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###### Exercises

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Exercises 1 In order to change the sentence into an inequality, we should look at the individual parts of the sentence. First we have "the sum of a number y and 9" which can be expressed algebraically as: y+9. We are then told that this sum is "at least -1," which is the same thing as saying that it is greater than or equal to -1, the minimum value is at least -1. Altogether we have: Sum of a number y and 9 is at least -1y+9 ≥ -1 | |

Exercises 2 We are told that r is either greater than 0 or less than or equal to -8. This means we need to create an "or" compound inequality, one inequality to represent each situation. The first will be: a number r is more than 0. r > 0 The second inequality will be: a number r is less than or equal to -8.r ≤ -8 Combining the two gives us the compound inequality: r>0 or r≤-8. | |

Exercises 3 We need to use an inequality to express a distance away from a central point, this requires us to use an absolute value inequality. The most basic form of an absolute value equation can be written as: ∣x−a∣=b, where a is the central point and b is the distance from the central point. Here we need an inequality when our central point is 10 and the distance of point k from 10 must be less than 3. A number k is less than 3 units away from 10.∣k−10∣ <3 | |

Exercises 4 Solving an inequality is similar to solving an equation. We will isolate the variable using inverse operations, then graph our solution on a number line. 2x−5≥-9LHS+5≥RHS+52x≥-4LHS⋅2≥RHS⋅2x≥-8 This inequality tells us that all values greater than or equal to -8 will satisfy the inequality. Notice that x can equal -8, which we show with a closed circle on the number line. | |

Exercises 5 Solving an inequality is similar to solving an equation. We will isolate the variable using inverse operations, then graph our solution on a number line. Remember that if we multiply or divide an inequality by a negative, we have to reverse the inequality symbol. -4s<6s+1LHS−6s<RHS−6s-10s<1Divide by -10 and flip inequality signs>-101 This inequality tells us that all values greater than -101 will satisfy the inequality. Notice that s cannot equal -101, which we show with an open circle on the number line. | |

Exercises 6 Solving an inequality is similar to solving an equation. We will isolate the variable using inverse operations, then graph our solution on a number line. 4p+3≥2(2p+1)Distribute 24p+3≥4p+2LHS−4p≥RHS−4p3≥2 Since the inequality 3≥2 is always true, there are infinitely many solutions and the solution set is all real numbers. | |

Exercises 7 First, let's split the compound inequality into separate inequalities: Compound Inequality:-7<2cFirst Inequality:-7<2cSecond Inequality:2c−1<10−1−1<10 Notice that compound inequalities written in this way are equivalent to compound inequalities that involves the word and. -7<2c−1and2c−1<10First 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. -7<2c−1LHS+1<RHS+1-6<2cLHS/2<RHS/2-3<cRearrange inequalityc>-3 This inequality tells us that all values greater than -3 will satisfy the inequality.Note that the point on -3 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. 2c−1<10LHS+1<RHS+12c<11LHS/2<RHS/2c<211Calculate quotientc<5.5 This inequality tells us that all values less than 5.5 will satisfy the inequality.Note that the point on 5.5 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>-3 as -3<c. First solution set:Second solution set:Intersecting solution set:-3<c-3<c<5.5-3<c<5.5 Finally, we'll graph the solution set to the compound inequality on a number line. | |

Exercises 8 First, let's split the compound inequality, -2≤4−3a≤13, into two separate inequalities: -2≤4−3aand4−3a≤13. Now we can solve them separately and later compare their solution sets.First inequality We can solve inequalities just like ordinary equations. However, if we multiply or divide an inequality by a negative, we have to reverse the inequality sign. -2≤4−3aLHS−4≤RHS−4-6≤-3aDivide by -3 and flip inequality sign2≥aRearrange inequalitya≤2 The first inequality is satisfied by all values less than or equal to 2. Note that a can equal 2 as the inequality sign contains or equal to.Second inequality Now we can solve the second inequality. Again, if we multiply or divide an inequality by a negative, we have to reverse the inequality sign. 4−3a≤13LHS−4≤RHS−4-3a≤9Divide by -3 and flip inequality signa≥-3 The second inequality is true for numbers greater than or equal to -3. Notice that a can equal -3 as the inequality sign contains or equal to.Combining solution sets Finally, we can combine the obtained solution sets. The first inequality describes all values to the left of 2, including 2.The second inequality describes all values to the right of -3, including -3.The intersection of these sets, -3≤a≤2, describes the solutions of the compound inequality. | |

Exercises 9 To solve the compound inequality, we have to solve each of the inequalities separately. Since the word between the individual inequalities is "or," the solution set for the compound inequality consists of the sets of the individual solutions.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. -5<2−hLHS−2<RHS−2-7<-hMultiply by -1 and flip inequality sign7>hRearrange inequalityh<7 This inequality tells us that all values less than 7 will satisfy the inequality.Note that the point on 7 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. 6h+5>71LHS−5>RHS−56h>66LHS/6>RHS/6h>11 This inequality tells us that all values greater than 11 will satisfy the inequality.Note that the point on 11 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:h<7h<7 or h>11h<7 or h>11 Finally, we'll graph the solution set to the compound inequality. | |

Exercises 10 We are asked to find and graph the solution set for all possible values of n in the given inequality. ∣2q+8∣>4 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 4 away from the midpoint in the positive direction or a distance greater than 4 away from the midpoint in the negative direction. 2q+8 >4 or 2q+8<-4 Let's isolate q in both of these cases before graphing the solution set.Case 1 2q+8>4LHS−8>RHS−82q>-4LHS/2>RHS/2q>-2 This inequality tells us that all values greater than -2 will satisfy the inequality.Case 2 2q+8<-4LHS−8<RHS−82q<-12LHS/2<RHS/2q<-6 This inequality tells us that all values less than -6 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: q<-6 or q>-2 q<-6 q<-6 or q>-2Graph The graph of this inequality includes all values less than -6 or greater than -2. We show this with open circles on the endpoints. | |

Exercises 11 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. First, let's isolate the absolute value expression. Remember that if we multiply or divide an inequality by a negative, we have to reverse the inequality symbol. -2∣y−3∣−5≥-4LHS+5≥RHS+5-2∣y−3∣≥1Divide by -2 and flip inequality sign∣y−3∣≤-21 Thus, we can state the following. ∣y−3∣≥0⇒∣y−3∣ ≰ -21 This means there is no value of y that will satisfy the inequality. Thus, there is no solution. | |

Exercises 12 We are asked to find and graph the solution set for all possible values of a in the given inequality. 4∣-3b+5∣−9<7 To do this, let's isolate the absolute value expression first. 4∣-3b+5∣−9<7LHS+9<RHS+94∣-3b+5∣<16LHS/4<RHS/4∣-3b+5∣<4 Now, we can create a compound inequality by removing the absolute value. In this case, the solution set is any number less than 4 away from the midpoint in the positive direction and any number less than 4 away from the midpoint in the negative direction. Absolute value: Compound: ∣-3b+5∣<4 -4< -3b+5 <4 This compound inequality means that the distance -3b+5 is greater than -4 and less than 4. -3b+5>-4and-3b+5<4 Let's isolate b in both of these cases before graphing the solution set.Case 1 -3b+5>-4LHS−5>RHS−5-3b>-9Divide by -3 and flip inequality signb<3 This inequality tells us that all values less than 3 will satisfy the inequality.Case 2 -3b+5<4LHS−5<RHS−5-3b<-1Divide by -3 and flip inequality signb>31 This inequality tells us that all values greater than 31 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: 31<Intersecting solution set: 31<b<3bb<3Graph The graph of this inequality includes all values from 31 to 3, not inclusive. We show this by using open circles on the endpoints. | |

Exercises 13 Before we begin constructing the inequality, let's just review a few of the key words used in this exercise.Expenses: Money spent to run your business. Revenue: All the money earned at your business. Profit: Your revenue minus any of your expenses (actual money you get to keep). You start a business and in the first month you want to make a profit of at least $250. That means that your profit has to be greater than or equal to $250. We can write the right-hand side of the inequality as follows. …≥250 To write the left-hand side of the inequality, remember that profit equals revenue minus expenses. Knowing the expenses in the first month are $155, we can write x−155≥250, where x represents the revenue. Let's solve this inequality for x to find its value. x−155≥250LHS+155≥RHS+155x−155+155≥250+155Add termsx≥405 You need to have a revenue of $405 or more in order to have a profit of at least $250 in the first month. | |

Exercises 14 We know the required width of the chain which is 0.3 inches. This is the midpoint of the range. We are asked to find the acceptable widths of the chain which is at most 0.0003 inches from the required width. This will be the maximum distance from the midpoint. Let's name the acceptable width as x and write the absolute value inequality. ∣x−0.3∣≤0.0003 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. -0.0003≤x−0.3≤0.0003 Finally, we can isolate x by adding 0.3 to all sides of the inequality. 2.9997≤x≤3.0003 The acceptable widths are between 2.9997 inches and 3.0003 inches. | |

Exercises 15 Before we can think about the possible solution sets for x, let's simplify our inequality a little bit so that x is as isolated as possible. ax+b<cx+dLHS−b<RHS−bax<cx+d−bLHS−cx<RHS−cxax−cx<d−bFactor out xx(a−c)<d−b We cannot go any further just yet. What if a−c=0? What if a−c is a negative number? There are actually four cases we need to consider for this inequality. First case: 1Second case: 1Third case: 1Fourth case: 1a=c and d>ba=c and d<ba>ca<c Let's look at these cases one at a time.First case In the first case, we should consider when a=c and d>b. When a=c we have a−c=0. x(a−c)<d−b(a−c)=0x⋅0<d−bZero Property of Multiplication0<d−b With the assumption that d>b we must have that d−b is always a positive number. Therefore, we have 0<"a positive number" which is always true. The inequality would then hold true for all real numbers.Second case In the second case, we should consider when a=c and d<b. When a=c, we have a−c=0. x(a−c)<d−b(a−c)=0x⋅0<d−bZero Property of Multiplication0<d−b With the assumption that d<b we must have that d−b is always a negative number. Therefore, we have 0<"a negative number" which is never true. The inequality would then have no solution.Third case In the third case, we should consider when a>c. When a>c, we must have that a−c will always be a positive number in which case we can simply divide and keep the inequality symbol as it is. x(a−c)<d−bLHS/(a−c)<RHS/(a−c)x<a−cd−b Our solution set is then all values such that x<a−cd−b.Fourth case In the fourth case, we should consider when a<c. When a<c, we must have that a−c will always be a negative number. This means we would need to reverse the inequality symbol when we divide to completely isolate x. x(a−c)<d−bDivide by (a−c) and flip inequality signx>a−cd−b Our solution set is then all values such that x>a−cd−b. | |

Exercises 16 The given graph shows a closed circle at -3 with shading to the left. It also has an open circle at 2 with shading to the right. This solution set represents values that are less than or equal to -3 or greater than 2. However, we want the numbers that are not in the solution set. These numbers can be stated with the following phrase. Greater than -3 and Less than or equal to 2. The algebraic symbols that represent greater than and less than or equal to are > and ≤, respectively. If we let x represent the numbers that are not the solutions, we can write the compound in equality as the following. First inequality: -3<Second inequality: Compound inequality: -3<xx≤2x≤2 The graph for the numbers that are not solutions can be shown as the following.Notice that -3 is represented with an open circle because it was a solution in the original graph. Similarly, 2 is represented with a closed circle because it was not a solution in the original graph. | |

Exercises 17 The given graph shows closed circles at -4 and -1 with shading between the two points. This solution set represents values that are Greater than or equal to -4 and Less than or equal to -1. However, we want the numbers that are not in the solution set. These numbers can be stated with the following phrase. Less than -4 or Greater than -1. The algebraic symbols that represent less than and greater than are < and >, respectively. If we let x represent the numbers that are not the solutions, we can write the compound in equality as the following. First inequality: Second inequality: Compound inequality: x<-4x<-4 or x>-1x<-4 or x>-1 The graph for the numbers that are not solutions can be shown as the following.Notice that, -4 and -1 are represented with open circles because they were solutions in the original graph. | |

Exercises 18 aThe first thing we need to do is find the taxable sales amount and then we can find what percent of that amount is $7.50. Since the sales tax only kicks in after $175, and the receipt shows that the suit cost $295, we have that amount taxable amount was: 295−175=$120. We can now use a percent proportion to find the sales tax percentage rate. wholepart=100percentSubstitute values1207.50=100x Solve for x Write as a decimal0.0625=100xLHS⋅100=RHS⋅1006.25=xRearrange equation x=6.25bTo find the possible prices p of coats that the shopper can afford, the general inequality we need to look at is: p+Sales Tax≤$430. The sales tax rate is 6.25% which can be rewritten as the decimal value 0.0625. Also, since the tax is only applied to clothing purchases over $175, we can express the total sales tax for the purchase algebraically as: 0.0625(p−175), because we only need to multiply the tax rate by the amount that exceeds 175. Therefore, our inequality becomes: p+0.0625(p−175)≤430, which we can now solve for the possible values of p. p+0.0625(p−175)≤430 Solve for p Distribute 0.0625p+0.0625p−10.9375≤430Add terms1.0625p−10.9375≤430LHS+10.9375≤RHS+10.93751.0625p≤440.9375LHS/1.0625≤RHS/1.0625 p≤415 This means that when taxes are included, the shopper can afford any winter coats that cost less than or equal to $415.cWe want to compare for which prices p of clothing would it be cheaper to buy things in the state with a flat 5% sales tax rate compared to the sales tax rate of 6.25% on purchases over $175. We can do this with the following inequality: p+0.05p<p+0.0625(p−175). In the state represented by the left-hand side, we have that any item will have a total cost comprised of the item's price plus 5% of that item's price. In the state represented by the right-hand side, we have that any item will have a total cost comprised of the item's price plus 6.25% of any amount above $175. Let's solve for the possible values of p now. p+0.05p<p+0.0625(p−175) Solve for p LHS−p<RHS−p0.05p<0.0625(p−175)Distribute 0.06250.05p<0.0625p−10.9375LHS+10.9375<RHS+10.93750.05p+10.9375<0.0625pLHS−0.05p<RHS−0.05p10.9375<0.0125pLHS/0.0125<RHS/0.0125875<pRearrange inequality p>875 Any item costing more than $875 will be cheaper overall if purchased in the state with a flat 5% sales tax rate. Three example prices that this would apply to are: $945, $1200 and $853215.99. |

##### 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 Review
- Cumulative Assessment