1. Key Words and Introduction to Expressions

2. Writing expressions Here are some examples of algebraic expressions:
a number n plus 7
5 – n
5 minus a number n 2n
2 lots of the number n or 2 × n 6 n
6 divided by a number n
4n + 5
4 lots of a number n plus 5
Explain that algebra is very much like a language. It follows special rules, like the rules of grammar in a language. We have to keep to these rules so that any mathematician in the world can understand it.
Algebra is very important in mathematics because it describes the relationships between numbers.
Explain that there is a difference between an unknown and a variable. An unknown usually has a unique value which we can work out given enough information. A variable can have many different values.
We can use any letter in the alphabet to stand for unknowns or variables but some letters are used more than others. For example, we often use a, b, n, x or y. But we try not to use o (because it looks like 0).
Explain that in algebra we do not need to write the multiplication sign, ×, and so 2 lots of n is written as 2n. 3 lots of a, or 3 times a would be written as 3a. 5 lots of t, or 5 times t would be written as 5t.
Give some examples of possible values for 2n. If n is worth 5 then 2n is equal to 10 (not 25). If n is worth 20 then 2n is worth 40.
When we divide in algebra we write the number we are dividing by underneath, like a fraction. In this example, if n was worth 2, 6/n would be equal to 3.
Tell pupils that n³ is read as ‘n cubed’ or ‘n to the power of 3’. Give some examples, if n is worth 2 then n³ is 2 × 2 × 2, 8.
a number n multiplied by itself twice or n × n × n n3
3 × (n + 4) or 3(n + 4)
a number n plus 4 and then times 3.

3. Writing expressions
Miss Green is holding n number of cubes in her hand:
Write an expression for the number of cubes in her hand if:
She takes 3 cubes away.
n – 3
Discuss the examples on the slides and give other examples from around the classroom.
For example,
Suppose Joe has p number of pencils in his pencil case. If Harry, sitting next to him, gives him two more pencils, what expression could we write for the number of pencils in his pencil case? (p + 2)
We don’t know what p is (without counting) but we can still write an expression for the number of pencils in the case.
If p, the original number of pencils in the pencil case was 9, Joe would now have 11. If p was 15, Joe would now have 17.
Now, Joe gives Harry back his pencils, so he has p pencils again. Suppose he shares his pencils equally between himself and two of his friends. How many pencils will they have each? (p ÷ 3)
Tell pupils that if they are not sure whether or not an expression works, they should try using numbers in place of the letters to check. For example,
If Joe had 18 pencils and shared them between himself and his 2 friends they would get 6 each, 18 ÷ 3, so our expression works.
Suppose Mary has p pencils and Julia has q pencils. How many pencils do they have altogether? (p + q)
She doubles the number of cubes she is holding.
2 × n or 2n

4. Equivalent expression match

5. Simplifying Algebraic Expressions

6. Like terms
An algebraic expression is made up of terms and operators such as +, –, ×, ÷ and ( ).
A term is made up of numbers and letter symbols but not operators.
For example,
3a + 4b – a + 5 is an expression.
Explain that we can think of algebra as the language of mathematics.
3a, 4b, a and 5 are terms in the expression.
3a and a are called like terms because they both contain a number and the letter symbol a.

7. Collecting together like terms
Remember, in algebra letters stand for numbers, so we can use the same rules as we use for arithmetic.
In arithmetic,
= 4 × 5
In algebra,
a + a + a + a = 4a
In arithmetic, a number plus the same number plus the same number plus the same number = 4 × the number.
Emphasize that a can be any number.
In this example, we are using algebra to generalize the result.
The a’s are like terms.
We collect together like terms to simplify the expression.

8. Collecting together like terms
Remember, in algebra letters stand for numbers, so we can use the same rules as we use for arithmetic.
In arithmetic,
(7 × 4) + (3 × 4) = 10 × 4
In algebra,
7 × b + 3 × b = 10 × b or
In arithmetic, 7 × a number plus 3 × the same number = 10 × the number.
Pupils may want an ‘answer’ to 7b + 3b = 10b. Explain that what this is telling us is that any number multiplied by 7 plus the same number multiplied by 3 is always equal to the number multiplied by 10.
Emphasize that b can be any number.
In this example, we are using algebra to generalize a result rather than to give an answer to a specific problem.
7b + 3b = 10b
7b, 3b and 10b are like terms.
They all contain a number and the letter b.

9. Collecting together like terms
Remember, in algebra letters stand for numbers, so we can use the same rules as we use for arithmetic.
In arithmetic,
2 + (6 × 2) – (3 × 2) = 4 × 2
In algebra,
x + 6x – 3x = 4x
In arithmetic, a number plus 6 × the same number minus 3 × the same number = 4 × the number.
Discuss the algebraic equivalent of this.
Emphasize that x can be any number.
Again, we are using algebra to generalize a result rather than to give an answer to a specific problem.
x, 6x, 3x and 4x are like terms.
They all contain a number and the letter x.

10. Collecting together like terms
When we add or subtract like terms in an expression we say we are simplifying an expression by collecting together like terms.
An expression can contain different like terms.
For example,
3a + 2b + 4a + 6b
= 3a + 4a + 2b + 6b
Explain the meanings of each key word and phrase.
In the example 3a + 2b + 4a + 6b, explain that it is helpful to write like terms next to each other. (Remember that we can add terms in any order.)
Stress that we cannot simplify 7a + 8b any further. We can’t combine a’s and b’s. This says 7 times one number plus eight times another number.
= 7a + 8b
This expression cannot be simplified any further.

11. Collecting together like terms
Simplify these expressions by collecting together like terms.
1) a + a + a + a + a = 5a
2) 5b – 4b = b
3) 4c + 3d + 3 – 2c + 6 – d =
4c – 2c + 3d – d
= 2c + 2d + 9
Whenever possible make comparisons to arithmetic by substituting actual values for the letters. If it is true using numbers then it is true using letters. For example, in 1) we could say that is equivalent to 5 × 7.
For example 1) and example 2), stress again that in algebra we don’t need to write the number 1 before a letter to multiply it by 1. 1a is just written as a and 1b is just written as b.
For example 3), explain that when there are lots of terms we can write like terms next to each other so that they are easier to collect together. The numbers without any letters are added together separately.
In example 4) stress that n² is different from n. They cannot be collected together. 4n – 3n is n and n² stays as it is.
If we can’t collect together any like terms, as in example 5), we write ‘cannot be simplified’.
4) 4n + n2 – 3n =
4n – 3n + n2 =
n + n2
5) 4r + 6s – t
Cannot be simplified

12. Algebraic perimeters
Remember, to find the perimeter of a shape we add together the lengths of each of its sides.
Write algebraic expressions for the perimeters of the following shapes: 2a 3b
Perimeter =
2a + 3b + 2a + 3b
= 4a + 6b
For the first example, remind pupils that we need to add together the length of each side. The lengths of two of the sides have not been written on. Since this is a rectangle we can deduce that the length of the side opposite the side of length 2a is also 2a and the length of the side opposite the side of length 3b is also 3b.
Ask pupils,
How could the longer side be represented by 2a and the shorter side by 3b?
Deduce that the letter a must represent a bigger number than the letter b. 5x 4y x
Perimeter =
4y + 5x + x + 5x
= 4y + 11x

13. Using Index Notation

14. Using index notation Simplify: x + x + x + x + x = 5x
x to the power of 5
Simplify:
x × x × x × x × x
= x5
This is called index notation.
Similarly,
Start by asking pupils to simplify x + x + x + x + x. This is 5 lots of x, which we have seen is written as 5x.
Now, ask pupils how they might simplify x × x × x × x × x. Impress upon pupils the difference between this, and the previous expression, as they are often confused.
If x is equal to 2, for example, x + x + x + x + x equals 10, and x × x × x × x × x equals 32.
Some pupils may suggest writing xxxxx. This is not strictly incorrect, however, it should be discouraged in favour of using index notation.
When we write a number or term to the power of another number it is called index notation. The power, or index, is the raised number, in this case 5. The plural of index is indices. The number or letter that we are multiplying successive times, in this case, x, is called the base.
x² is read as ‘x squared’ or ‘x to the power of 2’.
x³ is read as ‘x cubed’ or ‘x to the power of 3’.
x4 is read as ‘x to the power of 4’.
x × x
= x2
x × x × x
= x3
x × x × x × x
= x4

15. Using index notation
We can use index notation to simplify expressions.
For example,
3p × 2p =
3 × p × 2 × p =
6p2
q2 × q3 =
q × q × q × q × q = q5
Discuss each example briefly.
In the last example 2t × 2t the use of brackets may need further clarification. We must put a bracket around the 2t since both the 2 and the t are squared. If we wrote 2t², then only the t would be squared.
Give a numerical example, if necessary. If t was 3 then 2t would be equal to 6. We would then have 6², 36. If we wrote 2t², that would mean 2 × 32 or 2 × 9 which is 18.
Remember the order of operations - BIDMAS. Brackets are worked out before indices, but indices are worked out before multiplication.
3r × r2 =
3 × r × r × r =
3r3
2t × 2t =
(2t)2 or
4t2

16. Expressions of the form (xm)n
When a term is raised to a power and the result raised to another power, the powers are multiplied.
For example,
(a5)3 =
a5 × a5 × a5
= a( )
= a15
= a(3 × 5)
In general,
(xm)n = xmn

17. Expressions of the form (xm)n
Rewrite the following without brackets.
1) (2a2)3 =
8a6
2) (m3n)4 =
m12n4
3) (t–4)2 =
t–8
4) (3g5)3 =
27g15
5) (ab–2)–2 =
a–2b4
6) (p2q–5)–1 =
p–2q5
You may wish to ask pupils to complete this exercise individually before talking through the answers.
The zero index is introduced in the last question and discussed on the next slide.
7) (h½)2 = h
8) (7a4b–3)0 = 1

18. Any number or term divided by itself is equal to 1.
The zero index
Any number or term divided by itself is equal to 1.
Look at the following division:
y4 ÷ y4 = 1
But using the rule that xm ÷ xn = x(m – n)
y4 ÷ y4 = y(4 – 4) = y0
That means that
y0 = 1
Stress that this rule is only true for non-zero values of x. 00 is undefined.
In general, for all x  0,
x0 = 1

19. Index laws xm × xn = x(m + n) (xm)n = xmn x0 = 1 (for x = 0) x1 = x
Here is a summary of the index laws.
xm × xn = x(m + n)
(xm)n = xmn
x0 = 1 (for x = 0)
x1 = x
x = x 1 2
Discuss the general form of each result where x is any number and m and n are integers.

20. Multiplying Algebraic Terms

21. Multiplying terms together
In algebra we usually leave out the multiplication sign ×.
Any numbers must be written at the front and all letters should be written in alphabetical order.
For example,
4 × a = 4a
1 × b = b
We don’t need to write a 1 in front of the letter.
b × 5 =
Ask pupils why they think we try not to use the multiplication symbol ×.
One reason is that it is easily confused with the letter, x.
Another reason is that when we use algebra we try to write things as simply as possible, only writing what is absolutely necessary. It’s simpler to write 2n than 2 x n.
It is also unnecessary to write a 1 in front of a letter to multiply it by 1. Multiplying by 1 has no effect so we can leave it out altogether. 5b
We don’t write b5.
3 × d × c =
3cd
We write letters in alphabetical order.
6 × e × e =
6e2

22. Brackets Look at this algebraic expression: 4(a + b)
What do you think it means?
Remember, in algebra we do not write the multiplication sign, ×.
This expression actually means:
4 × (a + b) or
Discuss the meaning of 4(a + b). Take suggestions from pupils and correct any misconceptions.
The number outside the brackets multiplies every term inside the brackets.
Remind pupils again that, in algebra, letters stand for numbers and that algebra follow the same rules as arithmetic.
Link this expression with the multiplication method we met at the beginning of the lesson.
a could be equal to 30, for example, and b could be equal to 2. This expression would then be equivalent to 4 × 32.
Remember, a and b could each be any number.
The expression 4(a + b) is equivalent to 4a + 4b. They are different ways of writing the same thing.
Explain that since algebra follows the same rules as arithmetic we can use the grid method to help us multiply out brackets in algebra.
(a + b) + (a + b) + (a + b) + (a + b)
= a + b + a + b + a + b + a + b
= 4a + 4b

23. Expanding Algebraic Expressions

24. Expanding brackets then simplifying
Sometimes we need to multiply out brackets and then simplify.
For example,
2(5 – x)
We need to multiply the bracket by 2. 10
– 2x
Show pupils that it is not necessary to construct a grid to multiply out a bracket. If required pupils can use lines, as shown here in orange, to make sure that every term inside the bracket is multiplied by the term outside the bracket.

25. Expanding brackets then simplifying
We need to multiply the bracket by –1 and collect together like terms. 4
– 5n
+ 3
In this example, we have 4 and then –(5n – 3).
This is equivalent to –1 × (5n – 3).
Repeat that a minus sign in front of a bracket has the effect of changing the sign of every term inside the bracket. 5n becomes – 5n and – 3 becomes + 3.
Let’s write the like terms next to each other.
When we collect the like terms together we have 7 – 5n.
= – 5n
= 7 – 5n

26. Expanding brackets then simplifying
2(3n – 4) + 3(3n + 5)
We need to multiply out both brackets and collect together like terms. 6n
– 8
+ 9n
+ 15
In this example, we have two sets of brackets. The first set is multiplied by 2 and the second set is multiplied by 3.
We don’t need to use a grid as long as we remember to multiply every term inside the bracket by every term outside it.
Talk through the multiplication of (3n – 4) by 2 and (3n + 5) by 3.
Let’s write the like terms next to each other.
When we collect the like terms together we have 6n + 9n which is 15n and – = 7.
= 6n + 9n –
= 15n + 7

27. Expanding brackets then simplifying
5(3a + 2b) – 2(2a + 5b)
We need to multiply out both brackets and collect together like terms.
15a
+ 10b
– 4a
–10b
Here is another example with two sets of brackets. The first set is multiplied by 5 and the second set is multiplied by minus 2.
Talk through the multiplication of (3a + 2b) by 5.
Next stress that we need to multiply (2a + 5b) by negative 2. Talk through this.
Let’s write the like terms next to each other.
When we collect together the like terms we have 15a – 4a which is 11a. The b terms cancel out.
= 15a – 4a + 10b – 10b
= 11a