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NCERT Solutions for Exercise 3.4 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

NCERT Solutions for Exercise 3.4 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Edited By Ramraj Saini | Updated on Nov 13, 2023 03:03 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.4

NCERT Solutions for class 10 maths ex 3.4 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 3.4 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.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.4
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 3 Exercise 3.4
  3. Assess NCERT Solutions for Class 10 Maths chapter 3 exercise 3.4
  4. More About NCERT Solutions for Class 10 Maths Exercise 3.4
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 3.4
  6. NCERT Solutions of Class 10 Subject Wise
  7. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 3.4 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables
NCERT Solutions for Exercise 3.4 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

These class 10 maths ex 3.4 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. 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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Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excrcise: 3.4

Q1(i) Solve the following pair of linear equations by the elimination method and the substitution method :

x + y =5 \ \textup{and} \ 2x - 3y = 4

Answer:

Elimination Method:

Given, equations

\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)

Now, multiplying (1) by 3 we, get

\\3x +3 y =15............(3)

Now, Adding (2) and (3), we get

\\2x-3y+3x +3 y =4+15

\Rightarrow 5x=19

\Rightarrow x=\frac{19}{5}

Substituting this value in (1) we, get

\frac{19}{5}+y=5

\Rightarrow y=5-\frac{19}{5}

\Rightarrow y=\frac{6}{5}

Hence,

x=\frac{19}{5}\:and\:y=\frac{6}{5}

Substitution method :

Given, equations

\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)

Now, from (1) we have,

y=5-x.......(3)

substituting this value in (2)

2x-3(5-x)=4

\Rightarrow 2x-15+3x=4

\Rightarrow 5x=19

\Rightarrow x=\frac{19}{5}

Substituting this value of x in (3)

\Rightarrow y=5-x=5-\frac{19}{5}=\frac{6}{5}

Hence,

x=\frac{19}{5}\:and\:y=\frac{6}{5}

Q1(ii) Solve the following pair of linear equations by the elimination method and the substitution method :

3x + 4 y = 10 \ \textup{and} \ 2x - 2y = 2

Answer:

Elimination Method:

Given, equations

\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)

Now, multiplying (2) by 2 we, get

\\4x -4 y =4............(3)

Now, Adding (1) and (3), we get

\\3x+4y+4x -4 y =10+4

\Rightarrow 7x=14

\Rightarrow x=2

Putting this value in (2) we, get

2(2)-2y=2

\Rightarrow 2y=2

\Rightarrow y=1

Hence,

x=2\:and\:y=1

Substitution method :

Given, equations

\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)

Now, from (2) we have,

y=\frac{2x-2}{2}=x-1.......(3)

substituting this value in (1)

3x+4(x-1)=10

\Rightarrow 3x+4x-4=10

\Rightarrow 7x=14

\Rightarrow x=2

Substituting this value of x in (3)

\Rightarrow y=x-1=2-1=1

Hence,

x=2\:and\:y=1

Q1(iii} Solve the following pair of linear equations by the elimination method and the substitution method: (iii) 3x - 5y -4 = 0\ \textup{and} \ 9x = 2y + 7

Answer:

Elimination Method:

Given, equations

\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7

\\\Rightarrow 9x - 2y -7=0........(2)

Now, multiplying (1) by 3 we, get

\\9x -15 y -12=0............(3)

Now, Subtracting (3) from (2), we get

9x-2y-7-9x+15y+12=0

\Rightarrow 13y+5=0

\Rightarrow y=\frac{-5}{13}

Putting this value in (1) we, get

3x-5(\frac{-5}{13})-4=0

\Rightarrow 3x=4-\frac{25}{13}

\Rightarrow 3x=\frac{27}{13}

\Rightarrow x=\frac{9}{13}

Hence,

x=\frac{9}{13}\:and\:y=-\frac{5}{13}

Substitution method :

Given, equations

\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7

\\\Rightarrow 9x - 2y -7=0........(2)

Now, from (2) we have,

y=\frac{9x-7}{2}.......(3)

substituting this value in (1)

3x-5\left(\frac{9x-7}{2} \right )-4=0

\Rightarrow 6x-45x+35-8=0

\Rightarrow -39x+27=0

\Rightarrow x=\frac{27}{39}=\frac{9}{13}

Substituting this value of x in (3)

\Rightarrow y=\frac{9(9/13)-7}{2}=\frac{81/13-7}{2}=\frac{-5}{13}

Hence,

x=\frac{9}{13}\:and\:y=-\frac{5}{13}

Q1(iv) Solve the following pair of linear equations by the elimination method and the substitution method :(iv) \frac{x}{2} + \frac{2y}{3} = -1\ \textup{and} \ x - \frac{y}{3} = 3

Answer:

Elimination Method:

Given, equations

\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)

Now, multiplying (2) by 2 we, get

\\2x - \frac{2y}{3} =6............(3)

Now, Adding (1) and (3), we get

\\\frac{x}{2}+\frac{2y}{3}+2x-\frac{2y}{3}=-1+6

\Rightarrow \frac{5x}{2}=5

\Rightarrow x=2

Putting this value in (2) we, get

2-\frac{y}{3}=3

\Rightarrow \frac{y}{3}=-1

\Rightarrow y=-3

Hence,

x=2\:and\:y=-3

Substitution method :

Given, equations

\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)

Now, from (2) we have,

y=3(x-3)......(3)

substituting this value in (1)

\frac{x}{2}+\frac{2(3(x-3))}{3}=-1

\Rightarrow \frac{x}{2}+2x-6=-1

\Rightarrow \frac{5x}{2}=5

\Rightarrow x=2

Substituting this value of x in (3)

\Rightarrow y=3(x-3)=3(2-1)=-3

Hence,

x=2\:and\:y=-3

Q2(i) Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \frac{1}{2 } if we only add 1 to the denominator. What is the fraction?

Answer:

Let the numerator of the fraction be x and denominator is y,

Now, According to the question,

\frac{x+1}{y-1}=1

\Rightarrow x+1=y-1

\Rightarrow x-y=-2.........(1)

Also,

\frac{x}{y+1}=\frac{1}{2}

\Rightarrow 2x=y+1

\Rightarrow 2x-y=1..........(2)

Now, Subtracting (1) from (2) we get

x=3

Putting this value in (1)

3-y=-2

\Rightarrow y=5

Hence

x=3\:and\:y=5

And the fraction is

\frac{3}{5}

Q2(ii) Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer:

Let the age of Nuri be x and age of Sonu be y.

Now, According to the question

x-5=3(y-5)

\Rightarrow x-5=3y-15

\Rightarrow x-3y=-10.........(1)

Also,

x+10=2 (y+10)

\Rightarrow x+10=2y+20

\Rightarrow x-2y=10........(2)

Now, Subtracting (1) from (2), we get

y=20

putting this value in (2)

x-2(20)=10

\Rightarrow x=50

Hence the age of Nuri is 50 and the age of Nuri is 20.

Q2(iii) Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer:

Let the unit digit of the number be x and 10's digit be y.

Now, According to the question,

x+y=9.......(1) 24323

Also

9(10y+x)=2(10x+y)

\Rightarrow 90y+9x=20x+2y

\Rightarrow 88y-11x=0

\Rightarrow 8y-x=0.........(2)

Now adding (1) and (2) we get,

\Rightarrow 9y=9

\Rightarrow y=1

now putting this value in (1)

x+1=9

\Rightarrow x=8

Hence the number is 18.

Q2(iv) Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Answer:

Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.

Now, According to the question,

x+y=25..........(1)

And

50x+100y=2000

\Rightarrow x+2y=40.............(2)

Now, Subtracting(1) from (2), we get

y=15

Putting this value in (1).

x+15=25

\Rightarrow x=10

Hence Meena received 10 50 Rs notes and 15 100 Rs notes.

Q2(v) Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:

Let fixed charge be x and per day charge is y.

Now, According to the question,

x+4y=27...........(1)

And

x+2y=21...........(2)

Now, Subtracting (2) from (1). we get,

4y-2y=27-21

\Rightarrow 2y=6

\Rightarrow y=3

Putting this in (1)

x+4(3)=27

\Rightarrow x=27-12=15

Hence the fixed charge is 15 Rs and the per day charge is 3 Rs.

More About NCERT Solutions for Class 10 Maths Exercise 3.4

In the elimination method from NCERT solutions for Class 10 Maths exercise 3.4 we pick a variable which we need to eliminate then we multiply both the equations with non-zero constants such that the coefficient of the variable we picked becomes numerically equal but with opposite signs (one negative the other positive) so that they get cancelled when we add the equation.

Thus, we are left with the other variable which we can solve easily and get the value of the remaining variable. Once we get the value of the other variable, we substitute it to any of the equations to get the value of the first variable.

In this example, we will eliminate ‘y’ from the equation

x + 2y = 4 ..........(1)

2x - y = 3 ............(2)

We multiply equation (I) with 1 and equation (II) with 2 to make the coefficient of ‘y’ numerically equal.

x + 2y = 4 ..........(3)

4x - 2y = 6 ............(4)

Now add these two equations we get.

5x = 10

x = 10/2 = 2

Now substituting the value of ‘x’ on equation (I) we get.

(2) + 2y = 4

2y = 2

y = 1

So, we have ‘x’ as 2 and ‘y’ as 1.

The elimination method can only be applied if the ratios of the coefficient are different. It doesn’t solve the equations if the ratios are the same.

In the above example, the ratio of equation (I) is 1:2 and the ratio of Equation (II) is 2: -1 which is different. 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 3.4

  • Class 10 Maths chapter 3 exercise 3.4 introduces us to a more convenient, easier and time-efficient method for solving a pair of linear equations by elimination method.
  • The elimination from NCERT solutions for Class 10 Maths exercise 3.4 method is quite conceptual and this method makes dealing with higher-level questions easier.
  • Apart from solving through the algebraic method and graphical method the elimination method of Class 10 Maths chapter 3 exercise 3.4 is quite an easier and time-efficient method to solve the pair of linear equations.
  • In some exams, the question asks only the value of one variable so by using the elimination method of Class 10th Maths chapter 3 exercise 3.4 we can eliminate the second variable and end up with the value of the first (required) variable only.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. Choose the convenient and time efficient method to solve linear a pair of linear equations?

  1. Graphical Method 

  2. Elimination method 

Elimination method as it is time-efficient and not lengthy.

2. What is the basic concept of elimination method according to NCERT solutions for Class 10 Maths 3 exercise 3.4?

We multiply both the equations with some non-zero constant to eliminate one of the variables by adding both of the new equations.

3. Why do we multiply both the equations with some non-zero variable?

In order to make the coefficient of one of the variables numerically equal but with opposite signs so that they get cancelled when we add them together.

4. Does substitution play any role in elimination method?

Yes, once we find the value of the first variable, we substitute its value in any of the equation to get the value of the second equation.

5. Why graphical method is not always useful?

It doesn’t provide us the desired point if the values are irrational.

6. Can we apply elimination method in the equation?

It doesn’t provide us the desired point if the values are irrational.

7. Can we apply elimination method in the equation?

It doesn’t provide us the desired point if the values are irrational.

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sai raja 01 September,2024
show the previous year exams ?

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

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as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

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best of luck!

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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.

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