Lesson Notes By Weeks and Term v5 - Grade 8

Lesson Video

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Performance objectives

• Simplify algebraic expressions by combining like terms and using the distributive property.
• Solve linear equations of the form ax + b = c.
• Solve linear equations involving brackets.
• Verify solutions to linear equations by substitution.
• Apply algebraic equations to solve simple word problems.

Lesson summary

Algebraic expressions and equations are fundamental tools in mathematics that allow us to represent and solve problems involving unknown quantities. Understanding these concepts is crucial for success in higher-level mathematics, as well as in various real-life situations. From budgeting your pocket money to calculating distances or even understanding the spread of a disease, algebraic thinking is essential. In South Africa, understanding finances, particularly with fluctuating exchange rates and price increases, requires the ability to manipulate and solve equations. This week, we will focus on simplifying algebraic expressions and solving simple linear equations.

Lesson notes

2.1 Algebraic Expressions: An algebraic expression is a combination of variables, constants, and mathematical operations (addition, subtraction, multiplication, division). Variables are symbols (usually letters like x, y, or z) that represent unknown quantities. Constants are fixed numerical values.

Example: 3x + 2y - 5 is an algebraic expression. Here, x and y are variables, 3 and 2 are coefficients (the numbers multiplying the variables), and -5 is a constant. 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).

Example 1: 5x and -2x are like terms (both have x to the power of 1).

Example 2: 3y² and 7y² are like terms (both have y to the power of 2).

Example 3: 4x and 4x² are not like terms (x is to the power of 1 in the first term and to the power of 2 in the second term).

Example 4: 6 and -9 are like terms (both are constants).

Combining Like Terms: To combine like terms, add or subtract their coefficients while keeping the variable part the same.

Example: 5x + 2x - 3x = (5 + 2 - 3)x = 4x 2.3 The Distributive Property: The distributive property states that a(b + c) = ab + ac. This means that we multiply the term outside the parentheses by each term inside the parentheses. This is essential when dealing with expressions containing brackets.

Example 1: 2(x + 3) = 2 x + 2 * 3 = 2x + 6 Example 2: -3(y - 4) = -3 y + (-3) * (-4) = -3y + 12 Example 3: 5(2x + 1) - 2(x - 3) = (5 2x + 5 1) + (-2 x + (-2) * (-3)) = 10x + 5 - 2x + 6 = (10x - 2x) + (5 + 6) = 8x + 11 2.4 Algebraic Equations: An algebraic equation is a 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.

Example: 2x + 3 = 7 is an algebraic equation. 2.5 Solving Linear Equations: A linear equation is an equation where the highest power of the variable is

1. To solve a linear equation, we want to isolate the variable on one side of the equation. We do this by performing the same operation on both sides of the equation to maintain the equality.

Steps for Solving Linear Equations: Simplify both sides: Combine like terms and use the distributive property if necessary.

Isolate the variable term: Add or subtract the same number from both sides to get the variable term alone on one side.

Isolate the variable: Multiply or divide both sides by the coefficient of the variable to get the variable by itself.

Verify the solution: Substitute the solution back into the original equation to check if it makes the equation true.

Example 1: Solve for x: x + 5 = 9 Subtract 5 from both sides: x + 5 - 5 = 9 - 5 Simplify: x = 4 Verification: Substitute x = 4 back into the original equation: 4 + 5 =

9. This is true, so the solution is correct.

Example 2: Solve for y: 2y - 3 = 7 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 Verification: Substitute y = 5 back into the original equation: 2(5) - 3 = 10 - 3 =

7. This is true, so the solution is correct.

Example 3: Solve for z: 3(z + 2) = 15 Use the distributive property: 3 z + 3 * 2 = 15 Simplify: 3z + 6 = 15 Subtract 6 from both sides: 3z + 6 - 6 = 15 - 6 Simplify: 3z = 9 Divide both sides by 3: 3z / 3 = 9 / 3 Simplify: z = 3 Verification: Substitute z = 3 back into the original equation: 3(3 + 2) = 3(5) =

1

5. This is true, so the solution is correct. Guided Practice (With Solutions)

Question 1: Simplify the expression: 4x + 7 - 2x + 3 Solution: Identify like terms: 4x and -2x are like terms, and 7 and 3 are like terms.

Combine like terms: (4x - 2x) + (7 + 3)

Simplify: 2x + 10

Commentary: The key here is to correctly identify and group the like terms before performing the addition and subtraction.

Question 2: Solve for a: a - 8 = -2 Solution: Add 8 to both sides: a - 8 + 8 = -2 + 8 Simplify: a = 6 Verification: Substitute a = 6 into the original equation: 6 - 8 = -

2. This is true.

Commentary: Remember to perform the same operation on both sides of the equation to maintain balance.

Question 3: Solve for b: 5b + 10 = 25 Solution: Subtract 10 from both sides: 5b + 10 - 10 = 25 - 10 Simplify: 5b = 15 Divide both sides by 5: 5b / 5 = 15 / 5 Simplify: b = 3 Verification: Substitute b = 3 into the original equation: 5(3) + 10 = 15 + 10 =

2

5. This is true.

Commentary: Follow the order of operations in reverse to isolate the variable. First, deal with addition/subtraction, then multiplication/division.

Question 4: Solve for c: 2(c - 1) = 8 Solution: Use the distributive property: 2c - 21 = 8 Simplify: 2c - 2 = 8 Add 2 to both sides: 2c - 2 + 2 = 8 + 2 Simplify: 2c = 10 Divide both sides by 2: 2c / 2 = 10 / 2 Simplify: c = 5 Verification: Substitute c = 5 into the original equation: 2(5 - 1) = 2(4) =

8. This is true.

Commentary: Remember to distribute correctly before attempting to isolate the variable.