)$

The binomial kept the minus, but the trinomial's middle sign became a plus.

Factorise completely:

5 x^{3} + 40

Worked answer

Take the common factor out first: $5 x^{3} + 40 = 5 (x^{3} + 8)$ .
Now $x^{3} + 8$ is a sum of two cubes, since $8 = 2^{3}$ .
The rule gives $(x + 2) (x^{2} - 2 x + 4)$ .
Nothing inside factorises further.
So the full answer is $5 (x + 2) (x^{2} - 2 x + 4)$ .

Answer: $5 (x + 2) (x^{2} - 2 x + 4)$

Taking the $5$ out first is what made the sum of two cubes visible.

Traps

Grouping four terms into two pairs and stopping as soon as each pair has a common factor out. Check the two remaining brackets match exactly. If they do not, reorder the terms and try again.
Copying the plus into the middle, so $x^{3} + 8$ gives $(x + 2) (x^{2} + 2 x + 4)$ . The trinomial's middle sign is the opposite of the binomial's, so it is $(x + 2) (x^{2} - 2 x + 4)$ .
Trying to factorise $x^{2} + 2 x + 4$ further, as though it were a perfect square. That trinomial does not factorise here. Once the cube rule is used, it is finished.
Leaving the common factor in, so the sum of two cubes stays hidden. Take the common factor out first. Then $5 x^{3} + 40$ becomes $5 (x^{3} + 8)$ and the rule works.

Learn and practise “Grouping and cubes” in the app →

Simplify algebraic fractions

Factorise the top and the bottom first, and a fraction with letters in it gets a lot smaller.

algebraic fraction: A fraction with a letter in the top or the bottom, like $\frac{x + 1}{x - 3}$ .
cancel: To divide the top and the bottom by the same factor, so both get smaller.
difference of two squares: One square subtracted from another, of the form $a^{2} - b^{2}$ .
undefined: A fraction has no value when its bottom is $0$ . We say it is undefined there.

An algebraic fraction is simplified by cancelling, and you can only cancel a factor. So factorise the top and the bottom first. Until they are in brackets, there is usually nothing you are allowed to cancel.

A whole bracket cancels; a single term never does. The $x$ in $\frac{x + 3}{x + 5}$ is not a factor of either part, so that fraction is already as simple as it gets.

Watch for a sign flip. $(b - a) = - (a - b)$ , so a bracket that looks wrong way round can still cancel. It leaves $- 1$ behind, not $1$ .

If the bottom is a sum of two cubes or a difference of two cubes, use the cube rule on it. That produces a binomial and a trinomial, and one of them often matches the top exactly.

The bottom of a fraction may never be $0$ , because the fraction is undefined there. In $\frac{x + 1}{x - 3}$ the value $x = 3$ is not allowed. Say so as part of your answer.

Rules to remember

$\frac{k a}{k b} = \frac{a}{b}$ , as long as $k$ is not $0$
$(b - a) = - (a - b)$

Examples

Simplify:

\frac{x ^{2} - 9}{x + 3}

Worked answer

The top is a difference of two squares: $x^{2} - 9 = (x + 3) (x - 3)$ .
The fraction is now $\frac{( x + 3 ) ( x - 3 )}{x + 3}$ .
$(x + 3)$ is a whole factor of both, so it cancels.
What is left is $x - 3$ , and $x$ may not be $- 3$ .

Answer: $x - 3$

Factorising is what produced a bracket that was allowed to cancel.

Simplify:

\frac{x ^{2} + 5 x + 6}{x ^{2} - 4}

Worked answer

Factorise the top: $x^{2} + 5 x + 6 = (x + 2) (x + 3)$ .
Factorise the bottom: $x^{2} - 4 = (x + 2) (x - 2)$ .
$(x + 2)$ sits in both, so cancel it.
The answer is $\frac{x + 3}{x - 2}$ .

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

Both halves had to be factorised before anything could be seen to match.

Simplify:

\frac{x ^{2} - 2 x + 4}{x ^{3} + 8}

Worked answer

The bottom is a sum of two cubes, since $8 = 2^{3}$ .
Factorise it: $x^{3} + 8 = (x + 2) (x^{2} - 2 x + 4)$ .
That trinomial is exactly the top of the fraction.
Cancel it, and only $1$ is left on top.
So the answer is $\frac{1}{x + 2}$ .

Answer: $\frac{1}{x + 2}$

The cube rule handed back the very trinomial that was sitting on top.

Traps

Cancelling a letter that appears on the top and bottom, even though it is not a common factor. Only cancel whole factors, never a single term. $\frac{x + 3}{x + 5}$ cannot be simplified at all.
Cancelling $(a - b)$ against $(b - a)$ as though they were the same bracket. They differ by a sign, because $(b - a) = - (a - b)$ . Cancelling them leaves $- 1$ .
Leaving a trinomial as it is and cancelling nothing but a number in front. Factorise it first: $x^{2} + 5 x + 6 = (x + 2) (x + 3)$ . A bracket is what cancels.
Cancelling the $x$ on top against the $x^{3}$ inside $x^{3} + 8$ . $x^{3} + 8$ is a sum, not a product. Nothing cancels until you have factorised it.

Learn and practise “Simplify algebraic fractions” in the app →

Operations with fractions

Adding, subtracting, multiplying and dividing fractions that have letters in them.

numerator: The top of a fraction.
denominator: The bottom of a fraction. It may never be $0$ .
reciprocal: A fraction turned upside down. The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$ .
lowest common denominator: The simplest expression that every denominator in the sum divides into.
compound fraction: A fraction that has another fraction inside its top or its bottom.

To multiply, factorise everything first. Then cancel any factor on a top against the same factor on a bottom, even across the two fractions. Multiply what is left. Cancelling first keeps the numbers small.

To divide, turn the second fraction upside down and multiply by that reciprocal. Only the fraction you are dividing by flips. The first one is left exactly as it was.

To add or subtract, the denominators must match. Factorise each denominator, then build the lowest common denominator from the brackets you see. Rewrite each fraction over it, and only the numerators change.

Subtraction needs a bracket. The minus sign applies to every term of the second numerator, so write that numerator inside brackets before you take it away. Losing that bracket is the most common slip in this work.

A compound fraction is done from the inside out. Turn the top into one fraction and the bottom into one fraction. Then you have one fraction divided by another, which you already know how to do.

Rules to remember

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$

Examples

Simplify:

\frac{x + 2}{x - 1} \times \frac{x ^{2} - 1}{x ^{2} - 4}

Worked answer

Factorise: $x^{2} - 1 = (x - 1) (x + 1)$ .
Factorise: $x^{2} - 4 = (x + 2) (x - 2)$ .
$(x - 1)$ is on a bottom and a top, so cancel it.
$(x + 2)$ cancels the same way.
What is left is $\frac{x + 1}{x - 2}$ .

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

Cancelling before multiplying meant no big expanding was ever needed.

Simplify:

\frac{3}{x - 2} - \frac{1}{x + 1}

Worked answer

The lowest common denominator is $(x - 2) (x + 1)$ .
The numerators become $3 (x + 1)$ and $1 (x - 2)$ .
Subtract, keeping the bracket: $3 (x + 1) - (x - 2)$ .
That comes to $3 (x + 1) - (x - 2) = 2 x + 5$ .
So the answer is $\frac{2 x + 5}{( x - 2 ) ( x + 1 )}$ .

Answer: $\frac{2 x + 5}{( x - 2 ) ( x + 1 )}$

The bracket kept the minus on both terms of the second numerator.

Simplify:

\frac{1 - \frac{1}{x}}{x - 1}

Worked answer

Tidy the top into one fraction: $1 - \frac{1}{x} = \frac{x - 1}{x}$ .
The whole thing is now $\frac{x - 1}{x} \div (x - 1)$ .
Dividing means multiplying by the reciprocal $\frac{1}{x - 1}$ .
Cancel $(x - 1)$ , and $\frac{1}{x}$ is left.

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

The inside had to become a single fraction before the division could happen.

Traps

Turning the first fraction upside down instead of the second one when dividing. Only the fraction you divide by flips. The first fraction stays exactly as it is.
Adding the tops and the bottoms, so $\frac{1}{x} + \frac{1}{y}$ becomes $\frac{2}{x + y}$ . You need a common denominator first. Here it is $x y$ , which gives $\frac{y + x}{x y}$ .
Subtracting only the first term of the second numerator. Put that numerator in a bracket: $3 (x + 1) - (x - 2) = 2 x + 5$ .
Cancelling an $x$ from the inner fraction against one outside it. Turn the compound fraction into a single division first. Only then can whole factors cancel.

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