NCERT Solutions for Exercise 3.1 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables
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NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.1
NCERT Solutions for class 10 maths ex 3.1 Pair of Linear Equations in two variables is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 3.1 class 10 deals with the graphical and algebraic representations of linear equations in two variables and their different type of solutions such as a unique solution, no solutions or infinitely many solutions, using the five methods namely the Graphical method, substitution method, cross Multiplication method, elimination method and determinant method.
These class 10 maths ex 3.1 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problems. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.
- Pair of Linear Equations in Two Variables Exercise 3.2
- Pair of Linear Equations in Two Variables Exercise 3.3
- Pair of Linear Equations in Two Variables Exercise 3.4
- Pair of Linear Equations in Two Variables Exercise 3.5
- Pair of Linear Equations in Two Variables Exercise 3.6
- Pair of Linear Equations in Two Variables Exercise 3.7
Access Exercise 3.1 Class 10 Maths Answers
Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.1
Answer: Let x be the age of Aftab and y be the age of his daughter
Now, According to the question,
$(x-7)=7\times(y-7)$
$\Rightarrow x-7=7y-49$
$\Rightarrow x-7y=-42.....(1)$
Also,
$(x+3)=3(y+3)$
$\Rightarrow x+3=3y+9$
$\Rightarrow x-3y=6.....(2)$
Now, let's represent both equations graphically,
From (1), we get
$y=\frac{x+42}{7}$
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,
$y=\frac{x-6}{3}$
So, Putting different values of x we get corresponding values of y
| X | 0 | 3 | 6 |
| Y | -2 | -1 | 0 |
GRAPH:
Answer:
Let the price of one Bat be x and the price of one ball be y,
Now, According to the question,
$3x+6y=3900.....(1)$
$x+3y=1300.....(2)$
From(1) we have
$y=\frac{3900-3x}{6}$
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,
$y=\frac{1300-x}{3}$
By putting different values of x, we get different corresponding values of y.So
| X | 100 | 400 | -200 |
| Y | 400 | 300 | 500 |
GRAPH:
Answer:
Let, x be the cost of 1kg apple and y be the cost of 1kg grapes.
Now, According to the question,
On a day:
$2x+y=160.....(1)$
After One Month:
$4x+2y=300.....(2)$
Now, From (1) we have
$y=160-2x$
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,
$y=\frac{300-4x}{2}=150-2x$
Putting different values of x we get corresponding values of y, so,
| X | 50 | 60 | 70 |
| Y | 50 | 30 | 10 |
Graph:
More About NCERT Solutions for Class 10 Maths Exercise 3.1
10th class maths exercise 3.1 answers – Consists of 3 simple questions in word problem format. In exercise 3.1 Class 10 Maths, If the system of equations is consistent and independent, then they have a unique solution and their condition is a1/a2 not equal to b1/b2 . Also if the system of equations is inconsistent and independent, then there will be no solution and their condition is a1/a2 = b1/b2 not equal to c1/c2 . Similarly, If the system of equations is consistent and dependent, then they have an infinite number of solutions and their condition is a1/a2 = b1/b2 = c1/c2 . The NCERT solutions for Class 10 Maths exercise 3.1 mainly focused on the graphical representation of the system of equations. All the three questions related to the graphical representation are given in exercise 3.1 Class 10 Maths. Also students can get access of Pair of linear equations in two variables notes to revise all the concepts discussed in this chapter.
Benefits of NCERT Solutions for Class 10 Maths Exercise 3.1
• NCERT solutions for Class 10 Maths is considered the best material for solving Class 10 Maths chapter 3 exercise 3.1 also assists us in solving and revising all of the questions in these exercises, as well as gaining a better understanding of the pair of linear equations in two variables.
• We can gain practise by solving all of the questions based on graphical representations of linear equations in two variables in the NCERT solution for Class 10th Maths chapter 3 exercise 3.1 and get a general idea about the topic, and we can get more marks in the examination if we practise it thoroughly.
• Exercise 3.1 Class 10 Maths, in grasping the fundamental notion of graphical and algebraic representations of linear equations in two variables, as well as the many types of solutions and their graphs, which are key concepts in the chapter.
Also see-
NCERT Solutions of Class 10 Subject Wise
Subject Wise NCERT Exemplar Solutions
Frequently Asked Questions (FAQs)
Q: What are the coefficients of the equation 7x - 5y = 10 ?
A:
The concepts related to coefficients of the equations are discussed in ex 3.1 class 10 comprehensively. Students can practice these questions and answer. as per this question, the coefficient of x is 7 and the coefficient of y is -5.
Q: What is the constant of the equation 7x - 5y = 10 ?
A:
The concepts related to coefficients of the equations are discussed in class 10 maths ex 3.1 comprehensively. Students can practice these questions and answer. As per this question, the constant of the equation 7x - 5y = 10 is 10.
Q: How many solutions are there for linear equations in two variables if the equations are consistent and dependent ?
A:
Concepts related to consistent and dependent equations are discussed in 10th class maths exercise 3.1 answers . according to this concept If the equations are consistent and dependent, then linear equations in two variables have infinite number solutions.
Q: How many methods are there in solving the linear equations in two variables?
A:
Mathematically, there are five methods in solving the linear equations in two variables. Practice problems enumerated in class 10 maths ex 3.1 to get deeper understanding of these methods.
Q: Find the value of y in 5x - y = 10 if x = 1 .
A:
Putting x = 1 in 5x - y = 10
5(1) - y = 10
-y = 10 - 5
Thus y = -5
Q: Solve 4x-10=2x-4
A:
On interchanging their sides,
4x - 2x = 10 - 4
2x = 6
Thus x = 3
Q: What is the substitution method according to NCERT solutions for Class 10 Maths chapter 3 exercise 3.1 ?
A:
Substituting the value of one variable in the other equation to get the value of another variable is known as the substitution method. Practice problems discussed in class 10 ex 3.1 to get deeper understanding about the substitution method to get solution of an equation.
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