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# NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.4) Exercise 3.4

## NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.4

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Ncert Solution Class 10 Maths provides solutions for Mathematics Book textbooks. You can download the Class 10 Maths NCERT Solutions Chapter 3 Exercise 3.4 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.

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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## Access NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

Exercise 3.4

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 3.4 - Free PDF Download

Get the Class 10 Maths Free NCERT Solutions Chapter 3 Exercise 3.4 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 3.4.

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 3 Questions & Solutions (2 Short Answers, 1 Long Answer) Exercise 3.2 7 Questions & Solutions (5 Short Answers, 2 Long Answers) Exercise 3.3 3 Questions & Solutions (2 Short Answers, 1 Long Answer) Exercise 3.5 4 Questions & Solutions (4 Short Answers) Exercise 3.6 2 Questions & Solutions (2 Long Answers) Exercise 3.7 8 Questions & Solutions (1 Short Answer, 7 Long Answers)

### 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.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.4) Exercise 3.4

1. What we are going to study in chapter 3?

NCERT solutions for class 10 maths chapter 3 discusses the idea of a pair of linear equations in two variables, a graphical method of solving a pair of linear equations in two variables, algebraic methods of solving a pair of linear equations, substitution, elimination, cross-multiplication method and equations reducible to a pair of linear equations in two variables, various types of equations. The chapter concludes by summarizing points that help you to revise the chapter and its concepts quickly.

2. How to reduce the fear of Maths?

Some students face difficulties when studying math. Thus, with the aid of its expert teachers, Vedantu has made the subject easier to understand. The best way to learn is to move smartly forward and download the NCERT Solutions PDF now. Vedantu's NCERT Solutions is one of the most important parts of Class 8 Maths study materials. These solutions have been developed with the utmost care by qualified and professional teachers to make your exams easier.

3. How many questions are there in Exercise 3.4 of Class 10 Mathematics Chapter 3?

Chapter 3 of Class 10 Mathematics is a Pair of Linear Equations in Two Variables. This is one of the most important chapters that are to be focused on while preparing for the exams. Exercise 3.4 is based on “Algebraic Methods Of Solving A Pair Of Linear Equations”. The exercise includes two questions. The first question has four parts, while the second one has five parts to be solved using the substitution and elimination method.

4. How many examples are based on Exercise 3.4 of Class 10 Mathematics?

The examples provided in the NCERT book for Class 10 Maths have equal importance as the questions given in various exercises have. The questions based on the examples can also appear in the exams. There are four examples in Chapter 3 that are based on Exercise 3.4. Students should practice these examples as well because they are important and they help students develop a better understanding of all the concepts.

5. Is it tough to score higher marks in Class 10 Maths?

Scoring well in any subject can be made easy if students are determined to do so. Understanding all the concepts that are taught in Class 10 Mathematics is important to score higher marks in your exams. However, understanding the concepts might seem difficult to some students. This problem can be solved by practising everything that has been taught in school. Regular practice of all exercises in each chapter will make scoring well in Maths very easy.

6. Where can I find NCERT Solutions for Class 10 Maths Exercise 3.4?

The solution of Exercise 3.4 can be found on Vedantu for the students who need help in solving questions provided in the Class 10 Maths NCERT book. The NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.4 has step-by-step solutions for all the exercise questions. Students can also refer to the official website of Vedantu to access study material related to the chapter.

7. What are the most important formulas that I need to remember in Class 10 Maths Chapter 3?

Chapter 3 Pair of Linear Equations in Two Variables in Class 10 Maths NCERT includes some formulas that are the basis for solving any question in the exercise, such as in Cross Multiplication method, below is the important formula.

x=b1c2−b2c1 / a1b2−a2b1

and y=a2c1−a1c2 / a1b2−a2b1

These formulas vary depending on the method used for solving the given linear equations. All the formulas in the entire chapter are equally important to remember. You can also access study materials like notes from Vedantu’s app. All the resources are free of cost.

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