Quadrants And Inequalities: Finding Solution Regions

Oct 21, 2025 by SLV Team 53 views

Hey guys! Let's dive into a fun math problem today that combines inequalities and coordinate geometry. We're going to figure out in which quadrants the solution region of a system of inequalities lies. This might sound a bit complex, but trust me, we'll break it down step by step. Understanding how inequalities define regions on a coordinate plane is super useful, not just in math class, but also in real-world applications like optimization problems and modeling different scenarios. So, grab your thinking caps, and let's get started!

Understanding the Problem

The problem we're tackling involves two inequalities: $y < 2x + 1$ and $y > rac{1}{2}x - 1$ . Our mission is to determine which quadrants on the xy-coordinate plane contain the region where both of these inequalities are true. Remember, a quadrant is one of the four regions of the plane divided by the x and y axes. Each quadrant has a specific combination of positive and negative x and y values:

Quadrant I: x > 0, y > 0
Quadrant II: x < 0, y > 0
Quadrant III: x < 0, y < 0
Quadrant IV: x > 0, y < 0

To solve this, we'll need to visualize the regions defined by each inequality and then find their intersection. This intersection represents the solution region for the system of inequalities. We need to consider each inequality separately and then combine their solutions. Think of it like this: each inequality creates a boundary line, and we need to figure out which side of each line satisfies the inequality. Where those sides overlap, that's our solution region. It's like finding the common ground between two conditions.

Graphing the Inequalities

First, let's graph the lines corresponding to the inequalities. To do this, we treat the inequalities as equations: $y = 2x + 1$ and $y = rac{1}{2}x - 1$ . These are both linear equations, so they represent straight lines on the graph. To graph a line, we need at least two points. Let's find some points for each line:

For $y = 2x + 1$ :

When x = 0, y = 2(0) + 1 = 1. So, the point (0, 1) is on the line.
When x = 1, y = 2(1) + 1 = 3. So, the point (1, 3) is on the line.

For $y = rac{1}{2}x - 1$ :

When x = 0, y = (1/2)(0) - 1 = -1. So, the point (0, -1) is on the line.
When x = 2, y = (1/2)(2) - 1 = 0. So, the point (2, 0) is on the line.

Now, plot these points on the xy-plane and draw the lines. Remember, because our original inequalities are strict inequalities (using < and >), we'll draw dashed lines to indicate that the points on the lines themselves are not part of the solution. If the inequalities were ≤ or ≥, we'd use solid lines. The dashed lines are like boundaries that the solution region approaches but doesn't include.

Determining the Solution Region

Next, we need to figure out which side of each line represents the solution to the inequality. We can do this by testing a point that is not on the line. A convenient point to test is often the origin (0, 0). Let's test it for each inequality:

For $y < 2x + 1$ :

Substitute x = 0 and y = 0: 0 < 2(0) + 1, which simplifies to 0 < 1. This is true!
Since (0, 0) satisfies the inequality, the solution region for this inequality is the area below the line $y = 2x + 1$ .

For $y > rac{1}{2}x - 1$ :

Substitute x = 0 and y = 0: 0 > (1/2)(0) - 1, which simplifies to 0 > -1. This is also true!
Since (0, 0) satisfies this inequality as well, the solution region for this inequality is the area above the line $y = rac{1}{2}x - 1$ .

Now, we need to find the region where both inequalities are satisfied. This is the area where the two solution regions overlap. Shade the region below the line $y = 2x + 1$ and above the line $y = rac{1}{2}x - 1$ . The overlapping area is the solution region for the system of inequalities. It's like finding the sweet spot where both conditions are met. This overlapping region is crucial for solving the problem.

Identifying Quadrants in the Solution Region

Now that we've identified the solution region, we need to determine which quadrants it lies in. By looking at the graph, we can see that the solution region extends into:

Quadrant I: In this quadrant, both x and y are positive. The solution region clearly extends into this quadrant.
Quadrant II: In this quadrant, x is negative, and y is positive. The solution region also extends into this quadrant.
Quadrant III: In this quadrant, both x and y are negative. A portion of the solution region falls within Quadrant III.

Quadrant IV, where x is positive and y is negative, does not contain any part of the solution region. To confirm this visually, imagine drawing lines from the solution region to the x and y axes. If the lines intersect both axes with the signs corresponding to a specific quadrant, then the solution region extends into that quadrant.

Analyzing the Given Options

The problem provides us with four statements about the quadrants:

(1) I (x > 0, y > 0) (2) II (x < 0, y > 0) (3) III (x < 0, y < 0) (4) IV (x > 0, y < 0)

Based on our analysis, the solution region lies in Quadrants I, II, and III. Therefore, statements (1), (2), and (3) are correct. Statement (4) is incorrect because the solution region does not extend into Quadrant IV.

The options presented are in the form:

(A) (1), (2), and (3) ONLY are correct. (B) ... (Other options)

Since we've determined that statements (1), (2), and (3) are the only correct ones, the answer is (A).

Final Answer and Key Takeaways

So, the final answer is (A) (1), (2), and (3) ONLY are correct.

Let's recap the key steps we took to solve this problem:

Graphing the Inequalities: We transformed the inequalities into equations and graphed the corresponding lines. Remember to use dashed lines for strict inequalities (< and >) and solid lines for inclusive inequalities (≤ and ≥).
Determining the Solution Region: We tested a point (usually the origin) in each inequality to determine which side of the line represented the solution. Then, we identified the overlapping region where both inequalities were satisfied.
Identifying Quadrants: We visually inspected the solution region and determined which quadrants it extended into.
Analyzing Options: We compared our findings with the given statements and selected the option that accurately described the quadrants containing the solution region.

This problem highlights the importance of understanding how to graph inequalities and how they define regions on the coordinate plane. It's a fundamental concept in algebra and calculus, so mastering it will definitely help you in your future math adventures! Keep practicing, guys, and you'll become pros at solving these types of problems. Remember, math is like a puzzle – challenging, but super rewarding when you crack the code!