NCERT Solutions for class 10 Maths Chapter 3: Pair Of Linear Equations In Two Variables

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NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables are provided in this article. A pair of linear equations in two variables having a solution is known as a consistent pair of linear equations. Equivalent pair of linear equations has infinitely many distinct common solutions, such a pair of solutions is known as a dependent pair of linear equations in two variables.

Class 10 Maths Chapter 3 Linear Equations in Two Variables belongs to Unit 2 Algebra which has a weightage of 20 marks in the CBSE Class 10 Maths Examination. The NCERT solutions of the chapter include questions related to the Substitution method, Elimination method, and Cross-multiplication method.

Download PDF: NCERT Solutions for Class 10 Mathematics Chapter 3

NCERT Solutions for Class 10 Mathematics Chapter 3

NCERT Solutions

Important Topics in Class 10 Maths Chapter 3

Linear Equations are the equations in which the powers of all the involved variables are one.

The general form of a linear equation in two variables is ax + by + c = 0, where a and b cannot be simultaneously zero.

The solution of a linear equation in two variables is generally a pair of values, one for x and the other for y, which makes the two sides of the equation equal.

For example: If 3x + y = 6, then (0,6) is one of its solutions as it satisfies the equation.

Graph of Linear Equation in two variables: A linear equation in two variables can be geometrically represented as a straight line.

A pair of linear equations in two variables can be represented as shown below –

\(a_1x + b_1y+c_1=0\\ a_2x + b_2y+c_2=0\)

The solution for a consistent pair of linear equations can be found using various methods.

i) Elimination method

ii) Substitution Method

iii) Cross-multiplication of solving linear equations

iv) Graphical method

NCERT Solutions For Class 10 Maths Chapter 3 Exercises:

The detailed solutions for all the NCERT Solutions for Pair of Linear Equations in Two Variables under different exercises are as follows:

Related Topics:

Difference between linear and non-linear equations	Pair of Linear Equations in Two Variables Important Questions	Pair of Linear Equations in Two Variables Formula

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Geometry	Math Formula	Maths Study Material
NCERT Solutions for Class 10 Maths	Class 10 Maths Notes	Mensuration
Difference between in Maths	Calculus	Math MCQs

CBSE X Related Questions

1.
The natural number 1 is :
- a prime number.
- a composite number.
- prime as well as composite.
- neither prime nor composite.
View Solution
2.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
View Solution
3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
View Solution
4.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.
View Solution
5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
View Solution
6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
View Solution

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NCERT Solutions for class 10 Maths Chapter 3: Pair Of Linear Equations In Two Variables

NCERT Solutions for Class 10 Mathematics Chapter 3

Important Topics in Class 10 Maths Chapter 3

NCERT Solutions For Class 10 Maths Chapter 3 Exercises:

CBSE X Related Questions

1.
The natural number 1 is :

2.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

Comments

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NCERT Solutions for class 10 Maths Chapter 3: Pair Of Linear Equations In Two Variables

NCERT Solutions for Class 10 Mathematics Chapter 3

Important Topics in Class 10 Maths Chapter 3

NCERT Solutions For Class 10 Maths Chapter 3 Exercises:

CBSE X Related Questions

1.The natural number 1 is :

2.D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

3.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The natural number 1 is :

2.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)