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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
Chapter 6, Problem 7

Graph each inequality. y>2x−1

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1
Identify the inequality given: \(y > 2x - 1\). This represents all points \((x, y)\) where the \(y\)-value is greater than \(2x - 1\).
First, graph the boundary line \(y = 2x - 1\). This is a straight line with slope \(2\) and \(y\)-intercept \(-1\).
Since the inequality is strict (\(>\), not \(\geq\)), draw the boundary line as a dashed line to indicate points on the line are not included in the solution.
Choose a test point not on the line, commonly \((0,0)\), and substitute into the inequality: check if \(0 > 2(0) - 1\) which simplifies to \(0 > -1\). Since this is true, shade the region of the graph that contains \((0,0)\).
The shaded region represents all solutions to the inequality \(y > 2x - 1\). This completes the graph of the inequality.

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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 (for ≤ or ≥) or dashed (for < 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 of a Line

The slope-intercept form, y = mx + b, expresses a line where m is the slope and b is the y-intercept. It helps quickly graph the line by starting at (0, b) and using the slope to find other points. For y > 2x - 1, the slope is 2 and the y-intercept is -1.
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Graphing Lines in Slope-Intercept Form

Testing Points to Determine the Solution Region

After graphing the boundary line, select a test point not on the line (often (0,0)) to check if it satisfies the inequality. If the test point makes the inequality true, shade the region containing that point; otherwise, shade the opposite side. This confirms the correct solution area.
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