Lesson Notes By Weeks and Term v5 - Grade 8

Lesson Video

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Performance objectives

• Simplify expressions by combining like terms.
• Solve linear equations with one variable.
• Substitute numerical values into expressions.
• Solve simple word problems with equations.
• Verify solutions to equations.

Lesson summary

Algebraic expressions and equations form the bedrock of more advanced mathematics. This week, we'll be focusing on manipulating and solving simple equations – a crucial skill for solving problems in everyday life. Think about budgeting your pocket money, calculating the cost of airtime, or even determining the number of bricks needed to build a small structure. All of these involve understanding and applying algebraic concepts. This is why learning about algebraic expressions and equations is so important! In South Africa, having strong mathematical skills opens doors to various career paths and enables informed decision-making in personal and professional contexts.

Lesson notes

2.1 Algebraic Expressions: An algebraic expression is a combination of numbers, variables (represented by letters), and operations (+, -, ×, ÷). A variable represents an unknown value.

Examples: 3x + 5 2y - 7z a² + 4ab - c 2.2 Like Terms: Like terms are terms that have the same variable raised to the same power. Only like terms can be combined (added or subtracted). The number in front of the variable is called the coefficient.

Examples: 3x and 5x are like terms. 2y² and -7y² are like terms. 4a and 4a² are NOT like terms (different powers of 'a'). 3ab and -2ba are like terms (ab is the same as ba).

Combining Like Terms: To combine like terms, simply add or subtract their coefficients.

Example 1: Simplify: 4x + 7x - 2x Solution: (4 + 7 - 2)x = 9x Example 2: Simplify: 5a + 3b - 2a + b Solution: (5a - 2a) + (3b + b) = 3a + 4b 2.3 Algebraic Equations: An algebraic equation is a statement that two algebraic expressions are equal. It always contains an equals sign (=).

Examples: x + 5 = 10 2y - 3 = 7 a² + b² = c² 2.4 Solving Equations: "Solving" an equation means finding the value of the variable that makes the equation true. We use inverse operations to isolate the variable on one side of the equation.

The golden rule of solving equations is: Whatever you do to one side of the equation, you must do to the other side.

Example 1: Solve for x: x + 5 = 10 Solution: Subtract 5 from both sides: x + 5 - 5 = 10 - 5 Simplify: x = 5 Example 2: Solve for y: 2y - 3 = 7 Solution: Add 3 to both sides: 2y - 3 + 3 = 7 + 3 Simplify: 2y = 10 Divide both sides by 2: 2y / 2 = 10 / 2 Simplify: y = 5 Example 3: Solve for a: 3a + 4 = 19 Solution: Subtract 4 from both sides: 3a + 4 - 4 = 19 - 4 Simplify: 3a = 15 Divide both sides by 3: 3a / 3 = 15 / 3 Simplify: a = 5 Example 4 (Incorporating integers): Solve for m: m - 7 = -2 Solution: Add 7 to both sides: m - 7 + 7 = -2 + 7 Simplify: m = 5 Example 5 (Incorporating integers): Solve for p: 2p + 5 = -1 Solution: Subtract 5 from both sides: 2p + 5 - 5 = -1 - 5 Simplify: 2p = -6 Divide both sides by 2: 2p / 2 = -6 / 2 Simplify: p = -3 2.5 Verification: To verify that your solution is correct, substitute the value you found for the variable back into the original equation. If the equation is true, your solution is correct.

Example: For the equation x + 5 = 10, we found x =

5. Verification: Substitute x = 5 into the equation: 5 + 5 = 10 Since 10 = 10, our solution is correct. Guided Practice (With Solutions)

Question 1: Simplify the expression: 6x - 2y + 4x + 5y Solution: Identify like terms: (6x and 4x) and (-2y and 5y)

Combine like terms: (6x + 4x) + (-2y + 5y) = 10x + 3y Answer: 10x + 3y

Commentary: We grouped the 'x' terms and the 'y' terms and then added their coefficients.

Question 2: Solve for z: z - 8 = 3 Solution: Add 8 to both sides: z - 8 + 8 = 3 + 8 Simplify: z = 11 Answer: z = 11

Commentary: We used the inverse operation of subtraction (addition) to isolate 'z'.

Question 3: Solve for k: 4k = 20 Solution: Divide both sides by 4: 4k / 4 = 20 / 4 Simplify: k = 5 Answer: k = 5

Commentary: We used the inverse operation of multiplication (division) to isolate 'k'.

Question 4: Solve for m: 2m + 1 = 9 Solution: Subtract 1 from both sides: 2m + 1 - 1 = 9 - 1 Simplify: 2m = 8 Divide both sides by 2: 2m / 2 = 8 / 2 Simplify: m = 4 Answer: m = 4

Commentary: We first isolated the term with 'm' and then isolated 'm' itself.

Question 5: Solve for x: x + 3 = -5 Solution: Subtract 3 from both sides: x + 3 - 3 = -5 - 3 Simplify: x = -8 Answer: x = -8

Commentary: Remember to apply the rules of integer arithmetic when working with negative numbers. Independent Practice (Questions Only)

Simplify: 9a + 2b - 5a - b Simplify: 3x² + 7x - x² + 2 Solve for p: p + 6 = 15 Solve for q: q - 4 = 7 Solve for r: 5r = 35 Solve for s: 2s + 3 = 11 Solve for t: 3t - 5 = 10 Solve for x: x + 8 = 2 Solve for y: 2y - 4 = -6 Solve for z: -3z = 12