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###### Communicate Your Answer

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Communicate Your Answer 3 | |

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###### Exercises

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Exercises 1 Any sentence that has one of the symbols <,>,≤, or ≥ relates quantities that are not equal. We call this type of sentence an "inequality." | |

Exercises 2 When a value is a solution to an inequality, a true statement is made when the value is substituted into the inequality. If it does not make a true statement, it is not a solution. To determine if 5 is a solution to x+3>8, we will substitute it for x and see if a true statement is made. x+3>8x=55+3>?8Add terms8≯8 We get 8≯8 because 8 is not greater than itself. Rather, 8 equals 8. Thus, 5 is not in the solution set. | |

Exercises 3 The solution set to an inequality is graphed on a number line.We place a circle on the number to which the variable is related. If the inequality symbol is < or >, the circle is an open circle because the variable cannot equal that value. If the symbol is ≤ or ≥, the symbol is a closed circle because the variable can equal that value. Next, we shade the region on the number line either to the right or the left of the circle to show the solution set. If the symbol is < or ≤, we shade to the left of the circle because the values to the left become smaller. If the symbol is > or ≥, we shade to the right of the circle because the values to the right become larger. Consider the inequality. x<7 We will place an open circle at 7 and shade the region to the left because x is less than 7.Now consider a second inequality. x≥3 We will place a closed circle at 3 and shade the region to the right because x is greater than 3. | |

Exercises 4 To begin, we will make sense of what each phrase means and which inequality symbol it is associated with.NameGiven StatementPhrase meansSymbolInequality Aw is greater than or equal to -7greater than or equal to≥w≥-7 Bw is no more than -7less than or equal to≤w≤-7 Cw is no less than -7greater than or equal to≥w≥-7 Dw is at least -7greater than or equal to≥w≥-7 Notice that Phrase B is the only one that cannot be expressed with the inequality w≥-7. Thus, B does not belong. The two different inequalities are: w≥-7 and w≤-7. | |

Exercises 5 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "greater than," which is represented by the symbol >. …>… On the left-hand side of the inequality we have "a number x," which we will express with the variable x. x>… On the right-hand side of the inequality, we have "3." We can now form our final answer. a number x is greater than 3x > 3 | |

Exercises 6 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case we have less than or equal to, which is represented by the symbol ≤. …≤… On the left-hand side of the inequality we have a number n plus 7, which we will express as a sum. n+7≤… On the right-hand side of the inequality we have 9. We can now form our final answer. a number n plus 7 is less than or equal to 9n + 7≤9 | |

Exercises 7 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "no more than," which is represented by the symbol ≤. …≤… On the left-hand side of the inequality we have 15. 15≤… On the right-hand side of the inequality, we have "a number divided by 5," which we will express as a fraction. We can now form our final answer. 15 is no more than a number t divided by 515≤5t | |

Exercises 8 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "less than," which is represented by the symbol <. …<… On the left-hand side of the inequality we have "three times a number w" which we will express as a product. 3×w<… On the right-hand side of the inequality, we have "18." We can now form our final answer. Three times a number w is less than 183×w<18 Notice that 3×w can be written as 3w. Writing a complete inequality, we have: 3w<18 | |

Exercises 9 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "more than," which is represented by the symbol >. …>… On the left-hand side of the inequality we have "one-half of a number y," which we will express as: 21y>… On the right-hand side of the inequality, we have "22." We can now form our final answer. One-half of a number y is more than 2221y>22 | |

Exercises 10 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "less than," which is represented by the symbol <. …<… On the left-hand side of the inequality we have "three." 3<… On the right-hand side of the inequality, we have "the sum of a number s and 4." We can now form our final answer. Three is less than the sum of a number s and 43<s+4 | |

Exercises 11 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "at least," which is represented by the symbol ≥. …≥… On the left-hand side of the inequality we have "thirteen." 13≥… On the right-hand side of the inequality, we have "the difference of a number v and 1." We can now form our final answer. Thirteen is at least the difference of a number v and 113≥v−1 | |

Exercises 12 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "no less than," which is represented by the symbol ≥. …≥… On the left-hand side of the inequality we have "four." 4≥… On the right-hand side of the inequality, we have "the quotient of a number x and 2.1," which we will represent as a fraction. We can now form our final answer. Four is no less than the quotient of a number x and 2.14≥2.1x | |

Exercises 13 To begin, we will determine which of the inequality symbols corresponds with the statement. It is given that the weight of the second fish is at least 0.5 pounds more than the weight of the first fish. The phrase at least is symbolically represented by ≥. …≥… If we let w represent the weight of the second fish, we can complete the left-hand side of our inequality. w≥… We know that the weight w is at least 0.5 pounds more than the weight of the first fish. The first fish weighs 1.2 pounds. We can now complete the inequality. w≥1.2+0.5⇔w≥1.7 | |

Exercises 14 It is given that the maximum capacity of the pool is 600 people. This means 600 or fewer people can be in the pool at one time. It is also given that there are currently 430 people in the pool. We can determine the number of additional people that can enter the pool by subtracting 430 from 600. If we let p be the number of additional people that can enter, we have the following inequality. p≤600−430⇔p≤170 | |

Exercises 15 If 2 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. r+4>8r=22+4>?8Add terms6≯8 The statement 6>8 says that 6 is greater than 8. As this is always a false statement, 2 is not a solution to the given inequality. | |

Exercises 16 If -3 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. 5−x<8x=-35−(-3)<?8a−(-b)=a+b8≮8 The statement 8<8 says that 8 is less than 8. As this is always a false statement, -3 is not a solution to the given inequality. | |

Exercises 17 If -6 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. 3s≤19s=-63(-6)≤?19a(-b)=-a⋅b-3⋅6≤?19Multiply-18≤19 The statement -18≤19 says that -18 is less than or equal to 19. As this is always a true statement, -6 is a solution to the given inequality. | |

Exercises 18 If 7 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. 17≥2yy=717≥?2(7)Multiply17≥14 The statement 17≥14 says that 17 is greater than or equal to 14. As this is always a true statement, 7 is a solution to the given inequality. | |

Exercises 19 If 3 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. -1>-2xx=3-1>?-23Calculate quotient-1>-1.5 The statement -1>-1.5 says that -1 is less than -1.5. As this is always a true statement, 3 is a solution to the given inequality. | |

Exercises 20 If 2 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. z4≥3z=224≥?3Calculate quotient2≱3 The statement 2≥3 says that 2 is greater than or equal to 3. As this is always a false statement, 2 is not a solution to the given inequality. | |

Exercises 21 If -5 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. 14≥-2n+4n=-514≥?-2(-5)+4-a(-b)=a⋅b14≥?10+4Add terms14≥14 The statement 14≥14 says that 14 is greater than or equal to 14. As this is always a true statement, -5 is a solution to the given inequality. | |

Exercises 22 If 10 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. -5÷(2s)<-1s=10-5÷(2⋅10)<?-1Multiply-5÷(20)<?-1a÷b=ba20-5<?-1 Simplify fraction ba=b/5a/54-1<?-1Put minus sign in front of fraction -41≮-1 The statement -41<-1 says that -1 is greater than -41. As this is always a false statement, 10 is not a solution to the given inequality. | |

Exercises 23 If 5 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. 20≤2z10+20z=520≤?2⋅510+20Multiply20≤?1010+20Calculate quotient20≤?1+20Add terms20≤21 The statement 20≤21 says that 20 is less than or equal to 21. As this is always a true statement, 5 is a solution to the given inequality. | |

Exercises 24 If 8 is a solution to the inequality, it will make a true statement when substituted for the variable. If it does not make a true statement, it is not a solution. 63m−2>3m=863⋅8−2>?3Multiply624−2>?3Calculate quotient4−2>?3Subtract term2≯3 The statement 2>3 says that 2 is greater than 3. As this is always a false statement, 8 is not a solution to the given inequality. | |

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Exercises 27 After substituting 8 into the inequality and simplifying, we get -1<-4. It is then concluded that 8 is a solution to the inequality. However, it should be: -1≮-4. The less negative a number is, the greater it is. By graphing -1 and -4 on a number line, we can determine which is greater.The further a number is to the right, the greater it is. Hence, we can conclude that -1 is greater than -4. Therefore, the error is that we have drawn the wrong conclusion. It should be that 8 is not a solution to the inequality. | |

Exercises 28 After substituting 8 into the inequality and simplifying, we get: 6≤6. The sign ≤ means that the number on the left-hand side is less than or equal to the number on the right-hand side. Since 6 is equal to 6, this inequality is correct. Therefore, 8 is in the solution set. | |

Exercises 29 The inequality x≥2 tells us that all values of x are greater than or equal to 2. This means the solution set lies on and to the right of 2 on a number line. Notice that because the sign is greater than or equal to, x=2 is a solution. We show this with a closed circle on the number line at 2. | |

Exercises 30 The inequality z≤5 tells us that all values of z are less than or equal to 5. This means that the solution set lies on and to the left of 5 on a number line. Because the inequality sign is "less than or equal to," z=5 is a solution. We show this with a closed circle on the number line at 5. | |

Exercises 31 The inequality -1>t tells us that -1 is greater than all values of t. This means that the solution set lies to the left of -1 on a number line. Notice that the inequality sign is strict. This indicates that t=-1 is not a solution, so we mark an open circle on the number line at -1. | |

Exercises 32 The inequality -2<w tells us that -2 is less than all values of w. This means the solution set lies to the right of -2 on a number line. Notice that the inequality symbol indicates that w=-2 is not a solution, so we mark an open circle on the number line at -2. | |

Exercises 33 The inequality v≤-4 tells us that all values of v are less than or equal to -4. This means the solution set lies on and to the left of -4 on a number line. Notice that v=-4 is a solution, so we mark a closed circle on the number line at -4. | |

Exercises 34 The inequality s<1 tells us that all values of s are less than 1. This means that the solution set lies to the left of 1 on a number line. Notice that the inequality sign indicates that s=1 is not a solution, so we will mark an open circle on the number line at 1. | |

Exercises 35 The inequality 41<p tells us that 41 is less than all values of p. This means the solution set lies to the right of 41 on a number line. Notice that the inequality sign indicates that p=41 is not a solution, so we will mark an open circle on the number line at 41. | |

Exercises 36 An absolute value always results in a positive number. We can simply remove the absolute value and write 5 instead of ∣5∣. Now we have the following inequality: r≥-5 To graph this, we place a closed dot on -5 on a number line. Since we are looking for all values of r that are greater than or equal to -5, we shade everything to the right of this dot. | |

Exercises 37 The solution set {x∣x<7} can be read as the sentence "the set of all x such that x is less than 7." This can be written as the inequality: x<7. To graph this inequality, we place an open dot on 7 on a number line. Since x is less than 7, the solution set includes all numbers that are below 7. Hence, we should shade everything to the left of the open dot. | |

Exercises 38 The solution set {n∣n≥-2} can be read as the sentence "the set of all n such that n is greater than or equal to -2." This can be written as the inequality: n≥-2. To graph this inequality, we place a closed dot on -2 on a number line. Since n is greater than or equal to -2, the solution set includes all numbers that are above or equal 7. Hence, we should shade everything to the right of the closed dot. | |

Exercises 39 The solution set {z∣1.3≤z} can be read as the sentence "the set of all z such that z is greater than or equal to 1.3." This can be written as the inequality: z≥1.3 To graph this inequality, we place a closed dot on 1.3 on a number line. Since z is greater than or equal to 1.3, the solution set includes all numbers that are above or equal to 1.3. Hence, we should shade everything to the right of the closed dot. | |

Exercises 40 The solution set {w∣5.2>w} can be read as the sentence "the set of all w such that w is less than 5.2." This can be written as the inequality: w<5.2. To graph this inequality, we place an open dot on 5.2 on a number line. Since w is less than 5.2, the solution set includes all numbers that are below 5.2. Hence, we should shade everything to the left of the open dot. | |

Exercises 41 Since the circle at 4 is closed, the graph describes all values less than or equal to 4. In other words, we have an inequality which is not strict and should include 4. If we let x represent these values, we can write the following inequality. x≤4 | |

Exercises 42 Since the circle at -2 is closed, the graph describes all values greater than or equal to -2. In other words, we have an inequality which is not strict and should include -2. If we let x represent these values, we can write the following inequality. x≥-2 | |

Exercises 43 Since the circle at 3 is open, the graph describes all values greater than but not equal to 3. In other words, we have a strict inequality and should exclude 3. If we let x represent these values, we can write the following inequality. x>3 | |

Exercises 44 Since the circle at -1 is open, the graph describes all values less than but not equal to -1. In other words, we have a strict inequality and should exclude -1. If we let x represent these values, we can write the following inequality. x<-1 | |

Exercises 45 The phrase no less than 76∘F indicates that the water temperature must be greater than or equal to 76∘F. The temperature is currently 74∘F, which is lower than the minimum expected. Let's calculate the difference between the minimum desired and the current temperature. 76∘F−74∘F=2∘F To reach the minimum expected, the temperature must be raised by 2∘F. Let x be the number of degrees the temperature needs to increase. We can write an inequality to illustrate the situation. x≥2 This inequality represents all numbers greater than or equal to 2. Notice that x can equal 2, which we show with a closed circle on a number line.This corresponds to option C. | |

Exercises 46 First of all, we can conclude that the minimum weight for each kind of truck has to be greater than 0, because a weight of zero would mean there isn't a truck at all. This means that the lower boundary, no matter how many axles it has, is: 0<w. Now we can continue and analyze each vehicle separately.2 axles For a vehicle with 2 axles, the maximum total weight including its contents is 40000 lb. This means that we have an upper boundary of: w≤40000. Combining this with the lower boundary, we get the following inequality: 0<w≤40000. To graph this, we place an open dot at 0, a closed dot at 40000, and shade everything in between.3 axles For a vehicle with 3 axles, the maximum total weight including its contents is 60000 lb. This means that: w≤60000. Combining this with the lower boundary, we get the following inequality: 0<w≤60000. To graph this, we place an open dot at 0, a closed dot at 60000, and shade everything in between.4 axles For a vehicle with 4 axles, the maximum total weight including its contents is 80000 lb. This means that: w≤80000. Combining this with the lower boundary, we get the following inequality: 0<w≤80000. To graph this, we place an open dot at 0, a closed dot at 80000, and shade everything in between. | |

Exercises 47 It is given that the length of the world's longest natural arch is 400 feet. This means that the length of any other arch is less than 400 feet. If we let ℓ represent the length of any other arch, we can express this statement as an inequality. ℓ<400 ft Knowing that 1 foot = 12 inches, we can form a conversion factor. We will use it later to convert 400 feet into inches. 1 ft12 in By multiplying 400 ft by this conversion factor, we can convert the full arch length from feet to inches. 400 ft⋅1 ft12 in Simplify a⋅cb=ca⋅b1 ft400 ft⋅12 inCross out common factors1ft400ft⋅12 inSimplify quotient1400⋅12 in1a=a400⋅12 inMultiply 4800 in By replacing 400 feet with 4800 inches, we have our final answer. ℓ<4800 in To graph this inequality, we will place an open circle at 4800 as ℓ cannot equal 4800. Since ℓ is smaller than 4800, we have to shade everything to the left of this circle. | |

Exercises 48 Let's assume that the working week for this student consists of 5 days. Let h be the number of hours the student works each day. We can write an expression for the total number of working hours per week. 5h We also know that a student doesn't work more than 25 hours per week. Therefore, our expression must be less than or equal to 25. 5h≤25 Let's isolate h to determine the inequality that describes how many hours the student can work per day. 5h≤25LHS/5≤RHS/5h≤5 The student can work 5 hours or less per day. | |

Exercises 49 The principal of an elementary school orders that there must be no more than 31 students per class. There are already 23 students registered to study in class 1A. How many more students can be signed up for this class? | |

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##### Other subchapters in Solving Linear Inequalities

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Solving Inequalities Using Addition or Subtraction
- Solving Inequalities Using Multiplication or Division
- Solving Multi-Step Inequalities
- Quiz
- Solving Compound Inequalities
- Solving Absolute Value Inequalities
- Chapter Review
- Chapter Test
- Cumulative Assessment