View
58
Download
9
Category
Preview:
DESCRIPTION
S1–S4 Mathematics. A1 Introduction to algebra. A1 Introduction to algebra. Contents. A. A1.2 Collecting like terms. A. A1.3 Multiplying terms and expanding brackets. A. A1.1 Writing expressions. A1.4 Dividing terms. A. A1.5 Factorizing expressions. A. A1.6 Substitution. A. - PowerPoint PPT Presentation
Citation preview
© Boardworks Ltd 2006 1 of 60
A1 Introduction to algebra
S1–S4 Mathematics
© Boardworks Ltd 2006 2 of 60
Contents
A1 Introduction to algebra
A
A
A
A
A
A
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
© Boardworks Ltd 2006 3 of 60
+ 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
© Boardworks Ltd 2006 4 of 60
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.
© Boardworks Ltd 2006 5 of 60
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.
© Boardworks Ltd 2006 6 of 60
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
© Boardworks Ltd 2006 7 of 60
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.
© Boardworks Ltd 2006 8 of 60
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.
© Boardworks Ltd 2006 9 of 60
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
© Boardworks Ltd 2006 10 of 60
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.
© Boardworks Ltd 2006 11 of 60
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:
© Boardworks Ltd 2006 12 of 60
Equivalent expression match
© Boardworks Ltd 2006 13 of 60
Contents
A
A
A
A
A
A
A1.2 Collecting like terms
A1.4 Dividing terms
A1.5 Factorizing expressions
A1.6 Substitution
A1 Introduction to algebra
A1.1 Writing expressions
A1.3 Multiplying terms and expanding brackets
© Boardworks Ltd 2006 14 of 60
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.
© Boardworks Ltd 2006 15 of 60
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
© Boardworks Ltd 2006 16 of 60
Collecting together like terms
7 × b + 3 × b = 10 × b
Remember, in algebra letters stand for numbers, so we can use the same rules as we use for arithmetic.
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
© Boardworks Ltd 2006 17 of 60
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,
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
© Boardworks Ltd 2006 18 of 60
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
= 7a + 8b
This expression cannot be simplified any further.
© Boardworks Ltd 2006 19 of 60
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
Collecting together like terms
n + n2
© Boardworks Ltd 2006 20 of 60
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
© Boardworks Ltd 2006 21 of 60
Algebraic pyramids
© Boardworks Ltd 2006 22 of 60
Algebraic magic square
© Boardworks Ltd 2006 23 of 60
Contents
A
A
A
A
A
A
A1.3 Multiplying terms and expanding brackets
A1.4 Dividing terms
A1.5 Factorizing expressions
A1.6 Substitution
A1 Introduction to algebra
A1.2 Collecting like terms
A1.1 Writing expressions
© Boardworks Ltd 2006 24 of 60
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.
© Boardworks Ltd 2006 25 of 60
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
© Boardworks Ltd 2006 26 of 60
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
© Boardworks Ltd 2006 27 of 60
Grid method for multiplying numbers
© Boardworks Ltd 2006 28 of 60
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
© Boardworks Ltd 2006 29 of 60
Using the grid method to expand brackets
© Boardworks Ltd 2006 30 of 60
Look at this algebraic expression:
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
© Boardworks Ltd 2006 31 of 60
Look at this algebraic expression:
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
© Boardworks Ltd 2006 32 of 60
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
© Boardworks Ltd 2006 33 of 60
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) =
© Boardworks Ltd 2006 34 of 60
Expanding brackets and simplifying
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) =
© Boardworks Ltd 2006 35 of 60
We need to multiply out both brackets and collect together like terms.
15a + 10b – 2a – 5ab
= 15a – 2a + 10b – 5ab
= 13a + 10b – 5ab
Expanding brackets and simplifying
5(3a + 2b) – a(2 + 5b) =
Expand and simplify: 5(3a + 2b) – a(2 + 5b)
© Boardworks Ltd 2006 36 of 60
Algebraic multiplication square
© Boardworks Ltd 2006 37 of 60
Pelmanism: Equivalent expressions
© Boardworks Ltd 2006 38 of 60
Algebraic areas
© Boardworks Ltd 2006 39 of 60
Contents
A
A
A
A
A
A
A1.4 Dividing terms
A1.5 Factorizing expressions
A1.6 Substitution
A1 Introduction to algebra
A1.2 Collecting like terms
A1.1 Writing expressions
A1.3 Multiplying terms and expanding brackets
© Boardworks Ltd 2006 40 of 60
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
© Boardworks Ltd 2006 41 of 60
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
© Boardworks Ltd 2006 42 of 60
Hexagon puzzle
© Boardworks Ltd 2006 43 of 60
Contents
A
A
A
A
A
A
A1.5 Factorizing expressions
A1.6 Substitution
A1 Introduction to algebra
A1.4 Dividing terms
A1.2 Collecting like terms
A1.1 Writing expressions
A1.3 Multiplying terms and expanding brackets
© Boardworks Ltd 2006 44 of 60
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.
© Boardworks Ltd 2006 45 of 60
Factorizing expressions
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)
© Boardworks Ltd 2006 46 of 60
Factorizing expressions
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
© Boardworks Ltd 2006 47 of 60
Factorizing expressions
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
© Boardworks Ltd 2006 48 of 60
Factorization
© Boardworks Ltd 2006 49 of 60
Contents
A
A
A
A
A
AA1.6 Substitution
A1 Introduction to algebra
A1.5 Factorizing expressions
A1.4 Dividing terms
A1.2 Collecting like terms
A1.1 Writing expressions
A1.3 Multiplying terms and expanding brackets
© Boardworks Ltd 2006 50 of 60
Work it out!
4 + 3 × 8
= 28
5
= 19
43
= 133
0.6
= 5.8
–7
= –17
© Boardworks Ltd 2006 51 of 60
Work it out!
2
7 × 6
= 21
9
= 31.5
22
= 77
0.4
= 1.4
–3
= –10.5
© Boardworks Ltd 2006 52 of 60
Work it out!
2 + 63
= 15
9
= 87
12
= 150
0.2
= 6.04
–4
= 22
© Boardworks Ltd 2006 53 of 60
Work it out!
2( + 8)7
= 30
18
= 52
69
= 154
3.6
= 23.2
–13
= –10
© Boardworks Ltd 2006 54 of 60
Substitution
What does substitution
mean?
In algebra, when we replace letters in an expression or equation with numbers we call it substitution.
© Boardworks Ltd 2006 55 of 60
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
© Boardworks Ltd 2006 56 of 60
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
© Boardworks Ltd 2006 57 of 60
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
© Boardworks Ltd 2006 58 of 60
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
© Boardworks Ltd 2006 59 of 60
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
© Boardworks Ltd 2006 60 of 60
Noughts and crosses – substitution
Recommended