Problem 23 - Algebra 2

Question

Find the coordinates of the vertices of the figure formed by each system of inequalities. $$ \begin{array}{l}{x+y \leq 9} \\ {x-2 y \leq 12} \\ {y \leq 2 x+3}\end{array} $$

Step-by-Step Solution

Answer

The only vertex satisfying all inequalities is (10, -1).

Step 1: Identify the System of Inequalities

We start by listing the system of inequalities given:\[\begin{align*}x + y &\leq 9, \x - 2y &\leq 12, \y &\leq 2x + 3.\end{align*}\]

Step 2: Convert Inequalities to Equations

For each inequality, convert it to an equation by replacing the inequality sign with an equality sign:1. \(x + y = 9\)2. \(x - 2y = 12\)3. \(y = 2x + 3\).

Step 3: Find Points of Intersection

Find points where each pair of lines intersect by solving the equations simultaneously.1. Solve \(x + y = 9\) and \(x - 2y = 12\): - Rearrange and substitute: \(x = 12 + 2y\); substituting in \(x + y = 9\) gives: \[ 12 + 2y + y = 9 \quad \Rightarrow \quad 3y = -3 \quad \Rightarrow \quad y = -1 \] - Substitute \(y = -1\) back: \(x = 12 + 2(-1) = 10\). - First intersection point: \((10, -1)\).2. Solve \(x + y = 9\) and \(y = 2x + 3\): - Substitute: \(y = 2x + 3\) in \(x + y = 9\): \[ x + 2x + 3 = 9 \quad \Rightarrow \quad 3x = 6 \quad \Rightarrow \quad x = 2 \] - Substitute \(x = 2\) back: \(y = 2(2) + 3 = 7\). - Second intersection point: \((2, 7)\).3. Solve \(x - 2y = 12\) and \(y = 2x + 3\): - Substitute: \(y = 2x + 3\) in \(x - 2y = 12\): \[ x - 2(2x + 3) = 12 \quad \Rightarrow \quad x - 4x - 6 = 12 \quad \Rightarrow \quad -3x = 18 \quad \Rightarrow \quad x = -6 \] - Substitute \(x = -6\) back: \(y = 2(-6) + 3 = -9\). - Third intersection point: \((-6, -9)\).

Step 4: Verify and Identify Feasible Region

Check the vertices from Step 3 to see which satisfy all the original inequalities.- Vertex \((10, -1)\): - **Check:** - \(10 + (-1) = 9\leq 9\) - \(10 - 2(-1) = 12 \leq 12\) - \((-1) \leq 2(10) + 3\) holds. - It satisfies all inequalities.- Vertex \((2, 7)\): - **Check:** - \(2 + 7 \leq 9\) fails.- Vertex \((-6, -9)\): - **Check:** - \(-6 + (-9)=-15 \leq 9\) - \(-6 - 2(-9)=12\) - \(-9 \leq 2(-6) + 3\) fails.Only the vertex \((10, -1)\) satisfies all inequalities.

Key Concepts

Graphical Solution

A graphical solution involves plotting each of the inequalities on a coordinate plane. Understanding the visual representation of inequalities is key to solving them.
Each inequality divides the plane into a region that satisfies the inequality and one that does not. When representing inequalities graphically, the equal parts form boundary lines. For example:

• For the inequality \(x + y \leq 9\), the line \(x + y = 9\) is a boundary.
• Similarly, \(x - 2y \leq 12\) has the boundary line \(x - 2y = 12\).
• Lastly, \(y \leq 2x + 3\) has the boundary line \(y = 2x + 3\).

To solve it graphically, draw each boundary on the graph. Use shading to show where the inequality holds. The overlapping area, where all shadings intersect, is the solution set or feasible region.

Intersection Points

Intersection points are where two lines cross on a graph. These points are crucial because they can represent potential solutions to systems of inequalities. Let's look at how to find them:
When two lines intersect, they have the same \(x\) and \(y\) coordinates simultaneously. Hence, solving the equations of both lines at the same time helps locate these points.

• For example, \(x + y = 9\) and \(x - 2y = 12\) intersect at \((10, -1)\).
• Similarly, \(x + y = 9\) and \(y = 2x + 3\) intersect at \((2, 7)\).
• Lastly, \(x - 2y = 12\) and \(y = 2x + 3\) intersect at \((-6, -9)\).

These coordinates highlight potential vertices of the feasible region. But only those within all shaded regions serve as solutions.

Feasible Region

The feasible region is the section of the graph where all inequalities overlap. In simpler terms, it's the set of all solutions that satisfy everyone inequality in the system. This region is usually a polygon or an open region on the graph.
To locate the feasible region, graph each inequality and determine where they coincide. The vertices of the feasible region are points where the boundary lines of these inequalities intersect.

• First, draw the lines for the equations that were derived from the inequalities.
• Then, identify where each solution set overlaps as this will form the feasible region.
• Finally, check if the points where the lines intersect fall within the initial shaded regions.

For this exercise, only the intersection point \((10, -1)\) was a valid vertex of the feasible region as it satisfied all the initial inequalities.

Inequality Equations

Inequality equations are simply mathematical expressions where one side is not necessarily equal to the other but instead is less than, greater than, less than or equal to, or greater than or equal to the other side. This is how:

• In the expression \(x + y \leq 9\), the sum of \(x\) and \(y\) is less than or equal to 9.
• Similarly, \(x - 2y \leq 12\) suggests that \(x\) minus twice \(y\) mustn't exceed 12.
• Finally, \(y \leq 2x + 3\) indicates that \(y\) must be at most twice \(x\) summed with 3.

To determine their graphical solution, these inequalities are converted into equations for plotting. Reinterpreting them with an equality symbol allows these lines to serve as boundaries on the graph, crucial for discovering the feasible region.