Activity 2.2.1.
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Gradually add a mass to the \(1\) kg mass until the beam balances.
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Represent the added mass with a letter and relate the weights on the two sides of the beam balance.
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Discuss and share with other groups.
Section 2.2 Linear Equations
Introduction to Linear Equations.
Have you ever tried to balance a scale? If you add weight to one side, you must do the same to the other side to keep it balanced. This idea is the foundation of linear equations.
What is a Linear Equation?
A linear equation is an equation where the highest power of the variable is 1. It has a simple form, such as:
\begin{equation*}
2x + 3 = 7
\end{equation*}
Here, \(x\) is the unknown number, called a variable. To find \(x\text{,}\) we solve the equation step by step:
\begin{equation*}
2x = 7 - 3
\end{equation*}
\begin{equation*}
2x = 4
\end{equation*}
\begin{equation*}
x = \frac{2}{4}
\end{equation*}
\begin{equation*}
x = 2
\end{equation*}
Why are Linear Equations Important?
Linear equations help us solve real-world problems, such as:
-
Calculating expenses: If a book costs \(b\) dollars and you buy \(4\) books, the total cost is \(4b\text{.}\) If the total cost is \($40\text{,}\) you can find the price of one book by solving \(4b = 40\text{.}\)
-
Finding unknown ages: If a person’s age is \(x\text{,}\) and their younger sibling is \(3\) years younger, we can write this as \(x - 3.\)
-
Distance and speed calculations: If a car travels at \(s\) km per hour for \(t\) hours, the distance covered is \(s \times t\) km.
You are going to learn how to:
-
Understand equations – Recognizing how equations express equality.
-
Solve for unknowns – Finding the value of variables step by step.
-
Use real-life applications – Applying equations to everyday situations.
-
Graph linear equations – Representing equations on a coordinate plane.
Linear equations are like puzzles waiting to be solved. With practice, you will see how they can make problem-solving easier and fun.
Subsection 2.2.1 Forming Equations in One Unknown
\begin{equation*}
{\text{Key Takeaway.}}
\end{equation*}
\begin{equation*}
\text{Equating the masses on both sides leads to the formation of an equation.}
\end{equation*}
Example 2.2.1.
Walibora bought a shirt and a pair of shorts. The cost of the pair of shorts was \(150\) shillings less than that of the shirt. If the two items cost \(950\) shillings. Write an equation representing the information using letter \(x\) as the cost of a shirt.
Solution.
The cost of a shirt and a pair of shorts can be represented using the expression shown below:
\begin{equation*}
x + (x - 150) = 950
\end{equation*}
Simplifying the equation, we get:
\begin{equation*}
2x - 150 = 950
\end{equation*}
Represents the cost of a shirt and a pair of shorts.
Checkpoint 2.2.2.
Checkpoint 2.2.3.
Subsection 2.2.2 Solving Linear Equations in One Unknown.
Activity 2.2.2. Work in Groups.
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On one side of a beam balance, place five matchboxes of the same weight.
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On the other side, place two matchboxes.
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To the side with two matchboxes, gradually add grains of sand until the beam balances.
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From both sides, remove two matchboxes and observe whether the beam balances.
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Represent the mass of the added sand with a letter and form an equation involving the letter.
-
Discuss and share your results with other groups.
\begin{equation*}
\text{Key Takeaway.}
\end{equation*}
-
\begin{equation*} \text{Subtracting a number from both sides of the equation.} \end{equation*}
-
\begin{equation*} \text{Adding a number to both sides of the equation.} \end{equation*}
-
\begin{equation*} \text{Dividing both sides of the equation by a number.} \end{equation*}
-
\begin{equation*} \text{Multiplying both sides of the equation by a number.} \end{equation*}
Example 2.2.4.
Solve the following equation:
\begin{equation*}
x + 4 = 7
\end{equation*}
Solution.
Subtract \(4\) from both sides.
Example 2.2.5.
Example 2.2.6.
Solution.
Multiply both sides by 4
\begin{equation*}
\frac{1}{4}h \times 4 = 3 \times 4
\end{equation*}
The \(4\) on the left side of the equation cancels out and we are left with
\begin{equation*}
h = 12
\end{equation*}
\begin{equation*}
\frac{4h}{4} = 12
\end{equation*}
\begin{equation*}
h = 12
\end{equation*}
Checkpoint 2.2.7.
Checkpoint 2.2.8.
Checkpoint 2.2.9.
Subsection 2.2.3 Application of Linear Equations in One Unknown
Example 2.2.10.
Dorine and Andrew are to share \(10\) kg of sugar so that Dorine gets \(2\) kg more than what Andrew receives. Find the weight received by each of them.
Solution.
Let Andrew’s share be \(x\) kg.
\begin{equation*}
2x + 2 = 10.
\end{equation*}
\begin{equation*}
2x = 8.
\end{equation*}
\begin{equation*}
x = 4
\end{equation*}
Checkpoint 2.2.11.
Exercise Here.
Example 2.2.12.
Makanda and Magut sell exercise books at \(40\) shillings per book. Peter bought some exercise books from Makanda’s shop and from Magut’s shop. In Magut’s shop, he bought \(30\) more books than he bought from Makanda’s shop. If he spent a total of \(4400\) shillings buying the books, find the number of books bought from each shop.
Solution.
Let the number of books bought from Makanda’s shop be \(x\text{.}\)
\begin{equation*}
= 40x + 1200.
\end{equation*}
\begin{equation*}
= 80x + 1200.
\end{equation*}
\begin{equation*}
80x = 4400 - 1200.
\end{equation*}
\begin{equation*}
80x = 3200.
\end{equation*}
\begin{equation*}
x = 40.
\end{equation*}
Number of books bought from Makanda’s shop is \(40\) and from Magut’s shop is \(40 + 30\) = \(70\text{.}\)
