1. 1 Topic 4.5.2 Borders of Regions

2. 2 California Standard: 6.0 Students graph a linear equation and compute the x - and y - intercepts (e.g., graph 2 x + 6 y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2 x + 6 y < 4). What it means for you: You’ll learn how to show the different types of inequality on a graph. Key Words: inequality strict inequality border region Topic 4.5.2

3. 3 Borders of Regions In Topic 4.5.1 you were dealing with regions defined by strict inequalities — the ones involving a sign. This Topic shows you how to graph inequalities involving  and  signs too. Topic 4.5.2 –2 y + 3 x > 6 y < – x + 3 y > 0.5 x + 2 4 y – 3 x  12 y  2 x + 32 x – y  –2

4. 4 Borders of Regions Regions Can Have Different Types of Borders The region defined by a strict inequality doesn’t include points on the border line, and you draw the border line as a dashed line. Topic 4.5.2 For example, the region defined by y > – x + 3 doesn’t include any points on the line y = – x + 3. 2 4 6 y –2 –4 –6 –4 0 –20246 x y > – x + 3

5. 5 Borders of Regions Topic 4.5.2 Regions defined by inequalities involving a  or  sign do include points on the border line. For example, the region defined by y  – x + 3 includes all the points on the line y = – x + 3. 2 4 6 y –2 –4 –6 –4 0 –20246 x y  – x + 3 In this case, you draw the border line as a solid line.

6. 6 Borders of Regions Example 1 Graph 2 x – y = –2 and show the three regions of the plane that include all the points on this line. Solution follows… Solution Topic 4.5.2 –6–4–20246 0 2 4 6 y x –4 –6 2 x – y  –2 Set of points on the line — all these points satisfy the equation 2 x – y = –2 and both the inequalities. Set of points above and on the line — all these points satisfy the inequality 2 x – y  –2. Set of points below and on the line — all these points satisfy the inequality 2 x – y  –2. 2 x – y  –2

7. 7 Borders of Regions Sketching Regions Inclusive of the Border Line The method for sketching the region is the same as the one in Topic 4.5.1, except now there’s an extra step for showing the border type — using a dashed line, or a solid line. Topic 4.5.2 = dashed line  or  = solid line

8. 8 Borders of Regions Example 2 Sketch the region of the coordinate plane defined by y  – x – 5. Solution follows… Solution Topic 4.5.2 First plot the border line. The border line equation is y = – x – 5. Table of values for sketching the line: xy ( x, y ) 0 y = – x – 5 = –(0) – 5 = –5(0, –5) –3 y = – x – 3 = –(–3) – 5 = –2(–3, –2) So the border line goes through the points (0, –5) and (–3, –2), as shown. y –6–4–20246 0 2 4 6 x –4 –6 Solution continues…

9. 9 Borders of Regions Example 2 Sketch the region of the coordinate plane defined by y  – x – 5. Solution (continued) Topic 4.5.2 Identify the border type: it’s solid, since the sign is . y –6–4–20246 0 2 4 6 x –4 –6 Test whether the point (0, 0) satisfies the inequality — substitute x = 0 and y = 0 into the inequality. y  – x – 5  0  –0 – 5 0  –5 — this is false. Therefore (0, 0) doesn’t lie in the region y  – x – 5 — so shade the region that doesn’t contain (0, 0). y  – x – 5

10. 10 Borders of Regions Guided Practice Solution follows… In Exercises 1–4, show whether the given point is in the solution set of –2 x + 3 y  –15. Topic 4.5.2 1. (0, 0) 2. (6, –1) 3. (4, –7) 4. (–3, –5) –2(4) + 3(–7)  –15  –29  –15. This is a true statement, so (4, –7) is a solution. –2(0) + 3(0)  –15  0  –15. This is a false statement, so (0, 0) is not a solution. –2(–3) + 3(–5)  –15  –9  –15. This is a false statement, so (–3, –5) is not a solution. –2(6) + 3(–1)  –15  –15  –15. This is a true statement, so (6, –1) is a solution.

11. 11 Borders of Regions Guided Practice Solution follows… Topic 4.5.2 5. 4 x + 3 y  9 6. y  2 x + 3 7. –2 y  x 8. 2 x + y  4 In each of Exercises 5–8 use a set of axes spanning from –6 to 6 on the x - and y -axes, and shade the region defined by the inequality. 5678 y  2 x + 3 –2 y  x 2 x + y  4 4 x + 3 y  9

12. 12 Borders of Regions Example 3 Graph the solution set of 4 y – 3 x  12. Solution follows… Solution Topic 4.5.2 First form the border-line equation: Table of values for sketching the line: xy ( x, y ) 0 y = ¾ x + 3 = ¾(0) + 3 = 3(0, 3) 4 y = ¾ x + 3 = ¾(4) + 3 = 6(4, 6) So the border line goes through the points (0, 3) and (4, 6). Solution continues… 4 y – 3 x = 12 4 y = 3 x + 12  y = x + 3 3 4  y –6–4–20246 0 2 4 6 x –4 –6

13. 13 Borders of Regions Example 3 Graph the solution set of 4 y – 3 x  12. Solution (continued) Topic 4.5.2 Test whether the point (0, 0) satisfies the inequality — substitute x = 0 and y = 0 into the inequality. Therefore (0, 0) doesn’t lie in the region 4 y – 3 x  12 — so shade the region that doesn’t contain (0, 0). 4 y – 3 x  12 4(0) – 3(0)  12 0  12 — This is false. The border line is solid, since the sign is . y –6–4–20246 0 2 4 6 x –4 –6 4 y – 3 x  12

14. 14 Borders of Regions Example 4 Sketch the region of the coordinate plane defined by 2 y < 2 x + 6. Solution follows… Solution Topic 4.5.2 First form the border line equation. Table of values for sketching the line: xy ( x, y ) 0 y = x + 3 = 0 + 3 = 3(0, 3) 3 y = x + 3 = 3 + 3 = 6(3, 6) So the border line goes through the points (0, 3) and (3, 6). Solution continues… 2 y = 2 x + 6  y = x + 3 The border line is dashed, since the sign is <. y –6–4–20246 0 2 4 6 x –4 –6

15. 15 Borders of Regions Example 4 Sketch the region of the coordinate plane defined by 2 y < 2 x + 6. Solution (continued) Topic 4.5.2 Therefore (0, 0) lies in the region 2 y < 2 x + 6 — so shade the region containing (0, 0). Test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. 2 y < 2 x + 6 2(0) < 2(0) + 6 0 < 6 — This is a true statement. y –6–4–20246 0 2 4 6 x –4 –6 2 y < 2 x + 6

16. 16 Borders of Regions Guided Practice Solution follows… In each of Exercises 9–14, use a set of axes spanning from –6 to 6 on the x - and y -axes. For each exercise, shade the region defined by the inequality. Topic 4.5.2 9. y  x 10. y – 2 < 0 11. x + 2 > 0 12. x – 4  0 13. y + 3  0 14. y  –3 x 91112141013 y  xy  x y – 2 < 0 x + 2 > 0 x – 4  0 y + 3  0 y  –3 x

17. 17 Borders of Regions Guided Practice Solution follows… In each of Exercises 15–18, use a set of axes spanning from –6 to 6 on the x - and y -axes. For each exercise, shade the region defined by the inequality. Topic 4.5.2 15. y  2 x – 5 16. x + 4 y < 4 17. y > 2 x + 6 18. 4 x – 3 y  8 15161718 y  2 x – 5 y > 2 x + 6 4 x – 3 y  8 x + 4 y < 4

18. 18 Borders of Regions Guided Practice Solution follows… 19. Show whether (5, 4) is in the solution set of 4 x – 3 y  8. Topic 4.5.2 1920 20. Show whether (–4, 2) is in the solution set of 2 x + y > –6. (5, 4) (5, 4) is in the solution set of 4 x – 3 y  8. (–4, 2) is not in the solution set of 4 x – 3 y  8. (–4, 2)

19. 19 Independent Practice Solution follows… Borders of Regions Topic 4.5.2 In Exercises 1–4, show whether the given point is in the solution set of the given line. 1. (–1, 3); 2 x + y < 1 2. (–2, –8); x – y  6 3. (0.5, 0.25); x + 2 y  1 4. (1, –5); x + 2 y > –9 2(–1) + 3 < 1  1 < 1. This is a false statement, so (–1, 3) is not a solution. –2 – (–8)  6  6  6. This is a true statement, so (–2, –8) is a solution. 0.5 + 2(0.25)  1  1  1. This is a true statement, so (0.5, 0.25) is a solution. 1 + 2(–5) > –9  –9 > –9. This is a false statement, so (1, –5) is not a solution.

20. 20 Independent Practice Solution follows… no Borders of Regions Topic 4.5.2 In Exercises 5–6, determine if the given point is in the solution set shown by the shaded region of the graph. 5. (4, 6) 6. (–1, 1) yes y = x + 2 2 1 (–1, 1) (4, 6)

21. 21 Independent Practice Solution follows… yes Borders of Regions Topic 4.5.2 In Exercises 7–8, determine if the given point is in the solution set shown by the shaded region of the graph. 7. (3, –2) 8. (–2, 3) yes –6–4–20246 0 2 4 6 y x –4 –6 y = –2 (3, –2) (–2, 3)

22. 22 Independent Practice Borders of Regions Topic 4.5.2 Graph the solution set in Exercises 9–14. 9. x > 4 10. y  –2 11. x – 3 y  9 12. x + y < 5 13. x + y  –1 14. y  2 x – 7 Solution follows… 91011141213 x > 4 y  –2 x + y < 5 x + y  –1 y  2 x – 7 x – 3 y  9

23. 23 Independent Practice Solution follows… Borders of Regions Topic 4.5.2 Graph the solution set in Exercises 15–20. 15. x + 2 y < 2 16. y  2 x + 2 17. y  3 x 18. y > –2 x + 1 19. x + y  3 20. 2 x + y  –3 151617201819 x + 2 y < 2 y  2 x + 2 y  3 x y > –2 x + 1 x + y  3 2 x + y  –3

24. 24 Round Up Remember — graphs of inequalities including signs will always have a dashed line, and graphs including  and  signs will always have a solid line. Borders of Regions Topic 4.5.2