which system of linear inequalities has the point (3

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29-03-2025

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Which System of Linear Inequalities Contains the Point (3, 1)?

Determining which system of linear inequalities contains a specific point, like (3, 1), involves testing the point in each inequality of the system. If the point satisfies all inequalities, it lies within the solution region of that system. Let's explore this process with some examples.

Understanding Linear Inequalities

Before we dive into examples, let's refresh our understanding of linear inequalities. A linear inequality is similar to a linear equation (like y = 2x + 1), but instead of an equals sign (=), it uses an inequality symbol (<, >, ≤, ≥). These symbols represent "less than," "greater than," "less than or equal to," and "greater than or equal to," respectively.

The solution to a linear inequality is a region on a graph, rather than a single line. This region represents all points (x, y) that satisfy the inequality. A system of linear inequalities involves multiple inequalities, and the solution is the intersection of all the solution regions of the individual inequalities.

Examples: Identifying the Correct System

Let's consider a few example systems of linear inequalities and see if the point (3, 1) satisfies them.

Example 1:

• y ≤ x + 1
• y > -x + 2

To check if (3, 1) is a solution:

1. Inequality 1 (y ≤ x + 1): Substitute x = 3 and y = 1. We get 1 ≤ 3 + 1, which simplifies to 1 ≤ 4. This is true.

2. Inequality 2 (y > -x + 2): Substitute x = 3 and y = 1. We get 1 > -3 + 2, which simplifies to 1 > -1. This is also true.

Since (3, 1) satisfies both inequalities in Example 1, it lies within the solution region of this system.

Example 2:

• y < x - 2
• y ≥ 2x - 5

Let's test (3, 1):

1. Inequality 1 (y < x - 2): 1 < 3 - 2 simplifies to 1 < 1. This is false.

Because (3, 1) does not satisfy the first inequality, it is not part of the solution region for Example 2. We don't even need to check the second inequality. If a point fails one inequality in a system, it fails the entire system.

Example 3:

• x + y ≤ 5
• x - y ≥ 1

Let's test (3,1):

1. Inequality 1 (x + y ≤ 5): 3 + 1 ≤ 5 simplifies to 4 ≤ 5. This is true.

2. Inequality 2 (x - y ≥ 1): 3 - 1 ≥ 1 simplifies to 2 ≥ 1. This is true.

Since (3, 1) satisfies both inequalities, it lies in the solution region of Example 3.

How to Solve This Type of Problem

To determine if a given point lies within the solution region of a system of linear inequalities, follow these steps:

1. Substitute the coordinates: Plug the x and y values of the point into each inequality in the system.

2. Evaluate each inequality: Simplify the expression to see if the inequality is true or false.

3. Check all inequalities: The point is only a solution if it satisfies all the inequalities in the system. If it fails even one, it's not in the solution region.

By systematically testing the given point in each inequality, you can effectively identify which system of linear inequalities contains that point. Remember that graphing the inequalities can also be a helpful visual aid. You can plot the inequalities and visually see if the point falls within the overlapping shaded region.