Linear Inequalities in Two Variables

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.

Browse more Topics Under Linear Inequalities

• Linear Inequalities in One Variable
• 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:

1. ax + by = c
2. ax + by > c
3. 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:

1. b > 0
2. 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.

1. x + y ≥ 5
2. 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.