You must be familiar with the concepts of Algebra and must have solved equations based on them in your previous classes. Here are some algebraic expressions that you must have dealt with.

8y –10 = 6, 5x = 40, etc.

Now, in Grade 8, you will come across slightly more complicated algebraic equations. The major difference that you will see is the use of the equal (=) sign. This sign is used in algebraic equations, but not in expressions such as 7x, 4y – 3. Another difference is that algebraic equations have only one variable, with the highest power as 1.

Example: 2x – 4 = 6 is a linear equation, while x² + 1 is not as the power of the variable x is 2.

You will learn more about linear equations in your next lessons.

In an equation, the variables on either side of the equal sign have the same value. When solving such equations, it is important to maintain this balance for obtaining the right solution.

Since you have already studied algebraic equations with one variable, you must know that RHS (Right Hand Side) denotes the expression on the right side of '=', and LHS (Left Hand Side) denotes the expression on the left side of '='.

After solving the equation, you will find that the values on the LHS are equal to those on the RHS. This will happen only for certain variables in the equation. If LHS = RHS, then your solution is correct. Substitute the answer you obtained with the variable in the equation to find if your solution is correct. This is a great way to check your work yourself.

 

There are several types of algebraic equations that you can solve using different techniques. Some equations have variables on one side only, while others have variables on both sides of the equal sign. Now, let us study an equation that has numbers on one side and a variable on the other side of ‘=’.

Example 1: Find the solution for 2x – 3 = 7.

Technique 1: Add 3 on both sides of the equation to balance it and then solve.

2x – 3 + 3 = 7 + 3

2x = 10

x = 10 / 2

x = 5

Technique 2: Take 3 to the right side of the equation and then solve.

2x = 7 + 3

2x = 10

x = 10 / 2

x = 5

Technique 3: Divide both sides by 2 and then solve.

2x = 10

2x / 2 = 10 / 2

x = 5

To check for the correctness of your solution, put the value of x in the original equation and see if LHS = RHS.

2x – 3 = 7

(2*5) – 3 = 7

10 – 3 = 7

Here, LHS = RHS. Thus, the value x = 5 is correct.

Let us solve different types of linear equations to understand them better.

Example 2: If the sum of two numbers is 64 and one number is 10 more than the other, what are the numbers?

This problem is like a puzzle. You can solve it by writing the given information in the form of a linear equation.

Assume the unknown number to be x. Now, according to the problem, one number is 10 more than the other, and the sum of both is 64.

You can write this condition in the equation form as (10 + x) + x = 64.

To solve this equation, you can move the number 10 to RHS.

x + x = 64 – 10

2x = 54

x = 54 / 2

x = 27

Example 3: Rima has three times as many 5-rupee coins as she has 2-rupee coins. The total amount with her is Rs. 68. Determine the amounts she has in the forms of 5-rupee and 2-rupee coins.

Assume the unknown variable to be x.

Then, 

amount in the form of 5-rupee coins = Rs. 5 * 3x = Rs. 15x, and

amount in the form of 2-rupee coins = Rs. 2 * x = Rs. 2x.

Form an equation using the given information and solve it as shown below.

15x + 2x = 68

17x = 68

x = 68 / 17

x = 4

Total amount in the form of 5-rupee coins = 15*4 = Rs. 60

Total amount in the form of 2-rupee coins = 2*4 =Rs. 8

To check for correctness, substitute the obtained values of variables in the initial equation and solve it. 

Total money that Rima has = Rs. 60 + Rs. 8 = Rs. 68

Some equations have variables on both sides of the equal sign, i.e., LHS and RHS.

Example 4: Solve for x: 3x – 2 = x + 4

3x – 2 = x + 4

3x = x + 4 + 2

3x = x + 6

3x – x = 6

2x = 6

x = 6 / 2

x = 3

To check for correctness, put the obtained value of x in the original equation and solve.

3(3) – 2 = 3 + 4

9 – 2 = 7

7 = 7 (LHS = RHS)

Example 5: Tina is twice as old as Rohan. Five years ago, Tina's age was three times that of Rohan. What are their present ages?

If Rohan's present age is x, Tina's present age is 2x.

Five years ago, Tina's age was 2x – 5, and Rohan's age was x – 5. Also, Tina’s age was three times that of Rohan.

Using the information about five years ago, you will get the following equation. 

2x – 5 = 3(x – 5)

Solve it as shown below.

2x – 5 = 3x – 15

2x – 5 + 15 = 3x

2x – 10 = 3x

10 = 3x – 2x

x =10

Rohan’s present age = x = 10 years

Tina’s present age = 2x = 2 * 10 = 20 years

Example 6: Solve for x: 5x – 2(2x – 7) = 2(3x – 1) + 7 / 2

Simplify LHS and RHS as shown.

LHS: 5x – 2(2x – 7) = 5x – 4x – 14 = x – 14

RHS: 2(3x – 1) + 7 / 2 = 6x – 2 + 7 / 2 = 6x – 4 / 2 + 7 / 2 = 6x – 3 / 2

Now, write the simplified equation as shown and solve further.

x – 14 = 6x – 3 / 2

14 = 6x – x – 3 / 2

14 = 5x – 3 / 2

5x =14 – 3 / 2

5x = 28 – 3 / 2

5x = 25 / 2

x = 25 / 2 * 1 / 5

x = 5 / 2

Example 7: The present ages of Ami and Ari are in the ratio of 4:5. Eight years from now, the ratio of their ages will be 5:6. Find their present ages.

Let the present age of Ami be 4x and that of Ari be 5x.

Then, after eight years, Ami's age will be 4x + 8 and Ari's age will be 5x + 8.

The ratio of their ages after eight years can be written as shown:

4x + 8 / 5x + 8 = 5:6

6(4x + 8) = 5(5x + 8)

24x + 48 = 25x + 40

48 – 40 = 25x – 24x

Finally, you get x = 8.

Ami's age after eight years = 4x + 8 or 4(8) + 8 = 32 + 8 = 40 years

Ami's present age = 4x or 4 * 8 = 32 years

Ari's age after eight years = 5x + 8 or 5 * 8 + 8 = 40 + 8 = 48 years

Ari's present age = 5x or 5*8 = 40 years

• An algebraic equation has variables in it. You can solve it by finding the value of the variable.

• In an algebraic equation, the value on LHS always equals that on the RHS. 

• You can check the correctness of your solution by substituting the obtained value with the variable. If your answer is correct, you will get LHS = RHS.

• To solve equations with variables on both sides of the equal sign, bring all variables to one side and numbers to the other side.

• To solve complex equations, first, simplify them.

1. How many sums are there in the NCERT Solutions for Class 8 Maths Chapter 2- Linear Equations in One Variable?

Ans. There are a total of 65 exercise sums in the chapter Linear Equations In One Variable for Class 8.

2. What is meant by ‘linear equations in one variable’ in mathematics?

Ans: Algebraic equations that have linear expressions in one variable only are called linear equations in one variable.

3. Can I get Class 8 Maths Chapter 2, Linear Equations In One Variable, online?

Ans. Yes, you can get this chapter online on the MSVgo App. Join MSVgo and study on the go, anywhere, anytime.

4. Can I download the solutions of the NCERT chapter Linear Equations In One Variable NCERT for free?

Ans. Yes, you can download the solutions of Linear Equations In One Variable for Grade 8 NCERT for free from the MSVgo App. Just download the app and get access to 15,000+ videos and 10,000+ questions bank for your Champ.

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