Problem 39 - Algebra and Trigonometry

Question

19–40 Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned} x^{2}+y^{2} &<9 \\ x+y &>0 \\ x & \leq 0 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Vertices: (0, 0) and \((-3/\sqrt{2}, 3/\sqrt{2})\); Solution set is bounded.

Step 1: Graph the first inequality

The inequality \(x^2 + y^2 < 9\) represents the interior of a circle centered at the origin \((0, 0)\) with a radius of 3. Start by drawing a dashed circle with radius 3 centered at the origin. The dashed boundary indicates that points on the circle are not included in the solution.

Step 2: Graph the second inequality

The inequality \(x + y > 0\) can be rewritten as \(y > -x\). This represents the half-plane above the line \(y = -x\). Draw the line \(y = -x\) with a dashed line to indicate that points on this line are not part of the solution, and shade the region above it.

Step 3: Graph the third inequality

The inequality \(x \leq 0\) represents the left half of the coordinate plane (the half-plane where \(x\) is negative or zero). Shade the region to the left of the \(y\)-axis including the axis itself.

Step 4: Find the solution region

The solution to the system of inequalities is the region where all shaded areas from Steps 1-3 overlap. This region is the intersection of the interior of the circle, the half-plane above the line \(y = -x\), and the left half-plane \(x \leq 0\).

Step 5: Identify the vertices

The vertices of the solution region are the points of intersection of the boundaries of these inequalities that lie within the solution set. Intersection points include \((0, 0)\) and points on the circle where \(x + y = 0\) and \(x \leq 0\). Calculate points on the circle by solving \(x^2 + y^2 = 9\) and \(x + y = 0\) to get \((-3/\sqrt{2}, 3/\sqrt{2})\) and \((3/\sqrt{2}, -3/\sqrt{2})\). Only the one with \(x \leq 0\) is included, which is \((-3/\sqrt{2}, 3/\sqrt{2})\).

Step 6: Determine if the solution set is bounded

A set is bounded if it can be contained within a circle of finite radius. This solution set is within a circle of radius 3, and hence it is bounded.

Key Concepts

Graphing Inequalities

Graphing inequalities involves shading areas on a graph where the solutions to the inequality are found. In this exercise, the first inequality is a circle, described by \(x^2 + y^2 < 9\), which represents a circle centered at the origin with a radius of 3. When graphing this, use a dashed circle to show that points directly on the circle are not included in the solution. The second inequality, \(x + y > 0\), or equivalently \(y > -x\), requires shading the half-plane above the line \(y = -x\). A dashed line indicates that solutions do not include points on the line. Lastly, \(x \leq 0\) means shading all areas to the left of the y-axis including the line itself. The graphical representation of an inequality shows all possible solutions that satisfy the inequality conditions.

Remember, using dashed lines is crucial to represent strict inequalities, ensuring correct visualization of the solution set.

Solution Region

The solution region for a system of inequalities is where the shaded regions for each individual inequality overlap. In our exercise, this region is a common area satisfying \(x^2 + y^2 < 9\), \(x + y > 0\), and \(x \leq 0\). To find this overlapping region on the graph, identify where the shaded areas from each inequality intersect. This involves visualizing:

• The interior of a circle with radius 3
• The half-plane above the line \(y = -x\)
• The entire space to the left of the y-axis

The triple intersection of these areas gives all possible solutions for the system of inequalities. This part of the plane is where all initial constraints are true simultaneously.

When solving graphically, it's beneficial to systematically draw and overlay each piece to ensure you capture the correct solution region.

Bounded Solution

A solution is bounded if it can fit within a closed region—a finite area, such as a circle. In this system of inequalities, the solution set is contained within the interior of a circle of radius 3. Thus, it is bounded. A bounded solution implies there are limits to how far solutions extend in the geometric space. This notion is crucial as bounded sets are contained, preventing solutions from stretching infinitely.

This property can be graphically confirmed by noting any finite boundary that completely encloses the solution region, such as the dashed circle indicating the area contained by \(x^2 + y^2 < 9\). For students, visual checkpoints like this help determine boundaries without calculating extensive coordinates.

Vertices of Inequalities

Vertices in a solution region are points where the boundary lines of different inequalities intersect. They are the corner points of the solution space. In the context of our exercise, finding these vertices involves looking where the boundaries of each inequality meet.

The point \((0,0)\) is a clear vertex, being the intersection of the line \(x+y=0\) and the y-axis. Another potential vertex is where the circle \(x^2 + y^2 = 9\) intersects with the line \(x + y = 0\). Solving these gives points like \((-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}})\) and \((\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}})\), of which only the point with a negative x-coordinate fits within the half-plane \(x \leq 0\). Calculating these coordinates enhances understanding by showing geometrical relationships between mathematical expressions.