2006 Education Pearson Education Inc Publishing as Pearson
© 2006 Education, Pearson Education, Inc. Publishing as Pearson Addison-Wesley Copyright © 2012 Pearson Inc. 1
2 2. 1 2. 2 2. 3 2. 4 2. 5 Linear Functions and Equations Linear Functions and Models Equations of Lines Linear Equations Linear Inequalities Absolute value Equations and Inequalities Copyright © 2012 Pearson Education, Inc.
2. 4 ♦ ♦ Linear Inequalities Understand basic terminology related to inequalities Solve linear inequalities symbolically Solve linear inequalities graphically and numerically Solve compound inequalities Copyright © 2012 Pearson Education, Inc.
Terminology Related to Inequalities • • Inequalities result whenever the equals sign in an equation is replaced with any one of the symbols ≤, ≥, <, >. Examples of inequalities include: • x – 5 > 2 x + 3 2 • x ≤ 1 – 2 x 2 • xy – x < x • 5 > 1 Copyright © 2012 Pearson Education, Inc. 4
Linear Inequality in One Variable • • • A linear inequality in one variable is an inequality that can be written in the form ax + b > 0 where a ≠ 0. (The symbol may be replaced by ≤, ≥, <, > ) Examples of linear inequalities in one variable: • 5 x + 4 ≤ 2 + 3 x simplifies to 2 x + 2 ≤ 0 • 1(x – 3) + 4(2 x + 1) > 5 simplifies to 7 x + 2 > 0 Examples of inequalities in one variable which are not linear: 2 • x < 1 Copyright © 2012 Pearson Education, Inc. 5
Interval Notation • The solution to a linear inequality in one variable is typically an interval on the real number line. See examples of interval notation below. Copyright © 2012 Pearson Education, Inc. 6
What Happens When Both Sides of An Inequality Are Multiplied By A Negative Number? • Note that 3 < 5, but if both sides are multiplied by 1, in order to produce a true statement the > symbol must be used. 3<5 but 3> 5 So when both sides of an inequality are multiplied (or divided) by a negative number the direction of the inequality must be reversed. Copyright © 2012 Pearson Education, Inc. 7
Solving Linear Inequalities Symbolically • The procedure for solving a linear inequality symbolically is the same as the procedure for solving a linear equation, except when both sides of an inequality are multiplied (or divided) by a negative number the direction of the inequality is reversed. Example of Solving a Linear Equation Symbolically Solve 2 x + 1 = x 2 2 x x = 2 1 3 x = 3 x=1 Example of Solving a Linear Inequality Symbolically Solve 2 x + 1 < x 2 2 x x < 2 1 3 x < 3 x>1 Note that we divided both sides by 3 so the direction was reversed. In interval notation the solution set is (1, ∞). Copyright © 2012 Pearson Education, Inc. 8
Example S T E P 1 Solve S Solution T E P 3 [ 2, 15, 1] by [ 2, 15, 1] S T E P 2 graphically. Note that the graphs intersect at the point (8. 20, 7. 59). The graph of y 1 is above the graph of y 2 to the right of the point of intersection or when x > 8. 20. Thus in interval notation the solution set is (8. 20, ∞). Copyright © 2012 Pearson Education, Inc. 9
Comments on Previous Problem S T E P 1 S T Solve E P 3 Note that the inequality above becomes y 1 > y 2 since in Step 1 we let y 1 equal the left-hand side of the inequality and y 2 equal the right hand side. To write the solution set of the inequality we are looking for the values of x for which the graph of y 1 is above the graph of y 2. These are values of x to the right of 8. 20, that is values larger than 8. 20. Thus the solution is x > 8. 20 which is (8. 20, ∞) in interval notation. Copyright © 2012 Pearson Education, Inc. 10
Example [ 10, 1] by [ 10, 1] S T E P 1 Solve Solution S T E P 3 S T E P 2 Note that the graphs intersect at the point ( 1. 36, 2. 72). The graph of y 1 is above the graph of y 2 to the left of the point of intersection or when x < 1. 36. When x ≤ 1. 36 the graph of y 1 is on or above the graph of y 2. Thus in interval notation the solution set is ( ∞, 1. 36]. Copyright © 2012 Pearson Education, Inc. 11
Comments on Previous Problem S T E P 1 Note that the inequality above becomes y 1 ≥ y 2 since in Step 1 we let y 1 equal the left-hand side of the inequality and y 2 equal the right hand side. S T Solve E P 3 To write the solution set of the inequality we are looking for the values of x for which the graph of y 1 is on or above the graph of y 2. These values of x are values less than or equal to 1. 36 which is ( ∞, 1. 36] in interval notation. Copyright © 2012 Pearson Education, Inc. 12
Example S T E P 1 Solve S Solution T E P 2 Note that the inequality above becomes y 1 ≥ y 2 since in Step 1 we let y 1 equal the lefthand side and y 2 equal the right hand side. To write the solution set of the inequality we are looking for the values of x in the table for which y 1 is the same or larger than y 2. Note that when x = 1. 3, y 1 is less than y 2; but when x = 1. 4, y 1 is larger than y 2. By the numerically. Intermediate Value Property, there is a value of x between 1. 4 and 1. 3 such that y 1 = y 2. In order to find an approximation of this value, make a new table in which x is incremented by 0. 01 (x is incremented by 0. 1 in the table to the left. ) Copyright © 2012 Pearson Education, Inc. 13
Example continued To write the solution set of the inequality we are looking for the values of x in the table for which y 1 is the same as or larger than y 2. Note that when x is approximately 1. 36, y 1 equals y 2 and when x is smaller than 1. 36 y 1 is larger than y 2 so the solutions can be written x ≤ 1. 36 or ( ∞, 1. 36] in interval notation. Copyright © 2012 Pearson Education, Inc. 14
Example The goal is to isolate the variable x in the middle of the three-part inequality • Suppose the Fahrenheit temperature x miles above the ground level is given by T(x) = 88 – 32 x. Determine the altitudes where the air temp is from 300 to 400. Solution We must solve the inequality 30 < 88 – 32 x < 40. Direction reversed – Divided each side of an inequality by a negative. Between 1. 5 and 1. 8215 miles above ground level, the air temperature is between 30 and 40 degrees Fahrenheit. Copyright © 2012 Pearson Education, Inc. 15
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