Algebra
Tidying Up Terms
Multiplying Terms
Algebra
Algebra
Solving Simple Equations ( x+1=5 )
Removing Brackets & Simplifying
Harder Equations ( 2x+1=9 )
Solving Equations ( brackets)
Solving Equations (terms either side)
Inequalities
Starter Questions
Starter Questions
6cm
2cm
2cm
3cm
Learning Intention
Learning Intention
Success Criteria
Success Criteria
1.
Understand the term ‘like
Understand the term ‘like
terms’.
terms’.
1. To explain how to gather
‘algebraic like terms’.
Algebra
Algebra
2.
2.
Gather like terms for simple
Gather like terms for simple
expressions.
expressions.
Multiplying Terms
First Row :
+
+
=
2c
c
4c
+
+
= 7c
2nd Row :
+
+
=
2d
2d
3d
+
+
= 7d
Tidying Terms
3rd Row :
+
+
=
3f
f
2f
+
+
= 6f
In Total we have
7c + 7d + 6f
CANNOT
TIDY UP
ANYMORE
WE CAN ONLY TIDY UP “LIKE TERMS”
Tidying Terms
Tidying Terms
WE CAN ONLY TIDY UP “LIKE TERMS”
Tidy up the following:
Q1.
2x + 4x + 5y -3y + 18 =
Q2.
4a + 3b + 5a + 6 – b =
6x + 2y + 18
9a + 2b + 6
Learning Intention
Learning Intention
Success Criteria
Success Criteria
1.
Understand the key steps of
Understand the key steps of
multiplying terms.
multiplying terms.
1. To explain how to multiply
out algebraic terms.
Algebra
Algebra
2.
2.
Apply multiplication rules
Apply multiplication rules
for simple expressions.
for simple expressions.
Multiplying Terms
Algebra
Algebra
Simplifying Algebraic Expressions
Reminder !
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Algebra
Algebra
Simplifying Algebraic Expressions
Reminder !
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Learning Intention
Learning Intention
Success Criteria
Success Criteria
1.
Understand the key steps in
Understand the key steps in
removing brackets.
removing brackets.
1. To explain how to multiply
out simply algebraic
brackets.
Algebra
Algebra
2.
2.
Apply multiplication rules
Apply multiplication rules
for integers numbers when
for integers numbers when
removing brackets.
removing brackets.
Removing brackets
3(b + 5) =
3b
+ 15
Example 1
4(w - 2) =
4w
- 8
Example 2
Removing a
Removing a
Single Bracket
Single Bracket
2(y - 1) =
2y
- 2
Example 3
7(w - 6) =
7w
- 42
Example 4
Removing a
Removing a
Single Bracket
Single Bracket
8(x + 3) =
8x
+ 24
Example 5
4(3 -2m) =
12
- 8m
Example 6
Removing a
Removing a
Single Bracket
Single Bracket
7 + 3(4 - y) =
7 +
12
Example 7
9 - 3(8 - y) =
9
- 24
Example 8
Removing a
Removing a
Single Bracket
Single Bracket
- 3y
=
1
9
-
3
y
Tidy Up
+ 3y
Tidy Up
=
-
1
5
+
3
y
4(m - 3) - (m + 2) =
4m - 12
- m - 2
Example 9
Example 10
= 3m - 14
Tidy Up
7(y - 1) - 2(y + 4) =
7y - 7
- 2y - 8
= 5y - 15
Tidy Up
Removing Two Single Brackets
Removing Two Single Brackets
Learning Intention
Learning Intention
Success Criteria
Success Criteria
2.
2.
Solving simple algebraic
Solving simple algebraic
equations.
equations.
1.
To solve simple equations
using the
‘Balancing Method’.
‘Balancing Method’.
1.
1.
Know the process of the
Know the process of the
‘Balancing Method’.
‘Balancing Method’.
Equations
Equations
Solving Equations
Balancing Method
Balancing Method
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?
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Balancing Method
Balancing Method
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+
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=
8
k
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Balancing Method
Balancing Method
P
P
4
P
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4
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m
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b
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=
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2
P
+
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P
+
1
2
What symbol should we
use for the scales ?
=
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-4
2
P
=
P
+
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Subtract P
from each side
-P
-P
P
=
8
Balancing Method
Balancing Method
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Equations
Equations
Write down the opposite of the following :
+
opposite is
x
-
÷
opposite is
opposite is
opposite is
-
+
÷
x
Solving Equations
x + 3 =
20
Example 1
x =
2
0
-
3
1
7
(
-
3
)
x =
Simple Equations
Simple Equations
24 – 8 =
24 - x =
8
Example 2
24 =
8
+
x
x
(
+
x
)
Simple Equations
Simple Equations
(
-
8
)
16 =
x
x
= 16
4x =
20
Example 3
x =
2
0
÷
4
5
(
÷
4
)
x =
Simple Equations
Simple Equations
8x =
28
Example 4
x =
2
0
÷
8
3
.
5
(
÷
8
)
x =
Simple Equations
Simple Equations
Learning Intention
Learning Intention
Success Criteria
Success Criteria
2.
2.
Solving harder algebraic
Solving harder algebraic
equations by using rule
equations by using rule
repeatedly.
repeatedly.
1.
To solve harder equations
using the rule
‘Balancing Method’.
‘Balancing Method’.
repeatedly.
repeatedly.
1.
1.
Know the process of
Know the process of
‘Balancing Method’.
‘Balancing Method’.
Equations
Equations
Harder Equations
Balancing Method
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
.
If a child’s ticket costs £
2
. How much
for an adult ticket ?
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Balancing Method
Balancing Method
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e
-6
-6
5
a
=
3
a
+
1
0
Subtract 3a
from each side
-3a
-3a
2
a
=
1
0
a
a
2
2
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b
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2
a
=
5
A
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£
5
1
Balancing Method
Balancing Method
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Equations
Equations
Write down the opposite of the following :
+
opposite is
x
-
÷
opposite is
opposite is
opposite is
-
+
÷
x
Level E Harder Equations
2x + 4 =
22
2x
Example 1
=
1
8
Equations
Equations
x =
9
(
-
4
)
(
÷
2
)
9x - 5 =
40
9x
Example 2
=
4
5
Equations
Equations
x =
5
(
+
5
)
(
÷
9
)
Starter Questions
Starter Questions
Q1.
Solve for x
(a)
x + 3 = 8
(b)
2x – 14 = 50
Q2.
Is this statement true
(x – 1) – 3(x + 1) = -2x
Learning Intention
Learning Intention
Success Criteria
Success Criteria
1.
To show how to solve
equations that have
bracket terms.
1.
Be able to multiply out
Be able to multiply out
brackets and solve
brackets and solve
equations.
equations.
Equations and brackets
Equations and brackets
5(x - 3) =
25
5x
Example 1
- 15
=
2
5
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Equations and brackets
Equations and brackets
5x =
2
5
+
1
5
x =
4
0
÷
5
=
4
0
=
8
(
+
1
5
)
(
÷
5
)
3(g - 1) =
9
3g
Example 2
- 3
=
9
Multiply out the bracket first
and then solve.
Equations and brackets
Equations and brackets
3g =
9
+
3
g =
1
2
÷
3
=
1
2
=
4
(
+
3
)
(
÷
3
)
Q1.
Solve for x
(a)
x + 7 = 29
(b)
2x – 5 = 21
Q2.
Is this statement true
(x + 1) – 2(x + 1) = -x
Learning Intention
Learning Intention
Success Criteria
Success Criteria
1.
To show how to solve
equations with terms on
both sides.
1.
Be able to solve equations
Be able to solve equations
with terms on both sides.
with terms on both sides.
Equations and brackets
Equations and brackets
6x - 3 =
x + 7
5x
Example 1
- 3
=
7
Equations and brackets
Equations and brackets
5x =
7
+
3
x =
1
0
÷
5
=
1
0
2
(
-
x
)
(
+
3
)
(
÷
5
)
x =
8y + 1 =
5y + 7
3y
Example 2
+ 1
=
7
Equations and brackets
Equations and brackets
3y =
6
y =
2
(
-
5
y
)
(
-
1
)
(
÷
2
)
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Presentation Transcript
Algebra Tidying Up Terms Multiplying Terms Removing Brackets & Simplifying Solving Simple Equations ( x+1=5 ) Harder Equations ( 2x+1=9 ) Solving Equations ( brackets) Solving Equations (terms either side) Inequalities
Starter Questions 1. A TV has been reduced in the sale by 20%. It was 250. How much is it now. 2. 30% of 300 3. Tidy up the expression 2cm 3cm 5y + 3d +4d + 10y 6cm 2cm 4. Explain how do find the area and perimeter of the shape.
Algebra Multiplying Terms Learning Intention Success Criteria 1. Understand the term like terms . 1. To explain how to gather algebraic like terms . 2. Gather like terms for simple expressions.
Tidying Terms + + + + = First Row : 2c c 4c = 7c + + + = 2nd Row : + 2d 2d 3d = 7d
Tidying Terms + + + + = 3rd Row : 3f f 2f = 6f CANNOT TIDY UP ANYMORE In Total we have 7c + 7d + 6f WE CAN ONLY TIDY UP LIKE TERMS
Tidying Terms WE CAN ONLY TIDY UP LIKE TERMS Tidy up the following: 6x + 2y + 18 Q1. 2x + 4x + 5y -3y + 18 = Q2. 4a + 3b + 5a + 6 b = 9a + 2b + 6
Algebra Multiplying Terms Learning Intention Success Criteria 1. Understand the key steps of multiplying terms. 1. To explain how to multiply out algebraic terms. 2. Apply multiplication rules for simple expressions.
Algebra Simplifying Algebraic Expressions Reminder ! We can only add and subtract like terms x x x x + + + = 4x 9 7 p p = 2p 3 6 a b a b + + + = + 3x 2 7 a b 10 6 + 2 x = + 9 6w 1 x w 2 + = 2 2 x
Algebra Simplifying Algebraic Expressions Reminder ! Multiplying terms 7 a = 7a 5 w = 5w b b = b 8m 2 ( NOT 2b ) = 2 2 4 m m ( NOT 8m )
Algebra Removing brackets Learning Intention Success Criteria 1. Understand the key steps in removing brackets. 1. To explain how to multiply out simply algebraic brackets. 2. Apply multiplication rules for integers numbers when removing brackets.
Removing a Single Bracket Example 1 3(b + 5) =3b + 15 Example 2 4(w - 2) =4w - 8
Removing a Single Bracket Example 3 2(y - 1) = 2y - 2 Example 4 7(w - 6) =7w- 42
Removing a Single Bracket Example 5 8(x + 3) = 8x+ 24 Example 6 12 - 8m 4(3 -2m) =
Removing a Single Bracket Tidy Up Example 7 - 3y 7 + 3(4 - y) = 7 + 12 = 19 - 3y Tidy Up Example 8 9 - 3(8 - y) = 9 - 24 + 3y = -15 + 3y
Removing Two Single Brackets Tidy Up Example 9 4(m - 3) - (m + 2) =4m - 12 - m - 2 = 3m - 14 Tidy Up Example 10 7(y - 1) - 2(y + 4) = 7y - 7 - 2y - 8 = 5y - 15
Equations Solving Equations Learning Intention Success Criteria 1. To solve simple equations using the Balancing Method . 1. Know the process of the Balancing Method . 2. Solving simple algebraic equations.
Balancing Method 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 4 4 How can we go about solving this using What instrument measures balance balance ?
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
Balancing Method What symbol should we use for the scales ? Let s solve it using maths. 4 Let P be the weight of a big bag. P P P 4 4 4 We know that a small bag = 4 2P + 4 P + 12 = Subtract 4 from each side Subtract P from each side -4 2P= P + 8 -P -P P = 8 -4
Balancing Method It would be far too time consuming to draw out the balancing scales each time. We will now learn how to use the rules for solving equations.
Equations Solving Equations The method we use to solve equations is The Balancing Method Write down the opposite of the following : +opposite is x + x - - opposite is opposite is opposite is
Simple Equations Example 1 x + 3 = 20 x = 20 - 3 x = (- 3) 17
Simple Equations Example 2 24 - x =8 24 =8 + x (+ x) 24 8 = 16 = x x = 16 x (- 8)
Simple Equations Example 3 4x = 20 x = 20 4 5 x = ( 4)
Simple Equations Example 4 8x = 28 x = 20 8 3.5 x = ( 8)
Equations Harder Equations Learning Intention Success Criteria 1. To solve harder equations using the rule Balancing Method . repeatedly. 1. Know the process of Balancing Method . 2. Solving harder algebraic equations by using rule repeatedly.
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 2 2 2 a 2 2 a 2 2 If a child s ticket costs 2. How much for an adult ticket ? 2 a a a a a 2 2 Let a be the price of an adult ticket. We know that a child price = 2
Balancing Method Subtract 6 from each side from each side Subtract 3a For balance we have 2 a 2 2 2 a 2 2 a 2 2 1 5a + 6 3a + 16 = 2 a a a a a 2 2 -6 5a= 3a + 10 -3a -3a 2a= 10 a = 5 -6 Divide each side by 2 Adult ticket price is 5
Balancing Method It would be far too time consuming to draw out the balancing scales each time. We will now learn how to use the rules for solving equations.
Equations Level E Harder Equations The rule we use to solve equations is The Balancing Method Write down the opposite of the following : +opposite is x + x - - opposite is opposite is opposite is
Equations Example 1 2x + 4 = 22 2x x = 9 = 18 (- 4) ( 2)
Equations Example 2 9x - 5 = 40 9x x = 5 = 45 (+ 5) ( 9)
Starter Questions Q1. Solve for x (a) x + 3 = 8 (b) 2x 14 = 50 Q2. Is this statement true (x 1) 3(x + 1) = -2x
Equations and brackets Learning Intention Success Criteria 1. To show how to solve equations that have bracket terms. 1. Be able to multiply out brackets and solve equations.
Equations and brackets Multiply out the bracket first and then solve. Example 1 5(x - 3) = 25 5x - 15 = 25 5x = 25 + 15 x = 40 5 = 40 = 8 (+15) ( 5)
Equations and brackets Multiply out the bracket first and then solve. Example 2 3(g - 1) = 9 3g - 3 = 9 3g = 9 + 3 g =12 3 = 12 = 4 (+3) ( 3)
Starter Questions Q1. Solve for x (a) x + 7 = 29 (b) 2x 5 = 21 Q2. Is this statement true (x + 1) 2(x + 1) = -x
Equations and brackets Learning Intention Success Criteria 1. To show how to solve equations with terms on both sides. 1. Be able to solve equations with terms on both sides.
Equations and brackets Example 1 x + 7 6x - 3 = 5x - 3 = 7 5x = 7 + 3 x = 10 5 x = (- x) = 10 (+3) ( 5) 2
Equations and brackets Example 2 5y + 7 8y + 1 = 3y + 1 = 7 3y = 6 y = 2 (- 5y) (-1) ( 2)
