Class 8 Maths Chapter 2 Notes - Linear Equations in One Variable

Introduction to Linear Equations in One Variable

An expression is a mathematical phrase that can contain numbers, variables, and operators like +, –, ×, ÷. For example, 3x + 9 is a mathematical expression. Equations are similar to expressions, except that, equations contain an ‘equal to’ operator along with some other expressions. The expression on the left of the equality sign is the left-hand side (L.H.S.). The expression on the right of the equality sign is the right-hand side (R.H.S.).

Linear equation in one variable is an equation of the form ax + b = 0. Here, a and b are integers and x is a variable. It has only one solution. For example: 8y-11 = 0 is a linear equation.

While solving an equation, we find the value of the variable at which the LHS is equal to the RHS of the equation. There are two methods to solve the linear equations, let's take a look at them.

Methods of Linear Equations

Systematic/Inverse method

In this method, whatever is done on the left-hand side (L.H.S.) of the equation, the same is to be done on the right-hand side (R.H.S.). Both sides of the equation can be added, subtracted, multiplied, or divided with the same number to get the solution.

Example:

Transposition method

In this method, any term of the equation can be taken from one side to the other with a change in its sign. This does not affect the solution of the equation. The plus sign of a term changes to a minus sign, and vice versa. Similarly, the multiplication sign of a term changes to the division sign and vice versa.

Example:

Now, let us see different forms of equations.

Forms of Equations

Variable is on One Side of the Equation

In these type of equations, the variable is only on one side of the equation.
For example, in the equation 2x – 5 = 55, the variable is only on one side of the equation.

Variable is on both the Sides of the Equation

In these type of equations, the variable is on both the sides of the equation.
For example, in the equation 5y + 11 = 6 (y – 9), variables are on both the sides of the equation.

Let us see, how we can solve linear equations.

Solving Linear Equations

Steps of Solving an Equation

If there is any number being multiplied with a bracket, simplify them first.

If there are any fractions on one side of the equation, use the LCM method to make the denominators equal and then do cross multiplication, if needed.

Figure out which terms are containing constants and which terms are containing variables.

Bring all the terms containing the variables on one side of the equation and the terms containing the constants on the other side.

Perform the algebraic operations on both sides of the equation to get the value of the variable.

An example is shown below.

Hence, z = 83/3 is the solution of the equation.
Linear equations can be used to solve real-world application problems.

Want to download NCERT solutions for class 8 maths linear equations? Get access to free PDF below:

You may also need this
NCERT Solutions for class 8 Maths Chapter 2
NCERT Solutions for class 8 Maths Chapter 2 Exercise 2.1
NCERT Solutions for class 8 Maths Chapter 2 Exercise 2.2
NCERT Solutions for class 8 Maths Chapter 2 Exercise 2.3
NCERT Solutions for class 8 Maths Chapter 2 Exercise 2.4
NCERT Solutions for class 8 Maths Chapter 2 Exercise 2.5
NCERT Solutions for class 8 Maths Chapter 2 Exercise 2.6

Real-World Application Problems of Linear Equations

Let us see some examples to understand this.

The ratio of the present ages of Ananya and Supriya is 2:7. Five years later, the ratio will be 2:3. What is the present age of Ananya?

According to the ratio, the present ages of Ananya and Supriya will be 2x and 7x years, respectively.

Five years later, age of Ananya = 2x + 5

Five years later, age of Supriya = 7x + 5

According to the question,

Now, 2x

So, the present age of Ananya is 7.5 years.

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Rohan and his friends covered a distance downstream in 15 hours and the same distance upstream in 30 hours by a boat. What will be speed of the boat in still water if the speed of the stream is 5 km/hr?

Let the speed of the boat in still water be x km/hr.

Speed of the stream = 5 km/hr
Downstream speed = (x + 5) km/hr
Upstream speed = (x – 5) km/hr
By using the formula,
Distance = speed × time
Distance covered downstream = (x + 5) × 15
Distance covered upstream = (x – 5) × 30

Now,
(x + 5) × 15 = (x – 5) × 30
15x + 750 = 30x – 150
30x – 15x = 750 + 150
15x = 900
x = 60
Hence, the speed of the boat in still water is 60 km/hr.

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