1. Systems of Inequalities

2. Graphing a Linear Inequality in Two Variables 1.Replace the inequality symbol with an equal sign and graph the corresponding linear equation. Draw a solid line if the original inequality contains a symbol. Draw a dashed line if the original inequality contains a symbol. 2.Choose a test point in one of the half-planes that is not on the line. Substitute the coordinates of the test point into the inequality 3.If a true statement results, shade the half-plane containing this test point. If a false statement results, shade the half-plane not containing this test point.

3. Graph: 3x – 5y < 15. Solution Step 1 Replace the inequality symbol by = and graph the linear equation. We need to graph 3x – 5y = 15. We can use intercepts to graph this line. We set y= 0 to find We set x = 0 to find the x-intercept:the y-intercept: 3x – 5y = 153x – 5y = 15 3x – 5 0 = 153 – 0 5y = 15 3x = 15-5y = 15 x = 5y = -3 Text Example

4. Solution -3-2123456 7 5 4 3 2 1 -3 -4 -5 -2 3x – 5y = 15 The x-intercept is 5, so the line passes through (5, 0). The y-intercept is -3, so the line passes through (0, -3). The graph is indicated by a dashed line because the inequality 3x – 5y < 15 contains a < symbol, rather than <. The graph of the line is shown below. Text Example cont.

5. Solution Step 2 Choose a test point in one of the half-planes that is not on the line. Substitute its coordinates into the inequality. The line 3x – 5y = 15 divides the plane into three parts – the line itself and two half-planes. The points in one half-plane satisfy 3x – 5y > 15. The points in the other half-plane satisfy 3x – 5y < 15. We need to find which half-plane is the solution. To do so, we test a point from either half-plane. The origin, (0, 0), is the easiest point to test. 3x – 5y < 15 This is the given inequality. Is 3 0 – 5 0 < 15? Test (0, 0) by substituting 0 for x and y. 0 – 0 < 15 0 < 15, true Text Example cont.

6. Solution Step 3 If a true statement results, shade the half-plane containing the test point. Because 0 is less than 15, the test point (0, 0) is part of the solution set. All the points on the same side of the line 3x - 5y = 15 as the point (0, 0) are members of the solution set. The solution set is the half-plane that contains the point (0, 0), indicated by shading this half-plane. The graph is shown using green shading and a dashed blue line. -3-2123456 7 5 4 3 2 1 -3 -4 -5 -2 Text Example cont.

7. Graph: x 2 + y 2 < 9. -5-4-3-212345 5 4 3 2 1 -2 -3 -4 -5 Solution Step 1 Replace the inequality symbol by = and graph the nonlinear equation. We need to graph x 2 + y 2 = 9. The graph is a circle of radius 3 with its center at the origin. The graph is shown below as a solid circle because equality is included in the < symbol. Text Example

8. Solution Step 2 Choose a test point in one of the regions that is not on the circle. Substitute its coordinates into the inequality. The circle divides the plane into three parts – the circle itself, the region inside the circle, and the region outside the circle. We need to determine whether the region inside or outside the circle is the solution. To do so, we will use the test point (0, 0) from inside the circle. x 2 + y 2 < 9 This is the given inequality. Is 0 2 + 0 2 < 9? Test (0, 0) by substituting 0 for x and 0 for y. 0 + 0 < 9 0 < 9, true Text Example cont.

9. Solution Step 3 If a true statement results, shade the region containing the test point. The true statement tells us that all the points inside the circle satisfy x 2 + y 2 < 9. The graph is shown using green shading and a solid blue circle. -5-4-3-212345 5 4 3 2 1 -2 -3 -4 -5 Text Example cont.

10. Example Graph: 2x+y 4 Solution: First graph 2x+y=8 x+y=4

11. Example cont. Graph: 2x+y 4 Solution: Test point (0,0) 2x+y<8 2(0) + 0 = 0 <8 true

12. Example cont. Graph: 2x+y 4 Solution: Test point (0,0) x+y>4 0+0=0 >4 false

13. Example cont. Graph: 2x+y 4 Solution:

14. Graph the solution set: x – y < 2 -2 < x < 4 y < 3 Solution We begin by graphing x - y < 2, the first given inequality. The line x – y = 2 has an x-intercept of 2 and a y-intercept of -2. The test point (0, 0) makes the inequality x – y < 2 true, and its graph is shown below. -5-4-3-212345 5 4 3 2 1 -2 -3 -4 -5 Text Example

15. Solution Now let's consider the second given inequality -2 < x < 4. Replacing the inequality symbols by =, we obtain x = -2 and x = 4, graphed as vertical lines. The line of x = 4 is not included. Using (0, 0) as a test point and substituting the x-coordinate, 0, into -2 < x < 4, we obtain the true statement - 2 < 0 < 4. We therefore shade the region between the vertical lines. -5-4-31235 5 4 3 2 1 -2 -3 -4 -5 Text Example cont.

16. Solution Finally, let's consider the third given inequality, y < 3. Replacing the inequality symbol by =, we obtain y = 3, which graphs as a horizontal line. Because (0, 0) satisfies y < 3 (0 < 3 is true), the graph consists of the half- plane below the line y = 3. -5-4-31235 5 4 3 2 1 -2 -3 -4 -5 Text Example cont.

17. Systems of Inequalities