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###### Exercises

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Exercises 1 Before we draw our graphs, let's take a look at what these compound inequalities really mean.First compound inequality In the first compound inequality, x is being contained between two values. -6≤x≤-4-6≤x AND x≤-4 This graph is shown below.Second compound inequality In the second inequality, the solutions for x are separated into two, distinct inequalities. x≤-6 OR x≥-4 This graph is shown below.Comparison The two graphs are almost complete opposites. They both represent an infinite number of solutions and there is only overlap at two distinct points, -6 and -4. | |

Exercises 2 In order to decide which inequality doesn't belong, let's graph each one and see if that helps us out.As we can see, the first three inequalities are comprised of two distinct solution sets. The fourth inequality is also an "or" compound inequality but the directions of the inequality symbols create overlapping sets and the solution set is actually just all real numbers. Therefore, the fourth inequality, a<6 or a>-9, is the one that doesn't belong. | |

Exercises 3 Let's first look at the shaded part of the number line. When a graph is shaded between two points, it represents an and compound inequality. This is because the value of the variable must be greater than (or greater than or equal to) the lesser point, and less than (or less than or equal to) the greater point.Let's now consider one at a time the endpoints of the shaded region.Lesser Point The graph is shaded to the right of -3, and the circle is open. This means the solutions to the inequality are greater than -3. If we let x be one of those solutions, we can write an inequality to illustrate this situation. x>-3Greater Point The graph is also shaded to the left of 2 and the circle is closed. This tells us that x is less than or equal to 2. x≤2Compound Inequality Notice that the solution set is sandwiched between the two points. This tells us that we have an and compound inequality. Rearranging x>-3 will allow us to visualize this sandwich. x>-3⇔-3<x By combining the two individual inequalities, we will obtain a compound inequality. -3 is less than x and x is less than or equal to 2. First inequality:Second inequality:Compound inequality: -3<x x≤2 -3<x≤2 | |

Exercises 4 The given graph shows open circles at 7 and 14 with shading between the two points. This solution set represents values that are Greater than 7 and Less than 14. The algebraic symbols that represent "greater than" and "less than" are > and <, respectively. If we let x represent a value on the number line, the graph can be expressed with the following compound inequality. x>7andx<14 We can express this "and" compound inequality in another form as well. To help with visualization, let's rewrite x>7 as 7<x. First inequality:7<xSecond inequality:xCompound inequality: 7<x<14<14 | |

Exercises 5 The given graph shows closed circles at -7 and -4 with shading to the left of -7 and to the right of -4. This solution set represents values that are Less than or equal to -7 or Greater than or equal to -4. The algebraic symbols that represent "less than or equal to" and "greater than or equal to" are ≤ and ≥, respectively. If we let x represent a value on the number line, the graph can be expressed with the following compound inequality. x≤-7orx≥-4 | |

Exercises 6 The given graph shows a closed circle at 4 with shading to the left. It also has an open circle at 6 with shading to the right. This solution set represents values that are Less than or equal to 4 or Greater than 6. The algebraic symbols that represent less than or equal to and greater than are ≤ and >, respectively. If we let x represent a value on the number line, the graph can be expressed with the following compound inequality. x≤4orx>6 | |

Exercises 7 To begin, we will decide the symbols that correspond with the given phrases. More than corresponds with the symbol >, while less than corresponds with <. We can now write the inequalities. A number p is less than 6 p<6 and greater than 2and p>2 Let's rewrite the above compound inequality as a single inequality. 2<p<6 To graph the above on a number line, we will place an open dot on both 2 and 6, because p cannot equal these values, and we will shade the region between them. | |

Exercises 8 To begin, we will decide the symbols that correspond with the given phrases. Less than or equal to corresponds with the symbol ≤, while greater than corresponds with >. We can now write the inequalities. n is less than or equal to -7n≤-7 or greater than 12 or n>12 To graph the first inequality, we will place a closed dot at -7, because n can equal -7. Then, we will shade the region to the left as n must be less than or equal to -7. To graph the second inequality, we will place an open dot at 12, because n cannot equal 12. We will shade the region to the right because n is greater than 12. | |

Exercises 9 To begin, we will decide the symbols that correspond with the given phrases. More than corresponds with the symbol >, while at most corresponds with ≤. We can now write the inequalities. A number m is more than -732 or at most -10.m>-732 or m≤-10 To graph the first inequality, we will place an open circle at -732 because m cannot equal -732. Then we will shade the region to the right as m must be more than -732. To graph the second inequality, we will place a closed circle at -10 because m can equal -10. We will shade the region to the left because m is at most -10. | |

Exercises 10 To begin, we will decide the symbols that correspond with the given phrases. No less than corresponds with the symbol ≥, while fewer than corresponds with <. We can now write the inequalities. r is no less than -1.5 r≥-1.5 and fewer than 9.5and r<9.5 When we have an "and" inequality, we are expressing a singular interval. Therefore, we can simplify this compound inequality a little bit and have: -1.5≤r<9.5. To graph these inequalities, we will place a closed circle on -1.5 and an open circle on 9.5. This is because r can equal -1.5 but cannot equal 9.5. We will shade the region between them. | |

Exercises 11 To begin, we will make sense of the given information. The given diagram shows that the mollusks usually are found between -2500 feet and -100 feet. If we let d represent the depth at which mollusks are found, it follows that d must be greater than or equal to -2500 and d must be less than or equal to -100. d≥-2500andd≤-100 To graph these inequalities, we will place a closed circle on both -2500 and -100, because d can equal these values, and we will shade the region between them. | |

Exercises 12 aFor this exercise, we need only focus on the information that pertains to the Low-Elevation Forest. It is given that this region is above 1700 feet to 2500 feet. If we let e represent the elevation, it follows that e is greater than 1700 and e is less than or equal to 2500. Symbolically, we can write e>1700 and e≤2500.bFor this exercise, we will combine the information for the Subalpine and Alpine regions. The Subalpine region covers from above 4000 feet to 6500 feet, while the Apline region covers from 6500 to the peak at 14410 feet. Together, we can see that these regions cover from above 4000 feet to 14410 feet. If we let e represent the elevation, it follows that e is greater than 4000 and e is less than or equal to 14410. Symbolically, we can write this as e>4000ande≤14410. | |

Exercises 13 First, let's split the compound inequality into separate inequalities. Compound Inequality:6<xFirst Inequality:6<xSecond Inequality: x+5≤11+5+5≤11 Notice that compound inequalities written in this way are equivalent to compound inequalities that involves the word "and." 6<x+5andx+5≤11 Let's solve the inequalities separately.First inequality 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, we must flip the inequality sign. 6<x+5LHS−5<RHS−51<xRearrange inequalityx>1 This inequality tells us that all values greater than 1 will satisfy the inequality.Note that the point on 1 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. x+5≤11LHS−5≤RHS−5x≤6 This inequality tells us that all values less than or equal to 6 will satisfy the inequality.Note that the point on 6 is closed because it is included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. To help visualize the algebraic expression, we will write x>1 as 1<x. First solution set:1<xSecond solution set:xIntersecting solution set: 1<x≤6≤6 Finally, we'll graph the solution set to the compound inequality on a number line. | |

Exercises 14 First, let's split the compound inequality into separate inequalities. Compound Inequality:24>-3rFirst Inequality:24>-3rSecond Inequality: -3r≥-9≥-9 Notice that compound inequalities written in this way are equivalent to compound inequalities that involves the word "and." 24>-3rand-3r≥-9 Let's solve the inequalities separately.First inequality 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, we must flip the inequality sign. 24>-3rDivide by -3 and flip inequality sign-8<rRearrange inequalityr>-8 This inequality tells us that all values greater than -8 will satisfy the inequality.Note that the point on -8 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. -3r≥-9Divide by -3 and flip inequality signr≤3 This inequality tells us that all values less than or equal to 3 will satisfy the inequality.Note that the point on 3 is closed because it is included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. To help visualize the algebraic expression, we will write r>-8 as -8<r. First solution set:-8<rSecond solution set:rIntersecting solution set: -8<r≤3≤3 Finally, we'll graph the solution set to the compound inequality on a number line. | |

Exercises 15 If we solve each inequality separately, we will find two solution sets. The union of those sets, the set containing values from one set or from the other, is the solution to the compound inequality.First Inequality We solve inequalities the same way we would solve equations. We just need to keep in mind that multiplying or dividing by a negative number reverses the sign of the inequality. We can start isolating by subtracting 8 from both sides of the inequality. v+8<3LHS−8<RHS−8v<-5 All possible values of v that are less than -5 will satisfy the inequality. Note that -5 is not included in the solution set.Second Inequality In this case, we can isolate v dividing by -8 on both sides of the inequality. -8v<-40Divide by -8 and flip inequality signv>5 The second inequality is satisfied for all values of v greater than 5. Again, note that 5 is not included in the solution set.Comparing Solution Sets Now we must compare our solution sets. From the first inequality, v<-5, we know that the solution sets consist of all numbers to the left of -5 on the number line, without including -5 itself. We will graph this using an open circle at -5.From the second inequality v>5, we know the solution sets include all values to the right of 5, not including 5. We will graph this using an open circle at 5 as well.The union of these solution sets is two intervals, v<-5 or v>5. | |

Exercises 16 If we solve each inequality separately, we will find two solution sets. The union of those sets, the set containing values from one set or from the other, is the solution to the compound inequality.First Inequality We solve inequalities the same way we would solve equations. We just need to keep in mind that multiplying or dividing by a negative number reverses the sign of the inequality. We can start isolating by subtracting 3 from both sides of the inequality. -14>w+3LHS−3>RHS−3-17>wRearrange inequalityw<-17 All possible values of w that are less than -17 will satisfy the inequality. Note that -17 is not included in the solution set.Second Inequality In this case, we can isolate w dividing by 3 on both sides of the inequality. 3w≥-27LHS/3≥RHS/3w≥-9 The second inequality is satisfied for all values of w greater than or equal to -9. In this case, -9 is included in the solution set.Comparing Solution Sets Now we must compare our solution sets. From the first inequality, w<-17, we know that the solution sets consist of all numbers to the left of -17 on the number line, without including -17 itself. We will graph this using an open circle at -17.From the second inequality w≥-9, we know the solution sets include all values to the right of -9, including -9. We will graph this using a closed circle at -9.The union of these solution sets is two intervals, w<-17 or w≥-9. | |

Exercises 17 If we solve each inequality separately, we will find two solution sets. The union of those sets, the set containing values from one set or from the other, is the solution to the compound inequality.First Inequality We solve inequalities the same way we would solve equations. We just need to keep in mind that multiplying or dividing by a negative number reverses the sign of the inequality. We can start isolating by subtracting 3 from both sides of the inequality. 2r+3<7LHS−3<RHS−32r<4LHS/2<RHS/2r<2 All possible values of r that are less than 2 will satisfy the inequality. Note that 2 is not included in the solution set.Second Inequality Same as the previous case, we can start isolating r subtracting 9 from both sides of the inequality. -r+9≤2LHS−9≤RHS−9-r≤-7Divide by -1 and flip inequality signr≥7 The second inequality is satisfied for all values of r greater than or equal to 7. In this case, 7 is included in the solution set.Comparing Solution Sets Now we must compare our solution sets. From the first inequality, r<2, we know that the solution sets consist of all numbers to the left of 2 on the number line, without including 2 itself. We will graph this using an open circle at 2.From the second inequality r≥7, we know the solution sets include all values to the right of 7, including 7. We will graph this using a closed circle at 7.The union of these solution sets is two intervals, r<2 or r≥7. | |

Exercises 18 First, let's split the compound inequality into separate inequalities. Compound Inequality:-6<3nFirst Inequality:-6<3nSecond Inequality:3n+9<21+9+9<21 Notice that compound inequalities written in this way are equivalent to compound inequalities that involves the word "and." -6<3n+9and3n+9<21First inequality 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. -6<3n+9LHS−9<RHS−9-15<3nLHS/3<RHS/3-5<pRearrange inequalityn>-5 This inequality tells us that all values greater than -5 will satisfy the inequality.Note that the point on -5 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. 3n+9<21LHS−9<RHS−93n<12LHS/3<RHS/3n<4 This inequality tells us that all values less than 4 will satisfy the inequality.Again, note that the point on 4 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. To help visualize the algebraic expression, we will write n>-5 as -5<n. First solution set:-5<nSecond solution set:nIntersecting solution set: -5<n<4<4 Finally, we'll graph the solution set to the compound inequality on a number line. | |

Exercises 19 Sometimes, it can be helpful to write a compound inequality as two individual inequalities. This lets us solve each one separately. To begin, let's write the compound inequality. -12<21(4x+16)<18⇕-12<21(4x+16)and21(4x+16)<18 Now we can solve them separately and later compare their solution sets.First inequality We can solve inequalities just like ordinary equations. Notice that, by multiplying both sides of both inequalities by 2, we can clear the fraction of the variable term and then proceed to solve for x. Let's start with the first inequality. -12<21(4x+16)LHS⋅2<RHS⋅2-24<4x+16LHS−16<RHS−16-40<4xLHS/4<RHS/4-10<xRearrange inequalityx>-10 The first inequality is satisfied by all values greater than -10. Note that x cannot equal -10.Second inequality Now we can solve the second inequality. Again, we will start by multiplying by 2 on both sides to eliminate the fraction. 21(4x+16)<18LHS⋅2<RHS⋅24x+16<36LHS−16<RHS−164x<20LHS/4<RHS/4x<5 The second inequality is true for all values less than 5. Notice that x cannot equal 5.Combining solution sets Finally, we can combine the obtained solution sets. The first inequality, x>-10, describes all values greater than -10, but not including -10. We can represent this using an open circle at -10 and shading the values the the right.The second inequality, x<5, describes all values less than 5, not including 5. For this, we will use an open circle again, this time at 5 and shade the values to the left.The intersection of these solution sets, -10<x<5, describes the solutions of the compound inequality. | |

Exercises 20 If we solve each inequality separately, we will find two solution sets. The union of those sets, the set containing values from one set or from the other, is the solution to the compound inequality.First Inequality We solve inequalities the same way we would solve equations. We just need to keep in mind that multiplying or dividing by a negative number reverses the sign of the inequality. We can start isolating by dividing both sides of the inequality by 7. 35<7(2−b)LHS/7=RHS/75<2−bLHS−2<RHS−23<-bMultiply by -1 and flip inequality sign-3>bRearrange inequalityb<-3 All possible values of b that are less than -3 will satisfy the inequality. Note that -3 is not included in the solution set.Second Inequality In this case, we can isolate b multiplying both sides of the inequality by 3. 31(15b−12)≥21LHS⋅3≥RHS⋅315b−12≥63LHS+12≥RHS+1215b≥75LHS/15≥RHS/15b≥5 The second inequality is satisfied for all values of b greater than or equal to 5. In this case, 5 is included in the solution set.Comparing Solution Sets Now we must compare our solution sets. From the first inequality, b<-3, we know that the solution sets consist of all numbers to the left of -3 on the number line, without including -3 itself. We will graph this using an open circle at -3.From the second inequality b≥5, we know the solution sets include all values to the right of 5, including 5. We will graph this using a closed circle at 5.The union of these solution sets is two intervals, b<-3 or b≥5. | |

Exercises 21 To locate the error, we can analyze each step of the work to ensure it is mathematically correct. From the first line to the second, we notice that the student attempted to subtract 3. This was done from the middle and from the right-hand side, but not from the left. This is the error. Inverse operations need to be applied to all sides of the compound inequality. We will solve and graph the inequality correctly. 4<-2x+3<9LHS−3<MHS−3<RHS−31<-2x<6Divide by -2 and flip inequality sign-21>x>-3Rearrange equation-3<x<-21 To graph the solution -3<x<-21 we will place open circles at both -3 and -21, and shade the region between them. | |

Exercises 22 To find the error, we will analyze the work step by step. We can see that the inequalities were solved correctly. That means, the error must be in the graph. We will analyze the solutions and the graph.First solution The solution x>5 means that x can only take values that are larger than 5. The graph for this should show an open circle at 5 and shading to the right, since values on a number line increase as you move to the right. The given graph does not show this.Second solution The solution x<-10 means that x can only take values less than -10. The graph for this should show an open circle at -10 and shading to the left, since values on a number line decrease as you move to the left. The given graph does not show this.Correct graph The error is that the solutions were not graphed correctly. Let's graph the correct solution set! | |

Exercises 23 To begin, we will make sense of the given information. It is given that the temperature inside an iceberg ranges from -20 degrees Celsius to -15 degrees Celsius. If we let t be the temperature, t must be greater than or equal to -20 and t must also be less than or equal to -15. With symbols, we can write this as t≥-20 and t≤-15 To graph these inequalities, we will place a closed circle on both -20 and -15, because t can equal these values, and then shade the region between them. | |

Exercises 24 Let's start be considering the heights of skiers the company does provide skis for. It is recommended that the length of the skis should be 1.16 times the height of the skier. If we let x represent the height of the skier, we can then write an expression for the recommended length of the skis. 1.16xForming inequalities It is also given that the length of the skis offered by the company range from 150 cm to 220 cm. This means that 150 cm is the minimum length and 220 cm is the maximum length offered. Thus, skis with length less than 150 or greater than 220 are not offered. We can write a compound inequality for the lengths that are not provided. 1.16x<150 or 1.16x>220 We will find the heights of skiers for which the company does not offer skis by solving the above compound inequality.Solutions We will solve the inequality by dividing by 1.16 on both sides, and approximating to the nearest hundredth.1.16x<150 or 1.16x>220 x<129.3103… or x>189.6551… x<129.31 or x>189.66 Skiers with heights less than 129.31 cm or greater than 189.66 cm cannot obtain skis from this shop. | |

Exercises 25 First, let's split the compound inequality into separate inequalities. Compound Inequality:First Inequality:Second Inequality:22<-3c+4<1422<-3c+422<-3c+4<14 Notice that compound inequalities written in this way are equivalent to compound inequalities that involves the word "and." 22<-3c+4and-3c+4<14 Let's solve the inequalities separately.First inequality 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, we must flip the inequality sign. 22<-3c+4LHS−4<RHS−418<-3cDivide by -3 and flip inequality sign-6>cRearrange inequalityc<-6 This inequality tells us that all values less than -6 will satisfy the inequality.Note that the point on -6 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. -3c+4<14LHS−4<RHS−4-3c<10Divide by -3 and flip inequality signc>-310 This inequality tells us that all values greater than -310 will satisfy the inequality.Note that the point on -310 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. To help visualize the algebraic expression, we will write c>-310 as -310<c. First solution set:Second solution set:Intersecting solution set:-310<c-310<c<-6 -310<c<-6× Since a number cannot be greater than -310 and less than -6 simultaneously, the compound inequality does not have solution. | |

Exercises 26 If we solve each inequality separately, we will find two solution sets. The union of those sets, the set containing values from one set or from the other, is the solution to the compound inequality.First Inequality We solve inequalities the same way we would solve equations. We just need to keep in mind that multiplying or dividing by a negative number reverses the sign of the inequality. We can start isolating by adding 1 to both sides of the inequality. 2m−1≥5LHS+1≥RHS+12m≥6LHS/2≥RHS/2m≥3 All possible values of m that are greater than or equal to 3 will satisfy the inequality. Note that 3 is included in the solution set.Second Inequality In this case, we can isolate m dividing by 5 on both sides of the inequality. 5m>-25LHS/5>RHS/5m>-5 The second inequality is satisfied for all values of m greater than -5. In this case, -5 is not included in the solution set.Comparing Solution Sets Now we must compare our solution sets. From the first inequality, m≥3, we know that the solution sets consist of all numbers to the right of 3 on the number line, including 3 itself. We will graph this using a closed circle at 3.From the second inequality m>-5, we know the solution sets include all values to the right of -5, without including -5. We will graph this using an open circle at -5.The union of these solution sets is m>-5. | |

Exercises 27 To solve the compound inequality, we have to solve each of the inequalities separately. The intersection of these solution sets is the solution set for the compound inequality.First inequality 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. -y+3≤8LHS−3≤RHS−3-y≤5Multiply by -1 and flip inequality signy≥-5 This inequality tells us that all values greater than or equal to -5 will satisfy the inequality.Note that the point on -5 is closed because it is included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. y+2>9LHS−2>RHS−2y>7 The second inequality is satisfied for all values of y that are greater than 7.Note that the point on 7 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets, which is y>7. Finally, we'll graph the solution set to the compound inequality. | |

Exercises 28 If we solve each inequality separately, we will find two solution sets. Since the compound inequality includes the word "or," the union of those sets, is the solution to the compound inequality.First Inequality We solve inequalities the same way we would solve equations. We just need to keep in mind that multiplying or dividing by a negative number reverses the sign of the inequality. We can start isolating by adding 8 to both sides of the inequality. x−8≤4LHS+8≤RHS+8x≤12 All possible values of x that are less than or equal to 12 will satisfy the inequality. Note that 12 is included in the solution set.Second Inequality In this case, we can isolate x subtracting 3 from both sides of the inequality. 2x+3>9LHS−3>RHS−32x>6LHS/2>RHS/2x>3 The second inequality is satisfied for all values of x greater than 3. In this case, 3 is not included in the solution set.Comparing Solution Sets Now we must compare our solution sets. From the first inequality, x≤12, we know that the solution set consists of all numbers to the left of 12 on the number line, including 12 itself. We will graph this using a closed circle at 12.From the second inequality, x>3, we know the solution set includes all values to the right of 3, without including 3. We will graph this using an open circle at 3.The union of these solution sets is all real numbers | |

Exercises 29 If we solve each inequality separately, we will find two solution sets. The union of those sets, the set containing values from one set or from the other, is the solution to the compound inequality.First Inequality We solve inequalities the same way we would solve equations. We just need to keep in mind that multiplying or dividing by a negative number reverses the sign of the inequality. We can start isolating by subtracting n from both sides of the inequality. 2n+19≤10+nLHS−n≤RHS−nn+19≤10LHS−19≤RHS−19n≤-9 All possible values of n that are less than pr equal to -9 will satisfy the inequality. Note that -9 is included in the solution set.Second Inequality In this case, we can isolate n adding 2n to both sides of the inequality. -3n+3<-2n+33LHS+2n<RHS+2n-n+3<33LHS−3<RHS−3-n<30Multiply by -1 and flip inequality signn>-30 The second inequality is satisfied for all values of n greater than -30. In this case, -30 is not included in the solution set.Comparing Solution Sets Now we must compare our solution sets. From the first inequality, n≤-9, we know that the solution sets consist of all numbers to the left of -9 on the number line, including -9 itself. We will graph this using a closed circle at -9.From the second inequality, n>-30, we know the solution sets include all values to the right of -30, without including -30. We will graph this using an open circle at -30.The union of these solution sets is all real numbers. | |

Exercises 30 To solve the compound inequality, we have to solve each of the inequalities separately. The intersection of these solution sets is the solution set for the compound inequality.First inequality 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−18<4x−23LHS+23<RHS+233x+5<4xLHS−3x<RHS−3x5<xRearrange inequalityx>5 This inequality tells us that all values greater than 5 will satisfy the inequality.Note that the point on 5 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. x−16<-22LHS+16<RHS+16x<-6 The second inequality is satisfied for all values of x that are less than -6.Note that the point on -6 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. Since there is not intersection of the solution sets, a number cannot be both less than -6 and greater than 5, there is no solution for the compound inequality. | |

Exercises 31 For this exercise, we will solve both inequalities and then compare their solutions. Then, we will determine which symbol the first inequality must have to ensure that the solution set of the compound inequality only has one value. First Inequality: Second Inequality: 4(x−6) ≤ 2(x−10)5(x+2)≥2(x+8) We will begin by solving the second inequality. 5(x+2)≥2(x+8) Solve for x Distribute 55x+10≥2(x+8)Distribute 25x+10≥2x+16LHS−2x≥RHS−2x3x+10≥16LHS−10≥RHS−103x≥6LHS/3≥RHS/3 x≥2 The solution to the second inequality is x≥2. We will now solve the first inequality. Notice that we ignore what's in the box for the moment being. 4(x−6) ≤ 2(x−10) Solve for x Distribute 44x−24 ≤ 2(x−10)Distribute 24x−24 ≤ 2x−20LHS−2x ≤ RHS−2x2x−24 ≤ -20LHS+24 ≤ RHS+242x ≤ 4LHS/2 ≤ RHS/2 x ≤ 2 Solving the first inequality gives x ≤ 2. Recall that we have already calculated the solution to the second inequality. x≥2 This means the second inequality is true only when x takes values 2 or larger. We will use the table below to show the number of solutions the compound inequality would have based on the different symbols for the first inequality.x ≤ 2Number of solutions x<2None x>2Infinitely many x≤2One x≥2Infinitely many If the solution to the first inequality is x≤2, and the solution to the second one is x≥2, the compound inequality has exactly one solution, x=2. | |

Exercises 32 First, we should acknowledge that there are infinitely many possible stories that can be represented by the given graph. What's provided here is just one possibility. A runner is training for a half-marathon. Her weekly training plan involves runs of different lengths, ranging from 6 miles to 10 miles. If we let x represent the number of miles she runs on any given day then x≥6 and x≤10 describes her daily runs in number of miles. To graph this compound inequality we place closed circles at 6 and 10, because x can equal those values, and we shade the region between. | |

Exercises 33 To begin writing the inequalities, we can choose any two sides to add together. We will arbitrarily choose to add 7 and x for the first inequality. The sum of these should be larger than the third side which we can write as 7+x>5. We will write the other two inequalities in a similar way. The table below shows all three inequalities.#Inequality 17+x>5 2x+5>7 35+7>x To determine if it's possible for x to equal 1, we solve all three inequalities.#InequalitySolution 17+x>5x>-2 2x+5>7x>2 35+7>xx<12 The table above shows that x must be greater than -2, x must be greater than 2, and x must be less than 12. All of these apply at the same time which means that the solution from Inequality 2 overrides the solution to Inequality 1. Thus, the possible range of values for x is between 2 and 12, exclusive: 2<x<12. Hence, it is not possible for x to equal 1. Our friend is incorrect. | |

Exercises 34 aIn 2012, we locate the greatest profit at 90 million and, in 2009, we have the least profit at 50 million. This means that between 2006 and 2016 profit ranged from 50 million to 90 million. If we let x represent the profit for any year, then x≥50 and x≤90 describes the range of the annual profits between 2006 and 2013. This can also be combined and rewritten as one inequality: 50≤x≤90.bWe know that the minimum and maximum profits P between 2006 to 2013 were 50 and 90 million which we can write as the compound inequality: 50≤P≤90. If we replace P with the expression R−C we get: 65≤R−C≤90. It is given that C=125. Substituting this value in the compound inequality and solving for R will give us the range of revenue from 2006 to 2013. 50≤R−C≤90C=12550≤R−125≤90LHS+125≤MHS+125≤RHS+125175≤R≤215 The solution 175≤R≤215 means that the revenue from 2006 to 2013 ranges from 175 million to 215 million. Thus, a revenue of 160 million is not possible. | |

Exercises 35 An absolute value is always a non-negative number because it measures the expression's distance from zero on a number line. In this exercise, the equation ∣∣∣∣9d∣∣∣∣=6 means that the value of the expression 9d is 6 steps from zero, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣∣∣∣9d∣∣∣∣=6⇒9d=6 and 9d=-6 We need to solve both of these cases so that we can find both values that satisfy the equation.First Equation Let's start with the first equation. 9d=6Multiply by 9d=54 The solution for the first equation is the point d=54.Second Equation Next, we will solve the second equation. 9d=-6Multiply by 9d=-54 The point d=-54 is the solution for the second equation. Now, let's graph the solutions.Graph Both d=54 and d=-54 are solutions to the given equation. When shown on a number line graph, we have the following. | |

Exercises 36 An absolute value represents distance and, thereby, is always non-negative. ∣x∣=x and ∣-x∣=x No matter what value you substitute for p, the equation will never equal -21. 7∣5p−7∣≠-21 Hence, there are no solutions to this equation. | |

Exercises 37 When the absolute values of two expressions are equal, either the expressions are equal or the opposite of the expressions are equal. For example, if we have the equation, ∣ax+b∣=∣cx+d∣ then we can have either ax+b=cx+dorax+b=-(cx+d). To solve the given absolute value equation, we will write two equations like above when we remove the absolute value.Solving the Equations Let's remove the absolute value and solve the equations. ∣r+2∣=∣3r−4∣r+2≥0: r+2=(3r−4)r+2<0: r+2=-(3r−4)(I)(II)r1+2=3r1−4r2+2=-(3r2−4)(I)(II) Solve for r1 and r2 (II): Remove parentheses and change signsr1+2=3r1−4r2+2=-3r2+4(I), (II): LHS−2=RHS−2r1=3r1−6r2=-3r2+2(I): LHS−3r1=RHS−3r1-2r1=-6r2=-3r2+2(I): LHS/(-2)=RHS/(-2)r1=3r2=-3r+2(II): LHS+3r2=RHS+3r2r1=34r2=2(II): LHS/4=RHS/4r1=3r2=42(II): Calculate quotient r1=3r2=0.5 After solving an absolute value equation, it is necessary to check for extraneous solutions.Checking the Solutions In order to check extraneous solutions, we substitute the found solutions into the given equation and determine if a true statement is made. ∣r+2∣=∣3r−4∣r=3∣3+2∣=?∣3⋅3−4∣ Simplify equation Multiply∣3+2∣=?∣9−4∣Add terms∣5∣=?∣5∣∣5∣=5 5=5 r=3 is not extraneous. ∣r+2∣=∣3r−4∣r=0.5∣0.5+2∣=?∣3⋅0.5−4∣ Simplify equation Multiply∣0.5+2∣=?∣1.5−4∣Add and subtract terms∣2.5∣=?∣-2.5∣∣2.55∣=2.552.5=?∣-2.5∣∣-2.5∣=2.5 2.5=2.5 r=0.5 is also not extraneous.Graph Solutions for the given equation are the points which are r=3 and r=0.5. In order to graph the solutions, we plot them on the number line. | |

Exercises 38 When the absolute values of two expressions are equal, either the expressions are equal or the opposite of the expressions are equal. For example, if we have the equation, ∣ax+b∣=∣cx+d∣ then we can have either ax+b=cx+dorax+b=-(cx+d). To solve the given absolute value equation, we will write two equations like above when we remove the absolute value.Positive Case Let's start with positive case. 21w−6=w+7LHS+6=RHS+621w=w+13LHS−w=RHS−w-21w=13LHS⋅(-2)=RHS⋅(-2)w=-26 The solution for the positive case is the point w=-26.Negative Case Next, we will solve the absolute value equation for the negative case. 21w−6=-(w+7)Distribute -121w−6=-w−7LHS+6=RHS+621w=-w−1LHS+w=RHS+w23w=-1LHS⋅32=RHS⋅32w=-32 The point w=-32 is the solution for the negative case.Checking the Solutions After solving an absolute value equation, it is necessary to check for extraneous solutions. In order to check extraneous solutions, we substitute the found solutions into the given equation and determine if a true statement is made. ∣∣∣∣21w−6∣∣∣∣=∣w+7∣w=-26∣∣∣∣21⋅(-26)−6∣∣∣∣=?∣-26+7∣ Simplify equation a(-b)=-a⋅b∣-13−6∣=?∣-26+7∣Add and subtract terms∣-19∣=?∣-19∣∣-19∣=19 19=19 w=-26 is not extraneous. ∣∣∣∣21w−6∣∣∣∣=∣w+7∣w=-32∣∣∣∣21⋅(-32)−6∣∣∣∣=?∣∣∣∣-32+7∣∣∣∣ Simplify equation Multiply fractions∣∣∣∣-62−6∣∣∣∣=?∣∣∣∣-32+7∣∣∣∣ba=b/2a/2∣∣∣∣-31−6∣∣∣∣=?∣∣∣∣-32+7∣∣∣∣a=33⋅a∣∣∣∣-31−318∣∣∣∣=?∣∣∣∣-32+321∣∣∣∣Add and subtract terms∣∣∣∣-319∣∣∣∣=?∣∣∣∣319∣∣∣∣∣∣∣∣-319∣∣∣∣=319319=?∣∣∣∣319∣∣∣∣∣∣∣∣319∣∣∣∣=319 319=319 w=-32 is also not extraneous.Graph Both w=-26 and w=-32 are the solutions for the absolute value equation. We will graph the solutions by plotting them on the number line. | |

Exercises 39 We have been asked to find and interpret the mean absolute deviation. This requires us to find the mean, the absolute deviations, and then find the mean of those deviations. Let's start by calculating the mean, the sum of all of the data divided by the number of data points. Sum of the Data: Number of Data Points: Mean: 70101070=7 Next we need to find all of the absolute deviations, which are the distances between each data point and the mean value.Data Point, x∣7−x∣Absolute Deviation 1∣7−1∣6 1∣7−1∣6 2∣7−2∣5 5∣7−5∣2 6∣7−6∣1 8∣7−8∣1 10∣7−10∣3 12∣7−12∣5 12∣7−12∣5 13∣7−13∣6 Now that we know all of the absolute deviations, we need to add them all together. Then, we can find the average deviation by dividing the sum of the deviations by the number of deviations. Sum of the Deviations: Number of Deviations: Mean: 40101040=4 Therefore, the mean absolute deviation is 4. Such a low value implies that the data is clustered close together. | |

Exercises 40 We have been asked to find and interpret the mean absolute deviation. This requires us to find the mean, the absolute deviations, and then find the mean of those deviations. Let's start by calculating the mean, the sum of all of the data divided by the number of data points. Sum of the Data: Number of Data Points: Mean: 3001010300=30 Next we need to find all of the absolute deviations, which are the distances between each data point and the mean value.Data Point, x∣30−x∣Absolute Deviation 24∣30−24∣6 26∣30−26∣4 28∣30−28∣2 28∣30−28∣2 30∣30−30∣0 30∣30−30∣0 32∣30−32∣2 32∣30−32∣2 34∣30−34∣4 36∣30−36∣6 Now that we know all of the absolute deviations, we need to add them all together. Then, we can find the average deviation by dividing the sum of the deviations by the number of deviations. Sum of the Deviations: Number of Deviations: Mean: 28101028=2.8 Therefore, the mean absolute deviation is 2.8. Such a low value implies that the data is clustered close together. |

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##### 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 Absolute Value Inequalities
- Chapter Review
- Chapter Test
- Cumulative Assessment