Last updated on June 5th, 2025
Equations are significant components in mathematics. The equations use the “=’’ sign to indicate that both sides of an expression are equal. They are widely utilized in finding unknown values. In this topic, we will talk more about simple equations and their applications.
Simple equations are mathematical statements with an equal sign and at least one unknown quantity (variable) to solve for. As the term suggests, they are straightforward and can be solved without the use of complex methods.
For example: 2y +10 = 12 is a simple equation where y is the variable.
Here, we apply arithmetic operations to find the value of y
2y + 10 = 12
2y = 12 – 10
2y = 2
y = 1
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The transposition method involves shifting terms across the equal sign while changing their signs accordingly. For example, positive becomes negative and vice-versa. Also, arithmetic operations change when transposing terms.
For e.g., addition becomes subtraction, and division becomes multiplication (and vice versa). This technique helps us in finding the unknown variable by isolating it.
Take a look at this example for better understanding:
Find the value of p
3p – 3 = 12
3p = 12 + 3
3p = 15
p = 15/3
p = 5
Hence, p is 5.
A linear equation is another term for a simple equation that involves one or more variables. Solving linear equations is straightforward and can be done using simple methods like graphing. Linear equations can also be solved by transposing the terms or by balancing the LHS and RHS.
Note that, when a number is transposed, its preceding sign changes.
For example:
Prove LHS = RHS for 8x + y = 24, given x = 2 and y = 8
8 (2) + 8 = 24
16 + 8 = 24
24 = 24
Therefore, LHS = RHS.
Solving simple equations determines the values of the unknown variable in the given equation. In a simple equation, the LHS and RHS should be equivalent. There are different methods for solving simple equations, as mentioned below:
In a trial and error method, we substitute random values for the variable to check if it satisfies the equation LHS = RHS.
For example: x + 5 = 15
Where:
LHS = x + 5
RHS = 15
Let’s now perform the trial and error method by substituting values starting from 1 to check if LHS = RHS.
x | LHS (x + 5) | RHS | Is LHS = RHS? |
1 | 1 + 5 = 6 | 15 | No |
2 | 2 + 5 = 7 | 15 | No |
3 | 3 + 5 = 8 | 15 | No |
4 | 4 + 5 = 9 | 15 | No |
10 | 10 + 5 = 15 | 15 | Yes |
Therefore, LHS = RHS for x = 10
The systematic method also known as the balance method is used to balance the equation by performing the mathematical operations on both sides of the equation. This systematic approach compares both sides of the equation to a weighing balance. It maintains equality by adding or removing values from each side of the equation.
For example: y – 2 = 8
We add 2 to both sides to isolate y
y – 2 + 2 = 8 + 2
Thus, y = 10.
The transposition method simplifies the equation by shifting terms across the equal sign. Below is a table exhibiting both systematic and transposition methods for the same equation.
Adding or subtracting on both sides (systematic method) |
Transposing |
2x – 8 = 4 Adding 8 to both sides, 2x – 8 + 8 = 4 + 8 2x = 12 |
2x – 8 = 4 Transposing –8 from LHS to RHS, (When transposing, –8 becomes +8) 2x = 12 |
Therefore, we can apply the transposition method to solve simple equations.
Simple equations are widely used in problem-solving in various real-life situations. They are used in different fields beyond mathematics. Let’s look into some:
Simple equations help students find unknown variables easily without complex formulas. However, there are a few common mistakes that students should watch out for. Here’s a list of such mistakes along with steps to avoid them:
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The sum of Roy’s age and his father’s age is 66. If Roy is 20 years old, calculate his father’s age.
Roy’s father is 46 years old
Consider Roy’s father’s age to be y
The equation for the given problem: 20 + y = 66
Now, subtract 20 from both sides,
y = 66 – 20
y = 46
Therefore, Roy’s father is 46 years old.
A teacher asked her students: “If a number is added to 12, the result is 24. Find the number.”
The unknown number is 12.
Consider the unknown number as x
The equation for the given problem:
x +12 = 24
We find the value of x by subtracting 12 from both sides:
x = 24 – 12
x = 12
Therefore, the unknown number is 12.
A shopkeeper has 6 boxes of pencils. Each box contains the same number of pencils. If there are 30 pencils in total, how many pencils are in each box?
There are 5 pencils in each box.
Consider the number of pencils in each box to be y
6 × x = 30
We find x by dividing both sides by 6:
X = 30/ 6
X = 5
So, there are 5 pencils in each box.
Find the value of x in the equation: 5x – 10 = 40
x = 10
We have:
5x – 10 = 40
To isolate the term with x, we add 10 to both sides:
5x = 50
x = 50/5
x = 10
Assume a number is subtracted from 60 to 15. Find the number.
The number is 45.
Consider the number to be y
The equation for the given problem:
60 – x = 15
Solving for x:
x = 60 – 15
x = 45
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Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.