Transcript 2 - Bedfordmathsacademy

Algebraic pyramids
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Algebraic magic square
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Contents
A1 Algebraic expressions
A1.1 Writing expressions
A1.2 Collecting like terms
A1.3 Multiplying terms
A1.4 Dividing terms
A1.5 Factorising expressions
A1.6 Substitution
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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
We write letters in alphabetical order.
6 × e × e = 6e2
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Using index notation
Simplify:
x + x + x + x + x = 5x
Simplify:
x × x × x × x × x = x5
x to the power of 5
This is called index notation.
Similarly,
x × x = x2
x × x × x = x3
x × x × x × x = x4
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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
3r × r2 = 3 × r × r × r = 3r3
2t × 2t = (2t)2
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or
4t2
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Grid method for multiplying numbers
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Brackets
Look at this algebraic expression:
4(a + b)
What do do think it means?
Remember, in algebra we do not write the multiplication sign, ×.
This expression actually means:
4 × (a + b)
or
(a + b) + (a + b) + (a + b) + (a + b)
=a+b+a+b+a+b+a+b
= 4a + 4b
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Using the grid method to expand brackets
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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
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Expanding brackets then simplifying
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
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Expanding brackets then simplifying
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
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Expanding brackets then simplifying
Simplify
5(3a + 2b) – 2(2a + 5b)
We need to multiply out both brackets and collect together
like terms.
15a + 10b – 4a –10b
= 15a – 4a + 10b – 10b
= 11a
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Algebraic multiplication square
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Pelmanism: Equivalent expressions
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Algebraic areas
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Contents
A1 Algebraic expressions
A1.1 Writing expressions
A1.2 Collecting like terms
A1.3 Multiplying terms
A1.4 Dividing terms
A1.5 Factorising expressions
A1.6 Substitution
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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
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is written as
a+b
c
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Dividing terms
Like a fraction, we can often simplify expressions by
cancelling.
For example,
3
n
n3 ÷ n2 = 2
n
2
6p
6p2 ÷ 3p =
3p
1
1
1
n×n×n
=
n×n
6×p×p
=
3×p
=n
= 2p
1
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2
1
1
1
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Algebraic areas
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Hexagon Puzzle
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Contents
A1 Algebraic expressions
A1.1 Writing expressions
A1.2 Collecting like terms
A1.3 Multiplying terms
A1.4 Dividing terms
A1.5 Factorizing expressions
A1.6 Substitution
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Factorizing expressions
Some expressions can be simplified by dividing each term by
a common factor and writing the expression using 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 and mentally
divide 5x + 10 by 5.
(5x + 10) ÷ 5 = x + 2
This is written inside the bracket.
5(x + 2)
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Factorizing expressions
Writing 5x + 10 as 5(x + 2) is called factorizing the
expression.
Factorize 6a + 8
Factorize 12 – 9n
The highest common
factor of 6a and 8 is 2.
The highest common
factor of 12 and 9n is 3.
(6a + 8) ÷ 2 = 3a + 4
(12 – 9n) ÷ 3 = 4 – 3n
6a + 8 = 2(3a + 4)
12 – 9n = 3(4 – 3n)
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Factorizing expressions
Writing 5x + 10 as 5(x + 2) is called factorizing the
expression.
Factorize 3x + x2
The highest common
factor of 3x and x2 is x.
(3x +
x 2)
÷x=3+x
Factorize 2p + 6p2 – 4p3
The highest common factor
of 2p, 6p2 and 4p3 is 2p.
(2p + 6p2 – 4p3) ÷ 2p
= 1 + 3p – 2p2
3x + x2 = x(3 + x)
2p + 6p2 – 4p3
= 2p(1 + 3p – 2p2)
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Algebraic multiplication square
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Pelmanism: Equivalent expressions
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Contents
A1 Algebraic expressions
A1.1 Writing expressions
A1.2 Collecting like terms
A1.3 Multiplying terms
A1.4 Dividing terms
A1.5 Factorising expressions
A1.6 Substitution
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Work it out!
4 + 3 × 0.6
43
–7
8
5
===–17
133
5.8
28
19
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Work it out!
7 × 0.4
22
–3
6
9
2
====–10.5
31.5
1.4
21
77
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Work it out!
0.2
12
–4
3
9
2
+6
===6.04
150
22
15
87
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Work it out!
2( –13
3.6
18
69
7 + 8)
===23.2
–10
154
30
52
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Substitution
What does
substitution
mean?
In algebra, when we replace letters in an expression or
equation with numbers we call it substitution.
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Substitution
How can 4 + 3 ×
be written as an algebraic expression?
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
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Substitution
7×
can be written as
7n
2
2
7n
We can evaluate the expression
by substituting different
2
values for n.
When n = 4
7n
2
= 7×4÷2
= 28 ÷ 2
= 14
When n = 1.1
7n
2
= 7 × 1.1 ÷ 2
= 7.7 ÷ 2
= 3.85
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Substitution
2
+6
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 + 7 = 0.62 + 6
= 0.36 + 6
= 6.36
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Substitution
2(
+ 8)
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
= 41
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Substitution exercise
Here are five expressions.
1) a + b + c = 5 + 2 + –1 = 6
2) 3a + 2c = 3 × 5 + 2 × –1 = 15 + –2 = 13
3) a(b + c) = 5 × (2 + –1) = 5 × 1 = 5
4) abc = 5 × 2 × –1= 10 × –1 = –10
22 – –1
b2 – c
5)
=
=5÷5=1
a
5
Evaluate these expressions when a = 5, b = 2 and c = –1
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Noughts and crosses - substitution
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