# NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.7) Exercise 3.7

There is a fundamental connection between human intellect, logic, and mathematics. MathematicsMathematics fosters logical thinking and mental rigour, and it is an effective method for developing mental discipline. Furthermore, understanding Mathematics is required for learning other academic disciplines such as Physics, Social Studies, and even Music and Art. As a result, students begin learning mathematics from the very beginning of their education.

Mathematics serves as a foundation for other disciplines. Integration and Differentiation are used to explain various theories and laws in Physics, Chemistry, and Biology. It is used in the Social Sciences in the form of Statistics to conduct research. Statistics are also used in Business and Economics. As a result, even if students do not pursue a career in Mathematics after Class 10, branches of Mathematics are always used.  Furthermore, Logical Reasoning and Aptitude are tested in almost all competitive examinations. The curriculum prescribed for various competitive examinations frequently includes the syllabus of Mathematics up to class 10.

Knowledge of Mathematics can be useful for pursuing careers in a variety of fields. With the help of its principles, students can pursue careers in Computer Science, Engineering, Astronomy, Statistics, Data Science, Science and other fields. Students who wish to pursue these careers must focus on the syllabus of Class 10 Mathematics. The academic curriculum of Class 10 Mathematics is very important for clearing the students’ fundamental concepts. Class 10 Mathematics will help students throughout their lives. Therefore, one can appreciate the significance of Mathematics in all aspects of life.

Relevance of Mathematics in real life: When students can relate what they are learning in class to real-life applications, learning becomes easier and more enjoyable. The following are some real-world applications of mathematical concepts

• Algebraic expressions are widely used in computer science. It is also used in Chemistry, for example, the study of crystal symmetry employs Algebra concepts.
• Statistics and probability: Statistics is used in business all over the world today. Statistics is employed in Data Science as well. Probability is used in many fields, including weather Forecasting, Politics, and Sales Forecasting, to predict the occurrence of natural disasters or everyday events like traffic.
• Geometry is used in many professions, including Architecture, Interior Design, Civil Engineering, and Product Manufacturing. Logarithms are used to make complex calculations easier to understand.
• Logarithms are used in Programming Languages and Memory Space Computation. Mensuration: Mensuration is used in everyday life to measure things, calculate weight, and more.

To understand more practical applications of Mathematics, students must focus on Class 10 Mathematics syllabus.

Class 10 is a very important academic year in students’ academic careers. They appear for the board examinations for the first time in Class 10. Students must therefore work hard for their Class 10 examinations. Class 10 board examinations are very important, and the marks obtained in the board examinations often play a huge role in deciding which subject stream students are going to opt for further studies in Class 11. Class 10 board examinations are conducted every year by the Central Board of Secondary Education(CBSE) in the written mode. For Class 10 CBSE Board examinations students are recommended to read the NCERT books. The National Council of Educational Research and Training releases the NCERT books every year. NCERT books are very helpful to students in their preparation for the Class 10 board examinations. The NCERT textbooks are written in an easy language that is understandable to most students. Moreover, NCERT textbooks contain diagrams, facts, and figures to aid visual learning. The exercises mentioned at the end of each chapter are very important and must be solved by all students.

Importance of Class 10: Class 10 board examination results are used to determine admission to senior secondary schools of students’ choice. Board examination results are an important part of students’ academic profiles because they demonstrate their academic performance and achievements. In the contemporary scenario of higher education, many prestigious colleges and universities award extra points in consideration of exceptional Class 10 marks when admitting students. A good performance in Class 10 board examinations can facilitate a student’s career development. Furthermore, in order to enrol in any entrance examination or scholarship programme available after Class 10, a student must have obtained the minimum required percentage/grades in the Class 10 board examinations.

Class 10 Mathematics is also asked in various competitive examinations such as the Civil Service Examination organised by the Union Public Service Commission (UPSC) Civil Services Examinations. Some information on this examination is given below:

Union Public Service Commission: UPSC is India’s central agency that administers examinations such as the Civil Services Examination (CSE) to recruit candidates for top government positions such as IAS, IPS, and IFS. The UPSC recruits candidates for both civil and defence services. UPSC is a constitutional body. The Civil Services Examinations are conducted every year by the UPSC. The CSE is considered one of the toughest exams in India. Class 10 Mathematics is very helpful to prepare for the Aptitude test of the CSE. Moreover, all NCERT textbooks from Classes 6 to Classes 10 are read by the UPSC aspirants. Students preparing for this examination can take the help of the resources provided by Extramarks for their preparation. Extramarks provides resources for both Hindi medium and English medium students. All the study materials provided by Extramarks are cautiously verified.

Mathematics is a very important subject for Class 10. The syllabus of Class 10 Mathematics is vast and requires consistent effort. Mathematics is a very scoring subject, and with enough practice, students can score very well in Mathematics. To overcome their apprehension of Mathematics, students must practise questions regularly. They should start by solving the easy questions first. Mathematics in Class 10 introduces students to numerous new concepts. For instance, Trigonometry is taught to students for the very first time in Class 10.

To find the best solutions for Chapter 3 Pair of Linear Equations in Two Variables, students can refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 offered by Extramarks. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be downloaded from the website and mobile application of Extramarks. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be downloaded in PDF format.

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.7) Exercise 3.7

Examinations are given to students to assess their knowledge in each subject. Students must efficiently prepare for examinations, and it is recommended that they regularly revise the chapters of a subject. Students must solve sample papers and past years’ papers for all subjects on a regular basis. Solving sample papers and past years’ papers can help students get a sense of the types of questions that may appear in the examinations.

A lot of students find Class 10 Mathematics challenging. To make studying Class 10 Mathematics easier for them here are a few tips:

1. Students should attend Class 10 Mathematics classes regularly. Overlooking the knowledge imparted in one lecture can cause confusion in the next lecture. Attending regular lectures ensures continuity in learning the formulas and theorems which are very important. Students should particularly focus on the derivations of various formulas in Class 10. This will help them in solving questions in examinations when they are not able to recall a formula.
2. Students can make short notes in class. The short notes should include example questions taught during the class. Students can also refer to the NCERT revision notes offered by Extramarks for Class 10 Mathematics. Extramarks provides revision notes for all classes and all subjects. The notes provided by Extramarks are reliable and authentic, and they cover the whole syllabus holistically. Students are advised to write down all formulas for Class 10 Mathematics in a consolidated format. These short notes can be revised before the examination. Students must also revisit difficult questions before the examination.
3. After the class, students should always practise a few questions. By doing so they will be able to retain what was taught in the class. In case of any difficulty, they can refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7. They can download the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 from the Extramarks website and mobile application. For the benefit of the students, the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be downloaded in PDF format.
4. Students should regularly practise mock tests and sample papers to be adequately prepared for examinations. Every year before the board examinations, CBSE released sample papers for the students. These sample papers are very important and must be solved by the students, as they can give them an idea of the kinds of questions that are going to be asked.Moreover, the sample papers released by CBSE are designed according to the latest CBSE guidelines. If students face any challenges while solving the sample papers, they can refer to the solutions provided by Extramarks. For Class 10 Chapter 3, students can take help from the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7.  The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 are available on the website and mobile application of Extramarks. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 contain the most authentic and easy-to-understand solutions for Chapter 3.
5. Importance of Syllabus: Students are expected to carefully read each chapter and understand the meaning of each topic and subtopic. To effectively prepare the chapters, it is critical to understand the fundamentals of each chapter. Before beginning preparations for the Mathematics examination, students should become acquainted with the CBSE syllabus. The syllabus is required for developing a learning strategy for Mathematics chapters.
6. Lastly, Class 10 Mathematics should be studied very seriously as it forms the basis of Class 11 and Class 12 Mathematics. Mathematics will further also help them to understand the curriculums of Physics and Chemistry in Class 11 and Class 12.

### All Topics and Subtopics of NCERT Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

Students should focus on all the chapters and complete the whole syllabus to prepare well for the examinations. Each chapter should be revised thoroughly. Learners can find the NCERT Solutions for all Class 10 Mathematics chapters on the Extramarks website and mobile application. They should refer to those solutions for the practise of asking important questions. Important Points [Include Download PDF Button]

While solving the questions for Class 10 Chapter 3, students must revise all the formulas for this chapter. They must regularly revise the Elimination and Substitution methods. Students must also prepare brief notes of all the important formulas from this chapter to revise before the examination. Students can use the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 to make short notes for all the important formulas in this chapter.  The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be downloaded from the link given below.

The best way to excel in Mathematics is to practise as many questions as possible. Students gain confidence and avoid making negligent errors with plenty of practice. Students should practise answering questions of all types in order to perform better in the examination. To practise questions of Chapter 3 Exercise 3.7, students can refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7. To get the solutions, students can download the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 from the link given below. Moreover, the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be downloaded in PDF versions for the convenience of students.

The NCERT Solutions for all subjects are available on the Extramarks website and mobile application. The NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6, NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2, NCERT Solutions Class 1 for all subjects can be downloaded from the Extramarks website and mobile application. The NCERT solutions can be downloaded in PDF format. These NCERT solutions will help students solve and practise the subjects better. NCERT Solutions provided by Extramarks’ are trustworthy and compiled in collaboration with proficient mentors.

Moreover, Extramarks also provides solutions for the ICSE curriculum. The resources provided by Extramarks include revision notes, solutions, etc. All the solutions provided by Extramarks are credible and very beneficial for the preparation of students.

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

Students can access the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 from the website and mobile application of Extramarks. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 provided by Extramarks are accurate and reliable. For easy access, students are recommended to download the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 as a PDF. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 provided by Extramarks are written in a step-by-step manner for better student understanding. NCERT Solution Maths Class 10 Chapter 3 – An Overview

To effectively prepare for Class 10 Chapter 3 it is necessary that students solve all NCERT exercise questions. Students can utilise the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 to find the most accurate solutions for Exercise 3.7. The key features of the NCERT Solutions provided by Extramarks are:

1. The solutions provided by Extramarks are accurate and solved by outstanding instructors of Mathematics.
2. The Class 10 Maths Ch 3 Ex 3.7 Solutions are updated regularly as per the latest CBSE guidelines.
3. The NCERT Solutions of Chapter 3 contain solutions to the miscellaneous questions too.
4. Extramarks provides step-wise answers to all questions for enhancing the conceptual clarity of students.
5. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be easily downloaded in PDF format.
6. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 are very advantageous to all students. Students will not have to look anywhere else for the most accurate solutions.
7. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 are also useful for the preparation for competitive examinations.
8. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 offered by Extramarks are updated regularly to make them error-free.

### NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.7

The Chapter 3 of Class 10 Mathematics is Pair of Linear Equations in Two Variables. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 will help students to understand how to solve questions on this topic. All the solutions provided by Extramarks are compiled with reference to contributions made by skilled teachers of Mathematics. To find the best solutions for Chapter 3 Class 10 students must refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be found on the Extramarks website and mobile application. They are advised to download the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 in PDF format for offline access.

### Class 10th Maths Weightage Marks

In Class 10 Mathematics, some chapters carry more weightage than others from the perspective of examinations. Students must prepare accordingly. Class 10 Chapter 3 is a very important chapter. Students should solve all the exercise questions from this chapter including exercise 3.7 Maths Class 10. If students face any challenges in solving any of the questions given in the exercise then they can refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 provide solutions to all questions, even the miscellaneous questions. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be accessed via the Extramarks website and mobile application. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be accessed in PDF format.

Chapter 3 Pair of Linear Equations in Two Variables forms part of Unit 2 Algebra. Unit 2 holds 20 marks out of 80 marks in total. Therefore, students must study the entire unit thoroughly and practice all the NCERT exercises from this unit. To practise Exercise 3.7, students must refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 will guide students to solve all the questions from the exercise. The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 are available on the Extramarks website as well as its mobile application. Students can download the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 in PDF format to read without internet access too.

### Benefits of Exercise 3.7 Class 10 Maths NCERT Solutions PDF

There are numerous benefits to solving the NCERT exercises for Class 10 Mathematics. In Class 10 Maths, students must complete all of the questions from Exercise 3.7.Students should start by solving easy questions first, which require direct application of formulas. Students should then solve the difficult questions, which require them to use their analytical skills. If students face any challenges, they can refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7.The NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 can be downloaded in PDF format from the website and mobile application of Extramarks. With the help of the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7 students will be able to retain all the formulas from the chapter.

Q.1

$\begin{array}{l}\mathrm{The}\mathrm{ages}\mathrm{of}\mathrm{two}\mathrm{friends}\mathrm{Ani}\mathrm{and}\mathrm{Biju}\mathrm{differ}\mathrm{by}3\mathrm{years}.\\ \mathrm{Ani}‘\mathrm{s}\mathrm{father}\mathrm{Dharam}\mathrm{is}\mathrm{twice}\mathrm{as}\mathrm{old}\mathrm{as}\mathrm{Ani}\mathrm{and}\mathrm{Biju}\mathrm{is}\\ \mathrm{twice}\mathrm{as}\mathrm{old}\mathrm{as}\mathrm{his}\mathrm{sister}\mathrm{Cathy}.\mathrm{The}\mathrm{ages}\mathrm{of}\mathrm{Cathy}\mathrm{and}\\ \mathrm{Dharam}\mathrm{differ}\mathrm{by}30\mathrm{years}.\mathrm{Find}\mathrm{the}\mathrm{ages}\mathrm{of}\mathrm{Ani}\mathrm{and}\mathrm{Biju}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{ages}\mathrm{of}\mathrm{Ani}\mathrm{and}\mathrm{Biju}\mathrm{be}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{years}\mathrm{respectively}\mathrm{.}\\ \mathrm{If}\mathrm{Ani}\mathrm{is}\mathrm{older}\mathrm{than}\mathrm{Biju}\mathrm{then}\\ \mathrm{x}-\mathrm{y}=3 ...\left(1\right)\\ \mathrm{Ani}‘\mathrm{s}\mathrm{father}\mathrm{Dharam}\mathrm{is}\mathrm{twice}\mathrm{as}\mathrm{old}\mathrm{as}\mathrm{Ani}.\mathrm{So},\\ \mathrm{Dharam}‘\mathrm{s}\mathrm{age}= 2\mathrm{x}\\ \mathrm{The}\mathrm{ages}\mathrm{of}\mathrm{Cathy}\mathrm{and}\mathrm{Dharam}\mathrm{differ}\mathrm{by}30\mathrm{years}\mathrm{.}\\ \mathrm{Therefore}, \\ \mathrm{Cathy}‘\mathrm{s}\mathrm{age}= 2\mathrm{x}-\mathrm{30}\\ \mathrm{Biju}\mathrm{is}\mathrm{twice}\mathrm{as}\mathrm{old}\mathrm{as}\mathrm{Cathy}.\mathrm{Therefore},\\ \mathrm{y}=2\mathrm{ }\left(2\mathrm{x}-\mathrm{30}\right) \mathrm{ }...\left(2\right)\\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{from}\mathrm{equation}\left(2\right)\mathrm{in}\mathrm{equation}\left(1\right),\\ \mathrm{we}\mathrm{get}\\ \mathrm{x}-2\left(2\mathrm{x}-\mathrm{30}\right)=3\\ ⇒ \mathrm{ }\mathrm{x}-4\mathrm{x}+60=3\\ ⇒ \mathrm{ }-3\mathrm{x}=3-60=-57\\ ⇒ \mathrm{ }\mathrm{x}=19\\ \mathrm{From}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}\\ \mathrm{y}=19-3\mathrm{ }=16\\ \mathrm{Hence},\mathrm{Ani}\mathrm{is}19\mathrm{years}\mathrm{old}\mathrm{and}\mathrm{Biju}\mathrm{is}16\mathrm{years}\mathrm{old}\mathrm{.}\\ \mathrm{Again},\mathrm{if}\mathrm{Biju}\mathrm{is}\mathrm{older}\mathrm{than}\mathrm{Ani}\mathrm{then}\\ \mathrm{y}-\mathrm{x}=3 ...\left(3\right)\\ \mathrm{Again}, \mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{from}\left(2\right)\mathrm{in}\left(3\right),\mathrm{we}\mathrm{get}\\ 2\left(2\mathrm{x}-\mathrm{30}\right)-\mathrm{x}=3\\ ⇒ \mathrm{ }4\mathrm{x}-60-\mathrm{x}=3\\ ⇒ \mathrm{ }3\mathrm{x}=3+60=63\\ ⇒ \mathrm{ }\mathrm{x}=21\\ \mathrm{Putting}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{equation}\left(3\right),\mathrm{we}\mathrm{get}\\ \mathrm{y}=21+3\mathrm{ }=24\\ \mathrm{Hence},\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{Ani}\mathrm{is}\mathrm{of}21\mathrm{years}\mathrm{and}\mathrm{Biju}\mathrm{is}\mathrm{of}24\mathrm{years}\mathrm{.}\end{array}$

Q.2

$\begin{array}{l}\text{One says, “Give me a hundred, friend! I shall then}\\ \text{become twice as rich as you”. The other replies,}\\ \text{“If you give me ten, I shall be six times as rich as}\\ \text{you”. Tell me what is the amount of their}\\ \text{(respective) capital?}\\ \text{[From the Bijaganita of Bhaskara II]}\\ \text{[Hint:}\mathrm{x}+100=2\left(\mathrm{y}-100\right),\text{}\mathrm{y}+10=6\left(\mathrm{x}-10\right)\right]\end{array}$

Ans.

$\begin{array}{l}\text{Let their respective capitals are}₹\mathrm{x}\text{and}₹\mathrm{y}.\\ \text{According to question,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+100=2\left(\mathrm{y}-100\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-2\mathrm{y}=-300\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}}...\left(1\right)\\ \text{Also,}\\ \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}}\mathrm{y}+10=6\left(\mathrm{x}-10\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}-6\mathrm{x}=-70\text{\hspace{0.17em}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=6\mathrm{x}-70\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}}...\left(2\right)\\ \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}}\mathrm{x}-2\left(6\mathrm{x}-70\right)=-300\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-12\mathrm{x}=-300-140\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{-440}{-11}=40\\ \text{Putting this value of x 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}}\mathrm{y}=6×40-70=240-70=170\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Hence, their respective capitals are}₹40\text{and}₹170.\end{array}$

Q.3

$\begin{array}{l}\text{A train covered a certain distance at a uniform}\\ \text{speed. If the train would have been 10 km/h faster,}\\ \text{it would have taken 2 hours less than the scheduled}\\ \text{time. And, if the train were slower by 10 km/h; it}\\ \text{would have taken 3 hours more than the scheduled}\\ \text{time. Find the distance covered by the train.}\end{array}$

Ans.

$\begin{array}{l}\text{Let the distance covered by the train is x km and the}\\ \text{uniform speed of the train is y km/h. Then time taken}\\ \text{to travel this distance is}\frac{\mathrm{x}}{\mathrm{y}}\text{hour.}\\ \text{According to question,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\mathrm{y}}-2=\frac{\mathrm{x}}{\mathrm{y}+10}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{x}-2\mathrm{y}\right)\left(\mathrm{y}+10\right)=\mathrm{xy}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{xy}+10\mathrm{x}-2{\mathrm{y}}^{2}-20\mathrm{y}=\mathrm{xy}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\mathrm{x}-2{\mathrm{y}}^{2}-20\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}}...\left(1\right)\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\mathrm{y}}+3=\frac{\mathrm{x}}{\mathrm{y}-10}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{x}+3\mathrm{y}\right)\left(\mathrm{y}-10\right)=\mathrm{xy}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{xy}-10\mathrm{x}+3{\mathrm{y}}^{2}-30\mathrm{y}=\mathrm{xy}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-10\mathrm{x}+3{\mathrm{y}}^{2}-30\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}}...\left(2\right)\\ \text{Adding equations (1) and (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\mathrm{x}-2{\mathrm{y}}^{2}-20\mathrm{y}=0\text{\hspace{0.17em}}\\ \underset{¯}{-10\mathrm{x}+3{\mathrm{y}}^{2}-30\mathrm{y}=0\text{\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}\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}}^{2}-50\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}}\mathrm{y}\left(\mathrm{y}-50\right)=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}}\mathrm{y}=0\text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=50\end{array}$   

$\begin{array}{l}\text{Here, speed can not be taken as zero unit as then distance}\\ \text{travelled would be zero unit.}\\ \text{So,}\mathrm{y}=\text{50}\\ \text{Putting this value of y in equation (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\mathrm{x}-2{\text{(50)}}^{\text{2}}-20×50=0\text{\hspace{0.17em}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\mathrm{x}-5000-1000=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=600\\ \text{Hence, distance covered by the train is 600 km.}\end{array}$

Q.4 The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Ans.
Let the number of rows be x and the number of students in a row be y.
Total students in the class
= (Number of rows) × (Number of students in a row)
= xy
Given that if 3 students are extra in a row, then there would be 1 row less.
Therefore,
Total number of students = (x – 1)(y + 3)
or xy =(x-1)(y+3)=xy – y + 3x – 3
or 3x – y – 3 = 0 …(1)
It is also given that if 3 students are less in a row, then there would be 2 rows more.
Therefore,
Total number of students = (x+2)(y– 3)
or xy = (x+2)(y– 3) = xy + 2y – 3x – 6
or 3x – 2y + 6 = 0 …(2)
Subtracting equation (2) from equation (1),
– y + 2y = 3 + 6
or y= 9
By using equation (1)
3x – 9 = 3
or 3x = 12
or x = 4
Number of rows = x = 4
Number of students in each row = y = 9
Hence, number of students in the class = 9 × 4 = 36.

Q.5

$\begin{array}{l}\text{In a ΔABC,}\angle \text{C=3}\angle \text{B=2}\left(\angle \text{A+}\angle \text{B}\right)\text{. Find the}\\ \text{three angles.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}\text{that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}=3\angle \mathrm{B}=2\text{\hspace{0.17em}}\left(\angle \mathrm{A}+\angle \mathrm{B}\right)\\ ⇒\text{}3\angle \mathrm{B}=2\angle \mathrm{A}+2\angle \\ ⇒\text{}\angle \mathrm{B}=2\angle \mathrm{A}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\angle \mathrm{A}-\angle \text{B = 0}...\text{(1)}\\ \text{We know that the sum of measures of all angles of}\\ \text{a triangle is 180}°.\text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°\\ ⇒\angle \mathrm{A}+\angle \mathrm{B}+3\angle \mathrm{B}=180°\\ ⇒\angle \mathrm{A}+4\angle \mathrm{B}=180°\text{}...\text{(2)}\\ \text{Multiplying equation (1) by 4, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\angle \mathrm{A}-\text{4}\angle \text{B = 0}...\text{(3)}\\ \mathrm{Addi}\text{ng equations (2) and (3), we get}\\ \text{9}\angle \text{A = 180°}⇒\angle \mathrm{A}=20\text{°}\\ \text{From equation (2), we get}\\ \text{20°}+4\angle \mathrm{B}=180\text{°}\\ ⇒\text{}4\angle \mathrm{B}=160\text{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}=40\text{°}\\ \text{and}\angle \mathrm{C}=3\angle \mathrm{B}⇒\angle \mathrm{C}=120\text{°}.\\ \text{Therefore,}\angle \mathrm{A},\text{}\angle \mathrm{B}\text{and}\angle \mathrm{C}\text{are respectively}\\ \text{20°},\text{40° and 120°}.\end{array}$

Q.6 Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis

Ans.
The given equations are
5x – y = 5 or y = 5(x – 1) …(1)
3x – y = 3 or y = 3(x – 1) …(2)
We find the value of y when x is zero and the value of x when y is zero for both equations and write the corresponding values in tables as below.

 x 0 1 y = 5(x – 1) –5 0

(1)

 x 0 1 y = 3(x – 1) –3 0

(2)
Now, we draw graphs of the given equations as given below.

From the graph above, we find that the co-ordinates of the vertices of the triangle formed by lines and the y-axis are (1, 0), (0, –3) and (0, –5).

Q.7

$\begin{array}{l}\text{Solve the following pair of linear equations:}\\ \text{(i) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}px+qy = p-q\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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) ax+by = c}\\ \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}qx-py = p+q\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}bx+ay = 1+c}\\ \text{(iii)\hspace{0.17em}\hspace{0.17em}}\frac{\text{x}}{\text{a}}\text{–}\frac{\text{y}}{\text{b}}{\text{= 0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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)\hspace{0.17em}(a-b)x+(a+b)y = a}}^{\text{2}}{\text{-2ab-b}}^{\text{2}}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ax+by = a}}^{\text{2}}{\text{+b}}^{\text{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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(a+b)(x+y) = a}}^{\text{2}}{\text{+b}}^{\text{2}}\\ \text{(v)\hspace{0.17em}\hspace{0.17em}152x-378y = -74}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-378x+152y = -604}\end{array}$

Ans.

$\begin{array}{l}\text{(i) \hspace{0.17em}\hspace{0.17em}}\\ \text{The given equations can be written as:}\\ \text{\hspace{0.17em}}\mathrm{px}+\mathrm{qy}-\left(\mathrm{p}-\mathrm{q}\right)=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \mathrm{qx}-\mathrm{py}-\left(\mathrm{p}+\mathrm{q}\text{\hspace{0.17em}}\right)=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\\ \text{By cross-multiplication method, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{-\mathrm{pq}-{\mathrm{q}}^{2}-\left({\mathrm{p}}^{2}-\mathrm{pq}\right)}=\frac{\mathrm{y}}{-\mathrm{pq}+{\mathrm{q}}^{2}+{\mathrm{p}}^{2}+\mathrm{pq}}=\frac{1}{-{\mathrm{p}}^{2}-{\mathrm{q}}^{2}}\\ ⇒\frac{\mathrm{x}}{-{\mathrm{q}}^{2}-{\mathrm{p}}^{2}}=\frac{\mathrm{y}}{{\mathrm{q}}^{2}+{\mathrm{p}}^{2}}=\frac{1}{-{\mathrm{p}}^{2}-{\mathrm{q}}^{2}}\\ ⇒\mathrm{x}=1\text{and y}=-1\end{array}$

$\begin{array}{l}\text{(ii) \hspace{0.17em}\hspace{0.17em}}\\ \text{The given equations can be written as:}\\ \text{\hspace{0.17em}}\mathrm{ax}+\mathrm{by}-\mathrm{c}=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}}...\left(1\right)\\ \mathrm{bx}+\mathrm{ay}-\left(1+\mathrm{c}\text{\hspace{0.17em}}\right)=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}}...\left(2\right)\\ \text{By cross-multiplication method, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{-\mathrm{b}-\mathrm{bc}+\mathrm{ac}}=\frac{\mathrm{y}}{-\mathrm{bc}+\mathrm{a}+\mathrm{ac}}=\frac{1}{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ ⇒\frac{\mathrm{x}}{\mathrm{c}\left(\mathrm{a}-\mathrm{b}\right)-\mathrm{b}}=\frac{\mathrm{y}}{\mathrm{c}\left(\mathrm{a}-\mathrm{b}\right)+\mathrm{a}}=\frac{1}{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ ⇒\mathrm{x}=\frac{\mathrm{c}\left(\mathrm{a}-\mathrm{b}\right)-\mathrm{b}}{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\text{and \hspace{0.17em}\hspace{0.17em}y}=\frac{\mathrm{c}\left(\mathrm{a}-\mathrm{b}\right)+\mathrm{a}}{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \\ \text{(iii)}\\ \text{The given equations can be written as:}\\ \text{\hspace{0.17em}}\mathrm{bx}-\mathrm{ay}=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}}...\left(1\right)\\ \mathrm{ax}+\mathrm{by}-\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)=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}}...\left(2\right)\\ \text{By cross-multiplication method, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\mathrm{a}\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)}=\frac{\mathrm{y}}{\mathrm{b}\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)}=\frac{1}{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{a}\text{and \hspace{0.17em}\hspace{0.17em}y}=\mathrm{b}\end{array}$

$\begin{array}{l}\text{(iv)}\\ \text{The given equations can be written as:}\\ \text{\hspace{0.17em}}\left(\mathrm{a}-\mathrm{b}\right)\mathrm{x}+\left(\mathrm{a}+\mathrm{b}\right)\mathrm{y}-\left({\mathrm{a}}^{2}-2\mathrm{ab}-{\mathrm{b}}^{2}\right)=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}}...\left(1\right)\\ \left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\left(\mathrm{a}+\mathrm{b}\right)\mathrm{y}-\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)=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}}...\left(2\right)\\ \text{By cross-multiplication method, we have}\\ \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}}\frac{\mathrm{x}}{-\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)+\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^{2}-2\mathrm{ab}-{\mathrm{b}}^{2}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{y}}{-\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^{2}-2\mathrm{ab}-{\mathrm{b}}^{2}\right)+\left(\mathrm{a}-\mathrm{b}\right)\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)}=\frac{1}{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}-{\left(\mathrm{a}+\mathrm{b}\right)}^{2}}\\ \\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{\left(\mathrm{a}+\mathrm{b}\right)\left(-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}+{\mathrm{a}}^{2}-2\mathrm{ab}-{\mathrm{b}}^{2}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{y}}{-{\mathrm{a}}^{3}+2{\mathrm{a}}^{2}\mathrm{b}+{\mathrm{ab}}^{2}-{\mathrm{a}}^{2}\mathrm{b}+2{\mathrm{ab}}^{2}+{\mathrm{b}}^{3}+{\mathrm{a}}^{3}+{\mathrm{ab}}^{2}-{\mathrm{a}}^{2}\mathrm{b}-{\mathrm{b}}^{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}}=\frac{1}{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}-2\mathrm{ab}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{-2\mathrm{b}{\left(\mathrm{a}+\mathrm{b}\right)}^{2}}=\frac{\mathrm{y}}{4{\mathrm{ab}}^{2}}=\frac{1}{-2\mathrm{b}\left(\mathrm{a}+\mathrm{b}\right)}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\mathrm{a}+\mathrm{b}\text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\frac{-2\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\\ \\ \text{(v)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}152\mathrm{x}-378\mathrm{y}=-74\\ -378\mathrm{x}+152\mathrm{y}=-604\\ \text{The given equations can be written as:}\\ \text{\hspace{0.17em}}76\mathrm{x}-189\mathrm{y}+37=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}}...\left(1\right)\\ -189\mathrm{x}+76\mathrm{y}+302=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}}...\left(2\right)\\ \text{By cross-multiplication method, we have}\\ \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}}\frac{\mathrm{x}}{-37×76-189×302}=\frac{\mathrm{y}}{-189×37-76×302}=\frac{1}{{76}^{2}-{189}^{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}}\mathrm{x}=-\frac{37×76+189×302}{{76}^{2}-{189}^{2}}=2\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\frac{-189×37-76×302}{{76}^{2}-{189}^{2}}=1\end{array}$

Q.8 ABCD is a cyclic quadrilateral. Find its all angles.

Ans.

$\begin{array}{l}\text{We know that sum of the measures of the opposite}\\ {\text{angles of a cyclic quadrilateral is 180}}^{°}.\\ \mathrm{Therefore},\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{C}=180\\ ⇒\text{}4\mathrm{y}+20-4\mathrm{x}=180\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-\mathrm{y}=-40\text{}...\left(\mathrm{i}\right)\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}+\angle \mathrm{D}=180\\ ⇒\text{}3\mathrm{y}-5-7\mathrm{x}+5=180\\ ⇒\text{}-7\mathrm{x}\text{}+\text{}3\mathrm{y}\text{}=\text{}180\text{}...\left(\mathrm{ii}\right)\\ \text{Multiplying equation (i) by 3, we get}\\ \text{}3\mathrm{x}-3\mathrm{y}=-120\text{}...\text{(iii)}\\ \text{Adding equations (ii) and (iii), we get}\\ \text{}-7\mathrm{x}+3\mathrm{x}=60⇒\mathrm{x}=-15\\ \mathrm{Putting}\text{value of x in equation (i), we get}\\ \text{}-15\text{}-\mathrm{y}=-40\\ \text{}⇒\text{}\mathrm{y}=25\\ \mathrm{Thus},\\ \angle \mathrm{A}=4\mathrm{y}+20=4×25+20=120°\\ \angle \mathrm{B}=3\mathrm{y}-5=3×25-5=70°\\ \angle \mathrm{C}=-4\mathrm{x}=-4×\left(-15\right)=60°\\ \angle \mathrm{D}=-7\mathrm{x}+5=-7×\left(-15\right)+5=110°.\end{array}$

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