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Practice Problems for Solving Linear Inequalities Prepared by Richard Gill For Mth 04: Intermediate Algebra Tidewater Community College.

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Presentation on theme: "Practice Problems for Solving Linear Inequalities Prepared by Richard Gill For Mth 04: Intermediate Algebra Tidewater Community College."— Presentation transcript:

1 Practice Problems for Solving Linear Inequalities Prepared by Richard Gill For Mth 04: Intermediate Algebra Tidewater Community College

2 Practice problems for Lesson 5.3: Systems of Inequalities. Graph the solution set for each system of inequalities. First do the graph yourself, then see the solution on one of the following frames.

3 Practice problems. Graph the solution set for each system of inequalities. First do the graph yourself, then see the solution on one of the following frames.

4 x + y < 2 2x - y > 4 The solution to #1. Points in the red region solve x + y < 2. Points in the yellow region solve 2x – y > 4. Points in the double- shaded brown region solve both inequalities simultaneously. 2x – y = 4 x + y = 2

5 Solution to #2. Points that solve both inequalities lie in the region that is double-shaded. 3x – y > 6 x + 2y < 4 3x – y = 6 x + 2y = 4

6 Solution to #3. Points that solve both inequalities lie in the region that is double-shaded. 2x – y < 4 x + y > -1 x + y = 1 2x – y = 4y x

7 Solution to #4. Points that solve both inequalities lie in the region that is double-shaded. Therefore, no solution. 2x + 3y > 6 2x + 3y = 6

8 The solution to #5. Points in the yellow region solve y >(1/3)x –3. Points in the red region solve y < -(4/3)x + 3. Points in the brown region solve both inequalities. y >(1/3)x –3 y < -(4/3)x + 3 y = (1/3) x - 3 y = -(4/3)x + 3 x y

9 The solution to #6. Points in the yellow are solve y > -2x + 4. Points in the red are solve y < 2. Points in the brown are solve both inequalities simultaneously. y > -2x + 4 y < 2 y = 2 y = -2x + 4 x

10 y > -3 This is the solution for #7. Points in the red area satisfy y > -3. Points in the yellow area satisfy Points in the brown region satisfy both inequalities in the system. y =3

11 y>-2x+3 y < ½ x -2 This is the solution to #8. Points in the yellow region satisfy y > -2x + 3. Points in the red region satisfy y < ½ x – 2. Points in the brown region satisfy both inequalities simultaneously. y = -2x + 3 y = ½ x - 2 x y


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