Graphical Solution Of Inequalities: Find The Solution Set

Graphical Solution of Inequalities: Find the Solution Set

Hey there, math enthusiasts! Today, we're diving into the visual world of solving systems of inequalities graphically. It's a fantastic way to understand the regions where multiple conditions are met simultaneously. We'll be tackling a specific problem: graphing the system $\begin{array}{l} y \geq \frac{5}{2} x-8 \\ y \leq-x-1 \end{array}$. This process not only helps us visualize the solution but also allows us to pinpoint specific points that satisfy all the given inequalities. So, grab your pencils, rulers, and let's get ready to transform abstract inequalities into a concrete graphical representation!

Understanding the Components of Our System

Before we jump into graphing, let's break down what each inequality represents. Our system consists of two linear inequalities: $y \geq \frac{5}{2} x-8$ and $y \leq-x-1$. Each of these inequalities defines a region in the Cartesian plane. The solution set of the system will be the area where these two regions overlap. Think of it like finding the intersection of two shaded areas on a map; the intersection is where both conditions are true. Let's analyze the first inequality, $y \geq \frac{5}{2} x-8$. This is a linear inequality, and its boundary line is given by the equation $y = \frac{5}{2} x-8$. The inequality sign '$\\geq$' tells us that we need to include the line itself in our solution, and the region above the line.

• The Boundary Line: To graph $y = \frac{5}{2} x-8$, we can identify its slope and y-intercept. The slope ($m$) is $\frac{5}{2}$, meaning for every 2 units we move to the right on the x-axis, we move 5 units up on the y-axis. The y-intercept ($b$) is -8, which is the point where the line crosses the y-axis (0, -8). We can plot this point and then use the slope to find other points on the line. For example, starting from (0, -8), moving 2 units right and 5 units up brings us to (2, -3). Moving another 2 units right and 5 units up gives us (4, 2), and so on. We can also go in the opposite direction: moving 2 units left and 5 units down from (0, -8) gives us (-2, -13).
• The Shaded Region: Since the inequality is $y \geq \frac{5}{2} x-8$, we are looking for all points (x, y) where the y-coordinate is greater than or equal to the value of $\frac{5}{2} x-8$. This means we shade the region above the boundary line. To confirm this, we can pick a test point not on the line, for instance, (0, 0). Plugging this into the inequality: $0 \geq \frac{5}{2}(0) - 8$, which simplifies to $0 \geq -8$. This statement is true, so the region containing (0, 0) – which is above the line – is the correct region to shade.

Now, let's look at the second inequality: $y \leq-x-1$. Similar to the first, this is also a linear inequality. The boundary line is $y = -x-1$, and the inequality sign '$\\leq$' indicates that we include the line and shade the region below it.

• The Boundary Line: For $y = -x-1$, the slope ($m$) is -1 (meaning for every 1 unit right, we go 1 unit down), and the y-intercept ($b$) is -1 (the point (0, -1)). We can plot (0, -1) and then use the slope to find other points. For example, moving 1 unit right and 1 unit down from (0, -1) gives us (1, -2). Moving another unit right and down gives us (2, -3), and so on. Going left and up: moving 1 unit left and 1 unit up from (0, -1) gives us (-1, 0).
• The Shaded Region: Because the inequality is $y \leq -x-1$, we need points where the y-coordinate is less than or equal to the value of $-x-1$. This means we shade the region below the boundary line. Let's use (0, 0) as a test point again: $0 \leq -(0) - 1$, which simplifies to $0 \leq -1$. This statement is false. Therefore, the region containing (0, 0) is not part of the solution, and we shade the region below the line $y = -x-1$.

By understanding these individual components, we are now well-equipped to graph the entire system and identify the overlapping region that represents our solution set. The key is to draw each boundary line accurately and shade the correct region for each inequality. The final solution will be the area that gets shaded by both inequalities.

Graphing the System of Inequalities

Now that we've dissected each inequality, it's time to bring them together on a single set of axes. The goal is to visualize the region where both conditions, $y \geq \frac{5}{2} x-8$ and $y \leq-x-1$, are satisfied simultaneously. This overlapping region is the heart of our solution set. Let's meticulously plot each boundary line and shade the corresponding regions.

• Step 1: Graph the first boundary line, $y = \frac{5}{2} x-8$. As we discussed, the y-intercept is at (0, -8). The slope is $\frac{5}{2}$. From (0, -8), we can move 2 units to the right and 5 units up to find another point, say (2, -3). We can continue this pattern to plot more points like (4, 2). Connecting these points with a ruler gives us the line $y = \frac{5}{2} x-8$. Since our inequality is $y \geq \frac{5}{2} x-8$, the boundary line itself is included in the solution. Therefore, we will draw this line as a solid line, not a dashed one. Now, we shade the region above this line, as this is where y-values are greater than or equal to those on the line.

• Step 2: Graph the second boundary line, $y = -x-1$. The y-intercept is at (0, -1). The slope is -1. From (0, -1), we can move 1 unit to the right and 1 unit down to find another point, say (1, -2). Continuing this, we can find points like (2, -3). Connecting these points forms the line $y = -x-1$. Since the inequality is $y \leq -x-1$, the boundary line is included in the solution. Thus, this line will also be a solid line. For this inequality, we shade the region below this line, as this is where y-values are less than or equal to those on the line.

• Step 3: Identify the Solution Set. The solution set for the system of inequalities is the region on the graph that is shaded by both inequalities. This is the area where the shading from the first inequality (above $y = \frac{5}{2} x-8$) and the shading from the second inequality (below $y = -x-1$) overlap. Visually, this will be a wedge-shaped region bounded by the two lines.

It's important to be precise when drawing these lines. Using graph paper is highly recommended to ensure accuracy. The intersection point of the two boundary lines is a critical point, as it lies on both lines and thus satisfies both inequalities. Let's find this intersection point algebraically. We set the expressions for y equal to each other:

$\frac{5}{2} x-8 = -x-1$

To solve for x, first, let's get rid of the fraction by multiplying the entire equation by 2:

$2 \left(\frac{5}{2} x-8\right) = 2(-x-1)$

$5x - 16 = -2x - 2$

Now, gather the x terms on one side and the constant terms on the other. Add 2x to both sides:

$5x + 2x - 16 = -2x + 2x - 2$

$7x - 16 = -2$

Add 16 to both sides:

$7x - 16 + 16 = -2 + 16$

$7x = 14$

Divide by 7:

$x = \frac{14}{7}$

$x = 2$

Now that we have the x-coordinate of the intersection point, we can substitute it into either of the original equations to find the corresponding y-coordinate. Let's use the second equation, $y = -x-1$, as it's simpler:

$y = -(2) - 1$

$y = -2 - 1$

$y = -3$

So, the two boundary lines intersect at the point (2, -3). This point is a vertex of our solution region. If we look at our graph, we should see that the point (2, -3) lies on both solid lines, which is exactly what we expect for the intersection.

Stating the Coordinates of a Point in the Solution Set

We've successfully graphed the system and identified the overlapping shaded region. The final part of the problem asks us to state the coordinates of a point that lies within this solution set. This means we need to pick any point that falls within the area where both inequalities are satisfied. Remember, this area is bounded by the two solid lines, and it's the region above $y = \frac{5}{2} x-8$ and below $y = -x-1$.

There are infinitely many points in the solution set. We can choose any point that is clearly within the shaded overlap. Let's consider a few possibilities:

• The Intersection Point: The point (2, -3) lies on both boundary lines. Since both lines are solid (meaning the boundary is included), this point satisfies both inequalities. Therefore, (2, -3) is a valid point in the solution set. If we plug it back into the original inequalities:

  • For $y \geq \frac{5}{2} x-8$: $-3 \geq \frac{5}{2}(2) - 8 \implies -3 \geq 5 - 8 \implies -3 \geq -3$. This is true.
  • For $y \leq -x-1$: $-3 \leq -(2) - 1 \implies -3 \leq -2 - 1 \implies -3 \leq -3$. This is also true.

• A Point Clearly Inside the Overlap: Let's try to find a point that is not on the boundary lines but is definitely in the shaded region. Looking at our intersection point (2, -3), we can try a point with a slightly larger x-value and a y-value that is below the first line and above the second. Or, we could try a point with a similar x-value but a y-value between the two lines. Let's consider x = 1. At x=1, the first line gives $y = \frac{5}{2}(1) - 8 = 2.5 - 8 = -5.5$. The second line gives $y = -(1) - 1 = -2$. So, for x=1, any y-value between -5.5 and -2 (inclusive) would work. Let's pick $y = -3$. So, the point (1, -3) is a candidate. Let's check (1, -3):

  • For $y \geq \frac{5}{2} x-8$: $-3 \geq \frac{5}{2}(1) - 8 \implies -3 \geq 2.5 - 8 \implies -3 \geq -5.5$. This is true.
  • For $y \leq -x-1$: $-3 \leq -(1) - 1 \implies -3 \leq -1 - 1 \implies -3 \leq -2$. This is true. So, (1, -3) is another point in the solution set.

• Another Point: How about x = 3? At x=3, the first line gives $y = \frac{5}{2}(3) - 8 = 7.5 - 8 = -0.5$. The second line gives $y = -(3) - 1 = -4$. So, for x=3, we need y such that $-4 \leq y \leq -0.5$. Let's pick $y = -2$. The point (3, -2) is a candidate. Let's check (3, -2):

  • For $y \geq \frac{5}{2} x-8$: $-2 \geq \frac{5}{2}(3) - 8 \implies -2 \geq 7.5 - 8 \implies -2 \geq -0.5$. This is false. So (3, -2) is NOT in the solution set. This highlights the importance of checking points carefully!

Let's re-evaluate our choice for x=3. We need $y \geq -0.5$ and $y \leq -4$. This is impossible! This means that for x=3, there are no y-values that satisfy both conditions. This tells us that our solution region does not extend indefinitely to the right. The intersection point (2, -3) is crucial here. Let's look at the graph. The region opens downwards and to the left of the intersection point.

Let's try a point with an x-value less than 2, say x = 0. At x=0, the first line gives $y = \frac{5}{2}(0) - 8 = -8$. The second line gives $y = -(0) - 1 = -1$. So, for x=0, we need y such that $-8 \leq y \leq -1$. Let's pick $y = -4$. The point (0, -4) is a candidate. Let's check (0, -4): * For $y \geq \frac{5}{2} x-8$: $-4 \geq \frac{5}{2}(0) - 8 \implies -4 \geq -8$. This is true. * For $y \leq -x-1$: $-4 \leq -(0) - 1 \implies -4 \leq -1$. This is true. So, (0, -4) is indeed another valid point in the solution set.

To be absolutely sure, when stating a point in the solution set, it's best to pick one that is clearly within the visually identified overlapping region on your graph. The point (2, -3) is always a safe bet as it's the intersection of the boundaries. If you pick another point, like (1, -3) or (0, -4), make sure it visually falls within the double-shaded area.

Conclusion

Graphing systems of inequalities is a powerful tool for visualizing the set of all solutions. By carefully plotting the boundary lines (solid for $\geq$ or $\\leq$, dashed for $>$ or $<$) and shading the appropriate regions, we can determine the intersection of these regions, which represents the solution set of the system. We found that the boundary lines $y = \frac{5}{2} x-8$ and $y = -x-1$ intersect at the point (2, -3). The solution set for the system $\begin{array}{l} y \geq \frac{5}{2} x-8 \\ y \leq-x-1 \end{array}$ is the region above or on the line $y = \frac{5}{2} x-8$ AND below or on the line $y = -x-1$. As requested, we can state the coordinates of a point in the solution set. A clear example is the intersection point itself, (2, -3). Other points, such as (1, -3) or (0, -4), also lie within this solution space. Understanding these graphical methods is fundamental in various areas of mathematics, including linear programming and optimization problems.

For further exploration into the world of inequalities and their graphical representations, you can visit resources like Khan Academy's section on linear inequalities. They offer comprehensive explanations and practice exercises to solidify your understanding.