1. Algebra Collecting Like Terms Removing Brackets
Removing Brackets & Simplifying
Solving Simple Equations
Solving Equations with Brackets
Evaluating Expressions & Formulae
Constructing Formulae
17-Sep-18

2. Algebra Simplifying Algebraic Expressions Learning Intention
Success Criteria
We are learning to simplify terms.
Understand the techniques for simplifying terms.
2. Be able too simplify algebraic expressions.
17-Sep-18

3. Algebra - m + a 1 + 4g (NOT 5g ) 7 k2 (NOT 7k6)
Simplifying Algebraic Expressions
Collecting LIKE terms
- m + a
1 + 4g
(NOT 5g )
7 k2
(NOT 7k6)
17-Sep-18

4. Algebra 6 d 4 h e2 (NOT 2e ) 8 a2 (NOT 8a)
Simplifying Algebraic Expressions
Multiplying Terms
6 d
4 h e2
(NOT 2e ) 8 a2
(NOT 8a)
17-Sep-18

5. Algebra e 4 3 Simplifying Algebraic Expressions Dividing Terms
17-Sep-18

6. Algebra y2 5y 3y 15 y2 + 8y + 15 y 5 y 3 www.mathsrevision.com
What is the total area of the downstairs house?
Algebra
Simplifying Algebraic Expressions
Below is a floor plan of the down stairs of a house. y 5 y
Living
Room y2 5y
Tidy up !
Bedroom 3y 15 3
Bathroom
Kitchen y2
+ 8y
+ 15
17-Sep-18

7. Algebra Now try TJ3a Ex 1 Ch7 (page 54) www.mathsrevision.com
Removing Brackets
Now try TJ3a
Ex 1
Ch7 (page 54)
17-Sep-18

8. Starter Questions 2a 6h
100o
60o
17-Sep-18

9. Algebra www.mathsrevision.com Removing brackets Learning Intention
Success Criteria
1. We are learning how to multiply out simply algebraic brackets.
Understand the key steps in removing brackets.
2. Apply multiplication rules for positive and negative numbers when removing brackets.
17-Sep-18

10. Removing a Single Bracket
Example
3(b + 5) = 3b
+ 15
Example
4(w - 2) = 4w
- 8
17-Sep-18

11. Removing a Single Bracket
Example
8(x + 3) = 8x
+ 24
Example
4(3 -2m) = 12
- 8m
17-Sep-18

12. Created by Mr. Lafferty@mathsrevision.com
Removing a
Single Bracket
Example : Write down an expression
for the amount indicate below
in two difference ways.
y-2
y-2
y-2
5(y - 2)
= 5y
- 10
y-2
y-2
17-Sep-18
17-Sep-18
Created by Mr.

13. Removing a Single Bracket
Example
a(b + 5) = ab
+ 5a
Example
p(w - 2) = pw
- 2p
17-Sep-18

14. Removing a Single Bracket
Example
y(y - 1) = y2
- y
Example
7w(w - 3) =
7w2
- 21w
17-Sep-18

15. Removing a Single Bracket
Example
a(a2 + 3b) = a3
+ 3ab
Example
y2(3y -2c) =
3y3
- 2cy2
17-Sep-18

16. Removing a Single Bracket
Example
-(x + y) = -x
- y
Example
-5(a - 7) =
-5a
+ 35
17-Sep-18

17. Algebra (i) A = L x B www.mathsrevision.com = 4(y + 5) (ii) = 4(y + 5)
Removing brackets
Question : Find the area of the rectangle
(i) With brackets
(ii) Without brackets.
y + 5
(i) A = L x B
= 4(y + 5) 4
(ii) = 4(y + 5)
= 4y
+ 20
17-Sep-18

18. Algebra Now try TJ3a Ex 2 Ch7 (page 56) www.mathsrevision.com
Removing Brackets
Now try TJ3a
Ex 2
Ch7 (page 56)
17-Sep-18

19. Starter Questions
(x-6)cm
3cm
17-Sep-18

20. Algebra www.mathsrevision.com Removing brackets Learning Intention
Success Criteria
1. We are learning how to multiply out simply algebraic brackets and collecting like terms.
Understand the key steps in removing brackets and simplifying.
2. Apply multiplication rules for positive and negative numbers when removing brackets.
17-Sep-18

21. Removing a Single Bracket
Tidy Up
Example
7 + 3(4 - y) =
7 + 12
- 3y
= y
Tidy Up
Example
9 - 3(8 - y) = 9
- 24
+ 3y
= y
17-Sep-18

22. Removing a Single Bracket
Example 12 10
Find my Area
Find my Area x
(x - 2)
A = L x B
A = L x B
A = 10(x – 2)
A = 12x
17-Sep-18

23. Removing Two Single Brackets
Tidy Up
Example
4(m - 3) - (m + 2) =
4m - 12
- m - 2
= 3m - 14
Tidy Up
Example
7(y - 1) - 2(y + 4) =
7y - 7
- 2y - 8
= 5y - 15
17-Sep-18

24. Removing a Single Bracket
Example : Find area of the orange border 12
Area =Big Rec – Small Rec 10 x
(x - 2)
= 12x – 10(x - 2)
= 12x
-10x
+ 20
= 2x + 20
17-Sep-18

25. Removing Two Single Brackets
Example : Find an expression for the difference in
the areas.
(y - 5)
A = 4(y – 5) 7 4
A = 7(y - 4)
(y - 4)
Diff = 3 x 14 – 8
= 34
Calculate difference if y = 14
Difference = 7(y - 4) - 4(y – 5)
= 7y - 28
- 4y + 20
= 3y - 8
17-Sep-18

26. Removing Two Single Brackets
Example : Find an expression for the difference in
the areas.
(x + 1)
A = 7(x + 1) 9 7
A = 9(x + 3)
x + 3
Diff = 2 x
=28
Calculate difference if x = 4
Difference = 9(x + 3) - 7(x + 1)
= 9x + 27
– 7x - 7
= 2x + 20
17-Sep-18

27. Algebra Ab= 6x As= 4(x - 3) As= 6x - 4(x - 3) www.mathsrevision.com
Removing Brackets & Simplifying
Question : Find the area of the large rectangle, then the small
rectangle and hence find the area of the red section. x
x-3
Ab= 6x 6 4
As= 4(x - 3)
= 4x
- 12
As= 6x - 4(x - 3)
= 6x
- 4x
+ 12
= 2x + 12
17-Sep-18

28. Algebra Now try TJ3a Ex 3 Ch7 (page 57) www.mathsrevision.com
Removing Brackets
Now try TJ3a
Ex 3
Ch7 (page 57)
17-Sep-18

29. Starter Questions
17-Sep-18

30. Equations www.mathsrevision.com Revision of Level 2 Learning Intention
Success Criteria
To revise Level 2
using the rule
‘Balancing Method’
1. Understand the process of the
‘Balancing Method’
2. Solving simple algebraic equations.
17-Sep-18

31. Balancing Method www.mathsrevision.com
Kirsty goes to the shops every week to buy some potatoes.
She always buys the same total weight.
One week she buys 2 large bags and 1 small bag.
The following week she buys 1 large bag and 3 small bags.
If a small bags weighs 4 kgs. How much does a large bag weigh? 4 4
What instrument measures balance
How can we go about solving this using balance ?
17-Sep-18

32. Balancing Method Take a small bag away from each side.4 4 4 4
Take a big bag away
from each side.
We can see that a big bag
is equal to
4 + 4 =
8 kg
17-Sep-18

33. Balancing Method Lets solve it using maths. Let P be the weigh
What symbol should we use for the scales ?
Balancing Method
Lets solve it using maths. 4 P P P
Let P be the weigh
of a big bag. 4 4 4
We know that a small bag = 4
Subtract 4 from each side
2P + 4 =
P + 12 -4 -4
Subtract P from each side 2P =
P + 8 -P -P P = 8
17-Sep-18

34. We know that a child price = £2
Balancing Method
Group of 5 adults and 3 children go to the local swimming.
Another group of 3 adults and 8 children also go swimming.
The total cost for each group is the same.
A child’s ticket costs £2. 2 a
If a child’s ticket costs £2. How much for an adult ticket ? a 2 a 2
Let a be the price
of an adult ticket. a 2
We know that a child price = £2
17-Sep-18

35. Balancing Method For balance we have 5a + 6 = 3a + 16 5a = 3a + 10
Subtract 3a from each side
Subtract 6 from each side
For balance we have 2 a
5a + 6 =
3a + 16 a 2 -6 -6 2 2 2 1 5a =
3a + 10 a a
-3a
-3a
Divide each side by 2 2a = 10 a = 5
Adult ticket price is £5
17-Sep-18

36. Balancing Method It would be far too time consuming to draw out
the balancing scales each time.
We will now review how to use the rules for solving equations.
17-Sep-18

37. Equations x – 5 = 11 x + 8 = 37 x = 16 x = 29 www.mathsrevision.com
Revision of Level 2
Examples
x – 5 = 11
x + 8 = 37
+ 5
+ 5
- 8
- 8
x = 16
x = 29
17-Sep-18

38. Equations 6x = 72 2x = 11 6 6 2 2 x = 12 x = 5.5 www.mathsrevision.com
Revision of Level 2
Examples
6x = 72
2x = 11 6 6 2 2
x = 12
x = 5.5
17-Sep-18

39. ½x = 5 ¾x = 9 Equations x = 10 3x = 36 3 3 x = 12
Revision of Level 2
Examples
½x = 5
¾x = 9
(x 2)
(x 4)
x = 10
3x = 36 3 3
x = 12
17-Sep-18

40. Equations 4x – 2 = 34 2x + 9 = 31 4x = 36 2x = 22 4 4 2 2 x = 9 x = 11
Revision of Level 2
Examples
4x – 2 = 34
2x + 9 = 31
+ 2
+ 2
- 9
- 9
4x = 36
2x = 22 4 4 2 2
x = 9
x = 11
17-Sep-18

41. Algebra Now try TJ3a Ex 4 Ch7 (page 58) www.mathsrevision.com
Removing Brackets
Now try TJ3a
Ex 4
Ch7 (page 58)
17-Sep-18

42. Starter Questions
17-Sep-18

43. Equations www.mathsrevision.com Equations with Brackets
Learning Intention
Success Criteria
1. Remember rule for Multiply out brackets.
1. We are learning how to solve equations with brackets and fractions .
2. Apply ‘Balancing Method’ to solve fractional equations.
17-Sep-18

44. Equations 3(x – 2) = 33 3x - 6 = 33 3x = 39 3 3 x = 13
Each group has
2 minutes come up with a method for solving this type of equation
Equations
Equations with Brackets
Multiply out the brackets first and then
Balancing Method
Example
3(x – 2) = 33
3x - 6 = 33
+ 6
+ 6
3x = 39 3 3
x = 13
17-Sep-18

45. Equations 5(3x + 2) – 2(4x - 3) = 2x + 36 15x + 10 – 8x + 6 = 2x + 36
Equations with Brackets
Example
5(3x + 2) – 2(4x - 3) = 2x + 36
15x
+ 10
– 8x
+ 6
= 2x + 36
Tidy up terms !
7x = 2x + 36
- 16
- 16
7x = 2x + 20
- 2x
- 2x
5x = 20 5 5
17-Sep-18
x = 4

46. Equations x + 3 = 7 x + 6 = 14 x = 8 www.mathsrevision.com
Each group has
2 minutes come up with a method for solving this type of equation
Equations
Equations with Fractions
Multiply EVERY term to get rid of fractional term.
and then apply
‘Balancing Method’
Example
x + 3 = 7 1 2
(x 2)
x + 6 = 14
- 6
- 6
x = 8
17-Sep-18

47. Multiply EVERY term to get rid of fractional term.
Equations
Equations with Fractions
Multiply EVERY term to get rid of fractional term.
and then apply
‘Balancing Method’
Example
x - = 4 3 4 1 2
(x 4)
3x - 2 = 16
+ 2
+ 2
3x = 18 3 3
x = 6
17-Sep-18

48. Algebra Now try TJ3a Ex 5 Ch7 (page 59) www.mathsrevision.com
Removing Brackets
Now try TJ3a
Ex 5
Ch7 (page 59)
17-Sep-18

49. Starter Questions
(x+2)cm
17-Sep-18
7cm

50. Algebra www.mathsrevision.com
Evaluating Expressions – number for letter
Learning Intention
Success Criteria
1. Be able to substitute numbers for letters in an expression.
1. We are learning how to evaluate an expression.
2. Use previous knowledge of Number to evaluate a expression.
17-Sep-18

51. Algebra a = 3 ; b = 4 and c = -1 2a + c b2 – a2 2b2 = 2x3 + (-1)
Evaluating Expressions – number for letter
BODMAS
Given the following information find the values of :-
a = 3 ; b = 4 and c = -1
2a + c
b2 – a2
2b2
= 2x3 + (-1)
= 42 – 32
= 2x(4)2
= 2x3 - 1
= 16 – 9
= 2x16
= 6 - 1
= 7
= 32
= 5
17-Sep-18

52. Algebra 2a2 + 3b + 30c = 2x(3)2 + 3x4 + 30x(-1) = 18 + 12 - 30 = 0
Evaluating Expressions – number for letter
Given the following information find the values of :-
a = 3 ; b = 4 and c = -1
2a2 + 3b + 30c
= 2x(3)2 + 3x4 + 30x(-1) =
= 0
17-Sep-18

53. Algebra ¾(d + f2) d = 3 ; e = 4 and f = -1 2(d - f) √(16e)
Evaluating Expressions – number for letter
BODMAS
Given the following information find the values of :-
d = 3 ; e = 4 and f = -1
2(d - f)
¾(d + f2)
√(16e)
= 2(3 -(-1))
= ¾(3 + (-12))
= √(16x4)
= 2 x 4
= ¾(4)
= √64
= 8
= 3
= 8
17-Sep-18

54. Algebra Now try TJ3a Ex 6 Ch7 (page 60) www.mathsrevision.com
Removing Brackets
Now try TJ3a
Ex 6
Ch7 (page 60)
17-Sep-18

55. Starter Questions
17-Sep-18

56. Formulae www.mathsrevision.com Learning Intention Success Criteria
To explain how to work with formulae expressed in words.
Understand formulae expressed in words.
2. Solve problems given formulae in words.
17-Sep-18

57. Sometimes formulae are expressed in words.
A tile fitter charges £20 per hour.
How much will he charge for 8 hours of work.
C =
20xh
For h = 8
C = 20 x 8 =
£160
17-Sep-18

58. Formulae C = 30 + 3d For d = 7 C = 30 + 3 x 7 = £51
The cost for hiring a carpet cleaner is
£30 plus £3 a day
How much will it cost to hire for a week.
C = 30 + 3d
For d = 7
C = x 7 =
£51
17-Sep-18

59. Formulae p = s - b £2.01 Profit = 12 – 9.99 = Profit :
Subtract how much you paid for goods in the first place from the amount you actually sold them for.
How much profit did Daniel make if he bought a CD for £9.99 and sold it for £12
p = s - b
Profit = 12 – 9.99 =
£2.01
17-Sep-18

60. Have you updated your Learning Log ?
Algebra
Removing Brackets
Now try TJ3a
Ex 7
Ch7 (page 62)
Are you on Target ?
I can ?
17-Sep-18