Problem 44
Question
In Exercises \(41-46\), sketch the graph of the system of linear inequalities. $$ \left\\{\begin{array}{r} 2 x-6 y>6 \\ x \quad y \leq 0 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The graph of the system of linear inequalities would include the region below the line \(y = \frac{1}{3}x - 1\) and either in the forth quadrant (\(x \geq 0, y \leq 0\)) or in the second quadrant (\(x \leq 0, y \geq 0\)).
1Step 1: Rewrite the inequalities
Start by translating the inequalities into slope-intercept form (i.e. \(y = mx + b\)). Doing so, we have: \[2x - 6y > 6 \implies y < \frac{1}{3}x - 1\]For the second part: \(x y \leq 0 \) this implies that the solutions exist either in the fourth quadrant (\(x \geq 0, y \leq 0\)) or in the second quadrant (\(x \leq 0, y \geq 0\))
2Step 2: Sketch the lines
The line \(y = \frac{1}{3}x - 1\) is a straight line with a positive slope passing through the point (0, -1). Since the inequality is 'greater than', we exclude the line itself from solution, and as the coefficient of y in initial inequality is negative, the area of solution lies below the line (y < ). The line x=0 and y=0 are the x and y axis respectively.
3Step 3: Shading the relevant area
The region satisfying both inequalities lies below the line \(y = \frac{1}{3}x - 1\) and either in the forth quadrant or second quadrant. Using two different shadings for these regions, we find the intersection. After shading the intersection, we have the final graph of the system of linear inequalities.
Key Concepts
Graphing InequalitiesCoordinate PlaneQuadrants in Graphing
Graphing Inequalities
Graphing inequalities is a key concept when dealing with systems of linear inequalities. It involves not just plotting a line, but also shading the region where the inequalities hold true. When graphing an inequality like \( y < \frac{1}{3}x - 1 \), first start by graphing the equation \( y = \frac{1}{3}x - 1 \). Draw this line as a dashed line to indicate that the points on the line are not included in the solution set. Once you've graphed the line, determine which side of the line to shade by selecting a test point. A common choice is the origin, (0, 0), if it's not on the line. Substitute this point into the inequality to see if it satisfies it. If the inequality holds true, shade the area on the same side of the line as the test point. Otherwise, shade the opposite side.With systems of inequalities, where you're dealing with more than one inequality at once, find the region where the shaded areas of all inequalities overlap. This overlap area represents the solution set for the system. The method of shading helps visualize feasible solutions that satisfy each inequality in the system.
Coordinate Plane
A coordinate plane is a two-dimensional surface where each point is defined by a pair of numerical coordinates. These coordinates represent a point's distance from two fixed, perpendicular lines, often called the x-axis (horizontal) and y-axis (vertical). On this plane, you can graph equations, lines, and inequalities to visualize mathematical relationships.
The x-axis typically defines horizontal movement, whereas the y-axis defines vertical movement. Each point on the plane is denoted by a pair (x, y), where x is the horizontal value and y is the vertical value.
When graphing inequalities, it is crucial to use the coordinate plane to accurately depict the regions where the conditions specified by the inequalities are satisfied. Understanding how to navigate this plane and plot points correctly is foundational for solving systems of inequalities.
Quadrants in Graphing
The coordinate plane is divided into four regions known as quadrants. Each is labeled with Roman numerals:
- Quadrant I: Both x and y coordinates are positive.
- Quadrant II: The x-coordinate is negative, while the y-coordinate is positive.
- Quadrant III: Both x and y coordinates are negative.
- Quadrant IV: The x-coordinate is positive, while the y-coordinate is negative.
Other exercises in this chapter
Problem 43
In Exercises 35-46, solve the system by the method of substitution. $$ \left\\{\begin{array}{l} \frac{x}{4}+\frac{y}{2}=1 \\ \frac{x}{2}-\frac{y}{3}=1 \end{arra
View solution Problem 43
In Exercises \(43-48\), write the equations of the lines in slope-intercept form. What can you conclude about the number of solutions of the system? $$ \left\\{
View solution Problem 44
Find an equation of the line of slope \(m=-2\) passing through the intersection of the lines \(2 x+5 y=11\) and \(4 x-y=11\).
View solution Problem 44
In Exercises 35-46, solve the system by the method of substitution. $$ \left\\{\begin{aligned} -\frac{x}{6}+\frac{y}{12} &=1 \\ \frac{x}{2}+\frac{y}{8} &=1 \end
View solution