Lesson Objectives

Dear learner;

At the end of this lesson you will be able to:

• State different techniques to solve systems of Linear Equations in Two Variables
• Solve systems of Linear Equations in Two Variables.
• Use solutions to systems of Linear Equations in Two Variables in real life problems.

Key Terms

• system of linear equations
• one solution
• many solution
• no solution

Brainstorming Questions

1. A bakery sells two types of cakes: chocolate cakes and vanilla cakes. The bakery sold a total of 80 cakes on a particular day, with the total revenue amounting to birr 600. If a chocolate cake costs birr 8 and a vanilla cake costs birr 6, how many of each type of cake were sold that day?

Solution

To determine how many chocolate cakes and vanilla cakes were sold, we’ll set up and solve a system of linear equations based on the given information.

Define the Variables

Let

• $ x$ be the number of chocolate cakes sold.
• $y$ be the number of vanilla cakes sold.

Set Up the Equations

Total Cakes Sold: The total number of cakes sold is 80.

$x + y = 80$

Total Revenue: The revenue from selling chocolate cakes and vanilla cakes is 600 Birr. Chocolate cakes cost 8 Birr each, and vanilla cakes cost 6 Birr each.

$8x + 6y = 600$.

Solve the System of Equations

We have the following system of equations:

$\begin{cases}
x + y = 80 \\
8x + 6y = 600
\end{cases}$

Step 1: Solve for $ y \text{ in terms of } x $ from the first equation

$y = 80 – x$

Step 2: Substitute $ y $ into the second equation

$8x + 6(80 – x) = 600.$

Step 3: Distribute and simplify

$8x + 480 – 6x = 600.$

Step 4: Substitute $ x $ back into the first equation to find $ y$

$y = 80 – 60.$

Step 5: Solve for $ y$

$y = 20.$

2.1. Solving systems of linear equations in two variables

Dear learner,

An equation of the type 𝑎𝑥 + 𝑏𝑦 = 𝑐 where 𝑎, 𝑏 and 𝑐 are arbitrary constants and 𝑎 ≠ 0, 𝑏 ≠ 0, is called a linear equation in two variables.

Example 1:

$3x +2y =5, x -3y = 6 $etc. are linear equations

Definition:

• A system of linear equations in two variables is a set of two or more equations that involve the same two variables and can be solved simultaneously to find the unique point where the equations intersect on a coordinate plane.
• Some linear systems may not have a solution, while others may have an infinite number of solutions.
• For a linear system to have a unique solution, there must be at least as many equations as there are variables. However, even this does not guarantee a unique solution.

Example 2:

a). the equation: 3x +y = 6 has many solutions in the set of real numbes

b). $\begin{cases}
3x +2 y = 5 \\
2x -y = 1
\end{cases}$ has unique or one solution.

2.1.1 Solving systems of linear equations in two variables using tables

• Dear learner; Solving systems of linear equations in two variables using tables involves organizing the equations into a table format and systematically finding the values of the variables that satisfy both equations simultaneously. This method is particularly useful for visual learners and can provide a clear and structured approach to solving such systems.

Example 3:

Find the solution for the system of equations: $\begin{cases}
2x -y = -1 \\
x +2y = 7
\end{cases}$ using table method.

Solution:

To solve this system using tables, we can create a table with rows for $x, y$, and the equations. We then substitute different values for $x$ or $y$ into each equation at a time and calculate the corresponding values for the other variable. By filling out the table with these values, we can identify the solution where both equations are satisfied.

x	0	1	2	3
$2x – y= -1$	1	3	5	7
$x + 2y= 7$	$\frac{7}{2}$	3	$\frac{5}{2}$	2

By substituting the given values of $x$ into each equation, the value of $y$ is simultaneously similar at $x = 1$. Therefore, the solution set that satisfies both equations at the same time at (1, 3) means at $x = 1$ and $y = 3$.

Solution set = {(1,3)}.

2.1. 2 Solving systems of linear equations in two variables by substitution

• Dear learner; Solving systems of linear equations in two variables by substitution involves replacing one variable in one equation with an expression involving the other variable from the second equation. This method allows for finding the values of both variables that satisfy both equations simultaneously.

To solve systems of linear equations in two variables by substitution, follow these steps:

Step 1: Solve one of the two equations for one of the variables in terms of the other.
Step 2: Substitute the expression for this variable into the second equation, then solve for the remaining variable.
Step 3: Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
Step 4: Check the solution in both equations to ensure accuracy.

Example 4:

Solve the system of equations.

$\begin{cases}
2x – y = -1 \\
x + 2y = 7
\end{cases}$

Using substitution method.

Solution:

To solve the system of equations using the substitution method, follow these steps:

Given system of equations:

$\begin{cases}
2x – y = -1 & \text{(Equation 1)} \\
x + 2y = 7 & \text{(Equation 2)}
\end{cases}.$

Step 1: Solve One Equation for One Variable

Let’s solve Equation 2 for $ x$:

$x + 2y = 7.$

Rearrange to solve for $x$:

$x = 7 – 2y.$

Step 2: Substitute the Expression into the Other Equation

Now substitute $ x = 7 – 2y$ into Equation 1:

$\begin{cases}
2x – y &= -1 \\
2(7 – 2y) – y &= -1
\end{cases}$

Step 3: Solve for $ y$

Expand and simplify:

$\begin{aligned}
2 \cdot 7 – 2 \cdot 2y – y &= -1 \
14 – 4y – y &= -1
\end{aligned}$

Combine like terms:

$14 – 5y = -1$

Subtract 14 from both sides:

$-5y = -1 – 14$

$-5y = -15$

Divide by -5:

$y = \frac{-15}{-5} = 3.$

Step 4: Substitute $ y $ back into the expression for $ x$

Now substitute $ y = 3$ back into $ x = 7 – 2y$:

$x = 7 – 2(3)=7-6=1$

Thus, $x = 1$.

2.1. 3 Solving systems of linear equations in two variables by Addition (Elimination) method

• The Addition (Elimination) method is a technique used to solve systems of linear equations in two variables by adding or subtracting the equations to eliminate one variable.

Steps to solve a system of equations using the addition (elimination) method:

Step 1: Write both equations with 𝑥 and 𝑦 variables on the left side of the equal sign and constants on the right.
Step 2: Write one equation above the other, aligning corresponding variables. If one variable in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations to eliminate one variable. If not, multiply by a nonzero number to create opposite coefficients and then add the equations.
Step 3: Solve the resulting equation for the remaining variable.
Step 4: Substitute that value into one of the original equations and solve for the second variable.
Step 5: Check the solution by substituting the values into the other equation.

Example 5:

Solve the system of equations

$\begin{cases}
2x – y = -1 \\
x + 2y = 7
\end{cases}$

Using addition (elimination) method

To solve the system of equations using the addition (elimination) method, follow these steps:

Given System:

$\begin{cases}
2x – y = -1 \\
x + 2y = 7
\end{cases}$

1. Align the equations:

$\begin{aligned}
2x – y &= -1 \\
x + 2y &= 7
\end{aligned}$

2. Multiply one or both equations if necessary to make the coefficients of one of the variables either $ x$ or $y$ match in magnitude but with opposite signs. In this case, we will eliminate $y$ . To do this, multiply the first equation by 2:

$\begin{aligned}
2(2x – y) &= 2(-1) \\
4x – 2y &= -2
\end{aligned}$

The system now looks like:

$\begin{aligned}
4x – 2y &= -2 \\
x + 2y &= 7
\end{aligned}$

3. Add the two equations to eliminate $ y$:

$\begin{aligned}
(4x – 2y) + (x + 2y) &= -2 + 7 \\
4x + x &= 5 \\
5x &= 5 \\
x &= \frac{5}{5} = 1
\end{aligned}$

4. Substitute $ x = 1 $ back into one of the original equations to solve for $y$. Using the second equation:

$\begin{aligned}
x + 2y &= 7 \\
1 + 2y &= 7 \\
2y &= 7 – 1 \\
2y &= 6 \\
y &= \frac{6}{2} = 3
\end{aligned}$

5. The solution to the system of equations is:

$\begin{aligned}
x &= 1, \quad y = 3
\end{aligned}$

To summarize:

$\begin{aligned}
(x, y) = (1, 3)
\end{aligned}$

• Let p be the cost of one pen (in Birr).
• Let b be the cost of one book (in Birr).

We are given two pieces of information:

1. Three pens and two books cost Birr 118.
2. One pen and two books cost Birr 106.

We can translate these statements into the following system of linear equations:

$\begin{cases}
3p + 2b = 118 \\
p + 2b = 106
\end{cases}$

To find the values of $ p$ and $ b$, we can use the elimination method. Here’s the step-by-step solution:

Step 1: Set Up the Equations

$\text{Equation 1:} 3p + 2b = 118 $
$\text{Equation 2:} p + 2b = 106$

Step 2: Eliminate One Variable

To eliminate $b$, we can subtract Equation 2 from Equation 1.

First, let’s subtract Equation 2 from Equation 1:

$(3p + 2b) – (p + 2b) = 118 – 106.$

Simplify:

$\begin{aligned}
3p + 2b – p – 2b &= 12
\end{aligned}$

Combine like terms:

$\begin{aligned}
2p &= 12
\end{aligned}$

Solve for $p$:

$\begin{aligned}
p &= \frac{12}{2} = 6
\end{aligned}$

So, the cost of one pen$ p$ is Birr 6.

Step 3: Substitute $p$ into One of the Original Equations

Now that we have $p = 6$ , substitute $ p$ into Equation 2 to solve for $ b$:

$\begin{aligned}
p + 2b &= 106 \\
6 + 2b &= 106
\end{aligned}$

Substitute $ p=6$:

$\begin{aligned}
6 + 2b &= 106
\end{aligned}$

Solve for $ b$:

$\begin{aligned}
2b &= 106 – 6 \\
2b &= 100 \\
b &= \frac{100}{2} = 50
\end{aligned}$

So, the cost of one book $b$ is Birr 50.

Summary
The cost of one pen is Birr 6.
The cost of one book is Birr 50.

2.1. 4 Graphical method of solving system of linear equations in two variables

Dear learner,

The Graphical method of solving a system of linear equations in two variables involves plotting the equations on a graph and finding the point where the lines intersect, which represents the solution to the system.

To solve systems of linear equations using graphs, follow these steps:

Step 1: Graph the first equation.
Step 2: Graph the second equation using the same coordinate system.
Step 3: Determine whether the lines intersect, are parallel, or are the same line.
Step 4: Identify the solution to the system. If the lines intersect, identify the point of intersection.

Example 7:

Solve the following system of linear equations graphically.

Solution:

Thus, the solution to the system of linear equation is (2, 0), meaning x = 2 and y = 0.

b) Solve the following system of linear equations graphically.

When we draw the line of each component equation, we see that the lines are parallel. This means the lines do not intersect. Hence the system does not have a solution.

Example 8:

Solve the following system of linear equations graphically.

Solution:

When we draw the line of each component equation, we see that the lines coincide one over the other, which shows that the system has infinite solutions. That is, all points (ordered pairs) on the line are solutions of the system.