Problem 42
Question
In Exercises 27–62, graph the solution set of each system of inequalities or indicate that the system has no solution. $$\left\\{\begin{array}{l} x+y>3 \\ x+y<-2 \end{array}\right.$$
Step-by-Step Solution
Verified Answer
The system of inequalities has no solution, as the two inequalities represent non-overlapping areas in the graph.
1Step 1: Graphing the first inequality
Begin by graphing the line of the first inequality, \(x + y = 3\). It's a diagonal line, passing through points (3, 0) and (0, 3). However, because the inequality is '>', the solution will be the area above the line, not including the line itself.
2Step 2: Graphing the second inequality
Next, graph the line of the second inequality, \(x + y = -2\). It's a diagonal line passing through points (-2, 0) and (0, -2), and the solution for the inequality '<' is the area below the line.
3Step 3: Analyzing both Inequalities
Upon analyzing the two inequalities side by side, it can be noticed that the areas they represent do not overlap - one is above the line \(x + y = 3\) while the other one is below the line \(x + y = -2\). Therefore these equations don't share any common solutions.
Key Concepts
graphing inequalitiessolution setlinear inequalitiesoverlapping regions
graphing inequalities
Graphing inequalities involves drawing the graph of an equation, but with a twist. Here, you have to decide whether the solution includes the line itself or just the region above or below it.
When graphing the inequality, such as in the form \(x + y > 3\), you begin by plotting the line \(x + y = 3\) as if it were an equality. This line serves as the boundary line.
But, because our inequality is 'greater than' \((>)\), the solution set includes all points above this line. It's important to remember that we do not include the line itself in the solution, so we draw it as a dashed line to show it is not part of the solution set.
In contrast, for the inequality \(x + y < -2\), plot the line \(x + y = -2\) but focus on the area below this line for the solution. Again, use a dashed line to exclude the boundary line from the solution set.
When graphing the inequality, such as in the form \(x + y > 3\), you begin by plotting the line \(x + y = 3\) as if it were an equality. This line serves as the boundary line.
But, because our inequality is 'greater than' \((>)\), the solution set includes all points above this line. It's important to remember that we do not include the line itself in the solution, so we draw it as a dashed line to show it is not part of the solution set.
In contrast, for the inequality \(x + y < -2\), plot the line \(x + y = -2\) but focus on the area below this line for the solution. Again, use a dashed line to exclude the boundary line from the solution set.
solution set
The solution set of a system of inequalities is the region satisfying all given inequalities. It's essential to understand this because a solution meets the conditions of all inequalities in the system simultaneously.
Graphically, the solution set is often represented by a shaded region on the coordinate plane. If we have multiple inequalities, we look for regions where the shaded areas overlap, indicating shared solutions.
However, it's vital to analyze the possibility of no common solutions. In some cases, like in the given exercise, the inequalities might not overlap at all. This absence of overlap indicates that there is no solution set because no points satisfy both inequalities simultaneously.
Graphically, the solution set is often represented by a shaded region on the coordinate plane. If we have multiple inequalities, we look for regions where the shaded areas overlap, indicating shared solutions.
However, it's vital to analyze the possibility of no common solutions. In some cases, like in the given exercise, the inequalities might not overlap at all. This absence of overlap indicates that there is no solution set because no points satisfy both inequalities simultaneously.
linear inequalities
Linear inequalities are expressions that relate linear functions using inequality signs, such as 'greater than' or 'less than'. In the context of graphing, they form straight lines on the coordinate plane.
These lines set boundary values for the inequalities, dividing the plane into two regions. For example, \(x + y > 3\) and \(x + y < -2\) create boundaries at lines \(x + y = 3\) and \(x + y = -2\).
The trick here is determining which side of the line constitutes the solution. Determining this relies on the type of inequality symbol used. Greater than symbols point to areas above, while less than symbols highlight areas below the dividing line. Using test points from the coordinate plane can help verify which side of the line the solution is on.
These lines set boundary values for the inequalities, dividing the plane into two regions. For example, \(x + y > 3\) and \(x + y < -2\) create boundaries at lines \(x + y = 3\) and \(x + y = -2\).
The trick here is determining which side of the line constitutes the solution. Determining this relies on the type of inequality symbol used. Greater than symbols point to areas above, while less than symbols highlight areas below the dividing line. Using test points from the coordinate plane can help verify which side of the line the solution is on.
overlapping regions
When graphing a system of inequalities, one critical task is finding overlapping regions. These are areas that satisfy all the system's inequalities simultaneously.
This region is key because it represents the solution to the system of inequalities. In practice, you graph each inequality one by one and identify the region where all shaded areas overlap.
In some systems, you may observe that no overlapping region exists, as is the case in the original exercise. Here, with one inequality being the area above \(x + y = 3\) and the other below \(x + y = -2\), it's clear that the boundary conditions cannot meet. Hence, there's no common region or solution. Validating the existence of overlapping regions provides clarity on whether a system of inequalities has solutions.
This region is key because it represents the solution to the system of inequalities. In practice, you graph each inequality one by one and identify the region where all shaded areas overlap.
In some systems, you may observe that no overlapping region exists, as is the case in the original exercise. Here, with one inequality being the area above \(x + y = 3\) and the other below \(x + y = -2\), it's clear that the boundary conditions cannot meet. Hence, there's no common region or solution. Validating the existence of overlapping regions provides clarity on whether a system of inequalities has solutions.
Other exercises in this chapter
Problem 42
Writing in Mathematics What is a system of linear equations in three variables?
View solution Problem 42
Write the partial fraction decomposition of each rational expression. $$\frac{3 x-5}{x^{3}-1}$$
View solution Problem 43
How do you determine whether a given ordered triple is a solution of a system in three variables?
View solution Problem 43
Perform each long division and write the partial fraction decomposition of the remainder term. $$\frac{x^{5}+2}{x^{2}-1}$$
View solution