Lesson Notes By Weeks and Term v5 - Grade 8

Lesson Video

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Performance objectives

• Simplify algebraic expressions by combining like terms.
• Solve linear equations with one variable using inverse operations.
• Solve word problems that can be modelled with linear equations.
• Substitute numerical values into algebraic expressions.
• Identify variables and constants in an expression.

Lesson summary

Algebraic expressions and equations are fundamental building blocks in mathematics. They provide a powerful way to represent relationships between quantities and solve problems involving unknown values. In everyday life in South Africa, we use these concepts, often without even realizing it, when calculating costs, sharing resources fairly, or planning budgets. This week, we will focus on simplifying algebraic expressions and solving simple linear equations. These skills will allow you to tackle more complex problems in mathematics and other subjects like science and accounting.

Lesson notes

2.1 Algebraic Expressions An algebraic expression is a combination of variables, constants, and mathematical operations (addition, subtraction, multiplication, division, powers).

Variable: A symbol (usually a letter, like x, y, or a) that represents an unknown value.

Constant: A fixed numerical value (e.g., 5, -3, 1/2).

Coefficient: The numerical factor of a term containing a variable (e.g., in the term 3x, 3 is the coefficient).

Like Terms: Terms that have the same variable(s) raised to the same power (e.g., 2x and 5x are like terms, but 2x and 2x 2 are not).

Example 1: Identifying parts of an algebraic expression Consider the expression: 4x + 7 - 2y + x 2 - 3 Terms: 4x, 7, -2y, x 2 , -3 Variables: x, y Constants: 7, -3 Coefficients: 4 (for the term 4x), -2 (for the term -2y), 1 (for the term x 2 , as x 2 is the same as 1 x 2 ) 2.2 Simplifying Algebraic Expressions by Combining Like Terms To simplify an algebraic expression, we combine like terms. This means adding or subtracting the coefficients of terms with the same variable(s) raised to the same power.

Example 2: Simplifying an algebraic expression Simplify: 5a + 3b - 2a + b - 4 Step 1: Identify like terms.

Like terms with a: 5a and -2a Like terms with b: 3b and b Constant term: -4 Step 2: Combine like terms. (5a - 2a) + (3b + b) - 4 3a + 4b - 4 Final Answer: 3a + 4b - 4 Why does this work? It's based on the distributive property. Think of 5a + 3b - 2a + b as (5 a) + (3 b) + (-2 a) + (1 b). We can rearrange the terms because addition is commutative. Then, we're essentially saying we have 5 "a's" and take away 2 "a's", leaving us with 3 "a's". 2.3 Solving Linear Equations A linear equation is an equation where the highest power of the variable is

1. Solving a linear equation means finding the value of the variable that makes the equation true. We do this by isolating the variable on one side of the equation using inverse operations. Remember, whatever operation you perform on one side of the equation, you must perform on the other side to maintain equality.

Inverse Operations: Addition and subtraction are inverse operations. Multiplication and division are inverse operations.

Example 3: Solving a linear equation Solve for x: x + 5 = 12 Step 1: Isolate x by subtracting 5 from both sides of the equation. x + 5 - 5 = 12 - 5 x = 7 Final Answer: x = 7 Example 4: Solving a linear equation with multiplication Solve for y: 3y = 18 Step 1: Isolate y by dividing both sides of the equation by 3. 3y/3 = 18/3 y = 6 Final Answer: y = 6 Example 5: Solving a linear equation with multiple steps Solve for z: 2z - 4 = 6 Step 1: Isolate the term with z by adding 4 to both sides of the equation. 2z - 4 + 4 = 6 + 4 2z = 10 Step 2: Isolate z by dividing both sides of the equation by 2. 2z/2 = 10/2 z = 5 Final Answer: z = 5 2.4 Solving Word Problems with Linear Equations Many real-world problems can be represented and solved using linear equations. To solve a word problem, follow these steps: Read the problem carefully and identify the unknown value. Assign a variable to represent the unknown value. Translate the word problem into a linear equation. Look for keywords that indicate mathematical operations (e.g., "sum" means addition, "difference" means subtraction, "product" means multiplication, "quotient" means division). Solve the linear equation. Answer the question in the context of the word problem. Make sure your answer makes sense.

Example 6: Word Problem Thando has x number of apples. Sindi has twice as many apples as Thando. Together they have 15 apples. How many apples does Thando have?

Step 1: Define the variable. Let x be the number of apples Thando has. Sindi has 2x apples.

Step 2: Write the equation. x + 2x = 15 Step 3: Solve the equation. 3x = 15 x = 15/3 x = 5 Step 4: Answer the question. Thando has 5 apples. Guided Practice (With Solutions)

Question 1: Simplify the expression: 7x - 3 + 2x + 5 - x Solution: Step 1: Identify like terms: 7x, 2x, -x, -3, 5 Step 2: Combine like terms: (7x + 2x - x) + (-3 + 5)

Step 3: Simplify: 8x + 2 Final Answer: 8x + 2

Commentary: We grouped the 'x' terms together and the constant terms together to simplify the expression. This makes it easier to understand and use.

Question 2: Solve for m: m - 8 = -2 Solution: Step 1: Add 8 to both sides: m - 8 + 8 = -2 + 8 Step 2: Simplify: m = 6 Final Answer: m = 6

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

Question 3: Solve for p: 4p + 2 = 10 Solution: Step 1: Subtract 2 from both sides: 4p + 2 - 2 = 10 - 2 Step 2: Simplify: 4p = 8 Step 3: Divide both sides by 4: 4p/4 = 8/4 Step 4: Simplify: p = 2 Final Answer: p = 2

Commentary: This question required two steps. We first isolated the term with the variable and then isolated the variable itself.

Question 4: Sifiso buys a loaf of bread and a packet of polony for R

3

5. The bread costs R

1

2. How much does the polony cost?

Solution: Step 1: Define the variable: Let c be the cost of the polony.