Solving Inequalities Examples

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1 Solving Inequalities Examples 1. Joe and Katie are dancers. Suppose you compare their weights. You can make only one of the following statements. Joe s weight is less than Kate s weight. Joe s weight is the same as Kate s weight. Joe s weight is greater than Kate s weight. 2. Let j stand for Joe s weight and k stand for Katie s weight. Then you can use inequalities and an equation to compare their weights. j < k J = k j > k This is an illustration of the trichotomy property 3. Trichotomy Property For any two real numbers a and b, exactly one of the following statements is true. a < b a = b a > b 4. Addition and Subtraction Properties for Inequalities 1. If a > b, then a + c > b + c and a c > b c. 2. If a < b, then a + c < b + c and a c < b c. Adding the same number to each side of an inequality does not change the truth of the inequality. 5. The following numerical examples may be used to illustrate the Addition and Subtraction Properties. 5 < < < 8 6 > > > 19 These properties can be used to solve inequalities. Each solution set can be graphed on the number line.

2 6. Example Solve 9x + 7 < 8x 2. Graph the solution set. 9x + 7 < 8x 2 8x + 9x + 7 < 8x + 8x 2 x + 7 < 2 x ( 7) < 2 + ( 7) x < 9 To check inequalities, first check the boundary point for the variable, and see if the two sides are equivalent. A true equation should occur if the inequality sign is replaced by the equals sign. To make sure the direction of the inequality is correct, check a point on each side of the boundary point. Check 9( 9) + 7 = 8( 9) 2 74 = 74 Choose a point in the solution set. 9( 10) + 7 < 8( 10) 87 < -80 TRUE Choose a point outside the solution set. 9( 8) + 7 < 8( 8) 65 < 64 FALSE 7. Example Solve y + 6 > 3. Graph the solution set. y + 6 > 3 y > 3 6 y > 3 8. Example Solve 18 < t 7. Graph the solution set. 18 < t (7) < t 7 + (7) 11 < t 9. Example Solve 2m + 9 < m + 4. Graph the solution set. 2m + 9 < m + 4 m + 9 < 4 m < We know that 18 > 11 is a true inequality. If you multiply each side of this inequality by a positive number, the result is a true inequality. 18 > (3) > 11(3) 54 > 33 TRUE What happens if we multiply by a negative number?

3 11. Suppose you multiply each side of a true inequality by a negative number. 18 > 11 18( 2) > 11( 2) 36 > 22 FALSE! We must reverse the inequality symbol when we multiply or divide by a negative number. 12. Multiplication and Division Properties for Inequalities 1. If c is positive and a < b, then ac < bc and c a < c b. 2. If c is positive and a > b, then ac > bc and c a > c b. 3. If c is negative and a < b, then ac > bc and c a > c b. 4. If c is negative and a > b, then ac > bc and c a < c b. 13. The following examples may be used to illustrate the Multiplication and Division Properties. 16 > 8 5 < ( 5)( 4) > 2( 4) < > 8 8 < Example Solve 0.5y < 6. Graph the solution set. 0.5y < 6 ( 2)( 0.5y) > ( 2)(6) y > 12 The solution set can be written {y y > 12}. It is read the set of all numbers y such that y is greater than 12. This notation for solution sets is called set builder notation.

4 15. Example Solve x 11 x 3 3x x 11 4x x 4 x x 11. Graph the solution set The solution set can be written {x x } The symbols,, and can also be used when comparing numbers. The symbol means is not equal to. The symbol means is less than or equal to. The symbol means is greater than or equal to. 6x 18 means 6x > 18 or 6x < Example Solve 3x > 27. Graph the solution set. 3x > 27 x > 9 The solution set is {x x > 9} 18. Example Solve 3x > 27. Graph the solution set. 3x > 27 x < 9 The solution set is {x x < 9} 19. Example Solve 3y y 3 4 y 4 The solution set is {y y 4} 3y Graph the solution set. 4

5 20. Example Solve 4a + 16 < 2(a + 4). Graph the solution set. 4a + 16 < 2(a + 4) 4a + 16 < 2a 8 6a < 24 a < 4 The solution set is {a a < 4} 21. The absolute value of a number represents its distance from zero on the number line. You can use this idea to help solve absolute value inequalities. 22. Example Solve x < 3. Graph the solution set. x < 3 means the distance between x and 0 is less than 3 units. To make 3 true, you must substitute values for x that are less than 3 units from 0. All the numbers between 3 and 3 are less than three units from zero. The solution set is {x 3< x < 3}. 23. Example Solve x 2. Graph the solution set. To make this true, you must substitute values for x that are 2 or more units from 0. The solution set is {x x 2 or x 2}. 24. Example Solve 2x 5 > 9. Graph the solution set. The inequality 2x 5 > 9 says that 2x 5 is more than 9 units from 0. 2x 5 > 9 2x > 14 x > 7 OR 2x 5 < 9 2x < 4 x < 2 The solution set is {x x < 2 or x > 7}

6 25. Example Solve 8x 24. Graph the solution set. The inequality 8x 24 says that 8x is less than 24 units from 0. 8x 24 x 3 AND 8x 24 x 3 The solution set is {x 3 x 3} 26. Example Solve x + 2 > 5. Graph the solution set. The inequality x + 2 > 5 says that x + 2 is more than 5 units from 0. x + 2 > 5 x > 3 OR x + 2 < 5 x < 7 The solution set is {x x >3 or x < 7} 27. Example Solve 2x < 5. Graph the solution set. First rewrite the inequality by subtracting 4 from each side 2x + 3 < 1 The inequality 2x + 3 < 1 says that 2x + 3 is less than 1 units from 0. 2x + 3 < 1 2x < 2 x < 1 AND 2x + 3 > 1 2x > 4 x > 2 The solution set is {x 2 < x < 1}

7 28. Example Solve 3x 8 < 19. Graph the solution set. The inequality 3x 8 < 19 says that 3x 8 is less than 19 units from 0. 3x 8 < 19 3x < 27 x < 9 AND 3x 8 > 19 3x > x > The solution set is {x < x < 9} Some absolute value inequalities have no solutions. For example, 4x 9 < 7 is never true. Since the absolute value of a number is always positive or zero, there is not replacement for x that will make the sentence true. The inequality 4x 9 < 7 has no solution. Therefore, its solution set is Ø. 30. Some absolute value inequalities are always true. For example, 10x + 3 > 5 is always true. Since the absolute value of a number is always positive or zero, any replacement for x will make the sentence true. The solution set for 10x + 3 > 5 is the set of real numbers. 31. Example Solve 6x > 3. Graph the solution set. First rewrite the inequality by subtracting 5 from each side 6x + 2 > 2 The solution is all numbers since this statement is always true.

8 32. Example Solve 6x < 3. Graph the solution set. Rewrite the inequality by subtracting nine from each side 6x 8 < 6 There are no solutions since this statement is always false.

9 Name: Date: Class: Graph the solution set of each inequality. Solving Inequalities Worksheet 1. x > 3 2. a 0 3. p x < k n > 6 Solve each inequality. Graph the solution set. 7. 3x + 7 > x x < x 5 < x + 1 < x x x < (3x + 2) > 7x 2 State an absolute value inequality for each of the following. Then graph each inequality. 15. All numbers between 3 and All numbers less than 8 and greater than All numbers greater than 6 or less than All numbers less than or equal to 5, and greater than or equal to x > 3 or x < x < 6 and x > x 4 and x 4

10 Solve each inequality. Graph each solution set. 22. x < x x + 1 > x < x x < x < x > 0

11 Name: Date: Class: Solving Inequalities Worksheet Key Graph the solution set of each inequality. 1. x > 3 2. a 0 3. p x < k 10 k n > 6 n < 2 Solve each inequality. Graph the solution set. 7. 3x + 7 > 43 3x > 36 x > x x 49 x x < 44 3x < 36 x > x 5 < 0.1 x < 5.1

12 11. 5(3x + 2) > 7x 2 15x + 10 > 7x 2 8x > 12 x > x + 1 < x + 5 2x < 4 x < x x 2.32 x x < x + 3 < x < 2.85 x < State an absolute value inequality for each of the following. Then graph each inequality. 15. All numbers between 3 and 3. x < All numbers less than 8 and greater than 8. x > All numbers greater than 6 or less than 6. x > All numbers less than or equal to 5, and greater than or equal to 5. x x > 3 or x < 3 x > 3

13 20. x < 6 and x > 6 x < x 4 and x 4 x 4 Solve each inequality. Graph each solution set. 22. x < 9 x < 9 and x > x 2 x 2 or x x + 1 > 3 x + 1 < 3 x < 4 or x + 1 > 3 x > x < 6 3x > 6 x > 2 and 3x < 6 x < x 64 Since absolute values must be positive, all numbers work x < 15 Since absolute values must be positive, no numbers work Ǿ 28. 6x < 6 First rewrite inequality by subtracting 6 from each side. (6x + 25 < 8 Since absolute values must be positive, no numbers work Ǿ

14 x > 0 First rewrite inequality by subtracting 6 from each side. 3x > 6 Since absolute values must be positive, all numbers work.

15 Student Name: Date: Solving Inequalities Checklist 1. On questions 1 thru 6, did the student graph the inequality correctly? a. Yes (30 points) b. 5 out of 6 (25 points) c. 4 out of 6 (20 points) d. 3 out of 6 (15 points) e. 2 out of 6 (10 points) f. 1 out of 6 (5 points) 2. On questions 7 thru 14, did the student solve the inequality correctly? a. Yes (40 points) b. 7 out of 8 (35 points) c. 6 out of 8 (30 points) d. 5 out of 8 (25 points) e. 4 out of 8 (20 points) f. 3 out of 8 (15 points) g. 2 out of 8 (10 points) h. 1 out of 8 (5 points) 3. On questions 7 thru 14, did the student graph the inequality correctly? a. Yes (40 points) b. 7 out of 8 (35 points) c. 6 out of 8 (30 points) d. 5 out of 8 (25 points) e. 4 out of 8 (20 points) f. 3 out of 8 (15 points) g. 2 out of 8 (10 points) h. 1 out of 8 (5 points) 4. On questions 15 thru 21, did the student state a correct absolute value for each inequality? a. Yes (35 points) b. 6 out of 7 (30 points) c. 5 out of 7 (25 points) d. 4 out of 7 (20 points) e. 3 out of 7 (15 points) f. 2 out of 7 (10 points) g. 1 out of 7 (5 points) 5. On questions 15 thru 21, did the student graph the inequality correctly? a. Yes (35 points) b. 6 out of 7 (30 points) c. 5 out of 7 (25 points) d. 4 out of 7 (20 points) e. 3 out of 7 (15 points) f. 2 out of 7 (10 points) g. 1 out of 7 (5 points) 6. On questions 22 thru 29, did the student solve the inequality correctly? a. Yes (40 points) b. 7 out of 8 (35 points) c. 6 out of 8 (30 points) d. 5 out of 8 (25 points) e. 4 out of 8 (20 points) f. 3 out of 8 (15 points) g. 2 out of 8 (10 points) h. 1 out of 8 (5 points) Total Number of Points A 198 points and above B 176 points and above Any score below C needs remediation! C 154 points and above D 132 points and above F 131 points and below NOTE: The sole purpose of this checklist is to aide the teacher in identifying students that need remediation. Students who meet the C criteria are ready for the next level of learning.

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