Introduction Finding the solutions to a system of linear equations requires graphing multiple linear inequalities on the same coordinate plane. Most real-world applications dealing with linear inequalities are actually systems of linear inequalities where at least one of the conditions is that one of the variables must be greater than 0. 1 2.3.2: Solving Systems of Linear Inequalities
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IntroductionFinding the solutions to a system of linear equations requires graphing multiple linear inequalities on the same coordinate plane. Most real-world applications dealing with linear inequalities are actually systems of linear inequalities where at least one of the conditions is that one of the variables must be greater than 0.
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2.3.2: Solving Systems of Linear Inequalities
Key Concepts• A system of inequalities is two or more inequalities in
the same variables that work together.
• The solution to a system of linear inequalities is the intersection of the half planes of the inequalities. The solution will be the set of all points that make all the inequalities in the system true.
• Graph each inequality and include the correct shading.
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2.3.2: Solving Systems of Linear Inequalities
Key Concepts, continued• Look for the area where the shading of the
inequalities overlaps; this is the “solution”.
• Remember to add constraints to the system that aren’t explicitly stated in the problem context but that make sense for the problem.
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2.3.2: Solving Systems of Linear Inequalities
Common Errors/Misconceptions• incorrectly graphing one or more of the inequalities in
the system
• not seeing the solution or where the intersections occur
• forgetting to include constraints or boundaries that make sense for the context of the problem
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2.3.2: Solving Systems of Linear Inequalities
“Wise Notes” ≥ ≤ Solid Line > < Dashed Line
<- - - - - - - ->
If x is present, vertical line Ex: x < 3
If y is present, horizontal line Ex: y >3
Use (0,0) as test pointsIf true, shade with sign. If false, shade opposite sign
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2.3.2: Solving Systems of Linear Inequalities
Guided PracticeExample 1Solve the following system of inequalities graphically:
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2.3.2: Solving Systems of Linear Inequalities
Guided Practice: Example 1, continued1. Graph the line x + y = 10. Use a dashed line
because the inequality is non-inclusive (greater than).
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2.3.2: Solving Systems of Linear Inequalities
Guided Practice: Example 1, continued2. Shade the solution set. First pick a test
point. Choose a point that is on either side of the line.
Test point: (0, 0)
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2.3.2: Solving Systems of Linear Inequalities
Guided Practice: Example 1, continued3. Then, substitute that point into the
equality x + y > 10.If the test point makes the inequality true, shade the region that contains that point. If the test point makes the inequality false, shade on the opposite side of the line.
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 1, continued4. Since the point (0, 0) makes the inequality
false, shade the opposite side of the line.The shaded region represents the solutions for x + y > 10.
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 1, continued5. Graph the line 2x – 4y = 5 on the same
coordinate plane. Use a dashed line because the inequality is non-inclusive (greater than).
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 1, continued6. Shade the solution set.
First pick a test point. Choose a point that is on either side of the line.
Test point: (0, 0)
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 1, continued7. Then, substitute that point into the
inequality 2x – 4y = 5.If the test point makes the inequality true, shade the region that contains that point. If the test point makes the inequality false, shade on the opposite side of the line.
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 1, continued8. Since the point (0, 0) makes the inequality
false, shade the opposite side of the line.The second shaded region represents the solutions for 2x – 4y = 5.
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 1, continued9. Find the solutions to the system.
The overlap of the two shaded regions, which is darker, represents the solutions to the system:
A possible solution to this system is (14, 2) because it satisfies both inequalities.
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 1, continued
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 41. Create the system of inequalities given
the example below.Let x = the number of oil paintings he makes.Let y = the number of watercolor paintings he makes.It might be helpful to create a table:
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2.3.2: Solving Systems of Linear Inequalities
Oil (x) Watercolor (y) TotalPaint 8 6 24
Frame 2 3 12
Guided Practice: Example 4, continuedNow, think about what must always be true of creating the paintings: there will never be negative paintings. Add these two constraints to the system: x ≥ 0 and y ≥ 0.
2.3.2: Solving Systems of Linear Inequalities
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Guided Practice: Example 4, continued2. Graph the system on the same coordinate
plane.Start by graphing the first two inequalities.
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2.3.2: Solving Systems of Linear Inequalities
Guided Practice: Example 4, continuedNow apply the last two constraints: x ≥ 0 and y ≥ 0. This means the solution lies in the first quadrant.
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2.3.2: Solving Systems of Linear Inequalities
Guided Practice: Example 4, continuedThe solution is the darker shaded region; any points that lie within it are solutions to the system. The point (1, 1) is a solution because it satisfies both inequalities. The artist can create 1 oil painting and 1 watercolor painting given the time constraints he has. Or, he can create no oil paintings and 4 watercolor paintings, (0, 4). However, he cannot create 4 oil paintings and 1 watercolor painting, because the point (4, 1) only satisfies one inequality and does not lie in the darker shaded region.