algebra 2 unit 3.3
TRANSCRIPT
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UNIT 3.3 SOLVING SYSTEMS OFUNIT 3.3 SOLVING SYSTEMS OFLINEAR INEQUALITIES LINEAR INEQUALITIES
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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 = –62x + y = 2
3. (–4, 3)x – y = –1
x + 2y = 2
yes
no
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Solve systems of linear inequalities.
Learning Targets
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system of linear inequalities
Vocabulary
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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.
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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.
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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
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If you are unsure which direction to shade, use the origin as a test point.
Helpful Hint
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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.
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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.
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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.
The overlapping region is the solution region.
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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)< 610 < 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
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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.
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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
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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.
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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 ≥ 0y ≥ 0
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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
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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
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Systems of inequalities may contain more than two inequalities.
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Graph the system of inequalities, and classify the figure created by the solution region.
Example 3: Geometry Application
x ≥ –2
y ≥ –x + 1x ≤ 3
y ≤ 4
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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. The solution is a trapezoid.
Example 3 Continued
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Graph the system of inequalities, and classify the figure created by the solution region.
x ≤ 6
Check It Out! Example 3a
y ≥ –2x + 4
y ≤ x + 1
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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.
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Lesson Quiz: Part I1. 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
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Lesson Quiz: Part II2. 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.
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Lesson Quiz: Part II Continued
x + y ≤ 100
3x + 5y ≥ 400
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