Problem 48 - Algebra and Trigonometry

Question

\(47-50\) . Use a graphing calculator to graph the solution of the system of inequalities. Find the coordinates of all vertices, rounded to one decimal place. $$ \left\\{\begin{aligned} x+y & \geq 12 \\ 2 x+y & \leq 24 \\ x-y & \geq-6 \end{aligned}\right. $$

Step-by-Step Solution

Answer

Vertices: (12, 0), (6, 12), (3, 9).

Step 1: Graph the Region of Each Inequality

First, we need to individually graph each inequality. - **Inequality 1**: \( x + y \geq 12 \) can be rewritten as \( y \geq 12 - x \). Graph the line \( y = 12 - x \) with a solid line because the inequality includes \( \geq \), and shade the region above the line.- **Inequality 2**: \( 2x + y \leq 24 \) can be rewritten as \( y \leq 24 - 2x \). Graph the line \( y = 24 - 2x \) with a solid line and shade the region below the line.- **Inequality 3**: \( x - y \geq -6 \) can be rewritten as \( y \leq x + 6 \). Graph the line \( y = x + 6 \) with a solid line and shade the region below the line.

Step 2: Determine the Intersection Region

The solution to the system of inequalities is the region where all shaded areas overlap. Check where the shaded regions of \( y \geq 12 - x \), \( y \leq 24 - 2x \), and \( y \leq x + 6 \) intersect by using a graphing calculator.

Step 3: Find Vertices of the Intersection Region

Calculate the coordinates of the vertices by solving the equations formed by intersecting lines:- **Intersection of Line 1 and Line 2**: \[ x + y = 12 \] \[ 2x + y = 24 \] Subtract the first equation from the second: \[ (2x + y) - (x + y) = 24 - 12 \] \[ x = 12 \] Substitute \( x = 12 \) into the first equation: \[ 12 + y = 12 \Rightarrow y = 0 \] Vertex: \( (12, 0) \)- **Intersection of Line 2 and Line 3**: \[ 2x + y = 24 \] \[ x - y = -6 \] Add equations: \[ 3x = 18 \Rightarrow x = 6 \] Substitute \( x = 6 \) into \( x - y = -6 \): \[ 6 - y = -6 \Rightarrow y = 12 \] Vertex: \( (6, 12) \)- **Intersection of Line 1 and Line 3**: \[ x + y = 12 \] \[ x - y = -6 \] Add equations: \[ 2x = 6 \Rightarrow x = 3 \] Substitute \( x = 3 \) into \( x + y = 12 \): \[ 3 + y = 12 \Rightarrow y = 9 \] Vertex: \( (3, 9) \)

Step 4: List Vertices of the Feasible Region

The vertices of the feasible region, where all inequalities intersect, are found to be at the points \((12, 0)\), \((6, 12)\), and \((3, 9)\). These coordinates represent the corners of the intersection region in this system of inequalities.

Key Concepts

graphing inequalities

To solve a system of inequalities like the one at hand, the first step is graphing each individual inequality on a coordinate plane.
This method visually represents the solutions to each inequality and helps us identify the region where all three intersect. Let’s break down how to graph each inequality correctly:

• **Inequality 1:** We have the inequality \(x + y \geq 12\). This can be transformed for graphing purposes into \(y \geq 12 - x\). You would draw the line \(y = 12 - x\) as a solid line because the inequality is \(\geq\), indicating that points on the line are included in the solution. The region above this line satisfies the inequality, so you’d shade above the line.
• **Inequality 2:** The inequality \(2x + y \leq 24\) becomes \(y \leq 24 - 2x\) for graphing. The line \(y = 24 - 2x\) is drawn solid, and the region below this line is shaded, as it represents all the solutions to this inequality.
• **Inequality 3:** With \(x - y \geq -6\), rearrange this to \(y \leq x + 6\). Graph the line \(y = x + 6\) solidly, and shade the area below, as this contains the solutions for the third inequality.

Once each inequality has been graphed, the trick is to find the overlapping shaded region, which represents all the points that satisfy all three inequalities.

vertices of the region

Identifying the vertices of the overlapping region is a key goal when solving systems of inequalities. These vertices are the corner points of the feasible region and can be found through solving systems of equations formed by the intersection of the lines from graphing the inequalities.
After graphing, you'll notice that the shaded areas overlap, creating a polygon-shaped feasible region. The vertices of this polygon can be calculated by where two lines meet (intersect each other). Let's see how you can do this:

• **Vertex at (12, 0):** This point comes from the intersection of the lines defined by the equations \(x + y = 12\) and \(2x + y = 24\). By solving these equations, we found \((12, 0)\).
• **Vertex at (6, 12):** Achieving this by solving \(2x + y = 24\) and \(x - y = -6\), which yields the point \((6, 12)\).
• **Vertex at (3, 9):** This vertex is the intersection of \(x + y = 12\) and \(x - y = -6\), giving us the point \((3, 9)\).

By rounding these vertices to one decimal place, which are already integers in this example, you ensure that your final answer is accurate. They outline the shape of the feasible region produced by the system of inequalities.

intersection of lines

Finding the intersection of lines is critical in solving a system of inequalities because it helps determine the vertices of the overlapped, feasible region. Whenever you have equations of lines, like in our problem, those lines intersect at points that mark key solutions to the inequality system.
Here's how to find such intersection points:

• Algebraically, to find where two lines cross, set their equations equal to one another if they have one plane and solve for the variables. Inequalities such as \(x + y = 12\) and \(2x + y = 24\) can be solved simultaneously to find intersections.
• This involves either substitution or elimination methods, where subtraction, addition, or substitution of the equations facilitates finding exact points.
• For instance, using elimination on \(2x + y = 24\) and \(x + y = 12\) gives you \(x = 12\) and upon substitution, \(y = 0\). These cleaned algebraic results pinpoint intersections precisely like \((12, 0)\).

Such intersection points reveal themselves as vertices of your solution area, tying together methodical graphing with tangible algebra solutions.