Lesson Notes By Weeks and Term v5 - Grade 7

Lesson Video

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Performance objectives

• Simplify expressions by combining like terms.
• Evaluate expressions by substituting values.
• Solve simple one-step equations.
• Create equations from word problems.
• Check your solutions to equations.

Lesson summary

This week, we will delve deeper into the world of algebraic expressions and simple equations. This is a fundamental area of mathematics that builds upon what you learned in previous grades. Understanding algebra is crucial because it provides a powerful tool for representing and solving problems in various real-life scenarios. Think about calculating the cost of groceries, determining the distance travelled at a constant speed, or even figuring out the ingredients needed for a recipe if you want to make a bigger or smaller batch. Algebra provides the framework for these kinds of calculations.

Lesson notes

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

Variable: A letter (e.g., x, y, a, b) that represents an unknown number.

Constant: A number that stands alone (e.g., 5, -3, 1/2).

Coefficient: The number that multiplies a variable (e.g., in the term 3x, the coefficient is 3).

Like Terms: Terms that have the same variable(s) raised to the same power (e.g., 2x and 5x are like terms; 3y² and -y² are like terms). Note that 2x and 2x² are NOT like terms because the powers are different.

Simplifying Algebraic Expressions: Simplifying an algebraic expression means combining like terms. We can only add or subtract like terms.

Example 1: Simplify 3x + 2y - x + 5y.

Step 1: Identify like terms: 3x and -x are like terms; 2y and 5y are like terms.

Step 2: Group like terms together: (3x - x) + (2y + 5y).

Step 3: Combine the coefficients of the like terms: (3 - 1)x + (2 + 5)y. Remember that if a term only has the variable written, the coefficient is

1. So -x is the same as -1x.

Step 4: Simplify: 2x + 7y.

Example 2: Simplify 5a - 2b + 4 - a + 3b -

1. Step 1: Identify like terms: 5a and -a are like terms; -2b and 3b are like terms; 4 and -1 are like terms.

Step 2: Group like terms together: (5a - a) + (-2b + 3b) + (4 - 1).

Step 3: Combine the coefficients: (5 - 1)a + (-2 + 3)b + (4 - 1).

Step 4: Simplify: 4a + b +

3. Evaluating Algebraic Expressions: Evaluating an algebraic expression means finding its numerical value by substituting given values for the variables.

Example 3: Evaluate the expression 2x + 3y if x = 4 and y = -

2. Step 1: Substitute the given values for the variables: 2(4) + 3(-2).

Step 2: Perform the multiplication: 8 + (-6).

Step 3: Perform the addition: 8 - 6 =

2. Therefore, the value of the expression is

2. Example 4: Evaluate the expression a² - b if a = -3 and b =

5. Step 1: Substitute: (-3)² - 5 Step 2: Remember that (-3)² = (-3)(-3) =

9. So, 9 - 5 Step 3: Simplify: 4 2.2 Simple Equations An equation is a statement that two algebraic expressions are equal. It contains an equals sign (=).

Solving an Equation: Finding the value of the variable that makes the equation true. This value is called the solution.

Solving One-Step Equations: To solve a one-step equation, we need to isolate the variable on one side of the equation by performing the inverse operation. Remember, whatever operation you perform on one side of the equation, you must perform the same operation on the other side to keep the equation balanced.

Example 5: Solve x + 5 =

1

2. Step 1: The variable x is being added to

5. The inverse operation of addition is subtraction. Subtract 5 from both sides of the equation: x + 5 - 5 = 12 -

5. Step 2: Simplify: x =

7. Therefore, the solution is x =

7. Example 6: Solve y - 3 =

8. Step 1: The variable y is being subtracted by

3. The inverse operation of subtraction is addition.

Add 3 to both sides of the equation: y - 3 + 3 = 8 +

3. Step 2: Simplify: y =

1

1. Therefore, the solution is y =

1

1. Example 7: Solve 2z =

1

0. Step 1: The variable z is being multiplied by

2. The inverse operation of multiplication is division.

Divide both sides of the equation by 2: 2z / 2 = 10 /

2. Step 2: Simplify: z =

5. Therefore, the solution is z =

5. Example 8: Solve p / 4 =

3. Step 1: The variable p is being divided by

4. The inverse operation of division is multiplication.

Multiply both sides of the equation by 4: (p / 4) 4 = 3

4. Step 2: Simplify: p =

1

2. Therefore, the solution is p =

1

2. Formulating Equations from Word Problems: Translate the word problem into an algebraic equation. Identify the unknown quantity (and assign it a variable) and the relationship between the known and unknown quantities.

Example 9: Thando has some sweets. Her friend gives her 7 more sweets, and now she has 15 sweets in total. How many sweets did Thando have originally?

Step 1: Let 's' be the number of sweets Thando had originally.

Step 2: Write the equation: s + 7 =

1

5. Step 3: Solve the equation: s + 7 - 7 = 15 -

7. Step 4: Simplify: s =

8. Therefore, Thando originally had 8 sweets.

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

Example 10: Check if x = 7 is the solution to the equation x + 5 =

1

2. Step 1: Substitute x = 7 into the equation: 7 + 5 =

1

2. Step 2: Simplify: 12 =

1

2. Since the equation is true, x = 7 is the correct solution. Guided Practice (With Solutions)

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

Combine like terms: 3x + 7 Therefore, the simplified expression is 3x +

7. This step requires understanding of combining coefficients.

Question 2: Evaluate the expression 3a - 2b if a = 5 and b = -1.