Holt McDougal Algebra 2 3-3 Solving Systems of Linear Inequalities 3-3 Solving Systems of Linear...

Holt McDougal Algebra 2

3-3 Solving Systems of Linear Inequalities 3-3 Solving Systems of Linear Inequalities

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz



3-3 Solving Systems of Linear Inequalities

Warm Up

1. Graph 2x – y > 4.

Determine if the given ordered pair is a solution of the system of equations.

2. (2, –2)2y – x = –6

2x + y = 2

3. (–4, 3)x – y = –1

x + 2y = 2

yes

no



Solve systems of linear inequalities.

Objective



system of linear inequalities

Vocabulary



When a problem uses phrases like “greater than” or “no more than,” you can model the situation using a system of linear inequalities.

A system of linear inequalities is a set of two or more linear inequalities with the same variables. The solution to a system of inequalities is often an infinite set of points that can be represented graphically by shading.

When you graph multiple inequalities on the same graph, the region where the shadings overlap is the solution region.



Graph the system of inequalities.Example 1A: Graphing Systems of Inequalities

y ≥ –x + 2

y < – 3

For y < – 3, graph the dashed boundary line y = – 3, and shade below it. For y ≥ –x + 2, graph the solid boundary line y = –x + 2, and shade above it.

The overlapping region is the solution region.



Check Test a point from each region on the graph.

Region Point

Left (0, 0)

Right (5,–2)

Top (0, 3)

Bottom (0,–4)

Only the point from the overlapping (right) region satisfies both inequalities.

y ≥ –x + 2 y < x – 3 12

3 < (0)–3 12

–4 < (0)–3 12

–4 < –3

2 < –3

0 < (0)–3 12

0 < –3–2 < (5) –3 1

2–2 <– 1

2

0 ≥ –(0) + 2

–4 ≥ 2

–4 ≥ –(0) + 2

3 ≥ 2

3 ≥ –(0) + 2

–2 ≥ –3

0 ≥ 2

–2 ≥ –(5) + 2

x

xx

x



If you are unsure which direction to shade, use the origin as a test point.

Helpful Hint



Graph each system of inequalities.

Example 1B: Graphing Systems of Inequalities

y ≥ –1

y < –3x + 2

For y < –3x + 2, graph the dashed boundary line y = –3x + 2, and shade below it.

For y ≥ –1, graph the solid boundary line y = –1, and shade above it.



Example 1B Continued

Check Choose a point in the solution region, such as (0, 0), and test it in both inequalities.

y < –3x + 2 y ≥ –1

0 < –3(0) + 2

0 < 2

0 ≥ –1

0 ≥ –1

The test point satisfies both inequalities, so the solution region is correct.



Check It Out! Example 1a

Graph the system of inequalities.

2x + y > 1.5

x – 3y < 6

For x – 3y < 6, graph the dashed boundary line y = – 2, and shade above it.

1

3x

For 2x + y > 1.5, graph the dashed boundary line y = –2x + 1.5, and shade above it.



3-3 Solving Systems of Linear Inequalities Check Test a point from each region on the graph.

Region Point

Left (0, 0)

Right (4,–2)

Top (0, 3)

Bottom (0,–4)

Only the point from the overlapping (top) region satisfies both inequalities.

0 < 6

x

xx

x

x – 3y < 6 2x + y > 1.5

0 – 3(0)< 6

0 – 3(–4)< 6

0 – 3(3)< 6

4 – 3(–2)< 6

10 < 6

–9 < 6

–12 < 6

2(0) + 0 >1.5

0 > 1.5

2(0) – 4 >1.5

2(0) + 3 >1.5

2(4) – 2 >1.5 6 > 1.5

3 > 1.5

–4 > 1.5



Graph each system of inequalities.

y ≤ 4

2x + y < 1

For 2x + y < 1, graph the dashed boundary line y = –3x +2, and shade below it.

For y ≤ 4, graph the solid boundary line y = 4, and shade below it.

Check It Out! Example 1b




Check Choose a point in the solution region, such as (0, 0), and test it in both directions.

The test point satisfies both inequalities, so the solution region is correct.

y ≤ 4 2x + y < 1

0 ≤ 4

0 ≤ 4

2(0) + 0 < 1

0 < 1

Check It Out! Example 1b Continued



Example 2: Art Application

Lauren wants to paint no more than 70 plates for the art show. It costs her at least $50 plus $2 per item to produce red plates and $3 per item to produce gold plates. She wants to spend no more than $215. Write and graph a system of inequalities that can be used to determine the number of each plate that Lauren can make.



Example 2 Continued

Let x represent the number of red plates, and let y represent the number of gold plates.

The total number of plates Lauren is willing to paint can be modeled by the inequality x + y ≤ 70.

The amount of money that Lauren is willing to spend can be modeled by 50 + 2x + 3y ≤ 215.

The system of inequalities is . x + y ≤ 70

50 + 2x + 3y ≤ 215

x 0

y 0



Graph the solid boundary line x + y = 70, and shade below it.

Graph the solid boundary line 50 + 2x + 3y ≤ 215, and shade below it. The overlapping region is the solution region.

Example 2 Continued



Check Test the point (20, 20) in both inequalities. This point represents painting 20 red and 20 gold plates.

x + y ≤ 70 50 + 2x + 3y ≤ 215

20 + 20 ≤ 70

40 ≤ 70

50 + 2(20) + 3(20) ≤ 215

150 ≤ 215

Example 2 Continued



Check It Out! Example 2

Leyla is selling hot dogs and spicy sausages at the fair. She has only 40 buns, so she can sell no more than a total of 40 hot dogs and spicy sausages. Each hot dog sells for $2, and each sausage sells for $2.50. Leyla needs at least $90 in sales to meet her goal. Write and graph a system of inequalities that models this situation.



Let d represent the number of hot dogs, and let s represent the number of sausages.

The total number of buns Leyla has can be modeled by the inequality d + s ≤ 40.

The amount of money that Leyla needs to meet her goal can be modeled by 2d + 2.5s ≥ 90.

The system of inequalities is . d + s ≤ 40

2d + 2.5s ≥ 90

Check It Out! Example 2 Continued

d 0

s 0



Graph the solid boundary line d + s = 40, and shade below it.

Graph the solid boundary line 2d + 2.5s ≥ 90, and shade above it. The overlapping region is the solution region.




Check Test the point (5, 32) in both inequalities. This point represents selling 5 hot dogs and 32 sausages.

d + s ≤ 40 2d + 2.5s ≥ 90

5 + 32 ≤ 4037 ≤ 40

2(5) + 2.5(32) ≥ 90

90 ≥ 90




Systems of inequalities may contain more than two inequalities.



Graph the system of inequalities, and classify the figure created by the solution region.

Example 3: Geometry Application

x ≥ –2

y ≥ –x + 1

x ≤ 3

y ≤ 4



Graph the solid boundary line x = –2 and shade to the right of it. Graph the solid boundary line x = 3, and shade to the left of it.

Graph the solid boundary line y = –x + 1, and shade above it. Graph the solid boundary line y = 4, and shade below it. The overlapping region is the solution region.

Example 3 Continued




x ≤ 6

Check It Out! Example 3a

y ≥ –2x + 4

y ≤ x + 1



Check It Out! Example 3a Continued

Graph the solid boundary line x = 6 and shade to the left of it.Graph the solid boundary line, y ≤ x + 1 and shade below it. Graph the solid boundary line y ≥ –2x + 4, and shade below it. The overlapping region is the solution region. The solution is a triangle.




Check It Out! Example 3b

y ≤ 4

y ≤ –x + 8

y ≥–1

y ≤ 2x + 2



Graph the solid boundary line y = 4 and shade to the below it. Graph the solid boundary line y = –1, and shade to the above it.

Graph the solid boundary line y = –x + 8, and shade below it. Graph the solid boundary line y = 2x + 2, and shade below it. The overlapping region is the solution region.





The solution region is a four-sided figure, or quadrilateral. Notice that the boundary lines y = 4 and y = –1 are parallel, horizontal lines. The boundary lines y = –x + 8 and y = 2x + 2 are not parallel since the slope of the first is –1 and the slope of the second is 2.

A quadrilateral with one set of parallel sides is called a trapezoid. The solution region is a trapezoid.



Lesson Quiz: Part I

1. Graph the system of inequalities and classify the figure created by the solution region.

y ≤ x – 2

x ≤ 4

y ≥ –2x – 2

x ≥ 1

trapezoid



Lesson Quiz: Part II

2. The cross-country team is selling water bottles to raise money for the team. The price of the water bottle is $3 for students and $5 for everyone else. The team needs to raise at least $400 and has 100 water bottles. Write and graph a system of inequalities that can be used to determine when the team will meet its goal.



Lesson Quiz: Part II Continued

x + y ≤ 100

3x + 5y ≥ 400

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