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System of linear inequalities
 

System of linear inequalities

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    System of linear inequalities System of linear inequalities Presentation Transcript

    • Solving Systems of Linear Inequalities Warm Up Lesson Presentation Lesson Quiz
    • Warm Up1. Graph 2x – y > 4.Determine if the givenordered pair is asolution of the systemof equations. 2x + y = 22. (2, –2) 2y – x = –6 yes x – y = –1 3. (–4, 3) no x + 2y = 2
    • ObjectiveSolve systems of linear inequalities.
    • Vocabularysystem of linear inequalities
    • When a problem uses phrases like “greater than”or “no more than,” you can model the situationusing a system of linear inequalities.A system of linear inequalities is a set of two ormore linear inequalities with the same variables.The solution to a system of inequalities is often aninfinite set of points that can be representedgraphically by shading.When you graph multiple inequalities on the samegraph, the region where the shadings overlap is thesolution region.
    • Example 1A: Graphing Systems of Inequalities Graph the system of inequalities. y< –3 y ≥ –x + 2For y < – 3, graph thedashed boundary liney = – 3, and shade belowit.For y ≥ –x + 2, graph thesolid boundary liney = –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 y< 1 x–3 y ≥ –x + 2 2Left (0, 0) 0 < 1 (0)–3 2 0 ≥ –(0) + 2 0 < –3 x 0≥2 xRight (5,–2) –2 < 1 (5) –3 2 –2 ≥ –(5) + 2 –2 <– 1  2 –2 ≥ –3 Top (0, 3) 3 < 1 (0)–3 3 ≥ –(0) + 2 2 2 < –3 x 3≥2 Bottom (0,–4) –4 < 1 (0)–3 –4 ≥ –(0) + 2 2 –4 < –3  –4 ≥ 2 xOnly the point from the overlapping (right) regionsatisfies both inequalities.
    • Helpful HintIf you are unsure which direction to shade, usethe origin as a test point.
    • Example 1B: Graphing Systems of InequalitiesGraph each system of inequalities. y < –3x + 2 y ≥ –1For y < –3x + 2, graph thedashed boundary liney = –3x + 2, and shadebelow it.For y ≥ –1, graph the solidboundary line y = –1, andshade 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 ≥ –1 0<2  0 ≥ –1 The test point satisfies both inequalities, so thesolution region is correct.
    • Check It Out! Example 1a Graph the system of inequalities. x – 3y < 6 2x + y > 1.5For x – 3y < 6, graph the dashed 1boundary line y = x – 2, and 3shade above it.For 2x + y > 1.5, graph thedashed boundary liney = –2x + 1.5, and shade above it.The overlapping region is the solution region.
    • Check Test a point from each region on the graph. Region Point x – 3y < 6 2x + y > 1.5 0 – 3(0)< 6 2(0) + 0 >1.5Left (0, 0) 0<6 x 0 > 1.5 xRight (4,–2) 4 – 3(–2)< 6 2(4) – 2 >1.5 10 < 6 x 6 > 1.5 Top (0, 3) 0 – 3(3)< 6 2(0) + 3 >1.5 –9 < 6  3 > 1.5 Bottom (0,–4) 0 – 3(–4)< 6 2(0) – 4 >1.5 –12 < 6  –4 > 1.5 xOnly the point from the overlapping (top) regionsatisfies both inequalities.
    • Check It Out! Example 1bGraph each system of inequalities. y≤4 2x + y < 1For y ≤ 4, graph the solidboundary line y = 4, andshade below it.For 2x + y < 1, graphthe dashed boundary liney = –3x +2, and shadebelow it.The overlapping region is the solution region.
    • Check It Out! Example 1b ContinuedCheck Choose a point in the solution region,such as (0, 0), and test it in both directions. y≤4 2x + y < 1 0≤4 2(0) + 0 < 1 0≤4  0<1 The test point satisfies both inequalities, so thesolution region is correct.
    • Example 2: Art ApplicationLauren wants to paint no more than 70plates for the art show. It costs her at least$50 plus $2 per item to produce red platesand $3 per item to produce gold plates. Shewants to spend no more than $215. Writeand graph a system of inequalities that canbe used to determine the number of eachplate that Lauren can make.
    • Example 2 ContinuedLet x represent the number of red plates, and lety represent the number of gold plates.The total number of plates Lauren is willing to paintcan be modeled by the inequality x + y ≤ 70.The amount of money that Lauren is willing tospend can be modeled by 50 + 2x + 3y ≤ 215. x 0The system of inequalities is y 0 . x + y ≤ 70 50 + 2x + 3y ≤ 215
    • Example 2 ContinuedGraph the solid boundaryline x + y = 70, and shadebelow it.Graph the solid boundaryline 50 + 2x + 3y ≤ 215,and shade below it. Theoverlapping region is thesolution region.
    • Example 2 ContinuedCheck Test the point (20, 20) in both inequalities.This point represents painting 20 red and 20 goldplates. x + y ≤ 70 50 + 2x + 3y ≤ 215 20 + 20 ≤ 70 50 + 2(20) + 3(20) ≤ 215 40 ≤ 70  150 ≤ 215 
    • Check It Out! Example 2Leyla is selling hot dogs and spicy sausages atthe fair. She has only 40 buns, so she can sellno more than a total of 40 hot dogs and spicysausages. Each hot dog sells for $2, and eachsausage sells for $2.50. Leyla needs at least$90 in sales to meet her goal. Write and grapha system of inequalities that models thissituation.
    • Check It Out! Example 2 ContinuedLet d represent the number of hot dogs, and let srepresent the number of sausages.The total number of buns Leyla has can be modeledby the inequality d + s ≤ 40.The amount of money that Leyla needs to meether goal can be modeled by 2d + 2.5s ≥ 90. d 0 s 0The system of inequalities is . d + s ≤ 40 2d + 2.5s ≥ 90
    • Check It Out! Example 2 ContinuedGraph the solid boundaryline d + s = 40, and shadebelow it.Graph the solid boundaryline 2d + 2.5s ≥ 90, andshade above it. Theoverlapping region is thesolution region.
    • Check It Out! Example 2 ContinuedCheck Test the point (5, 32) in both inequalities.This point represents selling 5 hot dogs and 32sausages. d + s ≤ 40 2d + 2.5s ≥ 90 5 + 32 ≤ 40 2(5) + 2.5(32) ≥ 90 37 ≤ 40  90 ≥ 90 
    • Systems of inequalities may contain morethan two inequalities.
    • Example 3: Geometry ApplicationGraph the system of inequalities, and classifythe figure created by the solution region. x ≥ –2 x≤3 y ≥ –x + 1 y≤4
    • Example 3 ContinuedGraph the solid boundaryline x = –2 and shade to theright of it. Graph the solidboundary line x = 3, andshade to the left of it.Graph the solid boundaryline y = –x + 1, and shadeabove it. Graph the solidboundary line y = 4, andshade below it. Theoverlapping region is thesolution region.
    • Check It Out! Example 3aGraph the system of inequalities, and classifythe figure created by the solution region. x≤6 y≤ x+1 y ≥ –2x + 4
    • Check It Out! Example 3a ContinuedGraph the solid boundaryline x = 6 and shade to theleft of it.Graph the solid boundaryline, y ≤ x + 1 and shadebelow it.Graph the solid boundaryline y ≥ –2x + 4, and shadebelow it.The overlapping region isthe solution region. Thesolution is a triangle.
    • Check It Out! Example 3bGraph the system of inequalities, and classifythe figure created by the solution region. y≤4 y ≥–1 y ≤ –x + 8 y ≤ 2x + 2
    • Check It Out! Example 3b ContinuedGraph the solid boundaryline y = 4 and shade to thebelow it. Graph the solidboundary line y = –1, andshade to the above it.Graph the solid boundaryline y = –x + 8, and shadebelow it. Graph the solidboundary line y = 2x +2, and shade below it. Theoverlapping region is thesolution region.
    • Check It Out! Example 3b ContinuedThe solution region is a four-sided figure, or quadrilateral.Notice that the boundarylines y = 4 and y = –1 areparallel, horizontal lines. Theboundary lines y = –x + 8and y = 2x + 2 are notparallel since the slope of thefirst is –1 and the slope ofthe second is 2.A quadrilateral with one set of parallel sides is called atrapezoid. The solution region is a trapezoid.
    • Lesson Quiz: Part I1. Graph the system of inequalities and classify the figure created by the solution region. y≤ x–2 y ≥ –2x – 2 x≤4 x≥1 trapezoid
    • 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.
    • Lesson Quiz: Part II Continuedx + y ≤ 1003x + 5y ≥ 400