Problem 4
Question
For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{l} 3 x-y>6 \\ 2 x+y \leq 4 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
The solution set is the region between the dashed line of \(3x-y>6\) excluding the line, and the solid line of \(2x+y\leq4\) including the line, where they overlap.
1Step 1: Transform Inequalities into Equations
To graph the inequalities, we first convert them to equations. For the first inequality, \(3x - y > 6\), change it to \(3x - y = 6\). For the second inequality, \(2x + y \leq 4\), change it to \(2x + y = 4\). These equations will help us determine the boundary lines.
2Step 2: Graph the Boundary Lines
For the equation \(3x - y = 6\), we can find two points. Let \(x = 0 \Rightarrow y = -6\) and \(y = 0 \Rightarrow x = 2\). Draw a dashed line through (0, -6) and (2, 0).For the equation \(2x + y = 4\), we can find two points. Let \(x = 0 \Rightarrow y = 4\) and \(y = 0 \Rightarrow x = 2\). Draw a solid line through (0, 4) and (2, 0).
3Step 3: Test Points for Shading
To determine the appropriate shading area for each inequality, pick a test point not on the lines (e.g., the origin (0,0)). For \(3x - y > 6\): Place (0,0) into \(3(0) - 0 > 6\), which gives \(0 > 6\); this is false, so shade the side that does not include (0,0). For \(2x + y \leq 4\): Place (0,0) into \(2(0) + 0 \leq 4\), which gives \(0 \leq 4\); this is true, so shade the side that includes (0,0).
4Step 4: Identify the Solution Region
The solution to the system of inequalities is the area where the shaded regions overlap. Make sure the solution area includes the region below or on the solid line (since \(2x + y \leq 4\)) and above the dashed line (as \(3x - y > 6\)). This region represents all the points that satisfy both inequalities.
Key Concepts
Graphing InequalitiesSolution SetBoundary LinesTest Points
Graphing Inequalities
To start solving a system of inequalities, we first need to plot them on a graph. This helps us visually understand the regions that satisfy the given conditions. We begin by transforming each inequality into an equation by changing inequality signs to equals signs.
For instance, given the inequalities:
For instance, given the inequalities:
- First: \(3x - y > 6\)
- Second: \(2x + y \leq 4\)
- \(3x - y = 6\)
- \(2x + y = 4\)
Solution Set
The solution set for a system of inequalities is the set of all points that satisfy all individual inequalities at once. After graphing, it usually appears as the overlapping shaded area on our graph.
To determine this visually:
To determine this visually:
- Identify distinct regions created by boundary lines.
- Assess where the shaded areas for individual inequalities overlap.
- This overlapping region is our solution set, which fulfills every inequality in the system.
Boundary Lines
Boundary lines are key in graphing inequalities. They help us define the critical zones for each inequality on a coordinate plane.
Based on our transformed equations:
Based on our transformed equations:
- The line \(3x - y = 6\) is drawn as a dashed line to indicate that points on this line do not satisfy \(3x - y > 6\).
- The line \(2x + y = 4\) is drawn as a solid line, meaning points on this line satisfy \(2x + y \leq 4\).
- Dashed lines signify strict inequalities (\(<\) or \(>\)).
- Solid lines signify inclusive inequalities (\(\leq\) or \(\geq\)).
Test Points
To determine which side of the boundary line we should shade, using test points is a helpful strategy. This involves selecting a point on either side of the boundary and checking if it satisfies the inequality.
Choosing the origin (0,0) is often convenient when it is not on any of the boundary lines.
Choosing the origin (0,0) is often convenient when it is not on any of the boundary lines.
- For \(3x - y > 6\), plugging in (0,0) gives \(0 > 6\), which is false. Thus, we shade the opposite side.
- For \(2x + y \leq 4\), using (0,0) results in \(0 \leq 4\), which is true. Hence, we shade the side of the boundary line that contains (0,0).
Other exercises in this chapter
Problem 4
For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} 2 & 9 \\ 3 & 13 \end{array}\right] $$
View solution Problem 4
For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\r
View solution Problem 5
For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rrr} 3 & -2 & 1 \\ -1 & 4 & -7 \\ 0 & 5 & 9 \end{array}\right], \qua
View solution Problem 5
For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} -1 & 2 \\ 3 & 4 \end{array}\right] $$
View solution