Solving Inequalities: We already know that a graph of a linear inequality in one variable is a convenient way of representing the solutions of the inequality. In this article, we will look at the graphical solution of linear inequalities in two variables.

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## Graphical Solution for Solving Inequalities in Two Variables

We know that a line divides a Cartesian two-dimensional plane is into two parts called half-planes.

- A vertical line divides it into left and right half planes as shown below:

- A non-vertical line divides the plane into lower and upper half-planes as shown below:

Any point in the Cartesian plane will either:

- lie in the line or
- lie in half-plane I or
- lie in half-plane II

Let’s look at the relationship between the points in the plane and the inequalities ax + by < c or ax + by > c. An expression in two variables, ‘x’ and ‘y’ and a constant ‘c’, can have three possibilities:

- ax + by = c
- ax + by > c
- ax + by < c

where a ≠ 0 and b ≠ 0. Let’s look at each scenario:

### 1. ax + by = c

In this case, all points having coordinates (x, y) satisfying the equation ax + by = c lie on the line representing it. Conversely, all points which lie on the line represented by ax + by = c satisfy the equation.

### 2. ax + by > c

In this case, we will look at two further scenarios:

- b > 0
- b < 0

Let’s look at the first scenario where b > 0. As shown in the figure below, consider a point P(α, β) on the line ax + by = c, where b > 0.

Since the point P lies on the line, we have aα + bβ = c. Next, take a point Q(α, γ) in the half-plane II as shown above. Since the point Q lies above the point P, we have

γ > β

Multiplying both sides by ‘b’, we get

bγ > bβ

Adding ‘aα’ to both sides, we have

aα + bγ > aα + bβ

∴ aα + bγ > c … (since aα + bβ = c)

Hence, we can say that the point Q(α, γ) satisfies the inequality aα + bγ > c. We can also state that all points lying in half-plane II (b > 0), above the line ax + by = c, satisfy the inequality ax + by > c.

### Converse

Let the point P(α, β) be a point on the line ax + by = c and the point Q(α, γ) satisfying the inequality ax + by > c. Therefore, we have

aα + bγ > c

⇒ aα + bγ > aα + bβ … (since P(α, β) satisfies ax + by = c, we have aα + bβ = c)

⇒ bγ > bβ

⇒ γ > β … thereby indicating that the point Q(α, γ) lies in the half-plane II.

**To sum it up: **Any point in the half plane II satisfies ax + by > c, and conversely, any point satisfying the inequality ax + by > c lies in half plane II. Similarly, for b < 0, it can be proved that any point satisfying the inequality ax + by > c lies in the half-plane I and conversely.

### Here are some pointers to graphically represent and solving inequalities in two variables:

- Solution region is the region containing all solutions of the inequality.
- To identify the half-plane represented by the inequality, you can take a point (a, b) which does not lie on the line and check if it satisfies the inequality. If it does, then the inequality represents the half-plane containing the point and you can shade the region. If it doesn’t then the half-plane that does not contain the point should be shaded. For convenience, you can take the point (0, 0) to check if it satisfies the inequality.
- If the inequality is of the type ax + by ≥ c or ax + by ≤ c, then include the points on the line ax + by = c in the solution region. To do so, draw a dark line on the line representing ax + by = c.
- If the inequality is of the type ax + by > c or ax + by < c, then don’t includes the points on the line ax + by = c in the solution region. Draw a broken/dotted line in this case.

### Example 1

Question: Let’s solve the following inequality 40x + 20y ≤ 120 … (1), where x and y are whole numbers (not fractions or negative numbers).

Solution: To solve this, let’s start with substituting x = 0 in the LHS of the inequality (1). Therefore, we get

40x + 20y = 0 + 20y = 20y

Hence, we have

20y ≤ 120

⇒ y ≤ 6

So, for x = 0, the values of y can be 0, 1, 2, 3, 4, 5, and 6. The solutions can be written as (0, 0), (0, 1), (0,2), (0,3), (0,4), (0, 5) and (0, 6). Similarly,

For x = 1, the solutions are (1, 0), (1, 1), (1, 2), (1, 3), (1, 4)

For x = 2, the solutions are (2, 0), (2, 1), (2, 2)

For x = 3, the solution is (3, 0)

For x = 4, the value of y becomes negative and hence not valid.

These solutions can be represented as follows:

Let’s try to extend this solution to real numbers. For this, draw a graph for the equation 40x + 20y = 120. Now, let’s consider a point (0, 0) and check if it satisfies the inequality. We have,

40x + 20y ≤ 0

Substituting the values of ‘x’ and ‘y’, we get

40(0) + 20(0) ≤ 0

Or, 0 ≤ 0 which is true. Hence, we can conclude that the point (0, 0) satisfies the inequality. Look at the diagram given below:

In the diagram, you can see that the point (0, 0) lies in the half-plane I. Hence, the inequality represents this plane. Also, the points on the line satisfy the inequality. Hence, it is also a part of the shaded region.

## Solving Inequalities: More Examples for You

### A question on solving inequalities in two variables

Question: Solve 3x + 2y > 6 graphically in a two-dimensional plane.

Solution: To solve the inequality, let’s plot a graph of the equation 3x + 2y = 6 as shown below:

This line divides the plane into half-planes I and II. Next, we select a point (0, 0) and determine if it satisfies the given inequality. Hence, we have

3x + 2y > 6

⇒ 3(0) + 2(0) > 6

⇒ 0 > 6 which is FALSE.

Since (0, 0) lies in the half-plane I and it does not satisfy the inequality, half-plane I, is not the solution. Also, the inequality given is a ‘Strict inequality’. Hence, any point on the line represented by 3x + 2y = 0 does not satisfy the inequality either. Therefore, the solution of the inequality is the shaded region in the diagram above.

### A question on a system of linear inequalities in two variables

Question: Solve the following system of linear inequalities in two variables graphically.

- x + y ≥ 5
- x – y ≤ 3

Solution. To begin with, let’s draw a graph of the equation x + y = 5. Now, we determine if the point (0, 0), which is lying in the half-plane I, satisfies the inequality 1. We have,

x + y ≥ 5

⇒ 0 + 0 ≥ 5

Or, 0 ≥ 5 which is FALSE. Also, being a ‘Slack inequality, the points on the line represented by x + y = 5 satisfy the inequality 1. Hence, the solution lies in the half-plane II and includes the line.

Next, let’s draw a graph of the equation x – y = 3 on the same set of axes. Now, we determine if the point (0, 0), which is lying in the half-plane II, satisfies the inequality 2. We have,

x – y ≤ 3

⇒ 0 – 0 ≤ 3

Or, 0 ≤ 3 which is TRUE. Also, being a ‘Slack inequality, the points on the line represented by x – y = 3, satisfy the inequality 2. Hence, the solution lies in the half-plane II and includes the line. Look at the diagram below:

The solution of the system of linear inequalities in two variables x + y ≥ 5 and x – y ≤ 3 is the region common to the two shaded regions as shown above.

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