NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables: It is an important chapter of algebra where you will learn to solve the linear equation having two variables. In solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables, you will be getting the entire coverage of the solutions including the optional exercise. Many reallife situations can be formulated using Mathematical equations. For example, consider the statement "cost of 1 Kg Apple and 2Kg orange is 120 and the cost of 3 Kg Apple and 1 Kg orange is 210". The statement can be formulated using the Mathematical equation, for this consider the cost of Apple as x and that of orange as y. Then we can write two equations as x+2y=120 and 3x+y=210. CBSE NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables, introduces the multiple methods to solve the two linear equations. The problems related to this chapter are very interesting and are important in the board exam and also for many other competitive exams. CBSE NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables will help you in boosting your preparation for all type of examinations. Crossmultiplication method to solve the linear equations is also given in the NCERT Solutions of this chapter.
Solving a linear equation with two variables.
Representation of linear equation in a graph.
Solutions of linear equations using the graph.
Algebraic interpretation of linear equations.
Formation of linear equations using statements.
Let x be the age of Aftab and y be the age of his daughter
Now, According to the question,
Also,
Now, let's represent both equations graphically,
From (1), we get
So, Putting different values of x we get corresponding values of y
X 
0 
7 
7 
Y 
6 
7 
5 
And From (2) we get,
So, Putting different values of x we get corresponding values of y
X 
0 
3 
6 
Y 
2 
1 
0 
GRAPH:
Let the price of one Bat be x and the price of one ball be y,
Now, According to the question,
From(1) we have
By putting different values of x, we get different corresponding values of y.So
X 
100 
300 
100 
Y 
600 
500 
700 
Now, From (2), we have,
By putting different values of x, we get different corresponding values of y.So
X 
100 
400 
200 
Y 
400 
300 
500 
GRAPH:
Let, x be the cost of 1kg apple and y be the cost of 1kg grapes.
Now, According to the question,
On a day:
After One Month:
Now, From (1) we have
Putting different values of x we get corresponding values of y, so,'
X 
80 
60 
50 
Y 
0 
40 
60 
And From (2) we have,
Putting different values of x we get corresponding values of y, so,
X 
50 
60 
70 
Y 
50 
30 
10 
Graph:
Let the number of boys is x and the number of girls is y.
Now, According to the question,
Total number of students in the class = 10, i.e.
And
the number of girls is 4 more than the number of boys,i.e.
Different points (x, y) for equation (1)
X 
5 
6 
4 
Y 
5 
4 
6 
Different points (x,y) satisfying (2)
X 
5 
6 
7 
y 
1 
2 
3 
Graph,
As we can see from the graph, both lines intersect at the point (7,3). that is x= 7 and y = 3 which means the number of boys in the class is 7 and the number of girls in the class is 3.
Let x be the price of 1 pencil and y be the price of 1 pen,
Now, According to the question
And
Now, the points (x,y), that satisfies the equation (1) are
X 
3 
4 
10 
Y 
5 
10 
0 
And, the points(x,y) that satisfies the equation (2) are
X 
3 
8 
2 
Y 
5 
2 
12 
The Graph,
As we can see from the Graph, both line intersects at point (3,5) that is, x = 3 and y = 5 which means cost of 1 pencil is 3 and the cost of 1 pen is 5.
Give, Equations,
Comparing these equations with , we get
As we can see
It means that both lines intersect at exactly one point.
Given, Equations,
Comparing these equations with , we get
As we can see
It means that both lines are coincident.
Q2 On comparing the ratios , and , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:(iii)
Give, Equations,
Comparing these equations with , we get
As we can see
It means that both lines are parallel to each other.
Answer:
Give, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have exactly one solution and thus pair of linear equations is consistent.
Given, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have no solution and thus pair of linear equations is inconsistent.
Given, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have exactly one solution and thus pair of linear equations is consistent.
Given, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.
Given, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.
Given, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.
The points (x,y) which satisfies in both equations are
X 
1 
3 
5 
Y 
4 
2 
0 
Given, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have no solution and thus pair of linear equations is inconsistent.
Given, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have exactly one solution and thus pair of linear equations is consistent.
Now The points(x, y) satisfying the equation are,
X 
0 
2 
3 
Y 
6 
2 
0 
And The points(x,y) satisfying the equation are,
X 
0 
1 
2 
Y 
2 
0 
2 
GRAPH:
As we can see both lines intersects at point (2,2) and hence the solution of both equations is x = 2 and y = 2.
Given, Equations,
Comparing these equations with , we get
As we can see
It means the given equations have no solution and thus pair of linear equations is inconsistent.
Let be the length of the rectangular garden and be the width.
Now, According to the question, the length is 4 m more than its width.i.e.
Also Given Half Parameter of the rectangle = 36 i.e.
Now, as we have two equations, on adding both equations, we get,
Putting this in equation (1),
Hence Length and width of the rectangle are 20m and 16 respectively.
As we know that the condition for the intersection of lines , is ,
So Any line with this condition can be
Here,
As
the line satisfies the given condition.
Given the equation,
As we know that the condition for the lines , for being parallel is,
So Any line with this condition can be
Here,
As
the line satisfies the given condition.
Given the equation,
As we know that the condition for the coincidence of the lines , is,
So any line with this condition can be
Here,
As
the line satisfies the given condition.
Given, two equations,
And
The points (x,y) satisfying (1) are
X 
0 
3 
6 
Y 
1 
4 
7 
And The points(x,y) satisfying (2) are,
X 
0 
2 
4 
Y 
6 
3 
0 
GRAPH:
24151
As we can see from the graph that both lines intersect at the point (2,3), And the vertices of the Triangle are ( 1,0), (2,3) and (4,0). The area of the triangle is shaded with a green color.
Given, two equations,
Now, from (1), we have
Substituting this in (2), we get
Substituting this value of x in (3)
Hence, Solution of the given equations is x = 9 and y = 5.
Q1 Solve the following pair of linear equations by the substitution method (ii)
Given, two equations,
Now, from (1), we have
Substituting this in (2), we get
Substituting this value of t in (3)
Hence, Solution of the given equations is s = 9 and t = 6.
Given, two equations,
Now, from (1), we have
Substituting this in (2), we get
This is always true, and hence this pair of the equation has infinite solutions.
As we have
,
One of many possible solutions is .
Answer:
Given, two equations,
Now, from (1), we have
Substituting this in (2), we get
Substituting this value of x in (3)
Hence, Solution of the given equations is,
.
Given, two equations,
Now, from (1), we have
Substituting this in (2), we get
Substituting this value of x in (3)
Hence, Solution of the given equations is,
.
Given,
From (1) we have,
Putting this in (2) we get,
putting this value in (3) we get,
Hence
Q2 Solve and and hence find the value of ‘’ for which .
Given, two equations,
Now, from (1), we have
Substituting this in (2), we get
Substituting this value of x in (3)
Hence, Solution of the given equations is,
Now,
As it satisfies ,
Hence Value of m is 1.
(i) The difference between the two numbers is 26 and one number is three times the other. Find them.
Let two numbers be x and y and let the bigger number is y.
Now, According to the question,
And
Now, the substituting value of y from (2) in (1) we get,
Substituting this in (2)
Hence the two numbers are 13 and 39.
Let the larger angle be x and smaller angle be y
Now, As we know the sum of supplementary angles is 180. so,
Also given in the question,
Now, From (2) we have,
Substituting this value in (1)
Now, Substituting this value of x in (3), we get
Hence the two supplementary angles are
Let the cost of 1 bat is x and the cost of 1 ball is y.
Now, According to the question,
Now, From (1) we have
Substituting this value of y in (2)
Now, Substituting this value of x in (3)
Hence, The cost of one bat is 500 Rs and the cost of one ball 50 Rs.
Let the fixed charge is x and the per km charge is y.
Now According to the question
And
Now, From (1) we have,
Substituting this value of x in (2), we have
Now, Substituting this value in (3)
Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.
Now, Fair For 25 km :
Hence fair for 25km is 255 Rs.
Let the numerator of the fraction be x and denominator of the fraction is y
Now According to the question,
Also,
Now, From (1) we have
Substituting this value of y in (2)
Substituting this value of x in (3)
Hence the required fraction is
Let x be the age of Jacob and y be the age of Jacob's son,
Now, According to the question
Also,
Now,
From (1) we have,
Substituting this value of x in (2)
Substituting this value of y in (3),
Hence, Present age of Jacob is 40 years and the present age of Jacob's son is 10 years.
Q1 Solve the following pair of linear equations by the elimination method and the substitution method :
Elimination Method:
Given, equations
Now, multiplying (1) by 3 we, get
Now, Adding (2) and (3), we get
Substituting this value in (1) we, get
Hence,
Substitution method :
Given, equations
Now, from (1) we have,
substituting this value in (2)
Substituting this value of x in (3)
Hence,
Q1 Solve the following pair of linear equations by the elimination method and the substitution method :
Elimination Method:
Given, equations
Now, multiplying (2) by 2 we, get
Now, Adding (1) and (3), we get
Putting this value in (2) we, get
Hence,
Substitution method :
Given, equations
Now, from (2) we have,
substituting this value in (1)
Substituting this value of x in (3)
Hence,
Elimination Method:
Given, equations
Now, multiplying (1) by 3 we, get
Now, Subtracting (3) from (2), we get
Putting this value in (1) we, get
Hence,
Substitution method :
Given, equations
Now, from (2) we have,
substituting this value in (1)
Substituting this value of x in (3)
Hence,
Elimination Method:
Given, equations
Now, multiplying (2) by 2 we, get
Now, Adding (1) and (3), we get
Putting this value in (2) we, get
Hence,
Substitution method :
Given, equations
Now, from (2) we have,
substituting this value in (1)
Substituting this value of x in (3)
Hence,
Let the numerator of the fraction be x and denominator is y,
Now, According to the question,
Also,
Now, Subtracting (1) from (2) we get
Putting this value in (1)
Hence
And the fraction is
Let the age of Nuri be x and age of Sonu be y.
Now, According to the question
Also,
Now, Subtracting (1) from (2), we get
putting this value in (2)
Hence the age of Nuri is 50 and the age of Nuri is 20.
Let the unit digit of the number be x and 10's digit be y.
Now, According to the question,
24323
Also
Now adding (1) and (2) we get,
now putting this value in (1)
Hence the number is 18.
Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.
Now, According to the question,
And
Now, Subtracting(1) from (2), we get
Putting this value in (1).
Hence Meena received 10 50 Rs notes and 15 100 Rs notes.
Let fixed charge be x and per day charge is y.
Now, According to the question,
And
Now, Subtracting (2) from (1). we get,
Putting this in (1)
Hence the fixed charge is 15 Rs and per day charge is 3 Rs.
Given, two equations,
Comparing these equations with , we get
As we can see,
Hence, the pair of equations has no solution.
Given, two equations,
Comparing these equations with , we get
As we can see,
Hence, the pair of equations has exactly one solution.
By Cross multiplication method,
Given the equations,
Comparing these equations with , we get
As we can see,
Hence, the pair of equations has infinitely many solutions.
Given the equations,
Comparing these equations with , we get
As we can see,
Hence, the pair of equations has exactly one solution.
By Cross multiplication method,
Given equations,
As we know, the condition for equations to have an infinite solution is
So, Comparing these equations with, , we get
From here we get,
Also,
Now, Subtracting (2) from (1) we get
Substituting this value in (1)
Hence, .
Q2 (ii) For which value of k will the following pair of linear equations have no solution?
Given, the equations,
As we know, the condition for equations to have no solution is
So, Comparing these equations with, , we get
From here we get,
Hence, the value of K is 2.
Q3 Solve the following pair of linear equations by the substitution and crossmultiplication methods :
Given the equations
By Substitution Method,
From (1) we have
Substituting this in (2),
Substituting this in (3)
Hence .
By Cross Multiplication Method
Let the fixed charge be x and the cost of food per day is y,
Now, According to the question
Also
Now subtracting (1) from (2),
Putting this value in (1)
Hence, the Fixed charge is Rs 400 and the cost of food per day is Rs 30.
Let numerator of a fraction be x and the denominator is y.
Now, According to the question,
Also,
Now, Subtracting (1) from (2) we get,
Putting this value in (2) we get,
Hence, the fraction is
.
Let the number of right answer and wrong answer be x and y respectively
Now, According to the question,
And
Now, subtracting (2) from (1) we get,
Putting this value in (1)
Hence the total number of question is
Let the speed of the first car is x and the speed of the second car is y.
Let's solve this problem by using relative motion concept,
the relative speed when they are going in the same direction= x  y
the relative speed when they are going in the opposite direction= x + y
The given relative distance between them = 100 km.
Now, As we know,
Relative distance = Relative speed * time .
So, According to the question,
Also,
Now Adding (1) and (2) we get
putting this in (1)
Hence the speeds of the cars are 40 km/hour and 60 km/hour.
Let be the length of the rectangle and be the width,
Now, According to the question,
Also,
By Cross multiplication method,
Hence the length and width of the rectangle are 17 units and 9 units respectively.
Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:
Given Equations,
Let,
Now, our equation becomes
And
By Cross Multiplication method,
And Hence,
Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:
Given Equations,
Let,
Now, our equation becomes
And
By Cross Multiplication method,
So,
.
And hence
Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:
Given Equations,
Let,
Now, our equation becomes
And
By Cross Multiplication method,
And Hence,
Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:
Given Equations,
Let,
Now, our equation becomes
And
Multiplying (1) by 3 we get
Now, adding (2) and (3) we get
Putting this in (2)
Now,
Hence,
Q1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v)
Given Equations,
Let,
Now, our equation becomes
And
By Cross Multiplication method,
And Hence,
Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:
Given Equations,
Let,
Now, our equation becomes
And
By Cross Multiplication method,
And Hence,
Given Equations,
Let,
Now, our equation becomes
And
By Cross Multiplication method,
Now,
And,
Adding (3) and (4) we get,
Putting this value in (3) we get,
And Hence,
Given Equations,
Let,
Now, our equation becomes
And
Now, Adding (1) and (2), we get
Putting this value in (1)
Now,
And
Now, Adding (3) and (4), we get
Putting this value in (3),
Hence,
Let the speed of Ritu in still water be x and speed of current be y,
Let's solve this problem by using relative motion concept,
the relative speed when they are going in the same direction (downstream)= x +y
the relative speed when they are going in the opposite direction (upstream)= x  y
Now, As we know,
Relative distance = Relative speed * time .
So, According to the question,
And,
Now, Adding (1) and (2), we get
Putting this in (2)
Hence,
Hence Speed of Ritu in still water is 6 km/hour and speed of the current is 4 km/hour
Let the number of days taken by woman and man be x and y respectively,
The proportion of Work done by a woman in a single day
The proportion of Work done by a man in a single day
Now, According to the question,
Also,
Let,
Now, our equation becomes
And
By Cross Multiplication method,
So,
Let the speed of the train and bus be u and v respectively
Now According to the question,
And
Let,
Now, our equation becomes
And
By Cross Multiplication method,
And Hence,
Hence the speed of the train and bus are 60 km/hour and 80 km/hour respectively.
Let the age of Ani be , age of Biju be ,
Case 1: when Ani is older than Biju
age of Ani's father Dharam:
and
age of his sister Cathy :
Now According to the question,
Also,
Now subtracting (1) from (2), we get,
putting this in (1)
Hence the age of Ani and Biju is 19 years and 16 years respectively.
Case 2:
And
Now Adding (3) and (4), we get,
putting it in (3)
Hence the age of Ani and Biju is 21 years and 24 years respectively.
Let the amount of money the first person and the second person having is x and y respectively
Noe, According to the question.
Also
Multiplying (2) by 2 we get,
Now adding (1) and (3), we get
Putting this value in (1)
Thus two friends had 40 Rs and 170 Rs respectively.
Let the speed of the train be v km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Now As we Know,
Now, According to the question,
Now, Using equation (1), we have
Also,
Adding equations (2) and (3), we obtain:
Substituting the value of x in equation (2), we obtain:
Putting this value in (1) we get,
Hence the distance covered by train is 600km.
Let the number of rows is x and the number of students in a row is y.
Total number of students in the class = Number of rows * Number of students in a row
Now, According to the question,
Also,
Subtracting equation (2) from (1), we get:
Substituting the value of y in equation (1), we obtain:
Hence,
The number of rows is 4 and the Number of students in a row is 9.
Total number of students in a class
:
Hence there are 36 students in the class.
Q5 In a , . Find the three angles.
Given,
Also, As we know that the sum of angles of a triangle is 180, so
Now From (1) we have
Putting this value in (2) we have
Putting this in (3)
And
Hence three angles of triangles
Given two equations,
And
Points(x,y) which satisfies equation (1) are:
X 
0 
1 
5 
Y 
5 
0 
20 
Points(x,y) which satisfies equation (1) are:
X 
0 
1 
2 
Y 
3 
0 
3 
GRAPH:
As we can see from the graph, the three points of the triangle are, (0,3),(0,5) and (1,0).
Given Equations,
Now By Cross multiplication method,
Q7 Solve the following pair of linear equations: (ii)
Given two equations,
Now By Cross multiplication method,
Q7 Solve the following pair of linear equations: (iii)
Given equation,
Now By Cross multiplication method,
Q7 Solve the following pair of linear equations: (iv)
Given,
And
Now, Subtracting (1) from (2), we get
Substituting this in (1), we get,
.
Hence,
Q7 Solve the following pair of linear equations: (v)
Given Equations,
As we can see by adding and subtracting both equations we can make our equations simple to solve.
So,
Adding (1) and )2) we get,
Subtracting (2) from (1) we get,
Now, Adding (3) and (4) we get,
Putting this value in (3)
Hence,
Q8 ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.
As we know that in a quadrilateral the sum of opposite angles is 180 degrees.
So, From Here,
Also,
Multiplying (1) by 3 we get,
Now,
Subtracting, (2) from (3) we get,
Substituting this value in (1) we get,
Hence four angles of a quadrilateral are :
Chapter No. 
Chapter Name 
Chapter 1 
CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers 
Chapter 2 

Chapter 3 
Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables 
Chapter 4 
CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations 
Chapter 5 
NCERT solutions for class 10 chapter 5 Arithmetic Progressions 
Chapter 6 

Chapter 7 
CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry 
Chapter 8 
NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 
Chapter 9 
Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry 
Chapter 10 

Chapter 11 

Chapter 12 
Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles 
Chapter 13 
CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes 
Chapter 14 

Chapter 15 
NCERT solutions are the most important tool when you are appearing for board examinations. 90% paper of CBSE board examinations, directly come from the NCERT.
Now you have done the NCERT solutions for class 10 maths chapter 2 polynomials and learned the approach to solving questions in the step by step method.
After covering NCERTs you should target the past year papers of CBSE board examinations. The previous year papers will cover the rest 10% part of the class 10 board exams.
Keep working hard & happy learning!