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###### Exercises

Exercise name | Free? |
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Exercises 1 We can write the given verbal phrase, "a number d minus 2," as the following algebraic expression. A number d minus 2 d − 2 Now, we will determine which of the symbols, <,>,≤, or ≥ correspond with the statement. Less than is symbolically represented by <, which is called a strict inequality. Let's write the complete inequality. d−2<-1 | |

Exercises 2 To begin, we will determine which inequality symbol corresponds with the given statement. "At least" can also be expressed as "greater than or equal to," or symbolically represented by ≥. Since 10 comes before this statement, we can write it on the left-hand side of our inequality. 10≥… Since "the product of" indicates multiplication, we can write "the product of a number h and 5" as the following algebraic statement. the product of a number h and 5h × 5 Writing the complete inequality, we have the following. 10≥h×5 Whenever there is a multiplication of a variable and number, the number can be rewritten as a coefficient for the variable. This gives us the following inequality as our final answer. 10≥5h | |

Exercises 3 The inequality x>4 describes all the values of x that are to the right of 4 on a number line. Notice that x=4 is not a solution, so we mark this as an open circle on the number line at 4. | |

Exercises 4 The inequality y≤2 describes all the values of y that are to the left of 2 on a number line. Notice that y=2 is a solution, so we mark this as a closed circle on the number line at 2. | |

Exercises 5 The inequality -1≥z describes all the values of z that are to the left of -1 on a number line. Notice that z=-1 is a solution, so we mark this as a closed circle on the number line at -1. | |

Exercises 6 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. p+4<10LHS−4<RHS−4p<6 This inequality tells us that all values less than 6 will satisfy the inequality. Notice that p cannot equal 6, which we show with an open circle on the number line. | |

Exercises 7 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. r−4<-6LHS+4<RHS+4r<-2 This inequality tells us that all values less than -2 will satisfy the inequality. Notice that r cannot equal -2, which we show with an open circle on the number line. | |

Exercises 8 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 2.1≥m−6.7LHS+6.7≥RHS+6.78.8≥mRearrange inequalitym≤8.8 This inequality tells us that all values less than or equal to 8.8 will satisfy the inequality. Notice that m can be equal 8.8, which we show with a closed circle on the number line. | |

Exercises 9 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that, when you divide or multiply by a negative number, you must reverse the inequality sign. 3x>-21LHS/3>RHS/3x>-7 This inequality tells us that all values greater than -7 will satisfy the inequality. Notice that x cannot equal -7, which we show with an open circle on the number line. | |

Exercises 10 Inequalities can be solved in the same way as equations. By multiplying both sides by 5, we can isolate g. -4≤5gLHS⋅5≤RHS⋅5-4⋅5≤5g⋅55a⋅5=a-4⋅5≤gMultiply-20≤gRearrange inequalityg≥-20 This inequality tells us that all values greater than or equal to -20 will satisfy the inequality. Notice that g can equal -20, which we show with a closed circle on the number line. | |

Exercises 11 Inequalities can be solved in the same way as equations. By multiplying both sides by -34, we can isolate n. Because we are multiplying the inequality by a negative number, we have to reverse the inequality sign. -43n≤3Multiply by (-34) and flip inequality sign-34(-43)n≥-34⋅3ba⋅ab=1n≥-34⋅33a⋅3=an≥-4 This inequality tells us that all values greater than or equal to -4 will satisfy the inequality. Notice that n can equal -4 which we show with a closed circle on the number line. | |

Exercises 12 Inequalities can be solved in the same way as equations. By dividing both sides by -8, we can isolate s. Because we are dividing the inequality by a negative number, we have to reverse the inequality sign. -8s≥11Divide by -8 and flip inequality signs≤-88 This inequality tells us that all values less than or equal to -88 will satisfy the inequality. Notice that s can equal -88, which we show with a closed circle on the number line. | |

Exercises 13 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that, when you divide or multiply by a negative number, you must reverse the inequality sign. 36<2qLHS/2<RHS/218<qRearrange inequalityq>18 This inequality tells us that all values greater than 18 will satisfy the inequality. Notice that q cannot equal 18, which we show with an open circle on the number line. | |

Exercises 14 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that, when you divide or multiply by a negative number, you must reverse the inequality sign. -1.2k>6Divide by -1.2 and flip inequality signk<-5 This inequality tells us that all values less than -5 will satisfy the inequality. Notice that k cannot equal -5, which we show with an open circle on the number line. | |

Exercises 15 Remember, we can solve an inequality in the same way as an equation, by performing operations on both sides of the inequality to isolate the variable. Let's isolate x to solve the inequality. 3x−4>11LHS+4>RHS+43x>15LHS/3>RHS/3x>5 This inequality tells us that all values greater than 5 will satisfy the inequality. Notice that x cannot equal 5, which we show with an open circle on the number line. | |

Exercises 16 To solve the inequality, we should use inverse operations to isolate b. -4<2b+9LHS−9<RHS−9-13<2bLHS⋅2<RHS⋅2-26<2b⋅22a⋅2=a-26<bRearrange inequalityb>-26 This inequality tells us that all values greater than -26 will satisfy the inequality. Notice that b cannot equal -26 which we will show with an open circle on the number line. | |

Exercises 17 To solve the inequality, we should use inverse operations to isolate n. 7−3n≤n+3LHS+3n≤RHS+3n7≤4n+3LHS−3≤RHS−34≤4nLHS/4≤RHS/41≤nRearrange inequalityn≥1 This inequality tells us that all values greater than and equal to 1 will satisfy the inequality. Notice that n can equal 1 which we will show with a closed circle on the number line. | |

Exercises 18 To solve the inequality, we should use inverse operations to isolate s. We start with distributing 2 to each term inside the parentheses. 2(-4s+2)≥-5s−10Distribute 2-4s(2)+2(2)≥-5s−10Multiply-8s+4≥-5s−10LHS+8s≥RHS+8s4≥3s−10LHS+10≥RHS+1014≥3sLHS/3≥RHS/3314≥sRearrange inequalitys≤314 This inequality tells us that all values less than or equal to 314 will satisfy the inequality. Notice that s can equal 314 which we will show with a closed circle on the number line. | |

Exercises 19 To solve the inequality, we should use inverse operations to isolate t. We start with distributing 6 to each term inside the parentheses. 6(2t+9)≤12t−1Distribute 62t(6)+9(6)≤12t−1Multiply12t+54≤12t−1LHS−12t≤RHS−12t54≤-1 This is never true for any value of t because 54 will always be greater than -1. Therefore, there is no solution to this inequality. | |

Exercises 20 Let's solve the inequality by isolating r on one side. Start with distributing 3 to each term inside the parentheses. 3r−8>3(r−6)Distribute 33r−8>r(3)−6(3)Multiply3r−8>3r−18LHS−3r=RHS−3r-8>-18 This is always true for any value of r because -8 will always be greater than -18. Therefore, the solution is all real numbers. | |

Exercises 21 Let's start by writing the sentence as an inequality. More than is the same as greater than. Therefore, x is greater than -6. x>-6or-6<x The phrase at most is the same as saying not greater than or, equivalently, less than or equal to. This means we are saying x is less than or equal to 8. x≤8 Note that the exercise uses the word and between the two inequalities. This means that we can combine the two inequalities we wrote above. First Inequality: Second Inequality: Compound Inequality: -6<xx-6<x≤8≤8 To graph, we can focus on the two parts. For x>-6, we need an open circle on -6 to show that it is not included as a solution and then we shade to the right to cover the values greater than -6. For x≤8, we need a closed circle on 8 because the solution includes 8. We then need to shade to the left to cover the values that are less than 8. | |

Exercises 22 To make solving a little bit easier, we can separate the compound inequality into two cases. First case: Second case: 19≥3z+13z+1≥-5First case Let's isolate z in the first case so that we may easily graph it in the solution set. 19≥3z+1 Solve for z LHS−1≥RHS−118≥3zLHS/3≥RHS/36≥zRearrange inequality z≤6 Let's graph the solution set.Second case Now, let's isolate z in the second case so that we may easily graph it in the solution set as well. 3z+1≥-5 Solve for z LHS−1≥RHS−13z≥-6LHS/3≥RHS/3 z≥-2 Let's graph the this solution set.Recombining the solution sets Now we can graph the two cases on the same number line. Remember, we only care about the intersection of these two inequalities. The compound inequality is an "and" inequality and contains all possible values for z within one interval.The purple overlap of the two intervals, -2≤z≤6, is our solution set. | |

Exercises 23 To make solving a little bit easier, we can separate the compound inequality into two cases. First case: Second case: 4r<-5-2r−7≤3First case Let's isolate r in the first case so that we may easily graph it in the solution set. 4r<-5LHS⋅4<RHS⋅4r<-20 Let's graph the solution set.Second case Now, let's isolate r in the second case so that we may easily graph it in the solution set as well. -2r−7≤3 Solve for r LHS+7≤RHS+7-2r≤10Divide by -2 and flip inequality sign r≥-5 Let's graph the solution set.Recombining the solution sets The solution to the compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:r<-20r<-20 or r≥-5r<-20 or r≥-5 Finally, we'll graph the solution set to the compound inequality. | |

Exercises 24 We are asked to find and graph the solution set for all possible values of m in the given inequality. ∣m∣≥10 To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number greater than or equal to 10 or any number less than or equal to -10. m≤-10andm≥10 The solution to this type of compound inequality is the union of the solution sets. Let's recombine our cases back into one compound inequality. First solution set:Second solution set:Intersecting solution set:m≤-10m≥10m≤-10 or m≥10 Since m is already isolated, we can start to graph the solution set. The graph of this inequality includes all values less than -10 or greater than 10, inclusive. We show this by using closed circles on the endpoints. | |

Exercises 25 All absolute value expressions are greater than or equal to zero. It follows then that no absolute value expression can be less than or equal to a negative number. ∣k−9∣≥0⇒∣k−9∣≰-4 This means there is no value of k that will satisfy the inequality. Thus, there is no solution. | |

Exercises 26 We are asked to find and graph the solution set for all possible values of f in the given inequality. To do this, let's isolate the absolute value expression first. 4∣f−6∣≤12LHS/4≤RHS/4∣f−6∣≤3 Now, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number less than or equal to 3 away from the midpoint in the positive direction and any number less than or equal to 3 away from the midpoint in the negative direction. Absolute value inequality: Compound inequality: ∣f−6∣≤3 -3≤ f−6 ≤3 This compound inequality means that the distance f−6 is greater than or equal to -3 and less than or equal to 3. f−6≥-3andf−6≤3 Let's isolate f in both of these cases before graphing the solution set.Case 1 For f−6≤3: f−6≤3LHS+6≤RHS+6f≤9 This inequality tells us that all values less than or equal to 9 will satisfy the inequality.Case 2 For f−6≥-3: f−6≥-3LHS+6≥RHS+6f≥3 This inequality tells us that all values greater than or equal to 3 will satisfy the inequality.Solution Set The solution to this type of compound inequality is the overlap of the solution sets. Let's recombine our cases back into one compound inequality. First solution set:Second solution set:Intersecting solution set:f≤9 3≤f 3≤f≤9Graph The graph of this inequality includes all values from 3 to 9, inclusive. We show this by using closed circles on the endpoints. | |

Exercises 27 We are asked to find and graph the solution set for all possible values of b in the given inequality. 5∣b+8∣−7>13 To do this, let's isolate absolute value expression first. 5∣b+8∣−7>13LHS+7>RHS+75∣b+8∣>20LHS/5>RHS/5∣b+8∣>4 Now, we can create a compound inequality by removing the absolute value. In this case, the solution set is any number with a distance greater than 4 or less than -4. b+8 >4 or b+8<-4 Let's isolate b in both of these cases before graphing the solution set.Case 1 b+8>4LHS−8>RHS−8b>-4 This inequality tells us that all values greater than -4 will satisfy the inequality.Case 2 b+8<-4LHS−8<RHS−8b<-12 This inequality tells us that all values less than -12 will satisfy the inequality.Solution Set The solution to this type of compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:b>-4 b<-12 b<-12 or b>-4Graph The graph of this inequality includes all values less than -12 or greater than -4. We show this by keeping the endpoints open. | |

Exercises 28 We are asked to find and graph the solution set for all possible values of g in the given inequality. We will start by subtracting 1 on both sides to isolate the absolute value expression, ∣-3g−2∣+1<6⇔∣-3g−2∣<5 Now we will create a compound inequality by removing the absolute value. The less than symbol <, creates an "and" inequality because absolute value represents a distance and we need our distance from the center to be less than 5 away. -3g−2 >-5 and -3g−2<5 Let's isolate g in both of these cases before graphing the solution set.Case 1 Remember whenever we multiply or divide an inequality by a negative number, we reverse the direction of the inequality symbol. -3g−2>-5LHS+2>RHS+2-3g>-3Divide by -3 and flip inequality signg<1 This statement tells us that all values less than 1 will satisfy the inequality.Case 2 Again, remember that whenever we multiply or divide an inequality by a negative number, we reverse the direction of the inequality symbol. -3g−2<5LHS+2<RHS+2-3g<7Divide by -3 and flip inequality signg>-37 Write as Mixed Number Rewrite 7 as 6+1g>-36+1Write as a sum of fractionsg>-(36+31)Calculate quotientg>-(2+31)Add termsg>-(231)Remove parentheses g>-231 This statement tells us that all values greater than -231 will satisfy the inequality.Solution Set The solution to this type of compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set: g<1 g<1 and g>-231 g<1 and g>-231Graph The graph of this inequality includes all values which are less than 1 and greater than -231. As both are strict inequalities, we will use open circles. | |

Exercises 29 All absolute value expressions are greater than or equal to zero. It follows then that any absolute value expression is greater than all negative numbers. ∣9−2j∣+10≥2⇒∣9−2j∣≥-8 This means any value of j will satisfy the inequality. Thus, there are infinitely many solutions, and the solution set is the set of all real numbers. The graph of this inequality is a line containing all values on the number line. | |

Exercises 30 The required height of the guardrail is 106 cm. This is the midpoint of the range. We are asked to find the acceptable heights of the guardrail which is no more than 7 cm from the required height. This will be the maximum distance from the midpoint. Let's name the acceptable height as h and write the absolute value inequality. ∣h−106∣≤7 Now, because we know that our distance from the midpoint must be less than or equal to the given value, we can write an "and" compound inequality. -7≤h−106≤7 Finally, we can isolate h by adding 106 to all sides of the inequality. 99≤h≤113 The acceptable widths are between 99 cm and 113 cm. |

##### Other subchapters in Solving Linear Inequalities

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Writing and Graphing Inequalities
- 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 Test
- Cumulative Assessment