# NCERT Solutions for Class 10 Maths Chapter 3

The National Council for Educational Research and Training, or NCERT, is an organisation that creates a variety of educational resources for students and teachers in schools. It offers textbooks for all disciplines and classes. Numerous schools, particularly those connected to the Central Board of Secondary Education (CBSE), include NCERT texts in their curricula. The NCERT’s publications offer comprehensive knowledge of all academic disciplines and themes. For students to learn how to approach various types of questions, the NCERT book for Mathematics is a fantastic resource. It offers examples for students to better understand the appropriate format for responding to a question and the most accurate method of solving particular kinds of problems. In board exams, the method of solution is a crucial component of the answers because it helps students get a higher grade. There are numerous methods to answer questions, but not all of them may be deemed appropriate from an examination perspective. Therefore, it is crucial for students to apply a solution method that is precise and accurate. Students can learn which method to use for solving different types of questions by using the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3. Students can benefit from the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3.

Class 10 students may seek academic assistance from the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3. To perform well in their tests at the conclusion of the semester, Class 10 students must concentrate on the material presented in the NCERT. The exercises in NCERT are designed to prepare students for solving issues at a higher level. Before introducing pupils to the more difficult problems, the NCERT texts first teach them the fundamentals of an idea. The NCERT questions will help students strengthen their conceptual foundation of the chapter or course information, which will help them solve additional problems.Students can clarify their doubts pertaining to Exercise 3.3 by using the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3.Class 10 Maths Chapter 3 NCERT Solutions [Include Download PDF Button]

The Introduction to Pairs of Linear Equations in Two Variables and How to Interpret Them are some of the themes covered in the Class 10 Maths Chapter 3 Exercise 3.3. Students are expected to learn how to create a linear equation with two variables through this activity. They will learn how these equations are represented and integrated through this practice. However, in addition to the topics covered in NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3, this chapter also covers a number of significant topics, such as How to Solve Linear Equations in Two Variables using Algebraic Methods. Students will learn a lot about linear equations in this chapter, which will aid them in more difficult classes and academic levels. After the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3, the subsequent exercises include certain indispensable concepts that will aid students in responding to a variety of queries, both simple and challenging. Students would greatly benefit fromthe NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 in understanding the fundamentals of these ideas. Later chapters will allow them to perform and comprehend more effectively. If students want to study Mathematics in depth, they need to comprehend these themes very well. Students can get guidance from the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 if they get confused regarding any question.

They can use the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3 to find answers to their questions as well as identify and correct their mistakes. Students will benefit from strengthening concepts embedded in one section of the chapter because the formulas and concepts therein may be used to solve problems in other chapters and activities. On the website and Learning App of Extramarks, the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 are conveniently made available.

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

The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 provide an overview of the chapter and also present a brief conceptualisation of the organisation of information in the chapter. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 provide a lot of information on the chapter in addition to the solution to the exercise questions. The method of solution used and explained in the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 is similar to those used in the NCERT examples. This will help students in solving all the exercises in the chapter and also  the optional exercises and assignments. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 can be accessed by students through the Extramarks portal.

### Highlights of NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

The algebraic method of solution for a pair of linear equations in two variables is explained in this exercise. This exercise will be of help to students who want to understand the algebraic method of solution for a pair of linear equations in two variables. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 give a clear overview of the exercise and the concepts applied in the designated exercise. The method of substitution to solve linear equations in two variables algebraically is used. The NCERT Solutions for Class 10 Maths, Chapter 3 and Exercise 3.3, will provide students with answers to any questions they may have about the chapter or Exercise 3.3 Maths Class 10.The methods of solution presented will assist students in solving a variety of problems.This will also help them understand later exercises and their method of solution through the application of learned concepts. The application of this concept is vast, as it forms the basis for solving linear equations. Students can learn this by carefully reading NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3.

The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 will help students understand Exercise 3.3 Class 10 Maths, how these questions must be solved, and how to avoid making negligent errors while solving them.The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 are of great help to students who want to understand the logical structure of various methods of solutions for these types of questions. Students can adequately learn this through NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3.

### Advantages of Practising with Extramarks Class 10 Maths Chapter 3 Exercise 3.3 Solution

Visiting the Extramarks website for the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 has a number of advantages. The academic content on the Extramarks digital learning portal, including the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3, has been compiled by highly proficient subject matter experts of Extramarks.  As a result, these responses are reviewed for accuracy before being publicly released.  Instead of giving students the answers straightaway, Extramarks’ NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 will first give an overview of the exercise and the topics discussed therein. Students may study linear equations in two variables and comprehend how to use concepts like algebraic methods of solution: the cross multiplication approach, the elimination method, and the substitution method- to solve linear equations in two variables by using Extramarks’ NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3. Extramarks’ NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 are available on the website and the Learning App, making it available to students at their convenience.

The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 offer a clear approach to solving the problem with the right responses and a concise depiction of the procedure deployed to answer the presented question. This will help students learn and understand the importance of this exercise and chapter through explanations. One of the many study resources for all chapters and exercises in the prescribed textbook of Mathematics for Class 10 that are offered on Extramarks is the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3. For various levels of academic content particular to different classes, Extramarks further offers comprehensive NCERT Solutions, such as NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 are an excellent example of the detailed and accurate solutions offered by Extramarks. The aforementioned advantages of the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 are merely a fraction of the academic benefits offered by these learning assets. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 can also help students by assisting them in figuring out the logical sequence of steps while solving a particular type of mathematical problem.

### NCERT Solutions Class 10 Maths All Chapters

Students can find a wealth of study materials on the Extramarks website, including the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3.Study materials for partial help are also available. Students may discover NCERT Solutions for a number of classes and academic disciplines. Students pursuing education in senior secondary classes have access to NCERT Solutions Class 12 and NCERT Solutions Class 11. These solutions for students in senior secondary classes are present for subjects such as Physics, Biology, Mathematics, Chemistry, Business Studies, Economics, Hindi and English. Students in secondary classes can secure access to NCERT Solutions Class 9 and NCERT Solutions Class 10. For students in secondary classes, these solutions can be procured for a variety of disciplines such as Science, Mathematics, Social Science, Hindi and English. The Extramarks website and mobile application have NCERT Solutions Class 6, NCERT Solutions Class 7, and NCERT Solutions Class 8 available for middle school students. These solutions for students in middle school classes can be availed for academic subjects such as Science, Mathematics, Social Science, Hindi and English. NCERT Solution Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4 and NCERT Solutions Class 5 are available to students in schools who are pursuing primary education. For students in primary classes, these solutions can be accessed for the curriculum of diverse subjects such as Mathematics, EVS, Hindi and English. Students can look through the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 to experience the quality academic guidance provided by solutions crafted by Extramarks.

### NCERT Solutions Class 10 Maths Chapter 3 Exercises

The purpose of NCERT exercises is to evaluate students’ knowledge of the arguments expressed in a particular chapter. For board exams, NCERT Solutions are vital. If students are facing any challenges understanding the Chapter 3 Pair of Linear Equations in Two Variables, they can refer to the study materials for the academic curriculum of Mathematics for Class 10 like the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 on the Extramarks website and Learning App. Since this chapter is challenging for many students, they can use the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3, which provide answers to all of their questions. Students can utilise the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 to improve their academic performance and organise their test preparation. These are available online on the Extramarks website.

Students may find it hindering their learning process if they aren’t able to clarify their doubts related to various concepts.When trying to address a problem while studying, teachers and peers might be helpful. By using the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 and other resources made accessible on the Extramarks learning portal, students can also reduce their unanswered queries.  Students who have questions about the material can get clarification by clearing up any misunderstandings. Students must be able to comprehend a question approach in order to score well on an examination later and to be able to find solutions to problems that are similar to or the same as that question.  The faster students can decipher the question and find the answer, the easier it will be for them to respond effectively to a question during an examination.In order to succeed on each test, students must keep this in mind. Therefore, it is essential for students to practise NCERT questions in order to save time during examination preparation as well as on test day. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 can assist them in a myriad of ways with their learning.

### Important Topics under NCERT Solutions for Class 10 Maths Chapter 3

The Class 10 Maths Chapter 3 Exercise 3.3 contains questions and answers to various important topics and concepts. Pair of Linear Equations in Two Variables, Chapter 3, is the topic wherein students will learn how to use variables to solve problems and choose the correct response to questions with two variables by reading this chapter. These will aid students in resolving a variety of mathematical issues and boost their performance in board exams like the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3. Students will also learn about the General Structure of a Pair of Linear Equations, how to represent them, how to solve them algebraically and graphically, and how to identify linear equations based on their terms. This chapter will also help students answer more complex problems. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 will provide students with a lot of information about various topics. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 will be of great help to students who are striving to attain excellence in board examinations. the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 would prepare students to face any challenging questions during their examinations.

### Importance of Class 10 Maths Chapter 3 Linear Equations

With the help of the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3, students can feasibly retain the knowledge they acquired in this chapter. A pair of linear equations’ general form is taught to Class 10 students. Additionally, students will find how algebra and graphs can be utilised to express linear equations. On graph paper or a Cartesian plane, two lines are drawn to represent the equations using the graphical technique of representation. The nature of the solution to these equations is further illustrated by the interactions of these lines. Depending on the characteristics of their graph, equations can have consistent or inconsistent solutions. However, other topics are covered in detail, such as the algebraic techniques for solving a pair of linear equations. The chapter discusses three algebraic methods of solution: the cross multiplication approach, the elimination method, and the substitution method- to solve linear equations in two variables. This chapter teaches students how to understand a pair of linear equations and the nature of their algebraic solution, in addition to the method of solution for a pair of linear equations. To determine which pair of linear equations will be consistent or inconsistent, three cases are given in the chapter. This chapter also teaches about situations that can be algebraically represented using linear equations in two variables or mathematically through a graph. These scenarios might not be linear at first, but by trying to alter them, they can be reduced to a pair of linear equations. Students can look through the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 to acquire a general understanding of these topics, especially the ones Exercise 3.3 is based on.

By constantly responding to the questions and using the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3, students can overcome any lacunae in their knowledge and abilities. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 have been compiled in an internet-friendly format for viable accessibility by students. After reading the prescribed NCERT textbooks in-depth, students would have a complete comprehension of the concepts and topics, as well as know how to respond to various queries. By studying the NCERT textbooks, they will have the ability needed to solve new problems from different textbooks or to take examinations to have their skills assessed and any inadequacies discovered.

Q.1

$\begin{array}{l}\text{Solve the following pair of linear equations by the}\\ \text{substitution method.}\\ \text{(i)}\mathrm{x}+\mathrm{y}=14\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(ii)}\mathrm{s}-\mathrm{t}=3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-\mathrm{y}=4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{s}}{3}+\frac{\mathrm{t}}{2}=6\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(iii)}3\mathrm{x}-\mathrm{y}=3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(iv)}0.2\mathrm{x}+0.3\mathrm{y}=1.3\\ \text{}9\mathrm{x}-3\mathrm{y}=9\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0.4\mathrm{x}+0.5\mathrm{y}=2.3\\ \text{(v)}\sqrt{2}\mathrm{x}+\sqrt{3}\mathrm{y}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(vi)}\frac{3\mathrm{x}}{2}-\frac{5\mathrm{y}}{3}=-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{3}\mathrm{x}-\sqrt{8}\mathrm{y}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{2}=\frac{13}{6}\end{array}$

Ans.

$\begin{array}{l}\text{(i) The given pair of linear equations are:}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\mathrm{y}=14\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-\mathrm{y}=4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\\ \text{We express x in terms of y from equation (1) to get}\\ \text{}\mathrm{x}=14-\mathrm{y}\\ \text{We substitute this value of x in equation (2) to get}\\ \text{}14-\mathrm{y}-\mathrm{y}=4\\ \text{i.e.,}-2\mathrm{y}=4-14=-10\\ \text{i.e.,}\mathrm{y}=5\\ \text{Putting this value of y in equation (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-5=4\\ \text{ i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=9\\ \text{(ii) The given pair of linear equations are:}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{s}-\mathrm{t}=3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{s}}{3}+\frac{\mathrm{t}}{2}=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\\ \text{We express}\mathrm{s}\text{in terms of}\mathrm{t}\text{from equation (1) to get}\\ \text{}\mathrm{s}=\mathrm{t}+3\\ \text{We substitute this value of}\mathrm{s}\text{in equation (2) to get}\\ \text{}\frac{\mathrm{t}+3}{3}+\frac{\mathrm{t}}{2}=6\\ \text{i.e.,}\frac{5\mathrm{t}+6}{6}=6\\ \text{i.e.,}5\mathrm{t}+6=36\\ \text{i.e.,}\mathrm{t}=6\\ \text{Putting this value of t in equation (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{s}-6=3\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{s}=9\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}(iii) The given pair of linear equations are:}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}-\mathrm{y}=3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}9\mathrm{x}-3\mathrm{y}=9\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\\ \text{We express}\mathrm{y}\text{in terms of}\mathrm{x}\text{from equation (1) to get}\\ \text{}\mathrm{y}=3\mathrm{x}-3\\ \text{We substitute this value of}\mathrm{y}\text{in equation (2) to get}\\ \text{}9\mathrm{x}-9\mathrm{x}+9=9\\ \text{i.e.,}9=9\\ \text{This statement is true for all values of x and so we can not}\\ \text{obtain a specific value of x. We observe that both the given}\\ \text{equations are the same as one is derived from another.}\\ \text{Therefore, given equations have infinitely many solutions.}\\ \text{\hspace{0.17em}(iv) The given pair of linear equations are:}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0.2\mathrm{x}+0.3\mathrm{y}=1.3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0.4\mathrm{x}+0.5\mathrm{y}=2.3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\\ \\ \text{We express x in terms of y from equation (1) to get}\\ \text{}\mathrm{x}=\left(1.3-0.3\mathrm{y}\right)/\left(0.2\right)=\left(13-3\mathrm{y}\right)/2\\ \text{We substitute this value of x in equation (2) to get}\\ \text{0.2}\left(13-3\mathrm{y}\right)+0.5\mathrm{y}=2.3\\ \text{i.e.,}2.6-0.6\mathrm{y}+0.5\mathrm{y}=2.3\\ \text{i.e.,}-\text{0.1}\mathrm{y}=2.3-2.6=-0.3\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=3\\ \text{Putting this value of y in equation (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0.4\mathrm{x}+0.5×3=2.3\text{\hspace{0.17em}}\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\left(2.3-1.5\right)/0.4=8/4=2\end{array}$

$\begin{array}{l}\left(\mathrm{v}\right)\mathrm{The}\mathrm{given}\mathrm{pair}\mathrm{of}\mathrm{linear}\mathrm{equations}\mathrm{are}:\\ \\ \sqrt{2}\mathrm{x}+\sqrt{3}\mathrm{y}=0 ...\left(1\right) \\ \sqrt{3}\mathrm{x}-\sqrt{8}\mathrm{y}=0 ...\left(2\right)\\ \mathrm{We}\mathrm{express}\mathrm{x}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{y}\mathrm{from}\mathrm{equation}\left(1\right)\mathrm{to}\mathrm{get}\\ \mathrm{x}=\frac{\sqrt{3}}{\sqrt{2}}\mathrm{y}\\ \mathrm{We}\mathrm{substitute}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{equation}\left(2\right)\mathrm{to}\mathrm{get}\\ \sqrt{3}\frac{\sqrt{3}}{\sqrt{2}}\mathrm{y}-\sqrt{8}\mathrm{y}=0\\ \mathrm{i}.\mathrm{e}.,\frac{3-4}{\sqrt{2}}\mathrm{y}=0\\ \mathrm{i}.\mathrm{e}.,\mathrm{y}=0\\ \mathrm{Putting}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get}\\ \sqrt{3}\mathrm{x}=0\\ \mathrm{i}.\mathrm{e}., \mathrm{x}=0\\ \\ \left(\mathrm{vi}\right)\mathrm{The}\mathrm{given}\mathrm{pair}\mathrm{of}\mathrm{linear}\mathrm{equations}\mathrm{are}:\\ \\ \frac{3\mathrm{x}}{2}-\frac{5\mathrm{y}}{3}=-2 \\ \mathrm{or} 9\mathrm{x}-10\mathrm{y}\mathrm{ }=-12 ...\left(1\right)\mathrm{ }\\ \mathrm{and}\\ \frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{2}=\frac{13}{6} \\ \mathrm{or}2\mathrm{x}+3\mathrm{y}=13 ...\left(2\right)\\ \mathrm{We}\mathrm{express}\mathrm{x}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{y}\mathrm{from}\mathrm{equation}\left(1\right)\mathrm{to}\mathrm{get}\\ \mathrm{x}=\left(-12+10\mathrm{y}\right)/9\\ \mathrm{We}\mathrm{substitute}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{equation}\left(2\right)\mathrm{to}\mathrm{get}\\ 2\left(\frac{-12+10\mathrm{y}}{9}\right)+3\mathrm{y}=13\\ \mathrm{i}.\mathrm{e}.,-24+20\mathrm{y}+27\mathrm{y}=117\\ \mathrm{i}.\mathrm{e}., 47\mathrm{y}=117+24=141\\ \mathrm{i}.\mathrm{e}., \mathrm{y}=\frac{141}{47}=3\\ \mathrm{Putting}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get}\\ 2\mathrm{x}+9=13\\ \mathrm{i}.\mathrm{e}., \mathrm{x}=\frac{4}{2}=2\end{array}$

Q.2 Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

Ans.

$\begin{array}{l}\text{The given linear equations are:}\\ 2\mathrm{x}+3\mathrm{y}=11\text{}...\text{(1)}\\ 2\mathrm{x}-4\mathrm{y}=-24\text{\hspace{0.17em}}...\text{(2)}\\ \text{We express x in terms of y from equation (1) to get}\\ \text{}\mathrm{x}=\frac{11-3\mathrm{y}}{2}\\ \text{We substitute this value of x in equation (2) to get}\\ \text{}11-3\mathrm{y}-4\mathrm{y}=-24\\ \text{i.e., 11}-7\mathrm{y}=-24\\ \text{i.e., 7}\mathrm{y}=24+11=35\\ \text{i.e., \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=5\\ \text{Putting this value of y in equation (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+3×5=11\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}=11-15=-4\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=-2\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\mathrm{mx}+3\\ \text{or 5}=-2\mathrm{m}+3\\ \text{or}\mathrm{m}=\text{}-1\end{array}$

Q.3

$\begin{array}{l}\text{Form the pair of linear equations for the following}\\ \text{problems and find their solution by substitution method.}\\ \text{(i) The difference betrween two numbers is 26 and one}\\ \text{number is three times the other. Find them.}\\ \text{(ii) The larger of two supplementary angles exceeds}\\ \text{the smaller by 18 degrees. Find them.}\\ \text{(iii) The coach of a cricket team buys a 7 bats and 6 balls}\\ \text{for}₹\text{3800. Later she buys the 3 bats and 5 balls}\\ \text{for}₹\text{1750. Find the cost of each bat and each ball.}\\ \text{(iv) The taxi charges in a city consist of a fixed charge}\\ \text{together with the charge for the distance covered.}\\ \text{For\hspace{0.17em}\hspace{0.17em}a distance of 10 km, the charge paid is}₹\text{105 and}\\ \text{for a journey of 15 km, the charge paid is 155. What}\\ \text{are the\hspace{0.17em}\hspace{0.17em}fixed charges and the charge per km? How}\\ \text{much does\hspace{0.17em}\hspace{0.17em}a person have to pay for travelling a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}distance of 25 km?}\\ \text{(v) A fraction becomes}\frac{9}{11}\text{, if 2 is added to both the}\\ \text{numerator and the denominator. If, 3 is added to}\\ \text{both the numerator and the denominator it}\\ \text{becomes}\frac{5}{6}.\text{Find the fraction.}\\ \text{(vi) Five years hence, the age of Jacob will be three times}\\ \text{that of his son. Five years ago, Jacob’s age was seven}\\ \text{times that of his son. What are their present ages?}\end{array}$

Ans.
(i)
Let the larger number is y and the smaller number is x.
According to question,
y = 3x …(1)
and
y – x = 26 …(2)
We substitute the value of y from equation (1) in equation (2) and get
3x – x = 26
or 2x = 26
or x = 13
Putting this value of x in equation (1), we get
y = 39
Hence, the numbers are 13 and 39.

(ii)
Let the larger angle is y and the smaller angle is x.
According to question,
y – x = 18° …(1)
and
y + x = 180° …(2)
We write y in terms of x from (1) to get
y = 18° + x
Putting this value of y in equation (2), we get
18° + x + x = 180°
or 2x = 180° – 18° = 162°
or x = 81°
Putting this value of x in equation (1), we get
y = 18° + 81° = 99°
Hence, the two supplementary angles are 81° and 99°.

(iii)
Let the cost of one bat and one ball be ₹ x and ₹ y respectively.
According to question,
7x + 6y = ₹ 3800 …(1)
and
3x + 5y = ₹ 1750 …(2)
We write y in terms of x from (1) to get
y= (3800 – 7x)/6
Putting this value of y in equation (2), we get
3x + 5(3800 – 7x)/6 = 1750
or 18x – 35x = 1750 × 6 – 19000 = –8500
or x = 8500/17 = 500
Putting this value of x in equation (1), we get
7×500 + 6y = 3800
or y = (3800 – 3500)/6 = 50
Hence, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50.

(iv)
Let the fixed charge be ₹ x and charge per km be ₹ y.
According to question,
x + 10y = ₹ 105 …(1)
and
x + 15y = ₹ 155 …(2)
We write x in terms of y from (1) to get
x = 105 – 10y
Putting this value of x in equation (2), we get
105 – 10y + 15y = 155
or 5y = 155 – 105 = 50
or y = 10
Putting this value of y in equation (1), we get
x + 10×10 = 105
or x = 105 – 100 = 5
Hence, the fixed charge is ₹ 5 and charge per km is ₹ 10.
Charge for 25 km = 5 + 25 ×10 = ₹ 255

$\begin{array}{l}\left(\text{v}\right)\\ \text{Let the fraction be}\frac{\mathrm{x}}{\mathrm{y}}.\\ \text{According to question,}\\ \frac{\mathrm{x}+2}{\mathrm{y}+2}=\frac{9}{11}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}11\mathrm{x}-9\mathrm{y}=-4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \text{and}\\ \frac{\mathrm{x}+3}{\mathrm{y}+3}=\frac{5}{6}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{x}-5\mathrm{y}=-3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\\ \\ \text{From equation (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{5\mathrm{y}-3}{6}\\ \text{We put this value of}\mathrm{x}\text{in equation (1) and get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}11\frac{5\mathrm{y}-3}{6}-9\mathrm{y}=-4\text{\hspace{0.17em}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}55\mathrm{y}-33-54\mathrm{y}=-24\\ ⇒\mathrm{y}=-24+33=9\\ \text{Putting this value of y in equation (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}11\mathrm{x}-9×9=-4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}11\mathrm{x}=81-4=77\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=7\\ \text{Hence the fraction is}\frac{7}{9}.\end{array}$

(vi)
Let the age of Jacob be x and the age of his son be y.
According to question,
x + 5 = 3(y + 5)
or x – 3y = 10 …(1)
Also,
x – 5 = 7(y – 5)
or x – 7y = –30 …(2)
From equation (1), we find
x = 10 + 3y
We substitute this value of x in equation (2) and get
10 + 3y – 7y = –30
or –4y = –40
or y = 10
Putting this value of y in (1), we get
x – 3×10 = 10
or x = 40
Hence the present age of Jacob is 40 years and that of his son is 10 years.

## FAQs (Frequently Asked Questions)

### 1. How can students improve their Mathematics test results?

Many students inquire about the best method for doing well in Mathematics with their teachers, peers, schoolmates, and seniors. A few strategies that could assist pupils in performing better have been mentioned further to help students resolve this query.  One of the most important parts of preparation is to practice as many questions as possible. Students could improve their ability to accurately comprehend questions and discover answers as quickly as possible by practising various types of questions. Students should therefore practice Mathematics as often as they can to improve their comprehension. The strong and weak areas of each student should be identified by them in order to improve them. Students must be aware of the concepts they are lacking because these are the ones that will have a negative impact on their test results. The ability to create a better academic plan based on their strengths would enable students to devote more time to improving their weak areas while balancing this effort with other courses and chapters. Therefore, it is critical for students to identify their areas of weakness and put a lot of effort into overcoming them.  Students should solve sample papers or past years’ papers. Students will gain a better understanding of the format and pattern of examinations by studying and solving past years’ papers before their examinations. To identify the pattern of questions and recurring types of questions, they should analyse past years’ papers closely. It will assist students in getting better grades in Mathematics. Students should therefore solve past years’ papers.

Students must start off with easy questions and move their way up to the advanced level of questions. Starting with simple questions can help students gain confidence in their approach to answering questions and would also benefit in figuring out how to tackle more challenging problems later on. Chapters like Linear Equations in Two Variables and Trigonometry, which are entirely novel to students in Class 10, are undoubtedly complex to comprehend. Therefore, before moving on to the more challenging ones, students should clarify their understanding of the simpler themes to make their foundation of chapters and concepts stronger.  Before attempting to answer any questions, students must first comprehend the concepts studied in the previous exercise or chapter. Although Mathematics is not a heavily theoretical discipline, it is crucial to comprehend the rationale behind how the concepts and formulae are created, how they are put into practice and what their purposes are. Students won’t be able to properly analyse the question without this understanding. Therefore, understanding the topics is just as crucial as practising the questions.

Early preparation is crucial. Students must make an effort to learn and practice as soon as their classes start. This way they will get more time to practice, and as a result, their foundation knowledge will be adequate. They will have more time to revise the formulas and learn new methods of solving problems. Students’ academic schedules and study time must be with appropriate breaks. They must allow themselves rest periods in between study sessions. All of these steps are meant to positively impact students to experience considerable improvement in their academics.

### 2. How can students improve their performance with the assistance of the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3?

Students are aware that regular study is required to get ready for any examination. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 will help them with achieving compliance with a regular academic schedule. By studying and practising, students can reduce negligent errors they might make during tests. Students will grasp a concept more comprehensively after some practice, and after fully mastering a concept, it will be simpler for them to apply it to other problems from the same or different topics and chapters. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 are a good example of this. By carefully practising NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3, students can become aware of their mistakes when responding to various types of problems. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 could also be used for a holistic revision of the chapter before the examination. Their initial reaction can be inaccurate, but with enough practice and by using the right formula and technique, they will be able to correct it. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3 will be of help in this. If they take proper breaks to recuperate, their study time will be more productive. Being confident with their preparation might encourage students to continue studying on a routine basis.