1 Topic 4.5.3 Regions Defined by Two Inequalities Regions Defined by Two Inequalities.

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Topic 4.5.3Topic 4.5.3

Regions Defined by Two Inequalities

Regions Defined by Two Inequalities

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Regions Defined by Two InequalitiesRegions Defined by Two Inequalities

California Standard: 9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets.

What it means for you:You’ll graph two inequalities to show the solution set that satisfies both inequalities.

Key Words:• system of linear inequalities• region• point-slope formula

Topic4.5.3

3


The idea of graphing two different inequalities on one graph is really not as hard as it sounds.

When you’ve finished, your graph will show the region where all the points satisfy both inequalities.

Topic4.5.3

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

4


Regions Defined by More Than One Linear Inequality

A system of linear inequalities is made up of two or more linear inequalities that contain the same variables.

Topic4.5.3

For example, 3x + 2y > 6 and 4x – y < 5 are linear inequalities both containing the variables x and y.

An ordered pair (x, y) is a solution of a system of linear inequalities if it is a solution of each of the inequalities in the system.

For example, (1, –2) is a solution of the system of inequalities y < –x + 2 and 2y < 2x + 6.

The graph of two linear inequalities is the region consisting of all the points satisfying both inequalities.

5

= region defined by y < –x + 2


Example 1

Sketch the region satisfying both y < –x + 2 and 2y < 2x + 6.

Solution follows…

Solution

Topic4.5.3

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

Line 2y = 2x + 6

Line y = –x + 2

Set of points satisfying both y < –x + 2 and 2y < 2x + 6.

= region defined by 2y < 2x + 6

= region defined by y < –x + 2

and 2y < 2x + 6

6


Sketching Regions Defined by Two Linear Inequalities

To sketch the region defined by two linear inequalities, shade the regions defined by each inequality.

Topic4.5.3

The region defined by both inequalities is the area where your shading overlaps.

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

y < –x + 2 2y < 2x + 6

y < –x + 2 and 2y < 2x + 6

7


Example 2

Solution follows…

Solution

Topic4.5.3

First line:

Solution continues…

5y + 3x = –25

5y = –3x – 25

Graph the solution set satisfying 5y + 3x –25 and y – x –5.

y = – x – 53

5

x y (x, y)

0 y = – x – 5 = – (0) – 5 = –5 (0, –5)

5 y = – x – 5 = – (5) – 5 = –8 (5, –8)

3

5

3

5

3

5

3

5

The border line y = – x – 5 goes through the points (0, –5)

and (5, –8), and is a solid line.

3

5

8Solution continues…


Example 2

Solution (continued)

Topic4.5.3

Second line:

y – x = –5

y = x – 5


The border line y = x – 5 goes through the points (0, –5)and (3, –2), and is a solid line.

x y (x, y)

0 y = x – 5 = 0 – 5 = –5 (0, –5)

3 y = x – 5 = 3 – 5 = –2 (3, –2)

Now you can plot the two lines on the same set of axes.



Example 2


Topic4.5.3

First line:


The border line y = – x – 5

goes through the points (0, –5)and (5, –8), and is a solid line.

3

5

Second line:

The border line y = x – 5 goes through the points (0, –5) and (3, –2), and is a solid line.

y

–3 0 3 6 9

–3

0

3

x

–6

–9

y = x – 5

y = x – 53

5



Example 2


Topic4.5.3


Test whether the point (0, 0) satisfies each inequality.

5y + 3x –255(0) + 3(0) –25 0 –25

This is a true statement, so (0, 0) lies in the region 5y + 3x –25.

Shade the region above y = – x – 5.3

5

First inequality:y

–3 0 3 6 9

–3

0

3

x

–6

–9

y = x – 5

y = x – 53

5

5y + 3x –25

11


Example 2


Topic4.5.3


Test whether the point (0, 0) satisfies each inequality.

y – x –50 – 0 –5 0 –5

This is a false statement, so (0, 0) doesn’t lie in the region y – x –5.

Shade the region below y = –x – 5.

Second inequality:

The required region is the area where the shading overlaps.

y

–3 0 3 6 9

–3

0

3

x

–6

–9

y = x – 5

y = x – 53

5

and y – x –55y + 3x –25

12


Guided Practice

Solution follows…

In each of Exercises 1–2, use a set of axes spanning from –5 to 7 on the x-axis and –6 to 6 on the y-axis. For each exercise, shade the region containing all solution points for both inequalities.

Topic4.5.3

1. 2x + 3y < 6 and y – 2x < 2

2. y – x 4 and 2x + y 5

Test point: (0, 0)2x + 3y < 60 > 6 – TrueShade below the line

y – 2x < 20 < 2 – TrueShade below the line

Test point: (0, 0)y – x 40 4 – FalseShade above the line

2x + y 50 5 – TrueShade below the line

–4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

12

2x + 3y < 6 and y – 2x < 2

y – x 4 and 2x + y 5

13

3. y < –x + 4 and y < x

4. y < x + 2 and y > –2x + 5


Guided Practice

Solution follows…

In each of Exercises 3–4, use a set of axes spanning from –6 to 6 on the x- and y-axes. For each exercise, shade the region containing all solution points for both inequalities.

Topic4.5.3

Test point: (2, 1)y < –x + 41 < 2 – TrueShade below the line

y < x1 < 2 – TrueShade below the line

Test point: (0, 0)y < x + 20 < 2 – TrueShade below the line

y > –2x + 50 > 5 – FalseShade above the line

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

34

y < x + 2 and

y > –2x + 5

y < –x + 4 and y < x

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Guided Practice

Solution follows…

In each of Exercises 5–6, use a set of axes spanning from –6 to 6 on the x- and y-axes. For each exercise, shade the region containing all solution points for both inequalities.

Topic4.5.3

5. y < x and y < –x – 2

6. y 0.5x + 3 and y 2x + 2

Test point: (1, –1)y < x–1 < 1 – TrueShade below the line

y < –x – 2–1 < –3 – FalseShade below the line

Test point: (0, 0)y 0.5x + 3 0 3 – TrueShade below the line

y 2x + 20 2 – TrueShade below the line

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

56

y < x and y < –x – 2

y 0.5x + 3 and

y 2x + 2

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Identifying the Inequalities Defining a Region

To identify the inequalities defining a region, you first need to establish the border line equations.

Topic4.5.3

Then examine a point in the region to identify the inequalities — the method’s coming up next.

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Here’s the method:

Topic4.5.3

• Find the equations — use the point-slope formula to find the equations of each of the border lines.

• Identify the signs — choose a point in the region (but not on a line) and substitute its coordinates into each equation. Since the point does not lie on either of the lines, you will have two false statements. Replace the “=” in each equation with a “<” or “>” sign to make the statements true. If the line is solid, use “” or “.”

• Write the inequalities which define the region.

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Example 3

Find the inequalities whose simultaneous solution set is the shaded region shown below.

Solution follows…

Solution

Topic4.5.3

First line: Two points on this line are (0, 1) and (3, 2).

m1 = = 3 1 2 – 1

3 – 0

3 1

y – 2 = x – 1

3 1 y – 2 = (x – 3)

y – y1 = m(x – x1)

y = x + 1 3 1

2

4

6

y

0

–2

–4

–4 –2 0 2 4 6 8x

Line 2

Solution continues…

y = ⅓ x + 1Line 1

(0, 1)(3, 2)



Example 3



Topic4.5.3

Second line: Two points on this line are (3, 2) and (1, –2).

2

4

6

y

0

–2

–4

–4 –2 0 2 4 6 8xm1 = = = 2

–2 – 2 1 – 3 –2

–4

y = 2x – 4 y – 2 = 2x – 6

y – 2 = 2(x – 3)

y – y1 = m(x – x1)

y = 2x – 4Line 2

(1, –2)

y = ⅓ x + 1Line 1

(3, 2)



Example 3



Topic4.5.3

2

4

6

y

0

–2

–4

–4 –2 0 2 4 6 8x

y = 2x – 4Line 2

y = ⅓ x + 1Line 1

Choose a point in the shaded region, for example, (5, 4). Substitute this point into the two equations.

So the equations of the two border lines are

3 1

y = x + 1 and y = 2x – 4.

(5, 4)



Example 3



Topic4.5.3

2

4

6

y

0

–2

–4

–4 –2 0 2 4 6 8x

y = 2x – 4Line 2

y = ⅓ x + 1Line 1

(5, 4)Equation for line 1:3 1

y = x + 1

4 = + 1 3 5

3 8 4 =

4 > , so a > sign is needed to

make it true. However, the line is solid so the sign should be .

3 8

So the first inequality is y x + 1. 3 1

— this is a false statement.

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Example 3



Topic4.5.3

2

4

6

y

0

–2

–4

–4 –2 0 2 4 6 8x

y = 2x – 4Line 2

y = ⅓ x + 1Line 1

(5, 4)Equation for line 2: y = 2x – 4

4 < 6, so a < sign is needed to make it true. The line is dashed so the < sign is correct.The second inequality is y < 2x – 4.

— this is a false statement. 4 = 10 – 4 4 = 6

Therefore the inequalities defining the shaded region

are y x + 1 and y < 2x – 4. 3 1

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Guided Practice

Solution follows…

In Exercises 7–8, find inequalities whose simultaneous solution defines each of the shaded regions.

Topic4.5.3

7. 8.

Use test point: (–2, 2)–y + x = –2 + (–2) = –4 < –2 –y + x < –2 x + 3y = –2 + 3(2) = 4 < 9 x + 3y 9

Use test point: (1, 1)y = 1, x + 2 = 1 + 2 = 3 y > x + 2y = 1, –0.5x + 1 = –0.5 + 1 = 0.5 y > –0.5x + 1

–4 –2 0 2 4 60

2

4

6 y

x

–2

–4

–6

–4 –2 0 2 4 60

2

4

6 y

x

–2

–4

–6

y = x + 2 y = –0.5x + 1 –y + x = –2

x + 3y = 9(1, 1)(–2, 2)

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Guided Practice

Solution follows…

In Exercises 9–10, find inequalities whose simultaneous solution defines each of the shaded regions.

Topic4.5.3

9. 10.

Find the equation of each line, then use test point (1, 0) to find the inequalities.y 3x – 2 and y –2x + 3

Find the equation of each line, then use test point (–1, 0) to find the inequalities.

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14

y > – x – and y < –x

–4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

–4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

(1, 0) (–1, 0)

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In Exercises 1–4, use the graph below to determine if the given point is included in the solution set.

1. (0, 0)

2. (2, 6)

3. (6, 2)

4. (4, –2)

Independent Practice

Solution follows…

no

Regions Defined by Two InequalitiesRegions Defined by Two InequalitiesTopic4.5.3

no

no

yes

–6 –4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

2x – 3y = 6

5x + 4y = 12

(0, 0)

(2, 6)

(6, 2)

(4, –2)

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Solution follows…


5. Determine whether the point (–4, –3) lies in the solution region of both 3x – 4y 2 and x – 2y 1.

6. Determine whether the point (0, 0) lies in the solution region of both 3x – 4y 2 and x – 2y 1.

7. Determine whether the point (–3, 1) lies in the solution region of both 3x – 4y 2 and x – 2y 1.

3(–4) – 4(–3) 2 0 2 This is a true statement.–4 – 2(–3) 1 2 1 This is also a true statement, so the point is in the solution region of both inequalities.

3(0) – 4(0) 2 0 2 This is a true statement.0 – 2(0) 1 0 1 This statement is false, so the point is not in the solution region of both inequalities.

3(–3) – 4(1) 2 –13 2 This is a true statement.–3 – 2(1) 1 –5 1 This statement is false, so the point is not in the solution region of both inequalities.

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Solution follows…


8. On axes spanning from –3 to 9 on the x-axis and –6 to 6 on the y-axis, graph the solution set that satisfies both the inequalities y + 2x 4 and y 1.5x – 2.

–3 0 3 6 90

2

4

6y

x

–2

–4

–6

y + 2x 4 and y 1.5x – 2

27


Solution follows…


9. On axes spanning from –3 to 9 on the x-axis and –8 to 4 on the y-axis, graph the solution set that satisfies both the inequalities y > x – 6 and y > –x + 2.

–3 0 3 6 9

–2

0

2

4y

x

–4

–6

–8

y > x – 6 and y > –x + 2

28


Solution follows…


In each of Exercises 10–11, show on axes spanning from –5 to 7 on the x-axis and –6 to 6 on the y-axis the region defined by the set of inequalities:

10. 4y – 3x < 12, 2x + y < 0, and y 1

11. 3x + 5y > 10, x – y < 2, and y 2

–4 –2 0 2 4 60

2

4

6y

x

–2

–4

–6

1011

29


Solution follows…


12. The points (2, 6.6) and (7, 8.1) lie on a line bounding a region, but are not part of the region themselves. Find the equation of the line.

y = 0.3x + 6

13. The region in Exercise 12 is also bounded by the lines y = 0.3x and y = |1.5x – 6|.

If the points (2, 3), (5, 1.5), and (4, 4) all lie within the region, draw and shade the region on axes spanning from –2 to 12 on the x- and y-axes.

4

6

8

10

y

2

0

–2

12

–2 0 2 4 6 8 10x

12

(5, 1.5)(2, 3)

y = 0.3x + 6

y = |1.5x – 6|

y = 0.3x

(4, 4)

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Round UpRound Up

When you’re graphing a system of linear inequalities, don’t forget that you still have to pay attention to whether the lines should be solid or dashed.


Date post:	23-Dec-2015
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