NCERT Solutions for Exercise 3.1 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

# NCERT Solutions for Exercise 3.1 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Edited By Ramraj Saini | Updated on Nov 08, 2023 07:42 PM IST | #CBSE Class 10th

## 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.

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## 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:

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:

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

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### Frequently Asked Questions (FAQs)

1. What are the coefficients of the equation 7x - 5y = 10 ?

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.

2. What is the constant of the equation 7x - 5y = 10 ?

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.

3. How many solutions are there for linear equations in two variables if the equations are consistent and dependent ?

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.

4. How many methods are there in solving the linear equations in two variables?

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.

5. Find the value of y in 5x - y = 10 if x = 1 .

Putting x = 1 in 5x - y = 10

5(1) - y = 10

-y = 10 - 5

Thus y = -5

6. Solve 4x-10=2x-4

On interchanging their sides,

4x - 2x = 10 - 4

2x = 6

Thus x = 3

7. What is the substitution method according to NCERT solutions for Class 10 Maths chapter 3 exercise 3.1 ?

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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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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