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

### Solve the following system of inequalities graphically:

### Question 1: x ≥ 3, y ≥ 2

**Solution:**

For equation 1:Now draw a

solidline 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:

x y 3 0 3 5 Consider x ≥ 3

Lets, select origin point (0, 0)

⇒ 0 ≥ 3

⇒ 0 ≥ 3 (

this not is true)Hence, Solution region of the given inequality is the line x ≥ 3. where,

Origin is not included in the region

For equation 2:Now draw a

solidline 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:

x y 0 2 3 2 Consider y ≥ 2

Lets, select origin point (0, 0)

⇒ 0 ≥ 2

⇒ 0 ≥ 2 (

this not is true)Hence, Solution region of the given inequality is the line y ≥ 2. where,

Origin is not included in the regionThe graph will be as follows for Equation 1 and 2:

### Question 2: 3x + 2y ≤ 12, x ≥ 1, y ≥ 2

**Solution:**

For equation 1:Now draw a

solidline 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:

x y 0 6 4 0 Consider 3x + 2y ≤ 12

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 12

⇒ 0 + 0 ≤ 12 (

this is true)Hence, Solution region of the given inequality is the line 3x + 2y ≤ 12. where,

Origin is included in the region

For equation 2:Now draw a

solidline x = 1 in the graph (because (x = 1) is the part of the given equation)

x y 1 0 1 5 Consider x ≥ 1

Lets, select origin point (0, 0)

⇒ 0 ≥ 1

⇒ 0 ≥ 1 (

this not is true)Hence, Solution region of the given inequality is the line x ≥ 1. where,

Origin is not included in the region

For equation 3:Now draw a

solidline y = 2 in the graph (because (y = 2) is the part of the given equation)

x y 0 2 5 2 Consider y ≥ 2

Lets, select origin point (0, 0)

⇒ 0 ≥ 2

⇒ 0 ≥ 2 (

this not is true)Hence, Solution region of the given inequality is the line y ≥ 2. where,

Origin is not included in the regionThe graph will be as follows for Equation 1, 2 and 3:

### Question 3: 2x + y ≥ 6, 3x + 4y < 12

**Solution:**

For equation 1:Now draw a

solidline 2x + y = 6 in the graph (because (2x + y = 6) is the part of the given equation)

x y 0 6 3 0 Consider 2x + y ≥ 6

Lets, select origin point (0, 0)

⇒ 0 + 0 ≥ 6

⇒ 0 ≥ 6 (

this is not true)Hence, Solution region of the given inequality is the line 2x + y ≥ 6. where,

Origin is not included in the region

For equation 2:Now draw a

dottedline 3x + 4y = 12 in the graph (because (3x + 4y = 12) isNOTthe part of the given equation)

x y 0 3 4 0 Consider 3x + 4y < 12

Lets, select origin point (0, 0)

⇒ 0 + 0 < 12

⇒ 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 ≥ 4, 2x – y < 0

**Solution:**

For equation 1:Now draw a

solidline x + y = 4 in the graph (because (x + y = 4) is the part of the given equation)

x y 0 4 4 0 Consider x + y ≥ 4

Lets, select origin point (0, 0)

⇒ 0 + 0 ≥ 4

⇒ 0 ≥ 4 (

this is not true)Hence, Solution region of the given inequality is the line x + y ≥ 4. where,

Origin is not included in the region

For equation 2:Now draw a

dottedline 2x – y = 0 in the graph (because (2x – y = 0) isNOTthe part of the given equation)

x y 0 0 1 2 Consider 2x – y < 0

Lets, select point (3, 0)

⇒ 6 – 0 < 0

⇒ 0 > 6 (

this is not true)Hence, Solution region of the given inequality is the line 2x – 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 – y >1, x – 2y < – 1

**Solution:**

For equation 1:Now draw a

dottedline 2x – y =1 in the graph (because (2x – y =1) isNOTthe part of the given equation)

x y 0 -1 1 1 Consider 2x – y >1

Lets, select origin point (0, 0)

⇒ 0 – 0 > 1

⇒ 0 > 1 (

this is not true)Hence, Solution region of the given inequality is the line 2x – y >1. where,

Origin is not included in the region

For equation 2:Now draw a

dottedline x – 2y = – 1 in the graph (because (x – 2y = – 1) isNOTthe part of the given equation)

x y 1 1 -1 0 Consider x – 2y < – 1

Lets, select point (3, 0)

⇒ 0 – 0 < -1

⇒ 0 < -1 (

this is not true)Hence, Solution region of the given inequality is the line x – 2y < – 1.

Origin is not included in the regionThe graph will be as follows for Equation 1 and 2:

### Question 6: x + y ≤ 6, x + y ≥ 4

**Solution:**

For equation 1:Now draw a

solidline x + y = 6 in the graph (because (x + y = 6) is the part of the given equation)

x y 0 6 6 0 Consider x + y ≤ 6

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 6

⇒ 0 ≤ 6 (

this is true)Hence, Solution region of the given inequality is the line x + y ≤ 6. where,

Origin is included in the region

For equation 2:Now draw a

solidline x + y = 4 in the graph (because (x + y = 4) is the part of the given equation)

x y 0 4 4 0 Consider x + y ≥ 4

Lets, select origin point (0, 0)

⇒ 0 + 0 ≥ 4

⇒ 0 ≥ 4 (

this not is true)Hence, Solution region of the given inequality is the line x + y ≥ 4. where,

Origin is not included in the regionThe graph will be as follows for Equation 1 and 2:

### Question 7: 2x + y ≥ 8, x + 2y ≥ 10

**Solution: **

For equation 1:Now draw a

solidline 2x + y = 8 in the graph (because (2x + y = 8) is the part of the given equation)

x y 0 8 4 0 Consider 2x + y ≥ 8

Lets, select origin point (0, 0)

⇒ 0 + 0 ≥ 8

⇒ 0 ≥ 8 (

this is not true)Hence, Solution region of the given inequality is the line 2x + y ≥ 8. where,

Origin is not included in the region

For equation 2:Now draw a

solidline x + 2y = 10 in the graph (because (x + 2y = 10) is the part of the given equation)

x y 0 5 10 0 Consider x + 2y ≥ 10

Lets, select origin point (0, 0)

⇒ 0 + 0 ≥ 10

⇒ 0 ≥ 10 (

this not is true)Hence, Solution region of the given inequality is the line x + 2y ≥ 10. where,

Origin is not included in the regionThe graph will be as follows for Equation 1 and 2:

### Question 8: x + y ≤ 9, y > x, x ≥ 0

**Solution:**

For equation 1:Now draw a

solidline x + y = 9 in the graph (because (x + y = 9) is the part of the given equation)

x y 0 9 9 0 Consider x + y ≤ 9

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 9

⇒ 0 ≤ 9 (

this is true)Hence, Solution region of the given inequality is the line x + y ≤ 9. where,

Origin is included in the region

For equation 2:Now draw a

dottedline y = x in the graph (because (y = x) is NOT the part of the given equation)

x y 1 1 0 0 Consider y > x

Lets, select point (3, 0)

⇒ 0 > 3

⇒ 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

solidline x = 0 in the graph (because (x = 0) is the part of the given equation)

x y 0 0 0 5 Consider x ≥ 0

Lets, select point (3, 0)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)Hence, Solution region of the given inequality is the line x ≥ 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 ≤ 20, x ≥ 1, y ≥ 2

**Solution:**

For equation 1:Now draw a

solidline 5x + 4y = 20 in the graph (because (5x + 4y = 20) is the part of the given equation)

x y 0 5 4 0 Consider 5x + 4y ≤ 20

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 20

⇒ 0 ≤ 20 (

this is true)Hence, Solution region of the given inequality is the line 5x + 4y ≤ 20. where,

Origin is included in the region

For equation 2:Now draw a

solidline x = 1 in the graph (because (x = 1) is the part of the given equation)

x y 1 0 1 5 Consider x ≥ 1

Lets, select origin point (0, 0)

⇒ 0 ≥ 1

⇒ 0 ≥ 1 (

this not is true)Hence, Solution region of the given inequality is the line x ≥ 1. where,

Origin is not included in the region

For equation 3:Now draw a

solidline y = 2 in the graph (because (y = 2) is the part of the given equation)

x y 0 2 5 2 Consider y ≥ 2

Lets, select origin point (0, 0)

⇒ 0 ≥ 2

⇒ 0 ≥ 2 (

this not is true)Hence, Solution region of the given inequality is the line y ≥ 2. where,

Origin is not included in the regionThe graph will be as follows for Equation 1, 2 and 3:

### Question 10: 3x + 4y ≤ 60, x +3y ≤ 30, x ≥ 0, y ≥ 0

**Solution:**

For equation 1:Now draw a

solidline 3x + 4y = 60 in the graph (because (3x + 4y = 60) is the part of the given equation)

x y 0 15 20 0 Consider 3x + 4y ≤ 60

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 60

⇒ 0 ≤ 60 (

this is true)Hence, Solution region of the given inequality is the line 3x + 4y ≤ 60. where,

Origin is included in the region

For equation 2:Now draw a

solidline x +3y = 30 in the graph (because (x +3y = 30) is the part of the given equation)

x y 0 10 30 0 Consider x +3y ≤ 30

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 30

⇒ 0 ≤ 30 (

this is true)Hence, Solution region of the given inequality is the line x +3y ≤ 30. where,

Origin is included in the region

For equation 3:Now draw a

solidline x = 0 in the graph (because (x = 0) is the part of the given equation)

x y 0 0 0 5 Consider x ≥ 0

Lets, select point (3, 0)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)Hence, Solution region of the given inequality is the line x ≥ 0. where,

the point (3,0) is included in the region.

For equation 4:Now draw a

solidline y = 0 in the graph (because (y = 0) is the part of the given equation)

x y 0 0 5 0 Consider y ≥ 0

Lets, select point (0,3)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)Hence, Solution region of the given inequality is the line y ≥ 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 ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6

**Solution:**

For equation 1:Now draw a

solidline 2x + y = 4 in the graph (because (2x + y = 4) is the part of the given equation)

x y 0 4 2 0 Consider 2x + y ≥ 4

Lets, select origin point (0, 0)

⇒ 0 + 0 ≥ 4

⇒ 0 ≥ 4 (

this is not true)Hence, Solution region of the given inequality is the line 2x + y ≥ 4. where,

Origin is not included in the region

For equation 2:Now draw a

solidline x + y = 3 in the graph (because (x + y = 3) is the part of the given equation)

x y 0 3 3 0 Consider x + y ≤ 3

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 3

⇒ 0 ≤ 3 (

this is true)Hence, Solution region of the given inequality is the line x + y ≤ 3. where,

Origin is included in the region

For equation 3:Now draw a

solidline 2x – 3y = 6 in the graph (because (2x – 3y = 6) is the part of the given equation)

x y 0 -2 3 0 Consider 2x – 3y ≤ 6

Lets, select origin point (0, 0)

⇒ 0 – 0 ≤ 6

⇒ 0 ≤ 6 (

this is true)Hence, Solution region of the given inequality is the line 2x – 3y ≤ 6. where,

Origin is included in the regionThe graph will be as follows for Equation 1, 2 and 3:

### Question 12: x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0 , y ≥ 1

**Solution:**

For equation 1:Now draw a

solidline x – 2y = 3 in the graph (because (x – 2y = 3) is the part of the given equation)

x y 1 -1 3 0 Consider x – 2y ≤ 3

Lets, select origin point (0, 0)

⇒ 0 – 0 ≤ 3

⇒ 0 ≤ 3 (

this is true)Hence, Solution region of the given inequality is the line x – 2y ≤ 3. where,

Origin is included in the region

For equation 2:Now draw a

solidline 3x + 4y = 12 in the graph (because (3x + 4y = 12) is the part of the given equation)

x y 0 3 4 0 Consider 3x + 4y ≥ 12

Lets, select origin point (0, 0)

⇒ 0 + 0 ≥ 12

⇒ 0 ≥ 12 (

this is not true)Hence, Solution region of the given inequality is the line 3x + 4y ≥ 12. where,

Origin is not included in the region

For equation 3:Now draw a

solidline x = 0 in the graph (because (x = 0) is the part of the given equation)

x y 0 0 0 5 Consider x ≥ 0

Lets, select point (3, 0)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)Hence, Solution region of the given inequality is the line x ≥ 0. where,

the point (3,0) is included in the region.

For equation 4:Now draw a

solidline y = 1 in the graph (because (y = 1) is the part of the given equation)

x y 0 1 5 1 Consider y ≥ 1

Lets, select origin point (0, 0)

⇒ 0 ≥ 1

⇒ 0 ≥ 1 (

this is not true)Hence, Solution region of the given inequality is the line y ≥ 1. where,

Origin is not included in the regionThe graph will be as follows for Equation 1, 2. 3 and 4:

### Question 13: 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

**Solution:**

For equation 1:Now draw a

solidline 4x + 3y = 60 in the graph (because (4x + 3y = 60) is the part of the given equation)

x y 0 20 15 0 Consider 4x + 3y ≤ 60

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 60

⇒ 0 ≤ 60 (

this is true)Hence, Solution region of the given inequality is the line 4x + 3y ≤ 60. where,

Origin is included in the region

For equation 2:Now draw a

solidline y = 2x in the graph (because (y = 2x) is the part of the given equation)

x y 0 0 2 4 Consider y ≥ 2x

Lets, select point (3, 0)

⇒ 0 ≥ 6

⇒ 0 ≥ 6 (

this is not true)Hence, Solution region of the given inequality is the line y ≥ 2x. where,

the point (3,0) is not included in the region.

For equation 3:Now draw a

solidline x = 3 in the graph (because (x = 3) is the part of the given equation)

x y 3 0 3 5 Consider x ≥ 3

Lets, select origin point (0, 0)

⇒ 0 ≥ 3

⇒ 0 ≥ 3 (

this is not true)Hence, Solution region of the given inequality is the line x ≥ 3. where,

Origin is not included in the region

For equation 4:Now draw a

solidline x = 0 in the graph (because (x = 0) is the part of the given equation)

x y 0 0 0 5 Consider x ≥ 0

Lets, select point (3, 0)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)the point (3,0) is included in the region.

For equation 5:Now draw a

solidline y = 0 in the graph (because (y = 0) is the part of the given equation)

x y 0 0 5 0 Consider y ≥ 0

Lets, select point (0,3)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)Hence, Solution region of the given inequality is the line y ≥ 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 ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0

**Solution:**

For equation 1:Now draw a

solidline 3x + 2y = 150 in the graph (because (3x + 2y = 150) is the part of the given equation)

x y 0 75 50 0 Consider 3x + 2y ≤ 150

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 150

⇒ 0 ≤ 150 (

this is true)Hence, Solution region of the given inequality is the line 3x + 2y ≤ 150. where,

Origin is included in the region

For equation 2:Now draw a

solidline x + 4y = 80 in the graph (because (x + 4y = 80) is the part of the given equation)

x y 0 20 80 0 Consider x + 4y ≤ 80

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 80

⇒ 0 ≤ 80 (

this is true)Hence, Solution region of the given inequality is the line x + 4y ≤ 80. where,

Origin is included in the region

For equation 3:Now draw a

solidline x = 15 in the graph (because (x = 15) is the part of the given equation)

x y 15 0 15 5 Consider x ≤ 15

Lets, select origin point (0, 0)

⇒ 0 ≤ 15

⇒ 0 ≤ 15 (

this is true)Hence, Solution region of the given inequality is the line x ≤ 15. where,

Origin is included in the region

For equation 4:Now draw a

solidline x = 0 in the graph (because (x = 0) is the part of the given equation)

x y 0 0 0 5 Consider x ≥ 0

Lets, select point (3, 0)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)the point (3,0) is included in the region.

For equation 5:Now draw a

solidline y = 0 in the graph (because (y = 0) is the part of the given equation)

x y 0 0 5 0 Consider y ≥ 0

Lets, select point (0,3)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)Hence, Solution region of the given inequality is the line y ≥ 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 ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0

**Solution:**

For equation 1:Now draw a

solidline x + 2y = 10 in the graph (because (x + 2y = 10) is the part of the given equation)

x y 0 5 10 0 Consider x + 2y ≤ 10

Lets, select origin point (0, 0)

⇒ 0 + 0 ≤ 10

⇒ 0 ≤ 10 (

this is true)Hence, Solution region of the given inequality is the line x + 2y ≤ 10. where,

Origin is included in the region

For equation 2:Now draw a

solidline x + y = 1 in the graph (because (x + y = 1) is the part of the given equation)

x y 0 1 1 0 Consider x + y ≥ 1

Lets, select origin point (0, 0)

⇒ 0 + 0 ≥ 1

⇒ 0 ≥ 1 (

this is true)Hence, Solution region of the given inequality is the line x + y ≥ 1. where,

Origin is included in the region

For equation 3:Now draw a

solidline x – y = 0 in the graph (because (x – y = 0) is the part of the given equation)

x y 1 1 0 0 Consider x – y ≤ 0

Lets, select point (3, 0)

⇒ 3 – 0 ≤ 0

⇒ 3 ≤ 0 (

this is not true)Hence, Solution region of the given inequality is the line x – y ≤ 0. where,

the point (3,0) is not included in the region

For equation 4:Now draw a

solidline x = 0 in the graph (because (x = 0) is the part of the given equation)

x y 0 0 0 5 Consider x ≥ 0

Lets, select point (3, 0)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)the point (3,0) is included in the region.

For equation 5:Now draw a

solidline y = 0 in the graph (because (y = 0) is the part of the given equation)

x y 0 0 5 0 Consider y ≥ 0

Lets, select point (0,3)

⇒ 3 ≥ 0

⇒ 3 ≥ 0 (

this is true)the point (0,3) is included in the region.The graph will be as follows for Equation 1, 2. 3, 4 and 5: