Identifying Ordered Pairs That Satisfy Inequalities A Step-by-Step Guide

In mathematics, especially when dealing with inequalities, it's crucial to understand how to identify ordered pairs that satisfy a given set of conditions. This article delves into the process of determining which ordered pairs make a system of inequalities true. We will use a specific example to illustrate the concepts and techniques involved, ensuring a clear and thorough understanding. Understanding ordered pairs and inequalities is vital for various mathematical applications, from solving linear programming problems to modeling real-world constraints.

Understanding the Problem

To effectively identify ordered pairs that make inequalities true, it's essential to first grasp the problem at hand. Consider the following system of inequalities:

$ \begin{array}{l} y < 5x + 2 \ y \geq \frac{1}{2}x + 1 \end{array} $

We are given two inequalities: $y < 5x + 2$ and $y \geq \frac{1}{2}x + 1$. Our task is to determine which ordered pairs, from a given set, satisfy both inequalities simultaneously. An ordered pair $(x, y)$ satisfies an inequality if, upon substituting the values of $x$ and $y$ into the inequality, the resulting statement is true. This involves plugging in the x and y values from each ordered pair into both inequalities and checking if the inequalities hold. The solution set will consist of all ordered pairs that satisfy both inequalities, representing the region where the solution sets of the individual inequalities overlap.

Step-by-Step Solution

1. Understanding Ordered Pairs

An ordered pair is a set of two numbers written in the form $(x, y)$, where $x$ represents the horizontal coordinate and $y$ represents the vertical coordinate on a coordinate plane. For an ordered pair to satisfy an inequality, both its $x$ and $y$ values must make the inequality true when substituted. This means that the coordinate point represented by the ordered pair must lie within the solution region defined by the inequality. For a system of inequalities, an ordered pair must satisfy all inequalities in the system to be considered a solution. Let's analyze each ordered pair provided in the context of the given inequalities.

2. Testing the First Ordered Pair: (-1, 3)

The first ordered pair to consider is $(-1, 3)$. This means $x = -1$ and $y = 3$. We need to substitute these values into both inequalities and check if they hold true.

First Inequality: $y < 5x + 2$

Substitute $x = -1$ and $y = 3$ into the inequality:

$3 < 5(-1) + 2$

Simplify the expression:

$3 < -5 + 2$

$3 < -3$

This statement is false. Since 3 is not less than -3, the ordered pair $(-1, 3)$ does not satisfy the first inequality.

Second Inequality: $y \geq \frac{1}{2}x + 1$

Substitute $x = -1$ and $y = 3$ into the inequality:

$3 \geq \frac{1}{2}(-1) + 1$

Simplify the expression:

$3 \geq -\frac{1}{2} + 1$

$3 \geq \frac{1}{2}$

This statement is true. However, for the ordered pair to be a solution to the system of inequalities, it must satisfy both inequalities. Since $(-1, 3)$ does not satisfy the first inequality, we can conclude that it is not a solution to the system.

3. Testing the Second Ordered Pair: (0, 2)

The second ordered pair to test is $(0, 2)$. Here, $x = 0$ and $y = 2$. We will again substitute these values into both inequalities.

First Inequality: $y < 5x + 2$

Substitute $x = 0$ and $y = 2$ into the inequality:

$2 < 5(0) + 2$

Simplify the expression:

$2 < 0 + 2$

$2 < 2$

This statement is false. 2 is not less than 2, so the ordered pair $(0, 2)$ does not satisfy the first inequality.

Second Inequality: $y \geq \frac{1}{2}x + 1$

Substitute $x = 0$ and $y = 2$ into the inequality:

$2 \geq \frac{1}{2}(0) + 1$

Simplify the expression:

$2 \geq 0 + 1$

$2 \geq 1$

This statement is true. But as with the previous ordered pair, since $(0, 2)$ does not satisfy the first inequality, it is not a solution to the system.

General Strategy for Solving Systems of Inequalities

When solving systems of inequalities, the general strategy involves testing ordered pairs against each inequality to see if they hold true. Here’s a breakdown of the steps:

1. Understand the Inequalities: Clearly identify the inequalities in the system. Each inequality represents a region on the coordinate plane.
2. Choose Ordered Pairs: Select the ordered pairs you want to test. These pairs are potential solutions to the system.
3. Substitute Values: For each ordered pair, substitute the $x$ and $y$ values into each inequality.
4. Check for Truth: Determine if the resulting statements are true. An ordered pair satisfies an inequality if the statement is true after substitution.
5. Identify Solutions: An ordered pair is a solution to the system if it satisfies all inequalities in the system. If even one inequality is not satisfied, the ordered pair is not a solution.
6. Graphical Representation: (Optional but highly beneficial) Graph the inequalities on a coordinate plane. The region where the shaded areas of all inequalities overlap represents the solution set of the system. This visual aid can help verify algebraic solutions and provide a clearer understanding of the solution space.

Additional Ordered Pairs and Deeper Analysis

To further illustrate the process, let’s consider a few additional ordered pairs and analyze them in the context of the same system of inequalities:

$ \begin{array}{l} y < 5x + 2 \ y \geq \frac{1}{2}x + 1 \end{array} $

Example 1: Ordered Pair (1, 4)

Test the ordered pair $(1, 4)$, where $x = 1$ and $y = 4$.

First Inequality: $y < 5x + 2$

Substitute $x = 1$ and $y = 4$:

$4 < 5(1) + 2$

$4 < 5 + 2$

$4 < 7$

This statement is true.

Second Inequality: $y \geq \frac{1}{2}x + 1$

Substitute $x = 1$ and $y = 4$:

$4 \geq \frac{1}{2}(1) + 1$

$4 \geq \frac{1}{2} + 1$

$4 \geq \frac{3}{2}$

This statement is also true. Since $(1, 4)$ satisfies both inequalities, it is a solution to the system.

Example 2: Ordered Pair (-2, -8)

Test the ordered pair $(-2, -8)$, where $x = -2$ and $y = -8$.

First Inequality: $y < 5x + 2$

Substitute $x = -2$ and $y = -8$:

$-8 < 5(-2) + 2$

$-8 < -10 + 2$

$-8 < -8$

This statement is false since -8 is not less than -8.

Second Inequality: $y \geq \frac{1}{2}x + 1$

Substitute $x = -2$ and $y = -8$:

$-8 \geq \frac{1}{2}(-2) + 1$

$-8 \geq -1 + 1$

$-8 \geq 0$

This statement is also false. Since $(-2, -8)$ does not satisfy either inequality, it is not a solution to the system.

Example 3: Ordered Pair (0, 1)

Test the ordered pair $(0, 1)$, where $x = 0$ and $y = 1$.

First Inequality: $y < 5x + 2$

Substitute $x = 0$ and $y = 1$:

$1 < 5(0) + 2$

$1 < 0 + 2$

$1 < 2$

This statement is true.

Second Inequality: $y \geq \frac{1}{2}x + 1$

Substitute $x = 0$ and $y = 1$:

$1 \geq \frac{1}{2}(0) + 1$

$1 \geq 0 + 1$

$1 \geq 1$

This statement is also true. Thus, $(0, 1)$ is a solution to the system because it satisfies both inequalities.

Graphing Inequalities to Visualize Solutions

Graphing inequalities is a powerful way to visualize the solution set and understand the region in which ordered pairs will satisfy the system. Here’s how to graph the inequalities $y < 5x + 2$ and $y \geq \frac{1}{2}x + 1$:

Graphing $y < 5x + 2$

1. Treat as an Equation: First, treat the inequality as an equation: $y = 5x + 2$. This is a linear equation in slope-intercept form ($y = mx + b$), where $m$ is the slope and $b$ is the y-intercept.
2. Identify Slope and Y-Intercept: The slope $m$ is 5, and the y-intercept $b$ is 2. This means the line crosses the y-axis at the point $(0, 2)$.
3. Plot the Line: Plot the y-intercept $(0, 2)$. Use the slope to find another point. A slope of 5 can be interpreted as “rise over run,” so for every 1 unit you move to the right, you move 5 units up. From $(0, 2)$, move 1 unit to the right and 5 units up to find the point $(1, 7)$. Draw a line through these two points.
4. Dashed Line: Because the inequality is $y < 5x + 2$, not $y \leq 5x + 2$, the line should be dashed. A dashed line indicates that the points on the line itself are not included in the solution set.
5. Shade the Region: Since $y$ must be less than $5x + 2$, shade the region below the line. This shaded area represents all the ordered pairs that satisfy the inequality $y < 5x + 2$.

Graphing $y \geq \frac{1}{2}x + 1$

1. Treat as an Equation: Treat the inequality as an equation: $y = \frac{1}{2}x + 1$. This is another linear equation in slope-intercept form.
2. Identify Slope and Y-Intercept: The slope $m$ is $\frac{1}{2}$, and the y-intercept $b$ is 1. This line crosses the y-axis at the point $(0, 1)$.
3. Plot the Line: Plot the y-intercept $(0, 1)$. Use the slope of $\frac{1}{2}$ to find another point. For every 2 units you move to the right, move 1 unit up. From $(0, 1)$, move 2 units to the right and 1 unit up to find the point $(2, 2)$. Draw a line through these two points.
4. Solid Line: Because the inequality is $y \geq \frac{1}{2}x + 1$, the line should be solid. A solid line indicates that the points on the line are included in the solution set.
5. Shade the Region: Since $y$ must be greater than or equal to $\frac{1}{2}x + 1$, shade the region above the line. This shaded area represents all the ordered pairs that satisfy the inequality $y \geq \frac{1}{2}x + 1$.

Identifying the Solution Region

The solution region for the system of inequalities is the area where the shaded regions of both inequalities overlap. Any ordered pair that falls within this overlapping region satisfies both inequalities and is therefore a solution to the system. By visually inspecting the graph, you can quickly determine whether a given ordered pair is a solution.

Real-World Applications and Significance

The application of solving systems of inequalities extends beyond the classroom and into real-world scenarios. Understanding how to find ordered pairs that satisfy multiple conditions is crucial in various fields, including economics, engineering, and computer science.

Linear Programming

In linear programming, systems of inequalities are used to define constraints within which optimization problems are solved. For example, a business might use linear programming to determine the optimal production levels of different products, given constraints on resources like labor and materials. The solution set of these inequalities represents the feasible region, and the ordered pairs within this region represent possible production plans that satisfy all constraints.

Engineering Design

Engineers often use systems of inequalities to define the acceptable ranges for design parameters. For instance, in structural engineering, inequalities might represent the limits on the load that a bridge can safely bear. Ordered pairs representing different design configurations must satisfy these inequalities to ensure the structure's integrity.

Resource Allocation

Governments and organizations use systems of inequalities to allocate resources efficiently. For example, in healthcare, inequalities might define constraints on the distribution of medical supplies or personnel across different regions. The solution set helps decision-makers identify allocation strategies that meet the needs of all regions while staying within resource limits.

Computer Science

In computer science, systems of inequalities are used in areas like network design and optimization. Inequalities can represent constraints on network bandwidth, latency, and cost. Ordered pairs representing different network configurations are evaluated against these inequalities to find solutions that meet performance requirements while minimizing costs.

Conclusion

Identifying ordered pairs that make inequalities true is a fundamental skill in mathematics with wide-ranging applications. By understanding the steps involved in substituting values, checking for truth, and visualizing solutions graphically, you can effectively solve systems of inequalities. Whether it’s optimizing business processes, designing safe structures, allocating resources, or engineering computer networks, the ability to work with inequalities provides a powerful tool for problem-solving and decision-making.

Through the detailed examples and explanations provided in this article, you should now have a solid foundation for tackling various problems involving systems of inequalities. Remember to practice and apply these techniques to build your proficiency and confidence in this important mathematical area.