# NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities (Ex 6.3) Exercise 6.3

The Central Board of Secondary Education is an educational institution that operates in India. There are a huge number of schools that are affiliated with the board. CBSE is one of the most reputed as well as one of the largest educational boards that operate in India. The educational layout that is followed by CBSE and all CBSE schools is highly objective. The questions that are asked to students in the exams are highly objective, deeply conceptual, and straightforward. Students are advised by the top educators in the country to pay great attention to the concepts that are discussed in a chapter and have an in-depth understanding of the ideas to score well in the exams. Students must develop the skill of having very high precision and attention to detail for the pattern of the questions in order to be able to answer the questions that are asked by the CBSE. CBSE values traits like revising and regularly recapitulating rather than merely retaining information. The CBSE curriculum has a very particular application-based study pattern. Whatever the students learn in their courses, they must go through the ideas thoroughly and develop a solid understanding of the concepts and postulates. This is required because the pattern of the question paper requires students to know the subject matter well. The CBSE board accentuates the analytical portions of a topic over the theoretical portions of the topic. The questions asked are very brief and crisp and the answers demanded are very straightforward. Students must be confident while studying and extremely consistent with their revisions. These are the most efficient ways a student can score well in a subject like Mathematics. If a student does not have a deep understanding of the subject, then there is a high probability that they will lose a lot of marks. Experts at Extramarks, however, have asked their students not to stress over this. They emphasise the importance of consistency and regularity. CBSE students are graded based on the average of the marks they have achieved in all the subjects they have teaken. The educators at Extramarks reassure all the students that there is nothing to fear about the advanced level of difficulty of the CBSE board exam and the vast syllabus. Students can easily score well if they follow the guidance of their teachers.

The National Centre of Research and Educational Training NCERT is a central organisation. NCERT was established in1961 by the Indian government. The NCERT is in charge of all matters pertaining to the country’s educational development and academic prowess. NCERT provides great assistance to both the Central Government and the State Governments concerning any issues related to education or schooling. NCERT has put out their own set of guidelines, which they expect students to follow. NCERT has devised its syllabus, and NCERT even publishes its books. CBSE and NCERT work together symbiotically. CBSE asks all the schools that are affiliated with it to strictly follow the guidelines provided by NCERT. CBSE also asks its schools and all their students to consider the NCERT textbook as their primary textbook and study material. Following the NCERT textbook, students get a very clear understanding of the kind of questions that can be asked during the exam. NCERT wants its students to learn a concept well enough for them to apply it to real-life situations. NCERT, therefore, values cognition over memorisation. NCERT wants its students to understand the fundamentals of a concept very well. The paper is often felt to be extremely tricky and difficult because students who appear in the exam do not practise and revise what they have learned very much. NCERT is constantly trying to increase their reach, and therefore they have introduced many different amenities that a CBSE student can put to great use. NCERT has introduced amazing foreign exchange programmes to promote good education in their country while creating an amicable environment between students from different nations. NCERT has developed a journal in which students from the CBSE board can get their profound opinions published. NCERT, therefore, provides an amazing opportunity for students to use their highly regarded platform to build their careers and future.

Mathematics is one of the core subjects of any student who has just been promoted to Class 11 and has taken up Science as their primary stream of expertise. The concepts that a student is going to learn in Mathematics echo in other subjects like Physics and Chemistry all the time. For a student to score good marks in their exam they have to consistently put the effort into all the core Science subjects. CBSE is famous for the vastness of its syllabus and its advanced level of difficulty. Students often get overwhelmed by their exams under the board. This stress makes students make silly errors, which causes them to lose a large number of marks. Educators at Extramarks have emphasised that there is nothing for a student to feel stressed about. Extramarks assures these students that if they consistently study while steadily furthering the syllabus, regularly revise the portions of the syllabus that they have already learned, and give tests or solve questions regularly, they can score well in their exams. Many students change their board and transition to the CBSE board. CBSE suits the best students who are preparing themselves for competitive exams like JEE Mains, JEE Advanced, NEET, etc. The objective question paper pattern of CBSE exams completely corresponds to the multiple-choice questions asked in the competitive exams. Students who transition to CBSE from other boards generally take a considerable amount of time to get accustomed to the difficulty of the CBSE exams as well as the big syllabus. Educators have constantly told these students that the CBSE syllabus might be overwhelming but there are simple and easy tricks with which a student can score well.

Chapter 6 of the CBSE Class 11 Mathematics syllabus is Linear Inequalities. This is one of the most important chapters in the syllabus, and therefore teachers suggest that students allocate a significant amount of time to this chapter. Linear inequalities is one of the trickier chapters in the Class 11 CBSE Mathematics curriculum. Students often avoid the chapters that are trickier because, given the vast syllabus, these chapters take up a lot of time to finish. Top educators at Extramarks have advised against leaving such chapters. Therefore, Extramarks has come up with the NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3. Linear Inequality is a chapter from which a consequential number of questions are asked. Teachers emphasise the importance of having all the fundamental concepts of this chapter clear. Once the basic ideas are clear, students have generally found a way to enjoy the chapter. Top educators at Extramarks have suggested that for a student to score well in this Chapter they must practise problems from it regularly. Once the chapter is done, and a student has moved ahead with the syllabus, they must come back to this chapter and revise the problems from it. Therefore, Extramarks has released NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3. NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 are one of the most efficient NCERT Solutions available online.

The third exercise of Chapter 6 is based on finding solutions to a system in the milieu of Linear Inequality when there is more than one variable concerned. While solving, students come across many doubts and generally students wait for their teachers to solve their doubts. The Class 11 CBSE Mathematics is vast and students shouldn’t wait for solutions. Therefore, NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 come in handy for all students. NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 is provided by top educators at Extramarks. NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 provide all their answers in extreme detail making sure that they are easily understandable.

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## H2 – NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities (Ex 6.3) Exercise 6.3

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**H3 – Important Topics Covered in Exercise 6.3 of Class 11 Maths NCERT Solutions**

Linear inequalities is one of the most crucial chapters that a student has in their Class 11 CBSE curriculum. Class 11 Maths Chapter 6 Exercise 6.3 is among the trickier exercises in the chapter, and students ought to be careful while solving it. Students who are preparing for competitive exams like JEE Mains, JEE Advanced, NEET, etc. find that studying in a school that follows the CBSE curriculum helps them prepare for the competitive exams as well because both share an extremely similar question pattern. When students refer to NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3, while solving the chapter, they are also preparing for the competitive exam. NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3, abound with tricks and hacks with which students can solve a problem faster, more efficiently, and correctly. The way the solutions are provided in NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 makes sure that students can grasp the concept easily. The language is extremely magnanimous, which makes the process of studying easier and stress-free. Students from any kind of academic background can access the NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 anytime, so a student is always under the guidance of top educators.

**H3 – Access NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities**

Extramarks has released NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 for students who get stuck in the third exercise of the sixth chapter in their CBSE syllabus. Like NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 Extramarks releases solutions for every other exercise. The solutions have all the necessary information that CBSE wants its students to know. The solutions explain all the reasons and explanations behind complicated steps lucidly. They also have other information written in the bracket which is based on a particular axiom learned in the chapters. Whenever formulas are used in a particular step of a problem, the entire formula is mentioned alongside the step in the NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3. Students need to be extremely patient while solving the sums from Ex 6.3 Class 11.

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### H3 – NCERT Solutions for Class 11 Maths Chapters

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### H3 – NCERT Solution Class 11 Maths of Chapter 6 Exercise

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Maths Ex 6.3 Class 11 is a tricky exercise and teachers have asked students to be very careful while solving the problems making sure there are no mistakes.

**H3 – NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Exercise 6.3**

Linear Inequations is one of the most important chapters from the algebra section of the syllabus. Students must revise regularly and must use NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3 for help. Whenever a student is solving sums that they have solved before, they must keep an eye on the sums that they could not answer by themselves. When students redo these problems, they often can solve them. If a student cannot do a sum after solving the exercise before, Teachers ask the students not to fear situations. When a doubt arises for a student, it is not a testament to their performance but to the areas in which the student has doubts, which need to be addressed as soon as possible. Therefore, NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.3, helps students not just with their academic prowess but also with their exam strategies.

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**Q.1 ** Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{\ge}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\ge}\mathbf{2}$

**Ans.**

$\begin{array}{l}\mathrm{We}\text{are given}\\ \mathrm{x}\ge 3...\left(\mathrm{i}\right)\\ \mathrm{y}\ge 2...\left(\mathrm{ii}\right)\\ \mathrm{We}\text{draw the graph of linear equations x}=\text{3 and y}=\text{2 in xy-plane}\\ \text{as shown in figure. Solution of inequality}\left(\text{1}\right)\text{is represented by}\\ \text{the}\text{shaded region right the line\hspace{0.17em}\hspace{0.17em}x}=\text{3},\text{including the points on}\\ \text{the line and solution of inequality}\left(\text{2}\right)\text{is represented by the}\\ \text{shaded region above the line y}=\text{2},\text{including the points on the}\\ \text{line}.\text{\hspace{0.17em}The common region of these two lines is shaded and it is the}\\ \text{required solution region of the given system of inequalities}.\end{array}$

**Q.2 ** Solve the following system of inequalities graphically:

$\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\le}\mathbf{12}\mathbf{,}\mathbf{x}\mathbf{\ge}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\ge}\mathbf{1}$

**Ans.**

$\begin{array}{l}\mathrm{The}\text{given inequalities are:}\\ \text{3x}+\text{2y}\le \text{12 \u2026}\left(1\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}x}\ge \text{1 \u2026}\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}\ge \text{1 \u2026}\left(3\right)\\ \mathrm{The}\text{graphs of following lines are drawn in the figure:}\\ \text{3x}+2\mathrm{y}=1\text{2,\hspace{0.17em}\hspace{0.17em}x}=\text{1 and y}=\text{1.}\end{array}$ $\begin{array}{l}\text{The inequality}\left(1\right)\text{shows shaded region below the line}\\ \text{3x + 2y = 12, inequality}\left(2\right)\text{shows the shaded region right of}\\ \text{the line x = 1 and inequality}\left(3\right)\text{shows the shaded region}\\ \text{above the line y}=\text{1.Hence, the common region of all these}\\ \text{lines is the solution of the given system of the linear inequalities.}\end{array}$

**Q.3 **

**Ans.**

$\begin{array}{l}\text{The given inequalities are:}\\ \mathrm{x}+\mathrm{y}\ge 4\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ 2\mathrm{x}-\mathrm{y}\ge 0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \text{The graphs of linear equations}\mathrm{x}+\mathrm{y}=4\text{and}2\mathrm{x}-\mathrm{y}=0\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x + y* *= 6, including the points on the line.

On the same set of axes, we draw the graph of the equation 3x + 4y = 12 as shown in Figure.

Then we see that inequality (2) represents the shaded region below the line 3x + 4y = 12, including the points on the line.

In this way, the double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

**Q.4 **

**Ans.**

$\begin{array}{l}\text{The given inequalities are:}\\ \mathrm{x}+\mathrm{y}\ge 4\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\text{(1)}\\ 2\mathrm{x}-\mathrm{y}\ge 0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \text{The graphs of linear equations}\mathrm{x}+\mathrm{y}=4\text{and}2\mathrm{x}-\mathrm{y}=0\\ \text{are drawn in}\mathrm{xy}\text{-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line x + y* *= 4, including the points on the line.

On the same set of axes, we draw the graph of the equation 2x – y = 0 as shown in Figure.

Then we see that inequality (2) represents the shaded region above the line 2x – y =0, including the points on the line.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

**Q.5** Solve the following system of inequalities graphically:

$\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{>}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{<}\mathbf{-}\mathbf{1}$

**Ans.**

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given\hspace{0.33em}inequalities\hspace{0.33em}are:}\\ 2\text{x}-\text{y}>1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ \text{x}-2\text{y}<-1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations 2x}-\text{y}=1\text{and x}-2\text{y}=-1\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x – y* *= 1, excluding the points on the line.

On the same set of axes, we draw the graph of the equation 2x – y* *= 1 as shown in Figure.

Then we see that inequality (2) represents the shaded region above the line x – 2y = –1.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

**Q.6** Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le}\mathbf{6}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge}\mathbf{4}$

**Ans.**

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given\hspace{0.33em}inequalities\hspace{0.33em}are:}\\ \text{x}+\text{y}\le 6\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ \text{x}+\text{y}\ge 4\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations x}+\text{y}=6\text{and x}+\text{y}=4\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line x + y* *= 6, including the points on the line.

On the same set of axes, we draw the graph of the equation x + y* *= 4 as shown in Figure.

Then we see that inequality (2) represents the shaded region above the line x + y = 4, including the points on the line.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

**Q.7 ** Solve the following system of inequalities graphically:

$\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge}\mathbf{8}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\ge}\mathbf{10}$

**Ans.**

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given\hspace{0.33em}inequalities\hspace{0.33em}are:}\\ 2\text{x}+\text{y}\ge 8\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ \text{x}+2\text{y}\ge 10\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations 2x}+\text{y}=8\text{and x}+2\text{y}=10\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x + y* *= 8,

including the points on the line.

On the same set of axes, we draw the graph of the equation x + 2y* *= 10 as shown in Figure.

Then we see that inequality (2) represents the shaded region above the line x + 2y* *= 10,

including the points on the line.

In this way, the double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

**Q.8 ** Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le}\mathbf{9}\mathbf{,}\mathbf{y}\mathbf{>}\mathbf{x}\mathbf{,}\mathbf{x}\mathbf{\ge}\mathbf{0}$

**Ans.**

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given\hspace{0.33em}inequalities\hspace{0.33em}are:}\\ \text{x}+\text{y}\le 9\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ \text{y}>\text{x\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \text{x}\ge 0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(3\right)\\ \mathrm{The}\text{graphs of linear equations 2x}+\text{y}=8\text{and x}+2\text{y}=10\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line x + y* *= 9, including the points on the line.

On the same set of axes, we draw the graph of the equation y* *= x and x = 0 as shown in Figure.

Then we see that inequality (2) and (3) represent the shaded region above the line y* *= x and right of the line and

x = 0 excluding the points on the line respectively.

In this way, the triple shaded region, common to the above three shaded regions, is the required solution region of the given system of inequalities.

**Q.9 **

**Ans.**

$\begin{array}{l}\text{The given inequalities are:.}\\ 5\mathrm{x}+4\mathrm{y}\le 60\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{x}\ge 1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{y}\ge 2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(3\right)\\ \text{The graphs of linear equations}5\mathrm{x}+4\mathrm{y}=60,\mathrm{x}=1\text{and}\mathrm{y}=2\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line 5x + 4y* *= 60, including the points on the line.

On the same set of axes, we draw the graph of the equation x* *= 1 and y = 2 as shown in Figure.

Then we see that inequality (2) represents the shaded region above the line x* *= 1 on the line including the points on the line and (3) represents the shaded region above the line y = 2 on the line including the points on the line.

In this way, green region is the common region. Which is the required solution region of the given system of inequalities.

**Q.10 **

$\begin{array}{l}\mathbf{\text{Solve the following system of inequalities graphically:}}\\ \mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\le}\mathbf{60}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\le}\mathbf{30}\mathbf{,}\mathbf{x}\mathbf{\ge}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\ge}\mathbf{0}\end{array}$

**Ans.**

$\begin{array}{l}\text{The given inequalities are:}\\ 3\mathrm{x}+4\mathrm{y}\le 60\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ \text{\hspace{0.33em}}\mathrm{x}+3\mathrm{y}\le 30\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{x}\ge 0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(3\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{y}\ge 0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(4\right)\\ \text{The graphs of linear equations}3\mathrm{x}+4\mathrm{y}=60,\mathrm{x}+3\mathrm{y}=30\text{,}\\ \mathrm{x}=0\text{and}\mathrm{y}=0\text{are drawn in}\mathrm{xy}\text{-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line 3x + 4y* *= 60, including the points on the line.

On the same set of axes, we draw the graph of the equation x + 3y* *= 30, x* *= 0 and y = 0 as shown in Figure. Then we see that inequality (2), (3) and (4) represent the shaded region below the line x + 3y* *= 30,

x = 0 and y = 0 including the points on the line.

In this way, the green shaded region is the required solution region of the given system of inequalities.

**Q.11 **

**Ans.**

$\begin{array}{l}\text{The given inequalities are:.}\\ 2\mathrm{x}+\mathrm{y}\ge 4\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ \text{\hspace{0.33em}\hspace{0.33em}}\mathrm{x}+\mathrm{y}\le 3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ 2\mathrm{x}-3\mathrm{y}\le 6\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(3\right)\\ \text{The graphs of linear equations}2\mathrm{x}+\mathrm{y}=4,\mathrm{x}+\mathrm{y}=3,\\ \mathrm{and}\text{\hspace{0.33em}}2\mathrm{x}-3\mathrm{y}=6\text{\hspace{0.33em}are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.33em}}\left(1\right)\text{is false for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is above the line of equation}\\ \text{2x}+\text{y}=4\text{including the points on\hspace{0.33em}the line.}\\ \text{Inequality\hspace{0.33em}}\left(2\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below the line of equation}\\ \text{x}+\text{y}=3\text{including the points on\hspace{0.33em}the line.}\\ \text{Inequality\hspace{0.33em}}\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below the line of equation}\\ \text{2x}-\text{3y\hspace{0.33em}=\hspace{0.33em}6 including the points on\hspace{0.33em}the line.}\\ \text{The common shaded area of the given three inequalities is\hspace{0.33em}the solution area.}\end{array}$

**Q.12 **

**Ans.**

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given inequalities are:}\\ \text{\hspace{0.33em}\hspace{0.33em}x}-2\text{y}\le 3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(1\right)\\ 3\text{x}+4\text{y}\ge 12\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x}\ge 0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(3\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\ge 1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations x}-2\text{y}=3\text{,\hspace{0.33em}}3\text{x}+4\text{y}=12,\text{x}=0\\ \text{and y}=\text{1 are drawn in xy-plane.}\\ \text{Inequality\hspace{0.33em}}\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is left of the line of equation}\\ \text{x}-2\text{y}=3\text{including the points on\hspace{0.33em}}\mathrm{the}\text{\hspace{0.33em}}\mathrm{line}.\\ \text{Inequality\hspace{0.33em}}\left(2\right)\text{is false for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is above the line of equation}\\ 3\text{x}+4\text{y}=12\text{including the points on\hspace{0.33em}the\hspace{0.33em}line.}\\ \text{Inequality\hspace{0.33em}}\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is right of the line of equation}\\ \text{x}=0\text{including the points on the\hspace{0.33em}line.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.33em}}\left(4\right)\text{is false for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is above the line of equation}\\ \text{2x}-3\text{y}=6\text{including the points on \hspace{0.33em}}\mathrm{the}\text{\hspace{0.33em}}\mathrm{line}.\\ \text{Thus, the common green area of the given three inequalities is \hspace{0.33em}the solution area.}\end{array}$

**Q.13 **

**Ans.**

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ 4\mathrm{x}+3\mathrm{y}\le 60...\left(1\right)\\ \mathrm{y}\ge 2\mathrm{x}...\left(2\right)\\ \mathrm{x}\ge 3...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations 4}\mathrm{x}+3\mathrm{y}=60\text{,\hspace{0.17em}}\mathrm{y}=2\mathrm{x}\text{,x}=\text{3,}\mathrm{x}=0\\ \text{and y}=\text{0 are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.33em}}\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below of the line of equation}\\ \text{4x}+3\text{y}=60\text{including the points\hspace{0.33em}on the line.}\\ \text{Inequality\hspace{0.33em}}\left(2\right)\text{is true for the point}\left(1,3\right),\text{so its solution area\hspace{0.33em}is above the line of equation}\\ \text{y}=2\text{x including the points on\hspace{0.33em}the\hspace{0.33em}line.}\\ \text{Inequality\hspace{0.33em}}\left(3\right)\text{is false for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is right of the line of equation}\\ \text{x}=3\text{including the points on the\hspace{0.33em}line.}\\ \text{Inequality\hspace{0.33em}}\left(4\right)\text{is true for the values of x and y equal to zero\hspace{0.33em}or greater than zero.}\\ \text{So, its solution area is right of the x}-\mathrm{axis}\text{\hspace{0.33em}and above to the y-axis.}\\ \text{Thus, the green shaded area of the given three inequalities is\hspace{0.33em}the solution area.}\end{array}$

**Q.14 **

$\begin{array}{l}\text{Solve the following system of inequalities graphically:}\\ 3\text{x}+2\text{y}\le 150,\text{\hspace{0.33em}x}+4\text{y}\le 80,\text{x}\le 15,\text{x}\ge 0,\text{\hspace{0.33em}y}\ge 0\end{array}$

**Ans.**

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ 3\mathrm{x}+2\mathrm{y}\le 150...\left(1\right)\\ \mathrm{x}+4\mathrm{y}\le 80...\left(2\right)\\ \mathrm{x}\le 15...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations}3\mathrm{x}+2\mathrm{y}=150\text{,\hspace{0.17em}}\mathrm{x}+4\mathrm{y}=80\text{,\hspace{0.17em}x}=\text{15,}\\ \mathrm{x}=0\text{\hspace{0.17em}\hspace{0.17em}and y}=\text{0 are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.33em}}\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below of the line of equation}\\ \text{3x}+2\text{y}=150\text{including the points\hspace{0.33em}on the line.}\\ \text{Inequality\hspace{0.33em}}\left(2\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below the line of equation}\\ \text{x}+4\text{y}=80\text{including the points on\hspace{0.33em}the line.}\\ \text{Inequality\hspace{0.33em}}\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is left of the line of equation}\\ \text{x}=15\text{including the points on the\hspace{0.33em}}\mathrm{line}.\\ \text{Inequality\hspace{0.33em}}\left(4\right)\text{is true for the values of x and y equal to zero\hspace{0.33em}or greater than zero.}\\ \text{So, its solution area is right of the x}-\mathrm{axis}\text{\hspace{0.33em}and above to the y-axis.}\\ \text{Thus, the green shaded area of the given four inequalities is\hspace{0.33em}}\mathrm{the}\text{\hspace{0.33em}}\mathrm{solution}\text{\hspace{0.33em}}\mathrm{area}.\end{array}$

**Q.15 **

$\begin{array}{l}\text{Solve the following system of inequalities graphically:}\\ \text{}\mathrm{x}+2\mathrm{y}\le 10,\mathrm{x}+\mathrm{y}\ge 1,\mathrm{x}-\mathrm{y}\le 0,\mathrm{x}\ge 0,\mathrm{y}\ge 0\end{array}$

**Ans.**

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ \mathrm{x}+2\mathrm{y}\le 10...\left(1\right)\\ \mathrm{x}+\mathrm{y}\ge 1...\left(2\right)\\ \mathrm{x}-\mathrm{y}\le 0...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations}\mathrm{x}+2\mathrm{y}=10\text{, \hspace{0.17em}}\mathrm{x}+\mathrm{y}=1\text{,\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-\mathrm{y}=\text{0,}\\ \mathrm{x}=0\text{\hspace{0.17em}\hspace{0.17em}and y}=\text{0 are drawn in xy-plane.}\\ \text{Inequality\hspace{0.17em}}\left(1\right)\text{is true for the point}(0,0),\text{so its solution area}\\ \text{is below of the line of equation}\mathrm{x}+2\mathrm{y}=10\text{including the points}\\ \text{on the line.}\\ \text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(2\right)\text{is false for the point}(0,0),\text{so its solution area}\\ \text{is above the line of equation}\mathrm{x}+\mathrm{y}=1\text{including the points on}\\ \text{the line.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(3\right)\text{is true for the point}(2,1),\text{so its solution area}\\ \text{is above the line of equation}\mathrm{x}-\mathrm{y}=0\text{including the points on the}\\ \text{line.}\\ \text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(4\right)\text{is true for the values of x and y equal to zero}\\ \text{or greater than zero. So, its solution area is right of the}\mathrm{x}-\mathrm{axis}\\ \text{and above to the y-axis.}\\ \text{Thus, the blue shaded area of the given four inequalities is}\\ \text{the solution area.}\end{array}$

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