Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Solve linear equations and inequalities.
• Factorize quadratic expressions.
• Simplify algebraic fractions.
• Determine the equation of a straight line.
• Calculate the distance and midpoint between two points.

Lesson summary

This week's focus is a vital revision of concepts learned throughout the term, particularly those covered up to week

8. Mastering these fundamentals is essential for building a strong mathematical foundation, crucial not only for succeeding in Grade 10 but also for future academic and professional pursuits. Mathematics empowers us to understand and solve problems in everyday life, from budgeting and financial planning to analyzing data and making informed decisions. In South Africa, a solid grasp of mathematics is particularly important for participating in a growing economy that increasingly relies on quantitative skills.

Lesson notes

This week we will be revising important algebra and analytical geometry concepts. A. Linear Equations and Inequalities A linear equation is an equation of the form ax + b = 0, where a and b are constants, and x is the variable. An inequality uses symbols like , ≤, or ≥ instead of =. Solving involves isolating x.

Example 1: Solving a Linear Equation Solve for x: 3x + 5 = 14 Step 1: Isolate the term with x.

Subtract 5 from both sides: 3x + 5 - 5 = 14 - 5 3x = 9 Step 2: Solve for x.

Divide both sides by 3: 3x/3 = 9/3 x = 3 Therefore, the solution is x =

3. Example 2: Solving a Linear Inequality Solve for x: 2x - 7 > 5 Step 1: Isolate the term with x.

Add 7 to both sides: 2x - 7 + 7 > 5 + 7 2x > 12 Step 2: Solve for x.

Divide both sides by 2: 2x/2 > 12/2 x > 6 Therefore, the solution is x >

6. This means any value of x greater than 6 will satisfy the inequality. B. Factorizing Quadratic Expressions A quadratic expression is an expression of the form ax² + bx + c, where a, b, and c are constants and a ≠

0. Factorizing involves writing the expression as a product of two linear factors.

Example 1: Common Factor Factorize: 4x² + 8x Step 1: Identify the common factor. The common factor is 4x.

Step 2: Factor out the common factor. 4x² + 8x = 4x(x + 2) Therefore, the factored form is 4x(x + 2).

Example 2: Difference of Two Squares Factorize: x² - 9 Step 1: Recognize the pattern. This is in the form a² - b², where a = x and b =

3. Step 2: Apply the formula. a² - b² = (a + b)(a - b) x² - 9 = (x + 3)(x - 3) Therefore, the factored form is (x + 3)(x - 3).

Example 3: Trinomial Factorization Factorize: x² + 5x + 6 Step 1: Find two numbers that multiply to give c (6) and add to give b (5). These numbers are 2 and

3. Step 2: Write the factors. x² + 5x + 6 = (x + 2)(x + 3) Therefore, the factored form is (x + 2)(x + 3). C. Simplifying Algebraic Fractions Algebraic fractions are fractions where the numerator and/or denominator are algebraic expressions. Simplifying involves factoring and cancelling common factors.

Example: Simplify: ( x² - 4 ) / ( x² + 4x + 4 )

Step 1: Factorize the numerator and denominator. x² - 4 = (x + 2)(x - 2) x² + 4x + 4 = (x + 2)(x + 2)

Step 2: Rewrite the fraction with factored expressions. ( x² - 4 ) / ( x² + 4x + 4 ) = ( (x + 2)(x - 2) ) / ( (x + 2)(x + 2) )

Step 3: Cancel common factors. ( (x + 2)(x - 2) ) / ( (x + 2)(x + 2) ) = (x - 2) / (x + 2) Therefore, the simplified form is (x - 2) / (x + 2). D. Equation of a Straight Line The general form of the equation of a straight line is y = mx + c, where m is the gradient and c is the y-intercept. Finding the equation given two points (x₁, y₁) and (x₂, y₂): Calculate the gradient: m = (y₂ - y₁) / (x₂ - x₁)

Use the point-slope form: y - y₁ = m(x - x₁) Simplify to the form y = mx + c.

Example: Find the equation of the line passing through (1, 3) and (4, 9).

Calculate the gradient: m = (9 - 3) / (4 - 1) = 6 / 3 = 2 Use the point-slope form: y - 3 = 2(x - 1)

Simplify: y - 3 = 2x - 2 y = 2x + 1 Therefore, the equation of the line is y = 2x +

1. E. Distance and Midpoint Formulae Distance Formula: The distance d between two points (x₁, y₁) and (x₂, y₂) is given by: d = √((x₂ - x₁)² + (y₂ - y₁)²)

Midpoint Formula: The midpoint M of the line segment connecting two points (x₁, y₁) and (x₂, y₂) is given by: M = ((x₁ + x₂) / 2 , (y₁ + y₂) / 2)

Example: Find the distance and midpoint between the points (2, 1) and (5, 5).

Distance: d = √((5 - 2)² + (5 - 1)²) = √(3² + 4²) = √(9 + 16) = √25 = 5 Midpoint: M = ((2 + 5) / 2 , (1 + 5) / 2) = (7/2 , 6/2) = (3.5 , 3)* Therefore, the distance is 5 units and the midpoint is (3.5, 3). Guided Practice (With Solutions)

Question 1: Solve for x: 5x - 8 = 22 Solution: Add 8 to both sides: 5x = 30 Divide both sides by 5: x = 6 Therefore, x =

6. This isolates x by performing inverse operations in the correct order.

Question 2: Factorize: x² - 16 Solution: Recognize the difference of two squares: x² - 4² Apply the formula: (x + 4)(x - 4) Therefore, x² - 16 = (x + 4)(x - 4). This is a direct application of a standard factorization pattern.

Question 3: Simplify: ( x² + 3x + 2 ) / ( x + 1 )

Solution: Factorize the numerator: x² + 3x + 2 = (x + 1)(x + 2)

Rewrite the fraction: ( (x + 1)(x + 2) ) / ( x + 1 ) Cancel the common factor (x + 1): x + 2 Therefore, the simplified form is x +

2. Factorization is key to identifying and canceling common terms.

Question 4: Find the equation of the line passing through (0, 2) and (3, 8).

Solution: Calculate the gradient: m = (8 - 2) / (3 - 0) = 6 / 3 = 2 Since (0,2) is the y-intercept, c = 2 Therefore, the equation of the line is y = 2x + 2 This highlights the connection between the gradient, y-intercept, and the equation of a line.

Question 5: Find the distance between the points (-1, 4) and (2, 0).

Solution: Apply the distance formula: d = √((2 - (-1))² + (0 - 4)²) = √((3)² + (-4)²) = √(9 + 16) = √25 = 5 Therefore, the distance between the points is 5.