Linear Inequalities

Linear inequalities are mathematical expressions where two algebraic expressions are compared using inequality symbols like < (less than), > (greater than), ≤ (less than or equal to), or ≥ (greater than or equal to). They represent a range of possible values rather than a single solution. Understanding linear inequalities is crucial in various fields such as economics, engineering, and optimization problems, as they help in defining constraints and determining feasible regions.

This article will explore the domain of linear inequalities, exploring methods for solving them and graphically representing their solutions. Let's embark on this journey to deepen our understanding of linear inequalities, their solutions, and the visual depiction through graphing.

1.0What is Linear Inequality?

In mathematics, inequalities arise when we compare two mathematical expressions or numbers and find that they are not equal. These inequalities can take different forms, such as numerical or algebraic, and sometimes a combination of both. Linear inequalities specifically involve comparing at least one linear algebraic expression, typically a polynomial of degree 1, with another expression of degree less than or equal to 1. 

Linear inequalities compare two linear expressions using inequality symbols. There are five symbols used for this purpose:

• ≠ (Not equal)
• < (Less than)
• > (Greater than)
• ≤ (Less than or equal to)
• ≥ (Greater than or equal to)

It's important to note that if p < q, then p is strictly less than q. If p ≤ q, it means p is either strictly less than q or equal to q. The same logic applies to > and ≥ inequalities.

2.0Linear Inequalities Rules

Let's delve into the rules governing linear inequalities:

1. Addition and Subtraction: You can add or subtract the same number from both sides of the inequality without changing its direction. For instance, if 3x < 9, you can subtract 2 from both sides to get 3x - 2 < 7.
2. Multiplication and Division: You can multiply or divide each side of the inequality by a positive number without changing its direction. However, if you multiply or divide by a negative number, you must flip the direction of the inequality. For example, if 2x < 6, multiplying both sides by 3 yields 6x < 18, but if you multiply both sides by –2, you must reverse the inequality sign to get –4x > –12.
3. Combining Inequalities: When you have multiple inequalities connected by "and" or "or", you need to consider their combined effect. For "and" statements, the solution is where both inequalities are true, while for "or" statements, the solution is where at least one inequality is true.
4. Graphing: Linear inequalities can be graphed on the Cartesian coordinate plane. If it's in the form ax + by < c, you first graph the line ax + by = c, then decide which side of the line to shade based on the inequality sign.
5. Intersections and Unions: When dealing with systems of inequalities, the solution is often the intersection (where all inequalities overlap) or the union (where any inequality is true). Graphically, this corresponds to shaded regions that overlap or combine.

Following these rules ensure that you manipulate linear inequalities correctly and accurately determine their solutions.

3.0Solving System of Linear Inequalities

Solving multi-step linear inequalities in one variable closely resembles the process for solving multi-step linear equations. The process starts by separating the variable from the constants. It's crucial to adhere to the rules of inequalities, particularly when multiplying or dividing by negative numbers, which necessitates reversing the inequality sign.

Step 1: Simplify the inequality on each side according to the rules of linear inequality.

Step 2: Once the value is obtained, if the inequality is strict, then the solution for x is either less than or greater than the obtained value, depending on the problem's conditions. If the inequality is not strict, the solution for x is either less than or equal to, or greater than or equal to, the obtained value.

Let's illustrate this with an example:

2x + 3 > 7 

To solve this linear inequality, we follow these steps:

⇒ 2x > 7 – 3 

⇒ 2x > 4 

⇒ x > 2 

The solution to this inequality comprises all values of x for which the inequality x > 2 holds true, indicating all real numbers strictly greater than 2.

4.0Graphing Linear Inequalities

Graphing linear inequalities involves visually representing regions on the Cartesian coordinate plane where the inequality holds true. Here's a step-by-step guide to graphing linear inequalities:

1. Identify the Inequality: Start with the given linear inequality in the form ax + by < c, ax + by ≤ c, ax + by > c, or ax + by ≥c.
2. Graph the Boundary Line: Treat the inequality as an equation and graph the corresponding boundary line. To do this:

• Rewrite the inequality as an equation by replacing the inequality sign with an equal sign.
• Plot the boundary line by finding two points that satisfy the equation.
• Draw a straight line passing through these points. If the inequality is strict (< or >), use a dashed line. If it includes equality (≤ or ≥), use a solid line.

3. Determine the Shaded Region: To determine which side of the boundary line to shade:

• Choose a test point not on the boundary line. The origin (0, 0) is often convenient.
• Substitute the coordinates of the test point into the original inequality.
• If the inequality holds true for the test point, shade the region containing it. Otherwise, shade the opposite region. 

4. Label the Shaded Region: Once you've determined which side of the boundary line to shade, label it accordingly to denote the solution region.
5. Final Touches: Don't forget to label the axes to provide context for your graph. Additionally, a title or key can enhance understanding, especially in more complex graphs.

By following these steps meticulously, you'll effectively graph linear inequalities, visually illustrating mathematical relationships on the Cartesian plane.

A sample of Graph is given below:

5.0Linear Equations and Inequalities in Two Variables

Linear equations and inequalities in two variables extend the concepts of single-variable linear equations to involve multiple variables. These equations and inequalities typically take the form ax + by = c or ax + by < c, where a, b, and c are constants, and x and y are the variables. The solutions to these equations and inequalities represent points or regions on the Cartesian coordinate plane where the expressions hold true. Graphically, linear equations represent straight lines, while linear inequalities represent shaded regions on the plane. Understanding these concepts allows for the analysis and interpretation of relationships between two variables in various mathematical contexts.

6.0Solving Linear Inequalities with Variable on Both Sides

Let's tackle the linear inequality with one variable given by 3x – 15 > 2x + 11. Here's our approach:

⇒ –15 – 11 > 2x – 3x 

⇒ –26 > –x 

⇒ x > 26

Thus, the solution to this inequality is x > 26, indicating that all real numbers greater than 26 satisfy the given inequality.

7.0Solved Examples of Linear Inequalities

Question 1: Solve the following linear inequality for x: 2x + 5 ≤ 3x – 2

Solution: To solve the inequality, we'll first simplify it:

2x + 5 ≤ 3x – 2

Subtract 2x from both sides:

5 ≤ x - 2

Add 2 to both sides:

5 + 2 ≤ x

7 ≤ x 

So, the solution to the inequality is x ≥ 7.

This means all real numbers greater than or equal to 7 satisfy the given inequality.

Question 2: Solve the following linear inequality for x: 4x – 3 ≥ 2x + 9

Solution: 4x – 3 ≥ 2x + 9 

Subtract 2x from both sides: 

⇒ 4x – 2x – 3 ≥ 9 

⇒2x – 3 ≥ 9 

Add 3 to both sides:

⇒ 2x – 3 + 3 ≥ 9 + 3 

⇒ 2x ≥ 12

Divide both sides by 2:

$\Rightarrow x\ge \frac{12}{2}$

⇒x ≥ 6 

So, the solution to the inequality is x ≥ 6.

Question 3: Solve the following linear inequality for x: 3(x – 4) < 2(x + 6)

Solution: 3(x – 4) < 2(x + 6) 

Expand both sides:

⇒ 3x – 12 < 2x + 12 

Subtract 2x from both sides:

⇒ 3x – 2x – 12 < 12 

⇒ x – 12 < 12 

Add 12 to both sides:

⇒ x – 12 + 12 < 12 + 12 

⇒ x < 24 

So, the solution to the inequality is x < 24.

Question 4: Solve the following linear inequality for x : 5 – 2x ≥ 3x + 8 

Solution: 5 – 2x ≥ 3x + 8 

Add 2x to both sides:

⇒ 5 ≥ 3x + 2x + 8 

⇒ 5 ≥ 5x + 8 

Subtract 8 from both sides:

⇒ 5 – 8 ≥ 5x 

⇒ –3 ≥ 5x 

Divide both sides by 5 

$\Rightarrow \frac{-3}{5}\ge x$

$\Rightarrow \mathrm{x}\le -\frac{3}{5}$

So, the solution to the inequality is

$x\le -\frac{3}{5}$

Question 5: Solve the following linear inequality for x: 2(x + 3) – 4 ≤ 3(2x – 1)

Solution: 2(x + 3) – 4 ≤ 3(2x – 1) 

Expand both sides:

⇒ 2x + 6 – 4 ≤ 6x – 3 

⇒ 2x + 2 ≤ 6x – 3 

Subtract 2x from both sides:

⇒ 2x – 2x + 2 ≤ 6x – 2x – 3 

⇒ 2 ≤ 4x – 3 

Add 3 to both sides:

⇒ 2 + 3 ≤ 4x – 3 + 3 

⇒ 5 ≤ 4x 

Divide both sides by 4:

$\Rightarrow \frac{5}{4}\le x$

$\Rightarrow x\ge \frac{5}{4}$

So, the solution to the inequality is

$x\ge \frac{5}{4}$

Question 6: Solve the following linear inequality for $x:\frac{3}{2}x-4>\frac{1}{2}x+5$

Solution:

$\frac{3}{2}x-4>\frac{1}{2}x+5$

Subtract $\frac{x}{2}$ from both sides:

$\Rightarrow \frac{3}{2}x-\frac{1}{2}x-4>5$

⇒ x – 4 > 5

Add 4 to both sides:

⇒ x – 4 + 4 > 5 + 4 

⇒ x > 9 

So, the solution to the inequality is x > 9.

8.0Word Problems in Linear Inequalities

Question 1: A fruit vendor sells apples and oranges. Let x represent the number of apples and y represent the number of oranges. The vendor has at most 100 fruits to sell and wants to make at least Rs. 50 in sales. Each apple costs Rs. 2 and each orange costs Rs. 3. Write a linear inequality representing this situation.

Solution: Let's break down the problem:

• The total number of fruits sold can be represented as x + y.
• The total revenue from selling apples and oranges is 2x + 3y.
• The vendor wants to make at least Rs. 50, so the inequality is 2x + 3y ≥ 50.
• The vendor has at most 100 fruits to sell, so the inequality is x + y ≤ 100.

Thus, the linear inequalities representing the situation are: 

2x + 3y ≥ 50 

x + y ≤ 100

These inequalities describe the constraints on the number of apples and oranges the vendor can sell to meet their sales and quantity goals.

Question 2: A company produces two types of products: Product A and Product B. The company can produce at most 500 units of Product A and 300 units of Product B per day. Each unit of Product A requires 2 hours of machine time, while each unit of Product B requires 3 hours. The company has a total of 1200 machine hours available per day. If the profit from each unit of Product A is Rs. 10 and from each unit of Product B is Rs. 15, write a system of linear inequalities to represent the production constraints and maximize the daily profit.

Solution: Let x represent the number of units of Product A and y represent the number of units of Product B produced per day. Based on the given constraints:

1. Production limits: 

x ≤ 500 

y ≤ 300

2. Machine time constraints: 2x + 3y ≤ 1200

To maximize profit, we want to maximize the objective function: 

Profit = 10x + 15y

So, the system of linear inequalities representing the situation is: 

x ≤ 500 

y ≤ 300 

2x + 3y ≤ 1200

This system describes the production constraints and allows us to maximize the profit from Product A and Product B.