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Graphing Inequalities in Two Variables
11-6 Graphing Inequalities in Two Variables Course 3 Warm Up Problem of the Day Lesson Presentation
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables Warm Up Find each equation of direct variation, given that y varies directly with x. 1. y is 18 when x is 3. 2. x is 60 when y is 12. 3. y is 126 when x is 18. 4. x is 4 when y is 20. y = 6x y = x 1 5 y = 7x y = 5x
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables Problem of the Day The circumference of a pizza varies directly with its diameter. If you graph that direct variation, what will the slope be?
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Learn to graph inequalities on the coordinate plane.
Course 3 11-6 Graphing Inequalities in Two Variables Learn to graph inequalities on the coordinate plane.
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Insert Lesson Title Here
Course 3 11-6 Graphing Inequalities in Two Variables Insert Lesson Title Here Vocabulary boundary line linear inequality
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables A graph of a linear equation separates the coordinate plane into three parts: the points on one side of the line, the points on the boundary line, and the points on the other side of the line.
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables When the equality symbol is replaced in a linear equation by an inequality symbol, the statement is a linear inequality. Any ordered pair that makes the linear inequality true is a solution.
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Additional Example 1A: Graphing Inequalities
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 1A: Graphing Inequalities Graph each inequality. A. y < x – 1 First graph the boundary line y = x – 1. Since no points that are on the line are solutions of y < x – 1, make the line dashed. Then determine on which side of the line the solutions lie. (0, 0) Test a point not on the line. y < x – 1 0 < 0 – 1 ? Substitute 0 for x and 0 for y. 0 < –1 ?
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Additional Example 1A Continued
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 1A Continued Since 0 < –1 is not true, (0, 0) is not a solution of y < x – 1. Shade the side of the line that does not include (0, 0).
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Additional Example 1B: Graphing Inequalities
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 1B: Graphing Inequalities B. y 2x + 1 First graph the boundary line y = 2x + 1. Since points that are on the line are solutions of y 2x + 1, make the line solid. Then shade the part of the coordinate plane in which the rest of the solutions of y 2x + 1 lie. (0, 4) Choose any point not on the line. y ≥ 2x + 1 4 ≥ 0 + 1 ? Substitute 0 for x and 4 for y.
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Additional Example 1B Continued
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 1B Continued Since 4 1 is true, (0, 4) is a solution of y 2x + 1. Shade the side of the line that includes (0, 4).
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Additional Example 1C: Graphing Inequalities
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 1C: Graphing Inequalities C. 2y + 5x < 6 First write the equation in slope-intercept form. 2y + 5x < 6 2y < –5x + 6 Subtract 5x from both sides. y < – x + 3 5 2 Divide both sides by 2. Then graph the line y = – x + 3. Since points that are on the line are not solutions of y < – x + 3, make the line dashed. Then determine on which side of the line the solutions lie. 5 2
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Additional Example 1C Continued
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 1C Continued (0, 0) Choose any point not on the line. y < – x + 3 5 2 0 < 0 + 3 ? 0 < 3 ? Since 0 < 3 is true, (0, 0) is a solution of y < – x + 3. Shade the side of the line that includes (0, 0). 5 2
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 1A Graph each inequality. A. y < x – 4 First graph the boundary line y = x – 4. Since no points that are on the line are solutions of y < x – 4, make the line dashed. Then determine on which side of the line the solutions lie. (0, 0) Test a point not on the line. y < x – 4 0 < 0 – 4 ? Substitute 0 for x and 0 for y. 0 < –4 ?
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Try This: Example 1A Continued
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 1A Continued Since 0 < –4 is not true, (0, 0) is not a solution of y < x – 4. Shade the side of the line that does not include (0, 0).
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 1B B. y > 4x + 4 First graph the boundary line y = 4x + 4. Since points that are on the line are solutions of y 4x + 4, make the line solid. Then shade the part of the coordinate plane in which the rest of the solutions of y 4x + 4 lie. (2, 3) Choose any point not on the line. y ≥ 4x + 4 3 ≥ 8 + 4 ? Substitute 2 for x and 3 for y.
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Try This: Example 1B Continued
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 1B Continued Since 3 12 is not true, (2, 3) is not a solution of y 4x + 4. Shade the side of the line that does not include (2, 3).
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 1C C. 3y + 4x 9 First write the equation in slope-intercept form. 3y + 4x 9 3y –4x + 9 Subtract 4x from both sides. y – x + 3 4 3 Divide both sides by 3. 4 3 Then graph the line y = – x + 3. Since points that are on the line are solutions of y – x + 3, make the line solid. Then determine on which side of the line the solutions lie. 4 3
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Try This: Example 1C Continued
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 1C Continued (0, 0) Choose any point not on the line. y – x + 3 4 3 0 0 + 3 ? 0 3 ? Since 0 3 is not true, (0, 0) is not a solution of y – x + 3. Shade the side of the line that does not include (0, 0). 4 3
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Additional Example 2: Career Application
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 2: Career Application A successful screenwriter can write no more than seven and a half pages of dialogue each day. Graph the relationship between the number of pages the writer can write and the number of days. At this rate, would the writer be able to write a 200-page screenplay in 30 days? First find the equation of the line that corresponds to the inequality. In 0 days the writer writes 0 pages. point (0, 0) In 1 day the writer writes no more than 7 pages. 1 2 point (1, 7.5)
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Additional Example 2 Continued
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 2 Continued With two known points, find the slope. m = 7.5 – 0 1 – 0 7.5 1 = = 7.5 y 7.5 x + 0 The y-intercept is 0. No more than means . Graph the boundary line y = 7.5x. Since points on the line are solutions of y 7.5x make the line solid. Shade the part of the coordinate plane in which the rest of the solutions of y 7.5x lie.
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Additional Example 2 Continued
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 2 Continued (2, 2) Choose any point not on the line. y 7.5x 2 7.5 2 ? Substitute 2 for x and 2 for y. 2 15 ? Since 2 15 is true, (2, 2) is a solution of y 7.5x. Shade the side of the line that includes point (2, 2).
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Additional Example 2 Continued
Course 3 11-6 Graphing Inequalities in Two Variables Additional Example 2 Continued The point (30, 200) is included in the shaded area, so the writer should be able to complete the 200 page screenplay in 30 days.
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 2 A certain author can write no more than 20 pages every 5 days. Graph the relationship between the number of pages the writer can write and the number of days. At this rate, would the writer be able to write 140 pages in 20 days? First find the equation of the line that corresponds to the inequality. In 0 days the writer writes 0 pages. point (0, 0) In 5 days the writer writes no more than 20 pages. point (5, 20)
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Try This: Example 2 Continued
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 2 Continued 20 - 0 5 - 0 m = = 20 5 = 4 With two known points, find the slope. The y-intercept is 0. No more than means . y 4x + 0 Graph the boundary line y = 4x. Since points on the line are solutions of y 4x make the line solid. Shade the part of the coordinate plane in which the rest of the solutions of y 4x lie.
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Try This: Example 2 Continued
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 2 Continued (5, 60) Choose any point not on the line. y 4x 60 4 5 ? Substitute 5 for x and 60 for y. 60 20 ? Since 60 20 is not true, (5, 60) is not a solution of y 4x. Shade the side of the line that does not include (5, 60).
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Try This: Example 2 Continued
Course 3 11-6 Graphing Inequalities in Two Variables Try This: Example 2 Continued y 200 180 160 140 120 100 80 60 40\ 20 Pages x Days The point (20, 140) is not included in the shaded area, so the writer will not be able to write 140 pages in 20 days.
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables Lesson Quiz Graph each inequality. 1. y < – x + 4 2. 4y + 2x > 12 Tell whether the given ordered pair is a solution of each inequality. 3. y < x + 15 (–2, 8) 4. y 3x – 1 (7, –1) 1 3
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables 1 3 1. y < – x + 4
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables 2. 4y + 2x > 12
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Graphing Inequalities in Two Variables
Course 3 11-6 Graphing Inequalities in Two Variables Tell whether the given ordered pair is a solution of each inequality. 3. y < x + 15 (–2, 8) 4. y 3x – 1 (7, –1) yes no
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