1. Systems of Inequalities Created By Richard Gill and Jeannie Taylor For Mth 163: Precalculus 1 Funded by a Grant from the VCCS LearningWare Program

2. This presentation starts with a few review slides on graphing linear inequalities and a few review slides on graphing non-linear inequalities. To skip either or both review sessions, click the appropriate button below and you will bypass the review. You can certainly come back to the review later if you need to. At the end of the presentation there will be a set of exercises with answers and occasional complete solutions. Good luck. Click here to see the review on linear inequalities. Click here to see the review on non-linear inequalities. Click here to skip both review sessions.

3. A short review on graphing inequalities. In order to graph the inequality y > 3 – x first graph the equation y = 3 – x. This line will be the borderline between the points that make y > 3 – x and the points that make y < 3 – x. In y = mx + b form we have y = -x + 3. In this case we have a line whose slope is –1 and whose y- intercept is 3. Now we have to decide which side of the line satisfies y > 3 – x.

4. A short review on graphing inequalities. All we have to do is to choose one point that is off the line and test it in the original inequality. If the point satisfies the inequality then we are on the correct side of the line and we shade that side. If the point does not satisfy the line, we shade the other side. The most popular point to use in the shading test is (0, 0). THE TEST: substitute (0, 0) into y > 3 – x and see if you get a true statement. 0 > 3 - 00 > 3, which is false.

5. Since (0, 0) did not satisfy the inequality y > 3 – x we conclude that (0, 0) is on the wrong side of the tracks and we shade the other side. Our conclusion is that every point in the shaded area is part of the solution set for y > 3 – x. You can reinforce this idea by testing several points in the shaded area. (2, 2) 2 > 3 – 2 2 > 1 (0, 3) 3 > 3 – 0 3 > 3 (4, 1) 1 > 3 – 4 1 > -1 Each point that we pick in the shaded area generates a true statement.

6. A short review on graphing non-linear inequalities. This same shading technique can be used for shading nonlinear inequalities such as: Once again the first step is to graph the equality which will serve as the borderline between the points whose x-y coordinates make and the points whose x-y coordinates make We can start by solving for y and graphing the borderline as a dotted line. Do you know what kind of graph this equation will generate?

7. Take a bow if you thought the borderline was going to be a parabola. As we did with the linear inequality, we can determine which side to shade with a test point. But (0, 0) will not work as a test point this time. Why? (0, 0) does not work as a test point because it is not on either side of the curve. We need a test point that is on one side or the other.

8. Take a minute to test (2, 1) in the original inequality: (2, 1) does not solve the original inequality so we will shade on the other side of the curve.

9. So here is the region that solves the original inequality. You can use the label at the top or the one at the bottom. Why is the curve dotted? You are so shrewd. The curve is dotted to indicate that points on curve are not part of our solution set. When we don’t have the “equal to” option, the borderline is dotted.

10. Graphing Systems of Inequalities Finally, we will take up the mission of this presentation. Consider the following system on inequalities: We will solve each inequality individually and then look for points that satisfy both inequalities at the same time. As before, we graph the border and then decide which side of the curve to shade. The first function has a restriction on its domain. The contents of the root have to be greater than or equal to zero. What is the domain of the first function?

11. Since the contents of the root have to be non-negative, the domain will include values of x no larger than 3. What kind of shape will the graph of this equation take on? Any graph that can be written in the form will generate half of a horizontal parabola. Now we must decide which side of the curve to shade. Why don’t you take a minute to use (0, 0) as a test point and see if (0, 0) is on the side of the curve that satisfies

12. The test for (0, 0): (0, 0) passes the test, so (0, 0) is on the correct side of the curve and we shade in that direction. Next we turn our attention to the second inequality. Before you click, see if you can figure the border and the shading for

13. The border for the inequality is, of course, the world famous parabola (0, 2) passes the test so we shade in that direction. (0, 0) is not an appropriate test point but (0, 2) will suffice. Does (0, 2) satisfy

14. We now have the shading for each inequality in the system: We now put the two graphs together and look for the points that are shaded in both inequalities. In other words, we look for the points that solve the inequalities simultaneously.

15. The intersection of the two shadings is the solution for the system: One more example and you can try your luck with a few exercises.

16. First a few tips. You will frequently see systems of inequalities with some of the restrictions below. Try to visualize each one before you click. x > 0 y > 0 x > 0 and y > 0

17. So spend a few minutes with the system below before you proceed. Graph each borderline on the same axis, use a test point to shade each inequality and highlight the region that is shaded in each inequality. It will stick with you better if you try the work yourself before you proceed. If you have not already done so figure the shading before you click.

18. We can use (0, 0) as a test point for each inequality. This is not true so we shade on the other side of the curve.

19. We can use (0, 0) as a test point for each inequality. This is not true so we shade on the other side of the curve. Similarly… This is true so we shade on the side of the curve that contains (0, 0).

20. The area that gets shaded twice is our solution set. It contains all points that solve both inequalities simultaneously. More precisely, our solution set looks like this…

21. Each point in the shaded area solves each inequality simultaneously. One more example and then you can spend some time on the exercises.

22. See what you can do with the following system of inequalities. 1. Graph each borderline. 2. Shade each inequality. 3. Highlight the area that is common to each shading. will generate the first border. What kind of graph is this? Congratulations if you knew this was a circle centered at the origin with radius of 5.

23. In order to get the whole circle to show on Winplot or on a graphing calculator you have to solve for y and graph each root.

24. In order to shade, we go back to the first inequality. Using test point (0, 0) you can see that 0 + 0 > 25 so we will shade outside the circle.

25. In order to shade, we go back to the first inequality. Now we can focus on the absolute value inequalities. Rewrite as: Note that (0, 0) satisfies each of the linear inequalities. Using test point (0, 0) you can see that 0 + 0 > 25 so we will shade outside the circle.

26. x = 5 x = -5 y = 5 y = -5 With all four linear boundaries in place, can you see which regions will be shaded?

27. So the shaded area contains the points that solve: Note that points in the shaded region like (4, 4) and (-5, -3) satisfy each inequality in the system. x = 5

28. Now check out the following exercises. All of the answers are listed at the end of the presentation along with a few of the complete solutions.