The document provides information on solving linear equations with two unknowns. It discusses representing real-life situations as linear equations with two variables. It introduces the substitution and elimination methods for solving such equations. For the substitution method, one variable is isolated and substituted into the other equation to solve for the remaining variable. For the elimination method, the equations are manipulated to eliminate one variable, leaving an equation that can be solved for the remaining variable. Examples are provided to demonstrate both methods. The key learning is how to set up and solve linear equations with two unknowns that model real-world scenarios.

1. STRAND 2.0: ALGEBRA
Sub -strand 2.1 Algebraic expressions

2. Specific Learning Outcomes
•By the end of the sub- strand the learner should
be able to;
•factorize algebraic expressions in different
situations
•simplify algebraic fractions in different situations
•evaluate algebraic expressions by substituting
numerical values in different situations

3. Factorize Algebraic Expressions in Different
Situations
• Introduction to Factorization
• factorization is breaking down algebraic expressions into simpler factors.
• By breaking down an expression into simpler factors, we can rearrange them
more easily to solve equations.
• Example 1: Factorize
• 2x + 4
• Common Factor is 2 because it can be divided by both 2x and 4
• Factored: 2(x + 2)

4. Factorize
•3y + 12y + 21y
•Common factor is 3y because it can be
divided by the whole equation
•Factored 3y(1 + 4 + 7)

5. SIMPLIFY ALGEBRAIC FRACTIONS IN DIFFERENT
SITUATIONS
•Simplifying algebraic fractions involves making them
as simple as possible by canceling out common
factors in the numerator and the denominator.
•Here's how you can do it step by step:

6. • Step 1: Identify Common Factors. Look for common factors between
the terms in the numerator and the terms in the denominator. These
common factors can be letters or numbers.
• Step 2: Cancel Common Factors. If you find common factors, cancel
them out by dividing both the numerator and the denominator by
the common factor.
• Step 3: Check for Further Simplification. After canceling the common
factors, see if there are any more factors that can be canceled to
simplify the expression more.
• Step 4: Write down the simplified expression. with the remaining
terms in the numerator and denominator.

7. Example 1:
• Simplify the expression:
•
4𝑥2 − 8𝑥
2𝑥
• Solution: Step 1: Identify Common Factors There's a common factor of 2x between
the terms in the numerator and the denominator.
• Step 2: Cancel Common Factors 2x
•
4𝑥2−8𝑥
2𝑥
=
2𝑥×(2𝑥−4)
(2𝑥)
= 2x - 4
• Step 3: Check for Further Simplification No further simplification is possible.
• Step 4: Write the Simplified Expression The simplified expression is: 2x – 4
•
• By simplifying algebraic fractions, you make algebraic expressions easier to work with
and understand, which is important when you're solving equations or dealing with
mathematical problems.

8. EVALUATE ALGEBRAIC EXPRESSIONS BY SUBSTITUTING
NUMERICAL VALUES IN DIFFERENT SITUATIONS
• Step 1: Understand the Expression First, you need to understand the
algebraic expression. It might have letters and numbers in it.
• Step 2: Substitute the Values Replace the letters in the expression with the
actual numbers you're given. This is like putting the pieces of a puzzle
together.
• Step 3: Follow the Order If there are operations like addition, subtraction,
multiplication, or division in the expression, do them in the correct order
(remember BODMAS).
• Step 4: Calculate Perform the calculations in each part of the expression.
• Step 5: Write the Answer After doing all the calculations, write down the
final answer.

9. Evaluate the expression
• 3x - 2x + 5, with x = 4
• Solution: Step 1: Understand the Expression. We have an expression with x in
it.
• Step 2: Substitute the Values Replace x with 4:
• 3(4) - 2(4) + 5
• Step 3: Follow the Order ,Deal with the bracket first:
• 12 - 8 + 5
• Step 5: Add and subtract:
• 12 + 5 =17
• 17 - 8 = 9
• So, when x = 4, the expression becomes 9.

10. Quiz
• Factorize
• 5a + 10 b) 2y + 4y + 6y c) 3x + 6x + 9x
Simplify
• 19w + 10z + 5w 3z b) 2x – 6y – 4x + 5z -y c) 8w – p- 2w + 11p
Simplify
• 4𝑥2 − 8𝑥 b)
8𝑥
16𝑥
c)
3𝑥2 +9𝑥2+12𝑥2
6𝑥2

11. SPECIFIC LEARNING OUTCOMES
•By the end of the sub strand the learner should be able
to;
•form linear equations in two unknowns in real life
situations
•solve linear equations in two unknowns by Substitution
method in real life situations
•solve linear equations in two unknowns by elimination
method in real life situations
•apply linear equations in two unknowns in real life

12. SUB-STRAND: LINEAR EQUATIONS
• Introduction to Linear Equations in Two Unknowns
• Linear equations are straight line mathematical expressions.
• Linear equations can represent relationships between two unknown values.
• Linear equations can represent various real-life situations, from distances to
costs.
• Consider the following:
• The cost of two skirts and 3 blouses is 600.
• Expressing this as a linear equation would be
• Take the number of skirts to be s and number of blouses to be b
• 2s + 3b = 600

13. EXAMPLE
The cost of one skirt and two blouses is 350
Expressing this as a linear equation would be
S + 2b = 350

14. Alex has 3 books and 1 pen
• Expressing this as a linear equation would be
• Take b to be number of books and p to be number of
pens
• In total Alex has
• 3b + p
•Two ways to solve linear equations with two unknown are

15. SUBSTITUTION METHOD
Solving Linear Equations by Substitution Method
The substitution method involves isolating one
unknown in one equation and substituting it into
another.
Solve for the remaining unknown using the substituted
value.

16. EXAMPLE : SOLVE BY SUBSTITUTION
• 2s + 3b = 600 …………. (i)
• s + 2b = 350 …………. (ii)
• taking the second equation
• s + 2b = 350
• subtracting 2b from both sides
• s = 350 – 2b …………. (iii)

17. EXAMPLE : SOLVE BY SUBSTITUTION
Substituting the value of s in the equation
2(350 – 2b) + 3b = 600
700 – 4b + 3b = 600
b = 100
Substituting the value of y in the third equation
s = 350 – 2b
s = 350 – 2(100)
s = 350 – 200
s = 150

18. Solve Linear Equations in Two Unknowns by
Elimination Method
• The elimination method involves reducing equations to eliminate one unknown.
• Solve for the remaining unknown using the simplified equation.
• 2s + 3b = 600 …………. (i)
• s + 2b = 350 …………. (ii)
• multiply equation (ii) by 2 to get equation (iii)
• 2s + 4b = 700 ………... (iii)
• Subtracting equation (i) from (iii)
• 2s + 4b = 700 ……………. (iii)
• 2s + 3b = 600 ……………. (i)
• b = 100

19. BUT FROM EQUATION (II)
• s + 2b = 350
• but b = 100
• s + 2(100) = 350
• s + 200 = 350
• s = 350 – 200
• s = 150
• s = 150, b = 100
• we notice that despite using different methods we arrive at the same answer.

20. QUIZ
•Represent the following as linear equations
•Paul has 3 books and 10 pens.
•There are 20 items in the kitchen, 12 cups and 8
plates
•In a farm there are 9 cows and 15 goats