Lesson Notes By Weeks and Term v3 - Junior Secondary 3
Simultaneous linear equations
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Simultaneous linear equations
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Subject: General Mathematics
Class: Junior Secondary 3
Term: 3rd Term
Week: 5
Theme: Algebraic Processes
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Compile table of values for simultaneous linear functions Solve problems in volving linear simultaneous equations in two variables graphically Solve simultaneous linear equations in two variables using elimination method Apply elimination method to solve problems in volving real life activities Solve simultaneous linear equations in two variables using substitution method Apply substitution method to solve problems in volving real life activities.
2.1 Definition of a Linear Equation in Two Variables A linear equation in two variables is an equation that can be written in the form $ax + by = c$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero. The variables are typically $x$ and $y$. When plotted on a coordinate plane, a linear equation in two variables forms a straight line. 2.2 Definition of Simultaneous Linear Equations Simultaneous linear equations are a set of two or more linear equations that contain the same variables. The goal is to find the values of these variables that satisfy all the equations in the set concurrently. For JSS3, the focus is primarily on systems of two linear equations in two variables.
Example: $2x + y = 7$ (Equation 1) $x - y = 2$ (Equation 2) The solution to this system is a pair of values $(x, y)$ that makes both equations true. 2.3 Methods of Solving Simultaneous Linear Equations There are three primary methods for solving simultaneous linear equations in two variables: A. Graphical Method The graphical method involves plotting the lines represented by each equation on the same coordinate plane. The point where the two lines intersect is the solution to the system.
Steps:
1. Compile a table of values for each equation. To do this, choose a range of $x$-values (e.g., -2, -1, 0, 1, 2) and substitute each into the equation to find the corresponding $y$-values. It is often easier to rearrange the equation to make $y$ the subject (e.g., $y = mx + c$).
2. Plot the points from the table of values for each equation on a Cartesian coordinate plane.
3. Draw a straight line through the plotted points for each equation. Use a ruler to ensure accuracy.
4. Identify the point of intersection. The coordinates $(x, y)$ of this point represent the solution to the simultaneous equations.
5. Verify the solution by substituting the $x$ and $y$ values back into the original equations.
Worked Example (Graphical Method): Solve the following simultaneous equations graphically: 1. $x + y = 5$ 2. $2x - y = 4$ Step 1: Compile tables of values.
For Equation 1: $x + y = 5 \implies y = 5 - x$ | $x$ | $y = 5 - x$ | $(x, y)$ | | :-- | :---------- | :------- | | -1 | $5 - (-1) = 6$ | (-1, 6) | | 0 | $5 - 0 = 5$ | (0, 5) | | 1 | $5 - 1 = 4$ | (1, 4) | | 2 | $5 - 2 = 3$ | (2, 3) | | 3 | $5 - 3 = 2$ | (3, 2) | | 4 | $5 - 4 = 1$ | (4, 1) | For Equation 2: $2x - y = 4 \implies y = 2x - 4$ | $x$ | $y = 2x - 4$ | $(x, y)$ | | :-- | :----------- | :------- | | -1 | $2(-1) - 4 = -6$ | (-1, -6) | | 0 | $2(0) - 4 = -4$ | (0, -4) | | 1 | $2(1) - 4 = -2$ | (1, -2) | | 2 | $2(2) - 4 = 0$ | (2, 0) | | 3 | $2(3) - 4 = 2$ | (3, 2) | | 4 | $2(4) - 4 = 4$ | (4, 4) | Step 2 & 3: Plot the points and draw the lines. (Teacher should demonstrate plotting these points on a grid and drawing the lines using a ruler. Use graph paper or a blackboard with a grid.)
Step 4: Identify the point of intersection. Observe the graph. The two lines intersect at the point (3, 2).
Step 5: Verify the solution. Substitute $x = 3$ and $y = 2$ into both original equations: For Equation 1: $x + y = 5 \implies 3 + 2 = 5$ (True)
For Equation 2: $2x - y = 4 \implies 2(3) - 2 = 6 - 2 = 4$ (True) Since the values satisfy both equations, the solution $(x, y) = (3, 2)$ Use graph paper or a blackboard with a grid.)
Step 4: Identify the point of intersection. Observe the graph. The two lines intersect at the point (3, 2).
Step 5: Verify the solution. Substitute $x = 3$ and $y = 2$ into both original equations: For Equation 1: $x + y = 5 \implies 3 + 2 = 5$ (True)
For Equation 2: $2x - y = 4 \implies 2(3) - 2 = 6 - 2 = 4$ (True) Since the values satisfy both equations, the solution $(x, y) = (3, 2)$ is correct. B. Elimination Method The elimination method involves manipulating the equations (multiplying by constants) so that when the equations are added or subtracted, one of the variables is eliminated.
Steps:
1. Align the equations such that like terms (x-terms, y-terms, constants) are vertically aligned.
2. Choose a variable to eliminate. Determine if you can make the coefficients of one variable the same (or additive inverses) in both equations.
3. Multiply one or both equations by a suitable constant(s) to make the coefficients of the chosen variable numerically equal.
4. Add or subtract the equations to eliminate the chosen variable. If the coefficients have the same sign, subtract one equation from the other. If the coefficients have opposite signs, add the equations.
5. Solve the resulting single-variable equation for the remaining variable.
6. Substitute the value found in step 5 back into one of the original equations to find the value of the other variable.
7. Verify the solution by substituting both values into the other original equation.
Worked Example (Elimination Method): Solve the following simultaneous equations using the elimination method: 1. $3x + 2y = 12$ (Equation 1) 2. $2x - 3y = 5$ (Equation 2)
Step 1 & 2: Align and choose variable to eliminate. Equations are already aligned. Let's choose to eliminate $y$.
Step 3: Multiply equations. To make coefficients of $y$ equal (or additive inverses), multiply Equation 1 by 3 and Equation 2 by 2. $3 \times (3x + 2y = 12) \implies 9x + 6y = 36$ (Equation 3) $2 \times (2x - 3y = 5) \implies 4x - 6y = 10$ (Equation 4)
Step 4: Add or subtract equations. Since the coefficients of $y$ ($+6y$ and $-6y$) have opposite signs, add Equation 3 and Equation 4: $(9x + 6y) + (4x - 6y) = 36 + 10$ $9x + 4x + 6y - 6y = 46$ $13x = 46$ Step 5: Solve for the remaining variable. $13x = 46$ $x = \frac{46}{13}$ Step 6: Substitute back. Substitute $x = \frac{46}{13}$ into Equation 1 (or Equation 2, or 3, or 4): $3(\frac{46}{13}) + 2y = 12$ $\frac{138}{13} + 2y = 12$ $2y = 12 - \frac{138}{13}$ $2y = \frac{12 \times 13 - 138}{13}$ $2y = \frac{156 - 138}{13}$ $2y = \frac{18}{13}$ $y = \frac{18}{13 \times 2}$ $y = \frac{18}{26}$ $y = \frac{9}{13}$ So, the solution is $x = \frac{46}{13}$ and $y = \frac{9}{13}$.
Step 7: Verify the solution. Substitute $x = \frac{46}{13}$ and $y = \frac{9}{13}$ into Equation 2: $2x - 3y = 5$ $2(\frac{46}{13}) - 3(\frac{9}{13}) = \frac{92}{13} - \frac{27}{13}$ $= \frac{92 - 27}{13} = \frac{65}{13} = 5$ (True) The solution is correct. C. Substitution Method The substitution method involves expressing one variable in terms of the other from one equation, and then substituting this expression into the second equation.
Steps:
1. Choose one equation and express one variable in terms of the other. It is usually easiest to choose an equation where a variable has a coefficient of 1 or -1.
2. Substitute this expression into the other equation. This will result in a single-variable equation.
3. Solve the resulting single-variable equation for the variable.
4. Substitute the value found in step 3 back into the expression derived in step 1 to find the value of the second variable.
5. Verify the solution by substituting both values into the original equations.
Worked Example (Substitution Method): Solve the following simultaneous equations using the substitution method: 1. $2x + y = 7$ (Equation 1) 2. $x - y = 2$ (Equation 2) *Step 1: Express one variable in terms of the other equation. This will result in a single-variable equation.
3. Solve the resulting single-variable equation for the variable.
4. Substitute the value found in step 3 back into the expression derived in step 1 to find the value of the second variable.
5. Verify the solution by substituting both values into the original equations.
Worked Example (Substitution Method): Solve the following simultaneous equations using the substitution method: 1. $2x + y = 7$ (Equation 1) 2. $x - y = 2$ (Equation 2)
Step 1: Express one variable in terms of the other. From Equation 2, it is easy to express $x$ in terms of $y$: $x = 2 + y$ (Equation 3)
Step 2: Substitute the expression into the other equation.
Substitute Equation 3 into Equation 1: $2(2 + y) + y = 7$ Step 3: Solve the resulting single-variable equation. $4 + 2y + y = 7$ $4 + 3y = 7$ $3y = 7 - 4$ $3y = 3$ $y = 1$ Step 4: Substitute back to find the second variable. Substitute $y = 1$ into Equation 3: $x = 2 + y$ $x = 2 + 1$ $x = 3$ So, the solution is $x = 3$ and $y = 1$.
Step 5: Verify the solution. Substitute $x = 3$ and $y = 1$ into both original equations: For Equation 1: $2x + y = 7 \implies 2(3) + 1 = 6 + 1 = 7$ (True)
For Equation 2: $x - y = 2 \implies 3 - 1 = 2$ (True) The solution is correct.
Phase 1: Introduction and Recall (10 minutes)
Teacher Activity: Begin by revisiting linear equations in one variable and plotting points on a Cartesian plane. Ask students what a solution to an equation means. Introduce the idea that sometimes there are multiple unknowns and multiple relationships, leading to simultaneous equations.
Student Activity: Students answer questions on plotting points and identifying coordinates. They recall the meaning of a solution to a linear equation.
Phase 2: Graphical Method (25 minutes)
Teacher Activity: Explain that two linear equations form two lines, and their intersection is the common solution. Demonstrate step-by-step how to compile a table of values for a given linear function (e.g., $y = x + 3$). Emphasize choosing a good range of $x$-values (e.g., from -3 to 3). Demonstrate how to plot these points accurately on a grid and draw the straight line. Introduce a second equation (e.g., $y = -x + 1$) and guide students to compile its table of values. Demonstrate plotting the second line on the same grid and identifying the intersection point as the solution. Emphasize the need for clear labelling of axes, scales, and lines.
Student Activity: In pairs, students compile tables of values for given linear equations. Each student plots the points and draws the lines for a pair of simultaneous equations on graph paper. Students identify the point of intersection and state the solution. Students verify their graphical solution algebraically.
Phase 3: Elimination Method (25 minutes)
Teacher Activity: Explain the principle of elimination: making coefficients of one variable the same (or additive inverse) and then adding/subtracting equations to remove that variable. Work through a simple example with coefficients already suitable for elimination (e.g., $x+y=5$, $x-y=1$). Work through a more complex example where multiplication is needed for one or both equations (e.g., $2x+y=7$, $3x-2y=0$). Highlight how to choose which variable to eliminate and what to multiply by. Stress the importance of neatness and systematic working.
Student Activity: Students follow the teacher's examples, taking notes. Students attempt practice problems provided by the teacher in their notebooks. Students discuss in small groups how to identify the best variable to eliminate in different scenarios.
Phase 4: Substitution Method (25 minutes)
Teacher Activity: Explain the principle of substitution: isolating a variable in one equation and replacing it in the other. Work through an example where one variable is already isolated or easily isolated (e.g., $y = x+2$, $2x+y=11$). Work through an example where rearrangement is needed from the beginning (e.g., $3x+y=7$, $x+2y=9$). Compare the advantages and disadvantages of substitution vs. elimination for different types of equations.
Student Activity: Students follow the teacher's examples and actively participate in problem-solving steps. Students solve simultaneous equations using the substitution method from practice problems. Students choose a preferred method and justify their choice for a given problem.
Phase 5: Real-Life Application & Problem Solving (15 minutes)
Teacher Activity: Present word problems related to Nigerian contexts (e.g., buying items at the market, cost of transport). Guide students on how to translate these problems into simultaneous equations.
Student Activity: Students work in groups to formulate equations from word problems and solve them using any of the learned methods. They present their solutions and interpretation to the class.
Phase 6: Consolidation and Wrap-up (10 minutes)
Teacher Activity: Recap all three methods. Answer questions. Assign homework.
Student Activity: Students ask clarifying questions. Complete an exit ticket or quick assessment.
Materials: Whiteboard/Blackboard and markers/chalk Graph paper Rulers Pencils Worksheets with practice problems Real-life scenarios/word problems
Market Pricing in Nigeria: Scenario: A customer buys a certain quantity of oranges and mangoes at a local market in Ibadan. Another customer buys different quantities of the same fruits. Given the total cost for each customer, simultaneous equations can be used to determine the individual price of one orange and one mango. This is directly applicable to market transactions for common Nigerian foodstuffs like tomatoes, pepper, yam tubers, etc.
Integration: Students can be asked to simulate a market scenario, collect "data" (number of items and total cost), and then solve for unit prices. This connects mathematics to economics and entrepreneurship. Budgeting and Financial Planning for Nigerian Households: Scenario: A family sets aside a fixed amount for transport and food each week. They know that a bus trip costs a certain amount, and a meal costs another amount. If they track their weekly spending for transport and food, simultaneous equations can help them figure out the individual cost per bus trip and per meal, or to plan how many of each they can afford within their budget.
Integration: Students can discuss typical household expenses in Nigeria and use hypothetical data to formulate and solve equations, understanding the practical application of resource allocation and financial literacy. Resource Allocation in Small Businesses (e.g., Tailoring, Farming): Scenario: A tailor in Aba uses two types of fabric, A and B, to make different designs of traditional attire. Design 1 requires a specific amount of fabric A and B, while Design 2 requires different amounts. Given the total available quantities of fabric A and B, and the requirements for each design, simultaneous equations can determine how many pieces of Design 1 and Design 2 can be produced.
Integration: This helps students see how mathematics is used in small and medium-sized enterprises (SMEs) which are crucial to the Nigerian economy, fostering entrepreneurial thinking.