Problem 106
Question
In Exercises 106-109, determine whether each statement makes sense or does not make sense, and explain your reasoning. When graphing a linear inequality, I should always use \((0,0)\) as a test point because it's easy to perform the calculations when 0 is substituted for each variable.
Step-by-Step Solution
Verified Answer
The statement makes sense. Using (0,0) as a test point when graphing a linear inequality simplifies calculations, as long as the line doesn't pass through the origin. In that case, a different test point should be used.
1Step 1: Understanding the use of test points in graphing linear inequalities
When graphing a linear inequality like \(ax + by \geq c\), a test point is used to identify which region of the graph to shade that satisfies the inequality. Once a line is drawn to represent the equality \(ax + by = c\), any point not on the line can be used as a test point. If the ordered pair of the test point when substituted into the inequality results in a true statement, it means that the region of the graph containing that test point is the solution to the inequality. If the statement is false, the other region is the solution.
2Step 2: Application to the origin (0,0)
The origin \((0,0)\) is often used as the test point because calculations are simplified when 0 is substituted for each variable. If the origin is not contained in the line representing the equality (essentially if the line does not pass through the origin), it can be used as the test point in the inequality. For example, in the inequality \(2x + 3y \geq 1\), if we substitute \((0,0)\), we get \(0 \geq 1\), which is false. Hence, the region not containing the origin is the solution to the inequality.
3Step 3: Evaluation of the statement
Given these explanations, the statement 'When graphing a linear inequality, I should always use \((0,0)\) as a test point because it's easy to perform the calculations when 0 is substituted for each variable.' makes sense. It is true because using \((0,0)\) simplifies the calculations. However, it should also be noted that this approach works as long as the line does not pass through the origin. If it does, one must choose a different test point.
Key Concepts
Test Point MethodLinear Inequality ShadingInequality Solution Region
Test Point Method
The test point method is an essential tool when graphing linear inequalities. Understanding this technique enables students to easily determine which side of the boundary line represents the solution set.
To employ the test point method, begin by graphing the corresponding linear equation of the inequality, which acts as a boundary between two distinct regions on the graph. This line divides the coordinate plane into a 'solution region' and a 'non-solution region'. The next step is to select a test point that is not on the line. The origin (0,0) is a common choice due to the ease of computation when all variable values are zero.
Let's consider a linear inequality ax + by < c. By substituting the coordinates of the test point into the inequality, if the resulting statement is true, then the test point, and the side of the line it lies on, is part of the solution region. If the statement is false, the solution region is on the opposite side of the line.
However, the origin can only serve as a test point if the line does not pass through it. If the boundary line does cross the origin, another test point must be chosen to avoid ambiguity in the results.
To employ the test point method, begin by graphing the corresponding linear equation of the inequality, which acts as a boundary between two distinct regions on the graph. This line divides the coordinate plane into a 'solution region' and a 'non-solution region'. The next step is to select a test point that is not on the line. The origin (0,0) is a common choice due to the ease of computation when all variable values are zero.
Let's consider a linear inequality ax + by < c. By substituting the coordinates of the test point into the inequality, if the resulting statement is true, then the test point, and the side of the line it lies on, is part of the solution region. If the statement is false, the solution region is on the opposite side of the line.
However, the origin can only serve as a test point if the line does not pass through it. If the boundary line does cross the origin, another test point must be chosen to avoid ambiguity in the results.
Linear Inequality Shading
After determining the correct side of the boundary line using the test point method, the next step is to shade the solution region. This visual representation distinguishes the area where the inequality holds true.
To shade the graph correctly, use the results from the test point. If the test point makes the inequality true, then shade the region that includes the test point. Conversely, if the test point is not part of the solution, shade the opposite side.
Always use a light shading or patterning to ensure that any overlapping regions with other inequalities are visible and that all relevant points and lines remain clear.
To shade the graph correctly, use the results from the test point. If the test point makes the inequality true, then shade the region that includes the test point. Conversely, if the test point is not part of the solution, shade the opposite side.
Why Shading Matters
Shading is important because it provides a clear depiction of all the possible solutions to the inequality. Without this visual aid, identifying the range of solutions would be more challenging. It's also helpful when dealing with systems of inequalities, as the intersecting shaded areas reveal the common solution set for all inequalities involved.Always use a light shading or patterning to ensure that any overlapping regions with other inequalities are visible and that all relevant points and lines remain clear.
Inequality Solution Region
Identifying the inequality solution region is the ultimate goal when graphing a linear inequality. This region represents all the ordered pairs (x, y) that satisfy the inequality. It's essentially the 'answer' to the graphing process.
The solution region can be bounded or unbounded depending on the inequality. For bounded inequalities, like x + y , the solution region is confined to a specific area on the graph. For unbounded inequalities, the region extends infinitely in a direction or directions indicated by the shading.
When graphing inequalities, always be mindful of the inclusion or exclusion of the boundary line. Inequalities with a '<=' or '>=' signify that the boundary line is part of the solution and should be drawn in a solid line. If the inequality has just a '<' or '>', the line is not part of the solution and should be dashed to indicate its exclusivity.
The solution region can be bounded or unbounded depending on the inequality. For bounded inequalities, like x + y , the solution region is confined to a specific area on the graph. For unbounded inequalities, the region extends infinitely in a direction or directions indicated by the shading.
Bounded vs. Unbounded Regions
Bounded regions typically occur with '<=' or '>=' inequalities and form shapes like triangles or polygons, demonstrated by the boundaries intersecting. Unbounded regions, associated with '<' or '>' inequalities, go on indefinitely, represented by shading that extends to the edges of the graph and beyond.When graphing inequalities, always be mindful of the inclusion or exclusion of the boundary line. Inequalities with a '<=' or '>=' signify that the boundary line is part of the solution and should be drawn in a solid line. If the inequality has just a '<' or '>', the line is not part of the solution and should be dashed to indicate its exclusivity.
Other exercises in this chapter
Problem 103
If \(x=3, y=2,\) and \(z=-3,\) does the ordered triple \((x, y, z)\) satisfy the equation \(2 x-y+4 z=-8 ?\)
View solution Problem 105
Write an equation involving \(a, b,\) and \(c\) based on the following description: When the value of \(x\) in \(y=a x^{2}+b x+c\) is \(4,\) the value of \(y\)
View solution Problem 107
In Exercises 106-109, determine whether each statement makes sense or does not make sense, and explain your reasoning. When graphing \(3 x-4 y
View solution Problem 114
Write a system of inequalities whose solution set includes every point in the rectangular coordinate system.
View solution