# Class 11 NCERT Solutions – Chapter 6 Linear Inequalities – Exercise 6.3

Last Updated : 21 Feb, 2021

### Question 1: x â‰¥ 3, y â‰¥ 2

Solution:

For equation 1:

Now draw a solid line x = 3 in the graph (because (x = 3) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 3

Lets, select origin point (0, 0)

â‡’ 0 â‰¥ 3

â‡’ 0 â‰¥ 3 (this not is true)

Hence, Solution region of the given inequality is the line x â‰¥ 3. where, Origin is not included in the region

For equation 2:

Now draw a solid line y = 2 in the graph (because (y = 2) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 2

Lets, select origin point (0, 0)

â‡’ 0 â‰¥ 2

â‡’ 0 â‰¥ 2 (this not is true)

Hence, Solution region of the given inequality is the line y â‰¥ 2. where, Origin is not included in the region

The graph will be as follows for Equation 1 and 2:

### Question 2: 3x + 2y â‰¤ 12, x â‰¥ 1, y â‰¥ 2

Solution:

For equation 1:

Now draw a solid line 3x + 2y = 12 in the graph (because (3x + 2y = 12) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 3x + 2y â‰¤ 12

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 12

â‡’ 0 + 0 â‰¤ 12 (this is true)

Hence, Solution region of the given inequality is the line 3x + 2y â‰¤ 12. where, Origin is included in the region

For equation 2:

Now draw a solid line x = 1 in the graph (because (x = 1) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 1

Lets, select origin point (0, 0)

â‡’ 0 â‰¥ 1

â‡’ 0 â‰¥ 1 (this not is true)

Hence, Solution region of the given inequality is the line x â‰¥ 1. where, Origin is not included in the region

For equation 3:

Now draw a solid line y = 2 in the graph (because (y = 2) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 2

Lets, select origin point (0, 0)

â‡’ 0 â‰¥ 2

â‡’ 0 â‰¥ 2 (this not is true)

Hence, Solution region of the given inequality is the line y â‰¥ 2. where, Origin is not included in the region

The graph will be as follows for Equation 1, 2 and 3:

### Question 3: 2x + y â‰¥ 6, 3x + 4y < 12

Solution:

For equation 1:

Now draw a solid line 2x + y = 6 in the graph (because (2x + y = 6) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 2x + y â‰¥ 6

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ 6

â‡’ 0 â‰¥ 6 (this is not true)

Hence, Solution region of the given inequality is the line 2x + y â‰¥ 6. where, Origin is not included in the region

For equation 2:

Now draw a dotted line 3x + 4y = 12 in the graph (because (3x + 4y = 12) is NOT the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 3x + 4y < 12

Lets, select origin point (0, 0)

â‡’ 0 + 0 < 12

â‡’ 0 < 12 (this is true)

Hence, Solution region of the given inequality is the line 3x + 4y < 12. where, Origin is included in the region.

The graph will be as follows for Equation 1 and 2:

### Question 4: x + y â‰¥ 4, 2x â€“ y < 0

Solution:

For equation 1:

Now draw a solid line x + y = 4 in the graph (because (x + y = 4) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + y â‰¥ 4

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ 4

â‡’ 0 â‰¥ 4 (this is not true)

Hence, Solution region of the given inequality is the line x + y â‰¥ 4. where, Origin is not included in the region

For equation 2:

Now draw a dotted line 2x â€“ y = 0 in the graph (because (2x â€“ y = 0) is NOT the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 2x â€“ y < 0

Lets, select point (3, 0)

â‡’ 6 – 0 < 0

â‡’ 0 > 6 (this is not true)

Hence, Solution region of the given inequality is the line 2x â€“ y < 0. where, the point (3,0) is included in the region.

The graph will be as follows for Equation 1 and 2:

### Question 5: 2x â€“ y >1, x â€“ 2y < â€“ 1

Solution:

For equation 1:

Now draw a dotted line 2x â€“ y =1 in the graph (because (2x â€“ y =1) is NOT the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 2x â€“ y >1

Lets, select origin point (0, 0)

â‡’ 0 – 0 > 1

â‡’ 0 > 1 (this is not true)

Hence, Solution region of the given inequality is the line 2x â€“ y >1. where, Origin is not included in the region

For equation 2:

Now draw a dotted line x â€“ 2y = â€“ 1 in the graph (because (x â€“ 2y = â€“ 1) is NOT the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â€“ 2y < â€“ 1

Lets, select point (3, 0)

â‡’ 0 – 0 < -1

â‡’ 0 < -1 (this is not true)

Hence, Solution region of the given inequality is the line x â€“ 2y < â€“ 1. Origin is not included in the region

The graph will be as follows for Equation 1 and 2:

### Question 6: x + y â‰¤ 6, x + y â‰¥ 4

Solution:

For equation 1:

Now draw a solid line x + y = 6 in the graph (because (x + y = 6) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + y â‰¤ 6

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 6

â‡’ 0 â‰¤ 6 (this is true)

Hence, Solution region of the given inequality is the line x + y â‰¤ 6. where, Origin is included in the region

For equation 2:

Now draw a solid line x + y = 4 in the graph (because (x + y = 4) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + y â‰¥ 4

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ 4

â‡’ 0 â‰¥ 4 (this not is true)

Hence, Solution region of the given inequality is the line x + y â‰¥ 4. where, Origin is not included in the region

The graph will be as follows for Equation 1 and 2:

### Question 7: 2x + y â‰¥ 8, x + 2y â‰¥ 10

Solution:

For equation 1:

Now draw a solid line 2x + y = 8 in the graph (because (2x + y = 8) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 2x + y â‰¥ 8

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ 8

â‡’ 0 â‰¥ 8 (this is not true)

Hence, Solution region of the given inequality is the line 2x + y â‰¥ 8. where, Origin is not included in the region

For equation 2:

Now draw a solid line x + 2y = 10 in the graph (because (x + 2y = 10) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + 2y â‰¥ 10

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ 10

â‡’ 0 â‰¥ 10 (this not is true)

Hence, Solution region of the given inequality is the line x + 2y â‰¥ 10. where, Origin is not included in the region

The graph will be as follows for Equation 1 and 2:

### Question 8: x + y â‰¤ 9, y > x, x â‰¥ 0

Solution:

For equation 1:

Now draw a solid line x + y = 9 in the graph (because (x + y = 9) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + y â‰¤ 9

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 9

â‡’ 0 â‰¤ 9 (this is true)

Hence, Solution region of the given inequality is the line x + y â‰¤ 9. where, Origin is included in the region

For equation 2:

Now draw a dotted line y = x in the graph (because (y = x) is NOT the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y > x

Lets, select point (3, 0)

â‡’ 0 > 3

â‡’ 0 > 3 (this is not true)

Hence, Solution region of the given inequality is the line y > x. the point (3,0) is not included in the region.

For equation 3:

Now draw a solid line x = 0 in the graph (because (x = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 0

Lets, select point (3, 0)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line x â‰¥ 0. where, the point (3,0) is included in the region.

The graph will be as follows for Equation 1, 2 and 3:

### Question 9: 5x + 4y â‰¤ 20, x â‰¥ 1, y â‰¥ 2

Solution:

For equation 1:

Now draw a solid line 5x + 4y = 20 in the graph (because (5x + 4y = 20) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 5x + 4y â‰¤ 20

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 20

â‡’ 0 â‰¤ 20 (this is true)

Hence, Solution region of the given inequality is the line 5x + 4y â‰¤ 20. where, Origin is included in the region

For equation 2:

Now draw a solid line x = 1 in the graph (because (x = 1) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 1

Lets, select origin point (0, 0)

â‡’ 0 â‰¥ 1

â‡’ 0 â‰¥ 1 (this not is true)

Hence, Solution region of the given inequality is the line x â‰¥ 1. where, Origin is not included in the region

For equation 3:

Now draw a solid line y = 2 in the graph (because (y = 2) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 2

Lets, select origin point (0, 0)

â‡’ 0 â‰¥ 2

â‡’ 0 â‰¥ 2 (this not is true)

Hence, Solution region of the given inequality is the line y â‰¥ 2. where, Origin is not included in the region

The graph will be as follows for Equation 1, 2 and 3:

### Question 10: 3x + 4y â‰¤ 60, x +3y â‰¤ 30, x â‰¥ 0, y â‰¥ 0

Solution:

For equation 1:

Now draw a solid line 3x + 4y = 60 in the graph (because (3x + 4y = 60) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 3x + 4y â‰¤ 60

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 60

â‡’ 0 â‰¤ 60 (this is true)

Hence, Solution region of the given inequality is the line 3x + 4y â‰¤ 60. where, Origin is included in the region

For equation 2:

Now draw a solid line x +3y = 30 in the graph (because (x +3y = 30) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x +3y â‰¤ 30

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 30

â‡’ 0 â‰¤ 30 (this is true)

Hence, Solution region of the given inequality is the line x +3y â‰¤ 30. where, Origin is included in the region

For equation 3:

Now draw a solid line x = 0 in the graph (because (x = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 0

Lets, select point (3, 0)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line x â‰¥ 0. where, the point (3,0) is included in the region.

For equation 4:

Now draw a solid line y = 0 in the graph (because (y = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 0

Lets, select point (0,3)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line y â‰¥ 0. where, the point (0,3) is included in the region.

The graph will be as follows for Equation 1, 2. 3 and 4:

### Question 11: 2x + y â‰¥ 4, x + y â‰¤ 3, 2x â€“ 3y â‰¤ 6

Solution:

For equation 1:

Now draw a solid line 2x + y = 4 in the graph (because (2x + y = 4) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 2x + y â‰¥ 4

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ 4

â‡’ 0 â‰¥ 4 (this is not true)

Hence, Solution region of the given inequality is the line 2x + y â‰¥ 4. where, Origin is not included in the region

For equation 2:

Now draw a solid line x + y = 3 in the graph (because (x + y = 3) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + y â‰¤ 3

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 3

â‡’ 0 â‰¤ 3 (this is true)

Hence, Solution region of the given inequality is the line x + y â‰¤ 3. where, Origin is included in the region

For equation 3:

Now draw a solid line 2x â€“ 3y = 6 in the graph (because (2x â€“ 3y = 6) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 2x â€“ 3y â‰¤ 6

Lets, select origin point (0, 0)

â‡’ 0 â€“ 0 â‰¤ 6

â‡’ 0 â‰¤ 6 (this is true)

Hence, Solution region of the given inequality is the line 2x â€“ 3y â‰¤ 6. where, Origin is included in the region

The graph will be as follows for Equation 1, 2 and 3:

### Question 12: x â€“ 2y â‰¤ 3, 3x + 4y â‰¥ 12, x â‰¥ 0 , y â‰¥ 1

Solution:

For equation 1:

Now draw a solid line x â€“ 2y = 3 in the graph (because (x â€“ 2y = 3) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â€“ 2y â‰¤ 3

Lets, select origin point (0, 0)

â‡’ 0 – 0 â‰¤ 3

â‡’ 0 â‰¤ 3 (this is true)

Hence, Solution region of the given inequality is the line x â€“ 2y â‰¤ 3. where, Origin is included in the region

For equation 2:

Now draw a solid line 3x + 4y = 12 in the graph (because (3x + 4y = 12) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 3x + 4y â‰¥ 12

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ 12

â‡’ 0 â‰¥ 12 (this is not true)

Hence, Solution region of the given inequality is the line 3x + 4y â‰¥ 12. where, Origin is not included in the region

For equation 3:

Now draw a solid line x = 0 in the graph (because (x = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 0

Lets, select point (3, 0)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line x â‰¥ 0. where, the point (3,0) is included in the region.

For equation 4:

Now draw a solid line y = 1 in the graph (because (y = 1) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 1

Lets, select origin point (0, 0)

â‡’ 0 â‰¥ 1

â‡’ 0 â‰¥ 1 (this is not true)

Hence, Solution region of the given inequality is the line y â‰¥ 1. where, Origin is not included in the region

The graph will be as follows for Equation 1, 2. 3 and 4:

### Question 13: 4x + 3y â‰¤ 60, y â‰¥ 2x, x â‰¥ 3, x, y â‰¥ 0

Solution:

For equation 1:

Now draw a solid line 4x + 3y = 60 in the graph (because (4x + 3y = 60) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 4x + 3y â‰¤ 60

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 60

â‡’ 0 â‰¤ 60 (this is true)

Hence, Solution region of the given inequality is the line 4x + 3y â‰¤ 60. where, Origin is included in the region

For equation 2:

Now draw a solid line y = 2x in the graph (because (y = 2x) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 2x

Lets, select point (3, 0)

â‡’ 0 â‰¥ 6

â‡’ 0 â‰¥ 6 (this is not true)

Hence, Solution region of the given inequality is the line y â‰¥ 2x. where, the point (3,0) is not included in the region.

For equation 3:

Now draw a solid line x = 3 in the graph (because (x = 3) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 3

Lets, select origin point (0, 0)

â‡’ 0 â‰¥ 3

â‡’ 0 â‰¥ 3 (this is not true)

Hence, Solution region of the given inequality is the line x â‰¥ 3. where, Origin is not included in the region

For equation 4:

Now draw a solid line x = 0 in the graph (because (x = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 0

Lets, select point (3, 0)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line x â‰¥ 0. where, the point (3,0) is included in the region.

For equation 5:

Now draw a solid line y = 0 in the graph (because (y = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 0

Lets, select point (0,3)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line y â‰¥ 0. where, the point (0,3) is included in the region.

The graph will be as follows for Equation 1, 2. 3, 4 and 5:

### Question 14: 3x + 2y â‰¤ 150, x + 4y â‰¤ 80, x â‰¤ 15, y â‰¥ 0, x â‰¥ 0

Solution:

For equation 1:

Now draw a solid line 3x + 2y = 150 in the graph (because (3x + 2y = 150) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider 3x + 2y â‰¤ 150

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 150

â‡’ 0 â‰¤ 150 (this is true)

Hence, Solution region of the given inequality is the line 3x + 2y â‰¤ 150. where, Origin is included in the region

For equation 2:

Now draw a solid line x + 4y = 80 in the graph (because (x + 4y = 80) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + 4y â‰¤ 80

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 80

â‡’ 0 â‰¤ 80 (this is true)

Hence, Solution region of the given inequality is the line x + 4y â‰¤ 80. where, Origin is included in the region

For equation 3:

Now draw a solid line x = 15 in the graph (because (x = 15) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¤ 15

Lets, select origin point (0, 0)

â‡’ 0 â‰¤ 15

â‡’ 0 â‰¤ 15 (this is true)

Hence, Solution region of the given inequality is the line x â‰¤ 15. where, Origin is included in the region

For equation 4:

Now draw a solid line x = 0 in the graph (because (x = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 0

Lets, select point (3, 0)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line x â‰¥ 0. where, the point (3,0) is included in the region.

For equation 5:

Now draw a solid line y = 0 in the graph (because (y = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 0

Lets, select point (0,3)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line y â‰¥ 0. where, the point (0,3) is included in the region.

The graph will be as follows for Equation 1, 2. 3, 4 and 5:

### Question 15: x + 2y â‰¤ 10, x + y â‰¥ 1, x â€“ y â‰¤ 0, x â‰¥ 0, y â‰¥ 0

Solution:

For equation 1:

Now draw a solid line x + 2y = 10 in the graph (because (x + 2y = 10) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + 2y â‰¤ 10

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¤ 10

â‡’ 0 â‰¤ 10 (this is true)

Hence, Solution region of the given inequality is the line x + 2y â‰¤ 10. where, Origin is included in the region

For equation 2:

Now draw a solid line x + y = 1 in the graph (because (x + y = 1) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x + y â‰¥ 1

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ 1

â‡’ 0 â‰¥ 1 (this is true)

Hence, Solution region of the given inequality is the line x + y â‰¥ 1. where, Origin is included in the region

For equation 3:

Now draw a solid line x â€“ y = 0 in the graph (because (x â€“ y = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â€“ y â‰¤ 0

Lets, select point (3, 0)

â‡’ 3 – 0 â‰¤ 0

â‡’ 3 â‰¤ 0 (this is not true)

Hence, Solution region of the given inequality is the line x â€“ y â‰¤ 0. where, the point (3,0) is not included in the region

For equation 4:

Now draw a solid line x = 0 in the graph (because (x = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider x â‰¥ 0

Lets, select point (3, 0)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line x â‰¥ 0. where, the point (3,0) is included in the region.

For equation 5:

Now draw a solid line y = 0 in the graph (because (y = 0) is the part of the given equation)

we need at least two solutions of the equation. So, we can use the following table to draw the graph:

Consider y â‰¥ 0

Lets, select point (0,3)

â‡’ 3 â‰¥ 0

â‡’ 3 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line y â‰¥ 0. where, the point (0,3) is included in the region.

The graph will be as follows for Equation 1, 2. 3, 4 and 5:

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