Applications of Expressions, Equations and Inequalities

In this section, you will explore solving simultaneous linear equations in two variables using three key methods: elimination, substitution and the graphical method. You will begin by analysing two linear equations, learning how to eliminate one variable to solve the system algebraically. Substitution, where one equation is rearranged and substituted into the other, will also be practised.

You will then solve these equations graphically by plotting and identifying the point where the lines intersect. You will also apply your understanding to real- life problems, modelling these scenarios as simultaneous equations and solving them. The section emphasises practical applications, ensuring you can interpret your solutions and relate them to everyday contexts.

KEY IDEAS

• Elimination method: Add or subtract equations after manipulating coefficients to eliminate one variable, then solve for the remaining variable.

• Formulate equations: Translate the problem’s context into two linear equations involving the identified variables.

• Intersection point: The point where the two lines intersect. In simultaneous equations this is the solution, if it exists.

• Linear equation: is a mathematical equation that shows a straight-line relationship between two variables

• Simultaneous equation: is a set of two or more equations that share the same variables and are solved together.

• Substitution method: One equation is manipulated to define one variable in terms of the other. This new definition of the variable is then substituted into the second equation and solved.

In year one, you learned how to solve linear equations with one variable. Let’s revisit some examples as a quick review. This will lay the groundwork for better understanding of the concept of solving simultaneous equations.

Example 2.1

Solve 4x + 9 = 33

Solution

Given: 4x + 9 = 33 Subtract 9 from both sides to isolate the term with x :

4x + 9 − 9 = 33 − 9 4x + 0 = 24 4x = 24 Divide both sides by 4 to solve for x :

x = 24/4 x = 6 Therefore, the solution is x = 6 .

Example 2.2

Solve 7y − 12 = 2y + 8

Solution

Given: 7y − 12 = 2y + 8 Subtract 2y from both sides to get all y terms on one side:

7y − 2y − 12 = 2y − 2y + 8 5y − 12 = 8 Add 12 to both sides to isolate the term with y:

5y − 12 + 12 = 8 + 12 5y = 20 Divide both sides by 5 to solve for y:

y = 20/5 y = 4 Therefore, the solution is y = 4.

Example 2.3

Solve 3y/4 + 1/8 = 3

Solution

Given: 3y/4 + 1/8 = 3 Multiply the entire equation by 8 (LCM of 4 and 8) to remove the fractions:

8(3y_ 4 ) + 8(1_

8) = 8 × 3 6y + 1 = 24 Isolate y by subtracting 1 from both sides:

6y + 1 − 1 = 24 − 1 6y = 23 Divide both sides by 6 to solve for y:

y = 23/6 So, the solution is y = 23/6 .

Now that we’ve completed the revision, let’s explore how to solve simultaneous equations using the elimination method.

To grasp the elimination method for simultaneous equations, keep the following conditions in mind.

1. When solving simultaneous equations, you are looking for a solution which is true for both equations simultaneously, ie at the same time.

2. When we solve simultaneous linear equations with two unknowns, our goal is to find the value of each unknown that satisfies both equations at the same time.

3. Ensure that the coefficient (the number in front) of one of the unknowns is the same in both equations. This will help us eliminate that unknown.

4. We can eliminate this matching unknown by either adding or subtracting the two equations.

Perform the following activities on how to solve simultaneous equation using the Elimination Method

Activity 2.1: Solving simultaneous equation Solve x + y = 5 and x − y = 1 simultaneously.

Solution

Let:

Yellow shapes represent positive numbers.

Red shapes represent negative numbers.

x represent circular shapes.

y represent square shapes.

Constants be represented by triangular shapes.

If the variable is x and has a positive value, represent it with a yellow circle.

If x has a negative value, use a red circle.

For the variable y , use a yellow square for positive values and a red square for negative values.

For constants, represent positive values with a triangular shape and negative values with a red triangular shape Using the elimination method:

Step 1: Align the equations for clarity x + y = 5 (Equation 1) x − y = 1 (Equation 2)

Step 2: Use the specified shapes to represent the different parameters in the given equations: a yellow circle for a positive x , a yellow square for a positive y, a red square for a negative y, 5 yellow triangular shapes for positive constant in equation 1, and a yellow triangular shape for the constant in equation 2.

Step 3: Add the shapes to equations to eliminate y shape

Step 4: You will now share the 6 constants numbers between the two x variables.

x = 3

Step 5: Substitute the value of x = 3 (yellow circles) into either of the original equations to find y (yellow triangles). Using equation 1.

Now, you have 3 + y = 5

Step 6: Now taking one (circle) and one (triangle) from each side until no circles are left at the left hand side will lead to:

y = 2 Therefore, the solution is: x = 3, y = 2 Confirm that this is correct, but substituting into equation 2:

x − y = 1 3 − 2 = 1 is true, therefore, the solution is correct.

Activity 2.2: Solving simultaneous equation Solve the simultaneous equations, 2x + 3y = 12 and 4x − 3y = 6

Solution

Using the elimination method, follow these steps:

Step 1: Align the equations for clarity 2x + 3y = 12 (Equation 1) 4x − 3y = 6 (Equation 2)

Step 2: Look for a variable with the same magnitude coefficient. If the two coefficients have the same sign subtract the equations, if they are different, add them. In this case, we add the two equations to eliminate y:

(2x + 3y)+ (4x − 3y) = 12 + 6 2x + 4x = 18 6x = 18

Step 3: Now, divide both sides by 6 to find x x = 18/6 x = 3

Step 4: Substitute the value of x = 3 into either of the original equations to find y. We will use Equation 1:

2x + 3y = 12 2(3) + 3y = 12 6 + 3y = 12

Step 5: Isolate y by subtracting 6 from both sides to solve for y 6 − 6 + 3y = 12 − 6 3y = 6

Step 6: Divide both sides by 3 y = 6/3 y = 2 The solution to the simultaneous equations is: x = 3 , y = 2 To confirm that this is correct, substitute the values we have found into the equation we did not use. In this case, equation 2:

4(3) − 3(2) = 12 − 6 = 6 , as this is correct, we can be confident in our solution.

Activity 2.3: Solving simultaneous equation Solve the simultaneous equations, 2x + 3y = 12 and 4x − y = 5

Solution

Using the elimination method, follow these steps:

Step 1: Align the equations for clarity 2x + 3y = 12 (Equation 1) 4x− y = 5 (Equation 2)

Step 2: Match the coefficients To eliminate one variable, you can multiply (equation 2) by 3 to make the coefficients of y in both equations the same magnitude.

(Note, you could also have chosen to multiply (Equation 1) by 2 to make the coefficients of x the same, either one would give you the same answers.)

3(4x − y) = 3(5) This gives you:

12x − 3y = 15 (Equation 3) Now, we have:

2x + 3y = 12, (Equation 1) = The equation we have not touched yet 12x − 3y = 15 , (Equation 3) = The manipulated equation 2

Step 3: As the y coefficients have different signs; you will add Equation 1 and Equation 3 to eliminate y :

(2x + 3y) + (12x − 3y) = 12 + 15 14x = 27

Step 4: Now, divide both sides by 14 to find x :

x = 27/14

Step 5: Now, substitute x = 27/14 back into one of the original equations to find y. We will use Equation 1:

2(27_

14) + 3y = 12 54/14 + 3y = 12 54_ 14 − 54/14 + 3y = 12 − 54/14 3y = 12 − 54/14 3y = 57/7 (Divide both sides by 3) y = (57_ 7 ) ÷ 3 y = (57_ 7 ) × 1/3 = 57/21 y = 19/7 The solution to the simultaneous equations is: x = 27/14 , y = 19/7 To confirm that this is correct, substitute the values we have found into the equation we did not use. In this case, equation 2:

4(27/14) − 19/7 = 54/7 − 19/7 = 35/7 = 5 , as this is correct, we can be confident in our

solution.

For further practice, visit YouTube and search for tutorials on solving simultaneous equations using the elimination method. Watching these videos will help you better understand the steps and techniques involved. Make sure to practise what you learn.

Now that you have learned how to solve simultaneous equations using the elimination method, you can apply this knowledge to solve simultaneous equations using the substitution method with ease. Let’s move on to solving simultaneous linear equations in two variables using the substitution method.

The substitution method involves rearranging one of the equations for one variable, making it the subject of the equation, and then substituting that expression into the other equation. This transforms the system of equations into a single equation with one variable, making it easy to solve. Once we have the value for one variable, we can substitute this in to find the other unknown variable.

Follow these steps to solve simultaneous linear equations with two variables using the substitution method:

1. Rearrange one equation to solve and have one variable as the subject.

2. Substitute the expression for that variable into the other equation.

3. Simplify and solve for the remaining variable.

4. Substitute back to find the other variable.

5. Verify your solution by substituting both values back into the original equation. This step is optional, but it is good practice.

Let’s carry out the following tasks on solving simultaneous equations using the substitution method:

Activity 2.4: Solving simultaneous equation Solve the simultaneous equation 2x + y = 7 and 3x − y = 4.

Solution

Step 1: Choose an equation to rearrange to solve for one of the variables:

Choosing Equation 1: 2x + y = 7

Step 2: Solve for one variable (y):

y = 7 − 2x

Step 3: Substitute the new expression for y into Equation 2:

3x − (7 − 2x) = 4

Step 4: Simplify and solve for x:

3x − 7 + 2x = 4 5x − 7 = 4 5x = 11 x = 11/5

Step 5: Substitute x back into one of the original equations to find y. Using Equation 1:

2( 11/5 ) + y = 7 22/5 + y = 7 Subtract 22/5 from both sides of the equation to isolate y:

22_ 5 − 22/5 + y = 7− 22/5 y = 13/5 The solution to the simultaneous equations is: x = 11/5 , y = 13/5 Verification Equation 1: 2 (11/5 ) + 13/5 = 22/5 + 13/5 = 7 (True) Equation 2: 3 (11/5 ) − 13/5 = 33/5 − 13/5 = 4 (True)

Activity 2.5: Solving simultaneous equation Solve the simultaneous equation 2x + 3y = 12 and x − y = 1

Solution

2x + 3y = 12 (Equation 1) x − y = 1 (Equation 2)

Step 1: Choose an equation to solve for one variable. Choosing equation 2:

x − y = 1

Step 2: Solve for one variable (x ) x = y + 1

Step 3: Substitute the new expression for x into equation 1:

2(y + 1) + 3y = 12

Step 4: Simplify and solve for y:

2y + 2 + 3y = 12 5y + 2 = 12 5y = 10 y = 2

Step 5: Substitute y = 2 back into one of the original equations to find x.

Choosing equation 2:

x − 2 = 1 x = 3 Verify by plugging (x, y) into both original equations:

Equation 1: 2(3) + 3(2) = 6 + 6 = 12 (True) Equation 2: 3 - 2 = 1 (True)

The graphical method involves plotting linear equations on a coordinate plane to visually identify the solution to simultaneous equations. Each equation represents a straight line and the point where the lines intersect is the solution to the system of equations.

Follow the steps below to solve simultaneous equation by using graphical method:

Step 1: Write the equations in slope-intercept form:

a. Convert each equation into the form y = mx + c , where:

m is the slope (gradient) of the line.

c is the y-intercept (where the line crosses the y-axis).

Step 2: Plot the lines:

a For each equation, choose at least two values of x, and calculate the corresponding values of y.

b Plot these points on a coordinate plane and draw the line through them, extending the lines across the whole of your graph paper.

Step 3: Find the intersection point:

a After plotting both lines, observe where they intersect.

b The coordinates of this intersection point represent the solution to the system of equations.

Interpreting Graphs of Linear Equations in Two

Variables

1. If two lines intersect at a single point, the system has one unique solution.

This point is the common solution for both equations.

2. If two lines are parallel, they never intersect, meaning the system has no

solution (inconsistent system).

3. If two lines overlap (i.e., they are the same line), the system has infinitely many solutions, as all points on the line satisfy both equations.

Note that when two lines intersect, it shows where two conditions or relationships balance or agree.

Now that you have a basic understanding of how to solve simultaneous equations graphically, let’s engage in the following activities to deepen our grasp of these concepts:

Activity 2.6: Graphing simultaneous equation Solve − 2x + y = 1 and x + y = 4

Solution

Step 1: Rewrite each equation in slope-intercept form (i.e., y = mx + c ) For equation 1: − 2x + y = 1 Solve for y: y = 2x + 1 For equation 2: x + y = 4 Solve for y: y = − x + 4

Step 2: Create a table of values for each equation.

Choose values of x to calculate corresponding values of y for each equation.

For Equation 1: y = 2x + 1

Table 2.1: x and y coordinates for equ. 1 x y 0 1 1 3 2 5 For Equation 2: y = − x + 4

Table 2.2: x and y coordinates for equ. 2 x y 0 4 1 3 2 2

Step 3: Plot the points on a graph Using the tables from Step 2, plot the points on a graph:

For: y = 2x + 1 , plot the points (0, 1), (1, 3) and (2, 5).

For: y = − x + 4 , plot the points (0, 4), (1, 3) and (2, 2).

Step 4: Draw the lines Draw a straight line through the points for each equation, extending your lines across the whole of the graph paper. The lines represent the two equations as shown below:

Figure 2.1: Graph of a simultaneous equation

Step 5: Find the point of intersection The solution to the system of equations is the point where the two lines intersect. In this case, the lines intersect at the point (1, 3).

Step 6: Interpret the solution The coordinates of the intersection point (1, 3) represent the solution to the system of equations. This means that:

x = 1 and y = 3 These values satisfy both equations simultaneously.

Activity 2.7: Graphing simultaneous equation Solve the simultaneous equation 2x + y = 6 and x − y = 1 by using the graphical method

Solution:

Step 1: Rewrite each equation in slope-intercept form (i.e., y = mx + c ) For equation 1: 2x + y = 6 Solve for y: y = − 2x + 6 For equation 2: x − y = 1 Solve for y: y = x− 1

Step 2: Create a table of values for each equation.

Choose values of x to calculate corresponding values of y for each equation.

For y = − 2x + 6

Table 2.3: x and y coordinates for equ. 1 x y 0 6 1 4 2 2 For y = x− 1

Table 2.4: x and y coordinates for equ. 2 x y 0 -1 1 0 2 1

Step 3: Plot the points on a graph Using the tables from Step 2, plot the points on a graph:

For y = − 2x + 6, plot the points (0, 6), (1, 4) and (2, 2).

For y = x− 1 , plot the points (0, −1), (1, 0) and (2, 1).

Step 4: Draw the lines Draw a straight line through the points for each equation. The lines represent the two equations as shown below:

Figure 2.2: Graph of a simultaneous equation

Step 5: Find the point of intersection The solution to the system of equations is the point where the two lines intersect. In this case, the lines intersect at the point (2.3, 1.3).

Step 6: Interpret the solution The coordinates of the intersection point (2.3, 1.3) represent the solution to the system of equations. This means that:

x = 2.3 and y = 1.3 These values satisfy both equations simultaneously.

Activity 2.8: Graphing simultaneous equation Study the graph below carefully and use it to answer the questions that follow.

Figure 2.3: Graph of a simultaneous equation

1. What is the gradient of the line with the equation x – y = 2?

2. Which of the two lines has a negative gradient?

3. What is the coordinate of the point of intersection of the two lines?

4. Show that the point of intersection is truly the solution to the two systems of equations.

Solution

1. x – y = 2 Rewrite each equation in slope-intercept form (i.e., y = mx + c ) y = x − 2 Gradient = 1 as the coefficient of the x term in this form tells us the gradient.

2. The line x + y = 6 has a negative gradient, as y = − x + 6 . The negative coefficient of x indicates the negative gradient.

3. The coordinate of the point of intersection of the two lines is (4, 2)

4. (4, 2) means x = 4, y = 2 For x – y = 2 4 − 2 = 2 True For x + y = 6 4 + 2 = 6 True

To effectively tackle real-life problems involving simultaneous equations, use the following strategies:

1. Read the problem carefully to identify the quantities involved and what needs to be found.

2. Assign variables to the unknowns in the problem

3. Translate the problem into mathematical equations based on the relationships described in the problem.

4. Use methods such as substitution or elimination or graphical to solve the simultaneous equations.

5. Once you have found the values of your variables, interpret them in the context of the original problem.

6. For good practice and to confirm you have the correct solution, substitute your values back into the original equations to verify they satisfy both equations.

Activity 2.9: Real-life problem Sarah buys 3 apples and 2 bananas for Gh¢7.50. John buys 2 apples and 3 bananas for Gh¢7.00. How much does one apple and one banana cost?

Solution

You need to represent the situation with a system of simultaneous linear equations.

Let:

x represent the cost of one apple.

y represent the cost of one banana

Step 1: Formulate the system of equations from the problem Sarah buys 3 apples and 2 bananas for Gh¢7.50 3x + 2y = 7.50 (Equation 1) John buys 2 apples and 3 bananas for Gh¢7.00.

2x + 3y = 7 (Equation 2)

Step 2: To eliminate x , we will multiply the first equation by 2 and the second equation by 3, making the coefficients of x equal to 6 in both equations.

Multiply Equation 1 by 2:

2(3x + 2y) = 2(7.50) 6x + 4y = 15 (Equation 3) Multiply Equation 2 by 3:

2x + 3y = 7 3(2x + 3y) = 3(7) 6x + 9y = 21 (Equation 4)

Step 3: The coefficients of x are the same sign, so we subtract Equation 3 from Equation 4 to eliminate x .

(6x + 9y) − (6x + 4y) = 21 − 15 6x − 6x + 9y − 4y = 6 5y = 6

Step 4: Divide both sides by 6 to find y :

y = 6/5 = 1.20 The cost of one banana is Gh¢1.20

Step 5: Now that you have the cost of a banana, we can substitute y = 1.20 into one of the original equations to find x , the cost of an apple. Using equation 1:

3x + 2(1.20) = 7.50 3x + 2.40 = 7.50

Step 6: Subtract 2.40 from both sides of the equation to isolate x :

3x + 2.40 − 2.40 = 7.50 − 2.40 3x = 5.10

Step 7: Divide both sides by 3 to find x x = 5.10/3 = 1.70 The cost of one apple is Gh¢1.70.

Activity 2.10: Real-life problem A market woman in Accra sells plantains and tomatoes. She sells 2 plantains and 3 tomatoes for Gh¢12 and she sell 1 plantain and 1 tomato for Gh¢5. How much does 1 plantain cost and how much does 1 tomato cost?

Solution

Let:

x = cost of 1 plantain y = cost of 1 tomato 2x + 3y = 12 (2 plantains and 3 tomatoes for GH¢12): Equation 1 x + y = 5 (1 plantain and 1 tomato for Gh¢5): Equation 2

Step 1: Rearrange equation 2 for x to be the subject:

x = 5 − y

Step 2: Substitute x into Equation 1:

2(5 − y) + 3y = 12

Step 3: Simplify:

10 - 2y + 3y = 12 10 + y = 12 y = 2 The cost of one tomato is Gh¢2.00

Step 4: Substitute y back into any of the original equation. Using equation 2:

x + 2 = 5 x = 3 The cost of one plantain is Gh¢3.00

Step 5: Confirm your values work and you have the correct solution.

Activity 2.11: Real-life problem A farmer in the Northern Region sells maize and soybeans. He sells 2 bags of maize and 1 bag of soybeans for Gh¢150. If he sells 1 bag of maize for Gh¢40, how much does 1 bag of soybeans cost?

Solution

2x + y = 150 (2 bags of maize and 1 bag of soybeans for Gh¢150): Equation 1 x = 40 (1 bag of maize costs Gh¢40): Equation 2

Step 1: Substitute x = 40 into Equation 1:

2(40) + y = 150

Step 2: Simplify:

80 + y = 150

Step 3: Solve for y:

y = 150 - 80 y = 70 x = 40 (cost of 1 bag of maize) y = 70 (cost of 1 bag of soybeans) Therefore, 1 bag of soybeans costs Gh¢70.

Activity 2.12: Real-life problem In the local market of Chinderi, a town in the Oti Region, the price of a pineapple is Gh¢2.00 and the price of a mango is Gh¢3.00. Nana Okoegye Abraham, the chief of Chinderi, went to the market and bought a total of 10 fruits, spending Gh¢24.00.

How many pineapples and how many mangoes did he buy?

Solution

Let:

x represent the number of pineapples y represent the number of mangoes x + y = 10 (total fruits): Equation 1 2x + 3y = 24 (total cost): Equation 2

Step 1: Rearrange Equation 1 for x x = 10 − y

Step 2: Substitute x into Equation 2:

2(10 − y) + 3y = 24

Step 3: Simplify:

20 − 2y + 3y = 24 20 + y = 24

Step 4: Solve for y :

y = 4

Step 5: Substitute y back into Equation 1:

x + 4 = 10 x = 6 x = 6 (number of pineapples) y = 4 (number of mangoes) Therefore, Nana Okoegye Abraham bought 6 pineapples and 4 mangoes.

Using Software Demos to Graph Simultaneous

Equations Objective

You will learn how to use the Demos software on your phone to graph simultaneous equations with two variables. This activity will help you visualise solutions to equations and understand their graphical representations. Follow these steps to install and use the app with the guidance of your teacher.

Step 1: Download and Install Demos Software

For Android Users:

1. Open the Google Play Store on your Android phone.

2. In the search bar, type “Demos Graphing Calculator” and select the app from the results.

3. Tap Install to download the app.

4. Once installed, open the app and allow any necessary permissions.

For Apple Users (iOS):

1. Open the App Store on your iPhone or iPad.

2. In the search field, type “Demos Graphing Calculator” and select the app from the results.

3. Tap Get and, if prompted, confirm with your Apple ID or passcode.

4. Once installed, open the app and allow any necessary permissions.

Step 2: Prepare for Graphing

With your teacher’s guidance, you will:

1. Open the Demos app on your phone.

2. Enter the simultaneous equations as instructed by your teacher.

3. Use the app’s graphing features to view the intersection points and understand solutions visually.

Note: Remember to bring your phone to class fully charged or have it ready at home during designated learning time. If you have questions about downloading, installing or using the software, ask your teacher for help.

1. Solve these simultaneous equations using the elimination method.

a. 5x + 2y = 14 3x + y = 7

b. 2x + 3y = 12 x − 4y = − 6

c. 6x + y = 10 4x − y = 10

d. 7x + 3y = 20 2x − 3y = − 5

2. Solve these simultaneous equations using the substitution method.

a. x + 3y = 9 2x − y = 4

b. 2x + y = 5 3x − 4y = 10

c. x − 2y = 7 4x + y = 3

d. 5x − y = 11 x + y = 7

3. Solve these pairs of simultaneous equations graphically

a. 2x + 5y = 9 and 2x + 3y = 7

b. 3x + 4y = 23 and 2x – 4y = 2

c. 7x + 2y = 33 and 4x + 2y = 24

4. A farmer sells two types of crops: maize and beans. The farmer sells maize at Gh¢ 4.00 per bowl and beans at Gh¢ 6.00 per bowl. If he sold a total of 30 bowls for Gh¢150.00, how many bags of maize and beans did he sell?

5. Ama and Kofi run a small food stall in Accra, selling two types of dishes:

rice with chicken for Gh¢ 10 and jollof rice for Gh¢ 8. On a busy Saturday, they sold a total of 50 dishes and made Gh¢ 420 from their sales.

a Set up a system of simultaneous equations based on the information provided.

b Solve the system of equations to find out how many dishes of each type were sold that day.