**NCERT Solutions for Class 10 Maths Chapter 3 – ****Pair of Linear Equations in Two Variable**

CBSE prescribes NCERT textbooks for class 10. We have compiled ncert solutions for class 10 maths chapter 3 for you. Mathematics requires a clear understanding of the concepts, logic and a lot of practice. Hence, we are providing you ncert solutions for class 10 maths chapter 3. Our NCERT solutions for class 10 maths chapter 3 cover all the basic concepts provided in the NCERT textbook.

These ncert solutions for class 10 maths chapter 3 will help you to effectively prepare for your board examination. ncert solutions for class 10 maths chapter 3 are prepared by subject experts. In ncert solutions for class 10 maths chapter 3 the concepts are explained in details and all the doubts are instantly resolved by our subject matter experts during live doubt solving sessions. You will also be able to understand all the difficult concepts by referring to ncert solutions for class 10 maths chapter 3 and score good marks in your exams.

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**CBSE Class 10 Maths Chapter 3 NCERT Solutions – ****Pair of Linear Equations in Two Variable**

NCERT solutions for class 10 maths chapter 3 explains the concept of pair of linear equations in two variables, graphical method of solution of a pair of linear equations in two variable, algebraic methods of solving a pair of linear equations, substitution, elimination, cross-multiplication method and equations reducible to a pair of linear equations in two variable, different forms of equations which can be converted into a pair of linear equations. The chapter concludes with summarizing points that help you to revise the chapter and its concepts quickly.

**Sub-topics covered under Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variable**

- 3.1 -Introduction
- 3.2 – Graphical Method of Solution of a Pair of Linear Equations
- 3.3 – Algebraic Methods of Solving a Pair of Linear Equations
- 3.4 – Elimination and Substitution Method
- 3.5 – Cross-Multiplication Method
- 3.6 – Equations Reducible to a Pair of Linear Equations in Two Variables
- 3.7 – Summary

**NCERT Solutions for Class 10 Maths Chapter 3**

**Q.1 Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear. **

**Solution **

The given points are collinear then the area of triangle formed by these points will be 0 ∴ 12 [ x1(y2 − y1) + x2(y3 − y1) + x3(y1 − y2) ]

= 0

Here (x1, y1) = (x, y)

(x2, y2) = (1, 2)

(x3, y3) = (7, 0) ⟹ 21[x(2 − 0) + 1(0 − y) + 7(y − 2)] = 0 ⟹ 21[2x − y + 7y − 14] = 0 ⟹ 21[2x + 6y − 14] = 0

⟹ 2x + 6y − 14 = 0

⟹ x + 3y − 7 = 0

Hence this is the required relation between x and y.

**Q.2 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. **

**Solution **

Consider Aftab’s age as x and his daughter’s age as y.

Then , seven years ago,

Aftab’s age = x − 7

His daughter’s age = y − 7

According to the question,

x − 7 = 7(y − 7)

x − 7 = 7y − 49

x − 7y = − 49 + 7

x − 7y = − 42 …(i)

After three years,

Aftab’s age = x + 3

His daughter’s age = y + 3

According to the question,

x + 3 = 3(y + 3)

x + 3 = 3y + 9

x − 3y = 9 − 3

x − 3y = 6 …(ii)

Representing equation (i) and (ii) geometrically, we plot these equations by finding points on the lines representing these two equations

x − 7y = − 42 ⟹ x = 7y − 42

x = 7y − 42 42 35 49 y 12 11 13

x − 3y = 6 ⟹ x = 3y + 6

x = 3y + 6 42 36 48 y 12 10 14

From the graph we can see that two lines will intersect at a point.

x − 7y = − 42

x − 3y = 6

On subtracting the two equations, we get

4y = 48

or y = 12

Substituting value of y in (2),

x − 36 = 6

∴ x = 42

**Q. 3 The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. **

**Solution **

Let the cost of a bat be Rs. x and the cost of a ball be Rs. y.

Then algebraically representation of the question is –

3x + 6y = 3900 ⇒ x + 2y = 1300……(1)

x + 3y = 1300……..(2)

To represent it geometrically, we plot the two equations

x + 3y = 1300 ⇒ y = 1300 3

− x

x 100 1000 1300 y = 1300 3 − x

400 100 0

x + 2y = 1300 ⇒ y = 1300 2

− x

x 100 500 1300 y = 1300 2 − x

600 400 0

As we can see, the lines intersect at (1300, 0) which is the solution for this question.

**Q. 4 The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.**

**Solution **

x 65 55 45 35 y 20 40 60 80

Let the cost of 1 kg of apples be x and that of 1 kg be y

So the algebraic representation can be as follows:

2x + y = 160

4x + 2y = 300 ⇒ 2x + y = 150

The situation can be represented graphically by plotting these two equations.

2x + y = 160 ⇒ y = 160 − 2x

x 70 60 50 40 y = 160 − 2x 20 40 60 80

2x + y = 150 ⇒ y = 150 − 2x

x 65 55 45 35 y = 150 − 2x 20 40 60 80

We can see that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

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