Lesson Notes By Weeks and Term v5 - Grade 8

Lesson Video

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Performance objectives

• Simplify algebraic expressions by combining like terms.
• Expand algebraic expressions using the distributive property.
• Solve linear equations in one variable using inverse operations.
• Verify solutions to linear equations by substitution.
• Translate simple word problems into algebraic equations.

Lesson summary

Algebraic expressions and equations form the foundation for more advanced mathematical concepts. Understanding these concepts is crucial for success in higher grades and has wide-ranging applications in everyday life. In South Africa, this understanding can help with budgeting, understanding financial statements, and even making informed decisions about business ventures. For example, calculating profit margins for a small spaza shop requires understanding algebraic concepts. This week, we will be focusing on simplifying algebraic expressions and solving simple linear equations. These skills are vital for critical thinking and problem-solving abilities.

Lesson notes

Algebraic Expressions An algebraic expression is a mathematical phrase that contains variables, constants, and operations. A variable is a symbol (usually a letter) that represents an unknown value. A constant is a fixed number. Operations include addition, subtraction, multiplication, and division.

Like Terms: Like terms are terms that have the same variable(s) raised to the same power. Only like terms can be combined.

Simplifying Expressions: Simplifying an expression means rewriting it in a simpler form by combining like terms.

Example 1: Simplify the expression: `3x + 5y - 2x + y - 4` Step 1: Identify like terms. `3x` and `-2x` are like terms. `5y` and `y` are like terms. `-4` is a constant term.

Step 2: Combine like terms. `(3x - 2x) + (5y + y) - 4` `1x + 6y - 4` `x + 6y - 4` Therefore, the simplified expression is `x + 6y - 4`.

Distributive Property: The distributive property states that `a(b + c) = ab + ac`. This property is used to expand expressions where a term is multiplied by an expression in parentheses.

Example 2: Expand the expression: `2(x + 3)` Step 1: Distribute the 2 to each term inside the parentheses. `2 x + 2 * 3` Step 2: Simplify. `2x + 6` Therefore, the expanded expression is `2x + 6`.

Example 3: Expand and simplify the expression: `3(2a - 1) + 4a` Step 1: Distribute the 3 to each term inside the parentheses. `(3 2a) - (3 * 1) + 4a` `6a - 3 + 4a` Step 2: Combine like terms. `6a + 4a - 3` `10a - 3` Therefore, the simplified expression is `10a - 3`. Equations An equation is a statement that two expressions are equal. It contains an equals sign (=). A solution to an equation is a value for the variable that makes the equation true.

Linear Equations in One Variable: A linear equation in one variable is an equation that can be written in the form `ax + b = c`, where `a`, `b`, and `c` are constants and `x` is the variable.

Solving Linear Equations: To solve a linear equation, we use inverse operations to isolate the variable on one side of the equation. Inverse operations "undo" each other. For example, addition and subtraction are inverse operations, and multiplication and division are inverse operations.

Example 4: Solve the equation: `x + 5 = 12` Step 1: Isolate x by subtracting 5 from both sides of the equation. `x + 5 - 5 = 12 - 5` `x = 7` Therefore, the solution to the equation is `x = 7`.

Example 5: Solve the equation: `2x - 3 = 7` Step 1: Add 3 to both sides of the equation to isolate the term with x. `2x - 3 + 3 = 7 + 3` `2x = 10` Step 2: Divide both sides of the equation by 2 to isolate x. `2x / 2 = 10 / 2` `x = 5` Therefore, the solution to the equation is `x = 5`.

Example 6: Solve the equation: `(x/3) + 1 = 4` Step 1: Subtract 1 from both sides of the equation. `(x/3) + 1 - 1 = 4 - 1` `x/3 = 3` Step 2: Multiply both sides of the equation by 3. `(x/3) 3 = 3 * 3` `x = 9` Therefore, the solution to the equation is `x = 9`.

Verifying Solutions: To verify a solution, substitute the value of the variable back into the original equation. If the equation is true, then the solution is correct.

Example 7: Verify the solution `x = 7` for the equation `x + 5 = 12`. Substitute x = 7 into the equation: `7 + 5 = 12` `12 = 12` Since the equation is true, the solution `x = 7` is correct. Translating Word Problems To solve word problems, we need to translate the words into algebraic expressions and equations.

Example 8: Thabo has a certain number of apples. He gives 5 apples to his friend, and now he has 8 apples left. How many apples did Thabo have initially?

Step 1: Define a variable. Let `x` be the number of apples Thabo had initially.

Step 2: Write an equation. Thabo started with `x` apples, gave away 5, and has 8 left, so the equation is `x - 5 = 8`.

Step 3: Solve the equation. `x - 5 + 5 = 8 + 5` `x = 13` Therefore, Thabo initially had 13 apples. Guided Practice (With Solutions)

Question 1: Simplify the expression: `5a + 2b - 3a + 4b - 1` Solution: Step 1: Identify like terms. `5a` and `-3a` are like terms. `2b` and `4b` are like terms. `-1` is a constant term.

Step 2: Combine like terms. `(5a - 3a) + (2b + 4b) - 1` `2a + 6b - 1` Therefore, the simplified expression is `2a + 6b - 1`.

Question 2: Expand the expression: `-3(y - 2)` Solution: Step 1: Distribute the -3 to each term inside the parentheses. `(-3 y) - (-3 * 2)` `-3y + 6` Therefore, the expanded expression is `-3y + 6`.

Question 3: Solve the equation: `4x + 2 = 14` Solution: Step 1: Subtract 2 from both sides of the equation. `4x + 2 - 2 = 14 - 2` `4x = 12` Step 2: Divide both sides of the equation by 4. `4x / 4 = 12 / 4` `x = 3` Therefore, the solution to the equation is `x = 3`.

Question 4: Verify the solution `x = 3` for the equation `2x - 1 = 5`.

Solution: Substitute x = 3 into the equation: `2(3) - 1 = 5` `6 - 1 = 5` `5 = 5` Since the equation is true, the solution `x = 3` is correct.

Question 5: A farmer has a certain number of chickens. He sells 10 chickens at the market, and now he has 25 chickens left.