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Ch. 5 - Systems and Matrices
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 6, Problem 11

Graph each inequality. x + 2y ≤ 6

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Identify the inequality given: \(x + 2y \leq 6\). This represents a linear inequality in two variables.
Rewrite the inequality in slope-intercept form by isolating \(y\): subtract \(x\) from both sides to get \(2y \leq 6 - x\), then divide both sides by 2 to obtain \(y \leq \frac{6 - x}{2}\).
Graph the boundary line \(y = \frac{6 - x}{2}\). Since the inequality is \(\leq\), the boundary line should be solid, indicating points on the line satisfy the inequality.
Determine which side of the boundary line to shade by testing a point not on the line, such as the origin \((0,0)\). Substitute into the inequality: \(0 + 2(0) \leq 6\) which simplifies to \(0 \leq 6\), a true statement, so shade the region containing \((0,0)\).
Shade the region below or on the line \(y = \frac{6 - x}{2}\) to represent all solutions to the inequality \(x + 2y \leq 6\).

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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 first graphing the related linear equation as a boundary line. The inequality symbol determines whether the boundary is solid (≤ or ≥) or dashed (< or >). The solution region is the set of points that satisfy the inequality, typically shaded on one side of the boundary line.
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Slope-Intercept Form

Rearranging the inequality into slope-intercept form (y = mx + b) helps in graphing. For x + 2y ≤ 6, solving for y gives y ≤ (-1/2)x + 3. This form clearly shows the slope and y-intercept, making it easier to plot the boundary line accurately.
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Testing Points to Determine the Solution Region

After graphing the boundary line, select a test point not on the line (often the origin) to check if it satisfies the inequality. If the test point satisfies the inequality, shade the region containing that point; otherwise, shade the opposite side. This confirms the correct solution area.
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