NCERT notes for class 10 Mathematics, chapter 3 - pair of linear equations will help you learn with fun. Below notes are comprised of concepts, cross-word exercises, problems and much more to boost your talent. 
Every student needs to go through the below content before start their practice with the Linear equations chapter for class 10 at least two to three times. Notes are very helpful for the quick revision and have become an integral part of the studies.

NCERT Revision Notes for Class 10

Find below NCERT Notes for CBSE students for Class 10: Chapter 3 pair of linear equations in two variables
1. Two linear equations in the same two variables are called a 
Pair of linear equations is a pair of 2 equations consist two same variables.
General form is:  
$a_1$ x + $b_1$y = $c_1$ and 
$a_2$ x + $b_2$ y = $c_2$ 
Where $a_1, a_2, b_1,b_2,c_1$ and $c_2$ are real numbers where $a_1^2 + b_1^2$ $\neq$ 0 and $a_2^2 + b_2^2$ $\neq$ 0.

2. A set of 2 linear equations can be solved by below listed methods:
i) Graphical Method : The graph of each equation is a straight line.
There are few conditions to evaluate the solution set or we can say that solution for the given linear equation is based on below conditions:

 Situation  Graph of equations  Solution Type
Pair of linear equations is consistent  If lines intersect at a point  Unique solution 
Pair of equations is dependent (consistent) If lines coincide Infinitely many solutions
Pair of equations is inconsistent If lines are parallel No solution

ii) Algebraic Methods : One of the most popular methods to solve a pair of linear equations :
a) Substitution Method
b) Elimination Method
c) Cross-multiplication Method

  Pair of Linear Equations with two Variables

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations

Some of the practice questions based on NCERT exercises are solved below:
Question 1: 7 years ago, Ali was 7 times as old as his son. Also, 3 years from now, Ali will be 3 times as old as she will be. Find their present age.
Solution:
Step 1: Represent data in a table
Let say x and y be Ali and his son's present age.
   Ali   Ali's Son Expression 
Present Age   x  y 
 7 years ago  x - 7  y - 7 x-7 = 7(y-7)
 3 years from now x + 3 y + 3  x + 3 = 3(y + 3)

Step 2: Simplify expressions
Now we have 2 expressions,
(x - 7) = 7 (y – 7 )
x - 7 = 7y - 49 
Combine like terms, we get
x - 7y = - 49 + 7 
x - 7y = - 42  say equation (1)

Again, (x + 3) = 3 (y + 3) 
x + 3 = 3y + 9 
x - 3y = 9 - 3 
x - 3y = 6 say equation (2)

Step 3: Solve for x and y
Use elimination method, to find the value of x and y
Subtract equation (2) from equation (1), we have
- 7y + 3y = -48
or -4y = - 48
or y = 48/4 = 12
Substitute y = 12 in equation (2),
x - 3 $\times$ 12 = 6
x - 36 = 6
or x = 42
Therefore, Ali and his son's present ages are 42 and 12 respectively.

Question 2: Sudheer's cricket coach told him to buy 3 balls and 6 balls for Rs. 3900 for first match. For second match again, he spent Rs 1300 to buy another bat and 3 more balls of the same kind. What is cost of one bat and one ball.
Solution:
Step 1: Represent data in a table
Let x be the cost of one bat and y be the cost of one ball.
     Expression
Cost of one bat   x  
Cost of one ball  y  
Cost of 3 bats and 6 balls 3x + 6y 3x + 6y = 3900 
Cost of one bat and 2 balls x + 2y x + 2y = 1300

Step 2: Simplify expressions
Now we have 2 algebraic expressions,
3x + 6y = 3900 ...equation (1)
x + 2y = 1300  ...equation (2)

Step 3: Solve for x and y
Use substitution method, to find the value of x and y
Isolate equation (2) for variable x
x = 1300 - 2y, substitute the value in the equation (1), we get

3 ( 1300 - 2y) + 6y = 3900
3900 - 6y + 6y = 3900
or 3900 = 3900
Hence, the system of given linear equations has infinity many solutions.

Question 3: Verify if below system is consistent
x + y = 5 and 2x + 2y = 10 
Solution
Given equations are:  x + y = 5; 2x + 2y = 10 
To check if the system is consistent or not, first find the ratios of the coefficients of x , y and constant term, that is,
$\frac{a_1}{a_2}$ = 1/2 
$\frac{b_1}{b_2}$ = 1/2
$\frac{c_1}{c_2}$ = 5/10 = 1/2 
From above result, we can conclude that, $\frac{a_1}{a_2}$ =  $\frac{b_1}{b_2}$ =  $\frac{c_1}{c_2}$ = $\frac{1}{2}$

Above system of equations has coincident pair of lines, which represent it has infinite solutions. Hence, given set of linear equations is consistent. 

Practice Problems for Linear Equations

Practice Problem 1: Check if the pair of linear equations is are consistent/ inconsistent.
2x + y - 6 = 0, 4x - 2y - 4 = 0

Practice Problem 2: Find the dimensions of the rectangular garden whose length is four meter more than its breadth and its half perimeter is 36 meters.

Practice Problem 3: Solve x - y + 1 = 0 and 3x + 2y - 12 = 0.

Practice Problem 4: Solve: x – y = 26 and x = 3y