Algebraic Notation

How algebra writes multiplication, repeated addition, squares and cubes, division and brackets in short form, and how to read each short form back with × and ÷ signs.

Before you start

A letter (a variable) stands for a number. You have already met two short forms: $4 \times k$ is written $4k$ , and $k \div 6$ is written $\frac{k}{6}$ .

Order of operations: work out brackets first, then $\times$ and $\div$ from left to right, then $+$ and $-$ from left to right. So $2 \times (3 + 4) = 2 \times 7 = 14$ , but $2 \times 3 + 4 = 6 + 4 = 10$ .

Index notation: $5^{2} = 5 \times 5$ and $2^{3} = 2 \times 2 \times 2$ .

Leaving Out the Multiplication Sign

Between two numbers the $\times$ sign must stay: $3 \times 4$ is 12, and writing it as 34 would give a different number. Next to a letter there is no such danger, so algebra leaves the $\times$ sign out. This also stops the sign $\times$ being mistaken for the letter $x$ .

Numbers can be multiplied in any order ( $3 \times 5 = 5 \times 3$ ), and so can letters, which stand for numbers. So we choose one tidy way of writing every product: the number first, then the letters in alphabetical order.

Rule: writing a product

$a \times b$ is written $ab$ , and $3 \times p$ is written $3p$ .

The number is written first: $p \times 3$ is also written $3p$ , never $p3$ .

Letters are written in alphabetical order: $q \times p$ is written $pq$ , and $c \times a \times b$ is written $abc$ .

The numbers in a product are multiplied together: $2 \times k \times 5$ is written $10k$ .

$1 \times n$ is written $n$ , and $-1 \times x$ is written $-x$ .

Example 1: Products, both ways

Write (a) to (f) in algebraic notation, and write (g) and (h) with $\times$ signs.
(a) $7 \times m$ (b) $n \times 9$ (c) $y \times x$ (d) $1 \times t$ (e) $h \times (-1)$ (f) $2 \times q \times 6 \times p$
(g) $5ab$ (h) $-12mn$

Solution

(a) Leave out the $\times$ sign: $7m$ .

(b) Multiplication can be done in either order, so $n \times 9 = 9 \times n$ . With the number first, it is $9n$ .

(d) One lot of $t$ is $t$ , so the 1 is not written: $t$ .

(e) $h \times (-1) = -1 \times h$ , which is written $-h$ .

(f) Put the numbers together and the letters in order: $2 \times 6 \times p \times q$ . Then $2 \times 6 = 12$ , so it is $12pq$ .

(g) Put a $\times$ sign between each pair of neighbours: $5 \times a \times b$ .

(h) $-12mn = -12 \times m \times n$ .

(a) $7m$ (b) $9n$ (c) $xy$ (d) $t$ (e) $-h$ (f) $12pq$ (g) $5 \times a \times b$ (h) $-12 \times m \times n$

Repeated Addition

In Primary you learnt that $4 + 4 + 4 = 3 \times 4$ : adding the same number three times is multiplying it by 3. The same is true when the number is a letter.

Rule: repeated addition

$3y = y + y + y = 3 \times y$ .

In the same way, $5k = k + k + k + k + k = 5 \times k$ : the number in front says how many times the letter is added.

3y is not 3 + y

$3 + y$ means 3 more than $y$ ; $3y$ means 3 times $y$ . They are different: when $y = 10$ , $3 + y$ is 13 but $3y$ is 30.

Example 2: Repeated addition, both ways

(a) Write $k + k + k + k + k + k$ in algebraic notation.
(b) Write $4t$ as a sum, and as a product with a $\times$ sign.
(c) Write $w + w$ in algebraic notation.

Solution

(a) $k$ is added 6 times, which is $6 \times k$ , written $6k$ .

(b) $4t$ is 4 lots of $t$ : $t + t + t + t$ . As a product it is $4 \times t$ .

(a) $6k$ (b) $4t = t + t + t + t = 4 \times t$ (c) $2w$

Squares and Cubes of a Letter

Index notation works for letters exactly as it does for numbers. $5 \times 5$ is written $5^{2}$ ; in the same way $a \times a$ is written $a^{2}$ , read " $a$ squared". $a \times a \times a$ is written $a^{3}$ , read " $a$ cubed".

Rule: squares and cubes

$a^{2} = a \times a$ and $a^{3} = a \times a \times a$ .

The small raised number, the index, belongs only to the number, letter or bracket just before it. So $a^{2}b = a \times a \times b$ and $ab^{2} = a \times b \times b$ .

$3a^{2} = 3 \times a \times a$ : only $a$ is squared.

$(3a)^{2} = 3a \times 3a = 3 \times 3 \times a \times a = 9a^{2}$ : the brackets make the whole of $3a$ squared.

2a and a² are different

$2a = a + a = 2 \times a$ (add two $a$ s), but $a^{2} = a \times a$ (multiply two $a$ s). When $a = 5$ , $2a$ is 10 but $a^{2}$ is 25.

In the same way, $3a = a + a + a$ but $a^{3} = a \times a \times a$ .

Example 3: Writing with indices

Write each in algebraic notation.
(a) $m \times m$ (b) $k \times k \times k$ (c) $a \times a \times b$ (d) $x \times y \times y \times 5$ (e) $n \times m \times n \times m \times n$

Solution

(a) Two $m$ s multiplied: $m^{2}$ .

(b) Three $k$ s multiplied: $k^{3}$ .

(d) Number first: $5 \times x \times y \times y$ . There is one $x$ and two $y$ s, so it is $5xy^{2}$ .

(e) Put the same letters together, in alphabetical order: $m \times m \times n \times n \times n$ . Two $m$ s and three $n$ s give $m^{2}n^{3}$ .

(a) $m^{2}$ (b) $k^{3}$ (c) $a^{2}b$ (d) $5xy^{2}$ (e) $m^{2}n^{3}$

Example 4: Reading indices back

Write each with $\times$ signs and no indices.
(a) $p^{2}q$ (b) $4ab^{2}$ (c) $-x^{3}$ (d) $2c^{2}$ (e) $(2c)^{2}$

Solution

(a) The index 2 belongs to $p$ only: $p \times p \times q$ .

(b) The index 2 belongs to $b$ only: $4 \times a \times b \times b$ .

(d) The index belongs to $c$ only: $2 \times c \times c$ .

(e) The brackets make the whole of $2c$ squared: $2c \times 2c = 2 \times c \times 2 \times c$ . In short form this is $2 \times 2 \times c \times c = 4c^{2}$ , which is not the same as $2c^{2}$ in (d).

(a) $p \times p \times q$ (b) $4 \times a \times b \times b$ (c) $-1 \times x \times x \times x$ (d) $2 \times c \times c$ (e) $2 \times c \times 2 \times c$ , which is $4c^{2}$

Writing a Product in Its Shortest Form

A product of numbers and letters, however long, can be written as one short term. Because the order of multiplying does not matter, the factors can be rearranged before they are put together.

Steps: the shortest form of a product

1. Write every factor separately, with $\times$ signs ( $3x$ is $3 \times x$ ; $z^{2}$ is $z \times z$ ).

2. Multiply the numbers together, sign included.

3. Put the same letters together and write them with an index.

4. Write the number first, then the letters in alphabetical order.

Example 5: Shortest forms

Write each in its shortest form.
(a) $r \times r \times 3 \times s \times t$ (b) $3x \times 9y$ (c) $2y \times y \times 3z \times z^{2}$ (d) $4a \times (-5b) \times a$

Solution

(a) Number first: $3 \times r \times r \times s \times t$ . Two $r$ s make $r^{2}$ , so it is $3r^{2}st$ .

(b) $3x \times 9y = 3 \times x \times 9 \times y = 3 \times 9 \times x \times y$ . Then $3 \times 9 = 27$ , so it is $27xy$ .

Numbers: $2 \times 3 = 6$ . Letters: $y \times y = y^{2}$ , and $z \times z \times z = z^{3}$ (so $z \times z^{2} = z^{3}$ ).

So it is $6y^{2}z^{3}$ .

(d) $4 \times a \times (-5) \times b \times a = 4 \times (-5) \times a \times a \times b$ . Then $4 \times (-5) = -20$ , so it is $-20a^{2}b$ .

(a) $3r^{2}st$ (b) $27xy$ (c) $6y^{2}z^{3}$ (d) $-20a^{2}b$

Division Written as a Fraction

In algebra the $\div$ sign is rarely used. A division is written as a fraction, as in Primary, where $3 \div 4 = \frac{3}{4}$ .

Dividing by a number is the same as multiplying by its reciprocal: $12 \div 4 = 3$ , and $12 \times \frac{1}{4} = 3$ too. So a division can be read in two ways.

Rule: division

$\frac{a}{b} = a \div b = a \times \frac{1}{b}$ , where $b \neq 0$ .

So $\frac{x}{5} = x \div 5 = \frac{1}{5} \times x$ , which is also written $\frac{1}{5}x$ .

What stands below the fraction line cannot be 0, because dividing by 0 has no meaning: in $\frac{x}{y}$ , $y$ cannot be 0, and in $\frac{a}{b + c}$ , $b + c$ cannot be 0.

Example 6: Writing divisions as fractions

Write each in algebraic notation, as a fraction where there is a division.
(a) $x \div 4$ (b) $3 \div y$ (c) $2p \div q$ (d) $\frac{1}{5} \times m$ (e) $7p \div q \times 3$

Solution

(a) $x \div 4 = \frac{x}{4}$ .

(b) The number divided goes on top: $3 \div y = \frac{3}{y}$ .

(d) Multiplying by $\frac{1}{5}$ is dividing by 5: $\frac{1}{5} \times m = m \div 5 = \frac{m}{5}$ .

(e) $\times$ and $\div$ are worked from left to right. Dividing by $q$ is multiplying by $\frac{1}{q}$ : $7p \div q \times 3 = 7 \times p \times \frac{1}{q} \times 3$ .

Put the numbers together: $7 \times 3 \times p \times \frac{1}{q} = 21p \times \frac{1}{q} = \frac{21p}{q}$ .

(a) $\frac{x}{4}$ (b) $\frac{3}{y}$ (c) $\frac{2p}{q}$ (d) $\frac{m}{5}$ (e) $\frac{21p}{q}$

Example 7: Reading a fraction

Write each fraction in two ways, once with $\div$ and once with $\times$ .
(a) $\frac{x}{7}$ (b) $\frac{5m}{n}$ , where $n$ does not stand for 0

Solution

(a) $\frac{x}{7}$ is $x$ divided by 7: $x \div 7$ . Dividing by 7 is multiplying by $\frac{1}{7}$ , so it is also $\frac{1}{7} \times x$ .

(b) Everything above the line, $5m$ , is divided by $n$ : $5m \div n$ . Dividing by $n$ is multiplying by $\frac{1}{n}$ , so it is also $5m \times \frac{1}{n}$ .

(a) $x \div 7$ , or $\frac{1}{7} \times x$ (b) $5m \div n$ , or $5m \times \frac{1}{n}$

Brackets and the Fraction Line

Brackets keep a sum or a difference together, so that it is worked out first and then used as one number. $3(x + y)$ means $3 \times (x + y)$ : add $x$ and $y$ first, then multiply the whole sum by 3. The $\times$ sign in front of a bracket holding letters is left out, just as it is in front of a letter.

A fraction line also keeps things together: everything above the line is divided by everything below it. $\frac{3 + y}{5}$ means: add 3 and $y$ first, then divide the whole sum by 5.

Rule: brackets and the fraction line

$3(x + y) = 3 \times (x + y)$ . It is not $3x + y$ , in which only $x$ is multiplied by 3.

$\frac{3 + y}{5} = (3 + y) \div 5 = \frac{1}{5} \times (3 + y)$ . It is not $3 + \frac{y}{5}$ , in which only $y$ is divided by 5.

A number multiplying a bracket is written in front of it: $(x + y) \times 3$ is written $3(x + y)$ .

Example 8: Brackets and fractions, into algebraic notation

Write each in algebraic notation, as a fraction where there is a division.
(a) $4 \times (m + n)$ (b) $(p - 2) \times 5$ (c) $(x + 7) \div 3$ (d) $\frac{1}{4} \times (k - 1)$ (e) $a \div (b + c)$ (f) $(a - b) \times (a - b)$

Solution

(a) Leave out the $\times$ sign: $4(m + n)$ .

(b) The number goes in front of the bracket: $5(p - 2)$ .

(d) Multiplying by $\frac{1}{4}$ is dividing by 4: $(k - 1) \div 4 = \frac{k - 1}{4}$ .

(e) $a$ is divided by the whole of $b + c$ , so all of it goes below the line: $\frac{a}{b + c}$ .

(f) The same bracket multiplied by itself is its square: $(a - b)^{2}$ . The brackets show that the whole of $a - b$ is squared.

(a) $4(m + n)$ (b) $5(p - 2)$ (c) $\frac{x + 7}{3}$ (d) $\frac{k - 1}{4}$ (e) $\frac{a}{b + c}$ (f) $(a - b)^{2}$

Example 9: Brackets and fractions, read back

Write each with $\times$ or $\div$ signs.
(a) $2(a - b)$ (b) $\frac{y + 3}{5}$ , in two ways (c) $x + \frac{y}{2}$ (d) $5(p + q)^{2}$

Solution

(a) $2(a - b) = 2 \times (a - b)$ : the whole of $a - b$ is doubled.

(b) The whole of $y + 3$ is divided by 5: $(y + 3) \div 5$ , or $\frac{1}{5} \times (y + 3)$ . The bracket is needed in both: without it, in $y + 3 \div 5$ only 3 would be divided by 5, and in $\frac{1}{5} \times y + 3$ only $y$ would be multiplied by $\frac{1}{5}$ .

(c) Here only $y$ is above the line, so only $y$ is divided by 2: $x + y \div 2$ . (By the order of operations the division is done before the addition.)

(d) The index 2 belongs to the whole bracket, and 5 multiplies the result: $5 \times (p + q) \times (p + q)$ .

(a) $2 \times (a - b)$ (b) $(y + 3) \div 5$ , or $\frac{1}{5} \times (y + 3)$ (c) $x + y \div 2$ (d) $5 \times (p + q) \times (p + q)$

Words for Operations

Algebra is often described in words. Each operation has its own words. Some of these you have met already; the new ones are quotient, square and cube.

In words	In algebraic notation
the sum of $a$ and $b$ ; $a$ plus $b$	$a + b$
$a$ minus $b$ ; subtract $b$ from $a$	$a - b$
the difference between $a$ and $b$	$a - b$ (in the order the words give)
the product of $a$ and $b$ ; $a$ times $b$	$ab$
$a$ divided by $b$ ; the quotient when $a$ is divided by $b$	$\frac{a}{b}$
twice $a$ ; three times $a$	$2a$ ; $3a$
the square of $a$ ; $a$ squared	$a^{2}$
the cube of $a$ ; $a$ cubed	$a^{3}$

The words also fix the order of working. "The square of the sum of $a$ and $b$ " adds first and then squares, so it is $(a + b)^{2}$ . "The sum of the squares of $a$ and $b$ " squares first and then adds, so it is $a^{2} + b^{2}$ . When a sum or a difference is multiplied, divided or squared as a whole, it needs brackets, or must stand whole above or below a fraction line.

Take care with "subtract ... from": "subtract $b$ from $a$ " starts with $a$ , so it is $a - b$ .

Example 10: From words to algebraic notation

Write an algebraic expression, in its shortest form, for each of these.
(a) Subtract $11z$ from the product of $3x$ and $9y$ .
(b) Divide the difference between $8b$ and the square of $3a$ by $c$ .
(c) Subtract the square of the difference between $b$ and $c$ from the square of the difference between $a$ and $b$ .
(d) Subtract the quotient when $2p$ is divided by $3q$ from the square of $r$ .
(e) Divide the cube of $r$ by the sum of $5s$ and $4t$ .

Solution

(a) The product of $3x$ and $9y$ is $3x \times 9y = 3 \times 9 \times x \times y = 27xy$ . Subtracting $11z$ from it gives $27xy - 11z$ .

(b) The square of $3a$ is $(3a)^{2} = 3a \times 3a = 9a^{2}$ . The difference between $8b$ and $9a^{2}$ is $8b - 9a^{2}$ . All of it is divided by $c$ , so all of it goes above the line: $\frac{8b - 9a^{2}}{c}$ .

(c) The square of the difference between $a$ and $b$ is $(a - b)^{2}$ ; the square of the difference between $b$ and $c$ is $(b - c)^{2}$ . Subtract the second from the first: $(a - b)^{2} - (b - c)^{2}$ .

(d) The quotient when $2p$ is divided by $3q$ is $\frac{2p}{3q}$ . The square of $r$ is $r^{2}$ . Subtract the quotient from it: $r^{2} - \frac{2p}{3q}$ .

(e) The cube of $r$ is $r^{3}$ , and the sum of $5s$ and $4t$ is $5s + 4t$ . The whole sum goes below the line: $\frac{r^{3}}{5s + 4t}$ .

(a) $27xy - 11z$ (b) $\frac{8b - 9a^{2}}{c}$ (c) $(a - b)^{2} - (b - c)^{2}$ (d) $r^{2} - \frac{2p}{3q}$ (e) $\frac{r^{3}}{5s + 4t}$

Example 11: From algebraic notation to words

Write each expression in words.
(a) $2(x + y)$ (b) $a^{2} - b$ (c) $\frac{p - q}{r}$ (d) $xy^{3}$

Solution

(a) $2 \times (x + y)$ : $x$ and $y$ are added first, then the sum is doubled. "Twice the sum of $x$ and $y$ ."

(b) $a$ is squared, then $b$ is taken away. "Subtract $b$ from the square of $a$ ."

(d) $x \times y \times y \times y$ : only $y$ is cubed. "The product of $x$ and the cube of $y$ ."

(a) Twice the sum of $x$ and $y$ . (b) Subtract $b$ from the square of $a$ . (c) Divide the difference between $p$ and $q$ by $r$ . (d) The product of $x$ and the cube of $y$ .

Common Mistakes

Writing the number after the letter. $p \times 3$ is $3p$ , never $p3$ ; and $b \times a$ is $ab$ .
Writing $1x$ or $-1x$ . Write $x$ and $-x$ .
Reading $3y$ as $3 + y$ . $3y = y + y + y$ ; $3 + y$ is 3 more than $y$ .
Reading $a^{2}$ as $2a$ . $a^{2} = a \times a$ , but $2a = a + a$ .
Squaring only part of a term. $3a^{2} = 3 \times a \times a$ , but $(3a)^{2} = 3a \times 3a = 9a^{2}$ . "The square of $3a$ " is $9a^{2}$ , not $3a^{2}$ .
Leaving out a needed bracket. $3 \times (x + y)$ is $3(x + y)$ , not $3x + y$ ; and $(3 + y) \div 5$ is $\frac{3 + y}{5}$ , not $3 + \frac{y}{5}$ .
Leaving a $\div$ sign in an answer. Write $x \div 3$ as $\frac{x}{3}$ .
Leaving out the $\times$ sign between two numbers: $3 \times 4$ is 12, not 34. The $\times$ sign is left out only next to a letter, or in front of a bracket that holds letters.

Practice

Try each question before you open its solution. The questions get harder as you go.

Question 1

Write each in algebraic notation.
(a) $8 \times q$ (b) $b \times a \times 5$ (c) $n \times (-1)$ (d) $y \times y \times y \times 2$ (e) $h \div 6$

ShowHide solution

(a) Leave out the $\times$ sign: $8q$ .

(b) Number first, letters in alphabetical order: $5 \times a \times b = 5ab$ .

(d) Number first: $2 \times y \times y \times y$ . Three $y$ s multiplied make $y^{3}$ , so it is $2y^{3}$ .

(e) A division is written as a fraction: $\frac{h}{6}$ .

(a) $8q$ (b) $5ab$ (c) $-n$ (d) $2y^{3}$ (e) $\frac{h}{6}$

Question 2

Write each with $\times$ or $\div$ signs.
(a) $9pq$ (b) $a^{2}b$ (c) $\frac{x}{8}$ , in two ways (d) $3(m + n)$ (e) $\frac{k + 2}{7}$ , in two ways

ShowHide solution

(a) $9 \times p \times q$ .

(b) The index belongs to $a$ only: $a \times a \times b$ .

(d) The whole of $m + n$ is multiplied by 3: $3 \times (m + n)$ .

(e) The whole of $k + 2$ is divided by 7, so it keeps its bracket: $(k + 2) \div 7$ , or $\frac{1}{7} \times (k + 2)$ .

(a) $9 \times p \times q$ (b) $a \times a \times b$ (c) $x \div 8$ , or $\frac{1}{8} \times x$ (d) $3 \times (m + n)$ (e) $(k + 2) \div 7$ , or $\frac{1}{7} \times (k + 2)$

Question 3

Say whether each statement is true or false, and give a reason.
(a) $5t = t + t + t + t + t$
(b) $pq = qp$
(c) $r \times r \times r = r + r + r$ for every value of $r$
(d) $3 + y$ and $3y$ are the same
(e) $a^{2}b = a \times b \times b$

ShowHide solution

(a) $5t$ is 5 lots of $t$ , which is $t$ added 5 times. True.

(b) Two numbers can be multiplied in either order, so $p \times q = q \times p$ . True (it is written $pq$ ).

(c) $r \times r \times r = r^{3}$ , but $r + r + r = 3r$ . These are different: when $r = 2$ , $r^{3} = 8$ but $3r = 6$ . False.

(d) $3 + y$ is 3 more than $y$ , but $3y$ is 3 times $y$ . False.

(e) The index belongs to $a$ , so $a^{2}b = a \times a \times b$ . $a \times b \times b$ is $ab^{2}$ . False.

(a) True (b) True (c) False (d) False (e) False

Question 4

Write each in its shortest form.
(a) $m \times 4 \times m$ (b) $5a \times 2b$ (c) $x \times y \times x \times y \times x$ (d) $3p \times p^{2}$ (e) $2k \times (-3k)$

ShowHide solution

(a) $4 \times m \times m = 4m^{2}$ .

(b) $5 \times a \times 2 \times b = 5 \times 2 \times a \times b = 10ab$ .

(d) $p^{2} = p \times p$ , so $3p \times p^{2} = 3 \times p \times p \times p = 3p^{3}$ .

(e) $2 \times k \times (-3) \times k = 2 \times (-3) \times k \times k = -6k^{2}$ .

(a) $4m^{2}$ (b) $10ab$ (c) $x^{3}y^{2}$ (d) $3p^{3}$ (e) $-6k^{2}$

Question 5

Write each pair in full with $\times$ signs, then in its shortest form, and say whether the two are the same.
(a) $5a^{2}$ and $(5a)^{2}$
(b) $2xy^{2}$ and $(2xy)^{2}$

ShowHide solution

(a) $5a^{2} = 5 \times a \times a$ , already in shortest form: $5a^{2}$ .

$(5a)^{2} = 5a \times 5a = 5 \times a \times 5 \times a = 5 \times 5 \times a \times a = 25a^{2}$ .

$5a^{2}$ has the number 5 in front and $25a^{2}$ has 25: they are not the same.

(b) $2xy^{2} = 2 \times x \times y \times y$ , already in shortest form: $2xy^{2}$ .

$(2xy)^{2} = 2xy \times 2xy = 2 \times x \times y \times 2 \times x \times y = 2 \times 2 \times x \times x \times y \times y = 4x^{2}y^{2}$ .

In $2xy^{2}$ only $y$ is squared; in $(2xy)^{2}$ the 2, the $x$ and the $y$ are all squared. They are not the same.

(a) $5a^{2}$ and $25a^{2}$ : not the same. (b) $2xy^{2}$ and $4x^{2}y^{2}$ : not the same.

Question 6

Write each in algebraic notation, as a fraction. None of the letters stands for 0.
(a) $15a \div b$ (b) $\frac{1}{8} \times n$ (c) $x \div 3 \times y$ (d) $6p \div q \times 2$

ShowHide solution

(a) The number divided goes on top: $15a \div b = \frac{15a}{b}$ .

(b) Multiplying by $\frac{1}{8}$ is dividing by 8: $\frac{1}{8} \times n = n \div 8 = \frac{n}{8}$ .

(c) From left to right. Dividing by 3 is multiplying by $\frac{1}{3}$ : $x \div 3 \times y = x \times \frac{1}{3} \times y = xy \times \frac{1}{3} = \frac{xy}{3}$ .

(d) From left to right: $6p \div q = \frac{6p}{q}$ , and then $\times 2$ : $6 \times p \times \frac{1}{q} \times 2 = 12p \times \frac{1}{q} = \frac{12p}{q}$ .

(a) $\frac{15a}{b}$ (b) $\frac{n}{8}$ (c) $\frac{xy}{3}$ (d) $\frac{12p}{q}$

Question 7

(a) Which of these are the same as $4(a + b)$ ?
(i) $4 \times (a + b)$ (ii) $4a + b$ (iii) $(a + b) \times 4$ (iv) $4 \times a + b$
(b) Which of these are the same as $\frac{m + 6}{2}$ ?
(i) $(m + 6) \div 2$ (ii) $m + 6 \div 2$ (iii) $\frac{1}{2} \times (m + 6)$ (iv) $\frac{m}{2} + 6$

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(a) $4(a + b)$ means the whole of $a + b$ multiplied by 4. (i) is exactly that. (iii) is the same, since multiplication can be done in either order. In (ii) and (iv) only $a$ is multiplied by 4 (in (iv), $\times$ is done before $+$ ), so they are not the same.

(b) $\frac{m + 6}{2}$ means the whole of $m + 6$ divided by 2. (i) is exactly that, and (iii) is the same, since dividing by 2 is multiplying by $\frac{1}{2}$ . In (ii) only 6 is divided by 2 ( $\div$ is done before $+$ ), and in (iv) only $m$ is divided by 2, so they are not the same.

(a) (i) and (iii) (b) (i) and (iii)

Question 8

Write an algebraic expression, in its shortest form, for each of these.
(a) Add $7c$ to the product of $4a$ and $3b$ .
(b) The square of the sum of $p$ and $q$ .
(c) Divide the sum of $x$ and 5 by $y$ .
(d) Subtract the cube of $n$ from twice $m$ .
(e) Multiply the quotient when $r$ is divided by $2s$ by $5t$ .

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(a) $4a \times 3b = 4 \times 3 \times a \times b = 12ab$ . Add $7c$ : $12ab + 7c$ .

(b) Add first, then square the whole sum: $(p + q)^{2}$ .

(d) Twice $m$ is $2m$ ; the cube of $n$ is $n^{3}$ . Start with $2m$ and subtract: $2m - n^{3}$ .

(e) The quotient when $r$ is divided by $2s$ is $\frac{r}{2s} = r \times \frac{1}{2s}$ . Multiply by $5t$ : $r \times \frac{1}{2s} \times 5 \times t = 5rt \times \frac{1}{2s} = \frac{5rt}{2s}$ .

(a) $12ab + 7c$ (b) $(p + q)^{2}$ (c) $\frac{x + 5}{y}$ (d) $2m - n^{3}$ (e) $\frac{5rt}{2s}$

Question 9

Write each expression in words.
(a) $3(x - y)$ (b) $a^{2} + b^{2}$ (c) $\frac{m + n}{4}$ (d) $pq - r^{3}$

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(a) $3 \times (x - y)$ : subtract first, then multiply by 3. "Three times the difference between $x$ and $y$ ."

(b) Square each letter, then add. "The sum of the squares of $a$ and $b$ ."

(d) $p \times q$ , then take away $r \times r \times r$ . "Subtract the cube of $r$ from the product of $p$ and $q$ ."

(a) Three times the difference between $x$ and $y$ . (b) The sum of the squares of $a$ and $b$ . (c) Divide the sum of $m$ and $n$ by 4. (d) Subtract the cube of $r$ from the product of $p$ and $q$ .

Question 10

The floor of a void deck is covered with square tiles, each with sides of $s$ cm.
(a) Write the area of one tile in algebraic notation.
(b) Siti says the area is $2s$ cm². Explain her mistake.
(c) A storage box is a cube with edges of $s$ cm. Write its volume in algebraic notation.

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(a) Area of a square = side × side $= s \times s = s^{2}$ . The area is $s^{2}$ cm².

(b) $2s = s + s$ , which adds two sides; that is half the perimeter of the tile, in cm. The area multiplies the two sides: $s \times s = s^{2}$ , not $s + s$ .

(a) $s^{2}$ cm² (b) She added the sides ( $s + s = 2s$ ) instead of multiplying them ( $s \times s = s^{2}$ ). (c) $s^{3}$ cm³

Question 11

At a hawker stall, a plate of chicken rice costs $x$ dollars and a cup of iced tea costs $y$ dollars. Ravi buys 3 sets, each set being one plate of chicken rice and one cup of iced tea.
(a) Write the total cost with a $\times$ sign, then in algebraic notation.
(b) Hui Min writes the total cost as $(3x + y)$ dollars. What does her expression stand for, and why is it wrong?

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(a) One set costs $(x + y)$ dollars. Three sets cost $3 \times (x + y)$ dollars, which in algebraic notation is $3(x + y)$ dollars.

(b) $3x + y$ is $3 \times x + y$ : only $x$ is multiplied by 3. It stands for the cost of 3 plates of chicken rice and only 1 cup of iced tea. Ravi bought 3 cups of iced tea too, so the whole of $x + y$ must be multiplied by 3, and the bracket is needed.

(a) $3 \times (x + y)$ dollars, written $3(x + y)$ dollars (b) $(3x + y)$ dollars is the cost of 3 plates of chicken rice and 1 cup of iced tea; the bracket is missing.

Question 12

For sports day, a team of 5 students buys a banner for $40 and paint for $y$ dollars, and shares the total cost equally.
(a) Write the amount each student pays in algebraic notation, and then in two ways with $\div$ or $\times$ .
(b) Mei Ling writes $40 + \frac{y}{5}$ dollars. What does her expression stand for?

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(a) The total cost is $(40 + y)$ dollars. Shared equally among 5, each pays $(40 + y) \div 5$ dollars.

In algebraic notation the whole of $40 + y$ goes above the fraction line: $\frac{40 + y}{5}$ dollars.

Dividing by 5 is multiplying by $\frac{1}{5}$ , so this is also $\frac{1}{5} \times (40 + y)$ dollars.

(b) In $40 + \frac{y}{5}$ only $y$ is divided by 5. It stands for the whole $40 of the banner plus one fifth of the cost of the paint, which is more than each student pays.

(a) $\frac{40 + y}{5}$ dollars $= (40 + y) \div 5 = \frac{1}{5} \times (40 + y)$ dollars (b) The full cost of the banner and one fifth of the paint: only $y$ has been divided by 5.

Key Points

$a \times b$ is written $ab$ : no $\times$ sign, the number first, the letters in alphabetical order ( $3p$ , not $p3$ ).
$1 \times n$ is written $n$ , and $-1 \times x$ is written $-x$ .
$3y = y + y + y = 3 \times y$ , which is not $3 + y$ .
$a^{2} = a \times a$ , $a^{3} = a \times a \times a$ and $a^{2}b = a \times a \times b$ ; $2a = a + a$ is not $a^{2}$ .
An index belongs only to what is just before it: $3a^{2} = 3 \times a \times a$ , but $(3a)^{2} = 9a^{2}$ .
To write a product in its shortest form, multiply the numbers, gather each letter with an index, and write the number first.
$\frac{a}{b} = a \div b = a \times \frac{1}{b}$ , where $b \neq 0$ . An answer is written with a fraction, not $\div$ .
$3(x + y) = 3 \times (x + y)$ , not $3x + y$ .
$\frac{3 + y}{5} = (3 + y) \div 5 = \frac{1}{5} \times (3 + y)$ , not $3 + \frac{y}{5}$ .
The words fix the order: "the square of the sum of $a$ and $b$ " is $(a + b)^{2}$ ; "subtract $b$ from $a$ " is $a - b$ .