NCERT Solutions for Exercise 6.3 Class 11 Maths Chapter 6 - Linear Inequalities

# NCERT Solutions for Exercise 6.3 Class 11 Maths Chapter 6 - Linear Inequalities

Edited By Vishal kumar | Updated on Nov 07, 2023 06:48 AM IST

## NCERT Solutions for Class 11 Maths Chapter 6: Linear Inequalities Exercise 6.3- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 6: Linear Inequalities Exercise 6.3- In the section before exercise 6.3 Class 11 Maths, the NCERT presented the topics of linear inequalities in one and two variables and their graphical solutions. In NCERT solutions for exercise 6.3 Class 11 Maths chapter 6 the questions related to the system of two-variable linear inequalities is discussed. Solving the Class 11 Maths chapter 6 exercise 6.3 will give a good idea of the concepts discussed before Class 11th Maths chapter 6 exercise 6.3. Before practising the NCERT solutions for Class 11 Maths Chapter 6 Exercise 6.3 it's better to go through the solved examples given in the NCERT book prior to Class 11 Maths chapter 6 exercise 6.3.

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11th class maths exercise 6.3 answers are written by subject experts at Careers360 in great detail and step-by-step, ensuring that students can easily understand them. In addition to the text solutions of ex 6.3 class 11, PDF versions of the solutions are also available for students, and they are provided free of charge. This allows students to access and use them anytime and anywhere. The following exercises are also listed in the NCERT syllabus for Class 11 Mathematics.

**As per the new CBSE Syllabus for 2023-24, this chapter has been assigned a different number, and it is now referred to as Chapter 5.

## Question:1 Solve the following system of inequalities graphically:

$x \geq 3,\ y\geq 2$

Graphical representation of $x=3$ and $y=2$ is given in the graph below.

The line $x=3$ and $y=2$ divides plot in four regions i.e.I,II,III,IV.

For $x \geq 3$ ,

The solution to this inequality is region II and III including points on this line because points on the line also satisfy the inequality.

For $y \geq 2$ ,

The solution to this inequality is region IV and III including points on this line because points on the line also satisfy the inequality.

Hence, solution to $x \geq 3,\ y\geq 2$ is common region of graph i.e. region III.

Thus, solution of $x \geq 3,\ y\geq 2$ is region III.

This can be represented as follows:

The below green colour represents the solution

$3x +2y \leq 12,\ x \geq 1, \ y\geq 2$

Graphical representation of $x=1 \, \, ,3x+2y=12$ and $y=2$ is given in graph below.

For $x \geq 1$ ,

The solution to this inequality is region on right hand side of line $(x=1)$ including points on this line because points on the line also satisfy the inequality.

For $y \geq 2$ ,

The solution to this inequality is region above the line $(y=2)$ including points on this line because points on the line also satisfy the inequality.

For $3x+2y\leq 12$

The solution to this inequality is region below the line $(3x+2y= 12)$ including points on this line because points on the line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$2x +y \geq 6, 3x +4y\leq 12$

Graphical representation of $2x +y =6\, \, and\, \, 3x +4y=12$ is given in the graph below.

For $2x +y \geq 6$ ,

The solution to this inequality is region above line $(2x +y =6)$ including points on this line because points on the line also satisfy the inequality.

For $3x +4y\leq 12$ ,

The solution to this inequality is region below the line $( 3x +4y= 12)$ including points on this line because points on the line also satisfy the inequality.

Hence, the solution to these linear inequalities is the shaded region(ABC) as shown in figure including points on the respective lines.

This can be represented as follows:

$x + y \geq 4, 2x - y <0$

Graphical representation of $x +y =4\, \, and\, \, 2x -y=0$ is given in the graph below.

For $x + y \geq 4,$ ,

The solution to this inequality is region above line $(x +y =4)$ including points on this line because points on the line also satisfy the inequality.

For $2x - y <0$ ,

The solution to this inequality is half plane corresponding to the line $( 2x -y=0)$ containing point $(1,0)$ excluding points on this line because points on the line does not satisfy the inequality.

Hence, the solution to these linear inequalities is the shaded region as shown in figure including points on line $(x +y =4)$ and excluding points on the line $( 2x -y=0)$ .

This can be represented as follows:

$2x - y > 1, \ x -2y < -1$

Graphical representation of $x -2y =-1\, \, and\, \, 2x -y=1$ is given in graph below.

For $2x - y > 1,$

The solution to this inequality is region below line $( 2x -y=1)$ excluding points on this line because points on line does not satisfy the inequality.

For $\ x -2y < -1$ ,

The solution to this inequality is region above the line $(x -2y =-1)$ excluding points on this line because points on line does not satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure excluding points on the lines.

This can be represented as follows:

$x + y \leq 6, x + y \geq 4$

Graphical representation of $x + y = 6,\, \, and\, \, \, x + y = 4$ is given in the graph below.

For $x + y \leq 6,$

The solution to this inequality is region below line $( x+y=6)$ in cluding points on this line because points on the line also satisfy the inequality.

For $x + y \geq 4$ ,

The solution to this inequality is region above the line $( x+y=4)$ including points on this line because points on the line also satisfy the inequality.

Hence, the solution to these linear inequalities is shaded region as shown in figure including points on the lines.

This can be represented as follows:

$2x + y \geq 8 , x + 2y \geq 10$

Graphical representation of $2x + y = 8\, \, and\, \, x + 2y =10$ is given in graph below.

For $2x + y \geq 8 ,$

The solution to this inequality is region above line $(2x + y = 8)$ including points on this line because points on line also satisfy the inequality.

For $x + 2y \geq 10$ ,

The solution to this inequality is region above the line $( x + 2y =10)$ including points on this line because points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the lines.

This can be represented as follows:

$x + y \leq 9, y > x, x\geq 0$

Graphical representation of $x+y=9,x=y$ and $x=0$ is given in graph below.

For $x + y \leq 9$ ,

The solution to this inequality is region below line $(x+y=9)$ including points on this line because points on line also satisfy the inequality.

For $y > x$ ,

The solution to this inequality represents half plane corresponding to the line $(x=y)$ containing point $(0,1)$ excluding points on this line because points on line does not satisfy the inequality.

For $x\geq 0$ ,

The solution to this inequality is region on right hand side of the line $(x=0)$ including points on this line because points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure.

This can be represented as follows:

$5x+4y\leq20, \ x\geq 1, \ y\geq 2$

Graphical representation of $\, ,5x+4y=20,\, \, \, x=1\, \, and \, \, y=2$ is given in graph below.

For $5x+4y\leq20,$ ,

The solution to this inequality is region below the line $(5x+4y=20)$ including points on this line because points on line also satisfy the inequality.

For $\ x\geq 1,$ ,

The solution to this inequality is region right hand side of the line $(x=1)$ including points on this line because points on line also satisfy the inequality.

For $\ y\geq 2,$

The solution to this inequality is region above the line $(y=2)$ including points on this line because points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$3x + 4y \leq 60,\ x + 3y \leq 30, \ x \geq 0, \ y\geq 0$

Graphical representation of $3x+4y=60 \, \, ,x+3y=30\, \, \, ,x=0\, \, and\, \, y=0$ is given in graph below.

For $3x + 4y \leq 60$ ,

The solution to this inequality is region below the line $(3x+4y=60)$ including points on this line because points on line also satisfy the inequality.

For $\ x + 3y \leq 30$ ,

The solution to this inequality is region below the line $(x+3y=30)$ including points on this line because points on line also satisfy the inequality.

For $\ x \geq 0,$

The solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because points on line also satisfy the inequality.

For $\ y \geq 0,$

The solution to this inequality is region above the line $(y=0)$ including points on this line because points on line also satisfy the inequality.

Hence, the solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$2x +y \geq 4, \ x + y \leq 3, \ 2x - 3y \leq 6$

Graphical representation of $2x+y=4 \, \, ,x+y=3$ and $2x-3y=6$ is given in graph below.

For $2x +y \geq 4,$ ,

The solution to this inequality is region above the line $(2x+y=4)$ including points on this line because points on line also satisfy the inequality.

For $\ x + y \leq 3,$ ,

The solution to this inequality is region below the line $(x+y=3)$ including points on this line because points on line also satisfy the inequality.

For $\ 2x - 3y \leq 6,$

The solution to this inequality is region above the line $(2x-3y= 6)$ including points on this line because points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$x -2y \leq 3, 3x + 4y \geq 12, x \geq 0, y\geq 1$

Graphical representation of $x-2y=3 \, \, ,3x+4y=12\, \, \, ,x=0\, \, and\, \, y=1$ is given in graph below.

For $x -2y \leq 3$ ,

The solution to this inequality is region above the line $(x-2y=3)$ including points on this line because points on line also satisfy the inequality.

For $3x + 4y \geq 12$ ,

The solution to this inequality is region above the line $(3x+4y=12)$ including points on this line because points on line also satisfy the inequality.

For $\ x \geq 0,$

The solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because points on line also satisfy the inequality.

For $\ y \geq 1,$

The solution to this inequality is region above the line $(y=1)$ including points on this line because points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$4x + 3y \leq 60,\ y\geq 2x,\ x\geq 3,\ x,y\geq 0$

Graphical representation of $4x+3y=60 \, \, ,y=2x\, \, \,,x=3\, \, ,x=0\, \, and\, \, y=0$ is given in graph below.

For $4x + 3y \leq 60,$

The solution to this inequality is region below the line $(4x+3y=60)$ including points on this line because points on the line also satisfy the inequality.

For $y\geq 2x$ ,

The solution to this inequality is region above the line $(y=2x)$ including points on this line because points on the line also satisfy the inequality.

For $x\geq 3$ ,

The solution to this inequality is region right hand side of the line $(x=3)$ including points on this line because points on the line also satisfy the inequality.

For $\ x \geq 0,$

The solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because points on the line also satisfy the inequality.

For $\ y \geq 0,$

The solution to this inequality is region above the line $(y=0)$ including points on this line because points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$3x + 2y \leq 150, \ x +4y \leq 80,\ x\leq 15 \ y\geq 0, \ x\geq 0$

Graphical representation of $3x+2y=150 \, \, ,x+4y=80\, \, \,,x=15\, \, ,x=0\, \, and\, \, y=0$ is given in graph below.

For $3x + 2y \leq 150,$

The solution to this inequality is region below the line $(3x+2y=150)$ including points on this line because points on the line also satisfy the inequality.

For $x+4y\leq 80$ ,

The solution to this inequality is region below the line $(x+4y=80)$ including points on this line because points on the line also satisfy the inequality.

For $x\leq 15$ ,

The solution to this inequality is region left hand side of the line $(x=15)$ including points on this line because points on the line also satisfy the inequality.

For $\ x \geq 0,$

The solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because points on the line also satisfy the inequality.

For $\ y \geq 0,$

The solution to this inequality is region above the line $(y=0)$ including points on this line because points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$x+2y \leq 10, \ x +y \geq 1, \ x-y\leq 0, x\geq 0, \ y\geq 0$

Graphical representation of $x+2y=10 \, \, ,x+y=1\, \, \,,x-y=0\, \, ,x=0\, \, and\, \, y=0$ is given in graph below.

For $x+2y \leq 10,$

The solution to this inequality is region below the line $(x+2y=10)$ including points on this line because points on line also satisfy the inequality.

For $\ x +y \geq 1,$ ,

The solution to this inequality is region above the line $(x+y=1)$ including points on this line because points on line also satisfy the inequality.

For $\ x-y\leq 0,$ ,

The solution to this inequality is region above the line $(x-y=0)$ including points on this line because points on line also satisfy the inequality.

For $\ x \geq 0,$

The solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because points on line also satisfy the inequality.

For $\ y \geq 0,$

The solution to this inequality is region above the line $(y=0)$ including points on this line because points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

## More About NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.3

There are a total of fifteen questions in exercise 6.3 Class 11 Maths. Solving all these questions will improve the problem-solving skills of the concept solutions of linear inequalities in two variables graphically. Solved examples are given prior to the Class 11th Maths chapter 6 exercise 6.3 to understand the steps involved in solving the questions of the type discussed in Class 11 Maths chapter 6 exercise 6.3.

Also Read| Linear Inequalities Class 11th Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.3

• The content covered in exercise 6.3 Class 11 Maths is helpful in revising the concepts discussed in the chapter.

• Practising problems is important for maths subjects. And the Class 11 Maths chapter 6 exercise 6.3 helps for the same.

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## Key Features of NCERT 11th Class Maths Exercise 6.3 Answers

1. Comprehensive Coverage: The class 11 maths ex 6.3 answers provided cover all the exercises and problems in Chapter 6, Exercise 6.3 of the NCERT 11th Class Mathematics textbook.

2. Step-by-Step Solutions: The class 11 ex 6.3 answers are presented in a step-by-step format, making it easier for students to follow and understand the solution process.

3. Clarity and Accuracy: The class 11 maths chapter 6 exercise 6.3 solutions are written with clarity and accuracy, ensuring that students can grasp the concepts and methods required to solve mathematical problems.

4. Use of Proper Notation: The 11th class maths exercise 6.3 answers use appropriate mathematical notations and terminology, helping students become familiar with the language of mathematics.

5. Free Access: The ex 6.3 class 11 answers are typically made available for free, allowing students to access them without any cost, making it a valuable resource for self-study.

6. Supplementary Learning: These class 11 maths ex 6.3 answers can be used as a supplementary learning resource to reinforce classroom learning and aid in exam preparation.

7. Homework and Practice: Students can use these answers to check their work, practice problem-solving, and improve their overall performance in mathematics.

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## Subject Wise NCERT Exampler Solutions

1. Give examples of linear inequalities?

2x>5

2x+3y>5

2. Give examples for a system of linear inequalities

X>0,y>2

2x+y.6, 3x+y<12

3. The quadrant that represents both x and y coordinate greater than or equal to zero is………?

4. Give an application of linear inequalities?

In solving linear programming problems, linear inequalities are used.

5. How many questions are solved in the exercise 6.3 Class 11 Maths?

15 questions are solved in the NCERT solutions for Class 11 Maths chapter 6 exercise 6.3

6. What method is used to solve the linear inequalities given in the Class 11 Maths chapter 6 exercise 6.3?

Graphical method

7. Why do we solve Class 11 Maths chapter 6 exercise 6.3?

To get an exposure to the concepts of solving a system of linear inequalities in two variables using the graphical method

8. What is the importance of NCERT Class 11 Maths exercises?

As far as mathematics is concerned it is a subject that requires a good amount of practice. And the NCERT exercise helps for the same.

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