Lesson Notes By Weeks and Term v3 - Junior Secondary 2
Simple equations
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Simple equations
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Subject: General Mathematics
Class: Junior Secondary 2
Term: 2nd Term
Week: 3
Theme: Algebraic Processes
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This topic introduces learners to the fundamental concept of simple equations, a cornerstone of algebra. It builds upon prior knowledge of expressions, variables, constants, and basic arithmetic operations. Understanding simple equations is crucial as it equips learners with essential problem-solving skills applicable across various subjects and real-life scenarios. The ability to model real-world situations algebraically and solve for unknown quantities is a vital mathematical literacy skill.
Performance Objective (from curriculum): Students should be able to solve problems on simple equations.
Definition of an Equation: An equation is a mathematical statement that shows two expressions are equal. It is characterized by the presence of an equality sign (=). For example, `3x + 5 = 11`.
Definition of a Simple Equation: A simple equation (or linear equation in one variable) is an equation that contains only one variable, and the highest power of this variable is
1. The goal when solving a simple equation is to find the value of the unknown variable that makes the equation true.
Components of a Simple Equation: Variable: A symbol (usually a letter like `x`, `y`, `a`, `p`, etc.) that represents an unknown number.
Constant: A number whose value does not change.
Coefficient: A number that multiplies a variable. For example, in `3x`, `3` is the coefficient of `x`.
Expression: A combination of variables, constants, and mathematical operations (e.g., `3x + 5`, `11`). Equality Sign (=): Indicates that the expression on the left side has the same value as the expression on the right side.
The Principle of Balance: Solving an equation is like balancing a scale. To keep the scale balanced, whatever operation is performed on one side of the equation must also be performed on the other side. The objective is to isolate the variable on one side of the equation.
Inverse Operations: To isolate the variable, inverse operations are used: Addition is the inverse of subtraction. Subtraction is the inverse of addition. Multiplication is the inverse of division. Division is the inverse of multiplication.
Steps for Solving Simple Equations: Type 1: Equations involving addition or subtraction Concept: To isolate the variable, perform the inverse operation of the constant term on both sides of the equation.
Example 1: Solve `x + 7 = 15` To undo adding 7, subtract 7 from both sides. `x + 7 - 7 = 15 - 7` `x = 8` Example 2 (Nigerian Context): A food vendor in Ibadan sold 9 plates of Amala and had 12 plates remaining. How many plates did she have initially? Let `p` be the initial number of plates of Amala. `p - 9 = 12` To undo subtracting 9, add 9 to both sides. `p - 9 + 9 = 12 + 9` `p = 21` The vendor initially had 21 plates of Amala.
Type 2: Equations involving multiplication or division Concept: To isolate the variable, perform the inverse operation of the coefficient or the divisor on both sides.
Example 3: Solve `4y = 20` To undo multiplying by 4, divide both sides by 4. `4y / 4 = 20 / 4` `y = 5` Example 4 (Nigerian Context): A group of 5 students shared a bag of oranges equally, and each student received 8 oranges. How many oranges were in the bag initially? Let `o` be the initial number of oranges. `o / 5 = 8` To undo dividing by 5, multiply both sides by 5. `(o / 5) 5 = 8 5` `o = 40` There were 40 oranges in the bag initially.
Type 3: Equations involving a combination of operations (Two-step equations)
Concept: First, undo any addition or subtraction (to move constant terms away from the variable term). Second, undo any multiplication or division (to isolate the variable).
Example 5: Solve `2x + 3 = 11` Step 1: Subtract 3 from both sides. `2x + 3 - 3 = 11 - 3` `2x = 8` Step 2: Divide both sides by 2. `2x / 2 = 8 / 2` `x = 4` Example 6 (Nigerian Context): At a market in Kano, Aminu bought 3 metres of Ankara fabric and paid an extra N500 for tailoring service. His total bill was N
3
5
0
0. What was the cost per metre of the Ankara fabric? Let `c` be the cost per metre of Ankara fabric. `3c + 500 = 3500` Step 1: Subtract 500 from both sides. `3c 3` `2x = 8` Step 2: Divide both sides by 2. `2x / 2 = 8 / 2` `x = 4` Example 6 (Nigerian Context): At a market in Kano, Aminu bought 3 metres of Ankara fabric and paid an extra N500 for tailoring service. His total bill was N
3
5
0
0. What was the cost per metre of the Ankara fabric? Let `c` be the cost per metre of Ankara fabric. `3c + 500 = 3500` Step 1: Subtract 500 from both sides. `3c + 500 - 500 = 3500 - 500` `3c = 3000` Step 2: Divide both sides by 3. `3c / 3 = 3000 / 3` `c = 1000` The cost per metre of Ankara fabric was N
1
0
0
0. Type 4: Equations with variables on both sides Concept: Collect all terms containing the variable on one side of the equation and all constant terms on the other side. Then, proceed as with Type 3 equations.
Example 7: Solve `5x - 2 = 3x + 8` Step 1: Subtract `3x` from both sides to collect variables on the left. `5x - 3x - 2 = 3x - 3x + 8` `2x - 2 = 8` Step 2: Add 2 to both sides to collect constants on the right. `2x - 2 + 2 = 8 + 2` `2x = 10` Step 3: Divide both sides by 2. `2x / 2 = 10 / 2` `x = 5` Example 8 (Nigerian Context): Two mobile network providers, Glo and MTN, offer different data plans. Glo charges N100 per GB plus a N500 monthly access fee. MTN charges N150 per GB plus a N200 monthly access fee. For how many GB used would the monthly costs be equal? Let `g` be the number of GB used.
Glo cost: `100g + 500` MTN cost: `150g + 200` Set them equal: `100g + 500 = 150g + 200` Step 1: Subtract `100g` from both sides. `100g - 100g + 500 = 150g - 100g + 200` `500 = 50g + 200` Step 2: Subtract `200` from both sides. `500 - 200 = 50g + 200 - 200` `300 = 50g` Step 3: Divide both sides by 50. `300 / 50 = 50g / 50` `g = 6` The monthly costs would be equal for 6 GB of data usage.
Preparation: Ensure the classroom is conducive for learning. Prepare chalk/markers and a whiteboard/blackboard. Have pre-written examples and practice questions ready. Consider using a simple balance scale model (physical or drawn) to illustrate the concept of equality.
Introduction (5 minutes): Teacher Activity: Begin by reviewing the concepts of algebraic expressions, variables, and constants. Ask learners to identify these components in given expressions (e.g., `5x + 3`). Introduce the equality sign and explain how it transforms an expression into an equation, emphasizing that an equation seeks to find an unknown value. Use simple analogies like "a balanced seesaw" or "sharing items equally" to introduce the concept of balance in equations.
Learner Activity: Learners participate by answering questions on expressions, variables, and constants. They listen actively to the introduction of equations and contribute to analogies. Content Development & Explanation (20 minutes): Teacher Activity: Present the definition of a simple equation and its components. Explain the principle of inverse operations and the need to maintain balance by performing the same operation on both sides of the equation. Demonstrate step-by-step solutions for various types of simple equations (Type 1, 2, 3, and 4 as detailed in Section 2) using clear worked examples, including Nigerian context examples. Emphasize checking solutions by substituting the found value back into the original equation. Encourage questions and clarify any misconceptions immediately.
Learner Activity: Learners actively listen, take notes, and ask questions for clarification. They observe the step-by-step demonstrations and try to follow the logic. Some learners may be invited to attempt a step or check a solution on the board.
Guided Practice (15 minutes): Teacher Activity: Provide a few carefully selected problems for learners to solve under supervision. Walk around the classroom, observe their work, provide hints, and correct common errors. Encourage peer-to-peer discussion and problem-solving. Select a few learners to present their solutions on the board, explaining their steps.
Learner Activity: Learners attempt the guided practice problems individually or in pairs. They discuss their approaches with classmates and present their solutions to the class when called upon.
Conclusion & Wrap-up (5 minutes): Teacher Activity: Briefly summarize the key steps for solving simple equations. Reiterate the importance of checking solutions. Assign homework and prepare for the next lesson.
Learner Activity: Learners ask any final questions, note down homework, and review the day's lesson. The teacher will present these questions and guide learners through the solution process, allowing learners to attempt steps and provide answers.
Question 1: A farmer in Plateau State harvested some Irish potatoes. After selling 8 bags, he had 15 bags left. How many bags of Irish potatoes did he harvest initially?
Solution: Let `x` represent the initial number of bags of Irish potatoes. The problem can be translated into the equation: `x - 8 = 15` To isolate `x`, add 8 to both sides of the equation (inverse of subtraction): `x - 8 + 8 = 15 + 8` `x = 23`
Commentary: This is a Type 1 equation involving subtraction. The key is to apply the inverse operation (addition) to both sides to maintain balance and isolate the variable.
Question 2: Four children from a community in Rivers State contributed equally to buy a gift for their teacher. If the total cost of the gift was N2800, how much did each child contribute?
Solution: Let `y` represent the amount each child contributed. The problem can be translated into the equation: `4y = 2800` To isolate `y`, divide both sides of the equation by 4 (inverse of multiplication): `4y / 4 = 2800 / 4` `y = 700`
Commentary: This is a Type 2 equation involving multiplication. The inverse operation of division is applied to find the unknown quantity.
Question 3: A tailor in Kaduna collects payment for sewing clothes. For a particular order, she charged N500 per dress and an additional N1500 for special fabric sourcing. If the total bill for a customer was N4000, how many dresses did the customer order?
Solution: Let `d` represent the number of dresses ordered. The problem can be translated into the equation: `500d + 1500 = 4000` First, subtract 1500 from both sides (to undo addition): `500d + 1500 - 1500 = 4000 - 1500` `500d = 2500` Next, divide both sides by 500 (to undo multiplication): `500d / 500 = 2500 / 500` `d = 5`
Commentary: This is a Type 3 (two-step) equation. Learners must correctly apply the order of inverse operations: first addition/subtraction, then multiplication/division.
Question 4: Solve the equation: `3p + 7 = p + 15` Solution: The equation is: `3p + 7 = p + 15` Collect variable terms on one side. Subtract `p` from both sides: `3p - p + 7 = p - p + 15` `2p + 7 = 15` Collect constant terms on the other side.
Subtract 7 from both sides: `2p + 7 - 7 = 15 - 7` `2p = 8` Isolate `p`.
Divide both sides by 2: `2p / 2 = 8 / 2` `p = 4`
Commentary: This is a Type 4 equation with variables on both sides. The strategy is to consolidate variable terms to one side and constant terms to the other before isolating the variable.
Market Transactions and Budgeting: Application: Simple equations are used extensively in Nigerian markets and homes for managing money. For instance, a buyer wants to know how many kilograms of garri they can buy with a certain amount of money after setting aside funds for transport. Or a small business owner needs to calculate how many items to sell to reach a profit target after deducting fixed costs.
Example: "Madam Ngozi sells palm oil. If she wants to make a profit of N2,000 and each bottle of palm oil costs N800 to produce and sells for N1,000, how many bottles does she need to sell?" (Let `x` be bottles. Profit per bottle = N
2
0
0. Equation: `200x = 2000`).
Resource Allocation and Distribution: Application: In communities, simple equations can help in sharing resources (e.g., land, relief materials, borehole water supply schedules) fairly among a certain number of beneficiaries. It can also be used to calculate proportions for mixtures, such as in local food processing or building construction.
Example: "A community received 150 bags of rice as relief. If 15 bags were kept for community events, and the rest were distributed equally among `x` families, with each family receiving 5 bags, how many families received rice?" (Equation: `(150 - 15) / x = 5` or `135 / x = 5`).
Vocational Skills and Entrepreneurship: Application: Tailors calculate fabric requirements, carpenters determine wood lengths, mechanics estimate costs of repairs. Understanding equations helps them to plan, quote prices, and manage inventory efficiently.
Example: "A carpenter needs to cut a plank of wood 2.5 metres long into two pieces. One piece must be 50 cm longer than the other. What are the lengths of the two pieces?" (Convert to cm: 250 cm. Let `x` be the shorter piece.
Equation: `x + (x + 50) = 250`).