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1 Note that the “>” can be replaced by , <, or . Examples: Linear inequalities in one variable. 2x – 2 < 6x – 5 A linear inequality in one variable is.

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Presentation on theme: "1 Note that the “>” can be replaced by , <, or . Examples: Linear inequalities in one variable. 2x – 2 < 6x – 5 A linear inequality in one variable is."— Presentation transcript:

1 1 Note that the “>” can be replaced by , <, or . Examples: Linear inequalities in one variable. 2x – 2 < 6x – 5 A linear inequality in one variable is an inequality which can be put into the form ax + b > c can be written – 4x + (– 2) < – 5. 6x + 1  3(x – 5) 2x + 3 > 4 can be written 6x + 1  – 15. where a, b, and c are real numbers. 6.1 Solving Linear Inequalities in One Variable

2 2 The solution set for an inequality can be expressed in two ways. 1. Set-builder notation: 2. Graph on the real line: {x | x <  3} Example: Express the solution set of x <  3 in two ways. 1. Set-builder notation: 2. Graph on the real line: {x | x  4} Example: Express the solution set of x  4 in two ways. 0 1234-4-3-2 0 1234-4-3-2 Open circle indicate that the number is not included in the solution set. ° Closed circle indicate that the number is included in the solution set. 6.1 Solving Linear Inequalities in One Variable

3 3 A solution of an inequality in one variable is a number which, when substituted for the variable, results in a true inequality. Examples: Are any of the values of x given below solutions of 2x > 5? 2 is not a solution. 2.6 is a solution. x = 22(2) > 54 > 5 x = 2.62(2.6) > 55.2 > 5 The solution set of an inequality is the set of all solutions. 3 is a solution.x = 32(3) > 56 > 5 x = 1.52(1.5) > 53 > 51.5 is not a solution.False True ?? ?? ?? ?? 6.1 Solving Linear Inequalities in One Variable

4 4 The graph of a linear inequality in one variable is the graph on the real number line of all solutions of the inequality. 6.1 Solving Linear Inequalities in One Variable Verbal Phrase Inequality Graph All real numbers less than 2x < 2 All real numbers greater than -1x > -1 All real numbers less than or equal to 4 x < 4 All real numbers greater than or equal to -3 x > -3

5 5 If a > b and c is a real number, then a > b, a + c > b + c, and a – c > b – c have the same solution set. Addition and Subtraction Properties Example: Solve x – 4 > 7. x – 4 > 7 Add 4 to each side of the inequality. + 4 x > 11 Set-builder notation. {x | x > 11} If a < b and c is a real number, then a < b, a + c < b + c, and a – c < b – c have the same solution set. 6.1 Solving Linear Inequalities in One Variable

6 6 Example: Solve 3x  2x + 5. Subtract 2x from each side. 3x  2x + 5 x  5 – 2x Set-builder notation. {x | x  5} 6.1 Solving Linear Inequalities in One Variable

7 7 Multiplication and Division Properties If c b, ac < bc, and < have the same solution set. If c > 0 the inequalities a > b, ac > bc, and > have the same solution set. Example: Solve 4x  12. x  3 Divide by 4. 4 is greater than 0, so the inequality sign remains the same. 4x  12 6.1 Solving Linear Inequalities in One Variable

8 8 Example: Solve. (– 3) Multiply by – 3. – 3 is less than 0, so the inequality sign changes. 6.1 Solving Linear Inequalities in One Variable

9 9 Example: Solve x + 5 < 9x + 1. – 8x + 5 < 1 Subtract 9x from both sides. – 8x < – 4 Subtract 5 from both sides. x > Divide both sides by – 8 and simplify. Inequality sign changes because of division by a negative number. Solution set in set-builder notation. 6.1 Solving Linear Inequalities in One Variable

10 10 Example: Solve. Subtract from both sides. Add 2 to both sides. Multiply both sides by 5. 2030405060-20-10010 Solution set as a graph. 6.1 Solving Linear Inequalities in One Variable

11 11 Example: A cell phone company offers its customers a rate of $89 per month for 350 minutes, or a rate of $40 per month plus $0.50 for each minute used. Solve the inequality 0.50x + 40  89. 0.50x  49 x  24.5 Subtract 40. Divide by 0.5. Let x = the number of minutes used. The customer can use up to 24.5 minutes per month before the cost of the second plan exceeds the cost of the first plan. How many minutes per month can a customer who chooses the second plan use before the charges exceed those of the first plan? 6.2 Problem Solving

12 12 A compound inequality is formed by joining two inequalities with “and” or “or.” 6.3 Compound Inequalities

13 13 Example: Solve x + 2 – 8. x + 2 < 5 Solve the first inequality. The solution set of the “and” compound inequality is the intersection of the two solution sets. x < 3 {x | x > –1} {x | x < 3} Subtract 2. Solution set Solve the second inequality. 2x – 6 > – 8 2x > – 2 Add 6. Divide by 2. x > – 1 Solution set 0 1234-4-3-2 º ° 6.3 Compound Inequalities

14 14 Example: Solve 11 < 6x + 5 < 29. 6 < 6x < 24 Subtract 5 from each of the three parts. 1 < x < 4 Divide 6 into each of the three parts. Solution set. This inequality means 11 < 6x + 5 and 6x + 5 < 29. When solving compound inequalities, it is possible to work with both inequalities at once. 6.3 Compound Inequalities

15 15 Example: Solve. Multiply each part by – 2. Subtract 6 from each part. Multiplication by a negative number changes the inequality sign for each part. 0 1234-4-3-2 Solution set. 6.3 Compound Inequalities

16 16 Example: Solve x + 5 > 6 or 2x < – 4. Solve the first inequality. Since the inequalities are joined by “or” the solution set is the union of the solution sets. Solution set x + 5 > 62x < – 4 Solve the second inequality. { x | x > 1} 0 1234-4-3-2 x > 1 x < – 2 Solution set { x | x < – 2} ° ° 6.3 Compound Inequalities

17 17 6.4 Absolute Value and Inequalities There are 4 types of absolute value inequalities and equivalent inequalities |x| < a |x| > a

18 18 6.4 Absolute Value and Inequalities Translating Absolute Value Inequalities 1.The inequality |ax + b| < c is equivalent to -c < ax + b < c 2.The inequality |ax + b| > c is equivalent to ax + b < -c or ax + b > c

19 19 Example Solve |x - 4| < 3 -3 < x - 4 < 3 1 < x < 7 The solution set is {x| 1 < x < 7} and the interval is (1, 7) 6.4 Absolute Value and Inequalities º

20 20 Example Solve |4x - 1| < 9 -9 ≤ 4x - 1 ≤ 9 -8 ≤ 4x ≤ 10 -2 ≤ x ≤ 5/2 The interval solution is [-2, 5/2] 6.4 Absolute Value and Inequalities 

21 21 Solve |x| > a Solve |x + 1| > 2 x + 1 2 x 1 The solution interval is (-∞, -3) U (1, ∞) 6.4 Absolute Value and Inequalities º

22 22 Solve |x| ≥ a Solve |2x - 8| ≥ 4 2x – 8 ≤ -4 or 2x - 8 ≥ 4 2x ≤ 4 or 2x ≥ 12 x ≤ 2 or x ≥ 6 The solution interval is (-∞, 2] U [ 6, ∞) 6.4 Absolute Value and Inequalities


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