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

**Question 1. x + y < 5**

**Solution.**

Consider the equation x + y = 5. It passes through the points (0, 5) and (5, 0). The line x + y = 5 is represented by AB. Consider the inequality x + y < 5

Put x = 0, y = 0

0 + 0 = 0 < 5, which is true. So, the origin O lies in the plane x + y < 5

∴ Shaded region represents the inequality x + y < 5

**Question 2. 2x + y ≥ 6**

**Solution.**

Consider the equation 2x + y = 6

The line passes through (0, 6), (3, 0).

The line 2x + y = 6 is represented by AB.

Now, consider 2x + y ≥ 6

Put x = 0, y = 0

0 + 0 ≥ 6, which does not satisfy this inequality.

∴ Origin does not lie in the region of 2x + y ≥ 6.

The shaded region represents the inequality 2x + y ≥ 6

**Question 3. 3x + 4y ≤ 12**

**Solution.**

We draw the graph of the equation 3x + 4y = 12. The line passes through the points (4, 0), (0, 3). This line is represented by AB. Now consider the inequality 3x + 4y ≤ 12

Putting x = 0, y = 0 0 + 0 = 0 ≤ 12, which is true

∴ Origin lies in the region of 3x + 4y ≤ 12 The shaded region represents the inequality 3x + 4y ≤ 12

**Question 4. y + 8 ≥ 2x**

**Solution.**

Given inequality is y + 8 ≥ 2x

Let us draw the graph of the line, y+ 8 = 2x

The line passes through the points (4, 0), (0, -8).

This line is represented by AB.

Now, consider the inequality y + 8 ≥ 2x.

Putting x = 0, y = 0

0 + 8 ≥ 0, which is true

∴ Origin lies in the region of y + 8 ≥ 2x

The shaded region represents the inequality y + 8 ≥ 2x.

**Question 5. x – y ≤ 2**

**Solution.**

Given inequality is x – y ≤ 2

Let us draw the graph of the line x – y = 2

The line passes through the points (2, 0), (0, -2)

This line is represented by AB.

∴ Origin lies in the region of x – y ≤ 2

The shaded region represents the inequality x – y ≤ 2.

**Question 6. 2x – 3y > 6**

**Solution.**

We draw the graph of line 2x – 3y = 6.

The line passes through (3, 0), (0, -2)

AB represents the equation 2x – 3y = 6

Now consider the inequality 2x – 3y > 6

Putting x = 0, y = 0

0 – 0 > 6, which is not true

∴ Origin does not lie in the region of 2x – 3y > 6.

The shaded region represents the inequality 2x – 3y > 6

**Question 7. -3x + 2y ≥ -6.**

**Solution.**

Let us draw the line -3x + 2y = -6

The line passes through (2, 0), (0, -3)

The line AB represents the equation -3x + 2y = -6

Now consider the inequality -3x+ 2y ≥ -6

Putting x = 0, y = 0

0 + 0 ≥ -6, which is true.

∴ Origin lies in the region of -3x + 2y ≥ -6

The shaded region represents the inequality -3x + 2y ≥ – 6

**Question 8. 3y- 5x < 30**

**Solution.**

Given inequality is 3y – 5x < 30

Let us draw the graph of the line 3y – 5x = 30

The line passes through (-6, 0), (0, 10)

The line AB represents the equation 3y – 5x = 30

Now, consider the inequality 3y – 5x < 30

Putting x = 0, y = 0

0 – 0 < 30, which is true.

∴ Origin lies in the region of 3y – 5x < 30

The shaded region represents the inequality 3y – 5x < 30

**Question 9. y<- 2**

**Solution.**

Given inequality is y < -2 ………(1)

Let us draw the graph of the line y = -2

AB is the required line.

Putting y = 0 in (1), we have

0 < -2, which is not true.

The solution region is the shaded region below the line.

Hence, every point below the line (excluding the line) is the solution area.

**Question 10. x > -3**

**Solution.**

Let us draw the graph of x = -3

∴ AB represents the line x = -3

By putting x = 0 in the inequality x > -3

We get, 0 > -3, which is true.

∴ Origin lies in the region of x > -3.

Graph of the inequality x > -3 is shown in the figure by the shaded area