Problem 28 - Algebra and Trigonometry

Question

21-46 . Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$ \left\\{\begin{aligned} x &>2 \\ y &<12 \\ 2 x-4 y &>8 \end{aligned}\right. $$

Step-by-Step Solution

Answer

Vertices: (2, -1), (2, 12), (28, 12). The solution set is unbounded.

Step 1: Rewrite the Inequalities

Rewrite the given inequalities in a form that's easier to graph. The system given is: 1. \( x > 2 \)2. \( y < 12 \)3. \( 2x - 4y > 8 \). The third inequality can be rewritten in slope-intercept form as follows: 3. Rearrange terms: \( 2x - 8 > 4y \) 4. Solve for \( y \): \( y < \frac{1}{2}x - 2 \). Your system is now: 1. \( x > 2 \)2. \( y < 12 \)3. \( y < \frac{1}{2}x - 2 \).

Step 2: Graph the Inequalities

Graph each inequality on a coordinate plane:1. Draw the line \( x = 2 \). Shade the region to the right of this line, representing \( x > 2 \).2. Draw the line \( y = 12 \). Shade the region below this line, representing \( y < 12 \).3. Draw the line \( y = \frac{1}{2}x - 2 \). Shade the region below this line, representing \( y < \frac{1}{2}x - 2 \). Note that this line has a slope of \( \frac{1}{2} \) and a y-intercept of \(-2\).Your solution set is the area where all shaded regions overlap.

Step 3: Find the Intersection Points

Find the intersection points of the lines to identify the vertices of the solution area:1. **Intersection of \( x = 2 \) and \( y = \frac{1}{2}x - 2 \)**: Substitute \( x = 2 \) into \( y = \frac{1}{2}x - 2 \) to get \( y = \frac{1}{2}(2) - 2 = -1 \). So, the intersection point is \((2, -1)\).2. **Intersection of \( x = 2 \) and \( y = 12 \)**: Since \( x = 2 \), the point is \((2, 12)\).3. **Intersection of \( y = 12 \) and \( y = \frac{1}{2}x - 2 \)**: Set \( 12 = \frac{1}{2}x - 2 \), solve for \( x \): \[ 12 + 2 = \frac{1}{2}x \] \[ 14 = \frac{1}{2}x \] \[ x = 28 \]. So, the intersection point is \((28, 12)\).

Step 4: Determine if the Solution Set is Bounded

The vertices found are \((2, -1)\), \((2, 12)\), and \((28, 12)\). Visually inspect the graphed area formed by these vertices:- The region is open to the right since there are no constraints on such a boundary.- Therefore, the solution set is unbounded.

Key Concepts

Graphing Inequalities

When tackling a system of inequalities, graphing helps us visualize the solution space. Each inequality divides the plane into two halves: one where the inequality holds true, and one where it doesn't. Let's break it down using the example inequalities.

• The inequality \(x > 2\) is represented by a vertical line at \(x = 2\), and we shade the region to the right, as this is where \(x\) values are greater than 2.
• The inequality \(y < 12\) is shown with a horizontal line at \(y = 12\), shading below the line highlights the area where \(y\) values are less than 12.
• In the transformed form \(y < \frac{1}{2}x - 2\), we have a line with a slope of 1/2 and a y-intercept at -2. Shading below this line indicates the region where the inequality holds true.

By graphing these lines and combining the shaded regions, we find where all conditions are satisfied simultaneously. This overlap is the solution set of the system.

Intersection Points

Finding intersection points is key to identifying the vertices of the solution area. These points are where the boundary lines of different inequalities meet.

• First, for the intersection of \(x = 2\) and \(y = \frac{1}{2}x - 2\), substitute \(x = 2\) into the inequality to get \(y = \frac{1}{2}(2) - 2 = -1\). So, one vertex is at \((2, -1)\).
• For the lines \(x = 2\) and \(y = 12\), substituting directly gives us the point \((2, 12)\).
• Finally, to find where \(y = 12\) intersects \(y = \frac{1}{2}x - 2\), set these equal to each other: \(12 = \frac{1}{2}x - 2\). Solving for \(x\), we find \(x = 28\), leading to the point \((28, 12)\).

These points, once plotted, form the vertices of a polygonal region representing the solution set, and help in determining its bounds.

Bounded and Unbounded Sets

The concept of bounded and unbounded sets in a graph of inequalities refers to whether the solution area is contained within finite boundaries or extends infinitely in one or more directions.

• A bounded set is a closed polygon where all inequalities restrict the graph into a finite region. It means there's both a maximum and minimum limit on the graph's extent.
• An unbounded set, however, means the solution area extends indefinitely. In our case, the region formed by the intersection points \((2, -1)\), \((2, 12)\), and \((28, 12)\) is open to the right without any boundary.

Since no inequality limits the set's extent to the right, we conclude that this particular solution set is unbounded. Understanding whether a solution is bounded or unbounded can influence how we approach the problem and interpret the results.