Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Solve linear equations with fractions and brackets.
• Represent solutions to linear inequalities on a number line.
• Find the general term for linear and quadratic number patterns.
• Use the general term to find specific values in a sequence.
• Apply these concepts to solve real-world problems.

Lesson summary

This week builds upon your Grade 8 knowledge of equations, inequalities, and number patterns. We'll be focusing on solving more complex linear equations, understanding and representing inequalities on a number line, and identifying and extending various types of number patterns, including quadratic sequences. These skills are crucial for problem-solving in various fields, from managing your pocket money and understanding budgeting to analyzing data and predicting trends in economics or science. Think about calculating the best mobile data plan, saving for a new pair of sneakers, or understanding how quickly the price of petrol changes.

Lesson notes

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

1. Our goal is to isolate the variable (usually 'x') on one side of the equation. Remember, whatever operation you perform on one side of the equation, you MUST perform the same operation on the other side to maintain balance.

Key Steps: Simplify both sides: Remove brackets by expanding and combine like terms.

Isolate the variable term: Use addition or subtraction to get all the terms with 'x' on one side and all the constant terms on the other side.

Solve for the variable: Use multiplication or division to isolate 'x'.

Check your solution: Substitute your answer back into the original equation to ensure it is correct.

Example 1: Equation with brackets Solve: 2(x + 3) - 5 = x + 4 Step 1: Simplify 2x + 6 - 5 = x + 4 2x + 1 = x + 4 Step 2: Isolate the variable term 2x - x = 4 - 1 x = 3 Step 3: Solve for the variable x = 3 (already solved!)

Step 4: Check 2(3 + 3) - 5 = 3 + 4 2(6) - 5 = 7 12 - 5 = 7 7 = 7 (Correct!)

Example 2: Equation with fractions Solve: (x/2) + 1 = (x/3) + 2 Step 1: Find the Lowest Common Denominator (LCD): The LCD of 2 and 3 is

6. Step 2: Multiply both sides of the equation by the LCD: 6 [(x/2) + 1] = 6 [(x/3) + 2] 3x + 6 = 2x + 12 Step 3: Isolate the variable term: 3x - 2x = 12 - 6 x = 6 Step 4: Check (6/2) + 1 = (6/3) + 2 3 + 1 = 2 + 2 4 = 4 (Correct!) 2.2 Linear Inequalities A linear inequality is similar to a linear equation, but instead of an equals sign (=), it uses an inequality sign ( , ≤, ≥). Solving an inequality means finding all the values of the variable that make the inequality true. Important

Note: When you multiply or divide both sides of an inequality by a negative number, you MUST reverse the direction of the inequality sign.

Representing Solutions: Number Line: Use an open circle (o) for and a closed circle (●) for ≤ or ≥. Shade the region that represents the solutions.

Interval Notation: Use parentheses ( ) for and square brackets [ ] for ≤ or ≥.

Example 3: Solving and Representing an Inequality Solve and represent the solution on a number line: 3x - 2 > 7 Step 1: Isolate the variable term: 3x > 7 + 2 3x > 9 Step 2: Solve for the variable: x > 3 Step 3: Representation Number Line: Draw a number line. Place an open circle at

3. Shade the region to the right of

3. Interval Notation: (3, ∞)

Example 4: Inequality with a negative coefficient Solve and represent the solution on a number line: -2x + 4 ≤ 10 Step 1: Isolate the variable term: -2x ≤ 10 - 4 -2x ≤ 6 Step 2: Solve for the variable (remember to reverse the inequality sign!): x ≥ -3 Step 3: Representation Number Line: Draw a number line. Place a closed circle at -

3. Shade the region to the right of -

3. Interval Notation: [-3, ∞) 2.3 Number Patterns Number patterns are sequences of numbers that follow a specific rule. We will focus on linear and quadratic patterns. 2.3.1 Linear Patterns (Arithmetic Sequences) In a linear pattern, the difference between consecutive terms is constant (called the common difference, 'd').

The general term (nth term) is given by: T n = a + (n - 1)d Where: T n is the nth term a is the first term n is the term number d is the common difference Example 5: Finding the general term of a linear pattern Consider the pattern: 2, 5, 8, 11, ... a = 2 d = 5 - 2 = 3 Therefore, T n = 2 + (n - 1)3 = 2 + 3n - 3 = 3n - 1 2.3.2 Quadratic Patterns In a quadratic pattern, the second difference between consecutive terms is constant.

The general term (nth term) is given by: T n = an 2 + bn + c Where a, b, and c are constants. To find a, b, and c, we need to use a system of equations based on the first three terms of the sequence.

Example 6: Finding the general term of a quadratic pattern Consider the pattern: 2, 7, 14, 23, ...

First differences: 5, 7, 9 Second difference: 2 This is a quadratic sequence because the second difference is constant. 2a = 2 => a = 1 3a + b = 5 => 3(1) + b = 5 => b = 2 a + b + c = 2 => 1 + 2 + c = 2 => c = -1 Therefore, T n = n 2 + 2n - 1 Guided Practice (With Solutions)

Question 1: Solve for x: 3(x - 2) = 5x + 4 Solution: Step 1: Expand the brackets: 3x - 6 = 5x + 4 Step 2: Isolate x terms: 3x - 5x = 4 + 6 Step 3: Simplify: -2x = 10 Step 4: Divide by -2 (and flip the sign if it were an inequality): x = -5 Answer: x = -5 Question 2: Solve and represent on a number line: 2x + 5 ≤ 11 Solution: Step 1: Isolate x terms: 2x ≤ 11 - 5 Step 2: Simplify: 2x ≤ 6 Step 3: Divide by 2: x ≤ 3 Number line: Draw a number line. Place a closed circle at 3 and shade to the left.

Interval Notation: (-∞, 3] Question 3: Find the general term of the linear pattern: 7, 10, 13, 16, ...

Solution: Identify a and d: a = 7, d = 10 - 7 = 3 Apply the formula: T n = a + (n - 1)d = 7 + (n - 1)3 = 7 + 3n - 3 = 3n + 4 Answer: T n = 3n + 4 Question 4: Find the 10th term of the sequence defined by T n = 2n 2 - 1.