# CBSE Class 10 Maths Revision Notes Chapter 3

## Class 10 Mathematics Revision Notes for Chapter 3 – Pair of Linear Equations in Two Variables

In Class 10, Mathematics is one of the most crucial subjects. Students should practise it daily to score good marks and improve their overall percentage in the board examinations. Extramarks has created Class 10 Mathematics Chapter 3 Notes to help students understand the concepts better. They have covered all the important questions, topics, and concepts in Class 10 Chapter 3 Mathematics Notes to help students in preparation. You can rely on the notes for board exam preparation as they have been created according to the CBSE syllabus and NCERT books.

## Access Class 10 Mathematics Chapter 3 – Pair of Linear Equations in two Variables

### Linear Equation

An equation is defined as a statement that two mathematical expressions with one or more variables are equal.

Linear equations are those in which the powers of all the variables are one. A linear equation’s degree is always one.

The general form of representing the linear equations is ax + by + c = 0, where a and b cannot be zero simultaneously.

### Solution of an Equation

The solution of a linear equation in two variables can be given by a pair of values (x,y), one for ‘x’ and the other for ‘y’, that make the two sides of an equation equal.

Let us take an example:

If 2x + y = 4, then (0,4) is one of the solutions that satisfy the equation. It can have a number of solutions.

### Pair of Linear Equations in Two Variables

The pair of linear equations in two variables can be represented as follows:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

The nature of two straight lines in a plane is as follows:

• Both the lines intersect at exactly one point.
• They are parallel to each other.
• Both the lines coincide with each other.

### Graphical Method of Solutions

One can find the solutions to a pair of linear equations through the graphical method given below.

• Plot both the equations.
• Now find the point of intersection of the lines.

Let us take an example:

2x + y – 6 = 0 and 4x – 2y – 4 = 0.

The point of intersection for the two graphs will be (2,2). Take a look at the graph below.

### Substitution Method

We will explain the substitution method with an example.

7x – 15y = 2 (1)

x + 2y = 3 (2)

Solution:

Step 1: Pick any equation and write one variable in terms of the other. If we take equation 2, then

x + 2y = 3

And now we write it as x = 3 – 2y (3)

Step 2: Now substitute the value of x in equation (1). We get

7(3 – 2y) – 15y = 2

21 – 14y – 15y = 2

-29y = -19

Therefore,  y = 19/29

Step 3: By substituting the value of y in equation 3, we will get

x = 3 – 2 (19/29) = 49/29

Hence the solution is x = 49/29 and y = 19/29.

Verification: Substituting x = 49/29 and y = 19/29, we can verify that both the equations are satisfied.

Let’s study the substitution method step-by-step to better comprehend it:

Step 1: Find the value of one variable, let’s say y, in terms of the other variable, i.e., x, as per your convenience.

Step 2: Substitute the value of y in the other equation, and simplify it to a single equation in one variable, i.e., in terms of x, which can be solved.

Step 3: Now complete the process by substituting the value of x or y from Step 2 in the equation used in Step 1 to obtain the values of the other variable.

### Elimination Method

This method of eliminating one variable is sometimes more convenient than the substitution method. Let’s see how this method works.

Step 1: Start by multiplying both the equations by some suitable non-zero constants to obtain the coefficients of one variable (either x or y) numerically equal.

Step 2: Then, you have to add or subtract one equation from the other so that one variable gets eliminated.

Step 3: Now solve the equation in one variable (x or y) to obtain its value.

Step 4: Substitute the value of x or y in either equation to get the value of the other variable.

Let us take an example:

2x + 3y = 8                     (1)

4x + 6y = 7                     (2)

Multiply equation (1) by 2 and equation (2) by 1 to make the coefficients of x equal.

4x + 6y = 16                   (3)

4x + 6y = 7                     (4)

Now, by subtracting equation (4) from equation (3), we will get

(4x – 4x) + (6y – 6y) =16 -7

0 = 9, which is a false statement.

Hence the pair of equations has no solution.

### Cross Multiplication Method

Cross multiplication method is used to solve linear equations in two variables. We can understand the cross multiplication method by the following example.

Let’s suppose, a1x + b1y + c1 = 0 and a2x + b2x + c2 = 0

By using cross multiplication, we can get the values of x and y in the following ways:

x = b1 c2 – b2 c1 / b2 a1 – b1 a2

y = c1 a2 – c2 a1 / b2 a1 – b1 a2

Where

b2 a1 – b1 a2 is not equal to 0.

Solution:

x/ b1 c2 – b2c1

= y / c1 a2 – c2 a1

= 1/b2 a1 – b1 a2

You can use the diagram given below to understand the cross multiplication method better.

Two conditions can arise in this method:

Case 1:

If a1/a2 is not equal to b1/b2, then we will have a unique solution, and the two linear equations in two variables will be consistent.

Case 2:

If a1/a2 = b1/b2 = c1/c2, then there will be infinitely many solutions, and the lines will be coincident and thus dependent and consistent.

Case 3:

If a1/a2 = b1/b2 c1/c2, then there will be no solution, and the pair of linear equations will be inconsistent.

### 1. How can I access the past year's question papers for Class 10 exam preparation?

Candidates studying in Class 10 can access the CBSE past years’ question papers and study materials from the website of Extramarks for board exam preparation. In addition, CBSE sample papers are also available to help students in their preparation.

### 2. How should students prepare for the Class 10 Mathematics exam?

Students should focus more on quality than quantity. To score better in the examination, candidates should practise all formulas, numerical and CBSE extra questions to understand the types of questions asked in the examination.

### 3. Define the pair of linear equations in two variables.

A linear equation in two variables can be represented in the form of ax + by + c = 0, where a, b and c are real numbers, but a and b are not equal to zero. Whereas, in a pair of linear equations in two variables, we deal with two such equations.

### 4. Solve 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0

These are two given equations:

9x + 3y + 12 = 0              (1)

18x + 6y + 24 = 0            (2)

By comparing these two equations, we will get:

a1x + b1y + c1 = 0

And a2x + b2y + c2 = 0

We will get the following coefficients as follows:

a1  = 9, b1 = 3, c1 = 12

a2 = 18, b2 = 6, c2 = 24

Now,

(a1/a2) = 9/18 = ½

(b1/b2) = 3/6 = ½

(c1/c2) = 12/24 = ½

Hence, (a1/a2) = (b1/b2) = (c1/c2)

Therefore, the pair of equations in the question have infinite solutions, and the lines are coincident.