1. Chapter 3 Section 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Graphing Linear Inequalities in Two Variables 1 1 3 3 2 2 3.53.5 Graph ≤ or ≥ linear inequalities. Graph linear inequalities. Graph inequalities with a boundary through the origin.

3. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley An inequality that can be written as or where A, B, and C are real numbers and A and B are both not 0, is a linear inequality in two variables. The symbols ≤ and ≥ may replace in the definition. Graphing Linear Inequalities in Two Variables In Section 3.2, we graphed linear equations such as 2x + 3y = 6. Now we extend this work to linear inequalities in two variables, such as 2x + 3y ≤ 6. Slide 3.5 - 3

4. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1 Objective 1 Graph ≤ or ≥ linear inequalities. Slide 3.5 - 4

5. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph ≤ or ≥ linear inequalities. The inequality 2x + 3y ≤ 6 means that or Slide 3.5 - 5 The graph of 2x + 3y = 6 is a line. This boundary line divides the plane into two regions. The graph of the solutions of the inequality 2x + 3y < 6 will include only one of these regions. The required region is found by solving the inequality for y. From this last statement, ordered pairs in which y is less than or equal to will be solutions of the inequality. Ordered pairs in which y is equal to are on the boundary line, so pairs in which y is less than will be below that line.

6. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph ≤ or ≥ linear inequalities. (cont’d) Slide 3.5 - 6 Alternatively, a test point gives a quick way to find the correct region to shade. We choose any point not on the boundary line. Because (0,0) is easy to substitute into an inequality, it is often a good choice. We substitute 0 for x and y in the given inequality to see whether the resulting statement is true or false. Since the last statement is true, we shade the region that includes the test point (0,0). To indicate the solutions, shade the region below the line. The shaded region, along with the boundary line, is the desired graph.

7. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing a Linear Inequality Slide 3.5 - 7 Step 1: Graph the boundary. Graph the line that is the boundary of the region. Use the methods of Section 3.2. Draw a solid line if the inequality has ≤ or ≥ because of the equality portion of the symbol; draw a dashed line if the inequality has. Step 2: Shade the appropriate side. Use any point not on the line as a test point. Substitute for x and y in the inequality. If a true statement results, shade the side containing the test point. If a false statement results, shade the other side.

8. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Graphing a Linear Inequality Solution: The graph should include a dotted line since there is no equal sign in the equation. Slide 3.5 - 8 Graph 3x + 5y >15. Since the statement is false the region above the dotted line should be shaded.

9. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2 Objective 2 Graph linear inequalities. Slide 3.5 - 9

10. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Graphing a Linear Inequality with a Horizontal Boundary Line Slide 3.5 - 10 Graph y < 4. Solution: The graph should include a dotted line since there is no equal sign in the equation. Since the statement is true the region to the left of the dotted line should be shaded.

11. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3 Objective 3 Graph inequalities with a boundary through the origin. Slide 3.5 - 11

12. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a Linear Inequality with a Boundary Line through the Origin Slide 3.5 - 12 Graph x ≥ −3y. Solution: The graph should include a solid line since there is an equality in the equation. Test point (1,1) is used. Since the statement is true the region above the line should be shaded. When the line goes through the origin, (0,0) cannot be used as a test point.