E-Math notes · Chapter 5

Algebraic Expressions and Formulae: O-Level E-Math Notes

Algebra is arithmetic with the numbers swapped for letters. In this chapter you learn to read and write that shorthand, simplify it, expand and factorise it, work with algebraic fractions, find the nth term of a pattern and rearrange a formula. These notes follow topic N5 of the SEC G3 Mathematics syllabus (K310), which is the same content as the O-Level syllabus.

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Chapter
5

Figure 1 · Cross method,

a = 1

16 outcomes · 50 worked examples · every step shown ✦

Letters, notation and evaluation

A letter is a number in disguise.

Words into algebra

"Less than" turns the order round. " $5$ less than $x$ " starts from $x$ and takes $5$ away, so it is $x - 5$ . Writing $5 - x$ is one of the most common slips.

Patterns and the nth term

The gap is the number in front of $n$ .

By the end of this chapter

Read algebra notation, use letters for numbers, and evaluate expressions and formulae.
Translate real situations into expressions, and find the nth term of a pattern.
Add, subtract and simplify linear expressions, and use brackets and common factors.
Expand products, use the two square identities and the difference of two squares, and factorise quadratics.
Multiply, divide, add and subtract algebraic fractions.
Change the subject of a formula and find an unknown quantity in a formula.

Start here

Maths In Real Life

Three situations where algebra does real work. The pictures are made first; the maths is then drawn on top by script, so each graph matches its equation exactly.

$A garden path in an HDB estate paved in rows of square tiles, with a line graph of tiles against section number drawn over the picture$

Maths In Real Life 1

Paving a garden path

A paved path is built one section at a time, and each new section adds the same number of tiles. That steady growth is exactly what a linear nth term describes.

A garden path in an HDB estate is paved in sections. Section $1$ uses $6$ tiles, section $2$ uses $10$ tiles, section $3$ uses $14$ tiles and section $4$ uses $18$ tiles. Each new section adds the same number of tiles.

(a)Write the number of tiles in section $n$ as an expression in $n$ .
(b)How many tiles does section $30$ use?
(c)Which section uses exactly $126$ tiles?

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

\displaystyle 10-6=4

Each section adds the same number of tiles: the gap is the coefficient of

n

\displaystyle 4(1)+c=6

Section

1

has

6

tiles; find

c

\displaystyle T_n = 4 n + 2

(b)

\displaystyle 4 n + 2

\displaystyle = 4(30)+2

Substitute

n = 30

\displaystyle = 122

(c)

\displaystyle 4 n + 2 = 126

Equate the rule to

126

\displaystyle 4n=124

Subtract

2

from both sides.

\displaystyle n = 31

Final answer

(a)

T_n = 4 n + 2

(b)

122

(c)Section

31

$A flower pot falling from a high-rise balcony beside an HDB block, with a curve of distance fallen against time drawn over the picture$

Maths In Real Life 2

A pot dropped from a balcony

In a school design challenge, the distance fallen is estimated by a simple rule of the form $s = 5t^2$ . Rules like this are rearranged all the time: you know the drop and want the time.

In a design challenge, a flower pot dropped from a high balcony is modelled by $s = 5t^2$ , where $s$ is the distance fallen in metres and $t$ is the time in seconds.

(a)Find $s$ when $t = 3$ .
(b)Make $t$ the subject of the formula.
(c)How long does a pot take to fall $80$ m?

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

\displaystyle 5t^2

\displaystyle = 5(3)^2

Square first, then multiply by

5

\displaystyle = 45

(b)

\displaystyle s=5t^2

\displaystyle \frac{s}{5}=t^2

Divide both sides by

5

\displaystyle t = \pm \sqrt{\frac{s}{5}}

Take the square root of both sides. A square root can be positive or negative, so write

\pm

\displaystyle t = \sqrt{\frac{s}{5}}

Solve by taking both roots, then keep only the positive one: t > 0 because a time cannot be negative.

(c)

\displaystyle s=5t^2

\displaystyle 80=5t^2

Substitute

80

for

s

\displaystyle 16=t^2

Divide both sides by

5

\displaystyle t = 4

Final answer

(a)

45

(b)

t = \sqrt{\frac{s}{5}}

(c)

t = 4

seconds

$Two taps filling a clear water tank, with a graph of the combined fill rate against x drawn over the picture$

Maths In Real Life 3

Two taps filling a tank

When two machines work together, their rates add. Rates are fractions of a job per hour, so combining them means adding algebraic fractions.

Tap A fills an empty water tank in $x$ hours. Tap B, which is slower, fills the same tank in $(x + 2)$ hours. Tap A fills $\frac{1}{x}$ of the tank in one hour and Tap B fills $\frac{1}{x+2}$ of it.

(a)Write, as a single fraction, the part of the tank that both taps fill in one hour together.
(b)Find that fraction when $x = 4$ .

Show full solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

\displaystyle \frac{1}{x}+\frac{1}{x+2}

\displaystyle = \frac{x+2}{x(x+2)}+\frac{x}{x(x+2)}

Common denominator

x(x+2)

. Every denominator must be non-zero: x ≠ 0, x ≠ -2.

\displaystyle = \frac{(x+2)+x}{x(x+2)}

Add the tops.

\displaystyle = \frac{2 x + 2}{x \left(x + 2\right)}

The result holds for x ≠ 0, x ≠ -2.

(b)

\displaystyle \frac{2 x + 2}{x \left(x + 2\right)}

\displaystyle = \frac{2(4)+2}{4(4+2)}

Substitute

x = 4

\displaystyle = \frac{10}{24}

Work out the top and the bottom.

\displaystyle = \frac{5}{12}

Final answer

(a)

\frac{2 x + 2}{x \left(x + 2\right)}

(b)

\frac{5}{12}

Supports N5.1 · N5.2.a · N5.2.b · N5.2.c · N5.2.d · N5.2.e ...

Letters, notation and evaluating

A letter in algebra stands for a number. If the number can change, the letter is a variable; if it is one fixed number we have not found yet, it is an unknown. Because a letter is just a number in disguise, every rule you already know from arithmetic keeps working.

The Core Idea

A letter is a number in disguise.

So $3x$ means three times the number $x$ , and when you are told what $x$ is, you can always swap the letter back for the number.

Links To

Need the quick version first? Read the free E-Math revision page on algebra

Choosing letters. Pick a letter that cannot be mixed up with something else. Use $n$ for a number of items, $t$ for time, $h$ for a height. Never give the same letter two meanings in one problem, and avoid $l$ and $o$ because they look like $1$ and $0$ . When you write by hand, make $x$ and $\times$ look different.

Reading algebra shorthand
Shorthand	Meaning in full	Say it as
$ab$	$a \times b$	a times b
$\frac{a}{b}$	$a \div b$ or $a \times \frac{1}{b}$	a divided by b
$a^2$ , $a^3$ , $a^2b$	$a \times a$ , $a \times a \times a$ , $a \times a \times b$	a squared, a cubed, a squared times b
$3y$	$y + y + y$ or $3 \times y$	three lots of y
$3(x+y)$	$3 \times (x+y)$	three times the sum of x and y
$\frac{3+y}{5}$	$(3+y) \div 5$ or $\frac{1}{5} \times (3+y)$	the sum of 3 and y, divided by 5

SGSchoolKaki Tip

Algebra drops the $\times$ sign: $ab$ means $a \times b$ , and $2x$ means $2 \times x$ . But two numbers written side by side are NOT multiplied: $23$ is twenty-three, so write $2 \times 3$ when you mean multiplication.

Think About It

3y

the same as

y^3

? Test it with

y = 2

before you answer.

Show the worked answerHide the worked answer

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 3y = y+y+y

Three lots of

y

means adding

y

to itself three times.

\displaystyle y^3 = y \times y \times y

A cube multiplies

y

by itself three times.

\displaystyle 3(2) = 6

Test with

y = 2

: three lots of two.

\displaystyle 2^3 = 8

Test with

y = 2

: two times two times two.

Final answerNo. The first adds

y

to itself three times, which gives 6 when

y

is two. The second multiplies

y

by itself three times, which gives 8. Same letters, different meaning.

Evaluating an expression or a formula means replacing every letter with its given number and working out the result. Put each substituted number in brackets, especially a negative one: $a^2$ with $a = -3$ is $(-3)^2$ , not $-3^2$ . Then follow the order of operations: brackets, powers, multiply and divide, then add and subtract.

Example 5.1

Reading the shorthand

Write each expression in full, with every $\times$ and $\div$ sign shown and no shorthand.

(i) $5pq^2$
(ii) $\frac{2m+n}{7}$
(iii) $4(a-b)$
(iv) $6k$ as repeated addition

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle 5 \times p \times q \times q

q^2

means

q \times q

; the

5

multiplies everything.

(ii)

\displaystyle (2m+n) \div 7

The fraction bar means divide, and it also acts like brackets around the top.

(iii)

\displaystyle 4 \times (a-b)

A number touching a bracket multiplies the whole bracket.

(iv)

\displaystyle k+k+k+k+k+k

6k

means six lots of

k

Final answer

(i)

5 \times p \times q \times q

(ii)

(2m+n) \div 7

(iii)

4 \times (a-b)

(iv)

k+k+k+k+k+k

Your Turn 5.1

Write each expression in full, with every $\times$ and $\div$ sign shown and no shorthand.

(i) $3a^2b$
(ii) $\frac{x-4}{9}$
(iii) $5(m+n)$
(iv) $4w$ as repeated addition

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle 3 \times a \times a \times b

Write out every multiplication and division sign that the shorthand hides.

(ii)

\displaystyle (x-4) \div 9

Write out every multiplication and division sign that the shorthand hides.

(iii)

\displaystyle 5 \times (m+n)

Write out every multiplication and division sign that the shorthand hides.

(iv)

\displaystyle w+w+w+w

Write out every multiplication and division sign that the shorthand hides.

Final answer

(i)

3 \times a \times a \times b

(ii)

(x-4) \div 9

(iii)

5 \times (m+n)

(iv)

w+w+w+w

Example 5.2

Evaluating with negative numbers and fractions

Given that $a = -3$ , $b = 4$ and $c = \frac{1}{2}$ , evaluate each expression.

(i) $2a^2-b$
(ii) $ab+4c$
(iii) $\frac{a+b}{c}$
(iv) $(a-b)^2$

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle 2a^2-b

\displaystyle = 2(-3)^2-4

Replace each letter by its number, with

-3

inside brackets.

\displaystyle = 2(9)-4

Powers before multiplying: square the

-3

first.

\displaystyle = 14

(ii)

\displaystyle ab+4c

\displaystyle = (-3)(4)+4\left(\frac{1}{2}\right)

Substitute;

ab

means

a \times b

\displaystyle = -12+2

Multiply before adding.

\displaystyle = -10

(iii)

\displaystyle \frac{a+b}{c}

\displaystyle = \frac{-3+4}{\frac{1}{2}}

Work out the top first; the bar is a bracket.

\displaystyle = 1 \div \frac{1}{2}

Dividing by a half is the same as multiplying by

2

\displaystyle = 2

(iv)

\displaystyle (a-b)^2

\displaystyle = (-3-4)^2

The bracket is worked out before the power.

\displaystyle = (-7)^2

Multiply the negative by itself: a negative squared is positive.

\displaystyle = 49

Final answer

(i)

14

(ii)

-10

(iii)

2

(iv)

49

Your Turn 5.2

Given that $a = -2$ , $b = 5$ and $c = \frac{1}{3}$ , evaluate each expression.

(i) $3a^2-b$
(ii) $ab+6c$
(iii) $\frac{a+b}{c}$
(iv) $(a-b)^2$

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle 3a^2-b

\displaystyle = 3 \times \left(-2\right)^{2} - 5

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = 3 \times 4 - 5

Work out the powers (and roots) first.

\displaystyle = 12 - 5

Next, multiply and divide.

\displaystyle = 7

Finally, add and subtract.

(ii)

\displaystyle ab+6c

\displaystyle = \left(-2\right) \times 5 + 6 \times \frac{1}{3}

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = -10 + 6 \times \frac{1}{3}

Next, multiply and divide.

\displaystyle = -10 + 2

Next, multiply and divide.

\displaystyle = -8

Finally, add and subtract.

(iii)

\displaystyle \frac{a+b}{c}

\displaystyle = \frac{-2 + 5}{\frac{1}{3}}

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = \frac{3}{\frac{1}{3}}

Work out the bracket (or the top and bottom of the fraction) first.

\displaystyle = 9

Next, multiply and divide.

(iv)

\displaystyle (a-b)^2

\displaystyle = \left(-2 - 5\right)^{2}

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = \left(-7\right)^{2}

Work out the bracket (or the top and bottom of the fraction) first.

\displaystyle = 49

Work out the powers (and roots) first.

Final answer

(i)

7

(ii)

-8

(iii)

9

(iv)

49

Example 5.3

Evaluating a formula

A school kitchen uses the rule $T = 40m + 20$ , where $T$ is the cooking time in minutes for a roast of mass $m$ kilograms. Find $T$ when $m = 2.5$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle 40m+20

\displaystyle = 40(2.5)+20

Substitute

2.5

for

m

\displaystyle = 100+20

Multiply first, then add.

\displaystyle = 120

Final answer

120

minutes

Example 5.4

A formula with a power

A ball rolls down a slope. The distance $s$ metres it travels is $s = ut + \frac{1}{2}at^2$ . Find $s$ when $u = 12$ , $a = -4$ and $t = 3$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle ut+\frac{1}{2}at^2

\displaystyle = 12(3)+\frac{1}{2}(-4)(3)^2

Substitute all three letters; the negative goes inside brackets.

\displaystyle = 36+\frac{1}{2}(-4)(9)

Work out the power first:

3^2

\displaystyle = 36-18

\frac{1}{2}

-4

-2

, then multiply by the square.

\displaystyle = 18

Final answer

18

Your Turn 5.4

Using $s = ut + \frac{1}{2}at^2$ , find $s$ when $u = 10$ , $a = -6$ and $t = 4$ .

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle ut+\frac{1}{2}at^2

\displaystyle = 10 \times 4 + \frac{1}{2} \times \left(-6\right) \times 4^{2}

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = 10 \times 4 + \frac{1}{2} \times \left(-6\right) \times 16

Work out the powers (and roots) first.

\displaystyle = 40 + \frac{1}{2} \times \left(-6\right) \times 16

Next, multiply and divide.

\displaystyle = 40 + \left(-48\right)

Next, multiply and divide.

\displaystyle = -8

Finally, add and subtract.

Final answer

-8

Supports N5.4

Turning words into algebra

Algebra is how we write a situation in short form. Reading the words slowly, one phrase at a time, and building the expression from the inside out, is the whole skill. When a phrase hides a bracket, put the bracket in.

Everyday words and their algebra
The words	Algebra	Note
the sum of $x$ and $y$ ; $x$ plus $y$	$x + y$	order does not matter
$x$ more than $y$ ; $y$ increased by $x$	$y + x$	order does not matter
the difference between $x$ and $y$ ; $x$ minus $y$	$x - y$	order matters
$a$ less than $b$ ; $b$ decreased by $a$	$b - a$	the thing you start from comes first
the product of $x$ and $y$ ; $x$ times $y$	$xy$	order does not matter
$x$ divided by $y$ ; the quotient of $x$ and $y$	$\frac{x}{y}$	$y$ is on the bottom
half of $x$ ; $x$ shared equally between two	$\frac{x}{2}$	same as $\frac{1}{2}x$
the square of $x$ ; $x$ squared	$x^2$	only $x$ is squared

SGSchoolKaki Tip

"Less than" turns the order round. " $5$ less than $x$ " starts from $x$ and takes $5$ away, so it is $x - 5$ . Writing $5 - x$ is one of the most common slips.

Think About It

"Twice the sum of

x

and

3

" and "the sum of twice

x

and

3

" sound alike. Are they the same expression?

Show the worked answerHide the worked answer

Colour keyexpand, rewrite, substitute or solve

\displaystyle 2(x+3) = 2x+6

"Twice the sum": the bracket is doubled, so both terms are doubled.

\displaystyle 2(10+3) = 26

Test with

x = 10

for the first expression.

\displaystyle 2(10)+3 = 23

Test with

x = 10

for "the sum of twice

x

and

3

": only the

x

is doubled.

Final answerNo. The first one doubles everything, so the bracket is needed. The second doubles only the

x

and then adds on the three. Try

x

equal to ten: the first gives 26 and the second gives 23. A bracket marks the difference.

Example 5.5

Turning sentences into expressions

Ken has $x$ marbles. Write an expression for each statement.

(i)Mei has $4$ more than twice as many marbles as Ken. How many marbles does Mei have?
(ii)Sam has $5$ fewer marbles than Ken, and Tara has three times as many as Sam. How many does Tara have?
(iii)Ken wins $6$ more marbles and then shares all his marbles equally among $4$ friends. How many does each friend get?

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle 2x

"Twice as many as Ken" is

2 \times x

\displaystyle 2 x + 4

4

more than" that adds

4

(ii)

\displaystyle x-5

Sam:

5

fewer than Ken.

\displaystyle 3 \left(x - 5\right)

Tara has three times Sam's number, so Sam's number goes in a bracket.

(iii)

\displaystyle x+6

Ken's marbles after winning.

\displaystyle \frac{x + 6}{4}

x+6

equally: divide the whole amount by

4

Final answer

(i)

2 x + 4

(ii)

3 \left(x - 5\right)

(iii)

\frac{x + 6}{4}

Your Turn 5.5

Mala has $y$ stickers. Write an expression for each statement.

(i)Rina has $2$ fewer than three times as many stickers as Mala. How many does Rina have?
(ii)Mala buys $7$ more stickers and then doubles her total by trading. How many does she have now?
(iii)Mala gives away $8$ stickers and shares the rest equally among $5$ people. How many does each person get?

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle 3y

"Three times as many as Mala" is

3 \times y

\displaystyle 3 y - 2

Two fewer than that: subtract

2

(ii)

\displaystyle y+7

Mala's stickers after buying

7

more.

\displaystyle 2 \left(y + 7\right)

Doubling means the whole total is doubled, so

y + 7

goes in a bracket.

(iii)

\displaystyle y-8

What is left after giving away

8

\displaystyle \frac{y - 8}{5}

Share equally among

5

: divide the whole amount by

5

Final answer

(i)

3 y - 2

(ii)

2 \left(y + 7\right)

(iii)

\frac{y - 8}{5}

Example 5.6

A shopping problem with brackets

At a school concert, an adult ticket costs $x$ dollars and a child ticket costs $4$ dollars less. A group buys $a$ adult tickets and $c$ child tickets. Write an expression for the total cost in dollars, and find its value when $x = 12$ , $a = 3$ and $c = 2$ .

(i)Total cost in terms of $a$ , $c$ and $x$ .
(ii)Value when $x = 12$ , $a = 3$ and $c = 2$ .

Show solutionHide solution

Colour keyexpand, rewrite, substitute or solve

(i)

\displaystyle x-4

A child ticket costs

4

dollars less than

x

dollars.

\displaystyle ax

a

adult tickets at

x

dollars each.

\displaystyle c(x-4)

c

child tickets at

(x-4)

dollars each: the bracket keeps the whole price together.

\displaystyle a x + c \left(x - 4\right)

(ii)

\displaystyle a x + c \left(x - 4\right)

\displaystyle = 3(12)+2(12-4)

Substitute.

\displaystyle = 36+2(8)

Bracket first.

\displaystyle = 52

Final answer

(i)

a x + c \left(x - 4\right)

(ii)

52

dollars

Your Turn 5.6

A cinema charges $y$ dollars for a standard seat and $3$ dollars more for a premium seat. A family buys $s$ standard and $r$ premium seats. Write an expression for the total cost in dollars, and find its value when $y = 9$ , $s = 2$ and $r = 3$ .

(i)Total cost in terms of $s$ , $r$ and $y$ .
(ii)Value when $y = 9$ , $s = 2$ , $r = 3$ .

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle y+3

A premium seat costs

3

dollars more than a standard seat.

\displaystyle sy

s

standard seats at

y

dollars each.

\displaystyle r(y+3)

r

premium seats at

(y+3)

dollars each: the bracket keeps the whole price together.

\displaystyle r \left(y + 3\right) + s y

(ii)

\displaystyle r \left(y + 3\right) + s y

\displaystyle = 3\left(9 + 3\right) + 2 \times 9

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = 3\left(9 + 3\right) + 18

Next, multiply and divide.

\displaystyle = 3 \times 12 + 18

Work out the bracket (or the top and bottom of the fraction) first.

\displaystyle = 36 + 18

Next, multiply and divide.

\displaystyle = 54

Finally, add and subtract.

Final answer

(i)

r \left(y + 3\right) + s y

(ii)

54

dollars

Example 5.7

Reading an expression back into words

Write a sentence in words for each expression, where $n$ is the number of pencils in a box.

(i) $3(n+5)$
(ii) $\frac{n-2}{4}$
(iii) $n^2 + 1$

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle 3(n+5)

Look for the outermost operation first: here it is a multiplication of a bracket.

(ii)

\displaystyle \frac{n-2}{4}

The bar divides the whole top line.

(iii)

\displaystyle n^2+1

Work out the power before the

+1

Final answer

(i)Add five to the number of pencils, then multiply by three (three times the sum of

n

and five).

(ii)Take away two from the number of pencils and divide the result by four.

(iii)Square the number of pencils, then add one.

Supports N5.5

Patterns and the nth term

A sequence is a list of numbers that follows a rule. Each number is a term; $T_1$ is the first term, $T_2$ the second, and $T_n$ the nth term, the term in position $n$ . If you can write $T_n$ as an expression in $n$ , you can find any term without listing the ones before it.

The Core Idea

The gap is the number in front of $n$ .

When the gap between neighbouring terms is always $d$ , the nth term is $dn + c$ , where $c$ is the first term minus $d$ . Write the expression, then test it on the first two terms before you trust it.

Case table 1: what the gaps between terms tell you
What you see	Picture of $T_n$ against $n$	nth term	Example sequence
Gaps all the same and positive (a constant difference $d > 0$ )	rising line	$dn + (T_1 - d)$	$5, 8, 11, 14, \ldots$
Gaps all the same and negative ( $d < 0$ )	falling line	$dn + (T_1 - d)$	$30, 26, 22, 18, \ldots$
Gaps themselves grow by a constant amount (a constant second difference)	curve	$an^2 + bn + c$	$2, 5, 10, 17, \ldots$

The terms

5, 8, 11, 14

sit on one straight line. The gap in

T_n

for each step along is the gradient of that line.

SGSchoolKaki Tip

Always check your nth term with $n = 1$ and $n = 2$ . If both give the right terms, a straight-line rule is right; if either fails, go back to the gap.

Think About It

A sequence has gaps

4, 4, 4, \ldots

. What number is the coefficient of

n

in its nth term, and why?

Show the worked answerHide the worked answer

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle T_{n+1} - T_n = 4

Neighbouring terms always differ by the gap.

\displaystyle 4(n+1) - 4n = 4

A rule containing

4n

grows by exactly

4

for every step along in

n

Final answerIt is the gap itself, 4. Each step along in position adds the gap to the term, so the term grows by the gap for every extra

n

, and the expression must contain the gap times

n

Example 5.8

Finding the nth term of an increasing sequence

The sequence $5, 9, 13, 17, \ldots$ continues in the same way.

(i)Give the rule for the nth term, $T_n$ .
(ii)Find the $60$ th term.
(iii)Is $150$ a term of the sequence? Give a reason.

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(i)

\displaystyle 9-5=4

Gap between terms:

+4

each time, so

T_n

starts with

4n

\displaystyle 4(1)+c=5

Match the first term to find

c

\displaystyle 4(2)+1=9

Check:

n=2

gives the second term.

\displaystyle T_n = 4 n + 1

(ii)

\displaystyle 4 n + 1

\displaystyle = 4(60)+1

Substitute

60

for

n

\displaystyle = 241

(iii)

\displaystyle 4 n + 1 = 150

150

is a term, then

T_n = 150

for some whole number

n

\displaystyle 4n=149

Subtract

1

from both sides.

\displaystyle n = 37.25

Final answer

(i)

T_n = 4 n + 1

(ii)

241

(iii)No. Solving gives

n =

37.25, which is not a whole number, so that value is not a term of the sequence.

Your Turn 5.8

The sequence $7, 12, 17, 22, \ldots$ continues in the same way.

(i)Give the rule for the nth term, $T_n$ .
(ii)Find the $40$ th term.
(iii)Is $203$ a term of the sequence? Give a reason.

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(i)

\displaystyle 7,\ 12,\ 17,\ 22

The terms of the sequence.

\displaystyle 12 - 7 = 5,\ 17 - 12 = 5,\ 22 - 17 = 5

Find the gap between neighbouring terms (first differences).

\displaystyle T_n = 5n + c

The gap is the same every time (5), so the nth term starts with 5n.

\displaystyle 5(1) + c = 7

Substitute

n = 1

: the first term tells us what

c

must be.

\displaystyle c = 2

Solve for

c

\displaystyle T_n = 5 n + 2

So T_n = 5n + 2; check it on the second term.

(ii)

\displaystyle 5 n + 2

\displaystyle = 5 \times 40 + 2

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = 200 + 2

Next, multiply and divide.

\displaystyle = 202

Finally, add and subtract.

(iii)

\displaystyle 5 n + 2 = 203

\displaystyle 5 n = 201

Subtract

2

from both sides.

\displaystyle n = 40.2

Divide both sides by

5

Final answer

(i)

T_n = 5 n + 2

(ii)

202

(iii)No. Solving gives

n =

40.2, which is not a whole number, so that value is not a term of the sequence.

Example 5.9

A decreasing sequence

The sequence $47, 41, 35, 29, \ldots$ keeps the same rule.

(i)Find the nth term.
(ii)Which term equals $-31$ ?

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(i)

\displaystyle 41-47=-6

The gap is

-6

, so

T_n

starts with

-6n

\displaystyle -6(1)+c=47

The first term is

47

, so

c

makes up the difference.

\displaystyle T_n = 53 - 6 n

(ii)

\displaystyle 53 - 6 n = -31

Equate

T_n

-31

\displaystyle -6n=-84

Subtract

53

from both sides.

\displaystyle n = 14

Final answer

(i)

T_n = 53 - 6 n

(ii)Term number

14

Your Turn 5.9

The sequence $60, 53, 46, 39, \ldots$ keeps the same rule.

(i)Find the nth term.
(ii)Which term equals $-24$ ?

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(i)

\displaystyle 60,\ 53,\ 46,\ 39

The terms of the sequence.

\displaystyle 53 - 60 = -7,\ 46 - 53 = -7,\ 39 - 46 = -7

Find the gap between neighbouring terms (first differences).

\displaystyle T_n = -7n + c

The gap is the same every time (-7), so the nth term starts with -7n.

\displaystyle -7(1) + c = 60

Substitute

n = 1

: the first term tells us what

c

must be.

\displaystyle c = 67

Solve for

c

\displaystyle T_n = 67 - 7 n

So T_n = -7n + 67; check it on the second term.

(ii)

\displaystyle 67 - 7 n = -24

\displaystyle - 7 n = -91

Subtract

67

from both sides.

\displaystyle n = 13

Divide both sides by

-7

Final answer

(i)

T_n = 67 - 7 n

(ii)Term number

13

Example 5.10

A pattern of tiles

A row of tiles is built in stages. Stage $1$ uses $5$ tiles, stage $2$ uses $8$ tiles, stage $3$ uses $11$ tiles and stage $4$ uses $14$ tiles. Each stage adds the same number of tiles.

(i)Write the number of tiles in stage $n$ as an expression in $n$ .
(ii)How many tiles does stage $25$ use?
(iii)Which stage uses exactly $95$ tiles?

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(i)

\displaystyle 8-5=3

Each stage adds

3

tiles, so the rule has

3n

\displaystyle 3(1)+c=5

Stage

1

has

5

tiles; find

c

\displaystyle T_n = 3 n + 2

(ii)

\displaystyle 3 n + 2

\displaystyle = 3(25)+2

Substitute

n = 25

in the rule.

\displaystyle = 77

(iii)

\displaystyle 3 n + 2 = 95

Equate the rule to

95

\displaystyle 3n=93

Subtract

2

from both sides.

\displaystyle n = 31

Final answer

(i)

T_n = 3 n + 2

(ii)

77

(iii)Stage

31

Your Turn 5.10

Stage $1$ of a pattern uses $6$ sticks, stage $2$ uses $10$ , stage $3$ uses $14$ and stage $4$ uses $18$ .

(i)Write the number of sticks in stage $n$ as an expression in $n$ .
(ii)How many sticks does stage $30$ use?
(iii)Which stage uses exactly $110$ sticks?

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(i)

\displaystyle 6,\ 10,\ 14,\ 18

The terms of the sequence.

\displaystyle 10 - 6 = 4,\ 14 - 10 = 4,\ 18 - 14 = 4

Find the gap between neighbouring terms (first differences).

\displaystyle T_n = 4n + c

The gap is the same every time (4), so the nth term starts with 4n.

\displaystyle 4(1) + c = 6

Substitute

n = 1

: the first term tells us what

c

must be.

\displaystyle c = 2

Solve for

c

\displaystyle T_n = 4 n + 2

So T_n = 4n + 2; check it on the second term.

(ii)

\displaystyle 4 n + 2

\displaystyle = 4 \times 30 + 2

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = 120 + 2

Next, multiply and divide.

\displaystyle = 122

Finally, add and subtract.

(iii)

\displaystyle 4 n + 2 = 110

\displaystyle 4 n = 108

Subtract

2

from both sides.

\displaystyle n = 27

Divide both sides by

4

Final answer

(i)

T_n = 4 n + 2

(ii)

122

(iii)Stage

27

Example 5.11

A pattern with growing gaps

The sequence $3, 6, 11, 18, \ldots$ is made by a rule that is not a straight line: the gaps are $3, 5, 7, \ldots$

(i)Find the nth term.
(ii)Find the $12$ th term.

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Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle n^2

The terms grow faster and faster, so try

n^2

first:

1, 4, 9, 16

\displaystyle 3-1=2

Each term is

2

more than

n^2

; the rule is

n^2 + 2

\displaystyle T_n = n^{2} + 2

(ii)

\displaystyle n^{2} + 2

\displaystyle = 12^2+2

Square first, then add.

\displaystyle = 146

Final answer

(i)

T_n = n^{2} + 2

(ii)

146

Your Turn 5.11

The sequence $2, 5, 10, 17, \ldots$ also has growing gaps.

(i)Find the nth term.
(ii)Find the $15$ th term.

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(i)

\displaystyle 2,\ 5,\ 10,\ 17

The terms of the sequence.

\displaystyle 5 - 2 = 3,\ 10 - 5 = 5,\ 17 - 10 = 7

Find the gap between neighbouring terms (first differences).

\displaystyle 5 - 3 = 2,\ 7 - 5 = 2

Second differences: the gaps themselves grow by a constant amount, so the rule contains n².

\displaystyle a = 2 \div 2 = 1

The coefficient of n² is half the second difference.

\displaystyle 1,\ 1,\ 1,\ 1

Subtract n^{2} from each term of the sequence: this is what is left.

\displaystyle T_n - n^{2} = 1

What is left is the same number every time: a constant.

\displaystyle T_n = n^{2} + 1

Add the leftover rule to the n² term.

(ii)

\displaystyle n^{2} + 1

\displaystyle = 15^{2} + 1

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = 225 + 1

Work out the powers (and roots) first.

\displaystyle = 226

Finally, add and subtract.

Final answer

(i)

T_n = n^{2} + 1

(ii)

226

Supports N5.6 · N5.7 · N5.4

Adding, subtracting and simplifying linear expressions

Like terms have exactly the same letters with exactly the same powers. They can be added or subtracted: $5x + 2x = 7x$ is just "five lots of $x$ plus two lots of $x$ ". Unlike terms cannot be combined, so $3a + 3b$ stays as it is.

Like terms and unlike terms
Terms	Like or unlike?	Why
$5x$ and $-2x$	like	same letter, same power
$3a$ and $3b$	unlike	different letters
$4x^2$ and $4x$	unlike	same letter, different power
$ab$ and $-3ba$	like	$ba$ is the same as $ab$
$7$ and $-2$	like	plain numbers are like terms with each other

The Core Idea

Combine only like terms.

Collect the like terms, keeping the sign in front of each one, then add their coefficients.

SGSchoolKaki Tip

A minus sign in front of a bracket changes the sign of every term inside it: $-(x+3)$ is $-x-3$ , not $-x+3$ .

SGSchoolKaki Tip

Underline or circle each group of like terms in a different way. The sign belongs to the term on its right, so $-2b$ moves as $-2b$ .

Think About It

3x + 2y

: can it be written as one term, like

5xy

Show the worked answerHide the worked answer

\displaystyle 3(2)+2(1) = 8

Test with

x = 2

and

y = 1

3x + 2y

\displaystyle 5(2)(1) = 10

Test the same values in

5xy

: a different answer, so they are not equal.

Final answerNo.

x

and

y

are different letters, so the terms are unlike and cannot be merged into one term.

Example 5.12

Adding and subtracting like terms

Simplify $5a + 3b - 2a + 7b - 4$ .

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Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 5a+3b-2a+7b-4

\displaystyle = (5a-2a)+(3b+7b)-4

Group the like terms; keep each sign with its term.

\displaystyle = 3 a + 10 b - 4

Add the coefficients of each group.

Final answer

3 a + 10 b - 4

Example 5.13

Removing brackets with a minus sign

Simplify $(4x - 3y) - (2x - 5y)$ .

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\displaystyle (4x-3y)-(2x-5y)

\displaystyle = 4x-3y-2x+5y

The minus sign changes both signs inside the second bracket.

\displaystyle = (4x-2x)+(-3y+5y)

Collect the like terms.

\displaystyle = 2 x + 2 y

Final answer

2 x + 2 y

Example 5.14

Multiplying a bracket, including a negative multiplier

Simplify each expression.

(i) $-4(x-3)+2(3x+1)$
(ii) $\frac{12x-18}{6}$

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Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle -4(x-3)+2(3x+1)

\displaystyle = -4x+12+6x+2

Multiply each term inside by the number outside;

-4 \times -3

is positive.

\displaystyle = (-4x+6x)+(12+2)

Collect like terms.

\displaystyle = 2 x + 14

(ii)

\displaystyle \frac{12x-18}{6}

\displaystyle = \frac{12x}{6}-\frac{18}{6}

Divide each term on top by the number underneath.

\displaystyle = 2 x - 3

Final answer

(i)

2 x + 14

(ii)

2 x - 3

Your Turn 5.14

Simplify each expression.

(i) $-3(2y-5)+4(y+3)$
(ii) $\frac{20a-15}{5}$

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(i)

\displaystyle -3(2y-5)+4(y+3)

\displaystyle = -6 y + 15 + 4 y + 12

Expand each bracket term by term; mind every sign.

\displaystyle = 27 - 2 y

Add the coefficients of each group of like terms.

(ii)

\displaystyle \frac{20a-15}{5}

\displaystyle = \frac{20 a}{5} - \frac{15}{5}

Divide each term on the top by the number underneath.

\displaystyle = 4 a - 3

Simplify each fraction.

Final answer

(i)

27 - 2 y

(ii)

4 a - 3

Example 5.15

Fractions with numbers underneath

Simplify each expression, giving each answer as a single fraction.

(i) $\frac{x+1}{3}+\frac{2x-5}{4}$
(ii) $\frac{2x}{3}-\frac{4(x-1)}{5}$

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(i)

\displaystyle \frac{x+1}{3}+\frac{2x-5}{4}

\displaystyle = \frac{4(x+1)}{12}+\frac{3(2x-5)}{12}

Common denominator: the smallest number both

3

and

4

go into.

\displaystyle = \frac{4x+4+6x-15}{12}

Expand each top; keep the whole second top together.

\displaystyle = \frac{10 x - 11}{12}

(ii)

\displaystyle \frac{2x}{3}-\frac{4(x-1)}{5}

\displaystyle = \frac{10x}{15}-\frac{12(x-1)}{15}

Common denominator for

3

and

5

\displaystyle = \frac{10x-12x+12}{15}

The minus sign changes every sign in the bracket.

\displaystyle = \frac{12 - 2 x}{15}

Final answer

(i)

\frac{10 x - 11}{12}

(ii)

\frac{12 - 2 x}{15}

Your Turn 5.15

Simplify each expression, giving each answer as a single fraction.

(i) $\frac{x-2}{4}+\frac{3x+1}{5}$
(ii) $\frac{3x}{4}-\frac{2(x+3)}{3}$

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Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle \frac{x-2}{4}+\frac{3x+1}{5}

\displaystyle = \frac{5\left(x - 2\right)}{20} + \frac{4\left(3 x + 1\right)}{20}

Common denominator 20: the smallest number that every denominator divides into. Multiply each top by what its bottom is missing.

\displaystyle = \frac{5\left(x - 2\right) + 4\left(3 x + 1\right)}{20}

Write over the one denominator. Keep each top in a bracket, especially after a minus sign.

\displaystyle = \frac{5 x - 10 + 12 x + 4}{20}

Expand each top; a minus sign in front of a bracket changes every sign inside it.

\displaystyle = \frac{17 x - 6}{20}

Collect the like terms in the top.

(ii)

\displaystyle \frac{3x}{4}-\frac{2(x+3)}{3}

\displaystyle = \frac{9 x}{12} - \frac{8\left(x + 3\right)}{12}

Common denominator 12: the smallest number that every denominator divides into. Multiply each top by what its bottom is missing.

\displaystyle = \frac{9 x - 8\left(x + 3\right)}{12}

Write over the one denominator. Keep each top in a bracket, especially after a minus sign.

\displaystyle = \frac{9 x - 8 x - 24}{12}

Expand each top; a minus sign in front of a bracket changes every sign inside it.

\displaystyle = \frac{x - 24}{12}

Collect the like terms in the top.

Final answer

(i)

\frac{17 x - 6}{20}

(ii)

\frac{x - 24}{12}

Example 5.16

Checking an answer by substitution

Simplify $3(2x - 1) - 2(x - 4)$ . Then check your answer by evaluating both the original and your result at $x = 2$ .

(i)Simplify.
(ii)Value of the original expression at $x = 2$ .
(iii)Value of your simplified expression at $x = 2$ .

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle 3(2x-1)-2(x-4)

\displaystyle = 6x-3-2x+8

Multiply out; the second bracket flips its signs.

\displaystyle = (6x-2x)+(-3+8)

Collect like terms.

\displaystyle = 4 x + 5

(ii)

\displaystyle 3(2x-1)-2(x-4)

\displaystyle = 3(2(2)-1)-2(2-4)

Substitute into the original.

\displaystyle = 13

(iii)

\displaystyle 4 x + 5

\displaystyle = 4(2)+5

Substitute into the answer.

\displaystyle = 13

Final answer

(i)

4 x + 5

(ii)

13

(iii)

13

Your Turn 5.16

Simplify $4(x + 2) - 3(2x - 1)$ and check your answer with $x = 3$ .

(i)Simplify.
(ii)Original at $x = 3$ .
(iii)Your answer at $x = 3$ .

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle 4(x+2)-3(2x-1)

\displaystyle = 4 x + 8 - 6 x + 3

Expand each bracket term by term; mind every sign.

\displaystyle = 11 - 2 x

Add the coefficients of each group of like terms.

(ii)

\displaystyle 4(x+2)-3(2x-1)

\displaystyle = 4\left(3 + 2\right) - 3\left(2 \times 3 - 1\right)

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = 4\left(3 + 2\right) - 3\left(6 - 1\right)

Next, multiply and divide.

\displaystyle = 4 \times 5 - 3 \times 5

Work out the bracket (or the top and bottom of the fraction) first.

\displaystyle = 20 - 15

Next, multiply and divide.

\displaystyle = 5

Finally, add and subtract.

(iii)

\displaystyle 11 - 2 x

\displaystyle = 11 - 2 \times 3

Substitute each value for its letter; a negative number goes inside brackets.

\displaystyle = 11 - 6

Next, multiply and divide.

\displaystyle = 5

Finally, add and subtract.

Final answer

(i)

11 - 2 x

(ii)

5

(iii)

5

Example 5.17

Perimeter of a rectangle

A rectangular garden has length $(3x + 2)$ m and width $(x - 1)$ m. Write the perimeter as a simplified expression in $x$ .

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Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 2(3x+2)+2(x-1)

\displaystyle = 6x+4+2x-2

Perimeter is twice the length plus twice the width.

\displaystyle = (6x+2x)+(4-2)

Collect the like terms.

\displaystyle = 8 x + 2

Final answer

8 x + 2

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 5 of 19