Getting Used to Algebra
Algebra is where you use letters to
represent numbers
Aims:
• To get to grips with the algebra idea
• To solve simple algebraic problems
Sunday, 11 January 2015
Page 1
Level
3
4
5
6
Algebra
I understand
the role of
the = sign
• I can solve positive
equations with
unknowns on one
side.
•I can simplify
expressions with
positive terms.
•I can find equivalent
expressions.
•I can substitute
positive values into
simple expressions.
• I can solve linear equations
with unknowns on one side
including negative numbers.
•I can identify equivalent
expressions.
•I can explain the difference
between expressions.
•I can substitute positive and
negative values into
expressions.
•I understand the meaning of
brackets in expressions.
• I can solve linear
equations with
unknowns on both
sides including
with negative
numbers
• I can construct
and solve simple
linear equations
I can work
out the
missing
number in a
box.
7/8
I can solve
simultaneous
equations
I am starting the lesson on level _____________________
By the end of this lesson I want to be able to _____________________
Rules of algebra
a + b means the value of a added to the value of b
c – d means take away the value of d from the
value of c
3x
means 3 times the value of x
3x = 3 × x
pq
means p × q
e/
means e ÷ m
m
If a = 10 and b = 20 what is the
value of
a+b ?
10
30
200
If c = 100 and d = 70 what is the
value of
c–d ?
30
20
70
If x = 5 what is the value of
3x ?
15
35
8
If p = 2 and q = 3 what is the
value of
pq ?
23
5
6
If e = 20 and m = 5 what is the
value of
e/
m
15
?
100
4
Example
Alex has some sweets, we do not know
how many sweets Alex has….
so we can say ‘Alex has x sweets’
If Alex is given 5 more sweets, how many
sweets has he got?
x + 5
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Example
Bob has some toys, we do not know how
many toys he has….
so we can say ‘Bob has m toys’
If Bob buys 6 more toys, how many toys
has he now got?
m + 6
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Another Example
Bill catches y fish.
Ben takes 3 away from him.
How many fish does Bill now have?
y - 3
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A Third Example
Fred has x DVDs
Frank has y DVDs
How many DVDs do they have altogether?
x + y
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Questions
Use algebra to write:
1) 2 less than w w – 2
2) 3 more than d d + 3
3) 5 together with c c + 5
4) f more than g f + g
5) p less than q q - p
6) m less than 7 7 - m
Sunday, 11 January 2015
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Adding and Subtracting with
Letters
a + a + a = 3a
b + b + b + b + b = 5b
c + c – c + c = 2c
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Questions
1)
2)
3)
4)
5)
6)
7)
8)
a
c
p
v
b
n
h
g
+
+
+
+
+
a = 2a
c + c + c = 4c
p + p =p
v + v + v - v + v = 4v
b + b – b = 2b
n + n + n + n - n + n = 3n
h =0
g + g - g 2g
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Adding Expressions & Terms
2a + 4a = 6a
6a – 5a = a
7a – 4a + 10a = 13a
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Questions
1)
2)
3)
4)
5)
6)
7)
8)
9)
5c + 7c 12c
9d – 4d 5d
2s + 12s 14s
13a – 5a 8a
4e – e + 3e 6e
6e – 2e + 5e 9e
12e – 10e 2e
3h – 2h + h 2h
4r – r + 5r 8r
Sunday, 11 January 2015
10)
11)
12)
13)
14)
15)
16)
17)
18)
3w + 9w – 5w 7w
2g + 5g – 3g 4g
7f – 4f + 9f 12f
5b + 50b 55b
75j – 43j 32j
34p + 12p – 5p 41p
3m – m + m + 5m 8m
d – d + d - d0
4f + 10f - 13f f
Page 17
Going Negative
3a – 5a = -2a
5p + 4p – 12p = -3p
-5a + 2a – 7a = -10a
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Working with Algebra
Aims:
 To be able to collect like terms in
order to simplify algebraic expressions
 To be able to multiply terms together
and expand the brackets from an
expression
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Example 1
3a + 4b + 2a + 6b =
firstly collect like-terms…
3a + 2a + 4b + 6b =
5a + 10b
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Example 2
-3a + 2b – 4a + 6b
collect like-terms
-3a – 4a + 2b + 6b
-7a + 8b
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Example 3
-4a + 9b + 3a - 12b
collect like-terms
-4a + 3a + 9b – 12b
- a – 3b
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Aims
 To be able to simplify expressions
(including expressions with indices)
 To be able to expand brackets and
simplify
 To be able to understand the 3 laws of
indices
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Simplify each expression…
1) 5a – 4y – 11a + 2y
6) 3d – 5e – 4d + 2e
2) -4f + 6g – 7f – 3g
7) p + 8r – 7p + 2r + 3p
3) 4m + 9n – 6m - 14n
8) 9x + 3x – 8 - 2x + 3
4) -4t + 7y – 10t + 5y
9) a – 6b + 2a + 3b - 3a
5) 1 + 2r – 7 – 7r
10) 3g + 2h – 3h + 2g
-6a - 2y
-11f + 3g
-2m – 5n
-14t + 12y
-6 – 5r
Sunday, 11 January 2015
-d – 3e
-3p + 10r
10x – 5
-3b
5g – h
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Harder Simplification…
(remember, a, a2 and a3 are completely different terms)
1) 4a – 5a2 + 2a + 3a2 = 6a – 2a2
2) 5h2 + 2h3 – 10h2 – 3h3 = -5h2 –h3
3) 3x2 – 4x – 5x2 + x = – 3x-2x2
4) -4t2 + 7t – 10t2 + 5t3 = 7t -14t2 + 5t3
5) r2 + 2r – 7r2 – 7r2 – 2r = -13r2
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Multiplying Terms Together
When two terms in algebra are being
multiplied together, they are simply
written next to each other.
e.g.
and
3a means 3 x a
efg means e x f x g
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Multiplying Terms Together
Simplify:
2a x 4b = 8ab
10c x 2de = 20cde
3w x 4y x 5z = 60wyz
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Questions – simple multiplication
1) 7ab x 8pq 56abpq
2) 3f x 2h x 5a 30afh
3) 3abc x 4def 12abcdef
4) 5g x 25 125g
5) 5mn x 2pq x 4 40mnpq
6) 5f x g x h x 2j 10fghj
7) 3w x 4d x 2h x yz 24dhwyz
8) 7pqr x 4abc x 2h 56abchpqr
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Questions – reverse multiplication
10c = 70abc
1) 7ab x ___
2) 3f x ___
4g x 5h = 60fgh
d = 3abcd
3) 3abc x ___
4) 5g x ___
4 = 20g
p x 4 = 20mnp
5) 5mn x ___
6) 5f x ___
4h x 2g = 40fgh
7) 3w x ___
4 x 2v x yz = 24vwyz
8) 4bc x 10e
___ x 2ad = 80abcde
Sunday, 11 January 2015
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Dividing Algebraic Terms
Simplify:
8a ÷ 4 = 2a
10ab ÷ 2a =5b
20pq
pq
= 20
Sunday, 11 January 2015
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Powers
Aims:
To remember how to work out the HCF &
LCM from any given pair of numbers
To be able to understand how indices
(powers) work in algebra
To be able to manipulate the powers in an
expression in order to simplify it
Sunday, 11 January 2015
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Powers Rules
a x a = a2
a x a x a x a = a4
Rule 1:
When you multiply
powers of the same
letter or number you
add the indices…
But, what is a2 x a4?
It’s (a x a) x (a x a x a x a) =a6
What did you do with the powers?
You added them!
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Indices Questions
a3 x a6 x a2 = a11
c3 x c5 x c1 = c9
2y2 x 4y5 = 8y7
3m3 x 4m5 = 12m8
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Harder Indices Questions
Q1.
2a3b6 x a6b2 = 2a9b8
Q2. 3c5d4 x 5c2d4 = 15c7d8
Q3. 2ab4 x 4a2b3 = 8a3b7
Q4. 3mnp3 x 4mn2p5 = 12m2n3p8
Sunday, 11 January 2015
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Powers
a x a x a x a x a = a5
a x a x a = a3
Rule 2:
When you divide
powers of the same
letter or number you
subtract the indices…
But, what is a5  a3?
It’s (a x a x a x a x a) (a x a x a) =a2
What did you do with the powers?
You subtracted them!
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Indices Questions
a6  a2
= a4
c3  c5
= c-2
3y2 x 4y5 = 6y4
2y3
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Harder Indices Questions
Q1.
4a6 x 8a2
2a9
= 16a-1
2 x 4y5 = 20y3
30y
Q2.
6 x y4
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Powers
a x a x a = a3
Rule 3:
When you raise a
power by another
power (separated by
brackets) you multiply
the indices…
What do you think is…
(a3)2
= a 3 x a3 = a6
what did you do
with the powers
here?
multiplied them!
Sunday, 11 January 2015
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Indices Questions
Q1. (a2)4 x (a3)2
=
a14
Q2.
4(a2)3 = 4a6
Q3.
(4a2)3 = 64a6
Q4.
(5ab3)2 = 25a2b6
Sunday, 11 January 2015
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Have you really understood
indices??
We’ll see…
Simplify this…
( 2 a )  4 ( a )  (( a ) )
4
5
a a
7
Sunday, 11 January 2015
3
19
6
5
 (2a )
9
6
2
3
Page 40
Expanding the Brackets
Expand these expressions:
3(a + b) = 3a + 3b
4(y + 6) = 4y + 24
2(2a + 3b – 4c) = 4a + 6b – 8c
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Expand the brackets…
1) 4(2p – 12) = 8p - 48
2) 5(2a + 4b) = 10a + 20b
3) 3(4c – 4b) = 12c – 12b
4) 7(3e + 2f – 4g) = 21e + 14f – 28g
5) 9(6w + 2y – 7z) = 54w + 18y – 63z
6) 10(3b – 9m) = 30b – 90m
7) 6(-6j – 7m) =-36j – 42m
8) 5(4a – 3b) =20a – 15b
Sunday, 11 January 2015
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Factorising
Aims:
• To remember how to expand brackets,
including expressions with indices
• To learn what the process factorising is
and be able to apply it to any expression
Sunday, 11 January 2015
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Factorising
Reminder of how to expand brackets:
4(2m – 5) = 8m - 20
What is factorising then?
Q1. 8m – 20 = 4 ( 2m - 5 )
Sunday, 11 January 2015
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Factorising
Q2.
25a – 30b = 5 ( 2m - 5 )
Q3.
40a + 6a2 = 2a ( 20 + 3a)
Page 45
Sunday, 11 January 2015
Adding Bracketed Expressions
Aims:
 To be able to multiply out the brackets
from expressions…
 … and then collect like-terms and
simplify
Sunday, 11 January 2015
Page 46
Adding Bracketed Expressions
4(a + 5b) + 3(7a – b) =
4a + 20b + 21a – 3b =
25a + 17b
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Adding Bracketed Expressions
2(7a – 6b) + 6(3a + b) =
14a – 12b + 18a + 6b =
32a – 6b
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Questions
1) 7(2a – 4b) + 2(2a + b) 18a – 26b
2) 2(3y – w) + 3(2y + 5w) 12y + 13w
3) 4(p + q) + 5(2p – 3q) 14p – 11q
4) 6(2n + m) + 2(n – m) 14n + 4m
5) 2(4s + r) + 7(2s – 3r) 22s – 19r
Sunday, 11 January 2015
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Subtracting Bracketed
Expressions
3(2a – 4b) – 2(a + 2b) =
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Subtracting Bracketed
Expressions
3(2a – 4b) – 2(a + 2b) =
6a – 12b – 2a – 4b =
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Subtracting Bracketed
Expressions
3(2a – 4b) – 2(a + 2b) =
6a – 12b – 2a – 4b =
4a – 16b
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Subtracting Bracketed
Expressions
3(2p + 3q) – 4(p – 2q) =
Sunday, 11 January 2015
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Subtracting Bracketed
Expressions
3(2p + 3q) – 4(p – 2q) =
6p + 9q - 4p + 8q =
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Subtracting Bracketed
Expressions
3(2p + 3q) – 4(p – 2q) =
6p + 9q - 4p + 8q =
2p + 17q
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Invisible 1
7(m – 2n) – (m – 3n) =
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Invisible 1
7(m – 2n) – (m – 3n) =
7m – 14n – m + 3n =
Sunday, 11 January 2015
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Invisible 1
7(m – 2n) – (m – 3n) =
7m – 14n – m + 3n =
6m - 11n
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Questions
1) 3(4n + 5m) – 5(2n – 4m)
2) 5(n – 2m) – (m + 2n)
3) 3(2n + 6m) – 6(n – 2m)
4) 2(3n – m) – 7(n – 5m)
5) 8(5n – m) – 2(2n + m)
Sunday, 11 January 2015
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Questions
1) 3(4n + 5m) – 5(2n – 4m)
2) 5(n – 2m) – (m + 2n)
3) 3(2n + 6m) – 6(n – 2m)
4) 2(3n – m) – 7(n – 5m)
5) 8(5n – m) – 2(2n + m)
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