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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. 2.3.2: Solving Systems of Linear Inequalities

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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. 2.3.2: Solving Systems of Linear Inequalities

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**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. 2.3.2: Solving Systems of Linear Inequalities

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**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 2.3.2: Solving Systems of Linear Inequalities

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**Guided Practice Example 1**

Solve the following system of inequalities graphically: 2.3.2: Solving Systems of Linear Inequalities

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**Guided Practice: Example 1, continued **

Graph the line x + y = 10. 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, continued **

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, continued **

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, continued **

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, continued **

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, continued 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, continued **

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, continued **

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, continued **

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 1, continued**

2.3.2: Solving Systems of Linear Inequalities

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**Guided Practice Example 4**

An artist wants to analyze the time that he spends creating his art. He makes oil paintings and watercolor paintings. The artist takes 8 hours to paint an oil painting. He takes 6 hours to paint a watercolor painting. He has set aside a maximum of 24 hours per week to paint his paintings. The artist then takes 2 hours to frame and put the final touches on his oil paintings. He takes 3 hours to frame and put the final touches on his watercolor paintings. He has set aside a maximum of 12 hours per week for framing and final touch-ups. Write a system of inequalities that represents the time the artist has to complete his paintings. Graph the solution. 2.3.2: Solving Systems of Linear Inequalities

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**Guided Practice: Example 4, continued **

Create the system of inequalities. 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: Oil (x) Watercolor (y) Total Paint 8 6 64 Frame 2 3 12 2.3.2: Solving Systems of Linear Inequalities

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**Guided Practice: Example 4, continued**

Now, 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, continued **

Graph the system on the same coordinate plane. Start by graphing the first two inequalities. 2.3.2: Solving Systems of Linear Inequalities

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**Guided Practice: Example 4, continued**

Now apply the last two constraints: x ≥ 0 and y ≥ 0. This means the solution lies in the first quadrant. 2.3.2: Solving Systems of Linear Inequalities

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**✔ Guided Practice: Example 4, continued**

The 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. ✔ 2.3.2: Solving Systems of Linear Inequalities

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**Guided Practice: Example 4, continued**

2.3.2: Solving Systems of Linear Inequalities

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