Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Simplify algebraic expressions.
• Factorize various quadratic expressions.
• Solve linear equations and inequalities.
• Solve quadratic equations using factorization.
• Apply algebraic skills to solve word problems.

Lesson summary

This week's revision focuses on consolidating our understanding of algebraic expressions, including simplification, factorization, and solving equations. This is a crucial foundation for more advanced topics in mathematics and is also vital for problem-solving in various real-life scenarios. From budgeting personal finances to calculating measurements for DIY projects, a strong understanding of algebra is essential for navigating everyday life effectively. In South Africa, with varying economic conditions, these skills are particularly important for making informed financial decisions and managing resources effectively.

Lesson notes

2.1 Simplifying Algebraic Expressions Simplifying algebraic expressions involves combining like terms and applying the order of operations (BODMAS/PEMDAS: Brackets, Orders, Division & Multiplication, Addition & Subtraction). Like terms are terms that have the same variable raised to the same power.

Example 1: Simplify 3x + 5y - 2x + y.

Identify like terms: 3x and -2x are like terms; 5y and y are like terms.

Combine like terms: (3x - 2x) + (5y + y) = x + 6y Example 2: Simplify 2(x + 3) - 4(2x - 1).

Expand the brackets: 2x + 6 - 8x + 4 Identify like terms: 2x and -8x are like terms; 6 and 4 are like terms.

Combine like terms: (2x - 8x) + (6 + 4) = -6x + 10 Example 3: Simplify (4x² + 2x - 1) - (x² - 3x + 5) Distribute the negative sign into the second bracket: 4x² + 2x - 1 - x² + 3x - 5 Identify and group like terms: (4x² - x²) + (2x + 3x) + (-1 - 5)

Combine like terms: 3x² + 5x - 6 2.2 Factorization Factorization is the reverse process of expansion. It involves expressing an algebraic expression as a product of its factors. 2.2.1 Highest Common Factor (HCF): Identify the HCF of all the terms in the expression. Write the expression as a product of the HCF and the remaining factors.

Example: Factorize 6x + 9y. The HCF of 6x and 9y is 3. 6x + 9y = 3(2x + 3y) 2.2.2 Difference of Two Squares: Applies to expressions of the form a² - b². a² - b² = (a + b)(a - b)

Example: Factorize x² - 16. x² - 16 = x² - 4² = (x + 4)(x - 4) 2.2.3 Perfect Square Trinomials: Applies to expressions of the form a² + 2ab + b² or a² - 2ab + b². a² + 2ab + b² = (a + b)² a² - 2ab + b² = (a - b)²

Example: Factorize x² + 6x + 9. x² + 6x + 9 = x² + 2(x)(3) + 3² = (x + 3)² 2.2.4 General Trinomials (ax² + bx + c): Find two numbers that multiply to give 'ac' and add up to give 'b'.

Example 1: a = 1 Factorize x² + 5x +

6. Find two numbers that multiply to 6 and add to

5. These are 2 and 3. x² + 5x + 6 = (x + 2)(x + 3)

Example 2: a != 1 Factorize 2x² + 7x +

3. Find two numbers that multiply to (2)(3) = 6 and add up to

7. These are 6 and

1. Rewrite the middle term: 2x² + 6x + x + 3 Factor by grouping: 2x(x + 3) + 1(x + 3) (2x + 1)(x + 3) 2.3 Solving Linear Equations and Inequalities Linear Equations: An equation where the highest power of the variable is

1. The goal is to isolate the variable on one side of the equation.

Example: Solve 2x + 5 =

1

1. Subtract 5 from both sides: 2x = 6 Divide both sides by 2: x = 3 Linear Inequalities: Similar to equations but involve inequality signs ( , ≤, ≥). When multiplying or dividing by a negative number, remember to reverse the inequality sign.

Example: Solve 3x - 2 >

7. Add 2 to both sides: 3x > 9 Divide both sides by 3: x > 3 2.4 Solving Quadratic Equations by Factorization A quadratic equation is an equation of the form ax² + bx + c = 0, where a ≠

0. To solve by factorization: Rearrange the equation so that it is equal to zero. Factorize the quadratic expression. Set each factor equal to zero and solve for x.

Example: Solve x² - 5x + 6 =

0. Factorize the quadratic expression: (x - 2)(x - 3) = 0 Set each factor equal to zero: x - 2 = 0 or x - 3 = 0 Solve for x: x = 2 or x = 3 2.5 Word Problems Algebraic skills can be used to solve a variety of word problems. The key is to translate the words into mathematical expressions and equations.

Example: A tuck shop sells pies for R15 each and drinks for R8 each. A customer buys 'x' pies and 'y' drinks and pays R

1

0

1. Write an equation to represent this situation and if the customer buys 5 pies, how many drinks did they buy?

Equation: 15x + 8y = 101 If x = 5: 15(5) + 8y = 101 => 75 + 8y = 101 => 8y = 26 => y = 3.25 They bought 3 drinks (you can't buy a quarter of a drink!) and probably got some change back. Guided Practice (With Solutions)

Question 1: Simplify: 4(a - 2b) + 3(2a + b)

Solution: Expand the brackets: 4a - 8b + 6a + 3b Identify like terms: (4a + 6a) + (-8b + 3b)

Combine like terms: 10a - 5b

Commentary: This question reinforces the distributive property and combining like terms.

Question 2: Factorize: x² - 49 Solution: Recognize that this is a difference of two squares: x² - 7² Apply the formula a² - b² = (a + b)(a - b): (x + 7)(x - 7)

Commentary: This tests the ability to recognize and apply the difference of two squares factorization.

Question 3: Solve for x: 5x - 3 = 2x + 9 Solution: Subtract 2x from both sides: 3x - 3 = 9 Add 3 to both sides: 3x = 12 Divide both sides by 3: x = 4

Commentary: This focuses on solving linear equations using basic algebraic manipulation.

Question 4: Solve for x: x² + x - 12 = 0 Solution: Factorize the quadratic expression: (x + 4)(x - 3) = 0 Set each factor equal to zero: x + 4 = 0 or x - 3 = 0 Solve for x: x = -4 or x = 3

Commentary: This reinforces solving quadratic equations by factorization.

Question 5: The length of a rectangular garden is twice its width. If the perimeter of the garden is 36 meters, find the dimensions of the garden.

Solution: Let the width be 'w'. Then the length is '2w'.