Problem 20
Question
For Problems \(17-32\), indicate the solution set for each system of inequalities by shading the appropriate region. $$ \left(\begin{array}{l} 2 x-y \leq 4 \\ 2 x+y>4 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
The solution is the region below the line \( y = 2x - 4 \) and above the line \( y = -2x + 4 \).
1Step 1: Graph the first inequality
First, consider the inequality \( 2x - y \leq 4 \). To graph this, you'll first rewrite it in slope-intercept form by solving for \( y \): \( y \geq 2x - 4 \). This represents a straight line where the region below or on the line is shaded. Plot the line by identifying two points: when \( x = 0 \), \( y = -4 \) and when \( y = 0 \), \( x = 2 \). Draw the line solid to indicate that the region on the line is included, and shade below the line.
2Step 2: Graph the second inequality
For the inequality \( 2x + y > 4 \), rewrite it as \( y > -2x + 4 \). This time, plot the line by finding two points: when \( x = 0 \), \( y = 4 \) and when \( y = 0 \), \( x = 2 \). Since this is a strict inequality (\( > \)), draw a dashed line, indicating points on the line are not part of the solution. Shade above the line.
3Step 3: Identify the solution set
The solution to the system of inequalities is the region that satisfies both inequalities simultaneously. This means looking for the area on the graph where the shaded regions from Steps 1 and 2 overlap. Ensure the overlapping region accounts for the conditions correctly: solid line from first inequality and dashed from the second indicate different inclusions.
Key Concepts
Graphing InequalitiesSolution SetShaded Region
Graphing Inequalities
When graphing inequalities, you are essentially drawing lines on a coordinate plane to represent the constraints given by inequalities. Each inequality gets its own line, and different types of lines indicate different types of solutions. To graph an inequality, follow these steps:
- Convert the inequality into "y = mx + b" form, where "m" is the slope of the line and "b" is the y-intercept.
- If your inequality is "less than or equal to" (\(\leq\)), you'll draw a solid line. In this case, solutions are not just above or below the line, but also on the line itself.
- For a "greater than" (\(>\)) inequality, use a dashed line. This illustrates that solutions do not include the points on the line, only the region above or below it.
- For inequalities like \( y \leq mx + b \), shade below the line.
- For \( y \geq mx + b \), shade above the line.
Solution Set
The solution set of a system of inequalities is the region on the graph where all shaded areas from the individual inequalities overlap. Only this intersection of shaded regions represents solutions that simultaneously satisfy each inequality in the system.
Finding the solution set involves these steps:
- Plot each inequality on the same coordinate plane, following the graphing rules for solid or dashed lines as necessary.
- Shade the appropriate region for each inequality.
- Observe where the shaded sections overlap; this is your solution set.
- Dashed lines mean solutions hug but don't include the line.
- Solid lines include the line itself as well, within the solutions.
Shaded Region
The shaded region plays a crucial role in visualizing the solutions to inequalities. By shading, we indicate which part of the graph includes numbers that satisfy the particular inequality.Here's how to think about shading:
- The direction of shading depends on whether the inequality is greater-than or less-than. More specifically, you'll shade above the line if the inequality involves greater-than (\(y > mx + b\) or \(y \geq mx + b\)), and shade below for less-than (\(y < mx + b\) or \(y \leq mx + b\)).
- Using distinct shading for different inequalities can make seeing the overlap clearer and more intuitive.
- In systems of inequalities, only the areas where all shadings from each inequality overlap will give valid solutions.
Other exercises in this chapter
Problem 20
For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$
View solution Problem 20
For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{l} 2 x-y=9 \\ 7 x+4 y=1 \end{array}\right) $$
View solution Problem 21
For Problems \(1-28\), (a) graph each system so that approximate real number solutions (if there are any) can be predicted, and (b) solve each system using the
View solution Problem 21
For Problems \(11-30\), use Cramer's rule to find the solution set of each system. (Objective 2) $$ \left(\begin{array}{rr} 2 x-y+3 z & =-5 \\ 3 x+4 y-2 z & =-2
View solution