Lesson Notes By Weeks and Term v5 - Grade 9
Equations, inequalities and number patterns – Week 3 focus
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Equations, inequalities and number patterns – Week 3 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 9
Term: 2nd Term
Week: 3
Theme: General lesson support
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This week, we will delve deeper into equations, inequalities, and number patterns, building upon the foundational knowledge you acquired in previous grades. Understanding these concepts is crucial not only for academic success in Mathematics but also for navigating real-world situations involving budgeting, problem-solving, and logical reasoning. For example, understanding equations can help you calculate the best cell phone contract, inequalities can help you determine the affordability of different loan options, and number patterns can help you predict trends in data analysis.
2.1 Equations with Variables on Both Sides When solving equations with variables on both sides, the goal is to isolate the variable on one side of the equation. This is achieved by performing inverse operations to both sides to eliminate the variable from one side and combine like terms.
Remember the golden rule: whatever you do to one side, you must do to the other!
Example 1: Solve for x: 3x + 5 = x - 1 Step 1: Eliminate the variable from one side.
Subtract 'x' from both sides: 3x + 5 - x = x - 1 - x This simplifies to: 2x + 5 = -1 Step 2: Isolate the variable term.
Subtract 5 from both sides: 2x + 5 - 5 = -1 - 5 This simplifies to: 2x = -6 Step 3: Solve for x.
Divide both sides by 2: 2x / 2 = -6 / 2 Therefore, x = -3 Example 2 (Brackets): Solve for y: 2(y + 3) = 5y - 4 Step 1: Expand the brackets. Multiply 2 by each term inside the brackets: 2 y + 2 3 = 5y - 4 This simplifies to: 2y + 6 = 5y - 4 Step 2: Eliminate the variable from one side.
Subtract 2y from both sides: 2y + 6 - 2y = 5y - 4 - 2y This simplifies to: 6 = 3y - 4 Step 3: Isolate the variable term.
Add 4 to both sides: 6 + 4 = 3y - 4 + 4 This simplifies to: 10 = 3y Step 4: Solve for y.
Divide both sides by 3: 10 / 3 = 3y / 3 Therefore, y = 10/3 (or 3.33 recurring)
Example 3 (Fractions): Solve for m: (m/2) + 1 = (m/3) - 2 Step 1: Eliminate the fractions. Find the least common multiple (LCM) of the denominators (2 and 3), which is
6. Multiply every term in the equation by 6: 6 (m/2) + 6 1 = 6 (m/3) - 6 2 This simplifies to: 3m + 6 = 2m - 12 Step 2: Eliminate the variable from one side.
Subtract 2m from both sides: 3m + 6 - 2m = 2m - 12 - 2m This simplifies to: m + 6 = -12 Step 3: Isolate the variable.
Subtract 6 from both sides: m + 6 - 6 = -12 - 6 Therefore, m = -18 2.2 Compound Inequalities Compound inequalities involve two inequalities joined by "and" or "or". "And" Inequalities: Represent values that satisfy both inequalities. The solution is the intersection of the two inequalities.
Example: x > 2 AND x 3 Example 1 ("And"): Solve and represent graphically: -3 ≤ x + 2 8 Step 1: Solve each inequality separately. 2x - 1 2x x 8 => x > 4 Step 2: Represent on a number line. Draw a number line. Use an open circle at 2 and shade the region to the left. Use an open circle at 4 and shade the region to the right.
Interval Notation: (-∞, 2) ∪ (4, ∞) 2.3 Linear Number Patterns A linear number pattern has a constant difference between consecutive terms.
The general term is given by: T n = an + b where: T n is the nth term a is the common difference n is the term number b is a constant Example 1: Find the general term and the 10th term of the sequence: 2, 5, 8, 11, ...
Step 1: Find the common difference (a). 5 - 2 = 3, 8 - 5 = 3, 11 - 8 =
3. So, a =
3. Step 2: Substitute 'a' into the general term formula: T n = 3n + b Step 3: Find 'b'. Substitute n = 1 and T 1 = 2 into the formula: 2 = 3(1) + b 2 = 3 + b b = -1 Step 4: Write the general term: T n = 3n - 1 Step 5: Find the 10th term (T 10 ): T 10 = 3(10) - 1 = 30 - 1 = 29 2.4 Quadratic Number Patterns A quadratic number pattern has a constant second difference between consecutive terms. To find the nth term, T n = an 2 + bn + c, we need to find the values of a, b, and c. We use the method of constant differences.
Example 1: Find the general term of the sequence: 2, 7, 14, 23, ...
Step 1: Find the first differences: 7-2 = 5, 14-7 = 7, 23-14 = 9 Step 2: Find the second differences: 7-5 = 2, 9-7 =
2. The second difference is constant and equal to
2. Step 3: Use the following relationships: 2a = second difference => 2a = 2 => a = 1 3a + b = first difference between term 1 and term 2 => 3(1) + b = 5 => b = 2 a + b + c = first term => 1 + 2 + c = 2 => c = -1 Step 4: Write the general term: T n = n 2 + 2n - 1 Guided Practice (With Solutions)
Question 1: Solve for x: 4x - 3 = 2x + 7 Solution: Subtract 2x from both sides: 4x - 3 - 2x = 2x + 7 - 2x => 2x - 3 = 7 Add 3 to both sides: 2x - 3 + 3 = 7 + 3 => 2x = 10 Divide both sides by 2: 2x / 2 = 10 / 2 => x = 5
Commentary: We isolated x by performing inverse operations in the correct order.
Question 2: Represent the inequality -1 n = 3n + b Substitute n = 1 and T 1 = 7: 7 = 3(1) + b => b = 4 General term: T n = 3n + 4
Commentary: This is a linear sequence because the difference between consecutive terms is constant.
Question 4: Find the next two terms in the following number pattern: 1, 4, 9, 16, ...
Solution: Calculate the differences between the terms: 4-1 = 3, 9-4 = 5, 16-9 = 7 The differences are increasing by 2 each time, so this is a quadratic pattern. The next difference is 7+2 = 9, and the term after that is 9+2 =
1
1. Add the differences to the last term in the sequence: 16 + 9 = 25, 25 + 11 = 36 The next two terms are 25 and 36.