1

Topic 4.5.2

Borders of Regions

2

Borders of Regions

California Standard: 6.0 Students graph a linear equation and compute the x- and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 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

Topic4.5.2

3

Borders of Regions

In Topic 4.5.1 you were dealing with regions defined by strict inequalities — the ones involving a < or > sign.

This Topic shows you how to graph inequalities involving and signs too.

Topic4.5.2

–2y + 3x > 6y < –x + 3y > 0.5x + 2

4y – 3x 12 y 2x + 32x – y –2

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.

Topic4.5.2

For example, the region defined by y > –x + 3 doesn’t include any points on the line y = –x + 3.

2

4

6y

–2

–4

–6

–6 –40

–2 0 2 4 6x

y > –x + 3

5

Borders of RegionsTopic4.5.2

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

6y

–2

–4

–6

–6 –40

–2 0 2 4 6x

y –x + 3

In this case, you draw the border line as a solid line.

6

Borders of Regions

Example 1

Graph 2x – y = –2 and show the three regions of the plane that include all the points on this line.

Solution follows…

Solution

Topic4.5.2

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

2x – y –2

Set of points on the line — all these points satisfy the equation 2x – y = –2 and both the inequalities.

Set of points above and on the line — all

these points satisfy the inequality 2x – y –2.

Set of points below and on the line — all

these points satisfy the inequality 2x – y –2.

2x – y –2

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.

Topic4.5.2

< or > = dashed line or = solid line

8

Borders of Regions

Example 2

Sketch the region of the coordinate plane defined by y –x – 5.

Solution follows…

Solution

Topic4.5.2

First plot the border line. The border line equation is y = –x – 5. Table of values for sketching the line:

x y (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 –2 0 2 4 60

2

4

6

x

–2

–4

–6

Solution continues…

9

Borders of Regions

Example 2

Sketch the region of the coordinate plane defined by y –x – 5.Solution (continued)

Topic4.5.2

Identify the border type: it’s solid, since the sign is . y

–6 –4 –2 0 2 4 60

2

4

6

x

–2

–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

Borders of Regions

Guided Practice

Solution follows…

In Exercises 1–4, show whether the given point is in the solution set of –2x + 3y –15.

Topic4.5.2

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

Borders of Regions

Guided Practice

Solution follows…

Topic4.5.2

5. 4x + 3y 9

6. y 2x + 3

7. –2y x

8. 2x + 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

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

y 2x + 3

–2y x

2x + y 4

4x + 3y 9

12

Borders of Regions

Example 3

Graph the solution set of 4y – 3x 12.

Solution follows…

Solution

Topic4.5.2

First form the border-line equation:

Table of values for sketching the line:

x y (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…

4y – 3x = 12 4y = 3x + 12 y = x + 334

y

–6 –4 –2 0 2 4 60

2

4

6

x

–2

–4

–6

13

Borders of Regions

Example 3

Graph the solution set of 4y – 3x 12.

Solution (continued)

Topic4.5.2

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 4y – 3x 12 — so shade the region that doesn’t contain (0, 0).

4y – 3x 124(0) – 3(0) 12 0 12 — This is false.

The border line is solid, since the sign is . y

–6 –4 –2 0 2 4 60

2

4

6

x

–2

–4

–6

4y – 3x 12

14

Borders of Regions

Example 4

Sketch the region of the coordinate plane defined by 2y < 2x + 6.

Solution follows…

Solution

Topic4.5.2

First form the border line equation.

Table of values for sketching the line:x y (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…

2y = 2x + 6 y = x + 3

The border line is dashed, since the sign is <.

y

–6 –4 –2 0 2 4 60

2

4

6

x

–2

–4

–6

15

Borders of Regions

Example 4

Sketch the region of the coordinate plane defined by 2y < 2x + 6. Solution (continued)

Topic4.5.2

Therefore (0, 0) lies in the region 2y < 2x + 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.

2y < 2x + 62(0) < 2(0) + 6 0 < 6 — This is a true statement.

y

–6 –4 –2 0 2 4 60

2

4

6

x

–2

–4

–6

2y < 2x + 6

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.

Topic4.5.2

9. y x

10. y – 2 < 0

11. x + 2 > 0

12. x – 4 0

13. y + 3 0

14. y –3x

91112

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

141013

y xy – 2 < 0

x + 2 > 0x – 4 0

y + 3 0

y –3x

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.

Topic4.5.2

15. y 2x – 5

16. x + 4y < 4

17. y > 2x + 6

18. 4x – 3y 8

151617

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

18

y 2x – 5

y > 2x + 6

4x – 3y 8x + 4y < 4

18

Borders of Regions

Guided Practice

Solution follows…

19. Show whether (5, 4) is in the solution set of 4x – 3y 8.

Topic4.5.2

19

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

20

20. Show whether (–4, 2) is in the solution set of 2x + y > –6.

(5, 4)

(5, 4) is in the solution set of 4x – 3y 8.

(–4, 2) is not in the solution set of 4x – 3y 8.

(–4, 2)

19

Independent Practice

Solution follows…

Borders of RegionsTopic4.5.2

In Exercises 1–4, show whether the given point is in the solution set of the given line.

1. (–1, 3); 2x + y < 1

2. (–2, –8); x – y 6

3. (0.5, 0.25); x + 2y 1

4. (1, –5); x + 2y > –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

Independent Practice

Solution follows…

no

Borders of RegionsTopic4.5.2

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

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

y = x + 221

(–1, 1)

(4, 6)

21

Independent Practice

Solution follows…

yes

Borders of RegionsTopic4.5.2

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 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

y = –2(3, –2)

(–2, 3)

22

Independent Practice

Borders of RegionsTopic4.5.2

Graph the solution set in Exercises 9–14.

9. x > 4

10. y –2

11. x – 3y 9

12. x + y < 5

13. x + y –1

14. y 2x – 7

Solution follows…

9101114

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

1213

x > 4

y –2

x + y < 5x + y –1

y 2x – 7x – 3y 9

23

Independent Practice

Solution follows…

Borders of RegionsTopic4.5.2

Graph the solution set in Exercises 15–20.

15. x + 2y < 2

16. y 2x + 2

17. y 3x

18. y > –2x + 1

19. x + y 3

20. 2x + y –3

15161720

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

1819

x + 2y < 2

y 2x + 2y 3x y > –2x + 1x + y 3

2x + y –3

24

Round Up

Remember — graphs of inequalities including < and > signs will always have a dashed line, and graphs including and signs will always have a solid line.

Borders of RegionsTopic4.5.2