Problem 8
Question
For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{r} 3 x-y<3 \\ x+y \geq 1 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
Solution set is the region above the line \(y = 3x - 3\) and above/inclusive of the line \(y = -x + 1\).
1Step 1: Rewrite Inequalities for Graphing
First, rewrite each inequality in slope-intercept form (solve for \(y\)): 1. The first inequality, \(3x - y < 3\), can be written as \(y > 3x - 3\).2. The second inequality, \(x + y \geq 1\), can be rewritten as \(y \geq -x + 1\).
2Step 2: Graph the Boundary Lines
Draw the lines represented by the equations \(y = 3x - 3\) and \(y = -x + 1\). Use a dashed line for \(y = 3x - 3\) because the inequality is strict ('\(<\)'), and a solid line for \(y = -x + 1\) because the inequality includes equality ('\(\geq\)').
3Step 3: Shade the Solution Areas
For the inequality \(y > 3x - 3\), shade the area above the line \(y = 3x - 3\). For \(y \geq -x + 1\), shade the area above and including the line \(y = -x + 1\).
4Step 4: Find the Overlapping Region
The solution set is the region where the shaded areas overlap. This includes the solid boundary line \(y \geq -x + 1\) but not the line \(y = 3x - 3\). This overlapping region satisfies both inequalities.
Key Concepts
Graphing InequalitiesSolution SetSlope-Intercept FormBoundary Lines
Graphing Inequalities
Understanding how to graph inequalities is a crucial skill when solving systems of inequalities. The process begins with transforming each inequality into an equation. This helps in sketching boundary lines which serve as visual guides for where the inequality is valid. For instance, to graph the inequality \(3x - y < 3\), we start by rewriting it in terms of equality as \(y = 3x - 3\).
Once the boundary lines are defined, these lines are drawn on the coordinate plane. When graphing inequalities, remember:
Once the boundary lines are defined, these lines are drawn on the coordinate plane. When graphing inequalities, remember:
- Strict Inequality (\(<\) or \(>\)): Draw a dashed line to indicate that points on the line are not included in the solution.
- Inclusive Inequality (\(\leq\) or \(\geq\)): Use a solid line to show that the line itself is part of the solution set.
Solution Set
The solution set in a system of inequalities consists of all possible points that satisfy every inequality in the system. Once you've graphed each inequality with its corresponding shaded region on the coordinate plane, the solution set can be visually identified by finding where the shaded areas overlap.
For example, once we graph \(y > 3x - 3\) and \(y \geq -x + 1\), we look for the intersecting region. This region, often referred to as the feasible region, includes all points satisfying both inequalities simultaneously. If one of the inequalities has a solid line, those points on the line are included in the solution set as well.
Understanding the solution set is essential as it directly represents the set of solutions that fulfill the conditions given by all inequalities in the problem.
For example, once we graph \(y > 3x - 3\) and \(y \geq -x + 1\), we look for the intersecting region. This region, often referred to as the feasible region, includes all points satisfying both inequalities simultaneously. If one of the inequalities has a solid line, those points on the line are included in the solution set as well.
Understanding the solution set is essential as it directly represents the set of solutions that fulfill the conditions given by all inequalities in the problem.
Slope-Intercept Form
The slope-intercept form is a popular way to express the equation of a line, written as \(y = mx + b\). Here, \(m\) represents the slope of the line, and \(b\) is the y-intercept where the line crosses the y-axis. This form makes it straightforward to both graph the line and understand its direction and steepness.
By converting inequalities like \(3x - y < 3\) and \(x + y \geq 1\) into the slope-intercept form, graphing becomes more intuitive as it simplifies identifying how the line should be drawn.
Additionally, the slope-intercept form helps easily compare slopes of lines, which can determine relative steepness and the direction of the shading for inequalities.
By converting inequalities like \(3x - y < 3\) and \(x + y \geq 1\) into the slope-intercept form, graphing becomes more intuitive as it simplifies identifying how the line should be drawn.
Additionally, the slope-intercept form helps easily compare slopes of lines, which can determine relative steepness and the direction of the shading for inequalities.
Boundary Lines
Boundary lines play a vital role in graphing systems of inequalities. They are the lines that stem from converting inequalities into equalities. Deciding whether these lines are dashed or solid is key to accurately representing an inequality on a graph.
For example, the inequality \(y > 3x - 3\) uses a dashed boundary line \(y = 3x - 3\) to show that points precisely on the line are not included in the solution. Conversely, \(y \geq -x + 1\) has a solid boundary line to indicate inclusivity of points on that line.
The nature of boundary lines helps distinguish distinct conditions between inequalities, guiding not only the graphing but also understanding the regions they define.
For example, the inequality \(y > 3x - 3\) uses a dashed boundary line \(y = 3x - 3\) to show that points precisely on the line are not included in the solution. Conversely, \(y \geq -x + 1\) has a solid boundary line to indicate inclusivity of points on that line.
The nature of boundary lines helps distinguish distinct conditions between inequalities, guiding not only the graphing but also understanding the regions they define.
Other exercises in this chapter
Problem 8
For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} 5 & -1 \\ 3 & 4 \end{array}\right] $$
View solution Problem 8
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 9
For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{rr} 2 & -1 \\ 0 & -4 \\ -5 & 3 \end{array}\right], \quad B=\left[
View solution Problem 9
For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} -3 & 2 \\ -4 & 5 \end{array}\right] $$
View solution