# NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Mathematics is a subject which is vast to a great extent. Whether it is about numbers or Arithmetic, measurement of land and space or Geometry, shapes and their areas and volumes or menstruation, or the formation of Equations using constants and variables in Algebra; every field is just a subtopic in the huge panorama of Mathematics.

For engineering, computer sciences, finance, medicine, and so on, the importance of Mathematics cannot be ignored.

Derived from the Arabic word al-jabr, the term, Algebra means “the reunion of broken parts”.The art of manipulating equations and  formulas is known as Algebra. An equation is an expression which has an equality sign on both sides. Broadly, Equations generally have two elements : variables and constants. The 10 basic digits of Mathematics from 0 to 9 are counted as constants. Whereas any symbol other than these digits is known as a variable. The value of variables is not fixed.A “linear equation in one variable” is one that is framed using a single variable, whereas a “linear equation in two variables” is one that is framed using two variables with maximum power one.The basic structure of a linear equation in two variables is ax+by+c=0, where x and y are variables and can have any value, whereas X, y, z, etc. are constants, which have a definite and fixed value. A system where two such equations are dealt with is known as a “Pair of Linear Equations in Two Variables.

The National Council of Educational Research and Training (NCERT) includes this vital chapter in the book of Class 10 Mathematics. Here, one can learn different methods to solve these Linear Equations. The entire curriculum has been designed to meet the standards of students in class 10.The linear equations can be represented in two ways: graphically and algebraically. The equation ax+by+c=0 is an algebraic representation. Where the same equation can be presented on a graph using the X and Y axis.There are so many methods discussed in the chapter to solve the pair of linear equations like substitution method, cross-multiplication method, coefficient method etc. These methods can be used to solve the equations algebraically. However, if the equations are supposed to be solved graphically, then the solutions can be found by observing the lines drawn on the graph, Whether these lines are intersecting, parallel, or overlapping.

It needs fair practise and guidance to prepare this chapter. This can be done with the guidance of Extramarks. Extramarks is an authentic website which hires quality experts to prepare NCERT Solutions for every subject. Each and every piece of content is meticulously prepared here.This helps the students  prepare any topic in a better way. When it comes to Class 10, one cannot take any subject lightly. Every subject carries equal weightage and counts in the overall ranking of a student. As a result, it is essential that they thoroughly prepare each subject and chapter.When it comes to Maths, some students feel panic. They feel afraid of this giant subject. But this feeling can be turned into a friendly one if one learns this subject in a subtle way.

Extramarks is a platform which helps the students of Class 10 in their board exam preparation. Students can find NCERT Solutions Class 10 on the website of Extramarks. These solutions can be of immense help to them in preparation.

Chapter 3 of Mathematics is entitled Pair of Linear Equations in Two Variables. The chapter is full of interesting concepts. These concepts, if learnt with diligence and understanding, can be mastered. Extramarks helps the students of Class 10  learni these methods of solving linear equations in an easy way. Students of Class 10 who are looking for the solutions to exercise 3.2 can find the same on Extramarks under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. This will make them familiar with all the important concepts of this exercise.

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.2) Exercise 3.2

When a student enters Class 10, he or she enters a race to secure a good rank. To win this race, one needs to gear oneself up. The importance of Class 10 cannot be neglected. It is a period whose success will be carried by the students wherever they go. No matter how qualified one is, the marksheet of Class 10 will always be the first to secure a place in the portfolio. Thus, students of Class 10 should utilise their time in a wise manner while preparing for their board exams. This tough preparation needs dedication as well as guidance. Extramarks is one of the online destinations where one can seek this guidance.

The students of Class 10 can approach Extramarks’ NCERT Solutions while preparing for Mathematics Chapter 3. The second exercise of this chapter deals with solving the linear equations graphically or by comparing the coefficients of the equations. For students’ convenience, NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 on Extramarks are provided.Under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2, each concept of the topic is thoroughly discussed.It imparts the knowledge to students in a subtle way. This enhances their understanding of the concepts, too. The authenticity of these NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 is verified from time to time, with any required upgrades included. However, making notes from online sources requires a lot of time and energy. Extramarks NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 can be used to overcome this obstacle.Extramarks provides these NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 in a PDF format. The PDF of NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 can be accessed using Extramarks’ website or mobile app. The fact that this PDF of NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 can be downloaded by students using either the Extramarks website or mobile app adds to their delight.This saves them the effort of making notes. After downloading the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2, they can take a print of the same and invest the saved time in practising more questions and revising the concepts multiple times. Thus, with multiple revisions and a firm understanding, students can ace any Class 10 Maths Chapter 3 Exercise 3.2 question.

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

Linear Equation is defined as an equation of straight line. The form of  a linear equation is ax + by + c = 0 where a, b and c are the real numbers, and x and y are the two variables,

Here a and b are known as the coefficients, and c is the constant of the equation.

When two linear equations have the same two variables, they are known as a “pair of Linear Equations in two variables.” The representation of these is as follows:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Here the value of the coefficients may vary, but for both the equations, the x and y variables will have the same values.

The algebraic and the graphical methods to solve these Pair of linear equations can be learnt using Extramarks. Extramarks offers well represented solutions to Exercise 3.2 Class 10 Maths under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. The students can get the best quality of answers on NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 using Extramarks. These NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 aid in the better preparation of Class 10 students.

The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 are easy to access. All that Is needed is to go to the Extramarks website and click on NCERT Solutions Class 10. NCERT Solutions for Mathematics can be found there.Upon clicking NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2, one can easily access the authentic solutions to Exercise 3.2 of Class 10 Mathematics. All the different types of solving these equations are discussed in a minute manner, highlighting all the important steps to solving these equations.Following this step by step method of solving Pair of Linear Equations in Two Variables,as given on the website of Extramarks under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2, one can get better marks in the examination. Thus, the students are advised to opt for NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 while preparing for the same.

### NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2

CBSE is the entity which decides the curriculum for all the classes. The pattern of questions in the exams is modified by the Central Board of Secondary Education as per the requirements of time. As per the advancement of technology and progress, the upgrading of skills is also necessary. This is done by the CBSE by making the learning process more innovative and advanced. Rather than cramming, cognition is being promoted. This is the reason behind introducing assertion reason based questions. This emphasises the need for critical thinking.

The same is true for Chapter 3 of Class 10 math.The methods to solve the linear equations in two variables can be learnt with Extramarks NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. Furthermore, the NCERT Solutions for Class 10 Maths, Chapter 3 Exercise 3.2, elaborate each concept with such precision that students can attempt multiple-choice questions or assertion-reason type questions with ease.These NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 also include practise questions of this nature.Practice these questions from  Extramarks

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 boosts the students’ problem-solving skills. The updating of these NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 from time to time improves their utility.NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 also provide answers to such questions where statements are provided and linear equations are to be formed. NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 also covers the questions where linear equations are given and graphs are to be drawn.These NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 give a systematic response to every question in the best possible manner. Thus, the students should incorporate NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 in their exam strategy if they want to ameliorate their performance.

### Class 10 Maths Chapter 3 Exercise 3.2

Chapter 3 of Class 10 is a vast topic. It must be covered in detail with its theory and questions.Extramarks NCERT Solutions for Class 10 Maths, Chapter 3, Exercise 3.2, assist students in properly preparing for this topic.There are three types of pairs of linear equations. Firstly, there are pairs of linear equations with only one solution. Such equations are known as consistent equations. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 give a complete description of consistent equations. NCERT Solutions for Class 10 Maths, Chapter 3, Exercise 3.2, also contains questions on the subject.

Next come such Equations that can have an infinite number of solutions. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 describe such equations as inconsistent equations. At last come the equations with no common solution at all.These are defined as dependent equations under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2.

These NCERT Solutions for Class 10 Maths, Chapter 3 Exercise 3.2, also include previous years’ papers in Mathematics.Accessing the same from Extramarks NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2, students can become well versed in the questions and concepts related to chapter 3. Thus NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 can help smoothen the difficult journey of Board exams for students.

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 equips students with the necessary skills and an analytical perspective to begin the arduous journey of board exam preparation.These NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 are prepared using such lucid language that they can be understood by even the average student. As a result, students in Class 10 are advised to visit Extramarks for NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 if necessary.These NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 will help them accelerate their study time.Also, the availability of these NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 in a downloadable PDF format adds to the benefits of using Extramarks. This saves the time of students as well. No doubt there are other online sources as well. However, the features offered by Extramarks NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 enhance its value and give it an upper hand over others.

### NCERT Solutions for Class 10 Maths Chapter 3 All Other Exercises

When it comes to the preparation of Board classes one cannot ignore any topic or chapter. Class 10 board exams are a crucial phase in one’s life. The marks scored  this year carry extreme significance throughout one’s life. Whether it is an entrance exam for any reputable institution or an application for a job, the Class 10 marksheet secures the first place for itself to be checked.Thus,the students of Class 10 should make sure that they work hard and score excellent marks.

Mathematics is a topic that is both technical and scoring. If one has attempted all the questions in a stepwise manner with proper representation, then there is no scope for any unnecessary mark deduction in the mathematics exam. Thus, if the students pay some extra attention to this subject, it can bear sweet results. Students can secure more marks with diligence and improve their overall rank.

Extramarks is a platform where students can resolve their doubts and queries. It provides NCERT Solutions for almost all the classes. The solutions for primary and upper primary classes can be accessed on Extramarks under NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4, NCERT Solutions Class 5, NCERT Solutions Class 6,NCERT Solutions Class 7 and NCERT Solutions Class 8. Apart from this, NCERT Solutions Class 9 and NCERT Solutions Class 11 are also available. For board classes, students can click on NCERT Solutions Class 10 and NCERT Solutions Class 12. This is of enormous help to the students of Class 10 and Class 12 whose pressure of Board exams is reduced by Extramarks. Thus, the search by the students  for NCERT Solutions, no matter which class they belong to, can come to an end by accessing Extramarks. This speeds up their learning journey, and the aura of their knowledge improves greatly.Extramarks provides NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 in a pdf format on its website as well as mobile app. However, the solutions to the rest of the exercises can also be found on Extramarks under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. In total, there are 7 exercises in Chapter 3. Extramarks provides solutions to all these exercises. The link for these can be found under the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. These  solutions can be helpful in completing homework as well as during revision for pre-board or final board exams.

However, because not all exercises have the same mark distribution, these exercises should be prepared with smartwork.For example, from Exercise 3.1, only 1 long question can be expected, whereas Exercise 3.7 can command a maximum of 7 long questions. Thus, it becomes important to go through the pattern of examination and the distribution of marks as recommended by the Central Board of Secondary Education( CBSE). A full table discussing the number of questions to be asked is also available under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. Thus, NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 can be of immense help to the students of Class 10 during their peak revision.

Q.1 Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Ans.
(i) Let the number of girls and the number of boys be x and y respectively.
According to question,
x + y = 10
x – y = 4
For x + y = 10,
y = 10 – x

 x 4 6 8 y = 10 – x 6 4 2

For x – y = 4,
y = x – 4

 x 4 6 8 y = x – 4 0 2 4

The graphs of equations are drawn below which shows that the two lines intersect at (7, 3).
Hence the number of boys and the number of girls are 7 and 3 respectively.

(ii)
Let the cost of 1 pencil and the cost of 1 pen be ₹ x and ₹ y respectively.
According to the question,
5x + 7y = 50
and
7x + 5y = 46
For 5x + 7y = 50,
y = (50 – 5x)/7

 x 3 -4 10 y = (50 – 5x)/7 5 10 0

For 7x + 5y = 46,
y = (46 – 7x)/5

 x 8 3 –2 y = (46 – 7x)/5 –2 5 12

The graphs of equations are drawn below which shows that the two lines intersect at (3, 5).
Hence the cost of 1 pencil and the cost of 1 pen are ₹ 3 and ₹ 5 respectively.

Q.2

$\begin{array}{l}\text{On }\mathrm{c}\text{omparing the ratios }\frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{,}\frac{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{, }\frac{{\text{c}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{, find out whether the lines representing the}\\ \text{following pairs of linear equations intersect at a point, are parallel or coincident: }\\ \text{(i) 5x – 4y + 8 = 0}\\ \text{ 7x + 6y – 9 = 0}\\ \text{(ii) 9x + 3y + 12 = 0}\\ \text{ 18x + 6y + 24 = 0}\\ \text{(iii) 6x – 3y + 10 = 0}\\ \text{ 2x – y + 9 = 0}\end{array}$

Ans.

$\begin{array}{l}\text{(i) 5x-4y+8 = 0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7x+6y-9 = 0}\\ {\text{Comparing these equations with a}}_{\text{1}}{\text{x+b}}_{\text{1}}{\text{y+c}}_{\text{1}}\text{= 0}\\ {\text{and a}}_{\text{2}}{\text{x+b}}_{\text{2}}{\text{y+c}}_{\text{2}}\text{= 0, we get}\\ {\text{a}}_{\text{1}}{\text{=5,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{1}}{\text{=-4,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c}}_{\text{1}}\text{=8}\\ {\text{a}}_{\text{2}}{\text{=7,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{2}}{\text{=6, \hspace{0.17em}c}}_{\text{2}}\text{=-9}\\ \text{Also,}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{\text{5}}{\text{7}}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\text{=}\frac{\text{-4}}{\text{6}}\text{=}\frac{\text{-2}}{\text{3}}\\ \text{and so}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{ }\ne \frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\\ \text{Therefore, the given pairs of linear equations intersect at a point.}\end{array}$

$\begin{array}{l}\text{(ii) 9x+3y+12=0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}18x+6y+24=0}\\ {\text{Comparing these equations with a}}_{\text{1}}{\text{x+b}}_{\text{1}}{\text{y+c}}_{\text{1}}\text{= 0}\\ {\text{and a}}_{\text{2}}{\text{x+b}}_{\text{2}}{\text{y+c}}_{\text{2}}\text{= 0, we get}\\ {\text{a}}_{\text{1}}{\text{=9,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{1}}{\text{=3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c}}_{\text{1}}\text{=12}\\ {\text{a}}_{\text{2}}{\text{=18,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{2}}{\text{=6, c}}_{\text{2}}\text{=24}\\ \text{Also,}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{\text{9}}{\text{18}}\text{=}\frac{\text{1}}{\text{2}}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\text{=}\frac{\text{3}}{\text{6}}\text{=}\frac{\text{1}}{\text{2}}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{c}}_{\text{1}}}{{\text{\hspace{0.17em}c}}_{\text{2}}}\text{=}\frac{\text{12}}{\text{24}}\text{=}\frac{\text{1}}{\text{2}}\\ \text{and so}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\text{=}\frac{{\text{c}}_{\text{1}}}{{\text{\hspace{0.17em}c}}_{\text{2}}}\\ \text{Therefore, the given pairs of linear equations are coincident.}\end{array}$

$\begin{array}{l}\text{(iii) 6x-3y+10 = 0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2x-y+9 = 0}\\ {\text{Comparing these equations with a}}_{\text{1}}{\text{x+b}}_{\text{1}}{\text{y+c}}_{\text{1}}\text{=0}\\ {\text{and a}}_{\text{2}}{\text{x+b}}_{\text{2}}{\text{y+c}}_{\text{2}}\text{= 0, we get}\\ {\text{a}}_{\text{1}}{\text{= 6,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{1}}{\text{= -3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c}}_{\text{1}}\text{= 10}\\ {\text{a}}_{\text{2}}{\text{= 2,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{2}}{\text{= -1, c}}_{\text{2}}\text{= 9}\\ \text{Also,}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{\text{6}}{\text{2}}\text{= 3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\text{=}\frac{\text{-3}}{\text{-1}}\text{= 3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{c}}_{\text{1}}}{{\text{\hspace{0.17em}c}}_{\text{2}}}\text{=}\frac{\text{10}}{\text{9}}\\ \text{and so}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\ne \frac{{\text{c}}_{\text{1}}}{{\text{\hspace{0.17em}c}}_{\text{2}}}\\ \text{Therefore, the given pairs of linear equations are parallel.}\end{array}$

Q.3

$\begin{array}{l}\text{On comparing the ratios}\frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}\text{,}\frac{{\mathrm{b}}_{1}}{{\mathrm{b}}_{2}}\text{,\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{{\mathrm{c}}_{2}}\text{, find out whether}\\ \text{the following pair of linear equations are consistent,}\\ \text{or inconsistent.}\\ \\ \text{(i)}3\mathrm{x}+2\mathrm{y}=5;\text{}2\mathrm{x}-3\mathrm{y}=7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{(ii)}2\mathrm{x}-3\mathrm{y}=8;\text{}4\mathrm{x}-6\mathrm{y}=9\\ \text{(iii)}\frac{3}{2}\mathrm{x}+\frac{5}{3}\mathrm{y}=7;\text{\hspace{0.17em}\hspace{0.17em}}9\mathrm{x}-10\mathrm{y}=14\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{(iv) \hspace{0.17em}}5\mathrm{x}-3\mathrm{y}=11;\text{\hspace{0.17em}\hspace{0.17em}}-10\mathrm{x}+6\mathrm{y}=-22\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}(v)}\frac{4}{3}\mathrm{x}+2\mathrm{y}=8;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+3\mathrm{y}=12\end{array}$

Ans.

$\begin{array}{l}\text{(i)}3\mathrm{x}+2\mathrm{y}=5;\text{}2\mathrm{x}-3\mathrm{y}=7\\ \text{Comparing these equations with}{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}=0\\ \text{and}{\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}=0,\text{we get}\\ {\mathrm{a}}_{1}=3,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{1}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}=-5\\ {\mathrm{a}}_{2}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{2}=-3,\text{}{\mathrm{c}}_{2}=-7\\ \text{Also,}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{3}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=-\frac{2}{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}=\frac{5}{7}\\ \text{and so}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}\ne \frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}\\ \text{Therefore, the given pair of linear equations has a unique}\\ \text{solution and hence the two linear equations are consistent.}\end{array}$

$\begin{array}{l}\text{(ii)}2x-3y=8;\text{}4x-6y=9\\ \text{Comparing these equations with}{a}_{1}x+{b}_{1}y+{c}_{1}=0\\ \text{and}{a}_{2}x+{b}_{2}y+{c}_{2}=0,\text{we get}\\ {a}_{1}=2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{b}_{1}=-3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}_{1}=-8\\ {a}_{2}=4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{b}_{2}=-6,\text{}{c}_{2}=-9\\ \text{Also,}\\ \frac{{a}_{1}}{{a}_{2}}=\frac{2}{4}=\frac{1}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{b}_{1}}{\text{\hspace{0.17em}}{b}_{2}}=\frac{-3}{-6}=\frac{1}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{c}_{1}}{\text{\hspace{0.17em}}{c}_{2}}=\frac{8}{9}\\ \text{and so}\\ \frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{\text{\hspace{0.17em}}{b}_{2}}\ne \frac{{c}_{1}}{\text{\hspace{0.17em}}{c}_{2}}\\ \text{Therefore, the given linear equations are parallel and}\\ \text{hence the two linear equations are inconsistent}\text{.}\end{array}$

$\begin{array}{l}\text{(iii)}\frac{3}{2}\mathrm{x}+\frac{5}{3}\mathrm{y}=7;\text{\hspace{0.17em}\hspace{0.17em}}9\mathrm{x}-10\mathrm{y}=14\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Comparing these equations with}{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}=0\\ \text{and}{\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}=0,\text{we get}\\ {\mathrm{a}}_{1}=\frac{3}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{1}=\frac{5}{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}=-7\\ {\mathrm{a}}_{2}=9,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{2}=-10,\text{}{\mathrm{c}}_{2}=-14\\ \text{Also,}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{\frac{3}{2}}{9}=\frac{1}{6},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{\frac{5}{3}}{-10}=-\frac{1}{6},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}=\frac{-7}{-14}=\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}\ne \frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}\\ \text{Therefore, the given pair of linear equations has a unique}\\ \text{solution and hence the two linear equations are consistent.}\end{array}$

$\begin{array}{l}\text{(iv) \hspace{0.17em}}5\mathrm{x}-3\mathrm{y}=11;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-10\mathrm{x}+6\mathrm{y}=-22\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{Comparing these equations with}{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}=0\\ \text{and}{\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}=0,\text{we get}\\ {\mathrm{a}}_{1}=\text{\hspace{0.17em}}5,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{1}=-3,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}=-11\\ {\mathrm{a}}_{2}=-10,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{2}=6,\text{}{\mathrm{c}}_{2}=22\\ \text{Also,}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{5}{-10}=-\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{-3}{6}=-\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}=\frac{-11}{22}=-\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}\\ \text{Therefore, the given linear equations have infinitely many}\\ \text{solutions and hence they consistent.}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}(v)}\frac{4}{3}\mathrm{x}+2\mathrm{y}=8;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+3\mathrm{y}=12\\ \text{Comparing these equations with}{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}=0\\ \text{and}{\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}=0,\text{we get}\\ {\mathrm{a}}_{1}=\frac{4}{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{1}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}=-8\\ {\mathrm{a}}_{2}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{2}=3,\text{}{\mathrm{c}}_{2}=-12\\ \text{Also,}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{\frac{4}{3}}{2}=\frac{2}{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{2}{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}=\frac{-8}{-12}=\frac{2}{3}\\ \text{and so}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}\\ \text{Therefore, the given linear equations have infinitely many}\\ \text{solutions and hence they are consistent.}\end{array}$

Q.4

$\begin{array}{l}\text{Which of the following pairs of linear equations are consistent/inconsistent?}\\ \text{If consistent, obtain the solution graphically:}\\ \text{(i) x + y = 5, 2x + 2y = 10 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{(ii) x – y = 8, \hspace{0.17em} 3x – 3y = 16}\\ \text{(iii) 2x + y }–\text{6 = 0, 4x – 2y }–\text{ 4 = 0}\\ \text{(iv) 2x – 2y – 2 =\hspace{0.17em}0, 4x – 4y }–\text{ 5 = 0\hspace{0.17em}\hspace{0.17em}}\end{array}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}(i)}\mathrm{x}+\mathrm{y}=5,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+2\mathrm{y}=10\text{}\\ \text{Comparing these equations with}{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}=0\\ \text{and}{\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}=0,\text{we get}\\ {\mathrm{a}}_{1}=1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{1}=1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}=-5\\ {\mathrm{a}}_{2}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{2}=2,\text{}{\mathrm{c}}_{2}=-10\\ \text{Now,}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}=\frac{-5}{-10}=\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}\\ \text{Therefore, the given linear equations have infinitely many}\\ \text{solutions and hence they are consistent.}\\ \text{Graphs of the two equations coincide and hence each and}\\ \text{every point on this graph is a solution of these equations.}\end{array}$

$\begin{array}{l}\text{(ii)}\mathrm{x}-\mathrm{y}=8,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3}\mathrm{x}-3\mathrm{y}=16\\ \text{Comparing these equations with}{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}=0\\ \text{and}{\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}=0,\text{we get}\\ {\mathrm{a}}_{1}=1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{1}=-1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}=-8\\ {\mathrm{a}}_{2}=3,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{2}=-3,\text{}{\mathrm{c}}_{2}=-16\\ \text{Now,}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{1}{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{1}{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}=\frac{-8}{-16}=\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}\ne \frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}\\ \text{Therefore, the given linear equations are parallel}\\ \text{and hence they are inconsistent.}\end{array}$

$\begin{array}{l}\text{(iii)}2\mathrm{x}+\mathrm{y}-6=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}}4\mathrm{x}-2\mathrm{y}-4=0\\ \text{Comparing these equations with}{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}=0\\ \text{and}{\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}=0,\text{we get}\\ {\mathrm{a}}_{1}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{1}=1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}=-6\\ {\mathrm{a}}_{2}=4,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{2}=-2,\text{}{\mathrm{c}}_{2}=-4\\ \text{Now,}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{2}{4}=\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=-\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}=\frac{-6}{-4}=\frac{3}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}\ne \frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}\\ \text{Therefore, the given linear equations are consistent.}\\ \text{Now,}\\ 2\mathrm{x}+\mathrm{y}-6=0⇒\mathrm{y}=6-2\mathrm{x}\\ \text{Some points which satisfy this equation are written}\\ \text{in the following table.}\\ \begin{array}{cccc}\mathrm{x}& 0& 2& 3\\ \mathrm{y}=6-2\mathrm{x}& 6& 2& 0\end{array}\\ \text{Again,}\\ 4\mathrm{x}-2\mathrm{y}-4=0⇒\mathrm{y}=2\mathrm{x}-2\\ \text{Some points which satisfy this equation are written}\\ \text{in the following table.}\\ \begin{array}{cccc}\mathrm{x}& 0& 2& 3\\ \mathrm{y}=2\mathrm{x}-2& -2& 2& 4\end{array}\\ \\ \text{We get the following graphs of the given equations and}\\ \text{find that they intersect at (2, 2). Hence, x = 2 and y = 2.}\end{array}$

$\begin{array}{l}\text{(iv) \hspace{0.17em}}2\mathrm{x}-2\mathrm{y}-2=0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{x}-4\mathrm{y}-5=0\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{Comparing these equations with}{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{b}}_{1}\mathrm{y}+{\mathrm{c}}_{1}=0\\ \text{and}{\mathrm{a}}_{2}\mathrm{x}+{\mathrm{b}}_{2}\mathrm{y}+{\mathrm{c}}_{2}=0,\text{we get}\\ {\mathrm{a}}_{1}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{1}=-2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{1}=-2\\ {\mathrm{a}}_{2}=4,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}_{2}=-4,\text{}{\mathrm{c}}_{2}=-5\\ \text{Now,}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{2}{4}=\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}=\frac{1}{3}=\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}=\frac{-2}{-5}=\frac{2}{5}\\ \text{and so}\\ \frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{1}}{\text{\hspace{0.17em}}{\mathrm{b}}_{2}}\ne \frac{{\mathrm{c}}_{1}}{\text{\hspace{0.17em}}{\mathrm{c}}_{2}}\\ \text{Therefore, the given linear equations are parallel}\\ \text{and hence they are inconsistent.}\end{array}$

Q.5 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Ans.
Let the width of the garden be x and length be y.
According to the question, length is 4 m more than its width. Therefore,
y − x = 4 …(1)
Also, half the perimeter of a rectangular garden is 36 m.
Therefore,
x + y = 36 …(2)
Now,
y − x = 4 …(1)
or y = x + 4

 x –4 0 y = x + 4 0 4

Similarly,
x + y = 36 …(2)
or y = 36 – x

 x 28 20 y = 36 – x 8 16

Graphs of the equations (1) and (2) are drawn below and from there we observe that they intersect at (16, 20).
Therefore, x = 16 and y = 20. Hence, width of the garden is 16 m and length of the garden is 20 m.

Q.6 Given the linear equation 2x+ 3y– 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Ans.

$\begin{array}{l}\text{(i) Intersecting lines:}\\ \text{A line which will intersect the given line}2\mathrm{x}+3\mathrm{y}-8=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is}3\mathrm{x}+2\mathrm{y}-8=0\text{as}\\ \text{}\frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{2}{3}\text{and}\frac{{\mathrm{b}}_{1}}{{\mathrm{b}}_{2}}=\frac{3}{2}\text{impllies that}\frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}\ne \frac{{\mathrm{b}}_{1}}{{\mathrm{b}}_{2}}.\\ \text{(ii) P lines:}\\ \text{A line which is paralel to the given line}2\mathrm{x}+3\mathrm{y}-8=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is}4\mathrm{x}+6\mathrm{y}-7=0\text{as}\\ \text{}\frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{2}{4}=\frac{1}{2}\text{and}\frac{{\mathrm{b}}_{1}}{{\mathrm{b}}_{2}}=\frac{3}{6}=\frac{1}{2}\text{impllies that}\frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{1}}{{\mathrm{b}}_{2}}.\\ \left(\mathrm{iii}\right)\text{:}\\ \text{A line which is \hspace{0.17em}\hspace{0.17em}to the given line}2\mathrm{x}+3\mathrm{y}-8=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is}4\mathrm{x}+6\mathrm{y}-16=0\text{because}\\ \text{}\frac{{\mathrm{a}}_{1}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{b}}_{1}}{{\mathrm{b}}_{2}}=\frac{{\mathrm{c}}_{1}}{{\mathrm{c}}_{2}}=\frac{1}{2}.\end{array}$

Q.7 Draw the graphs of the equations
x – y + 1 = 0 and 3x + 2y – 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Ans.
We have,
x – y + 1 = 0
or y = x + 1
We have the following table.

 x 0 –1 y = x + 1 1 0

Again,
3x + 2y – 12 = 0
or y = (12–3x)/2
and so we have the following table.

 x 0 4 y = (12–3x)/2 6 0

Graphs of the given equations are drawn below and from there we find that coordinates of the vertices of the triangle formed by these lines and the x-axis are (-1, 0); (4, 0) and (2, 3).

## Please register to view this section

### 1. What are the benefits of using Extramarks NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 while preparing for Class 10 board exams?

Extramarks is a website by which quality content is always delivered. The students who seek guidance of Extramarks should be aware of the benefits they will be availing if they opt for Extramarks NCERT Solutions. The students who are preparing for Exercise 3.2 Class 10 Maths can seek help from Extramarks NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2. Here  are discussed some benefits of using Extramarks :NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 :

1. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 give a detailed and easy introduction to the chapter.
2. Each concept and topic under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 is discussed in an elaborative way.
3. All the solutions available on Extramarks under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 are prepared by top experts.
4. Wherever required, graphs are made for better concept clarity.
5. These NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 are upgraded from time to time as per the latest curriculum and trend of examination.
6. The past years’ papers section under NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 discusses the questions appeared in the exams in recent years from Exercise 3.2 Class 10 Maths which helps students in getting a better analysis of the latest trend of examination.
7. The practice questions available on Exercise 3.2 Class 10 Maths help students revise the concepts. Also the questions are framed following the latest style like assertion reason questions, multiple choice questions etc

### 2. How can one score good marks in Mathematics Class 10?

Class 10 is a crucial period that needs to be passed with sheer dedication and hard work. During the Mathematics exam, students should keep certain points in mind. First and foremost, approach the paper sequentially.Solve the questions in a stepwise manner, as representation matters a lot in math. Draw graphs and diagrams wherever necessary. Do not forget to write units wherever required. Following these steps, one can attempt the math exam in an effective manner. For more guidance, students can choose to opt for Extramarks. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 and solutions to other chapters of Mathematics available on Extramarks help students improve their weak areas as well as teach them the apt way to attempt the questions in the exam. Thus, using NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 one can learn multiple techniques to make one’s answer sheets effective and presentable. These NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 will help students improve their knowledge as well as their grades.