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Holt Algebra 2

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

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz

Holt Algebra 2

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

Holt Algebra 2


Solve systems of linear inequalities. Objective

Holt Algebra 2


system of linear inequalities

Vocabulary

Holt Algebra 2


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.

Holt Algebra 2


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.

Holt Algebra 2

3-3 Solving Systems of Linear Inequalities 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 1 2

3 < (0)–3 1 2

–4 < (0)–3 1 2

–4 < –3

2 < –3

0 < (0)–3 1 2

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

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ü ü

x

x x

x

ü

Holt Algebra 2


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

Helpful Hint

Holt Algebra 2


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.

Holt Algebra 2


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.

ü ü

Holt Algebra 2


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.

13x

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

Holt Algebra 2

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

ü

x ü

ü

x x

x

ü

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

Holt Algebra 2


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

The overlapping region is the solution region.

Holt Algebra 2


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

ü ü

Holt Algebra 2


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.

Holt Algebra 2


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

Holt Algebra 2


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

Holt Algebra 2


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 ü ü

Example 2 Continued

Holt Algebra 2


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.

Holt Algebra 2


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

Holt Algebra 2


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.


Holt Algebra 2


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 ≤ 40 37 ≤ 40

2(5) + 2.5(32) ≥ 90 90 ≥ 90 ü ü


Holt Algebra 2


Systems of inequalities may contain more than two inequalities.

Holt Algebra 2


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

Holt Algebra 2


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

Holt Algebra 2



x ≤ 6

Check It Out! Example 3a

y ≥ –2x + 4

y ≤ x + 1

Holt Algebra 2


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.

Holt Algebra 2



Check It Out! Example 3b

y ≤ 4

y ≤ –x + 8 y ≥–1

y ≤ 2x + 2

Holt Algebra 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.


Holt Algebra 2



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.

Holt Algebra 2


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

Holt Algebra 2


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.

Holt Algebra 2


Lesson Quiz: Part II Continued

x + y ≤ 100

3x + 5y ≥ 400

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