s1–s4 mathematics

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A1 Introduction to algebra

S1–S4 Mathematics


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A1.1 Writing expressions

A1.2 Collecting like terms

A1.4 Dividing terms

A1.5 Factorizing expressions

A1.6 Substitution

A1.3 Multiplying terms and expanding brackets


+ 9 = 17

Using symbols for unknowns

Look at this problem:

The symbol stands for an unknown number.

We can work out the value of .

= 8

because 8 + 9 = 17


Using symbols for unknowns

Look at this problem:

– = 5

The symbols stand for unknown numbers.and

In this example, and can have many values.

For example, 12 – 7 = 5 3.2 – –1.8 = 5or

and are called variables because their value can vary.


Using letter symbols for unknowns

In algebra, we use letter symbols to stand for numbers.

These letters are called unknowns or variables.

Sometimes we can work out the value of the letters and sometimes we can’t.

For example,

We can write an unknown number with 3 added on to it as

n + 3

This is an example of an algebraic expression.


Writing an expression

Suppose Jon has a packet of biscuits and he doesn’t know how many biscuits it contains.

He can call the number of biscuits in the full packet a.

If he opens the packet and eats 4 biscuits, he can write an expression for the number of biscuits remaining in the packet as:

a – 4


Writing an equation

Jon counts the number of biscuits in the packet after he has eaten 4 of them. There are now 22.

He can write this as an equation:

a – 4 = 22

We can work out the value of a:

a = 26

That means that there were 26 biscuits in the full packet.


Writing expressions

When we write expressions in algebra we don’t usually use the multiplication symbol ×.

For example,

5 × n or n × 5 is written as 5n.

The number must be written before the letter.

When we multiply a letter symbol by 1, we don’t have to write the 1.

For example,

1 × n or n × 1 is written as n.


Writing expressions

When we write expressions in algebra we don’t usually use the division symbol ÷. Instead we use a dividing line as in fraction notation.

For example,

When we multiply a letter symbol by itself, we use index notation.

For example,

n ÷ 3 is written asn3

n × n is written as n2.

n squared


Writing expressions

Here are some examples of algebraic expressions:

n + 7 a number n plus 7

5 – n 5 minus a number n

2n 2 lots of the number n or 2 × n

6n 6 divided by a number n

4n + 5 4 lots of a number n plus 5

n3 a number n multiplied by itself and by itself again or n × n × n

3 × (n + 4) or 3(n + 4)

a number n plus 4 and then times 3.


Writing expressions

Miss Green is holding n number of cubes in her hand:

She takes 3 cubes away.

n – 3

She doubles the number of cubes she is holding.

2 × n or 2n

Write an expression for the number of cubes in her hand if:


Equivalent expression match


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A1.4 Dividing terms


A1.6 Substitution





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.

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.


Collecting together like terms

Remember, in algebra letters stand for numbers, so we can use the same rules as we use for arithmetic.

In algebra,

a + a + a + a = 4a

The a’s are like terms.

We collect together like terms to simplify the expression.

In arithmetic,

5 + 5 + 5 + 5 = 4 × 5



7 × b + 3 × b = 10 × b


In algebra,

7b + 3b = 10b

7b, 3b and 10b are like terms.

They all contain a number and the letter b.

In arithmetic,

(7 × 4) + (3 × 4) = 10 × 4

or




In algebra,

x + 6x – 3x = 4x

x, 6x, 3x and 4x are like terms.

They all contain a number and the letter x.

In arithmetic,

2 + (6 × 2) – (3 × 2) = 4 × 2



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

= 7a + 8b

This expression cannot be simplified any further.


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 + 3 + 6

= 2c + 2d + 9

4) 4n + n2 – 3n = 4n – 3n + n2 =

5) 4r + 6s – t Cannot be simplified


n + n2


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

3bPerimeter = 2a + 3b + 2a + 3b

= 4a + 6b

5x

4y

5x

xPerimeter = 4y + 5x + x + 5x

= 4y + 11x


Algebraic pyramids


Algebraic magic square


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A1.4 Dividing terms


A1.6 Substitution





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 = 5b We don’t write b5.

3 × d × c = 3cd

6 × e × e = 6e2

We write letters in alphabetical order.


Using index notation

Simplify:

a + a + a + a + a = 5a

Simplify:

a × a × a × a × a = a5

a to the power of 5

This is called index notation.

Similarly,

a × a = a2

a × a × a = a3

a × a × a × a = a4


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

3r × r2 = 3 × r × r × r = 3r3

2t × 2t = (2t)2 or 4t2

Using index notation


Grid method for multiplying numbers


Look at this algebraic expression:

4(a + b)

What do you think it means?

Remember, in algebra we do not write the multiplication sign ×.

So this expression means:

4 × (a + b)or:

(a + b) + (a + b) + (a + b) + (a + b)

= a + b + a + b + a + b + a + b

= 4a + 4b

Brackets


Using the grid method to expand brackets



Expanding expressions with brackets

3y(4 – 2y)

This means 3y × (4 – 2y), but we do not usually write × in algebra.

To expand or multiply out this expression we multiply every term inside the bracket by the term outside the bracket.

3y(4 – 2y) = 12y – 6y2



Expanding expressions with brackets

–a(2a2 – 2a + 3)

When there is a negative term outside the bracket, the signs of the multiplied terms change.

–a(2a2 – 3a + 1) = –2a3 + 3a2 – a

In general, –x(y + z) = –xy – xz

–x(y – z) = –xy + xz

–(y + z) = –y – z

–(y – z) = –y + z


Expanding brackets then simplifying

Sometimes we need to multiply out brackets and then simplify.

For example, 3x + 2(5 – x)

We need to multiply the bracket by 2 and collect together like terms.

3x + 10 – 2x

= 3x – 2x + 10

= x + 10


Expanding brackets and simplifying

Expand and simplify: 4 – (5n – 3)

We need to multiply the bracket by –1 and collect together like terms.

4 – 5n + 3

= 4 + 3 – 5n

= 7 – 5n

4 – (5n – 3) =



Expand and simplify: 2(3n – 4) + 3(3n + 5)

We need to multiply out both brackets and collect together like terms.

6n – 8 + 9n + 15

= 6n + 9n – 8 + 15

= 15n + 7

2(3n – 4) + 3(3n + 5) =


We need to multiply out both brackets and collect together like terms.

15a + 10b – 2a – 5ab

= 15a – 2a + 10b – 5ab

= 13a + 10b – 5ab


5(3a + 2b) – a(2 + 5b) =

Expand and simplify: 5(3a + 2b) – a(2 + 5b)


Algebraic multiplication square


Pelmanism: Equivalent expressions


Algebraic areas


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A1.4 Dividing terms


A1.6 Substitution






Dividing terms

Remember, in algebra we do not usually use the division sign, ÷.

Instead we write the number or term we are dividing by underneath like a fraction.

For example,

(a + b) ÷ c is written asa + b

c


As with fractions, we can often simplify expressions by cancelling.

For example,

n3 ÷ n2 =n3

n2

=n × n × n

n × n

1

1

1

1

= n

6p2 ÷ 3p =6p2

3p

=6 × p × p

3 × p

2

1

1

1

= 2p

Dividing terms


Hexagon puzzle


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A1.6 Substitution


A1.4 Dividing terms





Factorizing expressions

Factorizing an expression is the opposite of expanding it.

a(b + c) ab + ac

Expanding or multiplying out

FactorizingOften:When we expand an expression we remove the brackets.When we factorize an expression we write it with brackets.



Expressions can be factorized by dividing each term by a common factor and writing this outside a pair of brackets.

For example, in the expression

5x + 10

the terms 5x and 10 have a common factor, 5.

We can write the 5 outside of a set of brackets

5(x + 2)

We can write the 5 outside of a set of brackets and mentally divide 5x + 10 by 5.

(5x + 10) ÷ 5 = x + 2

This is written inside the bracket.

5(x + 2)



Writing 5x + 10 as 5(x + 2) is called factorizing the expression.

Factorize 6a + 8

6a + 8 = 2(3a + 4)

Factorize 12 – 9n

12 – 9n = 3(4 – 3n)

The highest common factor of 6a and 8 is 2.

(6a + 8) ÷ 2 = 3a + 4

The highest common factor of 12 and 9n is 3.

(12 – 9n) ÷ 3 = 4 – 3n



Writing 5x + 10 as 5(x + 2) is called factorizing the expression.

3x + x2 = x(3 + x)2p + 6p2 – 4p3

= 2p(1 + 3p – 2p2)

The highest common factor of 3x and x2 is x.

(3x + x2) ÷ x = 3 + x

The highest common factor of 2p, 6p2 and 4p3 is 2p.

(2p + 6p2 – 4p3) ÷ 2p

= 1 + 3p – 2p2

Factorize 3x + x2 Factorize 2p + 6p2 – 4p3


Factorization


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AA1.6 Substitution



A1.4 Dividing terms





Work it out!

4 + 3 × 8

= 28

5

= 19

43

= 133

0.6

= 5.8

–7

= –17


Work it out!

2

7 × 6

= 21

9

= 31.5

22

= 77

0.4

= 1.4

–3

= –10.5


Work it out!

2 + 63

= 15

9

= 87

12

= 150

0.2

= 6.04

–4

= 22


Work it out!

2( + 8)7

= 30

18

= 52

69

= 154

3.6

= 23.2

–13

= –10


Substitution

What does substitution

mean?

In algebra, when we replace letters in an expression or equation with numbers we call it substitution.


How can be written as an algebraic expression?4 + 3 ×

Using n for the variable we can write this as 4 + 3n.

We can evaluate the expression 4 + 3n by substituting different values for n.

When n = 5 4 + 3n = 4 + 3 × 5= 4 + 15= 19

When n = 11 4 + 3n = 4 + 3 × 11= 4 + 33= 37

Substitution


can be written as7n

2

We can evaluate the expression by substituting different values for n.

7n

2

When n = 47n

2= 7 × 4 ÷ 2

= 28 ÷ 2

= 14

When n = 1.17n

2= 7 × 1.1 ÷ 2

= 7.7 ÷ 2

= 3.85

7 ×

2

Substitution


can be written as n2 + 6

We can evaluate the expression n2 + 6 by substituting different values for n.

When n = 4 n2 + 6 = 42 + 6= 16 + 6= 22

When n = 0.6 n2 + 6 = 0.62 + 6= 0.36 + 6= 6.36

2 + 6

Substitution


can be written as 2(n + 8)

We can evaluate the expression 2(n + 8) by substituting different values for n.

When n = 6 2(n + 8) = 2 × (6 + 8)= 2 × 14= 28

When n = 13 2(n + 8) = 2 × (13 + 8)= 2 × 21 = 42

2( + 8)

Substitution


Here are five expressions.

1) a + b + c

2) 3a + 2c

3) a(b + c)

4) abc

5) a

b2 – c

Evaluate these expressions when a = 5, b = 2 and c = –1.

= 5 + 2 + –1 = 6

= 3 × 5 + 2 × –1 = 15 + –2 = 13

= 5 × (2 + –1) = 5 × 1 = 5

= 5 × 2 × –1= 10 × –1 = –10

= 5 ÷ 5 = 15

22 – –1 =

Substitution exercise


Noughts and crosses – substitution

s1–s4 mathematics

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