Problem 36 - 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. $$\left\\{\begin{aligned} x+y &>12 \\ y &<\frac{1}{2} x-6 \\ 3 x+y &<6 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution set is bounded with vertices at \((12, 0)\), \((-3, 15)\), and \(\left(\frac{24}{7}, -\frac{18}{7}\right)\).

Step 1: Write the inequalities

First, we list the given system of inequalities: \(x + y > 12\), \(y < \frac{1}{2}x - 6\), \(3x + y < 6\). Our task is to graph these inequalities and find the solution set.

Step 2: Graph the first inequality

The first inequality is \(x + y > 12\). To graph \(x + y = 12\) as a line, choose points such that the equation holds, for instance: when \(x = 12, y = 0\) and when \(x = 0, y = 12\). Draw a dashed line through \((12, 0)\) and \((0, 12)\). Since the inequality is \(>\), shade the region above this line.

Step 3: Graph the second inequality

The second inequality is \(y < \frac{1}{2}x - 6\). Graph the line \(y = \frac{1}{2}x - 6\) using intercepts or slope. For this line, when \(x = 0, y = -6\) and when \(x = 12, y = 0\); draw a dashed line through these points. Since the inequality is \(<\), shade the region below this line.

Step 4: Graph the third inequality

The third inequality is \(3x + y < 6\). Graph the line \(3x + y = 6\). Using points: when \(x = 0, y = 6\) and when \(x = 2, y = 0\). Draw a dashed line through these points. Since the inequality is \(<\), shade the region below this line.

Step 5: Find intersections of lines

To find the vertices of the solution set, solve the system of equations at the line intersections: \(x + y = 12\), \(y = \frac{1}{2}x - 6\), and \(3x + y = 6\). Solve pairs of these equations to find points of intersection.

Step 6: Calculate vertices coordinates

1. Solve \(x + y = 12\) and \(y = \frac{1}{2}x - 6\) for intersection. Substituting \(y = \frac{1}{2}x - 6\) into \(x + y = 12\) gives \(x + \frac{1}{2}x - 6 = 12\), or \(\frac{3}{2}x = 18\), hence \(x = 12\). Then \(y = \frac{1}{2}(12) - 6 = 0\), leading to the vertex \((12, 0)\).2. Solve \(x + y = 12\) and \(3x + y = 6\) for intersection. Subtract \(x+y=12\) from \(3x+y=6\), yielding \(2x=-6\), hence \(x = -3\). With \(x = -3\), \(y = 15\) from \(x + y = 12\), resulting in \((-3, 15)\).3. Solve \(3x + y = 6\) and \(y = \frac{1}{2}x - 6\). Substitute \(y = \frac{1}{2}x - 6\) into \(3x + y = 6\), gives \(3x + \frac{1}{2}x - 6 = 6\), leading \(\frac{7}{2}x = 12\), hence \(x = \frac{24}{7}\). Substitute \(x = \frac{24}{7}\) into \(y = \frac{1}{2}x - 6\), results in \(y = -\frac{18}{7}\). This vertex is \(\left(\frac{24}{7}, -\frac{18}{7}\right)\).

Step 7: Determine if the solution set is bounded

The solution set is the overlapping shaded region from all the inequalities' graphs. Since the intersecting region forms a closed polygon defined by vertices \((12, 0)\), \((-3, 15)\), and \(\left(\frac{24}{7}, -\frac{18}{7}\right)\), the solution set is bounded.

Key Concepts

Graphing Inequalities

Graphing inequalities involves several steps to visually represent their solutions on a coordinate plane. To graph an inequality like \( x + y > 12 \), start by graphing the line \( x + y = 12 \) as if it were an equality. Choose any two points that satisfy the equation to plot the line. For instance, if you let \( x = 12 \), then \( y = 0 \), giving you the point \( (12, 0) \). Similarly, for \( x = 0 \), \( y = 12 \), resulting in the point \( (0, 12) \). Connect these points with a dashed line because the inequality is strict \((>)\) and doesn't include the boundary.

For the inequality itself, you need to shade one side of the line. Since \( x + y > 12 \), shade above the line to show that all points in this region satisfy the inequality. Remember, the approach is repeatable for any inequality: graph the boundary line and determine which side of the line should be shaded based on whether the solution is greater than or less than.

Coordinates of Vertices

In the context of inequalities, vertices are points where the boundary lines intersect. They form the corners of the solution set. To find the coordinates of these vertices, you need to solve pairs of equations from the lines of the inequalities. Let's illustrate this process with a couple of pairs:

- For lines \( x + y = 12 \) and \( y = \frac{1}{2}x - 6 \), substitute one equation into the other. This procedure will reveal common solutions. Substitute \( y = \frac{1}{2}x - 6 \) into \( x + y = 12 \) to solve for \( x \), and then solve for \( y \), giving \((12, 0)\).- Another intersection involves \( x + y = 12 \) and \( 3x + y = 6 \). Subtract the first equation from the second to ascertain \( x \) and consequently \( y \), resulting in \((-3, 15)\).- Finally, solving \( 3x + y = 6 \) and \( y = \frac{1}{2}x - 6 \) together will lead to \( \left(\frac{24}{7}, -\frac{18}{7}\right) \).

These vertices pinpoint where lines meet and define the edges of the solution's feasible region.

Bounded Solution Set

The solution set of a system of inequalities is found by locating the region on the graph where all the conditions are jointly satisfied. In the case of the system \( \{x + y > 12, y < \frac{1}{2}x - 6, 3x + y < 6\} \), the region where all shaded areas overlap represents the solution set.

A bounded solution set forms a closed shape on the graph, typically a polygon. For our specific problem, the vertices identified earlier, \((12, 0)\), \((-3, 15)\), and \(\left(\frac{24}{7}, -\frac{18}{7}\right)\), form a triangle. This closed region indicates a bounded solution set because it's completely contained within the plane defined by the boundaries of the intersecting inequalities.

A bounded set means there is a finite space that fulfills all the inequalities, unlike an unbounded set which would extend indefinitely in one or more directions.