Class 11 Maths Ncert Solutions Chapter 6 Ex 6.3 Linear Inequalities PDF

Class 11 Maths Ncert Solutions Ex 6.3

Class 11 Maths Ncert Solutions Chapter 6 Ex 6.3

Question 1

Solve the system of inequalities graphically:

= x 3 and y 2

Sol:

= x ≥ 3 ……………………(i)

= y ≥ 2……………………..(ii)

In order to draw the graph of the following inequalities, let us consider x = 3 and y = 2

In case of equation (i) the feasible region is the right hand side of the line including the points on the line x = 3 , whereas in case of equation (ii) the feasible region is above the line including the points on the line y = 2.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Linear Inequalites

 

 

Question 2

Solve graphically the system of inequalities:

3x + 2y 12, x 1 and y 2

Sol:

3x + 2y ≤ 12 …………… (i)

x ≥ 1 ……………………… (ii)

y ≥ 2 ……………………….(iii)

In order to draw the graph of the following inequalities, let us consider 3x + 2y = 12, x = 1 and y = 2

In case of equation (i) the feasible region is below the line including the points on the line 3x + 2y = 12. In case equation (ii) the feasible region is on the right hand side including the points on the line x = 1 and in case of equation (iii) the feasible region is above the line including the points on the line y = 2.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Linear Inequalities - Common Shaded Area

 

 

Question 3

Solve graphically the system of inequalities:

2x + y 6 and 3x + 4y 12.

Sol:

2x + y ≥ 6 …………………………. (i)

3x + 4y ≤ 12……………………… (ii)

In order to draw the graph of the following inequalities, let us consider 2x + y = 6 and 3x + 4y = 12.

In case of equation (i) the feasible region is above the line including the points on the line 2x + y  = 6 , whereas in case of equation (ii) the feasible region is below the line including the points on the line 3x + 4y = 12

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 4

Solve graphically the system of inequalities:

x + y 4 and 2x – y ˃ 0

Sol:

x + y ≥ 4 …………………………. (i)

2x – y ˃ 0……………………… (ii)

In order to draw the graph of the following inequalities, let us consider x + y = 4 and 2x – y = 0.

In case of equation (i) the feasible region is above the line including the points on the line x + y  = 4, whereas in case of equation (ii) the feasible region is the right hand side of  the line including the points on the line 2x – y = 0.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 5

Solve graphically the system of inequalities:

2x – y ˃ 1 and x – 2y ˂ -1

Sol:

2x – y ˃ 1 …………………………. (i)

x – 2y ˂ -1……………………… (ii)

In order to draw the graph of the following inequalities, let us consider 2x – y = 1 and x – 2y = -1.

In case of equation (i) the feasible region is below the line including the points on the line 2x – y = 1, whereas in case of equation (ii) the feasible region is the above   the line including the points on the line x – 2y = -1.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 6

Solve graphically the system of inequalities:

x + y 6 and x + y 4

Sol:

x + y ≤ 6 …………………………. (i)

x + y ≥ 4……………………… (ii)

In order to draw the graph of the following inequalities, let us consider x + y = 6 and x + y = 4.

In case of equation (i) the feasible region is below the line including the points on the line x + y = 6, whereas in case of equation (ii) the feasible region is the above   the line including the points on the line x + y = 4.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 7

Solve graphically the system of inequalities:

2x + y 8 and x + 2y 10

Sol:

2x + y ≥ 8 …………………………. (i)

x + 2y ≥ 10……………………… (ii)

In order to draw the graph of the following inequalities, let us consider 2x + y = 8 and x + 2y = 10.

In case of equation (i) the feasible region is above the line including the points on the line 2x + y = 8 , whereas in case of equation (ii) the feasible region is the above   the line including the points on the line x + 2y = 10.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 8

Solve graphically the system of inequalities:

x + y 9, y ˃ x and x 0

Sol:

x + y ≤ 9 …………………………. (i)

y ˃ x ……….……………………… (ii)

x ≥ 0 ………………………………… (iii)

In order to draw the graph of the following inequalities, let us consider x + y = 9, y = x and x = 0.

In case of equation (i) the feasible region is below the line including the points on the line x + y = 9. In case of equation (ii) the feasible region is above the line including the points on the line y =x  , whereas in case of equation (iii) the feasible region is the right hand side of the line  including the points on the line x = 0

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 9

Solve graphically the system of inequalities:

5x + 4y 20, x 1 and y 2

Sol:

5x + 4y ≤ 20 …………………………. (i)

x ≥ 1 ……….……………………… (ii)

y ≥ 2 ………………………………(iv)

In order to draw the graph of the following inequalities, let us consider 5x + 4y = 20, x = 1, x = 0 and y = 2

In case of equation (i) the feasible region is below the line including the points on the line 5x + 4y = 20. In case of equation (ii) the feasible region is on the right hand side of the line including the points on the line x = 1, where as in equation (iv) the feasible region is above line including the points on the line y = 2

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 10

Solve graphically the system of inequalities:

3x + 4y 60, x + 3y 30 and x 0

Sol:

3x + 4y ≤ 60 …………………………. (i)

x + 3y ≤ 30          ……………………… (ii)

x ≥ 0 …………………………………….(iii)

In order to draw the graph of the following inequalities, let us consider 3x + 4y = 60, x + 3y = 30 and x = 0

In case of equation (i) the feasible region is below the line including the points on the line 3x + 4y = 60. In case of equation (ii) the feasible region is the below the line including the points on the line x + 3y = 30, whereas in case of equation (iii) the feasible region is right and side of the line.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

linear inequalities

 

 

Question 11

Solve graphically the system of inequalities:

2x + y 4, x + y 3 and 2x -3y 6

Sol:

2x + y ≥ 4 …………………………. (i)

x + y ≤ 3 ……….……………………… (ii)

2x – 3y ≤ 6 ………………………………… (iii)

In order to draw the graph of the following inequalities, let us consider 2x +y = 4, x + y = 3 and 2x – 3y = 6.

In case of equation (i) the feasible region is above the line including the points on the line 2x + y = 4. In case of equation (ii) the feasible region is below the line including the points on the line x + y = 3, whereas in case of equation (iii) the feasible region is above line including the points on the line 2x- 3y = 6

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 12

Solve graphically the system of inequalities:

5x + 4y 3, 3x + 4y 12, x 0 and y 1

Sol:

x – 2y ≤ 3 …………………………. (i)

3x + 4y ≥ 12 ……….……………………… (ii)

x ≥ 0 ………………………………… (iii)

y ≥ 1 ………………………………(iv)

In order to draw the graph of the following inequalities, let us consider x – 2y = 3, 3x + 4y = 12, x = 0 and y = 1

In case of equation (i) the feasible region is above the line including the points on the line x – 2y = 3. In case of equation (ii) the feasible region is above the line including the points on the line 3x + 4y = 12, whereas in case of equation (iii) the feasible region is in the right hand side of the axis including the points on the line x = 0 and (iv) the feasible region is above line including the points on the line y = 1

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 13

Solve graphically the system of inequalities:

4x + 3y 60, y 2x, x 3 and x, y 0

Sol:

4x + 3y ≤ 60 …………………………. (i)

y ≥ 2x ……….……………………… (ii)

x ≥ 3 ………………………………… (iii)

x ≥ 0 and y ≥ 0 …………………(iv)

In order to draw the graph of the following inequalities, let us consider 4x + 3y = 60, y = 2x, x = 3 and x, y = 0

In case of equation (i) the feasible region is below the line including the points on the line 4x + 3y = 60. In case of equation (ii) the feasible region is above the line including the points on the line y = 2x, whereas in case of equation (iii) the feasible region is in the right hand side of the axis including the points on the line x = 3 and (iv) the feasible region is above the line and right hand side of the line including the points on the line y = 0 and x =0 respectively.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 14

Solve graphically the system of inequalities:

3x + 2y 150, x + 4y 80, x 15

Sol:

3x + 2y ≤ 150 …………………………. (i)

x + 4y ≤ 80 ……….……………………… (ii)

x ≤ 15 ………………………………… (iii)

In order to draw the graph of the following inequalities, let us consider 3x + 2y = 150, x + 4y = 80, x = 15.

In case of equation (i) the feasible region is below the line including the points on the line 3x + 2y = 150. In case of equation (ii) the feasible region is below the line including the points on the line x + 4y = 80 , whereas in case of equation (iii) the feasible region is in the left hand side of the axis including the points on the line x = 15.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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Question 15

Solve graphically the system of inequalities:

x + 2y 10, x + y 1, x – y  0

Sol:

x + 2y ≤ 10 …………………………. (i)

x + y ≥ 1 ……….……………………… (ii)

x – y ≤ 0 ………………………………… (iii)

In order to draw the graph of the following inequalities, let us consider x + 2y = 10, x + 4y = 1, x – y = 0.

In case of equation (i) the feasible region is below the line including the points on the line x + 2y = 10. In case of equation (ii) the feasible region is above the line including the points on the line x + y = 1, whereas in case of equation (iii) the feasible region is above the line including the points on the line x – y =0.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

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