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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblem 27
Chapter 6, Problem 27

Graph the solution set of each system of inequalities or indicate that the system has no solution.
{3x+6y≤62x+y≤8\(\begin{cases}\)3x + 6y \(\leq\) 6 \\2x + y \(\leq\) 8\(\end{cases}\)

Verified step by step guidance
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Step 1: Identify the system of inequalities to graph: \[\left\{ \begin{array}{l} 2x + 4y \leq 8 \\ 4x + y \leq 4 \end{array} \right.\]
Step 2: Convert each inequality into an equation to find the boundary lines: For the first inequality: \[2x + 4y = 8\] For the second inequality: \[4x + y = 4\]
Step 3: Find the intercepts for each boundary line to help graph them: - For \[2x + 4y = 8\]: - When \[x=0\], solve for \[y\]: \[4y = 8 \Rightarrow y = 2\] - When \[y=0\], solve for \[x\]: \[2x = 8 \Rightarrow x = 4\] - For \[4x + y = 4\]: - When \[x=0\], solve for \[y\]: \[y = 4\] - When \[y=0\], solve for \[x\]: \[4x = 4 \Rightarrow x = 1\]
Step 4: Graph the boundary lines using the intercepts found. Since the inequalities are \( \leq \), shade the region below or on each line. To determine which side to shade, pick a test point (usually the origin \(0,0\)) and check if it satisfies the inequality.
Step 5: The solution set to the system is the intersection of the shaded regions from both inequalities. This overlapping region represents all points \((x,y)\) that satisfy both inequalities simultaneously.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Graphing Linear Inequalities

Graphing linear inequalities involves plotting the boundary line of the corresponding equation and shading the region that satisfies the inequality. The boundary line is solid if the inequality includes equality (≤ or ≥) and dashed if it does not (< or >). This visual representation helps identify all possible solutions to the inequality.
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System of Inequalities

A system of inequalities consists of two or more inequalities considered simultaneously. The solution set is the intersection of the regions that satisfy each inequality individually. Graphing both inequalities on the same coordinate plane reveals the common shaded area representing all solutions.
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Finding Boundary Lines and Test Points

To graph inequalities, first convert each inequality to an equation to find the boundary line. Then, select a test point not on the line (commonly the origin) to determine which side of the line satisfies the inequality. This method ensures accurate shading of the solution region.
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Related Practice
Textbook Question

In Exercises 19–28, solve each system by the addition method. {y2−x=4x2+y2=4\(\begin{cases}\)y^2 - x = 4 \(\x\)^2 + y^2 = 4\(\end{cases}\)

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Textbook Question

In Exercises 25–35, solve each system by the method of your choice. This is a piecewise function, refer to textbook problem.

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Textbook Question

Write the partial fraction decomposition of each rational expression. x2/(x − 1)2 (x + 1)

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Textbook Question

In Exercises 25–35, solve each system by the method of your choice. This is a piecewise function, refer to textbook problem.

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Textbook Question

In Exercises 19–30, solve each system by the addition method. 3x - 4y = 11 2x + 3y = - 4

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Textbook Question

Solve each system in Exercises 25–26. {x+32−y−12+z+24=32x−52+y+13−z4=−256x−34−y+12+z−32=−52\(\begin{cases}\[\frac{x + 3}{2}\) - \(\frac{y - 1}{2}\) + \(\frac{z + 2}{4}\) = \(\frac{3}{2}\) \(\frac{x - 5}{2}\) + \(\frac{y + 1}{3}\) - \(\frac{z}{4}\) = - \(\frac{25}{6}\) \(\frac{x - 3}{4}\) - \(\frac{y + 1}{2}\) + \(\frac{z - 3}{2}\) = - \(\frac{5}{2}\]\end{cases}\)

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