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.
• Verify solutions by substituting them back into the original equation.
• Formulate and solve linear equations from simple word problems.

Lesson summary

Algebraic expressions and equations are fundamental tools for problem-solving in mathematics and in everyday life. They allow us to represent unknown quantities and relationships, leading to powerful techniques for calculating and predicting outcomes. In South Africa, understanding these concepts is crucial for learners as they prepare for further studies in mathematics, science, engineering, and economics, and for navigating everyday financial and logistical challenges. From calculating the best deals at the supermarket to understanding loan interest rates, algebra provides the necessary skills.

Lesson notes

2.1 Algebraic Expressions: Building Blocks An algebraic expression is a combination of variables (letters representing unknown values), constants (fixed numerical values), and mathematical operations (+, -, ×, ÷).

Terms: The individual parts of an expression separated by addition or subtraction signs are called terms. For example, in the expression `3x + 5y - 2`, the terms are `3x`, `5y`, and `-2`.

Like Terms: Like terms have the same variable raised to the same power. For example, `3x` and `-7x` are like terms, but `3x` and `3x²` are not. Similarly, `5y²` and `-2y²` are like terms, but `5y²` and `5y` are not. Constants are always considered like terms.

Simplifying Expressions: Simplifying an expression means combining like terms to write it in its most compact form. This is achieved by adding or subtracting the coefficients (the numerical part of the term) of the like terms.

Example 1: Simplify the expression `5x + 2y - 3x + 7y - 1`.

Step 1: Identify like terms. `5x` and `-3x` are like terms. `2y` and `7y` are like terms. `-1` is a constant term and is by itself in this expression.

Step 2: Combine like terms. `5x - 3x = (5 - 3)x = 2x` `2y + 7y = (2 + 7)y = 9y` Step 3: Write the simplified expression. `2x + 9y - 1` Example 2: Simplify the expression `2(x + 3) - 4x + 1`.

Step 1: Apply the distributive property. The distributive property states that `a(b + c) = ab + ac`. `2(x + 3) = 2 x + 2 * 3 = 2x + 6` Step 2: Rewrite the expression. `2x + 6 - 4x + 1` Step 3: Identify like terms. `2x` and `-4x` are like terms. `6` and `1` are like terms (constants).

Step 4: Combine like terms. `2x - 4x = (2 - 4)x = -2x` `6 + 1 = 7` Step 5: Write the simplified expression. `-2x + 7` 2.2 Linear Equations: Finding the Unknown A linear equation is an equation where the highest power of the variable is

1. The goal of solving a linear equation is to isolate the variable on one side of the equation to find its value. This is achieved using inverse operations.

Inverse Operations: The inverse operation of addition is subtraction, and vice versa. The inverse operation of multiplication is division, and vice versa.

Solving Equations: To solve an equation, perform the same operations on both sides of the equation to maintain equality.

Example 3: Solve the equation `x + 5 = 12`.

Step 1: Identify the operation being performed on the variable. `x` is being added to by

5. Step 2: Perform the inverse operation on both sides of the equation. The inverse of adding 5 is subtracting 5. `x + 5 - 5 = 12 - 5` Step 3: Simplify. `x = 7` Example 4: Solve the equation `3x - 2 = 10`.

Step 1: Isolate the term containing the variable. Add 2 to both sides. `3x - 2 + 2 = 10 + 2` `3x = 12` Step 2: Isolate the variable. Divide both sides by 3. `3x / 3 = 12 / 3` `x = 4` Example 5: Solve the equation `2(x - 1) = 8`.

Method 1: Distribute first Step 1: Apply the distributive property. `2 x - 2 * 1 = 8` `2x - 2 = 8` Step 2: Isolate the term containing the variable. Add 2 to both sides. `2x - 2 + 2 = 8 + 2` `2x = 10` Step 3: Isolate the variable. Divide both sides by 2. `2x / 2 = 10 / 2` `x = 5` Method 2: Divide first Step 1: Divide both sides by 2. `2(x - 1) / 2 = 8 / 2` `x - 1 = 4` Step 2: Isolate the variable. Add 1 to both sides. `x - 1 + 1 = 4 + 1` `x = 5` Verifying Solutions: To verify the solution, substitute the value of the variable back into the original equation. If the equation holds true, the solution is correct. For example, in Example 3, substitute `x = 7` into `x + 5 = 12`. `7 + 5 = 12`, which is true.

Therefore, `x = 7` is the correct solution. 2.3 Word Problems: Translating Words into Equations Many real-world problems can be represented using algebraic equations. The key is to identify the unknown quantity (assign it a variable) and translate the information given in the problem into a mathematical equation.

Example 6: Sindi has twice as many apples as John. Together they have 15 apples. How many apples does John have?

Step 1: Define the variable. Let `x` represent the number of apples John has.

Step 2: Express the other quantity in terms of the variable. Sindi has twice as many apples as John, so Sindi has `2x` apples.

Step 3: Formulate the equation. Together they have 15 apples, so `x + 2x = 15`.

Step 4: Solve the equation. `3x = 15` `x = 15 / 3` `x = 5` Step 5: Answer the question. John has 5 apples. Sindi has 2 5 = 10 apples. Guided Practice (With Solutions)

Question 1: Simplify the expression `7a - 3b + 2a + 5b - 4`.

Solution: Step 1: Identify like terms. `7a` and `2a` are like terms. `-3b` and `5b` are like terms. `-4` is a constant term.

Step 2: Combine like terms. `7a + 2a = (7 + 2)a = 9a` `-3b + 5b = (-3 + 5)b = 2b` Step 3: Write the simplified expression. `9a + 2b - 4` Question 2: Solve the equation `4y + 1 = 13`.

Solution: Step 1: Isolate the term containing the variable. Subtract 1 from both sides. `4y + 1 - 1 = 13 - 1` `4y = 12` Step 2: Isolate the variable. Divide both sides by 4.