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# Class 11 NCERT Solutions – Chapter 6 Linear Inequalities – Exercise 6.2

• Last Updated : 03 Mar, 2021

### Question 1: x + y < 5

Solution:

Now draw a dotted line x + y = 5 in the graph (because (x + y = 5) 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 + y < 5

Lets, select origin point (0, 0)

â‡’ 0 + 0 < 5

â‡’ 0 < 5 (this is true)

Hence, Solution region of the given inequality is the line x + y = 5. where, Origin is included in the region

The graph is as follows:

### Question 2: 2x + y â‰¥ 6

Solution:

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

The graph is as follows:

### Question 3: 3x + 4y â‰¤ 12

Solution:

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 true)

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

The graph is as follows:

### Question 4: y + 8 â‰¥ 2x

Solution:

Now draw a solid line y + 8 = 2x in the graph (because (y + 8 = 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 + 8 â‰¥ 2x

Lets, select origin point (0, 0)

â‡’ 0 + 8 â‰¥ 0

â‡’ 8 â‰¥ 0 (this is true)

Hence, Solution region of the given inequality is the line y + 8 â‰¥ 2x. where, Origin is included in the region

The graph is as follows:

### Question 5: x â€“ y â‰¤ 2

Solution:

Now draw a solid line x â€“ y = 2 in the graph (because (x â€“ 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 x â€“ y â‰¤ 2

Lets, select origin point (0, 0)

â‡’ 0 – 0 â‰¤ 2

â‡’ 0 â‰¤ 2  (this is true)

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

The graph is as follows:

### Question 6: 2x â€“ 3y > 6

Solution:

Now draw a dotted line 2x â€“ 3y = 6 in the graph (because (2x â€“ 3y = 6) 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€“ 3y > 6

Lets, select origin point (0, 0)

â‡’ 0 + 0 > 6

â‡’ 0 > 6 (this is not true)

Hence, Solution region of the given inequality is the line 2x â€“ 3y > 6. where, Origin is not included in the region

The graph is as follows:

### Question 7: â€“ 3x + 2y â‰¥ â€“ 6

Solution:

Now draw a solid line â€“ 3x + 2y = â€“ 6 in the graph (because (â€“ 3x + 2y = â€“ 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 â€“ 3x + 2y â‰¥ â€“ 6

Lets, select origin point (0, 0)

â‡’ 0 + 0 â‰¥ â€“ 6

â‡’ 0 â‰¥ â€“ 6 (this is true)

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

The graph is as follows:

### Question 8: 3y â€“ 5x < 30

Solution:

Now draw a dotted line 3y â€“ 5x = 30 in the graph (because (3y â€“ 5x = 30) 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 3y â€“ 5x < 30

Lets, select origin point (0, 0)

â‡’ 0 + 0 < 30

â‡’ 0 < 30 (this is true)

Hence, Solution region of the given inequality is the line 3y â€“ 5x < 30. where, Origin is included in the region

The graph is as follows:

### Question 9: y < â€“ 2

Solution:

Now draw a dotted line y = â€“ 2 in the graph (because (y = â€“ 2) 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 < â€“ 2

Lets, select origin point (0, 0)

â‡’ 0 < â€“ 2

â‡’ 0 < â€“ 2 (this is not true)

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

The graph is as follows:

### Question 10: x > â€“ 3

Solution:

Now draw a dotted line x = â€“ 3 in the graph (because (x = â€“ 3) 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€“ 3

Lets, select origin point (0, 0)

â‡’ 0 > â€“ 3

â‡’ 0 > â€“ 3 (this is true)

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

The graph is as follows:

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