PEARLY KIDS SCHOOL
SECOND TERM WEEKLY LESSON NOTES
WEEK 5
TEACHER: JAMES AHULOO
Week Ending: 7TH FEB Period: 4 TIMES Subject: Mathematics
2025
Duration: 60MINS Strand: Number
Sub Strand: Addition,
Class: B8 Class Size: 15 Subtraction Of Algebraic
Expressions
Content Standard: Indicator:
Lesson:
B8.2.1.1 Demonstrate the B8.2.2.1.2 Perform addition, subtraction,
ability to draw table of multiplication and division of algebraic 1 of 2
values for a linear relation expressions including fractions
Performance Indicator: Core Competencies:
Learners can perform addition, subtraction, Communication and Collaboration
multiplication and division of algebraic expressions (CC)
including fractions Critical Thinking and Problem
solving (CP)
References: Mathematics Curriculum Pg. 115-116
Phase/ Learners Activities Resources
Duration
PHASE 1: Revise with learners on the previous lesson.
STARTER
Share performance indicators with learners and
introduce the lesson.
PHASE 2: Guide learners to solve addition, subtraction, Counters,
NEW multiplication and division of algebraic expression using bundle and
LEARNING the PEDMAS strategy. loose straws
base ten cut
To use the PEMDAS strategy, follow these steps: square,
Bundle of
• Simplify any expressions inside parentheses first. sticks
• Evaluate any exponents next.
• Perform multiplication and division, from left to right.
• Perform addition and subtraction, from left to right.
Write an example on the board and task learners to
work in pairs.
Simplify 10x2 + (6x-4x) – (5x – 2x)2
Solution
To simplify the expression 10x2 + (6x-4x) - (5x - 2x)2
using the PEMDAS strategy, we follow the order of
operations as follows:
Simplify any expressions inside parentheses first:
(6x - 4x) = 2x
(5x - 2x)2 = (3x)2 = 9x2.
Now the expression becomes:
10x2 + 2x -
9x2 10x2 -
2
9x +2x =
x2 + 2x.
Therefore, the simplified expression is x^2 + 2x.
Example 2: solve (7y-5y)2 – 2(10y-8y) +4y
= (2y)2 – 2(2y) + 4y
= 4y2 – 4y + 4y // and +4y - 4y cancels out
= 4y2
Therefore, the simplified form of the expression is = 4y 2
Assessment
1. 3(5x+2x) – (4-5x)
2. (t + k) + (5t x 2)
3. (6m)2 – 4(2m x m) +2m
4. 2y-y(6y-2y) – (-2 x 2y)
Guide learners to solve problems based on
multiplication and division of algebraic fractions.
To solve problems based on multiplication and division
of algebraic fractions, follow these general steps:
1. Simplify each algebraic fraction by factoring out
any common factors in the numerator and
denominator.
2. To multiply algebraic fractions, multiply the
numerators together and multiply the denominators
together. Then, simplify the resulting fraction by
factoring out any common factors.
3. To divide algebraic fractions, invert the second
fraction and multiply it by the first. Then, simplify the
resulting fraction by factoring out any common factors.
(2𝑥 + 4𝑥) (𝑥 + 1)
2
Example 1: Multiply x
(𝑥+2)
Solution:
First, simplify each fraction. We can factor out a 2x
from the first fraction to get:
(2𝑥 + 4𝑥) 2𝑥(𝑥 + 2)
2
= = 2x
(𝑥+2)
For the second fraction, we can factor out an x from the
denominator to get:
(𝑥 + 1) (𝑥 + 1)
2−4𝑥) = 𝑥(𝑥−4)
(𝑥
Now we can multiply the two fractions together:
(𝑥 + 1)
2x *
𝑥(𝑥−4)
Multiplying the numerators gives us:
2x(x + 1) = 2x2 + 2x
Multiplying the denominators gives us:
x(x - 4) = x2 - 4x
So the final answer is:
(2𝑥 + 2𝑥)
2
= 2−4𝑥)
(
𝑥
We can simplify this by factoring out a 2x from the
numerator and a x from the denominator:
2𝑥(𝑥 + 2) 2(𝑥 + 2)
= =
𝑥(𝑥−4) 𝑥(𝑥−4)
(3𝑥 − 9𝑥) (2𝑥 + 8𝑥)
2 2
Example 2: Divide 2−4) ÷ (𝑥2−2𝑥)
(𝑥
Solution: First, simplify each fraction. We can factor out
a 3x from the numerator of the first fraction and factor
out a 2x from the numerator of the second fraction:
(3𝑥 − 9𝑥) 3𝑥(𝑥− 3)
2
=
(𝑥 −4) (𝑥−2)(𝑥+2)
2
(2𝑥 + 8𝑥) 2𝑥(𝑥+4)
2
=
(𝑥 −2𝑥) 𝑥(𝑥−2)
2
Now we can invert the second fraction and multiply it
by the first:
(3𝑥 −9𝑥) 𝑥(𝑥−2)
2
x
(𝑥 −4) 2𝑥(𝑥+4)
2
Multiplying the numerators gives us:
3x(x - 3)(x - 2)
Multiplying the denominators gives us:
2x(x + 4)(x - 2)(x + 2)
So the final answer is:
We can simplify this by cancelling out the (x - 2) factor
in the numerator and denominator:
Assessment
𝑎 𝑏
1) x
7 8
2)
𝑎 1
3) ÷ 𝑎𝑏 𝑎
4) x
8𝑟 5𝑟
PHASE 3: Use peer discussion and effective questioning to find
REFLECTION out from learners what they have learnt during the
lesson.
Take feedback from learners and summarize the lesson.
Week Ending: 7TH FEB Period: 4 TIMES Subject: Mathematics
2025
Duration: 60MINS Strand: Number
Sub Strand: Algebraic
Class: B7 Class Size: 15
Expressions
Content Standard: Indicator:
Lesson:
B8.2.1.1 Demonstrate the B8.2.2.1.3 Substitute values to evaluate
ability to draw table of algebraic expressions including fractions and 1 of 2
values for a linear relation use these to solve problems.
Performance Indicator: Core Competencies:
Learners can substitute values to evaluate algebraic Communication and Collaboration
expressions including fractions and use these to (CC)
solve problems Critical Thinking and Problem
solving (CP)
References: Mathematics Curriculum Pg. 119
Phase/ Learners Activities Resources
Duration
PHASE 1: Revise with learners on the previous lesson.
STARTER
Share performance indicators with learners and
introduce the lesson.
PHASE 2: Guide learners to substitute values to evaluate Counters,
NEW algebraic expressions including fractions and use bundle and
LEARNING loose straws
these to solve problems. base ten cut
square,
Take learners through the steps in substituting Bundle of
values into algebraic expressions. sticks
To substitute values to evaluate algebraic expressions
including fractions:
1. Identify the variables in the expression that you
want to substitute values for.
2. Replace each variable with the corresponding
value.
3. Simplify the expression by performing any
necessary arithmetic operations, such as addition,
subtraction, multiplication, and division.
Example, Evaluate the expression (3x - 2)/(x + 1) when
x = 4.
1. The variable in this expression is x.
2. We replace x with the value 4:
(3x - 2)/(x + 1) = (3(4) - 2)/(4 + 1)
3. Simplify the expression by performing the
arithmetic operations: (3(4) - 2)/(4 + 1) = (10/5) = 2
Therefore, when x = 4, the value of the expression (3x -
2)/(x + 1) is 2.
Example 2: Evaluate the expression when x =
5.
1.Identify the variable in the expression: x.
2.Replace x with the value 5:
= (2(5) + 3)/(5 - 4)
3. Simplify the expression by performing the arithmetic
operations: (2(5) + 3)/(5 - 4) = (13/1) = 13
Therefore, when x = 5, the value of the expression (2x
+ 3)/(x - 4) is 13.
Example 3: Evaluate the expression (5y - 2)/(2y +
1) when y = 3.
1. Identify the variable in the expression: y.
2. Replace y with the value -3:
(5y - 2)/(2y + 1) = (5(-3) - 2)/(2(-3) + 1)
3. Simplify the expression by performing the
arithmetic operations: (5(-3) - 2)/(2(-3) + 1) = (-17/-5)
= 3.4
Therefore, when y = -3, the value of the expression (5y
- 2)/(2y + 1) is 3.4.
Example 4: Evaluate the expression (4a2 - 3b)/(2a -
b) when a = 2 and b = 1.
1. Identify the variables in the expression: a and b.
2. Replace a with the value 2 and b with the value
1:
(4a2 - 3b)/(2a - b) = (4(2)2 - 3(1))/(2(2) - 1)
3. Simplify the expression by performing the
arithmetic operations: (4(2)2 - 3(1))/(2(2) - 1) = (13/3)
Therefore, when a = 2 and b = 1, the value of the
expression (4a^2 - 3b)/(2a - b) is 13/3.
PHASE 3: Use peer discussion and effective questioning to find
REFLECTION out from learners what they have learnt during the
lesson.
Take feedback from learners and summarize the lesson.
