Solving Inequalities Determine Solutions For Y < 5x + 2 And Y ≥ 1/2x + 1

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In mathematics, a system of inequalities represents a set of two or more inequalities that are considered simultaneously. The solution to a system of inequalities is the set of all points that satisfy all the inequalities in the system. Graphically, this solution is represented by the region where the shaded areas of the individual inequalities overlap. In this article, we will explore how to determine whether given points are solutions to the system of inequalities:

y < 5x + 2
y ≥ 1/2x + 1

We will analyze each point by substituting its coordinates into the inequalities and checking if the resulting statements are true. This process will help us understand the concept of solutions in the context of systems of inequalities and how they are identified.

Understanding Systems of Inequalities

Before diving into specific points, let's first understand the core concepts of systems of inequalities. A system of inequalities involves two or more inequalities that share the same variables. The solution to a system of inequalities is the set of all ordered pairs (x, y) that satisfy all inequalities in the system simultaneously. Graphically, each inequality represents a region in the coordinate plane, and the solution to the system is the intersection of these regions.

To determine if a point is a solution to a system of inequalities, we substitute the coordinates of the point into each inequality. If the point satisfies all inequalities, it is a solution to the system. If it fails to satisfy even one inequality, it is not a solution. This process is crucial for understanding the nature of solutions in systems of inequalities.

The given system of inequalities consists of two linear inequalities:

  1. y < 5x + 2
  2. y ≥ 1/2x + 1

The first inequality, y < 5x + 2, represents the region below the line y = 5x + 2. The line itself is not included in the solution because the inequality is strict (less than, not less than or equal to). The second inequality, y ≥ 1/2x + 1, represents the region above and including the line y = 1/2x + 1. The line is included because the inequality is non-strict (greater than or equal to).

The solution to the system is the region where these two shaded areas overlap. Points in this overlapping region satisfy both inequalities simultaneously. Now, let's evaluate each given point to see if it falls within this solution region.

Evaluating the Point (-1, 3)

To determine if the point (-1, 3) is a solution, we substitute x = -1 and y = 3 into both inequalities:

  1. For y < 5x + 2:

    3 < 5(-1) + 2
3 < -5 + 2
3 < -3

    This statement is false. Therefore, the point (-1, 3) does not satisfy the first inequality.

  2. Since the point does not satisfy the first inequality, we do not need to check the second inequality. A point must satisfy all inequalities in the system to be considered a solution.

Conclusion for (-1, 3)

Since the point (-1, 3) does not satisfy the inequality y < 5x + 2, it is not a solution to the system of inequalities.

Evaluating the Point (0, 2)

Next, we evaluate the point (0, 2) by substituting x = 0 and y = 2 into both inequalities:

  1. For y < 5x + 2:

    2 < 5(0) + 2
2 < 0 + 2
2 < 2

    This statement is false because 2 is not less than 2. Therefore, the point (0, 2) does not satisfy the first inequality.

  2. Since the point does not satisfy the first inequality, we do not need to check the second inequality.

Conclusion for (0, 2)

Because the point (0, 2) does not satisfy the inequality y < 5x + 2, it is not a solution to the system of inequalities.

Evaluating the Point (1, 2)

Now, let's evaluate the point (1, 2) by substituting x = 1 and y = 2 into both inequalities:

  1. For y < 5x + 2:

    2 < 5(1) + 2
2 < 5 + 2
2 < 7

    This statement is true, so the point (1, 2) satisfies the first inequality.

  2. For y ≥ 1/2x + 1:

    2 ≥ 1/2(1) + 1
2 ≥ 1/2 + 1
2 ≥ 3/2
2 ≥ 1.5

    This statement is also true, so the point (1, 2) satisfies the second inequality.

Conclusion for (1, 2)

Since the point (1, 2) satisfies both inequalities, it is a solution to the system of inequalities.

Evaluating the Point (2, -1)

We will now evaluate the point (2, -1) by substituting x = 2 and y = -1 into both inequalities:

  1. For y < 5x + 2:

    -1 < 5(2) + 2
-1 < 10 + 2
-1 < 12

    This statement is true, so the point (2, -1) satisfies the first inequality.

  2. For y ≥ 1/2x + 1:

    -1 ≥ 1/2(2) + 1
-1 ≥ 1 + 1
-1 ≥ 2

    This statement is false. Therefore, the point (2, -1) does not satisfy the second inequality.

Conclusion for (2, -1)

Since the point (2, -1) does not satisfy the inequality y ≥ 1/2x + 1, it is not a solution to the system of inequalities.

Evaluating the Point (2, 2)

Finally, let's evaluate the point (2, 2) by substituting x = 2 and y = 2 into both inequalities:

  1. For y < 5x + 2:

    2 < 5(2) + 2
2 < 10 + 2
2 < 12

    This statement is true, so the point (2, 2) satisfies the first inequality.

  2. For y ≥ 1/2x + 1:

    2 ≥ 1/2(2) + 1
2 ≥ 1 + 1
2 ≥ 2

    This statement is true, so the point (2, 2) satisfies the second inequality.

Conclusion for (2, 2)

Since the point (2, 2) satisfies both inequalities, it is a solution to the system of inequalities.

Summary of Solutions

After evaluating each point, we can summarize our findings:

  • (-1, 3): Not a solution
  • (0, 2): Not a solution
  • (1, 2): Is a solution
  • (2, -1): Not a solution
  • (2, 2): Is a solution

Conclusion

In this article, we have demonstrated how to determine whether a given point is a solution to a system of inequalities. By substituting the coordinates of each point into the inequalities and checking if the resulting statements are true, we can identify the solutions. In the given system of inequalities, y < 5x + 2 and y ≥ 1/2x + 1, the points (1, 2) and (2, 2) were found to be solutions, while the points (-1, 3), (0, 2), and (2, -1) were not. Understanding this process is crucial for solving systems of inequalities and visualizing their solutions graphically.