Solving Systems Of Inequalities A Step By Step Guide
In mathematics, a system of inequalities is a set of two or more inequalities involving the same variables. A solution to a system of inequalities is a point (an ordered pair) that satisfies all the inequalities in the system simultaneously. In other words, when the coordinates of the point are substituted into each inequality, the inequalities hold true. This article delves into the process of identifying solutions to systems of inequalities, using a specific example to illustrate the method. We will explore the graphical representation of inequalities, the concept of solution regions, and how to test points to determine if they satisfy a given system of inequalities. Understanding these concepts is crucial for solving a wide range of problems in algebra and calculus, particularly those involving optimization and constraint satisfaction.
Understanding Inequalities
Before diving into the specific problem, it's important to grasp the fundamental concepts of inequalities. An inequality is a mathematical statement that compares two expressions using symbols such as less than (<), greater than (>), less than or equal to (≤), or greater than or equal to (≥). Unlike equations, which have a single solution or a set of discrete solutions, inequalities often have a range of solutions. This range can be visualized graphically as a region in the coordinate plane.
Each inequality represents a boundary line and a region on one side of that line. The boundary line is determined by replacing the inequality symbol with an equal sign and graphing the resulting equation. The region that satisfies the inequality is then shaded. If the inequality includes an "equal to" component (≤ or ≥), the boundary line is solid, indicating that points on the line are part of the solution. If the inequality is strict (< or >), the boundary line is dashed, meaning that points on the line are not included in the solution.
When dealing with a system of inequalities, the solution set is the region where the shaded areas of all inequalities overlap. This overlapping region represents all the points that satisfy every inequality in the system. To determine if a specific point is a solution, we substitute its coordinates into each inequality and check if the inequalities hold true. This process is crucial for identifying solutions to systems of inequalities and is a fundamental skill in algebra.
The Given System of Inequalities
Let's consider the specific system of inequalities presented:
$ y ≤ (1/2)x - 3 $
$ y + 2x > 6 $
Our task is to determine which of the given points, A(5,-2), B(7,-8), C(2,-3), and D(4,1), is a solution to this system. To do this, we will substitute the coordinates of each point into both inequalities and check if both inequalities are satisfied.
This process involves careful substitution and evaluation. For each point, we replace 'x' and 'y' in the inequalities with the point's x-coordinate and y-coordinate, respectively. We then simplify the expressions and determine if the resulting statements are true. If both inequalities hold true for a particular point, that point is a solution to the system. If even one inequality is not satisfied, the point is not a solution.
Understanding how to substitute and evaluate points in inequalities is a critical skill for solving systems of inequalities. It allows us to move beyond graphical representations and identify solutions algebraically, which is particularly useful when dealing with complex systems or when precise solutions are required.
Testing Point A (5, -2)
To check if point A (5, -2) is a solution, we substitute x = 5 and y = -2 into the inequalities:
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For the first inequality, y ≤ (1/2)x - 3, we have:
-2 ≤ (1/2)(5) - 3
-2 ≤ 2.5 - 3
-2 ≤ -0.5
This statement is true, as -2 is indeed less than or equal to -0.5.
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For the second inequality, y + 2x > 6, we have:
-2 + 2(5) > 6
-2 + 10 > 6
8 > 6
This statement is also true, as 8 is greater than 6.
Since both inequalities are satisfied by point A (5, -2), we can conclude that point A is a solution to the system of inequalities.
This step-by-step evaluation highlights the importance of accurate substitution and simplification. By carefully replacing the variables with the coordinates of the point, we can determine whether the point lies within the solution region of the system. The fact that both inequalities hold true for point A confirms that it is a valid solution.
Testing Point B (7, -8)
Next, let's test point B (7, -8) by substituting x = 7 and y = -8 into the inequalities:
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For the first inequality, y ≤ (1/2)x - 3, we have:
-8 ≤ (1/2)(7) - 3
-8 ≤ 3.5 - 3
-8 ≤ 0.5
This statement is true, as -8 is less than or equal to 0.5.
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For the second inequality, y + 2x > 6, we have:
-8 + 2(7) > 6
-8 + 14 > 6
6 > 6
This statement is false, as 6 is not greater than 6.
Since the second inequality is not satisfied by point B (7, -8), point B is not a solution to the system of inequalities.
This example illustrates that a point must satisfy all inequalities in the system to be considered a solution. Even if a point satisfies one inequality, it is not a solution if it fails to satisfy any of the others. This underscores the importance of checking each inequality individually when determining solutions to systems of inequalities.
Testing Point C (2, -3)
Now, let's evaluate point C (2, -3) by substituting x = 2 and y = -3 into the inequalities:
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For the first inequality, y ≤ (1/2)x - 3, we have:
-3 ≤ (1/2)(2) - 3
-3 ≤ 1 - 3
-3 ≤ -2
This statement is true, as -3 is less than or equal to -2.
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For the second inequality, y + 2x > 6, we have:
-3 + 2(2) > 6
-3 + 4 > 6
1 > 6
This statement is false, as 1 is not greater than 6.
Since the second inequality is not satisfied by point C (2, -3), point C is not a solution to the system of inequalities.
This case further reinforces the principle that a solution to a system of inequalities must satisfy every inequality in the system. The failure of point C to satisfy the second inequality disqualifies it as a solution, even though it satisfies the first inequality. This careful evaluation process is essential for accurately identifying solutions.
Testing Point D (4, 1)
Finally, let's test point D (4, 1) by substituting x = 4 and y = 1 into the inequalities:
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For the first inequality, y ≤ (1/2)x - 3, we have:
1 ≤ (1/2)(4) - 3
1 ≤ 2 - 3
1 ≤ -1
This statement is false, as 1 is not less than or equal to -1.
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For the second inequality, y + 2x > 6, we have:
1 + 2(4) > 6
1 + 8 > 6
9 > 6
This statement is true, as 9 is greater than 6.
Since the first inequality is not satisfied by point D (4, 1), point D is not a solution to the system of inequalities.
In this instance, point D fails to satisfy the first inequality, making it ineligible as a solution to the system. This final example underscores the consistency of the requirement that all inequalities must be satisfied for a point to be considered a solution. The thorough evaluation of each point is crucial for accurate problem-solving.
Conclusion
In conclusion, by substituting the coordinates of each point into the given system of inequalities, we found that only point A (5, -2) satisfies both inequalities. Therefore, point A is the correct solution to the system. This process of substitution and evaluation is a fundamental method for solving systems of inequalities and determining whether a given point lies within the solution region.
Understanding how to work with inequalities and systems of inequalities is essential for various mathematical applications, including optimization problems, linear programming, and constraint satisfaction. The ability to accurately identify solutions and understand the graphical representation of inequalities is a valuable skill in both academic and practical contexts. By mastering these concepts, students and practitioners can effectively tackle a wide range of problems involving inequalities.
The correct answer is A. (5,-2)
