Solving Absolute Value Inequalities

Download for free
Find the solutions in the app
Android iOS
Exercises marked with requires Mathleaks premium to view it's solution in the app. Download Mathleaks app on Google Play or iTunes AppStore.
Sections
Communicate Your Answer
Exercise name Free?
Communicate Your Answer 4
Communicate Your Answer 5
Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Monitoring Progress 6
Monitoring Progress 7
Exercises
Exercise name Free?
Exercises 1 The important thing to remember in this situation is that the absolute value can never be negative. The most basic rule of absolute values can be shown as: ∣a∣=a and ∣-a∣=a. For this exercise, we can ignore everything inside the absolute value symbols and just look at the fact that we have the absolute value of something is greater than or equal to a negative number. ∣…∣≥-6 No matter what value of x we substitute into the expression 4x−2, whether it gives a positive or negative result, the absolute value will make it where the left-hand side is always a positive number. And a positive number will always be greater than a negative number. Hence, the solution is that the inequality is true for all real numbers.
Exercises 2 Before we can solve either of these inequalities, we need to remember that absolute values tell us the distance away from something. In an inequality, absolute values tell us if the distance is greater than or less than a given number. Let's look at how we solve each one and then compare.First inequality Let's think about what this inequality is telling us. We have: ∣w−9∣≤2 This can be translated to mean that: The distance from w−9 must be less than or equal to 2 units away. If we want the distance away from something to be less than some value, it means we want it to stay close to the center. We want to contain our solutions in one central set. This is thus an "and" compound inequality which we can write as -2≤w−9≤2 We can solve this by adding 9 to all three parts of the compound inequality giving us: 7≤w≤11Second inequality Once again, let's think about what the inequality is telling us. We have: ∣w−9∣≥2 This can be translated to mean that: The distance from w−9 must be greater than or equal to 2 units away. If we want the distance away from something to be greater than some value, it means we want it to be away from the center. We want to separate our solutions into two disjointed sets. This is then an "or" compound inequality: -2≥w−9 or 2≤w−9 We solve each of these inequalities separately and get: 7≥w or 11≤wThe difference When solving absolute value inequalities, you really need to think about the meaning of the absolute value in terms of distance. An inequality where the distance should be greater than the given value should be an "or" compound inequality. An inequality where the distance should be less than the given value should be an "and" compound inequality.
Exercises 3 We are asked to find and graph the solution set for all possible values of x in the given inequality. ∣x∣<3 To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number less than 3 away from the midpoint in the positive direction and any number less than 3 away from the midpoint in the negative direction. Absolute value inequality:   ∣x∣Compound inequality:   -3< x ​<3<3​ This compound inequality means that the distance x is greater than -3 and less than 3. x>-3andx<3 Since x is already isolated, we can graph the compound inequality. The graph of this inequality includes all values from -3 to 3, not inclusive. We show this by using open circles on the endpoints.
Exercises 4 We are asked to find and graph the solution set for all possible values of y in the given inequality. ∣y∣≥4.5 To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number with a distance greater than or equal to 4.5 away from the midpoint in the positive direction or a distance greater than or equal to 4.5 away from the midpoint in the negative direction. Combined solution set: y ≥4.5 or y≤-4.5​ Since y is already isolated, we can graph the solution set. The graph of this inequality includes all values less than or equal to -4.5 or greater than or equal to 4.5. We show this by keeping the the endpoints closed.
Exercises 5 We are asked to find and graph the solution set for all possible values of d in the given inequality. ∣d+9∣>3 To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number with a distance greater than 3 away from the midpoint in the positive direction or a distance greater than 3 away from the midpoint in the negative direction. d+9 >3 or d+9<-3​ Let's isolate d in both of these cases before graphing the solution set.Case 1 d+9>3LHS−9>RHS−9d>-6 This inequality tells us that all values greater than -6 will satisfy the inequality.Case 2 d+9<-3LHS−9<RHS−9d<-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:​d<-12 or  d>-6 d<-12 d<-12 or d>-6​Graph The graph of this inequality includes all values less than -12 or greater than -6. We show this by open circles on the endpoints closed.
Exercises 6 We are asked to find and graph the solution set for all possible values of h in the given inequality. ∣h−5∣≤10 To do this, 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 10 away from the midpoint in the positive direction and any number less than or equal to 10 away from the midpoint in the negative direction. Absolute value: Compound inequality: ​-10≤∣h−5∣≤10-10≤h−5≤10​ This compound inequality means that the distance from h−5 is greater than or equal to -10 and less than or equal to 10. h−5≥-10andh−5≤10 Let's isolate h in both of these cases before graphing the solution set.Case 1 h−5≤10LHS+5≤RHS+5h≤15 This inequality tells us that all values less than or equal to 15 will satisfy the inequality.Case 2 -10≤h−5LHS+5≤RHS+5-5≤h This inequality tells us that all values greater than or equal to -5 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: -5≤Intersecting solution set: -5≤​h≤15hh≤15​Graph The graph of this inequality includes all values from -5 to 15, inclusive. We show this by using closed circles on the endpoints.
Exercises 7 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. ∣2s−7∣≥0⇒∣2s−7∣>-1 This means any value of s will satisfy the inequality. Thus, there are infinitely many solutions, and the solution set is the set of all real numbers.
Exercises 8 We are asked to find and graph the solution set for all possible values of c in the given inequality. ∣4c+5∣>7 To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number with a distance greater than 7 away from the midpoint in the positive direction or a distance greater than 7 away from the midpoint in the negative direction. 4c+5>7 or 4c+5<-7​ Let's isolate c in both of these cases before graphing the solution set.Case 1 4c+5>7LHS−5>RHS−54c>12LHS/4>RHS/4c>3 This inequality tells us that all values greater than 3 will satisfy the inequality.Case 2 4c+5<-7LHS−5<RHS−54c<-12LHS/4<RHS/4c<-3 This inequality tells us that all values less than -3 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:​c<-3 or  c>3 c<-3 c<-3 or c>3​Graph The graph of this inequality includes all values less than -3 or greater than 3. We show this by open circles on the endpoints.
Exercises 9 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. ∣5p+2∣≥0⇒∣5p+2∣≮-4 This means there is no value of p that will satisfy the inequality. Thus, there is no solution.
Exercises 10 We are asked to find and graph the solution set for all possible values of n in the given inequality. ∣9−4n∣<5 To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number less than 5 away from the midpoint in the positive direction and any number less than 5 away from the midpoint in the negative direction. Absolute value: Compound inequality: ​  ∣9−4n∣<5  -5< 9−4n <5​ This compound inequality means that the distance from 9−4n is greater than -5 and less than 5. 9−4n>-5and9−4n<5 Let's isolate n in both of these cases before graphing the solution set.Case 1 9−4n>-5LHS−9>RHS−9-4n>-14Divide by -4 and flip inequality signn<414​Calculate quotientn<3.5 This inequality tells us that all values less than 3.5 will satisfy the inequality.Case 2 9−4n<5LHS−9<RHS−9-4n<-4Divide by -4 and flip inequality signn>1 This inequality tells us that all values greater than 1 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:​n<3.5 1<n 1<n<3.5​Graph The graph of this inequality includes all values from 1 to 3.5, not inclusive. We show this by using open circles on the endpoints.
Exercises 11 We are asked to find and graph the solution set for all possible values of t in the given inequality. ∣6t−7∣−8≥3 To do this, let's isolate absolute value expression first. ∣6t−7∣−8≥3LHS+8≥RHS+8∣6t−7∣≥11 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 or equal to 11 away from the midpoint in the positive direction or a distance greater than or equal to 11 away from the midpoint in the negative direction. 6t−7≥11 or 6t−7≤-11​ Let's isolate t in both of these cases before graphing the solution set.Case 1 6t−7≥11LHS+7≥RHS+76t≥18LHS/6≥RHS/6t≥3 This inequality tells us that all values greater than or equal to 3 will satisfy the inequality.Case 2 6t−7≤-11LHS+7≤RHS+76t≤-4LHS/6≤RHS/6t≤-64​ba​=b/2a/2​t≤-32​ This inequality tells us that all values less than or equal to -32​ 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:​ t≤-32​ or t≥3 t≤-32​ t≤-32​ or t≥3​Graph The graph of this inequality includes all values less than or equal to -32​ or greater than or equal to 3. We show this by keeping the the endpoints closed.
Exercises 12 First, let's isolate the absolute value expression for the given inequality. ∣3j−1∣+6>0LHS−6>RHS−6∣3j−1∣>-6 Remember that 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. Now, we can state the following expression. ∣3j−1∣≥0⇒∣3j−1∣>-6 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.
Exercises 13 We are asked to find and graph the solution set for all possible values of m in the given inequality. 3∣14−m∣>18 To do this, let's isolate the absolute value expression first. 3∣14−m∣>18LHS/3>RHS/3∣14−m∣>6 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 6 away from the midpoint in the positive direction or a distance greater than 6 away from the midpoint in the negative direction. 14−m>6 or 14−m<-6​ Let's isolate m in both of these cases before graphing the solution set.Case 1 14−m>6LHS−14>RHS−14-m>-8Multiply by -1 and flip inequality signm<8 This inequality tells us that all values less than 8 will satisfy the inequality.Case 2 14−m<-6LHS−14<RHS−14-m<-20Multiply by -1 and flip inequality signm>20 This inequality tells us that all values greater than 20 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:​ m<8 m<8 or m>20 m<8 or m>20​Graph The graph of this inequality includes all values less than 8 or greater than 20. We show this by open circles on the endpoints.
Exercises 14 First, let's substitute the absolute value expression of the given inequality. -4∣6b−8∣≤12Divide by -4 and flip inequality sign∣6b−8∣≥-3 Remember that 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. Thus, we can state the following. ∣6b−8∣≥0⇒∣6b−8∣≥-3 This means any value of b will satisfy the inequality. Thus, there are infinitely many solutions, and the solution set is the set of all real numbers.
Exercises 15 We are asked to find and graph the solution set for all possible values of w in the given inequality. 2∣3w+8∣−13≤-5 To do this, let's isolate the absolute value expression first. 2∣3w+8∣−13≤-5LHS+13≤RHS+132∣3w+8∣≤8LHS/2≤RHS/2∣3w+8∣≤4 Now, we can create a compound inequality by removing the absolute value. In this case, the solution set is any number less than or equal to 4 away from the midpoint in the positive direction and any number less than or equal to 4 away from the midpoint in the negative direction. Absolute value: Compound inequality: ​  ∣3w+8∣≤4  -4≤ 3w+8 ≤4​ This compound inequality means that the distance from 3w+8 is greater than or equal to -4 and less than or equal to 4. 3w+8≥-4and3w+8≤4 Let's isolate w in both of these cases before graphing the solution set.Case 1 3w+8≥-4Rearrange inequality-4≤3w+8LHS−8≤RHS−8-12≤3wLHS/3≤RHS/3-4≤w This inequality tells us that all values greater than or equal to -4 will satisfy the inequality.Case 2 3w+8≤4LHS−8≤RHS−83w≤-4LHS/3≤RHS/3w≤-34​ This inequality tells us that all values less than or equal to -34​ 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: ​-4≤w-4≤w≤-34​-4≤w≤-34​​Graph The graph of this inequality includes all values from -4 to -34​, inclusive. We show this by using closed circles on the endpoints.
Exercises 16 We are asked to find and graph the solution set for all possible values of u in the given inequality. -3∣2−4u∣+5<-13 To do this, let's isolate the absolute value expression first. -3∣2−4u∣+5<-13LHS−5<RHS−5-3∣2−4u∣<-18Divide by -3 and flip inequality sign∣2−4u∣>6 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 6 away from the midpoint in the positive direction or a distance greater than 6 away from the midpoint in the negative direction. 2−4u >6 or 2−4u<-6​ Let's isolate u in both of these cases before graphing the solution set.Case 1 2−4u>6LHS−2>RHS−2-4u>4Divide by -4 and flip inequality signu<-1 This inequality tells us that all values less than -1 will satisfy the inequality.Case 2 2−4u<-6LHS−2<RHS−2-4u<-8Divide by -4 and flip inequality signu>2 This inequality tells us that all values greater than 2 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:​ u<-1  u>2 u<-1 or u>2​Graph The graph of this inequality includes all values less than -1 or greater than 2. We show this with open circles on the endpoints.
Exercises 17 We are asked to find and graph the solution set for all possible values of f in the given inequality. 6∣-f+3∣+7>7 To do this, let's isolate the absolute value variable first. 6∣-f+3∣+7>7LHS−7>RHS−76∣-f+3∣>0LHS/6>RHS/6∣-f+3∣>0 Now, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number with a distance greater than 0 away from the midpoint in the positive direction or a distance greater than 0 away from the midpoint in the negative direction. -f+3 >0 or -f+3<0​ Let's isolate f in both of these cases before graphing the solution set.Case 1 -f+3>0LHS−3>RHS−3-f>-3Divide by -1 and flip inequality signf<3 This inequality tells us that all values less than 3 will satisfy the inequality.Case 2 -f+3<0LHS−3<RHS−3-f<-3Divide by -1 and flip inequality signf>3 This inequality tells us that all values greater than 3 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:​ f<3 f<3 or f>3 f<3 or f>3​Graph The graph of this inequality includes all values less than 3 or greater than 3. The only value that is not included in this solution set is 3 itself.
Exercises 18 We are asked to find and graph the solution set for all possible values of v in the given inequality. 32​∣4v+6∣−2≤10 To do this, let's isolate the absolute value expression first. 32​∣4v+6∣−2≤10LHS+2≤RHS+232​∣4v+6∣≤12LHS⋅23​≤RHS⋅23​∣4v+6∣≤18 Now, we can create a compound inequality by removing the absolute value. In this case, the solution set is any number less than or equal to 18 away from the midpoint in the positive direction and any number less than or equal to 18 away from the midpoint in the negative direction. Absolute value: Compound inequality: ​-18≤∣4v+6∣≤18-18≤4v+6≤18​ This compound inequality means that the distance from 4v+6 is greater than or equal to -18 and less than or equal to 18. 4v+6≥-18and4v+6≤18 Let's isolate v in both of these cases before graphing the solution set.Case 1 4v+6≥-18Rearrange inequality-18≤4v+6LHS−6≤RHS−6-24≤4vLHS/4≤RHS/4-6≤v This inequality tells us that all values greater than or equal to -6 will satisfy the inequality.Case 2 4v+6≤18LHS−6≤RHS−64v≤12LHS/4≤RHS/4v≤3 This inequality tells us that all values less 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: -6≤Intersecting solution set: -6≤​v≤3vv≤3​Graph The graph of this inequality includes all values from -6 to 3, inclusive. We show this by using open circles on the endpoints.
Exercises 19 We have been told that for an essay contest entries can have 500 words with an absolute deviation of at most 30 words. Let x be the number of words written. The absolute deviation is the difference between x and 500. We can state this as an inequality. ∣x−500∣≤30 Now, we will solve this absolute value inequality to find the acceptable numbers of words. To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set can be written as the following. Absolute value:  ∣x−500∣Compound:   -30≤ x−500 ​≤30≤30​ This compound inequality means that the distance from x−500 is greater than or equal to -30 and less than or equal to 30. x−500≥-30andx−500≤30 Let's isolate x in both of these cases before graphing the solution set.Case 1 x−500≥-30LHS+500≥RHS+500x≥470 This inequality tells us that all values greater than or equal to 470 will satisfy the inequality.Case 2 x−500≤30LHS+500≤RHS+500x≤530 This inequality tells us that all values less than or equal to 530 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: ​ 470≤xx≤530 470≤x≤530​ The acceptable number of words that can be written in the essay is between 470 and 530.
Exercises 20 To calculate the difference between the actual and normal body temperature of a camel, we subtract them. actual temperature−normal temperature​Writing the Inequality We are told the normal body temperature is 37∘C. Let x be the actual body temperature of a camel. x−37​ Note that the actual temperature can be lower or higher than the normal temperature. To keep the difference always nonnegative, we will use absolute value. ∣x−37∣​ We also know that the body temperature of a camel varies by up to 3∘C throughout a day. This means that the difference between actual and normal temperature must be less than or equal to 3∘C. ∣x−37∣≤3​Solving the Inequality The absolute value inequality we have written before can be expressed as a compound inequality using the word and.Absolute Value InequalityCompound Inequality ∣x−37∣≤3-3≤x−37andx−37≤3To solve the compound inequality, we need to solve its individual inequalities one at a time. Let's start by solving -3≤x−37. To do so, we will use inverse operations. -3≤x−37LHS+37≤RHS+3734≤xRearrange inequalityx≥34 Let's now solve x−37≤3. x−37≤3LHS+37≤RHS+37x≤40 The solution to this type of compound inequality is the intersection of the solution sets. First solution set: Second solution set: Intersecting solution set: ​34≤x34≤x≤4034≤x≤40​ The range of actual body temperature of a camel is between 34∘C and 40∘C.
Exercises 21 To solve absolute value inequalities, we must write the given inequality as two separate cases. If we have an inequality of the form ∣x+a∣<b then we can write this as the compound inequality x+a<b and x+a>-b Notice that in the work provided, the second case was not considered. The student should have also solved x−5>-20. We will solve the inequality correctly below. x−5>-20LHS+5>RHS+5x>-15 The solution set is thus both x<25 and x>-15⇔-15<x<25. To graph the solution, we will place open circles at 25 and -15 because x cannot equal neither 25 nor 15. Finally, we will shade the region between the circles.
Exercises 22 To solve absolute value inequalities, we must write the given inequality as two separate cases. If we have an inequality of the form ∣x+a∣>b then we can write this as the compound inequality x+a<b or x+a>-b Already here we can find a mistake. When splitting up the inequalities, the word "and" is used instead of "or". Let's make sure the inequalities were solved correctly.InequalitySubtract 4 x+4>13x>9 x+4<-13x<-17 They were in fact solved correctly so no problem there. The solution set is hence x<-17 or x>9. To graph the solution, we will place open circles at -17 and at 9 because x cannot equal either. Finally we shade the regions to the right of 9 and to the left of -17.
Exercises 23 We will choose x to be our variable. It is given that x is less than 6 units from 0. This means the absolute value of x is less than 6. Symbolically we can write this as ∣x∣<6. If we remove the absolute value, we get a compound inequality. Since the inequality sign is <, this will be written in the following form. -6<x<6 This is our solution, x must be greater than -6 but less than 6.
Exercises 24 In order to rewrite this sentence into an absolute value inequality, we should first remember the standard form for an absolute value equation: ∣x−midpoint∣=distance. We can look at the various pieces of the given sentence and determine how we can transform the general form for an absolute value equation into an inequality. If we let "a number" be x then we can fill in the numerical information as follows: ​A number…9 units from 3.∣x−3∣…9​ Now we need to use the phrase "is more than" to figure out the correct inequality symbol. If you recall, absolute value represents a distance. We need the distance from 3 to be more than 9 units. Another way to say this is: the distance from 3 should be greater than 9 units. Our inequality then becomes: ​A number is more than 9 units from 3.  ∣x−3∣>9​ Now we can solve this inequality for the possible values of x by separating the absolute value into two cases. Because we want the distance to be greater than 9, we will need to write an "or" inequality: First case: Second case: ​x−3>9x−3<-9​ We can solve these cases separately and then combine the results. Let's begin with the first case. x−3>9LHS+3>RHS+3x>12 Now let's solve the second case. x−3<-9LHS+3<RHS+3x<-6 Our final solution set is all values for x such that: x<-6orx>12
Exercises 25 It is given that half of a number x is at most 5 units from 14. Recall that absolute values are the distance between two values. Here, we are saying the 21​x is a distance from 14. This gives us the following expression inside the absolute value. ∣∣∣∣​21​x−14∣∣∣∣​​ We are also told that this distance is "at most 5." The phrase at most is an inclusive inequality phrase that means less than or equal to. This give us ∣∣∣∣​21​x−14∣∣∣∣​≤5​ as our absolute value inequality.Solving To solve this inequality, we will need to remove the absolute value and solve two different inequalities. Because the distance must be less than or equal to 5, we will have an "and" inequality. 21​x−14≤5and21​x−14≥-5 Now, we can solve inequality individually. 21​x−14≤5LHS+14≤RHS+1421​x≤19LHS⋅2≤RHS⋅2x≤38 The first inequality tells us that x must be less than or equal to 38. Next, we can find the lower bound of the solution set. 21​x−14≥-5LHS+14≤RHS+1421​x≥9LHS⋅2≥RHS⋅2x≥18 The second inequality tells us that x must also be greater than or equal to 18. When we combine our two solution sets, we have the following compound inequality. 18≤x≤38
Exercises 26 In order to rewrite this sentence into an absolute value inequality, we should first remember the standard form for an absolute value equation: ∣x−midpoint∣=distance from the midpoint. We can look at the various pieces of the given sentence and determine how we can transform the general form for an absolute value equation into an inequality. If we let "a number" be x then we can fill in the numerical information as follows: ​Twice a number…10 units from -1.∣2x−(-1)∣…10​ Now we need to use the phrase "no less than" to figure out the correct inequality symbol. If you recall, absolute value represents a distance. We need the distance from -1 to be no less than 10 units. Other ways to say this are: the distance from -1 should be at least 10 units or the distance from -1 should be greater than or equal to 10 units. Our inequality then becomes: ​Twice a number is no less than 10 units from -1.∣2x−(-1)∣≥10​ We can simplify this a little bit because subtracting a negative can be written as addition, (−)(−)=(+), making our final inequality: ∣2x+1∣≥10. Now we can solve this inequality for the possible values of x by separating the absolute value into two cases. Because we want the distance to be greater than or equal to 10, we will need to write an "or" inequality: First case: Second case: ​2x+1≥102x+1≤-10​ We can solve these cases separately and then combine the results. Let's begin with the first case. 2x+1≥10LHS−1≥RHS−12x≥9LHS/2≥RHS/2x≥29​ Now let's solve the second case. 2x+1≤-10LHS−1≤RHS−12x≤-11LHS/2≤RHS/2x≤-211​ Our final solution set is all values for x such that: x≥29​orx≤-211​
Exercises 27 The gaskets are only thrown out if they are not within 0.06 pounds of the mean weight of the batch. The first thing we need to do is calculate the mean weight of the batch so that we can know our central point of comparison. Mean=Number of valuesSum of values​Substitute valuesMean=50.58+0.63+0.65+0.53+0.61​Add termsMean=53​Use a calculatorMean=0.6 The mean weight of the batch is 0.6 pounds. We need the individual gasket weights w to be less than 0.06 pounds away from 0.6. We can solve for this range of values by writing an absolute value inequality: ∣w−0.6∣<0.06. This is an "and" compound absolute value inequality which can be rewritten as: -0.06<w−0.6<0.06. By the Addition Property of Inequalities, to find our solution set we only need to add 0.6 to all three sides of the compound inequality which isolates w in the middle. 0.54<w<0.66. The acceptable range of weights for the gaskets are those that are greater than 0.54 pounds and less than 0.66. The gasket that needs to be thrown out is the one weighing 0.53 pounds.
Exercises 28 Before we can find the greatest absolute deviation d, we need to calculate the mean of all the measurements taken during the experiment. Mean=Number of valuesSum of values​Substitute valuesMean=610.56+9.52+9.73+9.80+9.78+10.91​Add termsMean=660.30​Use a calculatorMean=10.05 Now, let's make a table to find the absolute deviation between the mean and each measurement.Measurement|Mean-Measurement|= 10.56∣10.05−10.56∣0.51 9.52∣10.05−9.52∣0.53 9.73∣10.05−9.73∣0.32 9.80∣10.05−9.80∣0.25 9.78∣10.05−9.78∣0.27 10.91∣10.05−10.91∣0.86 The greatest absolute deviation d is 0.86.
Exercises 29 Ultimately, we want to solve for the set of all possible values of x. We will do this by writing an absolute value inequality that looks like this: ∣Area of triangle−Area of Rectangle∣<2, because we need the difference of between the areas to be less than 2. First, we need to find the area of each figure as simplified as possible.Area of Triangle The area of a triangle is found using the formula: A=21​bh, where b is the base and h is the height. Let's substitute the values from the figure into our formula. A=21​bhb=4, h=x+6A=21​⋅4⋅(x+6)Area of Rectangle The area of a rectangle is found using the formula: A=lw, where l is the length and w is the width. Let's substitute the values from the figure into our formula. A=lwl=2, w=6A=2⋅6Absolute Value Inequality Now that we have our areas for the figures, we can substitute these expressions into the absolute value inequality from before: ∣∣∣∣​(21​⋅4⋅(x+6))−(2⋅6)∣∣∣∣​<2. We can simplify the expression inside the absolute value a bit before we solve for x. ∣∣∣∣​(21​⋅4⋅(x+6))−(2⋅6)∣∣∣∣​<2Multiply∣2(x+6)−12∣<2Distribute 2∣2x+12−12∣<2Subtract term∣2x∣<2 We can solve this by rewriting it as an "and" compound inequality and dividing all three sides by 2: -2<2x<2⇒-1<x<1 The values of x that lie in the solution set -1<x<1 will all satisfy the requirement that the difference between the areas of the figures must be less than 2.
Exercises 30 Ultimately, we want to solve for the set of all possible values of x. Because we need the difference between the perimeters to be less than or equal to 3, we can write an absolute value inequality that looks like: ∣Perimeter Rectangle​−Perimeter Square​∣≤3. First, we need to find the perimeter of each figure as simplified as possible.Perimeter of Rectangle The perimeter of a rectangle is found using the formula: P=2l+2w, where l is the length and w is the width of the rectangle. Let's substitute the values from the figure into our formula. P=2l+2wl=3, w=x+1P=2⋅3+2(x+1)MultiplyP=6+2(x+1)Distribute 2P=6+2x+2Add termsP=2x+8Perimeter of Square The perimeter of a square is found using the formula: P=4s, where s is the length of any side of a square. Let's substitute the values from the figure into our formula. P=4ss=xP=4⋅xMultiplyP=4xAbsolute Value Inequality Now that we have our perimeters for the figures, we can substitute these expressions into the absolute value inequality from before. ∣2x+8−4x∣≤3 We can simplify the expression inside the absolute value a bit before we solve for x. ∣2x+8−4x∣≤3Subtract terms∣-2x+8∣≤3 Now, let's write absolute value inequality and compound inequality as the following. Absolute value: ∣-2x+8∣Compound: -3≤-2x+8 ​≤3≤3​ Note that there is a third way to write this "and" type of compound inequality. -3≤-2x+8and-2x+8≤3 Let's isolate x in both of these cases before recombining them to form the final solution set.Case 1 -3≤-2x+8LHS−8≤RHS−8-11≤-2xDivide by -2 and flip inequality sign211​≥xCalculate quotient5.5≥xRearrange inequalityx≤5.5 This inequality is true when x is less than or equal to 5.5.Case 2 -2x+8≤3LHS−8≤RHS−8-2x≤-5Divide by 2 and flip inequality signx≥25​Calculate quotientx≥2.5 This inequality is true when x is greater than or equal to 2.5.Range of x The solution to this type of compound inequality is the intersection of the solution sets. First solution set:Second solution set:  2.5≤Intersecting solution set: 2.5≤​x≤5.5xx≤5.5​ The range of possible values of x is: 2.5≤x≤5.5
Exercises 31 The easiest way to compare the two given expressions is by rewriting the absolute value inequality as a compound inequality. Here we have ∣x+3∣≤8, which can be rewritten as an "and" compound inequality because the distance represented by the absolute value needs to be less than or equal to 8. The compound inequality then becomes: -8≤x+3 and x+3≤8 Now, we can compare these inequalities with other inequality given in the exercise, x+3≥-8. We can see that one of the inequalities from the compound inequality is the same as this, only rearranged, so the statement is true.
Exercises 32 The easiest way to compare the two given expressions is by rewriting the absolute value inequality as a compound inequality.Here we have ∣x+3∣>8, which can be rewritten as an "or" compound inequality because the distance represented by the absolute value needs to be greater than 8. The compound inequality then becomes: x+3>8 or x+3<-8. Now, we can compare these inequalities with other inequality given in the exercise, x+3>8. We can see that one of the inequalities from the compound inequality is the same as this, so the statement is true.
Exercises 33 The easiest way to compare the two given expressions is by rewriting the absolute value inequality as a compound inequality. Here we have ∣x+3∣≥8, which can be rewritten as an "or" compound inequality because the distance represented by the absolute value needs to be greater than or equal to 8. The compound inequality then becomes: x+3≥8 or x+3≤-8. Now, we can compare these inequalities with other inequality given in the exercise, x+3≥-8. We can see that this is a direct contradiction with the second member of the compound inequality.A contraditction Let's suppose that x=-12. In the compound inequality, this value satisfies the second inequality: -12+3≤-8⇒-9≤-8. However, with the inequality x+3≥-8, this is not a viable solution: -12+3≥?​-8⇒-9≱-8. Therefore, the statement is false.
Exercises 34 The easiest way to compare the two given expressions is by rewriting the absolute value inequality as a compound inequality. Here we have ∣x+3∣≥8, which can be rewritten as an "or" compound inequality because the distance represented by the absolute value needs to be greater than or equal to 8. The compound inequality then becomes: x+3≥8 or x+3≤-8. Now, we can compare these inequalities with other inequality given in the exercise, x+3≥8. We can see that one of the inequalities from the compound inequality is the same as this, so the statement is true.
Exercises 35 This is an easy mistake to make because it is almost true. Any nonzero value, positive or negative, substituted in for n will give a positive result because: ∣a∣=a and ∣-a∣=a​ However, what about 0? When n=0, what will ∣n∣ be? Well, ∣0∣=0. But, a number cannot be greater than itself so we have 0≯0. Therefore, the given inequality, ∣n∣>0, is true for all real numbers except 0.
Exercises 36 In order to create a polygon with an acceptable perimeter, we should solve the absolute value inequality first. Then we can know which perimeter values we can choose. We can rewrite the given absolute value inequality as an "and" compound inequality: ∣P−60∣≤12⇒-12≤P−60≤12 We can then use the Addition Property of Inequalities and add 60 to all three sides to get our solution set: 48≤P≤72. The perimeter of our polygon may be any values between 48 and 72, inclusive. Here is an example of a rectangle with a perimeter of 56.
Exercises 37 We are given two inequalities, let's look at them one at a time.First inequality The first inequality we are given is: ∣ax+b∣<c, where c<0. The constraint means that c can only be a negative value. We know that an absolute value, no matter what the expression inside simplifies to be, will always produce a positive result. Therefore, we can think about this absolute value inequality in general terms as: positive number < negative number. This will never be true, so there is no solution to this inequality.Second inequality The second inequality we are given is: ∣ax+b∣>c, where c<0. The constraint means that c can only be a negative value. We know that an absolute value, no matter what the expression inside simplifies to be, will always produce a positive result. Therefore, we can think about this absolute value inequality in general terms as: positive number > negative number. This will always be true, so the solution set is all real numbers.
Exercises 38 Before we can write the inequalities for the given graphs, we should remember two key things:An absolute value equation is written as ∣x−a∣=b, where a is the center point on the number line and b is distance to either endpoint from the center. How can we transform this into an inequality? Absolute value represents a distance from the center point. Do we need that distance to be greater than some value or less than some value?First graph The first graph is centered around the point where x=2. The closed dots are placed 3 spaces away from the center, 2−3=-1 and 2+3=5. From this, we can start writing our inequality: ∣x−2∣…3. To choose the inequality symbol, we need to decide whether the distance from the center should be greater than or equal to 3 or less than or equal to 3. In this case, the solution sets include points that are farther away from the center, so we need the distance to be greater than or equal to 3: ∣x−2∣≥3.Second graph The second graph is centered around the point where x=2. The open dots are placed 3 spaces away from the center, 2−3=-1 and 2+3=5. From this, we can start writing our inequality: ∣x−2∣…3. To choose the inequality symbol, we need to decide whether the distance from the center should be greater than 3 or less than 3. In this case, the solution set includes points that are closer to the center, so we need the distance to be less than to 3: ∣x−2∣<3.Third graph The third graph is centered around the point where x=2. The closed dots are placed 3 spaces away from the center, 2−3=-1 and 2+3=5. From this, we can start writing our inequality: ∣x−2∣…3. To choose the inequality symbol, we need to decide whether the distance from the center should be greater than or equal to 3 or less than or equal to 3. In this case, the solution set includes points that are closer to the center, so we need the distance to be less than or equal to to 3: ∣x−2∣≤3.Fourth graph The fourth graph is centered around the point where x=2. The open dots are placed 3 spaces away from the center, 2−3=-1 and 2+3=5. From this, we can start writing our inequality: ∣x−2∣…3. To choose the inequality symbol, we need to decide whether the distance from the center should be greater than 3 or less than 3. In this case, the solution sets include points that are farther away from the center, so we need the distance to be greater than to 3: ∣x−2∣>3.
Exercises 39 We are given two absolute value inequalities, let's look at them one at a time.First inequality First, let's rewrite the absolute value inequality as a compound inequality. ∣x∣<5 will become an "and" compound inequality because we need the solution sets where the values of x are less than 5 units away from 0: ∣x∣<5⇒-5<x and x<5. We can graph these intervals on a number line. The interval -5<x is represented with a blue line and x<5 with a red line. The purple segment represents the interval where these sets overlap.Because it is an "and" compound inequality, both conditions must be met and the overlap is the solution set. It is the "intersection" of these two intervals, the interval where the points belong to both sets.Second inequality First, let's rewrite the absolute value inequality as a compound inequality. ∣x∣>5 will become an "or" compound inequality because we need the solution sets where the values of x are greater than 5 units away from 0: ∣x∣>5⇒-5<x or x>5. We can graph these intervals on a number line. The interval -5>x is represented with a blue line and x>5 with a red line.Because it is an "or" compound inequality, the solution set includes all values that are contained in either set. It is the "union" of these two intervals.
Exercises 40 We have been given a compound inequality that is actually composed of two separate compound inequalities. First, let's rewrite and graph each part of the overall compound inequality, then combine the results.First half The first half of the compound inequality can be rewritten as an "and" compound inequality: ∣x−3∣<4⇒-4<x−3<4 Using the Addition Property of Inequalities, we can add 3 to all three sides and simplify this to be -1<x<7, which we can graph on a number line.Second half The second half of the compound inequality can be rewritten as an "or" compound inequality: ∣x+2∣>8⇒x+2>8orx+2<-8 Using the Subtraction Property of Inequalities, we can subtract 2 in each inequality and simplify this to be x>6 or x<-10, which we can graph on a number line.Combining both halves To find the solution set of both halves combined, we can graph both solution sets on the same number line.Since our overall compound inequality is an "and" inequality, we need the intersection of these two graphs. The overlapping interval between 6 and 7 is our set of solutions, 6<x<7. We can check our solution by substituting a few key values in for x to see if they satisfy both inequalities.x∣x−3∣<4Evaluate∣x+2∣>8Evaluate 6∣6−3∣<43<4∣6+2∣>88≯8 6.5∣6.5−3∣<43.5<4∣6.5+2∣>88.5>8 7∣7−3∣<44≮4∣7+2∣>89>8 The table of values shows us that the endpoints, 6 and 7, do not satisfy both inequalities, but the values between them do. Our solution is thus 6<x<7.
Exercises 41 In an ordered pair, the first number shows the x-coordinate and the second number shows the y-coordinate, (x,y). To graph A(1,3) we have to identify the point at which x=1 and y=3.
Exercises 42 In an ordered pair, the first number shows the x-coordinate and the second number shows the y-coordinate, (x,y). To graph B(0,-3) we have to identify the point at which x=0 and y=-3.
Exercises 43 In an ordered pair, the first number shows the x-coordinate and the second number shows the y-coordinate, (x,y). To graph C(-4,-2) we have to identify the point at which x=-4 and y=-2.
Exercises 44 In an ordered pair, the first number shows the x-coordinate and the second number shows the y-coordinate, (x,y). To graph D(-1,2) we have to identify the point at which x=-1 and y=2.
Exercises 45 We can fill out this table by substituting each value for x into the expression and simplifying.x01234 5x+15⋅0+15⋅1+15⋅2+15⋅3+15⋅4+1 Evaluate16111621
Exercises 46 We can fill out this table by substituting each value for x into the expression and simplifying.x-2-1012 -2x−3-2(-2)−3-2(-1)−3-2⋅0−3-2⋅1−3-2⋅2−3 Evaluate1-1-3-5-7