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Class 11 NCERT Solutions – Chapter 6 Linear Inequalities – Exercise 6.3
  • Difficulty Level : Hard
  • Last Updated : 21 Feb, 2021

Solve the following system of inequalities graphically:

Question 1: x ≥ 3, y ≥ 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:

xy
30
35

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 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:

xy
02
32

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 region

The 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 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:

xy
06
40

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 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:

xy
10
15

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 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:

xy
02
52

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 region

The 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 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:

xy
06
30

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 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:

xy
03
40

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 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:

xy
04
40

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 dotted line 2x – y = 0 in the graph (because (2x – 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:

xy
00
12

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 dotted line 2x – y =1 in the graph (because (2x – 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:

xy
0-1
11

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 dotted line x – 2y = – 1 in the graph (because (x – 2y = – 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:

xy
11
-10

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 region

The 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 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:

xy
06
60

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 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:

xy
04
40

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 region

The 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 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:

xy
08
40

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 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:

xy
05
100

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 region

The 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 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:

xy
09
90

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 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:

xy
11
00

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 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:

xy
00
05

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 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:

xy
05
40

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 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:

xy
10
15

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 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:

xy
02
52

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 region

The 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 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:

xy
015
200

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 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:

xy
010
300

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 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:

xy
00
05

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 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:

xy
00
50

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 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:

xy
04
20

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 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:

xy
03
30

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 solid line 2x – 3y = 6 in the graph (because (2x – 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:

xy
0-2
30

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 region

The 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 solid line x – 2y = 3 in the graph (because (x – 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:

xy
1-1
30

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 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:

xy
03
40

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 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:

xy
00
05

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 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:

xy
01
51

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 region

The 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 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:

xy
020
150

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 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:

xy
00
24

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 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:

xy
30
35

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 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:

xy
00
05

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 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:

xy
00
50

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 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:

xy
075
500

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 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:

xy
020
800

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 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:

xy
150
155

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 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:

xy
00
05

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 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:

xy
00
50

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 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:

xy
05
100

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 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:

xy
01
10

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 solid line x – y = 0 in the graph (because (x – 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:

xy
11
00

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 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:

xy
00
05

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 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:

xy
00
50

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:

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