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NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Ex 3.3

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NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.3

NCERT Solution Class 10 Maths provides solutions for Mathematics Book textbooks. You can download the Class 10 Maths NCERT Solutions Chapter 3 Exercise 3.3 from our website for free in PDF format. The NCERT Maths Class 10 PDF with the latest change are available on the Vedantu app and the website as per the latest CBSE syllabus.

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Class:

NCERT Solutions for Class 10

Subject:

Class 10 Maths

Chapter Name:

Chapter 3 - Pair of Linear Equations in Two Variables

Exercise:

Exercise - 3.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Vedantu's NCERT Solution not only include all probable types of questions that are relevant from a board exam perspective but also include sufficient and more well-solved examples and practice exercises. This gives you plenty of scope for practice. To obtain a better understanding of the exercise problems, download NCERT Solutions for Class 10 Maths Chapter 3. With the Vedantu learning app, you'll be able to take part in free conceptual videos and live masterclasses. You will also have access to all of the free PDFs for solutions and study materials. Students can also download Class 10 Science NCERT Solutions for free in PDF format only at Vedantu.

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NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Ex 3.3
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Access NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

Exercise 3.3

1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) $\mathbf{x}+\mathbf{y}=\mathbf{5}$ and $\mathbf{2x}\mathbf{3y}=\mathbf{4}$ 

Ans: Elimination method

The given equations are:

$x+y=5$          …… (i)

$2x-3y=4$   …… (ii)

Multiplying equation (ii) by $2$, we get

$2x+2y=10$   …… (iii)

Subtracting equation (ii) from equation (iii), we obtain

$5y=6$

$y=\dfrac{6}{5}$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$x=5-\dfrac{6}{5}$

$x=\dfrac{19}{5}$

Therefore, $x=\dfrac{19}{5}$ and $y=\dfrac{6}{5}$.

Substitution method:

From equation (i) we get

$x=5-y$      …… (v)

Substituting (v) in equation (ii), we get

$2\left( 5-y \right)-3y=4$

$-5y=-6$

$y=\dfrac{6}{5}$            …… (vi)

Substituting (vi) in equation (v), we obtain

$x=5-\dfrac{6}{5}$

$x=\dfrac{19}{5}$

Therefore, $x=\dfrac{19}{5}$ and $y=\dfrac{6}{5}$.


(ii) $\mathbf{3x}+\mathbf{4y}=\mathbf{10}$ and $\mathbf{2x}\mathbf{2y}=\mathbf{2}$

Ans: Elimination method

The given equations are:

$3x+4y=10$          …… (i)

$2x-2y=2$   …… (ii)

Multiplying equation (ii) by $2$, we get

$4x-4y=4$   …… (iii)

Adding equation (ii) and (iii), we obtain

$7x=14$

$x=2$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$6+4y=10$

$4y=4$

$y=1$

Therefore, $x=2$ and $y=1$.

Substitution method:

From equation (ii) we get

$x=1+y$      …… (v)

Substituting (v) in equation (i), we get

$3\left( 1+y \right)+4y=10$

$7y=7$

$y=1$            …… (vi)

Substituting (vi) in equation (v), we obtain

$x=1+1$

$x=2$

Therefore, $x=2$ and $y=1$.


(iii) $\mathbf{3x}\mathbf{5y}\mathbf{4}=\mathbf{0}$ and $\mathbf{9x}=\mathbf{2y}+\mathbf{7}$

Ans: Elimination method

The given equations are:

$3x-5y-4=0$          …… (i)

$9x=2y+7$

$9x-2y=7$   …… (ii)

Multiplying equation (i) by $3$, we get

$9x-15y-12=0$   …… (iii)

Subtracting equation (iii) from equation (ii), we obtain

$13y=-5$

$y=-\dfrac{5}{13}$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$3x+\dfrac{25}{13}-4=0$

$3x=\dfrac{27}{13}$

$x=\dfrac{9}{13}$

Therefore, $x=\dfrac{9}{13}$ and $y=-\dfrac{5}{13}$.

Substitution method:

From equation (i) we get

$x=\dfrac{5y+4}{3}$      …… (v)

Substituting (v) in equation (ii), we get

$9\left( \dfrac{5y+4}{3} \right)-2y-7=0$

$13y=-5$

$y=\dfrac{-5}{13}$            …… (vi)

Substituting (vi) in equation (v), we obtain

$x=\dfrac{5\left( \dfrac{-5}{13} \right)+4}{3}$

$x=\dfrac{9}{13}$

Therefore, $x=\dfrac{9}{13}$ and $y=\dfrac{-5}{13}$.


(iv) $\dfrac{x}{2}+\dfrac{2y}{3}=-1$ and $x-\dfrac{y}{3}=3$

Ans: Elimination method

The given equations are:

$\dfrac{x}{2}+\dfrac{2y}{3}=-1$

$3x+4y=-6$          …… (i)

$x-\dfrac{y}{3}=3$

$3x-y=9$   …… (ii)

Subtracting equation (ii) from equation (i), we obtain

$5y=-15$

$y=-3$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$3x+4\left( -3 \right)=-6$

$3x=6$

$x=2$

Therefore, $x=2$ and $y=-3$.

Substitution method:

From equation (ii) we get

$x=\dfrac{y+9}{3}$      …… (v)

Substituting (v) in equation (i), we get

$3\left( \dfrac{y+9}{3} \right)+4y=-6$

$5y=-15$

$y=-3$            …… (vi)

Substituting (vi) in equation (v), we obtain

$x=\dfrac{-3+9}{3}$

$x=2$

Therefore, $x=2$ and $y=-3$.


2. 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 $\mathbf{1}$ to the numerator and subtract $\mathbf{1}$ from the denominator, a fraction reduces to $\mathbf{1}$. It becomes $\dfrac{\mathbf{1}}{2}$ if we only add $\mathbf{1}$ to the denominator. What is the fraction?

Ans: Assuming the fraction be $\dfrac{x}{y}$.

Writing the algebraic representation using the information given in the question:

$\dfrac{x+1}{y-1}=1$

$x-y=-2$             …… (i)

$\dfrac{x}{y+1}=1$

$2x-y=1$    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

$x=3$              …… (iii)

Substituting the value of (iii) in equation (i), we get

$3-y=-2$

$y=5$

Therefore, $x=2$ and $y=-3$.

Hence the fraction is $\dfrac{3}{5}$.


(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?

Ans: Assuming the present age of Nuri be $x$ and the present age of Sonu be $y$.

Writing the algebraic representation using the information given in the question:

$\left( x-5 \right)=3\left( y-5 \right)$

$x-3y=-10$             …… (i)

$\left( x+10 \right)=2\left( y+10 \right)$

$x-2y=10$    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

$y=20$              …… (iii)

Substituting the value of (iii) in equation (i), we get

$x-60=-10$

$x=50$

Therefore, $x=50$ and $y=20$.

Hence Nuri’s present age is $50$ years and Sonu’s present age is $20$ years.


(iii) The sum of the digits of a two-digit number is $\mathbf{9}$. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Ans: Assuming the unit digit of the number be $x$ and the tens digit be $y$.

Therefore, the number is $10y+x$

The number after reversing the digits is $10x+y$.

Writing the algebraic representation using the information given in the question:

$x+y=9$             …… (i)

$9\left( 10y+x \right)=2\left( 10x+y \right)$

$-x+8y=0$    …… (ii)

Adding equation (i) and (ii), we obtain

$9y=9$

$y=1$              …… (iii)

Substituting the value of (iii) in equation (i), we get

$x=8$

Therefore, $x=8$ and $y=1$.

Hence the number is $10y+x=18$.


(iv) Meena went to bank to withdraw $\mathbf{Rs}\text{ }\mathbf{2000}$. She asked the cashier to give her $\mathbf{Rs}\text{ }\mathbf{50}$ and $\mathbf{Rs}\text{ }\mathbf{100}$ notes only. Meena got $\mathbf{25}$ notes in all. Find how many notes of $\mathbf{Rs}\text{ }\mathbf{50}$ and $\mathbf{Rs}\text{ }\mathbf{100}$ she received.

Ans: Assuming the number of $Rs\text{ }50$ notes be $x$ and the number of $Rs\text{ }100$ be $y$.

Writing the algebraic representation using the information given in the question:

$x+y=25$             …… (i)

$50x+100y=2000$    …… (ii)

Multiplying equation (i) by $50$, we obtain

$50x+50y=1250$     …… (iii)

Subtracting equation (iii) from equation (ii), we obtain

$50y=750$

$y=15$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$x=10$

Therefore, $x=10$ and $y=15$.

Hence Meena has $10$ notes of $Rs\text{ }50$ and $15$ notes of $Rs\text{ }100$.


(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid $\mathbf{Rs}\text{ }\mathbf{27}$ for a book kept for seven days, while Susy paid $\mathbf{Rs}\text{ }\mathbf{21}$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Ans: Assuming that the charge for first three days is $Rs\text{ }x$ and the charge for each day thereafter is $Rs\text{ y}$.

Writing the algebraic representation using the information given in the question:

$x+4y=27$             …… (i)

$x+2y=21$    …… (ii)

Subtracting equation (ii) from equation (i), we obtain

$2y=6$

$y=3$     …… (iii)

Subtracting equation (iii) from equation (i), we obtain

$x+12=27$

$x=15$              …… (iv)

Therefore, $x=15$ and $y=3$.

Hence, fixed charges are $Rs\text{ 15}$ and charges per day are $Rs\text{ 3}$.


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


Get the Class 10 Maths Free NCERT Solutions Chapter 3 Exercise 3.3 PDF. NCERT Solutions are highly helpful while doing your homework. The experienced Teachers prepare all the answers for Class 10 Maths. Detailed answers to all the questions in the NCERT TextBook Chapter 3, Maths Class 10 Pair of Linear Equations in Two Variables Exercise 3.3. 

Top academic experts at Vedantu have prepared the NCERT solutions provided here. These solutions will assist you in studying the question paper and in scoring the board exams well. In the form of PDF, you can also download these NCERT solutions and solve the problems offline when needed. As per the NCERT textbook, we offer step-by-step solutions to all exercise problems. Solving these problems will help you understand the chapter better and increase your marks. 

Practicing various types of questions will ensure that you understand the topic after studying Chapter 3. Free PDFs of NCERT Solutions are also available here and can be downloaded free of charge for online and offline use. Download solutions for NCERT now. You'll get 100% correct NCERT solutions for Maths Class 10 Chapter 3 solutions from our Maths Subject Experts. As per the NCERT guidelines, we also provide step-by-step solutions to the questions given in the math textbook. Download the Android Vedantu app here.


1. Solving pair of linear equations by Elimination method in two variables

In the Elimination method, either add or subtract the given linear equations which eliminate a variable and results in getting an equation in one variable. 

  • We can subtract the equations to eliminate one variable if the coefficients of any one of the variables are the same and the sign of the coefficients is the same.

  • Similarly, if the coefficients of any one of the variables are the same, but the sign of the coefficients are opposite, then we can add the linear equations to get the desired linear equation in one variable.


Steps in Elimination Method

  • Step 1 - Making the same coefficients: To make the coefficients of any one of the variables (either x or y) numerically equal, multiply both the given equations by some suitable non-zero constants.

  • Step 2 - Adding or Subtracting the Linear equations: After step 1, add or subtract one equation from the other in a way that one variable will be eliminated. Now, if you resulted in a linear equation in one variable, directly go to Step 3. Else;

  • If we obtain a true statement that includes no variable, then the original pair of equations will have infinitely many solutions.

  • If we obtain a false statement that includes no variable (2 ≠ 3), then the original pair of equations will have no solution, i.e., it is inconsistent.

  • Step 3 - Simplification: Simplify the equation in one variable (x or y) to get the variable value.

  • Step 4 - Substitution: Finally, Substitute the obtained variable value in any one of the given (original) equations to get the value of another unknown variable.


2. Solving pair of linear equations by Substitution method in two variables

The substitution method of solving linear equations is an algebraic method to solve simultaneous linear equations. As the name says, in the substitution method, the value of one variable from one equation is substituted in the second equation. By doing so, we get a pair of the linear equations transformed into one linear equation with only one variable, which can be simply solved.


Steps in Substitution Method

  • Step 1: Firstly, Simplify the given equations by expanding the parenthesis ( ).

  • Step 2: Transform one of the given linear equations for either variable x or variable y.

  • Step 3: Now, Substitute the variable value obtained in step 2 in the other equation.

  • Step 4: Solve the new equation obtained using basic arithmetic operations to get the variable value.

  • Step 5: Substitute the obtained variable value (from step 4) in any of the given equations and find out the value of the second variable.


Difference Between Substitution Method and Elimination method

As we discussed, the substitution method is the process of substituting and solving the equation to find the variable value, where the value of a variable is substituted in the other equation. And, the elimination method is the process of eliminating the variables in the equation so that the system of the equation can be with a single variable that can be easily solved.


We can conclude that the major difference between the Substitution and the Elimination method is that the substitution method involves replacing the variable with a value, whereas the elimination method involves removing the variable from the system of linear equations.


NCERT Solutions for Class 10 Maths Chapter 3 Exercises

Chapter 3 - Pair of Linear Equations in Two Variables Exercises in PDF Format

Exercise 3.1

7 Questions & Solutions

Exercise 3.2

2 Questions & Solutions


Benefits of NCERT Solutions for Class 10 Maths

To make your learning process easier, our professional and trained teachers have formulated all the solutions in a well-structured format. The following are the benefits that you would have if you use our free Class 10 Chapter 3 Maths Solutions: 

  • The NCERT PDF solution can be downloaded for offline use free of charge. 

  • Detailed and correct answers to all questions. 

  • The subject experts prepare each solution.

  • All solutions available are reliable and of high quality. 

  • All the responses comply with the CBSE guidelines. 

  • Vedantu expert teachers have prepared NCERT solutions in such a way to get maximum marks.

By downloading the free PDFs from Vedantu, you will obtain clarity on the problems you are facing with Class 10 Maths questions. 

These solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables will help you plan effectively for your board study. Class 10 maths chapter 3 solutions are prepared by subject experts. The concepts are explained in depth in NCERT solutions for class 10 maths chapter 3 and all doubts are instantly resolved by our subject matter experts during live doubt-solving sessions.


Other Related Links for CBSE Class 10 Maths Chapter 3


Chapter-wise NCERT Solutions Class 10 Maths


NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

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FAQs on NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Ex 3.3

1. What is the main objective of Chapter 3 Pair of Linear Equations in Two Variables in Class 10 Maths as per NCERT Solutions?

The primary objective of this chapter is to help students understand how to solve a pair of linear equations in two variables using graphical and algebraic methods, such as substitution, elimination, and cross-multiplication, in accordance with CBSE Class 10 syllabus (2025–26).

2. Which algebraic methods are explained for solving linear equations in Exercise 3.3 of Class 10 Maths NCERT Solutions?

In Exercise 3.3, students learn two main algebraic methods as per NCERT guidelines:

  • Elimination Method: Removing one variable by adding or subtracting equations.
  • Substitution Method: Substituting the value of one variable from one equation into the other equation.

3. How do you identify if a pair of linear equations has no solution using elimination or substitution?

If, after applying elimination or substitution, you reach a contradictory statement (such as 2 ≠ 3), then the system of equations is inconsistent and has no solution.

4. Why is understanding the step-by-step solution approach important in Class 10 Maths NCERT Solutions for chapter 3?

Following a stepwise approach in NCERT Solutions helps students:

  • Develop logical problem-solving skills
  • Clearly understand each algebraic method
  • Avoid calculation mistakes
  • Score better as per CBSE marking scheme

5. What are some real-life applications of solving pairs of linear equations introduced in Class 10 Chapter 3 NCERT Solutions?

Real-life scenarios where pairs of linear equations are used include:

  • Calculating ages based on conditions
  • Determining cost or price in business situations
  • Solving problems involving mixtures
  • Finding number of notes of different denominations as in banking problems
All such problems are included in NCERT solved examples for Exercise 3.3.

6. What precautions should students take to avoid errors while solving using the elimination method?

Students should:

  • Ensure variable coefficients are made exactly equal (by appropriate multiplication)
  • Pay attention to sign changes when adding/subtracting equations
  • Re-check substituted solutions in both equations

7. In what scenarios should the substitution method be preferred over the elimination method as per NCERT Solutions for Class 10 Chapter 3?

The substitution method is typically preferred when one equation can be easily rearranged to express one variable in terms of the other, making calculations simpler and reducing chances of arithmetic errors.

8. How can students confirm their solution for a pair of linear equations is correct in Class 10 Maths Chapter 3?

After finding the values of x and y, students must substitute both values back into the original equations to check if both equations are satisfied. If they do, the solution is correct as per NCERT standards.

9. What kind of conceptual traps can appear when solving ‘equations reducible to a pair of linear equations’ as per the NCERT Solutions?

Common traps include:

  • Mistaking equations with variables in denominators for non-linear equations
  • Incorrectly simplifying or cross-multiplying both sides
  • Not checking for extraneous (invalid) solutions introduced by multiplication

10. What role do solved examples in Exercise 3.3 play in understanding the chapter as per Vedantu's NCERT Solutions for Class 10 Maths?

Solved examples in Exercise 3.3:

  • Demonstrate correct application of methods
  • Highlight common student mistakes
  • Help reinforce conceptual understanding
  • Prepare students for similar exam-style questions

11. How can practicing all exercises and examples from NCERT Solutions for Class 10 Maths Chapter 3 help boost CBSE exam scores?

Solving every problem and example ensures thorough understanding of solved techniques, familiarizes students with various question types, and builds speed and accuracy—directly improving performance in CBSE 2025–26 board exams.

12. What is the fundamental difference between the substitution method and elimination method in the context of Class 10 Maths NCERT Solutions?

The substitution method replaces one variable in an equation with its equivalent from the other equation, while the elimination method adds or subtracts equations to eliminate one variable, both converting the two-variable system into a one-variable equation as per Chapter 3 guidelines.

13. Why is it important to learn multiple methods for solving linear equations as per the latest NCERT syllabus for Class 10 Maths?

Different exams or problems may favor one method over another in terms of speed or simplicity. Learning multiple methods provides flexibility, deeper conceptual understanding, and helps tackle a wider variety of question types in the CBSE exam.

14. What key formulas must be memorized for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables?

Students must know:

  • Cross-multiplication formulas for x and y
  • The standard form of a linear equation: ax + by + c = 0
  • The method-based steps for elimination and substitution
These are directly applicable to Exercise 3.3 as per the 2025–26 NCERT syllabus.

15. How do the solutions in Vedantu's Class 10 Maths Chapter 3 NCERT Solutions help students who find mathematics challenging?

Vedantu’s NCERT Solutions break down each step, provide explanations in simple language, highlight where students commonly make mistakes, and align with the CBSE approach—thus making it easier for all students to understand and excel in Maths.