Download presentation

Presentation is loading. Please wait.

Published byOscar Hart Modified over 8 years ago

1
§ 4.4 Linear Inequalities in Two Variables

2
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 4.4 Linear Inequalities in Two Variables Let’s consider what the graph of a linear inequality in two variables looks like. A line divides the plane into three parts – the points on the line, the points below the line, and the points above the line. The graph of a linear inequality that involves is a half plane. The graph of a linear equality that involves either or is a half-plane and a line.

3
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 4.4 Linear Inequalities in Two Variables 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 or 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.

4
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 4.4 Linear Inequalities in Two VariablesEXAMPLE SOLUTION Graph: x – y < 6. We set y = 0 to find the x-intercept: x – y = 6 1) Replace the inequality symbol by = and graph the linear equation. We need to graph x – y = 6. We can use intercepts to graph this line. We set x = 0 to find the y-intercept: x – 0 = 6 x = 6 x – y = 6 0 – y = 6 y = -6

5
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 4.4 Linear Inequalities in Two Variables The x-intercept is 6, so the line passes through the point (6,0). The y-intercept is -6, so the line passes through the point (0,-6). Using the intercepts, the graph of the line will be a dotted line. This is because the inequality x – y < 6 contains a < symbol, in which equality is not included. 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 x – y = 6 divides the plane into three parts – the line itself and two half-planes. The points in one half-plane satisfy x – y 6. CONTINUED

6
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 4.4 Linear Inequalities in Two Variables We need to find which half-plane belongs to the solution of x – y < 6. To do so, we test a point from either half-plane. The origin, (0,0), is the easiest point to test. CONTINUED x – y < 6 0 – 0 < 6 0 < 6 This is the given inequality Test (0,0) by substituting 0 for x and 0 for y Simplify ? true

7
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 4.4 Linear Inequalities in Two Variables 3) If a true statement results, shade the half-plane containing the test point. Since 0 is less than 6, the test point (0,0) is part of the solution set. All the points on the same side of the line x - y = 6 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 show using green shading and a dashed blue line in in the figure on the next page. CONTINUED

8
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 4.4 Linear Inequalities in Two VariablesCONTINUED

9
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 4.4 Linear Inequalities in Two VariablesEXAMPLE SOLUTION Graph the inequality without using a test point. First, we will replace the symbol with the = symbol and then graph the equation.

10
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 4.4 Linear Inequalities in Two Variables The line is solid because equality is included in. Because of the ‘greater than’ part of, we shade the half- plane above the horizontal line. CONTINUED

11
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 4.4 Linear Inequalities in Two Variables The solution set of a system of linear inequalities in two variables is the set of all ordered pairs that satisfy each inequality in the system. To graph a system of inequalities in two variables, begin by graphing each individual inequality on the same graph. Then find the region, if there is one, that is common to every graph in the system. This region of intersection gives the system’s solution set.

12
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 4.4 Linear Inequalities in Two VariablesEXAMPLE SOLUTION Graph the solution set of the system. Replacing each inequality symbol with an equal sign indicates that we need to graph x + y = 3 and x + y = -2. We can use intercepts to graph these lines. x + y > 3 x + y > -2 x + y = 3 The two x-intercepts are (3,0) and (-2,0). Replace y with 0x + 0 = 3 x = 3 x + y = -2 x + 0 = -2 x = -2Simplify

13
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 4.4 Linear Inequalities in Two Variables Graph x + y > 3. The blue line, x + y = 3, is dashed: Equality is not included in x + y > 3. Because (0,0) makes the inequality false, we shade the half-plane not containing (0,0) in green. x + y = 3 Replace x with 00 + y = 3 y = 3 x + y = -2 0 + y = -2 y = -2Simplify CONTINUED The two y-intercepts are (0,3) and (0,-2).

14
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 4.4 Linear Inequalities in Two Variables Add the graph of x + y > -2. The red line, x + y = -2, is dashed: Equality is not included in x + y > -2. Because (0,0) makes the inequality true, we shade the half-plane containing (0,0) using yellow horizontal shading. CONTINUED

15
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 4.4 Linear Inequalities in Two Variables The solution set of the system is graphed as the intersection (the overlap) of the two half-planes. This is the region in which the green vertical shading and the yellow horizontal shading overlap. Notice that the blue line is not part of the solution since it is not contained within both solution sets. CONTINUED

16
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 4.4 Linear Inequalities in Two VariablesEXAMPLE SOLUTION (a) An elevator can hold no more than 2000 pounds. If children average 80 pounds and adults average 160 pounds, write a system of inequalities that models when the elevator holding x children and y adults is overloaded. (b) Graph the solution set of the system of inequalities in part (a). (a) Since the number of children on the elevator will always be zero or more, we get the inequality:. Since the number of adults on the elevator will always be zero or more, we get the inequality:.

17
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 4.4 Linear Inequalities in Two Variables The number of pounds that children contribute to the weight in the elevator is (the average weight of a child) x (the number of children) = 80x. The number of pounds that adults contribute to the weight in the elevator is (the average weight of an adult) x (the number of adults) = 160y. Now, the sum of these two quantities cannot exceed 2000 pounds. Therefore, the inequality 80x + 160y > 2000 would tell us when the elevator is overloaded. So, the system of inequalities for this situation is: CONTINUED

18
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 4.4 Linear Inequalities in Two Variables (b) Now we can graph the inequalities. Each tick-mark on the y-axis represents 2 units. Each tick-mark on the x-axis represents 4 units. First I rewrite each of the inequalities with the = symbol. Now, we can find the intercepts for the third equation. CONTINUED 80x + 160y = 2000 We set y = 0 to find the x-intercept: We set x = 0 to find the y-intercept: 80x + 160y = 2000 80x + 160(0) = 200080(0) + 160y = 2000 80x + 0 = 20000 + 160y = 2000 x = 25y = 12.5

19
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 4.4 Linear Inequalities in Two Variables Therefore, the two intercepts for the third equation are: (25,0) and (0,12.5). CONTINUED Graph. The blue line, y = 0, is solid: Equality is included in. Because (0,1) makes the inequality true, we shade the half-plane containing (0,1) in green. 2 4

20
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 4.4 Linear Inequalities in Two VariablesCONTINUED Graph. The red line, x = 0, is solid: Equality is included in. Because (1,0) makes the inequality true, we shade the half-plane containing (1,0) in yellow. 2 4

21
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 4.4 Linear Inequalities in Two VariablesCONTINUED Graph 80x + 160y > 2000. The brown line, 80x + 160y = 2000, is dashed: Equality is not included in 80x + 160y > 2000. Because (0,0) makes the inequality false, we shade the half- plane not containing (0,0) in gray. 2 4 80x + 160y > 2000

22
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 4.4 Linear Inequalities in Two VariablesCONTINUED The region that all three graphs have in common is the solution. 2 4 80x + 160y > 2000

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google