Problem 35
Question
Graph the solution set of each system of inequalities or indicate that the system has no solution. $$ \left\\{\begin{array}{l} x \leq 2 \\ y \geq-1 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The solution to the system of inequalities is all the points to the left of the line \(x = 2\) (including the line) and above the line \(y = -1\) (including the line). The intersection forms a quadrant on the top left corner of the graph.
1Step 1: Graph the First Inequality
In the inequality \(x \leq 2\), the value of \(x\) is less than or equal to 2, which will result in a vertical boundary line at \(x = 2\). Since the inequality is \(\leq\), the line includes the points on the line. So we use a solid line to represent the boundary. All the area to the left of the line (including the line itself) represents the solution to the first inequality.
2Step 2: Graph the Second Inequality
Next, the inequality \(y \geq -1\) has \(y\) being greater than or equal to -1, which results in a horizontal boundary line at \(y = -1\). As the inequality symbol is \(\geq\), the boundary line includes the points on the line so its representation will also be a solid line. All the areas above the line (including the line itself) fulfill this condition and represent the solution set to the second inequality.
3Step 3: Find the Intersection of the Solution Sets
For the system of inequalities to hold a solution, there must be points that satisfy both inequalities at the same time. These points are represented by the overlap (intersection) of the two solution sets from step 1 and step 2. In this case, the intersection will be all the points that are at the left of the line at \(x = 2\) and at the same time above (and including) the line at \(y = -1\). Thus, the intersection forms a quadrant on the top left corner of the graph, with the solid lines included.
Key Concepts
Graphical RepresentationInequality Solution SetsIntersection of Inequalities
Graphical Representation
Graphing a system of inequalities is a visual method to find their solution set. This involves drawing each individual inequality on the same coordinate plane to see where they overlap.When you work with the inequality \( x \leq 2 \), you should plot a vertical line on \( x = 2 \). Since \( x \) can be equal to 2, use a solid line — which signals that all points on this line satisfy the inequality. Shading to the left of this line, including the line itself, marks all possible solutions for this inequality.Similarly, for the inequality \( y \geq -1 \), a horizontal line is drawn at \( y = -1 \). Again, your line will be solid because \( y \) can equal -1, meaning every point on this line is part of the solution set. Shade above the line to represent all the solutions for this inequality.Combining these graphical elements shows the possible region where the solutions for both inequalities exist.
Inequality Solution Sets
After plotting each inequality, the next step is identifying the solution set they represent. Each inequality's solution includes all values that make the inequality true.
- For \( x \leq 2 \), the solution set includes all points where \( x \) is 2 or less.
- For \( y \geq -1 \), it covers all points where \( y \) is -1 or more.
Intersection of Inequalities
The intersection of inequalities is where the magic of solving a system happens. It refers to the region where all the solution sets of individual inequalities overlap.For this set of inequalities: - The shaded region that is both to the left of the vertical line \( x = 2 \) and above the horizontal line \( y = -1 \) represents the complete solution set for the system. These overlapping or intersecting regions show the combined conditions that satisfy both inequalities at the same time. Always ensure both conditions in the inequalities are reflected in the graph.This intersection provides not only a visual guide but also confirms that the chosen points indeed satisfy both conditions of the system of inequalities. It's the visual representation of the shared solution, ensuring completeness and correctness in mathematical analysis.
Other exercises in this chapter
Problem 35
Group members should choose a particular field of interest. Research how linear programming is used to solve problems in that field. If possible, investigate th
View solution Problem 35
write the partial fraction decomposition of each rational expression. $$ \frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1} $$
View solution Problem 35
In Exercises \(29-42,\) solve each system by the method of your choice. $$ \left\\{\begin{array}{l} x^{3}+y=0 \\ x^{2}-y=0 \end{array}\right. $$
View solution Problem 35
In Exercises \(31-42,\) solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to
View solution