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

Exercise 6.3

1. Solve the following system of inequalities graphically: 

\[\mathrm{x}>\mathrm{=3}\], \[\mathrm{y}>\mathrm{=2}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\] and \[\left( 2 \right)\] respectively.

\[\text{x}>=\text{3}..............\left( 1 \right)\]

\[\text{y}>=\text{2}.............\left( 2 \right)\]

Graph of the inequalities \[\left( 1 \right)\] and \[\left( 2 \right)\] drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the right hand side region of the line \[\text{x=3}\]( Including the line \[\text{x=3}\]) and inequality \[\left( 2 \right)\] represents the region above the line \[\text{y=2}\]( Including the line \[y=2\]).

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

2. Solve the following system of inequalities graphically: 

\[\mathrm{3x+2y}<=\mathrm{12}\], \[\mathrm{x}>=\mathrm{1}\],\[\mathrm{y}>=\mathrm{2}\]

Ans: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\] respectively.

\[\text{3x+2y}<=\text{12}..............\left( 1 \right)\]

\[\text{x}>=\text{1}.............\left( 2 \right)\]

\[\text{y}>=\text{2}.............\left( 3 \right)\]

Graph of the inequalities \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\]  drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the below the line

region of the line \[\text{3x+2y=12}\]( Including the line \[\text{3x+2y=12}\]) .Inequality  

\[\left( 2 \right)\] represents the right hand side of the line \[\text{x=1}\] ( Including the line \[\text{x=1}\]) and inequality \[\left( 3 \right)\] represents the region above the line of the line \[\text{y=2}\] ( Including the line \[\text{y=2}\]).

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

3. Solve the following system of inequalities graphically: 

\[\mathrm{2x+y}>=\mathrm{6}\], 

\[\mathrm{3x+4y}<=\mathrm{12}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\] and \[\left( 2 \right)\] respectively.

\[\text{2x+y}>=\text{6}..............\left( 1 \right)\]

\[\text{3x+4y}<=\text{12}.............\left( 2 \right)\]

Graph of the inequalities \[\left( 1 \right)\] and \[\left( 2 \right)\] drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents above the line 

region of the line \[\text{2x+y=6}\]( Including the line represents \[\text{2x+y=6}\]) and 

inequality \[\left( 2 \right)\] represents the region below the line \[\text{3x+4y=12}\]( Including the line \[3x+4y=12\]). As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

4. Solve the following system of inequalities graphically: 

\[\mathrm{x+y}>=\mathrm{4}\], \[\mathrm{2x-y}>\mathrm{0}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\] and \[\left( 2 \right)\] respectively.

\[\text{x+y}>=\text{4}..............\left( 1 \right)\]

\[\text{2x-y}>\text{0}.............\left( 2 \right)\]

Graph of the inequalities \[\left( 1 \right)\] and \[\left( 2 \right)\] drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents above the line  region of the line \[x+y=4\]( Including the line represents \[x+y=4\]). 

We can say that the point \[\left( 1,0 \right)\] satisfy the inequality \[2\text{x-y}>\text{0}\].  

\[\left[ 2\left( 1 \right)-0=2>0 \right]\].

Therefore, we can say that the inequality \[\left( 2 \right)\] represents the region that  containing the point \[\left( 1,0 \right)\].(excluding the line \[2\text{x-y0}\] ).

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

5. Solve the following system of inequalities graphically: 

\[\mathrm{2x-y}>\mathrm{1}\], \[\mathrm{x-2y}<\mathrm{-1}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\] and \[\left( 2 \right)\] respectively.

\[\text{2x-y}>\text{1}..............\left( 1 \right)\]

\[\text{x-2y}<\text{-1}.............\left( 2 \right)\]

Graph of the inequalities \[\left( 1 \right)\] and \[\left( 2 \right)\] drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents below the line  region of the line \[2\text{x-y=1}\]( Excluding the line represents \[2\text{x-y=1}\]). 

Inequality \[\left( 2 \right)\] represents above the line region of the line \[\text{x-2y=-1}\]( Excluding the line represents \[\text{x-2y=-1}\]). 

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that excluding the points on the respective lines.

(Image will be uploaded soon)

6. Solve the following system of inequalities graphically: 

\[\mathrm{x+y}<=\mathrm{6}\], \[\mathrm{x+y}>=\mathrm{4}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\] and \[\left( 2 \right)\] respectively.

\[\text{x+y}<=\text{6}..............\left( 1 \right)\]

 \[\text{x+y}>=\text{4}............\left( 2 \right)\]

Graph of the inequalities \[\left( 1 \right)\] and \[\left( 2 \right)\] drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents below the line 

region of the line \[\text{x+y=6}\]( Including the line represents \[\text{x+y=6}\]) and 

inequality \[\left( 2 \right)\] represents the region above the line \[\text{x+y=4}\]( Including the line \[x+y=4\]). As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

7. Solve the following system of inequalities graphically: 

\[\mathrm{2x+y}>=\mathrm{8}\], \[\mathrm{x+2y}>=\mathrm{10}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\] and \[\left( 2 \right)\] respectively.

\[\text{2x+y}>=\text{8}..............\left( 1 \right)\]

 \[\text{x+2y}>=\text{10}............\left( 2 \right)\]

 Graph of the inequalities \[\left( 1 \right)\] and \[\left( 2 \right)\] drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents above the line  region of the line \[2\text{x+y=8}\]( Including the line represents \[2\text{x+y=8}\]) and 

inequality \[\left( 2 \right)\] represents the region below the line \[\text{x+2y=10}\]( Including the line \[\text{x+2y=10}\]). As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

8. Solve the following system of inequalities graphically: 

\[\mathrm{x+y}<=\mathrm{9}\], \[\mathrm{y}>\mathrm{x}\],\[\mathrm{x}>=\mathrm{0}\]

Ans: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\] respectively.

\[\text{x+y}<=\text{9}..............\left( 1 \right)\]

 \[\text{y}>\text{x}.............\left( 2 \right)\]

 \[\text{x}>=\text{0}............\left( 3 \right)\]

 Graph of the inequalities \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\]  drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the below the line

region of the line \[\text{x+y=9}\]( Including the line \[\text{x+y=9}\]) .

We can say that the point \[\left( 0,1 \right)\] satisfy the inequality \[2\text{x-y}>\text{0}\].  

 \[\text{y}>\text{x}\left[ 1>0 \right]\].

Therefore, inequality \[\left( 2 \right)\] represents the region that which lies at \[\left( 0,1 \right)\]. and inequality \[\left( 3 \right)\] represents the region right hand side of the line \[\text{x=0}\] or  ( Including \[\text{y}\]axis).

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

9. Solve the following system of inequalities graphically: 

\[\mathrm{5x+4y}<=\mathrm{20}\], \[\mathrm{x}>=\mathrm{1}\],\[\mathrm{y}>=\mathrm{2}\]

Ans: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\] respectively.

\[\text{5x+4y}<=\text{20}..............\left( 1 \right)\]

\[\text{x}>=\text{1}.............\left( 2 \right)\]

\[\text{y}>=\text{2}............\left( 3 \right)\]

Graph of the inequalities \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\]  drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the below the line

region of the line \[\text{5x+4y=20}\]( Including the line \[\text{5x+4y=20}\]) .

Inequality \[\left( 2 \right)\] represents the right hand side of the line \[\text{x}>=\text{1}\] (Including the line \[\text{x=1}\]) and inequality \[\left( 3 \right)\] represents the region right hand side of the line \[\text{x=0}\] or  ( Including \[\text{y}\]axis).

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

10. Solve the following system of inequalities graphically: 

\[\mathrm{3x+4y}<=\mathrm{60}\], \[\mathrm{x+3y}<=\mathrm{30}\],\[\mathrm{x}>=\mathrm{0}\], \[\mathrm{y}>=\mathrm{0}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\] respectively.

\[\text{3x+4y}>=\text{60}..............\left( 1 \right)\]

\[\text{x+3y}<=\text{30}.............\left( 2 \right)\]

\[\text{x}>=\text{0}.............\left( 3 \right)\]

\[\text{y}>=\text{0}............\left( 4 \right)\]

Graph of the inequalities \[\left( 1 \right)\]and \[\left( 2 \right)\] are drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the below the line region of the line \[\text{3x+4y=60}\]( Including the line \[\text{3x+4y=60}\]) .

Inequality \[\left( 2 \right)\] represents the below the line \[\text{x+3y=30}\] (Including the line \[\text{x+3y=30}\]) and inequality \[\left( 3 \right)\] represents the region right hand side of the line \[\text{x=0}\] or  ( Including \[\text{y}\]axis).

Since \[\text{x}>=\text{0}\] and \[\text{y}>=\text{0}\], it implies the common shaded region in first Quadrant including the points on the respective line.

11. Solve the following system of inequalities graphically:

\[\mathrm{2x+y>=4}\], \[\mathrm{x+y<=3}\],\[\mathrm{2x-3y<=6}\]

Solution: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\] respectively.

\[\text{2x+y>=4}..............\left( 1 \right)\]

\[\text{x+y<=3}.............\left( 2 \right)\]

\[\text{2x-3y<=6}............\left( 3 \right)\]

Graph of the inequalities \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\]  drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the region above the line \[\text{2x+y=4}\]( Including the line \[\text{2x+y=4}\]) .

Inequality \[\left( 2 \right)\] represents the region below the line \[\text{x+y=3}\] (Including the

line \[\text{x+y=3}\]) and inequality \[\left( 3 \right)\] represents the region above as the negative sign is present for $\text{y}$ of the line \[\text{2x-3y=6}\] ( Including \[\text{2x-3y=6}\] line).

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the various lines.

12. Solve the following system of inequalities graphically: 

\[\mathrm{x-2y}<=\mathrm{3}\], \[\mathrm{3x+4y}>=\mathrm{12}\],\[\mathrm{x}>=\mathrm{0}\], \[\mathrm{y}>=\mathrm{1}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\],\[\left( 3 \right)\] and \[\left( 4 \right)\] respectively.

\[\text{x-2y}<=\text{3}..............\left( 1 \right)\]

\[\text{3x+4y}>=\text{12}.............\left( 3 \right)\]

\[\text{x}>=\text{0}.............\left( 3 \right)\]

\[\text{y}>=\text{1}............\left( 4 \right)\]

Graphs of the inequalities\[\left( 1 \right)\],\[\left( 2 \right)\]and \[\left( 4 \right)\] are drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the above the line region of the line \[\text{x-2y=3}\]( Including the line \[\text{x-2y=3}\]) .

Inequality \[\left( 2 \right)\] represents the above the line \[\text{3x+4y=12}\] (Including the line \[\text{3x+4y=12}\])

and inequality \[\left( 4 \right)\] represents the region above the line \[\text{y=1}\] or  ( Including \[\text{y=1}\]).

Since \[\text{x}>=\text{0}\] it implies the common shaded region on right hand side of \[\text{y}\]axis.

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the respective lines.

13. Solve the following system of inequalities graphically: 

\[\mathrm{4x+3y}<=\mathrm{60}\], \[\mathrm{y}>=\mathrm{2x}\],\[\mathrm{x}>=\mathrm{3}\], \[\mathrm{y}>=\mathrm{0}\],\[\mathrm{x}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\],\[\left( 3 \right)\],\[\left( 4 \right)\]  and \[\left( 5 \right)\] respectively.

\[\text{4x+3y}<=\text{60}..............\left( 1 \right)\]

\[\text{y}>=\text{2x}.............\left( 3 \right)\]

\[\text{x}>=\text{3}.............\left( 3 \right)\]

\[\text{x}............\left( 4 \right)\]

\[\text{y}>=\text{0}............\left( 4 \right)\]

Graphs of the inequalities\[\left( 1 \right)\],\[\left( 2 \right)\]and \[\left( 4 \right)\] are drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the below the line

region of the line \[4x+3y=60\]( Including the line \[4x+3y=60\]) .

Inequality \[\left( 2 \right)\] represents the above the line \[\text{y=2x}\] (Including the line \[\text{y=2x}\])

and inequality \[\left( 3 \right)\] represents the region right hand side of the line \[\text{x=3}\] or  ( Including \[\text{x=3}\]).

Since \[\text{x}>=\text{0}\] it implies the common shaded region on the right hand side of \[\text{y}\]axis.

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the respective lines.

14. Solve the following system of inequalities graphically: 

\[\mathrm{3x+2y}<=\mathrm{150}\], \[\mathrm{x+4y<=80}\],\[\mathrm{x}<=\mathrm{15}\], \[\mathrm{y}>=\mathrm{0}\],\[\mathrm{x}>=\mathrm{0}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\],\[\left( 3 \right)\],\[\left( 4 \right)\]  and \[\left( 5 \right)\]    

respectively.

\[\text{3x+2y}<=\text{150}..............\left( 1 \right)\]

\[\text{x+4y}=\text{80}.............\left( 2 \right)\]

\[\text{x}<=\text{15}.............\left( 3 \right)\]

\[\text{y}>=\text{0}............\left( 4 \right)\]

\[\text{x}>=\text{0}............\left( 4 \right)\]

Graph of \[\left( 1 \right)\],\[\left( 2 \right)\]and \[\left( 3 \right)\] are drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the below the line

region of the line \[\text{3x+2y=150}\] ( Including the line \[\text{3x+2y=150}\]) .

Inequality \[\left( 2 \right)\] represents the below the line \[\text{x+4y=80}\] (Including the line \[\text{x+4y=80}\]) and inequality \[\left( 3 \right)\] represents the region left hand side of the line \[\text{x=15}\] or  ( Including \[\text{x=15}\]).

Since \[\text{x}>=\text{0}\] and \[\text{y}>=\text{0}\]and it implies the common shaded region in the first quadrant.

As a result, the solution of the given system of linear inequalities is the point (0,20)

15. Solve the following system of inequalities graphically: 

\[\mathrm{x+2y}<=\mathrm{10}\], \[\mathrm{x+y}>=\mathrm{1}\],\[\mathrm{x-y}<=\mathrm{0}\], \[\mathrm{y}>=\mathrm{0}\],\[\mathrm{x}>=\mathrm{0}\].

Ans: Let us assume the given inequalities as \[\left( 1 \right)\],\[\left( 2 \right)\],\[\left( 3 \right)\],\[\left( 4 \right)\]  and \[\left( 5 \right)\]    

 respectively.

\[\text{x+2y}<=\text{10}..............\left( 1 \right)\]

\[\text{x+y}>=\text{1}.............\left( 2 \right)\]

\[\text{x-y}<=\text{0}.............\left( 3 \right)\]

\[\text{x}>=\text{0}............\left( 4 \right)\]

\[\text{y}>=\text{0}............\left( 4 \right)\]

Graphs of \[\left( 1 \right)\],\[\left( 2 \right)\]and \[\left( 3 \right)\] are drawn below.

Therefore, we can see that the inequality \[\left( 1 \right)\] represents the below the line

region of the line \[\text{x+2y}<=\text{10}\] ( Including the line \[\text{x+2y}<=\text{10}\]) .

Inequality \[\left( 2 \right)\] represents the above the line \[\text{x+y=1}\] (Including the

line \[\text{x+y=1}\]) and inequality \[\left( 3 \right)\] represents the above the line \[\text{x-y=0}\] (Including the line \[\text{x-y=0}\])

Since \[\text{x}>=\text{0}\] and \[\text{y}>=\text{0}\]and it implies the common shaded region in the first quadrant.

As a result, the solution of the given system of linear inequalities is represented as follows by a common shaded region that includes the points on the respective lines.

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 6 Exercise

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

Opting for the NCERT solutions for Ex 6.3 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.3 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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Besides these NCERT solutions for Class 11 Maths Chapter 6 Exercise 6.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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