Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Solve linear equations in one variable.
• Represent and solve linear inequalities.
• Identify arithmetic sequences and find the general term.
• Apply these concepts to solve real-world problems.

Lesson summary

This week, we're diving into three fundamental areas of mathematics: equations, inequalities, and number patterns. These concepts are not just abstract ideas confined to the classroom; they are powerful tools that help us understand and solve real-world problems. Imagine you are saving up to buy a new pair of sneakers. You need to determine how much you need to save each week (equations). Or consider you want to ensure that your monthly expenses are less than your income (inequalities). Or perhaps you're planning a stokvel with friends, needing to predict how much money you'll all have after a certain number of contributions (number patterns).

Lesson notes

2.1 Equations An equation is a mathematical statement that two expressions are equal. It contains an equals sign (=). Solving an equation means finding the value(s) of the variable(s) that make the equation true.

Linear Equations: A linear equation is an equation where the highest power of the variable is

1. Examples: 2x + 3 = 7, (1/2)y - 5 =

0. Solving Linear Equations: The goal is to isolate the variable on one side of the equation. We do this by performing the same operation on both sides of the equation, maintaining the balance. Key operations include addition, subtraction, multiplication, and division.

Example 1: Solve for x: 3x + 5 = 14 Subtract 5 from both sides: 3x + 5 - 5 = 14 - 5 => 3x = 9 Divide both sides by 3: 3x / 3 = 9 / 3 => x = 3 Example 2: Solve for y: (y/2) - 1 = 4 Add 1 to both sides: (y/2) - 1 + 1 = 4 + 1 => y/2 = 5 Multiply both sides by 2: (y/2) 2 = 5 2 => y = 10 Example 3: Solve for z: 2(z + 1) = 8 Expand the brackets: 2 z + 2 1 = 8 => 2z + 2 = 8 Subtract 2 from both sides: 2z + 2 - 2 = 8 - 2 => 2z = 6 Divide both sides by 2: 2z / 2 = 6 / 2 => z = 3 Example 4: Equation with Fractions Solve for a: (a/3) + (a/2) = 5 Find a common denominator (in this case, 6): (2a/6) + (3a/6) = 5 Combine the fractions: (5a/6) = 5 Multiply both sides by 6: 5a = 30 Divide both sides by 5: a = 6 2.2 Inequalities An inequality is a mathematical statement that shows the relationship between two expressions that are not equal.

It uses inequality symbols such as: (greater than) ≤ (less than or equal to) ≥ (greater than or equal to)

Solving Linear Inequalities: Solving an inequality means finding the set of values that satisfy the inequality. The rules for solving inequalities are similar to those for solving equations, with one important difference: when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality sign.

Example 1: Solve for x: x + 3 x y ≥ 3 This means y can be any number greater than or equal to

3. Example 3: Solve for z: -3z ≤ 9 Divide both sides by -3 (and reverse the inequality sign): -3z / -3 ≥ 9 / -3 => z ≥ -3 This means z can be any number greater than or equal to -

3. Representing Inequalities on a Number Line: An open circle (o) indicates that the endpoint is not included in the solution (for ). A closed circle (•) indicates that the endpoint is included in the solution (for ≤ and ≥). For example, x 4x = 12 Divide both sides by 4: 4x / 4 = 12 / 4 => x = 3

Commentary: This is a straightforward linear equation. We isolate 'x' by performing inverse operations in the correct order (addition before division).

Question 2: Solve for y: -2y + 6 > 2 Solution: Subtract 6 from both sides: -2y + 6 - 6 > 2 - 6 => -2y > -4 Divide both sides by -2 (and reverse the inequality sign): -2y / -2 y < 2

Commentary: Remember to reverse the inequality sign when dividing by a negative number. The solution represents all numbers less than

2. Question 3: Determine the general term (T_n) for the sequence: 1, 4, 7, 10, ...

Solution: Identify the first term (a): a = 1 Identify the common difference (d): d = 4 - 1 = 3 Apply the formula: T_n = a + (n - 1)d: T_n = 1 + (n - 1)3 = 1 + 3n - 3 = 3n - 2 Therefore, the general term is T_n = 3n -

2. Commentary: We've identified this as an arithmetic sequence because the difference between consecutive terms is constant. Then we applied the general term formula.

Question 4: Solve for x: (2/3)x + 1 = 5 Solution: Subtract 1 from both sides: (2/3)x = 4 Multiply both sides by 3/2: x = 4 * (3/2) = 6

Commentary: Working with fractions requires finding a common denominator or, as in this case, multiplying by the reciprocal. Independent Practice (Questions Only)

Solve for a: 5a + 2 = 17 Solve for b: (b/4) - 3 = -1 Solve for c: 3(c - 2) = 9 Solve for d: -4d + 8 < 12 Solve for e: (e/2) + 5 ≥ 8 Determine the general term (T_n) for the sequence: 3, 7, 11, 15, ... Determine the general term (T_n) for the sequence: 20, 17, 14, 11, ... The general term of a sequence is T_n = 5n -

2. Find the 8th term (T_8).

Solve for x: (x/5) + (x/3) = 8 Solve for y: 2(y+3) - 5y = -6