Consider the system of inequalities and its graph.
[tex]\begin{array}{l}
y \leq-0.75 x \\
y \leq 3 x-2
\end{array}[/tex]
In which section of the graph does the actual solution to the system lie?
1
2
3
4
Answer (1)
The problem requires finding the region where both inequalities y ≤ − 0.75 x and y ≤ 3 x − 2 are satisfied.
Test a point in each section of the graph to see if it satisfies both inequalities.
The point (1, -1) in Section 4 satisfies both inequalities.
Therefore, the solution to the system lies in Section 4 .
Explanation
Understanding the Problem We are given a system of inequalities: y ≤ − 0.75 x and y ≤ 3 x − 2 . We need to determine in which section of the graph the solution to the system lies. The solution to the system of inequalities is the region where both inequalities are satisfied.
Analyzing the Inequalities Let's analyze the inequalities. The first inequality, y ≤ − 0.75 x , represents the region below the line y = − 0.75 x . The second inequality, y ≤ 3 x − 2 , represents the region below the line y = 3 x − 2 . The solution to the system is the intersection of these two regions.
Testing Points in Each Section To determine which section corresponds to the solution, we can test a point in each section to see if it satisfies both inequalities. Let's consider the following points:
Section 1: (1, 1) Section 2: (-1, 1) Section 3: (-1, -1) Section 4: (1, -1)
Now, let's test each point in the inequalities:
For (1, 1): 1 ≤ − 0.75 ( 1 ) R i g h t a rro w 1 ≤ − 0.75 (False) 1 ≤ 3 ( 1 ) − 2 R i g h t a rro w 1 ≤ 1 (True) Since the first inequality is false, (1, 1) is not in the solution region.
For (-1, 1): 1 ≤ − 0.75 ( − 1 ) R i g h t a rro w 1 ≤ 0.75 (False) 1 ≤ 3 ( − 1 ) − 2 R i g h t a rro w 1 ≤ − 5 (False) Since both inequalities are false, (-1, 1) is not in the solution region.
For (-1, -1): − 1 ≤ − 0.75 ( − 1 ) R i g h t a rro w − 1 ≤ 0.75 (True) − 1 ≤ 3 ( − 1 ) − 2 R i g h t a rro w − 1 ≤ − 5 (False) Since the second inequality is false, (-1, -1) is not in the solution region.
For (1, -1): − 1 ≤ − 0.75 ( 1 ) R i g h t a rro w − 1 ≤ − 0.75 (True) − 1 ≤ 3 ( 1 ) − 2 R i g h t a rro w − 1 ≤ 1 (True) Since both inequalities are true, (1, -1) is in the solution region.
Determining the Solution Region Since the point (1, -1) in Section 4 satisfies both inequalities, the solution to the system lies in Section 4.
Examples
Systems of inequalities are used in various real-world applications, such as linear programming, where they help optimize solutions under constraints. For example, a company might use a system of inequalities to determine the optimal production levels of two products, given limitations on resources like labor and materials. By graphing the inequalities, the company can identify the feasible region, representing all possible production combinations that satisfy the constraints. The solution to the system then helps the company maximize its profit while staying within its resource limitations.
