Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Application of Linear and quadratic equations to capital market etc.
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Application of Linear and quadratic equations to capital market etc.
Download the Lessonotes Mobile Nigeria 2025 app for faster lesson access on Android and iPhone.
Subject: General Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 2
Theme: Algebraic Process
This page supports the lesson note with a companion video and a short classroom-ready summary.
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Solve simultaneouslinear and quadraticequations Solve word problems on linear, quadratic and simultaneous linear and quadratic equation Solve problems on linearequation in volvingcapital markets
Interest Calculation): Mr. Okoro invests N500,000 in a savings bond at an annual simple interest rate of 8%. Calculate the interest earned and the total amount after 3 years.
Solution: Principal (P) = N500,000 Rate (R) = 8% Time (T) = 3 years Interest (I) = $\frac{PRT}{100}$ $I = \frac{500,000 \times 8 \times 3}{100}$ $I = 5,000 \times 8 \times 3$ $I = 120,000$ Interest earned = N120,
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0. Total Amount (A) = P + I $A = 500,000 + 120,000$ $A = 620,000$ Total amount after 3 years = N620,
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0. Example 2.4.2 (Finding Unknown Rate in Capital Market): A microfinance bank in Lagos offered Mrs. Eze a loan of N150,
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0. She repaid a total of N187,500 after 2.5 years. What was the simple interest rate charged by the bank?
Solution: Principal (P) = N150,000 Amount (A) = N187,500 Time (T) = 2.5 years Interest (I) = A - P = $187,500 - 150,000 = N37,500$. Using the formula $I = \frac{PRT}{100}$: $37,500 = \frac{150,000 \times R \times 2.5}{100}$ $37,500 = 1,500 \times R \times 2.5$ $37,500 = 3,750R$ $R = \frac{37,500}{3,750}$ $R = 10$ The simple interest rate was 10%. 2.
5. Cost, Revenue, and Profit Analysis (Business Application) Linear equations are fundamental in basic business analysis.
Total Cost (TC): $TC = FC + VC \times Q$ Where FC = Fixed Cost (costs that don't change with production volume, e.g., rent), VC = Variable Cost per unit (cost per item produced), Q = Quantity produced.
Total Revenue (TR): $TR = P \times Q$ Where P = Selling Price per unit, Q = Quantity sold.
Profit (Prof): $Prof = TR - TC$ Break-even Point: The point where Profit = 0, meaning $TR = TC$. Example 2.5.1 (Profit Analysis): A small-scale shoe manufacturer in Aba has fixed costs of N80,000 per month. The variable cost of producing one pair of shoes is N3,
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0. Each pair of shoes sells for N5,000. a) Write an equation for the total cost of producing $x$ pairs of shoes. b) Write an equation for the total revenue from selling $x$ pairs of shoes. c) How many pairs of shoes must be sold to break even?
Solution:** Let $x$ be the number of pairs of shoes. a)
Total Cost (TC): $TC = 80,000 + 3,000x$ b)
Total Revenue (TR): $TR = 5,000x$ c) Break-even point occurs when Profit = 0, i.e., $TR = TC$. $5,000x = 80,000 + 3,000x$ $5,000x - 3,000x = 80,000$ $2,000x = 80,000$ $x = \frac{80,000}{2,000}$ $x = 40$ The manufacturer must sell 40 pairs of shoes to break even. This section provides in-depth explanations and worked examples of the core concepts required for the lesson. 2.
1. Review of Linear Equations A linear equation is an algebraic equation in which each term has an exponent of 1, and when plotted, it forms a straight line. The standard form for a linear equation in one variable is $ax + b = 0$, where $a \neq 0$.
Solving Technique: Isolate the variable using inverse operations (addition/subtraction, multiplication/division).
Word Problem Translation: Identify the unknown quantity, assign a variable, and translate the problem statement into an algebraic equation. Example 2.1.1 (Linear Equation Word Problem): A trader at Onitsha Main Market sells a bag of rice for N25,
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0. His profit on each bag is N3,
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0. If his total sales for the day amounted to N350,000, how many bags did he sell? And what was his total profit?
Solution: Let $x$ be the number of bags of rice sold. Selling price per bag = N25,000 Total sales = N350,000 Equation for number of bags: $25,000x = 350,000$ $x = \frac{350,000}{25,000}$ $x = 14$ bags.
Total profit: Profit per bag = N3,500 Total profit = $14 \times 3,500 = N49,000$. 2.
2. Review of Quadratic Equations A quadratic equation is an algebraic equation of the second degree, meaning it contains at least one term in which the unknown variable is squared. The standard form is $ax^2 + bx + c = 0$, where $a \neq 0$.
Solving Techniques: Factorisation: Express the quadratic as a product of two linear factors.
Completing the Square: Manipulate the equation to form a perfect square trinomial.
General Formula (Quadratic Formula): $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Example 2.2.1 (Quadratic Equation Word Problem): The length of a rectangular plot of land in Abuja is 5 meters more than its width. If the area of the plot is 84 square meters, find the dimensions of the plot.
Solution: Let the width of the plot be $w$ meters. Then the length of the plot will be $(w + 5)$ meters. Area = Length $\times$ Width $84 = (w + 5)w$ $84 = w^2 + 5w$ $w^2 + 5w - 84 = 0$ Solving by factorisation: Find two numbers that multiply to -84 and add to 5 (which are 12 and -7). $(w + 12)(w - 7) = 0$ So, $w + 12 = 0$ or $w - 7 = 0$ $w = -12$ or $w = 7$. Since width cannot be negative, $w = 7$ meters. Length = $w + 5 = 7 + 5 = 12$ meters. Dimensions are 7 meters by 12 meters. 2.
3. Solving Simultaneous Linear and Quadratic Equations This involves finding the values of variables that satisfy both a linear equation and a quadratic equation at the same time. The most common and effective method is the substitution method. * Steps for Substitution Method:
1. Isolate a variable in the linear equation. This is usually the easiest step, as the linear equation is simpler.
2. Substitute the expression for this variable into the quadratic equation. This will result in a single quadratic equation in one variable.
3. Solve the resulting quadratic equation using any suitable method (factorisation, completing the square, or general formula) to find the values of that variable.
4. Substitute back the obtained values into the linear equation (or the isolated expression from step 1) to find the corresponding values of the other variable.
5. Check the solutions by substituting both pairs into both original equations. Example 2.3.1 (Simultaneous Linear and Quadratic Equations): Solve the following simultaneous equations: 1) $y = x + 1$ (Linear) 2) $y = x^2 - 5$ (Quadratic)
Solution:
1. Substitute (1) into (2): Since $y$ is already isolated in both, we can equate the right-hand sides. $x + 1 = x^2 - 5$
2. Rearrange into standard quadratic form: $x^2 - x - 5 - 1 = 0$ $x^2 - x - 6 = 0$
3. Solve the quadratic equation: By factorisation, find two numbers that multiply to -6 and add to -1 (which are -3 and 2). $(x - 3)(x + 2) = 0$ So, $x - 3 = 1$ (Linear) 2) $y = x^2 - 5$ (Quadratic)
Solution:
1. Substitute (1) into (2): Since $y$ is already isolated in both, we can equate the right-hand sides. $x + 1 = x^2 - 5$
2. Rearrange into standard quadratic form: $x^2 - x - 5 - 1 = 0$ $x^2 - x - 6 = 0$
3. Solve the quadratic equation: By factorisation, find two numbers that multiply to -6 and add to -1 (which are -3 and 2). $(x - 3)(x + 2) = 0$ So, $x - 3 = 0$ or $x + 2 = 0$ $x = 3$ or $x = -2$
4. Substitute $x$ values back into the linear equation (1) to find $y$: If $x = 3$: $y = 3 + 1 = 4$ First solution: $(x, y) = (3, 4)$ If $x = -2$: $y = -2 + 1 = -1$ Second solution: $(x, y) = (-2, -1)$ Example 2.3.2 (Word Problem leading to Simultaneous Linear and Quadratic): The sum of two numbers is 10, and the sum of their squares is
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8. Find the two numbers.
Solution: Let the two numbers be $x$ and $y$. 1)
Sum of two numbers is 10: $x + y = 10$ (Linear Equation) 2)
Sum of their squares is 58: $x^2 + y^2 = 58$ (Quadratic Equation) From equation (1), isolate $y$: $y = 10 - x$ (Equation 3)
Substitute Equation (3) into Equation (2): $x^2 + (10 - x)^2 = 58$ $x^2 + (100 - 20x + x^2) = 58$ $2x^2 - 20x + 100 = 58$ $2x^2 - 20x + 100 - 58 = 0$ $2x^2 - 20x + 42 = 0$ Divide by 2 to simplify: $x^2 - 10x + 21 = 0$ Solve the quadratic equation by factorisation (find two numbers that multiply to 21 and add to -10, which are -7 and -3): $(x - 7)(x - 3) = 0$ So, $x = 7$ or $x = 3$. Substitute these $x$ values back into Equation (3) to find $y$: If $x = 7$: $y = 10 - 7 = 3$ If $x = 3$: $y = 10 - 3 = 7$ The two numbers are 3 and 7. 2.
4. Application of Linear Equations to Capital Markets The capital market is a financial market where long-term funds are raised and invested, typically for periods longer than a year. It deals with instruments like stocks (equities), bonds, and debentures. Linear equations are primarily used in calculating simple interest, basic investment returns, and cost-revenue-profit analysis.
Key Terms: Principal (P): The initial amount of money borrowed or invested.
Interest (I): The cost of borrowing money or the income earned from lending/investing money.
Rate (R): The percentage at which interest is charged or earned per period (usually annually). It is often expressed as a decimal or a fraction (R/100).
Time (T): The duration for which the money is borrowed or invested, usually in years.
Amount (A): The total sum after interest is added to the principal ($A = P + I$).
Simple Interest Formula: $I = \frac{PRT}{100}$ (where R is a percentage, e.g., 5 for 5%) OR $I = PRT$ (where R is a decimal, e.g., 0.05 for 5%)
Note: For NERDC SS curriculum, using $I = PRT/100$ is common where R is the percentage value.* Amount Formula: $A = P + I$ $A = P + \frac{PRT}{100}$ $A = P(1 + \frac{RT}{100})$ These formulas can be rearranged using linear equation principles to find any unknown variable if the others are known. Example 2.4.1 (Simple Interest Calculation): Mr. Okoro invests N500,000 in a savings bond at an annual simple interest rate of 8%. Calculate the interest earned and the total amount after 3 years.
Solution: Principal (P) = N500,000 Rate (R) = 8% Time (T) = 3 years Interest (I) = $\frac{PRT}{100}$ $I = \frac{500,000 \times 8 \times 3}{100}$ $I = 5,000 \times 8 \times 3$ $I = 120,000$ Interest earned = N120,
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0. Total Amount (A) = P + I $A = 500,000 + 120,000$ $A = 620,000$ Total amount after 3 years = N620,000. *Example 2.4.2 This section outlines the activities for both the teacher and the students during the lesson. 3.
1. Introduction (10 minutes)
Teacher Activity: Initiate a brief review of solving linear and quadratic equations individually. Pose quick questions to assess prior knowledge (e.g., "Solve $2x - 7 = 3$" or "Factorise $x^2 - 4x + 3 = 0$"). Introduce the concept of solving simultaneous linear and quadratic equations by posing a scenario where two conditions must be met by the same variables (e.g., a path and a circular field, or intersecting functions). Briefly introduce the concept of capital markets and how mathematical equations can model financial situations in Nigeria (e.g., loans, investments, business profit). State the lesson objectives clearly, linking them to real-life relevance.
Student Activity: Respond to review questions. Listen attentively and participate in the introductory discussion. Note down the lesson objectives. 3.
2. Lesson Development (45 minutes)
Phase 1: Simultaneous Linear and Quadratic Equations (20 minutes)
Teacher Activity: Present a clear, step-by-step demonstration of solving a system of one linear and one quadratic equation on the whiteboard (e.g., Example 2.3.1). Emphasize the substitution method, detailing each step: isolating a variable, substituting, solving the resulting quadratic, and back-substituting. Guide students through the solution of a word problem that translates into simultaneous linear and quadratic equations (e.g., Example 2.3.2). Stress the importance of defining variables correctly. Circulate among students, providing support and clarification.
Student Activity: Observe the teacher's demonstration and take notes. Attempt to solve a similar problem in their notebooks or on individual whiteboards as the teacher guides them. Ask questions for clarification. Work in pairs to translate and set up equations for the word problem.
Phase 2: Applications to Capital Market (25 minutes)
Teacher Activity: Explain key capital market terms: Principal, Interest, Rate, Time, Amount. Use simple analogies related to savings or small loans. Introduce the simple interest formula ($I = \frac{PRT}{100}$) and the amount formula ($A = P(1 + \frac{RT}{100})$). Work through Example 2.4.1 (calculating interest and amount) and Example 2.4.2 (finding an unknown rate), explicitly showing how linear equations are derived and solved. Introduce the concepts of Fixed Cost, Variable Cost, Total Cost, Total Revenue, and Profit, demonstrating how linear equations model these business aspects (Example 2.5.1). Explain the break-even point. Encourage students to relate these concepts to local Nigerian businesses (e.g., market traders, small-scale manufacturers).
Student Activity: Define key terms in their notebooks. Copy and understand the formulas. Work through the examples with the teacher, asking questions where necessary. Participate in discussions on how these concepts apply to Nigerian business scenarios. 3.
3. Class Discussion and Wrap-up (10 minutes)
Teacher Activity: Facilitate a short discussion, inviting students to share their understanding and any challenges encountered. Summarize the key takeaways for the lesson, reiterating the practical value of algebraic equations. Assign independent practice questions as homework.
Student Activity: Contribute to the discussion, asking questions or sharing insights. Take note of assigned homework.
This topic has strong relevance to various aspects of Nigerian life: Personal Finance and Investment: Savings and Loans: Understanding simple interest calculations is crucial for managing personal savings accounts, evaluating small loans from cooperative societies (Esusu/Ajo), or microfinance banks. Students can apply linear equations to calculate the future value of their savings or the total cost of a loan over time.
Basic Investment Decisions: When considering informal investments or fixed deposits in Nigerian banks, understanding how interest accrues helps individuals project returns and make informed choices. For instance, comparing the returns from different investment vehicles over a specific period using the simple interest formula. Entrepreneurship and Small Business Management: Costing and Pricing: Small business owners in Nigeria, from market traders to artisans, can use linear equations to determine their total production costs (fixed + variable), set competitive prices, and calculate their profit margins.
Break-even Analysis: The concept of a break-even point is vital for any startup or existing business (e.g., a pure water sachet producer, a local baker). Using linear equations, they can determine the minimum sales volume required to cover all costs and avoid losses, which is critical for business sustainability in a competitive market.
Agriculture and Resource Management: Land Use Optimization: Farmers might use quadratic equations to optimize the layout of their farms for maximum yield within given perimeter constraints. For example, determining the dimensions of a rectangular plot to maximize area with a limited length of fencing.
Resource Allocation: In scenarios involving limited resources and competing demands, simultaneous equations can help model optimal allocation (though often simplified to linear for SS3). For example, allocating fertilizer types to different crops to maximize yield under budget constraints (can lead to linear programming in advanced stages).