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- 1. Inequalities
- 2. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line.
- 3. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line. Specifically, given two numbersL and R corresponding to two points on the real line as shown, L R– +
- 4. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line. Specifically, given two numbersL and R corresponding to two points on the real line as shown,we define the number to the right to be greater than thenumber to the left. L R– +
- 5. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line. Specifically, given two numbersL and R corresponding to two points on the real line as shown,we define the number to the right to be greater than thenumber to the left. Hence R is greater than L or that L issmaller than R . L R– +
- 6. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line. Specifically, given two numbersL and R corresponding to two points on the real line as shown,we define the number to the right to be greater than thenumber to the left. Hence R is greater than L or that L issmaller than R .– L < R +We write this as L < R or as R > L.
- 7. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line. Specifically, given two numbersL and R corresponding to two points on the real line as shown,we define the number to the right to be greater than thenumber to the left. Hence R is greater than L or that L issmaller than R .– L < R +We write this as L < R or as R > L.For example, 2 < 4, –3 < –2, 0 > –1 are true statements
- 8. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line. Specifically, given two numbersL and R corresponding to two points on the real line as shown,we define the number to the right to be greater than thenumber to the left. Hence R is greater than L or that L issmaller than R .– L < R +We write this as L < R or as R > L.For example, 2 < 4, –3 < –2, 0 > –1 are true statements and–2 < –5 , 5 < 3 are false statements.
- 9. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line. Specifically, given two numbersL and R corresponding to two points on the real line as shown,we define the number to the right to be greater than thenumber to the left. Hence R is greater than L or that L issmaller than R .– L < R +We write this as L < R or as R > L.For example, 2 < 4, –3 < –2, 0 > –1 are true statements and–2 < –5 , 5 < 3 are false statements.We use inequalities to represent segments of the real line, e.g.“all the numbers x that are greater than 5" as “5 < x".
- 10. InequalitiesWe recall that comparison of two numbers is based on theirlocations on the real line. Specifically, given two numbersL and R corresponding to two points on the real line as shown,we define the number to the right to be greater than thenumber to the left. Hence R is greater than L or that L issmaller than R .– L < R +We write this as L < R or as R > L.For example, 2 < 4, –3 < –2, 0 > –1 are true statements and–2 < –5 , 5 < 3 are false statements.We use inequalities to represent segments of the real line, e.g.“all the numbers x that are greater than 5" as “5 < x". 5 x– +
- 11. InequalitiesIn general, we write "a < x" for all In picture,
- 12. InequalitiesIn general, we write "a < x" for all the numbers x greater than abut excluding a. In picture, a a<x +– open dot
- 13. InequalitiesIn general, we write "a < x" for all the numbers x greater than abut excluding a. In picture, a a<x +– open dotIf we want all the numbers x greater than or equal to a whichincludes a, we write it as a < x.
- 14. InequalitiesIn general, we write "a < x" for all the numbers x greater than abut excluding a. In picture, a a<x +– open dotIf we want all the numbers x greater than or equal to a whichincludes a, we write it as a < x. In picture a a≤x – solid dot +
- 15. InequalitiesIn general, we write "a < x" for all the numbers x greater than abut excluding a. In picture, a a<x +– open dotIf we want all the numbers x greater than or equal to a whichincludes a, we write it as a < x. In picture a a≤x – solid dot +The numbers x such that a < x < b where a < b are all thenumbers x between a and b. a a<x≤b b– +
- 16. InequalitiesIn general, we write "a < x" for all the numbers x greater than abut excluding a. In picture, a a<x +– open dotIf we want all the numbers x greater than or equal to a whichincludes a, we write it as a < x. In picture a a≤x – solid dot +The numbers x such that a < x < b where a < b are all thenumbers x between a and b. A line segment as such is calledan interval. a a<x≤b b– +
- 17. InequalitiesIn general, we write "a < x" for all the numbers x greater than abut excluding a. In picture, a a<x +– open dotIf we want all the numbers x greater than or equal to a whichincludes a, we write it as a < x. In picture a a≤x – solid dot +The numbers x such that a < x < b where a < b are all thenumbers x between a and b. A line segment as such is calledan interval. a a<x≤b b– + –3 2Hence –3 ≤ x < 2 is
- 18. InequalitiesIn general, we write "a < x" for all the numbers x greater than abut excluding a. In picture, a a<x +– open dotIf we want all the numbers x greater than or equal to a whichincludes a, we write it as a < x. In picture a a≤x – solid dot +The numbers x such that a < x < b where a < b are all thenumbers x between a and b. A line segment as such is calledan interval. a a<x≤b b– + –3 2Hence –3 ≤ x < 2 isExpressions such as 2 < x > 3 or 2 < x < –3 do not have anysolution.
- 19. InequalitiesWe note the following algebra about inequalities.
- 20. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
- 21. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign,
- 22. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.
- 23. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
- 24. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign,
- 25. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.
- 26. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.Given that 6 < 12, if we multiply –1 to both sides
- 27. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.Given that 6 < 12, if we multiply –1 to both sides then (–1)6 > (–1)12 – 6 > –12 is true.
- 28. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.Given that 6 < 12, if we multiply –1 to both sides then (–1)6 > (–1)12 – 6 > –12 is true.Multiplying by –1 switches the relation of two numbers.
- 29. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.Given that 6 < 12, if we multiply –1 to both sides then (–1)6 > (–1)12 – 6 > –12 is true.Multiplying by –1 switches the relation of two numbers. – 0 + 6 < 12
- 30. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.Given that 6 < 12, if we multiply –1 to both sides then (–1)6 > (–1)12 – 6 > –12 is true.Multiplying by –1 switches the relation of two numbers. – 0 + –6 6 < 12
- 31. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.Given that 6 < 12, if we multiply –1 to both sides then (–1)6 > (–1)12 – 6 > –12 is true.Multiplying by –1 switches the relation of two numbers. – 0 + –12 < –6 6 < 12
- 32. InequalitiesWe note the following algebra about inequalities.Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.a. Adding or subtracting the same quantity to both sidesretains the inequality sign, i.e. if a < b, then a ± c < b ± c.Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.b. Multiplying or dividing a positive number c to both sidesretains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.Given that 6 < 12, if we multiply –1 to both sides then (–1)6 > (–1)12 – 6 > –12 which is true.Multiplying by –1 switches the relation of two numbers. – 0 + –12 < –6 6 < 12c. Multiplying or dividing by an negative number c reversesthe inequality sign, i.e. if c < 0 and a < b then ca > cb
- 33. InequalitiesFollowing are the steps for solving linear inequalities.
- 34. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.
- 35. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).
- 36. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).3. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around.
- 37. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).3. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around. (This rule can be avoided by keeping the x-term positive.)
- 38. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).3. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around. (This rule can be avoided by keeping the x-term positive.)Example A. a. Solve –3x + 2 < 11. Draw the solution.
- 39. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).3. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around. (This rule can be avoided by keeping the x-term positive.)Example A. a. Solve –3x + 2 < 11. Draw the solution.–3x + 2 < 11 move the x to the other side to make it positive
- 40. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).3. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around. (This rule can be avoided by keeping the x-term positive.)Example A. a. Solve –3x + 2 < 11. Draw the solution.–3x + 2 < 11 move the x to the other side to make it positive 2 – 11 < 3x
- 41. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).3. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around. (This rule can be avoided by keeping the x-term positive.)Example A. a. Solve –3x + 2 < 11. Draw the solution.–3x + 2 < 11 move the x to the other side to make it positive 2 – 11 < 3x –9 < 3x
- 42. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).3. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around. (This rule can be avoided by keeping the x-term positive.)Example A. a. Solve –3x + 2 < 11. Draw the solution.–3x + 2 < 11 move the x to the other side to make it positive 2 – 11 < 3x –9 < 3x divide by 3, no change with the inequality –9 3x 3 < 3 –3 < x
- 43. InequalitiesFollowing are the steps for solving linear inequalities.1. Simplify both sides of the inequalities.2. Gather the x-terms to one side and the number-terms to the other sides (use the “change side-change sign” rule).3. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around. (This rule can be avoided by keeping the x-term positive.)Example A. a. Solve –3x + 2 < 11. Draw the solution.–3x + 2 < 11 move the x to the other side to make it positive 2 – 11 < 3x –9 < 3x divide by 3, no change with the inequality –9 3x 3 < 3 –3 < x -3In picture, 0
- 44. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.
- 45. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,
- 46. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9
- 47. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
- 48. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3
- 49. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3 –3 –2–2 > x ≥ –3 or
- 50. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3 –3 –2–2 > x ≥ –3 orSigns and Inequalities
- 51. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3 –3 –2–2 > x ≥ –3 orSigns and InequalitiesOne of the most important association to inequalities isthe sign of a given expression.
- 52. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3 –3 –2–2 > x ≥ –3 orSigns and InequalitiesOne of the most important association to inequalities isthe sign of a given expression. Let f be an expression,following adjectives translate into the following inequalities.
- 53. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3 –3 –2–2 > x ≥ –3 orSigns and InequalitiesOne of the most important association to inequalities isthe sign of a given expression. Let f be an expression,following adjectives translate into the following inequalities.i. “f is positive” ↔ 0 < f or f > 0
- 54. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3 –3 –2–2 > x ≥ –3 orSigns and InequalitiesOne of the most important association to inequalities isthe sign of a given expression. Let f be an expression,following adjectives translate into the following inequalities.i. “f is positive” ↔ 0 < f or f > 0 “f is non–negative” ↔ 0 ≤ f or f ≥ 0
- 55. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3 –3 –2–2 > x ≥ –3 orSigns and InequalitiesOne of the most important association to inequalities isthe sign of a given expression. Let f be an expression,following adjectives translate into the following inequalities.i. “f is positive” ↔ 0 < f or f > 0 “f is non–negative” ↔ 0 ≤ f or f ≥ 0ii. “f is negative” ↔ f < 0 or 0 > f
- 56. Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.Draw the solution.5 < –3x + 1 ≤ 10 subtract 1 from all entries,–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,–6 –3x 9–3 > –3 ≥ –3 –3 –2–2 > x ≥ –3 orSigns and InequalitiesOne of the most important association to inequalities isthe sign of a given expression. Let f be an expression,following adjectives translate into the following inequalities.i. “f is positive” ↔ 0 < f or f > 0 “f is non–negative” ↔ 0 ≤ f or f ≥ 0ii. “f is negative” ↔ f < 0 or 0 > f “f is non–positive” ↔ f ≤ 0 or 0 ≥ f
- 57. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.
- 58. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 0
- 59. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12x > –4
- 60. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12 –4 xx > –4 or 0
- 61. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12 –4 xx > –4 or 0b. Let x be the price of a movie ticket. Translate the sentence“$20 minus the cost of two movie tickets is non–positive”into an inequality. Solve it and draw the solution.
- 62. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12 –4 xx > –4 or 0b. Let x be the price of a movie ticket. Translate the sentence“$20 minus the cost of two movie tickets is non–positive”into an inequality. Solve it and draw the solution.Two tickets cost $2x,
- 63. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12 –4 xx > –4 or 0b. Let x be the price of a movie ticket. Translate the sentence“$20 minus the cost of two movie tickets is non–positive”into an inequality. Solve it and draw the solution.Two tickets cost $2x, so “$20 minus the cost of two tickets”is (20 – 2x).
- 64. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12 –4 xx > –4 or 0b. Let x be the price of a movie ticket. Translate the sentence“$20 minus the cost of two movie tickets is non–positive”into an inequality. Solve it and draw the solution.Two tickets cost $2x, so “$20 minus the cost of two tickets”is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is20 – 2x ≤ 0
- 65. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12 –4 xx > –4 or 0b. Let x be the price of a movie ticket. Translate the sentence“$20 minus the cost of two movie tickets is non–positive”into an inequality. Solve it and draw the solution.Two tickets cost $2x, so “$20 minus the cost of two tickets”is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is20 – 2x ≤ 020 ≤ 2x10 ≤ x
- 66. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12 –4 xx > –4 or 0b. Let x be the price of a movie ticket. Translate the sentence“$20 minus the cost of two movie tickets is non–positive”into an inequality. Solve it and draw the solution.Two tickets cost $2x, so “$20 minus the cost of two tickets”is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is20 – 2x ≤ 020 ≤ 2x 10 x10 ≤ x or 0
- 67. InequalitiesExample B. Translate the sentence “3x + 12 is positive” intoinequalities and solve for x. Draw the solution.Being positive translates into the inequality3x + 12 > 03x > –12 –4 xx > –4 or 0b. Let x be the price of a movie ticket. Translate the sentence“$20 minus the cost of two movie tickets is non–positive”into an inequality. Solve it and draw the solution.Two tickets cost $2x, so “$20 minus the cost of two tickets”is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is20 – 2x ≤ 020 ≤ 2x 10 x10 ≤ x or 0So the price of a ticket has to be $10 or more.
- 68. InequalitiesIntersection and Union (∩ & U)
- 69. InequalitiesIntersection and Union (∩ & U)Let I be the interval consists of 1 ≤ x ≤ 3 as shown, 1 3 I:
- 70. InequalitiesIntersection and Union (∩ & U)Let I be the interval consists of 1 ≤ x ≤ 3 as shown, 1 3 I:and in the set notation we write I = {1 ≤ x ≤ 3 }.
- 71. InequalitiesIntersection and Union (∩ & U)Let I be the interval consists of 1 ≤ x ≤ 3 as shown, 1 3 I:and in the set notation we write I = {1 ≤ x ≤ 3 }.Let J = {2 < x < 4 } be another interval as shown. 2 4 J:
- 72. InequalitiesIntersection and Union (∩ & U)Let I be the interval consists of 1 ≤ x ≤ 3 as shown, 1 3 I:and in the set notation we write I = {1 ≤ x ≤ 3 }.Let J = {2 < x < 4 } be another interval as shown. 2 4 J:The common portion of the two intervals I and J shown here 1 2 3 I: 2 3 4 J:
- 73. InequalitiesIntersection and Union (∩ & U)Let I be the interval consists of 1 ≤ x ≤ 3 as shown, 1 3 I:and in the set notation we write I = {1 ≤ x ≤ 3 }.Let J = {2 < x < 4 } be another interval as shown. 2 4 J:The common portion of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 2 3
- 74. InequalitiesIntersection and Union (∩ & U)Let I be the interval consists of 1 ≤ x ≤ 3 as shown, 1 3 I:and in the set notation we write I = {1 ≤ x ≤ 3 }.Let J = {2 < x < 4 } be another interval as shown. 2 4 J:The common portion of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 2 3 I ∩ J:is called the intersection of I and J and it’s denoted as I ∩ J.
- 75. InequalitiesIntersection and Union (∩ & U)Let I be the interval consists of 1 ≤ x ≤ 3 as shown, 1 3 I:and in the set notation we write I = {1 ≤ x ≤ 3 }.Let J = {2 < x < 4 } be another interval as shown. 2 4 J:The common portion of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 2 3 I ∩ J:is called the intersection of I and J and it’s denoted as I ∩ J.In this case I ∩ J = {2 < x ≤ 3 }.
- 76. InequalitiesThe merge of the two intervals I and J shown here
- 77. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4
- 78. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4
- 79. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4 I U J:is called the union of I and J and it’s denoted as I U J.
- 80. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4 I U J:is called the union of I and J and it’s denoted as I U J.In this case I U J: = {1 ≤ x < 4 }.
- 81. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4 I U J:is called the union of I and J and it’s denoted as I U J.In this case I U J: = {1 ≤ x < 4 }.The Interval Notation
- 82. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4 I U J:is called the union of I and J and it’s denoted as I U J.In this case I U J: = {1 ≤ x < 4 }.The Interval NotationWe use the infinity symbol “∞” to mean “continues onward andsurpasses all numbers”.
- 83. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4 I U J:is called the union of I and J and it’s denoted as I U J.In this case I U J: = {1 ≤ x < 4 }.The Interval NotationWe use the infinity symbol “∞” to mean “continues onward andsurpasses all numbers”. We label the left of the real line with –∞ and to the right with ∞ as shown below when needed. –∞ 0 ∞
- 84. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4 I U J:is called the union of I and J and it’s denoted as I U J.In this case I U J: = {1 ≤ x < 4 }.The Interval NotationWe use the infinity symbol “∞” to mean “continues onward andsurpasses all numbers”. We label the left of the real line with –∞ and to the right with ∞ as shown below when needed. –∞ 0 ∞Infinity “∞” is not a number but we say that –∞ < x < ∞for all real number x.
- 85. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4 I U J:is called the union of I and J and it’s denoted as I U J.In this case I U J: = {1 ≤ x < 4 }.The Interval NotationWe use the infinity symbol “∞” to mean “continues onward andsurpasses all numbers”. We label the left of the real line with –∞ and to the right with ∞ as shown below when needed. –∞ 0 ∞Infinity “∞” is not a number but we say that –∞ < x < ∞for all real number x. In fact, the set of all real numbers R isR = {–∞ < x < ∞ }.
- 86. InequalitiesThe merge of the two intervals I and J shown here 1 2 3 I: 2 3 4 J: 1 2 3 4 I U J:is called the union of I and J and it’s denoted as I U J.In this case I U J: = {1 ≤ x < 4 }.The Interval NotationWe use the infinity symbol “∞” to mean “continues onward andsurpasses all numbers”. We label the left of the real line with –∞ and to the right with ∞ as shown below when needed. –∞ 0 ∞Infinity “∞” is not a number but we say that –∞ < x < ∞for all real number x. In fact, the set of all real numbers R isR = {–∞ < x < ∞ }. However, we do not write x ≤ ∞ because ∞ isnot a number in particular x can’t be ∞.
- 87. InequalitiesUsing the “∞” symbols, we write the line segment a ∞ or a ≤ x
- 88. InequalitiesUsing the “∞” symbols, we write the line segment a ∞ or a ≤ x as [a, ∞),
- 89. InequalitiesUsing the “∞” symbols, we write the line segment a ∞ or a ≤ x as [a, ∞),We use ] or [ to include the end points,
- 90. InequalitiesUsing the “∞” symbols, we write the line segment a ∞ or a ≤ x as [a, ∞), a ∞ or a < x as (a, ∞),We use ] or [ to include the end points, ) and ( to exclude theend points.
- 91. InequalitiesUsing the “∞” symbols, we write the line segment a ∞ or a ≤ x as [a, ∞), a ∞ or a < x as (a, ∞),–∞ a or x ≤ a as (–∞, a],–∞ a or x < a as (–∞, a),We use ] or [ to include the end points, ) and ( to exclude theend points.
- 92. InequalitiesUsing the “∞” symbols, we write the line segment a ∞ or a ≤ x as [a, ∞), a ∞ or a < x as (a, ∞),–∞ a or x ≤ a as (–∞, a],–∞ a or x < a as (–∞, a),Let a, b be two numbers such that a < b, we write the intervals a b or a ≤ x ≤ b as [a, b], a b or a < x < b as (a, b), a b or a ≤ x < b as [a, b), a b or a < x ≤ b as (a, b],We use ] or [ to include the end points, ) and ( to exclude theend points.
- 93. InequalitiesUsing the “∞” symbols, we write the line segment a ∞ or a ≤ x as [a, ∞), a ∞ or a < x as (a, ∞),–∞ a or x ≤ a as (–∞, a],–∞ a or x < a as (–∞, a),Let a, b be two numbers such that a < b, we write the intervals a b or a ≤ x ≤ b as [a, b], a b or a < x < b as (a, b), a b or a ≤ x < b as [a, b), a b or a < x ≤ b as (a, b],We use ] or [ to include the end points, ) and ( to exclude theend points. Note the interval notation (a, b) is the same as thecoordinate of a point so its interpretation depends on the context.
- 94. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.
- 95. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.
- 96. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.
- 97. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.
- 98. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.
- 99. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.
- 100. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.
- 101. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1
- 102. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U Jb. K ∩ I
- 103. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U J –3 1 0 Kb. K ∩ I
- 104. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U J –3 –2 1 0 Kb. K ∩ I
- 105. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U J –3 –2 1 –3 0 so K U J is 0 Kb. K ∩ I
- 106. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U J –3 –2 1 –3 0 so K U J is 0 KHence K U J is (–3, ∞) or {–3 < x}.b. K ∩ I
- 107. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U J –3 –2 1 –3 0 so K U J is 0Hence K U J is (–3, ∞) or {–3 < x}.b. K ∩ I –3 K 1 0
- 108. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U J –3 –2 1 –3 0 so K U J is 0Hence K U J is (–3, ∞) or {–3 < x}.b. K ∩ I –3 1 –4 –1 0 K I
- 109. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U J –3 –2 1 –3 0 so K U J is 0Hence K U J is (–3, ∞) or {–3 < x}.b. K ∩ I –3 1 –4 –1 0 K IThe common or overlapping portion is shown.
- 110. InequalitiesThe interval [a, a] consists of exactly one point {x = a}.The empty set which contains nothing is denoted as Φ = { }.The interval (a, a) or (a, a] or [a, a) = Φ.Example C. Given intervals I, J , and K, do the set operation.Draw. Put the answer in the interval and the inequality notation. –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1a. K U J –3 –2 1 –3 0 so K U J is 0Hence K U J is (–3, ∞) or {–3 < x}.b. K ∩ I –3 1 –4 –1 0 K IThe common or overlapping portion is shown.Hence K ∩ I is (–3, –1) or {–3 < x < –1}
- 111. Inequalities –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1c. (K ∩ J) U I
- 112. Inequalities –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1c. (K ∩ J) U IWe do the parenthesis first.
- 113. Inequalities –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1c. (K ∩ J) U IWe do the parenthesis first. –3 –2 1K∩J: 0
- 114. Inequalities –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1c. (K ∩ J) U IWe do the parenthesis first. –3 –2 1K∩J: 0 –2 1so K ∩ J is 0
- 115. Inequalities –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1c. (K ∩ J) U IWe do the parenthesis first. –3 –2 1K∩J: 0 –2 1so K ∩ J is 0Therefore (K ∩ J) U I is –4 –2 –1 1 0
- 116. Inequalities –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1c. (K ∩ J) U IWe do the parenthesis first. –3 –2 1K∩J: 0 –2 1so K ∩ J is 0Therefore (K ∩ J) U I is –4 –2 –1 1 0so (K ∩ J) U I = {–4 < x ≤ 1} or (–4, 1]
- 117. Inequalities –4 –1I: 0 J: x > –2 K: –3 < x ≤ 1c. (K ∩ J) U IWe do the parenthesis first. –3 –2 1K∩J: 0 –2 1so K ∩ J is 0Therefore (K ∩ J) U I is –4 –2 –1 1 0so (K ∩ J) U I = {–4 < x ≤ 1} or (–4, 1]Your turn. Find K ∩ (J U I ). Is this the same as (K ∩ J) U I?

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