Problem 26
Question
In Exercises 23-28, sketch the graph of the system of linear inequalities. $$ \left\\{\begin{array}{r} y \leq 10 \\ x+3 y \leq 15 \\ \geq x \\ y \geq 0 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The solution area on the graph is where all the four inequalities intersect. This area represents the solution to the given system of inequalities.
1Step 1: Graph the First Inequality
Draw a horizontal line passing through y = 10 on the y-axis. As the inequality is \(y \leq 10\), shade the area below the line.
2Step 2: Graph the Second Inequality
Rewrite the equation \(x + 3y \leq 15\) in intercept form to get \(y \leq \frac{15 - x}{3}\). Plot this line on the graph. Since the inequality is \(y \leq \frac{15 - x}{3}\), shade the area below this line.
3Step 3: Graph the Third Inequality
The inequality \(x \geq 0\) represents the area to the right of and including the y-axis. Shade this area on the graph.
4Step 4: Graph the Fourth Inequality
The inequality \(y \geq 0\) represents the area above and including the x-axis. Shade this area on the graph.
5Step 5: Identify the Solution Area
The solution area is the region that is shaded by all the four inequalities. This area represents all the possible (x, y) values that satisfy all the inequalities in the system.
Key Concepts
Inequality Shading TechniqueLinear Inequality GraphSolution Region Identification
Inequality Shading Technique
Understanding the inequality shading technique is fundamental when dealing with systems of linear inequalities. This technique involves shading the region of the coordinate plane that satisfies the inequality. Each linear inequality divides the plane into two regions: one that satisfies the inequality and one that does not. When confronting an inequality such as
\(y \leq 10\),
the process begins by drawing a boundary line, which in this case is a horizontal line at \(y = 10\). This line is solid, indicating that points on the line satisfy the inequality (since it includes the equality \(y = 10\)).
Shading is then applied to represent the solution set. For \(y \leq 10\), you would shade all the area below the line since all those points have a \(y\)-value less than or equal to 10. If the inequality were 'strict', such as \(y < 10\), a dashed line would be used, signifying that the line itself is not part of the solution set and only the area below it should be shaded. Applying this technique accurately to each inequality in a system is the key to finding the overall solution.
\(y \leq 10\),
the process begins by drawing a boundary line, which in this case is a horizontal line at \(y = 10\). This line is solid, indicating that points on the line satisfy the inequality (since it includes the equality \(y = 10\)).
Shading is then applied to represent the solution set. For \(y \leq 10\), you would shade all the area below the line since all those points have a \(y\)-value less than or equal to 10. If the inequality were 'strict', such as \(y < 10\), a dashed line would be used, signifying that the line itself is not part of the solution set and only the area below it should be shaded. Applying this technique accurately to each inequality in a system is the key to finding the overall solution.
Linear Inequality Graph
A linear inequality graph visually represents the solutions to a linear inequality. To graph an inequality like
\(x+3y \leq 15\),
you must first express it in a recognizable form, such as the intercept form, allowing you to find where the line intercepts the axes. In the intercept form, \(y \leq \frac{15 - x}{3}\), it's seen that the line crosses the \(y\)-axis at \(y = 5\) when \(x = 0\), and the \(x\)-axis at \(x = 15\) when \(y = 0\).
When drawing the graph, the line is plotted first. Since this is a '\(\leq\)' inequality, we use a solid line to encompass points lying exactly on the line. Next, we determine which side of the line to shade by choosing a test point not on the line. Often, \((0,0)\) is a convenient choice unless it lies on the graphed line. If the test point satisfies the inequality, the side containing that point is shaded. As with single inequalities, graphing systems requires considering all given inequalities simultaneously, and the graph becomes a powerful tool to visualize this.
\(x+3y \leq 15\),
you must first express it in a recognizable form, such as the intercept form, allowing you to find where the line intercepts the axes. In the intercept form, \(y \leq \frac{15 - x}{3}\), it's seen that the line crosses the \(y\)-axis at \(y = 5\) when \(x = 0\), and the \(x\)-axis at \(x = 15\) when \(y = 0\).
When drawing the graph, the line is plotted first. Since this is a '\(\leq\)' inequality, we use a solid line to encompass points lying exactly on the line. Next, we determine which side of the line to shade by choosing a test point not on the line. Often, \((0,0)\) is a convenient choice unless it lies on the graphed line. If the test point satisfies the inequality, the side containing that point is shaded. As with single inequalities, graphing systems requires considering all given inequalities simultaneously, and the graph becomes a powerful tool to visualize this.
Solution Region Identification
Solution region identification is the process of determining the common shaded area that satisfies all inequalities in the system. After graphing each inequality and applying the shading technique, the solution region is where the shaded areas overlap. Considering the given system of inequalities:
The solution region contains all the points that fulfill every single inequality in the system. In educational practice, to make sure the identified region is correct, one can pick a point from within the overlapping region and verify that it satisfies all inequalities. Accurately identifying this region is crucial, as it represents all the possible solutions to the system of inequalities.
- \(y \leq 10\)
- \(x+3y \leq 15\)
- \(x \geq 0\)
- \(y \geq 0\)
The solution region contains all the points that fulfill every single inequality in the system. In educational practice, to make sure the identified region is correct, one can pick a point from within the overlapping region and verify that it satisfies all inequalities. Accurately identifying this region is crucial, as it represents all the possible solutions to the system of inequalities.
Other exercises in this chapter
Problem 25
In Exercises 21-26, solve the system by the method of substitution. $$ \left\\{\begin{array}{rr} -6 x+1.5 y= & 6 \\ 8 x-2 y= & -8 \end{array}\right. $$
View solution Problem 25
In Exercises 19-26, solve the system by graphing. $$ \left\\{\begin{array}{r} x-2 y=4 \\ 2 x-4 y=8 \end{array}\right. $$
View solution Problem 26
In Exercises 23-28, use a system of linear equations to determine the number of each type of coin. 31 Nickels and quarters \(\$ 6.55\)
View solution Problem 26
In Exercises 25-28, solve the system by the method of elimination. $$ \left\\{\begin{array}{r} 7 x+10 y=0 \\ 21 x+30 y=0 \end{array}\right. $$
View solution