Problem 56
Question
Graph each compound inequality. y>x-4 \text { or } 3 x+2 y \geq 12
Step-by-Step Solution
Verified Answer
In order to graph the compound inequality \(y > x - 4\) or \(3x + 2y \geq 12\), we need to graph each inequality separately and then find the regions where either inequality is true. The equations of the boundary lines are \(y = x - 4\) and \(y = (-3/2)x + 6\). After graphing these lines, we find that the solution region for the first inequality is above the line \(y = x - 4\) (dashed line) and the solution region for the second inequality is above and to the right of the line \(y = (-3/2)x + 6\) (solid line). The final solution region is where either one of these inequalities is true - i.e., the regions above both lines.
1Step 1: Rewrite the inequalities as boundary line equations
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First, we will rewrite each inequality to find the equation of the boundary lines:
1. y > x - 4 becomes y = x - 4
2. 3x + 2y ≥ 12 becomes 2y = -3x + 12 or y = (-3/2)x + 6
2Step 2: Graphing the boundary lines
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Now, we will graph both boundary line equations on the same coordinate plane.
1. For the line y = x - 4, the slope is 1, and the y-intercept is -4. Draw a dashed line representing this equation (dashed because the inequality is strictly greater than, not equal to).
2. For the line y = (-3/2)x + 6, the slope is -3/2, and the y-intercept is 6. Draw a solid line representing this equation (solid because the inequality includes equal to, as well).
3Step 3: Shading the solution regions
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Now, we need to shade the regions where the inequalities are true:
1. For the inequality y > x - 4, pick a test point not on the boundary line, such as (0,0). Since 0 > 0 - 4, the inequality is true at the test point. Shade the region above the boundary line.
2. For the inequality 3x + 2y ≥ 12, again, pick a test point not on the boundary line, such as (0,0). Since 3(0) + 2(0) ≥ 12 is false, the inequality is not true at the test point. Shade the region above and to the right of the boundary line.
4Step 4: Final solution
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The problem asks us to graph the compound inequality y > x - 4 OR 3x + 2y ≥ 12. This means that any point that satisfies either of the inequalities will be part of the solution. By shading the regions for both inequalities, we can see that the region where either inequality is true will be the region above the line y = x - 4 and the region above and to the right of the line y = (-3/2)x + 6.
Key Concepts
Graphing InequalitiesBoundary LinesSolution RegionsInequality Shading
Graphing Inequalities
When we talk about graphing inequalities, we're looking at visualizing regions within a coordinate plane that satisfy specific conditions. Inequalities can involve two variables, commonly represented as x and y. Our goal is to find all of the x and y pairs (points on the graph) that make these inequalities true.
- To begin, we translate the inequalities into boundary line equations. For example, given the inequality \(y > x - 4\), we briefly change it into the equation \(y = x - 4\).
- We do the same for any additional inequalities, such as changing \(3x + 2y \geq 12\) into \(y = (-3/2)x + 6\).
Boundary Lines
Boundary lines are critical in graphing inequalities because they form the edges of the solution regions you will shade. The type of line you draw depends on the inequality sign:
- Dashed Lines: For inequalities that are "strict" (using symbols like > or <), the boundary line is dashed. It tells us that points on the line itself are not part of the solution. For example, this applies to \(y = x - 4\) when we are graphing \(y > x - 4\).
- Solid Lines: "Non-strict" inequalities (\(\leq\) or \(\geq\)) result in a solid boundary line. This indicates that the points on the line do belong to the solution region. Such is the case for \(y = (-3/2)x + 6\) when graphed from \(3x + 2y \geq 12\).
Solution Regions
By determining where the boundary lines lie on the graph, you can identify solution regions. These are the areas on the coordinate plane that satisfy the original inequality conditions. The process involves these steps:
- Choose a test point that is not on the boundary line, typically (0,0), to help determine which side of the line is part of the solution region.
- For example, if \( (0,0) \) satisfies the inequality when substituted back into the inequality, then the region containing \( (0,0) \) is the solution area.
- For \(y > x - 4\), the solution region is above the line \(y = x - 4\).
- For \(3x + 2y \geq 12\), since the inequality is false at \( (0,0) \), the solution region is above and to the right of the line \(y = (-3/2)x + 6\).
Inequality Shading
Shading is a simple yet powerful technique used to visually represent where solutions to inequalities lie within the coordinate plane. For compound inequalities, you must be precise about which regions to shade based on the test points:
- For the inequality \(y > x - 4\), after determining the correct region, shade the entire area above the line \(y = x - 4\) with a light shading to indicate that it's part of the solution.
- For \(3x + 2y \geq 12\), shade the region above and to the right of the line \(y = (-3/2)x + 6\). The solid line indicates that points on this boundary line are also included.
- Sometimes, inequalities are combined using "or" or "and." In "or" situations, as with this exercise, any point within either shaded region satisfies one of the original inequalities. Both shaded areas from the inequalities are part of the solution.
Other exercises in this chapter
Problem 55
Graph each compound inequality. \(y>-\frac{2}{3} x+1\) or \(-2 x+5 y \leq 0\)
View solution Problem 56
The following exercises contain absolute value equations, linear inequalities, and both types of absolute value inequalities. Solve each. Write the solution set
View solution Problem 57
The following exercises contain absolute value equations, linear inequalities, and both types of absolute value inequalities. Solve each. Write the solution set
View solution Problem 57
Graph each compound inequality. \(x \geq 5\) and \(y \leq-3\)
View solution