Download presentation

Presentation is loading. Please wait.

Published byKory Rose Modified over 7 years ago

1
© 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 7 Algebra: Graphs, Functions, and Linear Systems

2
© 2010 Pearson Prentice Hall. All rights reserved. 2 7.4 Linear Inequalities in Two Variables

3
© 2010 Pearson Prentice Hall. All rights reserved. Objectives 1.Graph a linear inequality in two variables. 2.Use mathematical models involving linear inequalities. 3.Graph a system of linear inequalities. 3

4
© 2010 Pearson Prentice Hall. All rights reserved. Linear Inequalities in Two Variables and Their Solutions If we change the symbol = in the equation Ax + By = C to >, <, ≥, or ≤, we obtain a linear inequality in two variables. For example, x + y < 2 and 3x – 5y ≥ 15 are linear inequalities in two variables. A solution of an inequality in two variables, x and y, is an ordered pair of real numbers such that when the x- coordinate is substituted for x and the y-coordinate is substituted for y in the inequality and we obtain a true statement. 4

5
© 2010 Pearson Prentice Hall. All rights reserved. The Graph of a Linear Inequality in Two Variables The graph of an inequality in two variables is the set of all points whose coordinates satisfy the inequality. 5

6
© 2010 Pearson Prentice Hall. All rights reserved. Graph: 3x – 5y ≥ 15. Solution: Step 1 We need to graph 3x – 5y = 15. We can use intercepts to graph this line. We set y = 0 to We set x = 0 to find the x-intercept. find the y-intercept. 3x – 5y = 15 3x – 5y = 15 3x – 5 · 0 = 15 3 · 0 – 5y = 15 3x = 15 −5y = 15 x = 5 y = −3 Example 1: Graphing a Linear Inequality in Two Variables 6

7
© 2010 Pearson Prentice Hall. All rights reserved. The x-intercept is 5, so the line passes through (5,0). The y-intercept is −3, so the line passes through (0,−3). Step 2 We choose (0,0) as a test point. 3x – 5y ≥ 15 3 · 0 – 5 · 0 ≥ 15 0 – 0 ≥ 15 0 ≥ 15 NOT TRUE! Example 1 continued 7

8
© 2010 Pearson Prentice Hall. All rights reserved. Step 3 Since the statement is false, we shade the half- plane that does not include the test point (0,0). Thus, the graph with the shading is the solution to the given inequality. Example 1 continued 8

9
© 2010 Pearson Prentice Hall. All rights reserved. Graph: Solution: Step 1 We need to graph Since the inequality > is given, we use a dashed line. Example 2: The Graph of a Linear Inequality in Two Variables 9

10
© 2010 Pearson Prentice Hall. All rights reserved. Step 2 We choose a test point not on the line, (1,1), which lies in the half-plane above the line. TRUE! Step 3 Since the statement is true, then we shade the half-plane that includes the test point (1,1). Example 2 continued 10

11
© 2010 Pearson Prentice Hall. All rights reserved. Graphing Linear Inequalities without Using Test Points For the vertical line x = a: If x > a, shade the half- plane to the right of x = a. If x < a, shade the half- plane to the left of x = a. For the horizontal line y = b: If y > b, shade the half-plane above y = b. If y < b, shade the half-plane below y = b. 11

12
© 2010 Pearson Prentice Hall. All rights reserved. Graph each inequality in a rectangular coordinate system:a. y ≤ −3b. x > 2 Solution: a. y ≤ −3b. x > 2 Example 3: Graphing Inequalities Without Using Test Points 12

13
© 2010 Pearson Prentice Hall. All rights reserved. Modeling with Systems of Linear Inequalities Just as two or more linear equations make up a system of linear equations, two or more linear inequalities make up a system of linear inequalities. A solution of a system of linear inequalities in two variables is an ordered pair that satisfies each inequalities in the system. 13

14
© 2010 Pearson Prentice Hall. All rights reserved. Graphing Systems of Linear Inequalities 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. 14

15
© 2010 Pearson Prentice Hall. All rights reserved. Example 5: Graphing a System of Linear Inequalities Graph the solution set of the system: x – y < 1 2x + 3y ≥ 12. Solution: Replacing each inequality symbol with an equal sign indicates that we need to graph x – y = 1 and 2x + 3y = 12. We can use intercepts to graph these lines. 15

16
© 2010 Pearson Prentice Hall. All rights reserved. Example 5 continued 16

17
© 2010 Pearson Prentice Hall. All rights reserved. Now we are ready to graph the solution set of the system of linear inequalities. Example 5 continued 17

Similar presentations

© 2023 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