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M098 Carson Elementary and Intermediate Algebra 3e Section 4.6 Objectives 1. 2. Determine whether an ordered pair is a solution for a linear inequality with two variables. Graph lines inequalities Vocabulary Prior Knowledge If an inequality is multiplied or divided by a negative number, the direction of the inequality must be reversed. To graph a line, either 2 points or a point and a slope are needed. To find the x-intercept, let y = 0. To find the y-intercept, let x = 0. New Concepts 1. Determine whether an ordered pair is a solution for a linear inequality. Substitute the x- and y-coordinates into the inequality. If the resulting statement is true, the ordered pair is a solution. If the statement is false, the ordered pair is not a solution. Example 1: 1 > -0 + 2 1>2 Is (0, 1) a solution for y > -x + 2? False statement so (0, 1) is NOT a solution. 2. Graph linear Inequalities. When an equation in one variable is solved the answer is a point on a line. 2x = 6 x=3 3 When the corresponding inequality is solved, there are many solutions. The point divides the line into two parts. We highlight a portion of the line to indicate the solutions. 2x > 6 x>3 ( 3 A ( is used on the line for > or < to show the point is NOT included in the solution. A [ is used if the problem includes a greater than or equal because the point is included. Similarly, when an equation in two variables is graphed, the result is a line that contains all of the solutions to the equation. V. Zabrocki 2011 page 1 M098 Carson Elementary and Intermediate Algebra 3e Section 4.6 y = 2x + 5 Now, when the corresponding inequality is graphed, the line will divide the plane into two pieces just like the point divided the line into two pieces previously. The solution will be all the points on one side of the line. If the line is included (greater than or equal to, or less than or equal to) a solid line is drawn. If the line is not included (< or >), a dotted line is drawn. Select a test point to decide which side of the line should be shaded. When possible, (0, 0) is the easiest point to use. Example 2: Graph y < - 3x + 2 The associated line is y = - 3x + 2, which is in slope-intercept form. The slope is -3. The y-intercept is (0, 2). The line will be dotted because the inequality does not include the line. The solution to the inequality will be all of the points on one side of the line. Pick a test point and substitute the x- and y-coordinates into the original inequality to determine which side to shade. Test Point: (0, 0) 0 < -3(0) + 2 0<2 This is a true statement so (0, 0) is one of many solutions to the inequality. Shade the side of the line that contains the test point (0, 0). Example 3: y 2 x 5 2 5 y-intercept: (0, 0) Slope: The line will be dotted. V. Zabrocki 2011 page 2 M098 Carson Elementary and Intermediate Algebra 3e Section 4.6 (0, 0) can not be used as the test point because it lies on the line. Pick ANY point not on the line. HINT: Pick a number for x that will be easy to work with. One possibility is (5, 1). 2 5 5 1 2 1 This is a false statement so (5, 1) is NOT a solution. Shade the opposite side of the line. Example 4: 3x – 2y ≥ 6 The associated line is 3x – 2y = 6, which is in standard form. Find the x- and y-intercepts to graph the line. x-intercept: (2, 0) y-intercept: (0, -3) Line is solid because of the ≥. Test Point: (0, 0) 3(0) – 2(0) ≥ 6 0>6 False statement – shade opposite side of the line. Example 5: x≤3 The associated equation is x = 3, which is a vertical line since it contains only an x. Draw a vertical line through the x-axis at 3. The line will be solid. Test Point: (0, 0) Because the inequality does not have a y variable, only the x-coordinate is used. 0≤3 True statement – shade the same side of the line that (0, 0) is on. V. Zabrocki 2011 page 3