Graphing Linear Inequalities

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1 Graphing Linear Inequalities Linear Inequalities in Two Variables: A linear inequality in two variables is an inequality that can be written in the general form Ax + By < C, where A, B, and C are real numbers and A and B are not both zero. The inequality symbol can be <, >,,. Example: Each of the following is a linear inequality in two variables. a. x + 5y > 8 b. 7 x 5y + c. 8x 3 4y d. 6y < 3 Example: Rewrite the linear inequality 7 x 6 y + 5in the general form Ax + By < C. Solution: In the general form Ax + By < C the x and y values are on the left side of the inequality and the constant is on the right side of the inequality. We will use the same techniques that we learned when solving linear equations in two variables. In this case all we need to do is use the addition property of inequality to subtract -6y from both sides of the equation. 7x 6y + 5 7x 6y 5 7x + 6y 5 In the last step, both sides of the inequality were multiplied by (-1). This was only done in order to reduce the number of negative signs in the inequality. Recall that when multiplying an inequality by a negative number, the direction of the inequality sign must be reversed. Example: Rewrite the linear equation < 3(7x 4) y + 3in the general form Ax + By < C. Solution: Again, we will use basic equation solving techniques. < 3(7x 4) y + 3 < 1x y + 3 < 1x y 9 11 < 1x y 1x y > 11 When switching the left hand and right hand expressions, don t forget to also switch the inequality sign. In other words, if Ax + By < C then, C > Ax + By

2 Solutions of Linear Inequalities in Two Variables: The solution to a linear inequality in two variables is an ordered pair that when substituted for the variables will make the inequality true. Example: Show that the ordered pair (-, 1) is a solution to the linear equation 5 x y < 8. Solution: Substitute the ordered pair (-, 1) into the inequality 5 x y < 8. 5x y < 8 5( ) (1) < 8 10 < 8 < 8 Because this is a true statement, the ordered pair is a solution to the inequality. Example: Determine if the ordered pair (-4, 5) is a solution to the linear inequality ( y 4) + 1 > x. Solution: Substitute the ordered pair (-4, 5) into the inequality ( y 4) + 1 > x. ( y 4) + 1 > x 3( 4) (5 4) + 1 > ( 4) (1) + 1 > 8 1 > > 13 Because this is NOT a true statement, the ordered pair is not a solution to the inequality. Notice that if the inequality were rewritten as ( y 4) + 1 x the statement would be true and the ordered pair would be a solution. This is demonstrated in the example below. ( y 4) + 1 x 3( 4) (5 4) + 1 ( 4) (1) Example: Determine if the ordered pair (, 5) is a solution to the linear inequality x 3y > 1. Solution: Substitute the ordered pair (, 5) into the inequality x 3y > 1. x 3y > 1 () 3(5) > > 1 19 > 1 The ordered pair (, 5) is not a solution to the inequality.

3 The Graph of a Linear Inequality: Just as with linear equations in two variables, a linear inequality in two variables will have an infinite number of solutions. The big difference between the two is that with a linear equation the solutions are all the ordered pairs which lie on the graph of the equation (the line) but with a linear inequality the solutions will be a region of the coordinate plane which is divided by the graph of the equation (line). Example: Given the graphs of the equation y = x + and the inequality y < x + determine if the points (, 4) and (4, -4) are solutions of each by inspecting the graph. Solution: It can also be determined whether or not the points are solutions by substituting them into the equation or inequality as demonstrated below. (, 4) (4, -4) (, 4) (4, -4) y < x + y < x + y = x + y = x + 4 < + 4 < 4 4 < < 6 4 = + 4 = 4 4 = = 6 Boundary Lines: Notice in the graph of the inequality above that the points on the line itself are NOT solutions to the inequality. This will always be the case when a less than (<) or greater than (>) symbol is used. In this case, we call the equation y = x + the boundary line for the graph of the inequality y < x +. This means that the equation y = x + forms a boundary between the solution region and non-solution region of the coordinate plane but is not a part of the solution region. The boundary equation is simply the equation created from the inequality by replacing the inequality symbol with an equal sign. Boundary lines may be graphed as a dashed line or a solid line. 1. Dashed Boundary Line: Less Than (<) or Greater Than (>). Solid Boundary Line: Less Than or Equal To ( ) or Greater Than or Equal To ( )

4 Procedure for Graphing Inequalities: 1. Graph the corresponding equation. This is the boundary line. Determine if the boundary line is a solid or dashed line.. Choose a test point in either half plane and substitute the coordinates of the point into the inequality. If the resulting inequality is true, that the graph of the inequality is the half-plane containing the test point. If not true, then the other half-plane is the solution. 3. Shade the half-plane that represents the solution to the inequality. Example: Graph the inequality y > x 3. Solution: The boundary equation is y = x 3. Graph this equation using previously learned techniques to obtain the following graph. Because the inequality is a (>) symbol the boundary line will be dashed. Select a test point somewhere in either half-plane. The point (0, 0) is always an easy point to work with so I generally use it unless it lies on the boundary line. y > x 3 0 > (0) 3 0 > 3 True Because the test point (0, 0) makes the inequality true, the half-plane containing (0, 0) will be the solution region to the inequality. Remember that any point in the shaded region is a solution to the inequality.

5 Example: Graph the inequality 1 y x +. 1 Solution: The boundary equation is y = x +. Graph this equation using previously learned techniques to obtain the graph to the right. Because the inequality is a ( ) symbol the boundary line will be solid. Select a test point somewhere in either half-plane. The point (0, 0) is always an easy point to work with so I generally use it unless it lies on the boundary line. 1 y x (0) + 0 Because the test point (0, 0) makes the inequality NOT true, the half-plane NOT containing (0, 0) will be the solution region to the inequality. Remember that any point in the shaded region is a solution to the inequality. Example: Graph the inequality x 3y < 1. Solution: The boundary equation is x 3y = 1 This can be graphed using the intercepts. The boundary line will be dashed because of the (<) symbol and the test point (0, 0) is a solution to the inequality so the top half of the coordinate plane is the shaded (solution) region.

6 Example: Graph the inequality + 3y 3 + 3( y 3) Solution: Before creating the boundary equation, simplify the inequality as much as possible. + 3y 3 + 3( y 3) + 3y 3 + 3y 9 6 x Therefore, the boundary equation is simply the vertical line: x = The boundary line will be solid because of the ( ) symbol and the test point (0, 0) is NOT a solution to the inequality so the left half of the coordinate plane is the shaded (solution) region. Example: Graph the inequality 4 y + x > 5y + x 10 Solution: Before creating the boundary equation, simplify the inequality as much as possible. 4 y + x > 5y + x 10 4 y > 5y 10 7 y > 14 y < Therefore, the boundary equation is simply the horizontal line: y = The boundary line will be dashed because of the (>) symbol and the test point (0, 0) is a solution to the inequality so the bottom half of the coordinate plane is the shaded (solution) region.

7 Applications: Example: Tickets to a swim meet cost $4 if purchased in advance. If purchased on the date of the meet, they cost $6. For the swim meet to make a profit, the total money collected from ticket sales must be at least $10. Write an inequality to represent this situation, and then graph the inequality and interpret the graph, including the intercepts. Solution: The inequality is 4a + 6d 10. The easiest way to graph this will be to find the intercepts using the boundary equation 4 a + 6d = 10. 4a + 6d 10 4(0) + 6d = 10 d = 0 (0,0) 4a + 6d 10 4a + 6(0) = 10 a = 30 (30,0) Any combination of ticket sales in the shaded region will be a solution to this inequality.

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