Inequalities in Two Variables 4.4 Inequalities in Two Variables Graphs of Linear Inequalities Systems of Linear Inequalities Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Graphs of Linear Inequalities When the equal sign in a linear equation is replaced with an inequality sign, a linear inequality is formed. Solutions of linear inequalities are ordered pairs. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Determine whether (1, 5) and (6, –2) are solutions of the inequality 3x – y < 5. Solution 3(1) – 5 5 –2 < 5 TRUE 3x – y < 5 3(6) – (–2) 5 20 < 5 FALSE 3x – y < 5 The pair (1, 5) is a solution of the inequality, but (6, –2) is not. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The graph of a linear equation is a straight line The graph of a linear equation is a straight line. The graph of a linear inequality is a half-plane, with a boundary that is a straight line. To find the equation of the boundary line, we simply replace the inequality sign with an equals sign. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Graph First graph the boundary y = x. Since the inequality is greater than or equal to, the line is drawn solid and is part of the graph of x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 y = x -4 -5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution (continued) Note that in the graph on the right each ordered pair on the half-plane above y = x contains a y x -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 y = x -4 -5 y- coordinate that is greater than the x- coordinate. It turns out that any point on the same side as (–2, 2) is also a solution. Thus, if one point in a half- plane is a solution, then all points in that half-plane are solutions. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution (continued) We finish drawing the solution set by shading the half-plane above y = x. The complete solution set consists of the shaded half-plane as well as the boundary itself. x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 y = x -4 -5 For any point here, y > x. For any point here, y = x. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Graph (3, 1) Since the inequality sign is < , points on the line y = 3 – 8x do not represent solutions of the inequality, so the line is dashed. Using (3, 1) as a test point, we see that it is not a solution. Thus the points in the other half-plane are solutions. x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 y = 3 – 8x -4 -5 (3, 1) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Steps for Graphing Linear Inequalities Replace the inequality sign with an equals sign and graph this line as the boundary. If the inequality symbol is < or >, draw the line dashed. If the inequality symbol is , draw the line solid. The graph of the inequality consists of a half-plane on one side of the line and, if the line is solid, the line as well. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Steps for Graphing Linear Inequalities a) If the inequality is of the form y < mx + b or shade below the line. If the inequality is of the form y > mx + b or shade above the line. b) If y is not isolated, either solve for y and graph as in part (a) or simply graph the boundary and use a test point. If the test point is a solution, shade the half-plane containing the point. If it is not a solution, shade the other half-plane. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Graph y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 y = (1/6)x – 1 -4 -5 The graph consists of the half-plane above the dashed boundary line y = (1/6)x – 1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Graph y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 The graph consists of the line x = –3 and the half-plane to the right of the line x = –3. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Systems of Linear Inequalities To graph a system of equations, we graph the individual equations and then find the intersection of the individual graphs. We do the same thing for a system of inequalities, that is, we graph each inequality and find the intersection of the individual graphs. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution First graph x + y > 3 in red. Graph the system y x -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 First graph x + y > 3 in red. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution Next graph in blue. (continued) y x 6 5 4 3 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution Now we find the intersection of the regions. (continued) x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 Solution set to the system Now we find the intersection of the regions. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution This is a system of inequalities Graph Red Blue -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 Solution set This is a system of inequalities Red Blue Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

A system of inequalities may have a graph that consists of a polygon and its interior. In Section 4.5 we will have use for the corners, or vertices (singular vertex), of such a graph. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example The graph of the system Blue Green (3, 5) Red -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 (3, 5) (3, –3 ) (–1, 1 ) is shown below and the vertices are labeled. These vertices can be found by finding the intersection points of the pairs of lines. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Linear equations 2x – 8 = 3(x + 5) A number in one variable Let’s look at 6 different types of problems that we have solved, along with illustrations of each type. Type Example Solution Linear equations 2x – 8 = 3(x + 5) A number in one variable Graph Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Linear inequalities –3x + 5 > 2 A set of numbers; Type Example Solution Linear inequalities –3x + 5 > 2 A set of numbers; in one variable an interval Graph Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Linear equations 2x + y = 7 A set of ordered Type Example Solution Linear equations 2x + y = 7 A set of ordered in two variables pairs; a line Graph Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Linear inequalities x + y ≥ 4 A set of ordered Type Example Solution Linear inequalities x + y ≥ 4 A set of ordered in two variables pairs; a half-plane Graph Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

System of x + y = 3, An ordered pair or Type Example Solution System of x + y = 3, An ordered pair or equations in 5x – y = –27 a (possibly empty) two variables set of ordered pairs Graph Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

two variables x + 7 ≥ 0 of a plane Type Example Solution System of 6x – 2y ≤ 12, A set of ordered inequalities in y – 3 ≤ 0, pairs; a region two variables x + 7 ≥ 0 of a plane Graph Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley