NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3 - Pair of Linear Equations in two variables

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3 - Pair of Linear Equations in two variables

Komal MiglaniUpdated on 30 May 2025, 03:07 PM IST

There are various methods for solving the linear equation. In the previous exercises graphical method and one of the algebraic methods for solving the equation have been discussed. There are many algebraic methods for solving equations, like substitution and elimination methods. Similar to the substitution method elimination method is also used to solve the linear equation. In this method, multiply the equation by numbers to make the coefficient of the equation equal. This will help to eliminate the value and get the solution.

This Story also Contains

  1. Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3
  2. Assess NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3
  3. Topics Covered in Chapter 3 Pair of Linear Equations in Two Variables: Exercise 3.3
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3 - Pair of Linear Equations in two variables
NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

These class 10 maths exercise 3.3 solutions are designed as per the students' demand, covering comprehensive, step-by-step solutions of every problem. Practice is necessary for all these questions to command the concepts, boost confidence and in-depth understanding of concepts. In the exercise, all the NCERT solutions are covered according to the syllabus of the NCERT. These solutions help the students to understand the concept in depth. This exercise included all the questions given in the NCERT Books. In exercise 3.3 of the NCERT, the chapter solutions are explained in detail.

Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

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Assess NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

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 the 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 the 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 the 10's digit be y.

Now, according to the question,

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

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 the 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.

Also Read:

Topics Covered in Chapter 3 Pair of Linear Equations in Two Variables: Exercise 3.3

  1. Elimination Method: In this method, multiply the equations if required to make the coefficient of the equation equal. Then, eliminate one of the variables by adding or subtracting the equations.
  2. Reducing Equations to a Linear Form: In some problems need to simplify the equation, or in some equations need to apply the substitution method to convert the equation into a standard linear equation. After reducing these equations, the elimination method is applied.
  3. Solving Word Problems: In this exercise, some questions are based on word problems. To solve these questions, these word problems are converted into a pair of linear equations for solving them.
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NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

Q: What is the basic concept of elimination method according to NCERT solutions for Class 10 Maths 3 exercise 3.3?
A:

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

Q: Choose the convenient and time efficient method to solve linear a pair of linear equations?
A:
  1. Graphical Method 

  2. Elimination method 

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

Q: Why do we multiply both the equations with some non-zero variable?
A:

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.

Q: Does substitution play any role in elimination method?
A:

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.

Q: Why graphical method is not always useful?
A:

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

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