Problem 48 - A Graphical Approach to Precalculus with Limits

Question

Graph the solution set of each system of inequalities by hand. $$\begin{aligned}&x+y \leq 4\\\&x-2 y \geq 6\end{aligned}$$

Step-by-Step Solution

Answer

Shade the region where the areas below \(x + y = 4\) and above \(x - 2y = 6\) overlap.

Step 1: Convert Inequalities to Equations

Begin by considering each inequality as an equation.1. Convert the inequality \(x + y \leq 4\) to the equation \(x + y = 4\).2. Convert \(x - 2y \geq 6\) to the equation \(x - 2y = 6\). These equations help us find the boundary lines for the inequalities.

Step 2: Graph the First Equation

Graph the line represented by \(x + y = 4\):1. Find the x-intercept by setting \(y = 0\): \[x + 0 = 4 \Rightarrow x = 4\]2. Find the y-intercept by setting \(x = 0\): \[0 + y = 4 \Rightarrow y = 4\]Plot the points (4,0) and (0,4) and draw a solid line through them, as the inequality is \(\leq\).

Step 3: Graph the Second Equation

Graph the line represented by \(x - 2y = 6\):1. Find the x-intercept by setting \(y = 0\): \[x - 2(0) = 6 \Rightarrow x = 6\]2. Find the y-intercept by setting \(x = 0\): \[0 - 2y = 6 \Rightarrow y = -3\]Plot the points (6,0) and (0,-3) and draw a solid line through them, as the inequality is \(\geq\).

Step 4: Determine Shaded Regions

For \(x + y \leq 4\), choose a test point not on the line (e.g., (0,0)):\[0 + 0 \leq 4\] is true, so shade the region below or on the line.For \(x - 2y \geq 6\), use the same test point:\[0 - 2(0) \geq 6\] is false, so do not shade the side containing (0,0); shade the opposite side, above this line.

Step 5: Identify and Shade the Solution Region

The solution set of the system is the region where the shaded areas of both inequalities intersect. This is the overlapping region on the graph, satisfying both \(x + y \leq 4\) and \(x - 2y \geq 6\). Shade this intersection area clearly.

Key Concepts

Graphing Inequalities

When graphing inequalities, you start by rewriting each inequality statement as an equation. This gives you a boundary line that assists in visualizing which region of the graph satisfies the inequality. For example, the inequality \(x + y \leq 4\) is first turned into the equation \(x + y = 4\). Similarly, \(x - 2y \geq 6\) becomes \(x - 2y = 6\). Processing it this way allows you to plot the lines on a coordinate plane directly. Once the lines are plotted, use solid lines for \(\leq\) or \(\geq\), indicating that points on the line fulfill the condition too. If the inequality were \( < \) or \( > \), you'd use a dashed line, showing that points on the line don't satisfy the inequality.

Intersection of Regions

Finding the intersections of regions means identifying where two or more shaded areas, which represent the solution sets of individual inequalities, overlap on your graph. This intersected region is the solution set to the system of inequalities. Shade each region as per the rules of the inequalities:

• The area that satisfies \(x + y \leq 4\) is plotted below or on the line \(x + y = 4\).
• The area fulfilling \(x - 2y \geq 6\) appears above or on the line \(x-2y = 6\).

The overall solution is the section where these shaded areas overlap. Only this section will satisfy both inequalities simultaneously.

Boundary Lines

Boundary lines in a system of inequalities define the limits of the solution region. These lines originate from converting inequalities to equations. For \(x+y\leq4\) and \(x-2y\geq6\), the boundary lines are \(x+y=4\) and \(x-2y=6\). To plot these:

• Find intercepts. For example, the x-intercept of \(x+y=4\) is found when \(y=0\), giving \((4,0)\). Similarly, the y-intercept is when \(x=0\), yielding \((0,4)\).
• Connect these points with a straight line. Do the same for \(x-2y=6\) with its intercepts: \((6,0)\) and \((0,-3)\).

Solid lines show that boundary is part of the solution; dashed lines exclude this boundary. Since these lines use \(\leq\) and \(\geq\), draw them solidly. This helps to build clarity when you're determining which regions to overlap for the solution.

Test Points (System of Inequalities)

Using test points will help you decide which side of the boundary line contains the solutions to an inequality. Choose a point that is not on the boundary line, generally the origin \((0, 0)\) if it's not on any line, to verify each inequality.If \((0,0)\) satisfies an inequality, shade the region inclusive of this point. For instance, with \(x+y\leq4\), \(0+0\leq4\) is true, so shade where \((0,0)\) lies. For \(x-2y\geq6\), \(0-2(0)\geq6\), which is false, indicating the region opposite of where \((0,0)\) is should be shaded.Test points simplify the process by providing a quick check to confirm which side of the boundary aligns with the inequality, helping you efficiently shade the correct area.