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

### Solve the following inequalities graphically in two-dimensional plane:

### Question 1: x + y < 5

**Solution:**

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

NOTthe 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:

xy0 5 5 5 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 regionThe 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:

xy0 6 3 0 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 regionThe 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:

xy0 3 4 0 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 regionThe 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)

xy0 -8 4 0 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 regionThe 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)

xy0 -2 2 0 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 regionThe 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

NOTthe part of the given equation)

xy0 -2 3 0 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 regionThe 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)

xy0 -3 2 0 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 regionThe 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

NOTthe part of the given equation)

xy0 10 -6 0 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 regionThe 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

NOTthe 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:

xy0 -2 5 -2 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 regionThe 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

NOTthe 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-3 0 -3 5 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 regionThe graph is as follows:

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