Inequality Graphs

The procedure for drawing the graph of inequality is similar to that of a linear graph.

Every point on the line is going to be a solution to the equations of the linear graphs above.

Boundaries within regions: 

The required boundary of a particular region is described by the equality sign =

However, if the inequality sign is:

1. \( \scriptsize < \: or \: > \): the points on the boundary region are to be excluded, and for this, broken lines are used to draw the graph.

The shaded region (in green), in the inequality graph \( y \scriptsize > x \) is above the line.

The shaded region (in green) in the inequality graph \( y \scriptsize < \: – x \) is below the line.

2. \( \scriptsize \leq \: or \: \geq \): the points on the boundary region are included, and for this, continuous (straight unbroken) lines are used.

The shaded region (in green), in the inequality graph \( y \scriptsize \geq x \) is above the line.

The shaded region (in green), in the inequality graph \( y \scriptsize \leq \: – x \) is below the line.

The additional point to note here will be the shading of regions in the graph of inequality. All the points in the shaded region are solutions to the inequality graphs.

Example 5.2.1:

1. Draw the graph of 3x + 2y < 12

Solution

• When x = 0, y = 6
• When y = 0, x = 4

The shaded region (in green), in the inequality graph with < sign is below the line.

Check:

Any point (x, y) in the shaded region will be a solution to the linear inequality.

For example: (-3, 4)

Substitute the values of x and y into 3x + 2y < 12

3(-3) + 2(4) < 12

-9 + 8 < 12

-1 < 12 ✔︎ This is true.

Any point in the shaded region will make the linear equation true.

The points on the boundary region are to be excluded, and for this, broken lines are used to draw the graph.

Let’s consider two points (2, 3) on the line (or boundary)

Substitute the values of x and y into 3x + 2y < 12

3(2) + 2(3) < 12

6 + 6 < 12

12 < 12 ❌ (This is incorrect, hence the broken lines)

Example 5.2.2:

Draw the graph of 4x + 3y ≥ 24

Solution

• When x = 0, y = 8
• When y = 0, x = 6

The shaded region (in green), in the inequality graph with ≥ sign is above the line.

Check:

Any point (x, y) in the shaded region will be a solution to the linear inequality.

For example: (7, 6)

Substitute the values of x and y into 4x + 3y ≥ 24

4(7) + 3(6) ≥ 24

28 + 18 ≥ 24

46 ≥ 24 ✔︎ This is true.

The points on the boundary region are included, and for this, continuous (straight unbroken) lines are used.

Let’s consider two points (3, 4) on the line (or boundary)

Substitute the values of x and y into 4x + 3y ≥ 24

4(3) + 3(4) ≥ 24

12 + 12 ≥ 24

24 ≥ 24 ✔︎ This is true (because 24 = 24)

Simultaneous Linear Inequalities in Two Variables:

Example 5.2.3:

Solution

Let us consider the inequality 4x – y ≤ 8 first. 

To get the boundary line, first, make y the subject 

i.e. 4x – y = 8

-y = 8 – 4x

y = -8 + 4x

When x = 0, y = -8. This gives points (0, – 8)

When y = 0, x = 2. This gives points (2, 0)

4x – y ≤ 8 can also be written as:

-y ≤ 8 – 4x

y ≥ 4x – 8

The shaded region (in light blue), in the inequality graph \( y \scriptsize \geq x \) is above the line

Mark out these two points on your axis and then join them together with a solid line.

Next, let’s consider the second inequality 3x + 2y > 12.

To get the boundary line which must be a broken line.

When x = 0, y = 6. This gives points (0, 6)

When y = 0, x = 4. This gives points (4, 0).

The shaded region (in pink), in the inequality graph \( y \scriptsize > x \) is above the line.

The solution lies within the region bounded by area ABC (the dark blue region). This is the only region where both equations are true.

Check:

Use points (2, 6) within the region bounded by area ABC.

4x – y ≤ 8  ……(1)
3x +2y > 12 ………(2)

4(2) – 6 ≤ 8  ……(1)

8 – 6 ≤ 8 

2 ≤ 8  ✔︎ This is true

3x +2y > 12 ………(2)

3(2) +2(6) > 12

6 + 12 > 12

18 > 12 ✔︎ This is true