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Chapter 1: Tools of Algebra 1-4: Solving Inequalities

Chapter 1: Tools of Algebra 1-4: Solving Inequalities. Essential Question: What is one important difference between solving equations and solving inequalities?. 1-4: Solving Inequalities. Inequalities are solved exactly the same as equations except for one key difference:

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Chapter 1: Tools of Algebra 1-4: Solving Inequalities

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  1. Chapter 1: Tools of Algebra1-4: Solving Inequalities Essential Question: What is one important difference between solving equations and solving inequalities?

  2. 1-4: Solving Inequalities • Inequalities are solved exactly the same as equations except for one key difference: • When multiplying or dividing by a negative number, you must reverse the inequality • Example: • 6 + 5 (2 – x) < 41

  3. 1-4: Solving Inequalities • Inequalities are solved exactly the same as equations except for one key difference: • When multiplying or dividing by a negative number, you must reverse the inequality • Example: • 6 + 5 (2 – x) < 41 (distribute) • 6 + 10 – 5x< 41

  4. 1-4: Solving Inequalities • Inequalities are solved exactly the same as equations except for one key difference: • When multiplying or dividing by a negative number, you must reverse the inequality • Example: • 6 + 5 (2 – x) < 41 (distribute) • 6 + 10 – 5x < 41 (combine like terms) • 16 – 5x < 41

  5. 1-4: Solving Inequalities • Inequalities are solved exactly the same as equations except for one key difference: • When multiplying or dividing by a negative number, you must reverse the inequality • Example: • 6 + 5 (2 – x) < 41 (distribute) • 6 + 10 – 5x < 41 (combine like terms) • 16 – 5x < 41 (subtract 16 from each side)-16 -16 • -5x < 25

  6. 1-4: Solving Inequalities • Inequalities are solved exactly the same as equations except for one key difference: • When multiplying or dividing by a negative number, you must reverse the inequality • Example: • 6 + 5 (2 – x) < 41 (distribute) • 6 + 10 – 5x < 41 (combine like terms) • 16 – 5x < 41 (subtract 16 from each side)-16 -16 • -5x < 25 (divide both sides by -5)-5 -5 (and flip the sign) • x > -5

  7. 1-4: Solving Inequalities • Graphing Inequalities • Regular equations were graphed simply by putting a point on a number line. • Because inequalities imply an infinite number of solutions, we graph them using a line • When the variable comes first, you can follow the arrow • If the inequality uses < or >, use an open circle • If the inequality uses < or >, use a closed circle • Think: If you do the extra work and underline the inequality, you have to do the extra work and fill in the circle.

  8. 1-4: Solving Inequalities • Graphing Inequalities • Example: • 3x – 12 < 3

  9. 1-4: Solving Inequalities • Graphing Inequalities • Example: • 3x – 12 < 3+ 12 +12 (add 12 to both sides) • 3x < 15

  10. 1-4: Solving Inequalities • Graphing Inequalities • Example: • 3x – 12 < 3 + 12 +12 (add 12 to both sides) • 3x < 153 3 (divide both sides by 3) • x < 5 • x comes first, which means: • Put an open circle at 5 (because the inequality is “<“) • Draw an arrow to the left

  11. 1-4: Solving Inequalities • “No Solutions” or “All Real Numbers” as solutions • If all variables get eliminated in a problem, it means that the solution is either “No Solution” or “All Real Numbers” • If the statement is false, there is “No Solution” • If the statement is true, “All Real Numbers” will solve • Example: • 2x – 3 > 2(x – 5)

  12. 1-4: Solving Inequalities • “No Solutions” or “All Real Numbers” as solutions • If all variables get eliminated in a problem, it means that the solution is either “No Solution” or “All Real Numbers” • If the statement is false, there is “No Solution” • If the statement is true, “All Real Numbers” will solve • Example: • 2x – 3 > 2(x – 5) • 2x – 3 > 2x – 10 (distribute)

  13. 1-4: Solving Inequalities • “No Solutions” or “All Real Numbers” as solutions • If all variables get eliminated in a problem, it means that the solution is either “No Solution” or “All Real Numbers” • If the statement is false, there is “No Solution” • If the statement is true, “All Real Numbers” will solve • Example: • 2x – 3 > 2(x – 5) • 2x – 3 > 2x – 10 (distribute)-2x -2x (subtract 2x from both sides) • -3 > -10

  14. 1-4: Solving Inequalities • “No Solutions” or “All Real Numbers” as solutions • If all variables get eliminated in a problem, it means that the solution is either “No Solution” or “All Real Numbers” • If the statement is false, there is “No Solution” • If the statement is true, “All Real Numbers” will solve • Example: • 2x – 3 > 2(x – 5) • 2x – 3 > 2x – 10 (distribute)-2x -2x (subtract 2x from both sides) • -3 > -10 (note: you could add 3 to both sides to see 0 > -7) • -3 is greater than -10, so “All Real Numbers” are solutions

  15. 1-4: Solving Inequalities • Real World Connection • Example: A band agrees to play for $200 plus 25% of the ticket sales. Find the ticket sales needed for the band to receive at least $500 • Cut the meat out of the problem • $200 + 25% of ticket sales > $500 • Let x = ticket sales (in dollars) • Write an equation and solve

  16. 1-4: Solving Inequalities • Real World Connection • Example: A band agrees to play for $200 plus 25% of the ticket sales. Find the ticket sales needed for the band to receive at least $500 • Cut the meat out of the problem • $200 + 25% of ticket sales > $500 • Let x = ticket sales (in dollars) • Write an equation and solve • 200 + 0.25x > 500

  17. 1-4: Solving Inequalities • Real World Connection • Example: A band agrees to play for $200 plus 25% of the ticket sales. Find the ticket sales needed for the band to receive at least $500 • Cut the meat out of the problem • $200 + 25% of ticket sales > $500 • Let x = ticket sales (in dollars) • Write an equation and solve • 200 + 0.25x > 500-200 -200 (subtract 200 from each side) • 0.25x > 300

  18. 1-4: Solving Inequalities • Real World Connection • Example: A band agrees to play for $200 plus 25% of the ticket sales. Find the ticket sales needed for the band to receive at least $500 • Cut the meat out of the problem • $200 + 25% of ticket sales > $500 • Let x = ticket sales (in dollars) • Write an equation and solve • 200 + 0.25x > 500-200 -200 (subtract 200 from each side) • 0.25x > 3000.25 0.25 (divide each side by 0.25) • x > 1200

  19. 1-4: Solving Inequalities • Compound Inequality • A pair of inequalities combined using the words and or or. • Solve the two inequalities separately • Inequalities that use “and” are going to meet in the middle • They have two ends • Inequalities that use “or” are going to go in opposite directions • Like oars on a boat

  20. 1-4: Solving Inequalities • Example: Compound Inequality using “And” • Graph the solution of 3x – 1 > -28 and 2x + 7 < 19

  21. 1-4: Solving Inequalities • Example: Compound Inequality using “And” • Graph the solution of 3x – 1 > -28 and 2x + 7 < 19

  22. 1-4: Solving Inequalities • Example: Compound Inequality using “And” • Graph the solution of 3x – 1 > -28 and 2x + 7 < 19

  23. 1-4: Solving Inequalities • Example: Compound Inequality using “And” • Graph the solution of 3x – 1 > -28 and 2x + 7 < 19

  24. 1-4: Solving Inequalities • Example: Compound Inequality using “And” • Graph the solution of 3x – 1 > -28 and 2x + 7 < 19

  25. 1-4: Solving Inequalities • Example: Compound Inequality using “And” • Graph the solution of 3x – 1 > -28 and 2x + 7 < 19

  26. 1-4: Solving Inequalities • Example: Compound Inequality using “Or” • Graph the solution of 4y – 2 > 14 or 3y – 4 < -13

  27. 1-4: Solving Inequalities • Example: Compound Inequality using “Or” • Graph the solution of 4y – 2 > 14 or 3y – 4 < -13

  28. 1-4: Solving Inequalities • Example: Compound Inequality using “Or” • Graph the solution of 4y – 2 > 14 or 3y – 4 < -13

  29. 1-4: Solving Inequalities • Example: Compound Inequality using “Or” • Graph the solution of 4y – 2 > 14 or 3y – 4 < -13

  30. 1-4: Solving Inequalities • Example: Compound Inequality using “Or” • Graph the solution of 4y – 2 > 14 or 3y – 4 < -13

  31. 1-4: Solving Inequalities • Example: Compound Inequality using “Or” • Graph the solution of 4y – 2 > 14 or 3y – 4 < -13

  32. 1-4: Solving Inequalities • Real World Connection • The ideal length of a bolt is 13.48 cm. The length can vary from the ideal length by at most 0.03 cm. A machinist finds one bolt that is 13.67 cm long. By how much should the machinist decrease the length so the bolt can be used? • Solution: • Minimum length < length after cut < maximum length • Let x be the amount cut from the bolt • Minimum length: • Maximum length: • Length after cut:

  33. 1-4: Solving Inequalities • Real World Connection • The ideal length of a bolt is 13.48 cm. The length can vary from the ideal length by at most 0.03 cm. A machinist finds one bolt that is 13.67 cm long. By how much should the machinist decrease the length so the bolt can be used? • Solution: • Minimum length < length after cut < maximum length • Let x be the amount cut from the bolt • Minimum length: 13.48 – 0.03 = 13.45 cm • Maximum length: • Length after cut:

  34. 1-4: Solving Inequalities • Real World Connection • The ideal length of a bolt is 13.48 cm. The length can vary from the ideal length by at most 0.03 cm. A machinist finds one bolt that is 13.67 cm long. By how much should the machinist decrease the length so the bolt can be used? • Solution: • Minimum length < length after cut < maximum length • Let x be the amount cut from the bolt • Minimum length: 13.48 – 0.03 = 13.45 cm • Maximum length: 13.48 + 0.03 = 13.51 cm • Length after cut:

  35. 1-4: Solving Inequalities • Real World Connection • The ideal length of a bolt is 13.48 cm. The length can vary from the ideal length by at most 0.03 cm. A machinist finds one bolt that is 13.67 cm long. By how much should the machinist decrease the length so the bolt can be used? • Solution: • Minimum length < length after cut < maximum length • Let x be the amount cut from the bolt • Minimum length: 13.48 – 0.03 = 13.45 cm • Maximum length: 13.48 + 0.03 = 13.51 cm • Length after cut: 13.67 – x

  36. 1-4: Solving Inequalities • Solution: • Minimum length < length after cut < maximum length • Let x be the amount cut from the bolt • Minimum length: 13.48 – 0.03 = 13.45 cm • Maximum length: 13.48 + 0.03 = 13.51 cm • Length after cut: 13.67 – x • 13.45 < 13.67 – x < 13.51

  37. 1-4: Solving Inequalities • Solution: • Minimum length < length after cut < maximum length • Let x be the amount cut from the bolt • Minimum length: 13.48 – 0.03 = 13.45 cm • Maximum length: 13.48 + 0.03 = 13.51 cm • Length after cut: 13.67 – x • 13.45 < 13.67 – x < 13.51-13.67 -13.67 -13.67 (subtract 13.67 from all parts) • -0.22 < -x < -0.16

  38. 1-4: Solving Inequalities • Solution: • Minimum length < length after cut < maximum length • Let x be the amount cut from the bolt • Minimum length: 13.48 – 0.03 = 13.45 cm • Maximum length: 13.48 + 0.03 = 13.51 cm • Length after cut: 13.67 – x • 13.45 < 13.67 – x < 13.51-13.67 -13.67 -13.67 (subtract 13.67 from all parts) • -0.22 < -x < -0.16 -1  -1  -1 (divide all parts by -1) • 0.22 > x > 0.16 (and flip all signs)

  39. 1-4: Solving Inequalities • Assignment • Page 29 • Problems 1 – 27 (odd problems)

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