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Exercises 1 When solving either an equation or an inequality, you must use a series of inverse operations to systematically isolate the variable. You must always follow the Addition, Subtraction, Multiplication, and Division Properties of Equality when solving equations. Similarly, you must always follow the Addition, Subtraction, Multiplication, and Division Properties of Inequality when solving inequalities. Basically, you follow the same process for both types of problems. The main difference between solving equations and inequalities is that, when dividing or multiplying by a negative in an inequality, you must reverse the symbol to compensate for the nature of negative values in comparisons. | |

Exercises 2 Without actually solving the inequality, we can notice a few key things about it that tell us that it has no solution.Both sides have positive 4x. The pieces that aren't the aforementioned 4x are: …8≤…-3.By the Properties of Inequality, we know that we can ignore everything that is the exact same thing on both sides and only look at the other information. The "other stuff," in this case, is that part that tells us: 8≤-3, which will never be true. So we can know that there is no solution to this inequality. | |

Exercises 3 We need to isolate b before we can try to find which graph matches our inequality. Let's do that now. 7b−4≤10LHS+4≤RHS+47b≤14LHS/7≤RHS/7b≤2 Now we have the inequality b≤2. Our solution set is then all real numbers that are less than or equal to 2. This corresponds to option B. | |

Exercises 4 We need to isolate p before we can try to find which graph matches our inequality. Let's do that now. 4p+4≥12LHS−4≥RHS−44p≥8LHS/4≥RHS/4p≥2 We have the inequality p≥2. Our solution set is then all real numbers that are greater than or equal to 2. This corresponds to option A. | |

Exercises 5 We need to isolate g before we can try to find which graph matches our inequality. Let's do that now. -6g+2≥20LHS−2≥RHS−2-6g≥18Divide by -6 and flip inequality signg≤-3 We have the inequality g≤-3. Our solution set is then all real numbers that are less than or equal to -3. This corresponds to option C. | |

Exercises 6 We need to isolate f before we can try to find which graph matches our inequality. Let's do that now. 3(2−f)≤15Distribute 36−3f≤15LHS−6≤RHS−6-3f≤9Divide by (-3) and flip inequality signf≥-3 Now we have the inequality f≥-3. Our solution set is then all real numbers that are less than or equal to -3. This corresponds to option D. | |

Exercises 7 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 2x−3>7LHS+3>RHS+32x>10LHS/2>RHS/2x>5 This inequality tells us that all values greater than 5 will satisfy the inequality. Notice that x cannot equal 5, which we show with an open circle on the number line. | |

Exercises 8 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 5y+9≤4LHS−9≤RHS−95y≤-5LHS/5≤RHS/5y≤-1 This inequality tells us that all values less than or equal to -1 will satisfy the inequality. Notice that y can equal -1, which we show with a closed circle on the number line. | |

Exercises 9 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. -9≤7−8vLHS−7≤RHS−7-16≤-8vDivide by -8 and flip inequality sign2≥vRearrange inequalityv≤2 This inequality tells us that all values less than or equal to 2 will satisfy the inequality. Notice that v can equal 2, which we show with a closed circle on the number line. | |

Exercises 10 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 2>-3t−10LHS+10>RHS+1012>-3tDivide by -3 and flip inequality sign-4<tRearrange inequalityt>-4 This inequality tells us that all values greater than -4 will satisfy the inequality. Notice that t cannot equal -4, which we show with an open circle on the number line. | |

Exercises 11 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. 2w+4>5LHS−4>RHS−42w>1LHS⋅2>RHS⋅2w>2 This inequality tells us that all values greater than 2 will satisfy the inequality. Notice that w cannot equal 2, which we show with an open circle on the number line. | |

Exercises 12 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 1+3m≤6LHS−1≤RHS−13m≤5LHS⋅3≤RHS⋅3m≤15 This inequality tells us that all values less than or equal to 15 will satisfy the inequality. Notice that m can equal 15, which we show with a closed circle on the number line. | |

Exercises 13 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. -8p+9>13LHS−9>RHS−9-8p>4Multiply by -8 and flip inequality signp<-32 This inequality tells us that all values less than -32 will satisfy the inequality. Notice that p cannot equal -32, which we show with an open circle on the number line. | |

Exercises 14 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 3+-4r≤3LHS−3≤RHS−3-4r≤3Multiply by -4 and flip inequality signr≥-12 This inequality tells us that all values greater than or equal to -12 will satisfy the inequality. Notice that r can equal -12, which we show with a closed circle on the number line. | |

Exercises 15 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≥-6(a+2)Distribute -66≥-6a−12LHS+12≥RHS+1218≥-6aDivide by -6 and flip inequality sign-3≤aRearrange inequalitya≥-3 This inequality tells us that all values greater than or equal to -3 will satisfy the inequality. Notice that a can equal -3, which we show with a closed circle on the number line. | |

Exercises 16 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. 18≤3(b−4)LHS/3≤RHS/36≤b−4LHS+4≤RHS+410≤bRearrange inequalityb≥10 This inequality tells us that all values greater than or equal to 10 will satisfy the inequality. Notice that b can equal 10 which we show with a closed circle on the number line. | |

Exercises 17 Let's isolate m to solve the inequality. 4−2m>7−3mLHS+3m>RHS+3m4+m>7LHS−4>RHS−4m>3 The solution m>3 means that any number greater than 3 will satisfy the inequality. | |

Exercises 18 Let's solve the inequality by isolating n on one side. 8n+2≤8n−9LHS−8n≤RHS−8n2≰-9 The inequality 2≤-9 is never true. Therefore, there is no solution for the given inequality. | |

Exercises 19 Let's solve the inequality by isolating d on one side. -2d−2<3d+8LHS+2d<RHS+2d-2<5d+8LHS−8<RHS−8-10<5dLHS/5<RHS/5-2<dRearrange inequalityd>-2 The solution to the inequality is d>-2. | |

Exercises 20 Let's solve the inequality by isolating f on one side. 8+10f>14−2fLHS+2f>RHS+2f8+12f>14LHS−8>RHS−812f>6LHS/12>RHS/12f>126ba=b/6a/6f>21 The solution to the inequality is f>21. | |

Exercises 21 Let's solve the inequality by isolating g on one side. 8g−5g−4≤-3+3gSubtract term3g−4≤-3+3gLHS−3g≤RHS−3g-4≤-3 The inequality -4≤-3 is always true. Therefore, the solution set is all real numbers. | |

Exercises 22 Let's solve the inequality by isolating w on one side. 3w−5>2w+w−7Add terms3w−5>3w−7LHS−3w>RHS−3w-5>-7 The inequality -5>-7 is always true. Therefore, the solution set for the inequality is all real numbers. | |

Exercises 23 Let's try to isolate the unknown l on one side of the inequality. 6(l+3)<3(2l+6)Distribute 66l+18<3(2l+6)Distribute 36l+18<6l+18LHS−18<RHS−186l<6lLHS/6<RHS/6l<l This is never true for any value of l because l will always be equal to l. Therefore, there is no solution to this inequality. | |

Exercises 24 Let's solve the inequality by isolating c on one side. 2(5c−7)≥10(c−3)Distribute 2 & 1010c−14≥10c−30LHS−10c≥RHS−10c-14≥-30 The inequality -14≥-30 is always true. Therefore, the solution set for the inequality is all real numbers. | |

Exercises 25 Let's solve the inequality by isolating t on one side. 4(21t−2)>2(t−3)Distribute 4 & 22t−8>2t−6LHS−2t>RHS−2t-8≯-6 The inequality -8>-6 is never true. Therefore, there is no solution for the inequality. | |

Exercises 26 Let's solve the inequality by isolating b on one side. 15(31b+3)≤6(b+9)Distribute 15 & 65b+45≤6b+54LHS−5b≤RHS−5b45≤b+54LHS−54≤RHS−54-9≤bRearrange inequalityb≥-9 The solution to the inequality is b≥-9. | |

Exercises 27 Let's try to isolate the unknown j on one side of the inequality. 9j−6+6j≥3(5j−2)Distribute 39j−6+6j≥15j−6Add terms15j−6≥15j−6LHS+6≥RHS+615j≥15jLHS/15≥RHS/15j≥j Any real number will always be equal to itself and, since inequality includes "or equal to," this will be true for all real numbers. | |

Exercises 28 Let's try to isolate the unknown h on one side of the inequality. 6h−6+2h<2(4h−3)Distribute 26h−6+2h<8h−6Add terms8h−6<8h−6LHS+6<RHS+68h<8hLHS/8≥RHS/8h<h This is never true for any value of h. Any value for h will be equal to h, not less than or greater than it. Therefore, there is no solution. | |

Exercises 29 In the first line of the shown solution, the denominator of the fraction was 4. In the second line the fraction had been removed. This means that the equation was multiplied by 4. Let's check if it was done correctly by doing it ourselves. 4x+6≥3LHS⋅4≥RHS⋅44(4x+6)≥4⋅3Distribute 44⋅4x+4⋅6≥4⋅3Multiplyx+24≥12 We can see that the 4 wasn't distributed properly. Both the fraction and the 6 on the left-hand side should have been multiplied by 4. We can continue solving by subtracting 24 on both sides to get the correct solution: x≥-12. | |

Exercises 30 The first step in solving this inequality is to use the Distributive Property. Let's check if it was done correctly. -2(1−x)≤2x−7Distribute -2-2⋅1+2⋅x≤2x−7Multiply-2+2x≤2x−7 So far, so good! Let's continue solving. -2+2x≤2x−7LHS−2x≤RHS−2x-2≤-7 The last inequality tells us that -2 is less than or equal to -7. However, this is not true, -2 is greater than -7. We can confirm this by just placing both values on a number line and checking which is further to the right:The conclusion drawn was that all real numbers are solutions. This is not true because -2≰-7. The correct conclusion is that there are no solutions. | |

Exercises 31 We see in the diagram provided that the current balance is $320, and that we need to keep $100 in the account. Therefore, the difference between 320 and 100 will give us the maximum amount we can withdraw. We will calculate the maximum number of $20 bills we can withdraw by dividing that difference by 20. 20320−100 Let n be the number of $20 bills to be withdrawn from the ATM. Since the above expression represents the maximum number of $20 bills we can withdraw, n must be less than or equal to the above expression. n≤20320−100 Simplifying the right-hand side of the inequality will give us the number of $20 bills we can withdraw. n≤20320−100Subtract termsn≤20220Calculate quotientn≤11 We can withdraw 11 or less $20 bills. | |

Exercises 32 To begin, we will make sense of the given information. It is given that the woodworker will sell the cabinet for 500 dollars and that the supplies cost 125 dollars. The profit the woodworker makes can be determined by subtracting the money spent from the money earned. We have 500−125=$375 The profit made is 375 dollars. It is also given that the woodworker wants to earn at least 25 dollars per hour. If we let x represent the number of hours worked, then 25x represents the amount earned for any arbitrary number of hours. This expression should be less than 375 which gives us the inequality 25x≤375 Solving the inequality will give the maximum number of hours the woodworker can work if they want to make 25 dollars per hour. 25x≤375LHS/25≤RHS/25x≤15 The solution x≤15 means the woodworker can work a maximum of 15 hours. | |

Exercises 33 To begin, let's recall the formula for the area of a rectangle. A=ℓw In the above equation, ℓ is the length of the rectangle and w is the width. It is given that the length of the rectangle is 12 feet and the width is 2x−3 feet. A=12(2x−3) It is also given that the area of the rectangle is greater than 60 square feet. A>60⇔12(2x−3)>60 Solving this inequality will give us the possible values for x. 12(2x−3)>60Distribute 1224x−36>60LHS+36>RHS+3624x>96LHS/24>RHS/24x>4 The solution x>4 means that x can take any value greater than 4. | |

Exercises 34 To begin, we can write an expression to determine the cost spent at each campground. Let x be the number of nights we camp. At Forest Park Campgrounds they charge a fee of $100 and $35 per night. 100+35x At Woodland Campground they charge a fee of $20 and $55 per night. 20+55x Let's calculate the cost of each campground for 4 nights by substituting 4 for x in each of the above equations.CampgroundExpressionx=4Evaluate Forest Park100+35x100+35(4)240 Woodland20+55x20+55(4)240 We can see that both campgrounds cost the same amount for 4 nights. Since Woodland Campground charges more per night, staying additional nights will be more expensive than at Forest Park Campgrounds. Therefore, our friend is incorrect. | |

Exercises 35 We want to know how many stories of the building the fire truck's ladder can reach. We know that each story is approximately 10 feet tall. If we let the number of stories be n then we can express the height of the story of the building, in feet, as: 10n. We also know that the bottom of the ladder must be placed at least 24 feet away from the wall. We have something that looks like the drawing below. Notice that we have simplified this a bit and let the ladder start at the tail end of the fire truck.As we can see, the triangle formed is a right triangle and we can use the Pythagorean Theorem to solve for the maximum height that the ladder can reach: W2+242=742 Note, we don't need to include an inequality at this time because we already know that the maximum height will be when the base is 24 because pulling the fire truck farther away from the wall will only decrease the height reached by the ladder. Let's solve for W now. W2+242=742Calculate powerW2+576=5476LHS−576=RHS−576W2=4900LHS=RHSW=70 ft The maximum height that the ladder can reach is 70 feet. Now, we also need to consider the fact that the ladder rests on top of an 8 ft tall fire truck. So we can find the number of stories n reached by the fire truck's ladder with the inequality: 10n≤70+8. Let's solve for n. 10n≤70+8Add terms10n≤78LHS/10≤RHS/10n≤7.8 The fire truck can reach up to 7.8 stories. However, we can only have a whole number value for stories of a building and we cannot round up in this situation. Hence, the fire truck can safe lives from a 7 story building. | |

Exercises 36 aTo begin, we should recognize that our budget, the amount we can spend, does not change. Additionally, it is not affected by the number of gallons of gas we buy. Looking at the graph, we can see that the line y=40 is constant. Thus, our budget is 40 dollars.bIn Part A, we established that the equation y=40 represented our budget. This implies that the equation y=3.55x+8 represents the amount spent on gasoline and a car wash. Notice that the quantity 3.55x will change depending on x, while the 8 is constant. We can assume that we will only purchase one car wash. Thus, the cost of a car wash is 8 dollars. It follows then that 3.55 represents the cost of one gallon of gas.cWe established in Part A that our budget is 40 dollars. The amount we actually spend can be less than or equal to 40, but not greater than 40. Using the given expression 3.55x+40, we can write the inequality 3.55x+40≤40.dOn the given graph, we can see that the lines intersect at the point (9,40). This point is significant. It shows that if we buy 9 gallons of gas, in addition to the car wash, we will spend exactly 40 dollars, our entire budget. If we purchase more than 9 gallons, notice that the graph of y=3.55x+8 surpasses y=40. This means that we cannot afford more than 9 gallons. We can estimate that the solution to the inequality in Part C is x≤9. | |

Exercises 37 In general, the process for solving this exercise involves us writing an inequality such that: (Area of Circle)−(Area of Square)≥9(π−2). This is because the shaded area is only the parts that are in the circle but not in the square and we want to know the values for r that make this area greater than or equal to 9(π−2).Area of the Circle The area of a circle is found by using the formula: A=πr2 In this case, we are only given that the radius is r and, therefore, we cannot reduce this any further.Area of the Square For this square, we are given that the radius r of the circle is the measurement from the center to the corners of the square. To find the side length s of the square, we can use the Pythagorean Theorem: r2+r2=s2 Let's solve for s in terms of r now. r2+r2=s2 Solve for s Add terms2r2=s2Rearrange equations2=2r2LHS=RHSs=±2rs>0 s=2r Notice, we don't need to concern ourselves with the possibility that r is negative because it is the side length of a square and cannot be a negative number. Now that we know the side length, we can solve for the area of the square using A=l⋅w. A=l⋅wl=2r, w=2rA=2r⋅2rRearrange factorsA=2⋅2⋅r⋅ra⋅a=aA=2r2Solving the Inequality Now we can combine everything we have to create our inequality which we can solve for all possible values of r. (Area of Circle)(πr2)−(Area of Square)−(2r2)12345≥9(π−2)≥9(π−2) Finally, let's solve for r! πr2−2r2≥9(π−2) Solve for r Factor out r2r2(π−2)≥9(π−2)LHS/(π−2)≥RHS/(π−2)r2≥9LHS≥RHSr≥±3r>0 r≥3 Once again, because we are looking at lengths of a figure, we only need to concern ourselves with the positive value for the square root. Therefore, the shaded region will have an area greater than or equal to 9(π−2) whenever r≥3. | |

Exercises 38 To begin we will recall how to calculate an average. In general, an average can be calculated by adding all of the values and dividing by the number of values. For this exercise, we want to take the average of five race times. Thus, we can calculate this average by adding the given times and dividing by 5. Let x represent the time of the race that has not yet been run. We have then that Average=525.5+24.3+24.8+23.5+x. It is given that the runner wishes that the average race time is no less than 24 minutes. In other words, the ideal average is greater than or equal to 24. We can then write the inequality 24≤525.5+24.3+24.8+23.5+x. Solving this inequality will tell us the necessary time for the remaining race. 24≤525.5+24.3+24.8+23.5+xAdd terms24≤598.1+xLHS⋅5≥RHS⋅5120≤98.1+xLHS−98.1≥RHS−98.121.9≤xRearrange inequalityx≥21.9 The solution x≥21.9 means the runner must run the last rate with a time greater than or equal to 21.9 minutes. | |

Exercises 39 In this exercise, we are asked to determine the value of a that will make the solution to the given inequality all real numbers. When a solution is "all real numbers" it means that the equation or inequality is true for any possible value of the variable. To begin, we will simplify the given inequality by distributing on the left side and combining like terms on the right. a(x+3)<5x+15−xDistribute aax+3a<5x+15−xSubtract termsax+3a<4x+15 After simplifying, our inequality becomes ax+3a<4x+15. Notice that the coefficient to the x term on the right-hand side is 4. Let's see what happens if we let a equal 4. ax+3a<4x+15a=44⋅x+3⋅4<4x+15Multiply4x+12<4x+15LHS−4x<RHS−4x12<15 The statement 12<15 is always true. Thus, the value a=4 ensures that the solution of the inequality is all real numbers. | |

Exercises 40 Let's use the Addition and Subtraction Properties of Inequality to isolate x as much as possible. We want to avoid using multiplication and division because we don't know if a is a negative number or not. 3x+8+2ax≥3ax−4a Gather all terms containing x on the LHS LHS−3ax≥RHS−3ax3x+8−ax≥-4aLHS−8≥RHS−8 3x−ax≥-4a−8Factor out xx(3−a)≥-4a−8 We want to find a value for a such that the solution set is all real numbers. If we let a=3, the left-hand side of the inequality will become 0, eliminating the x-variable. x(3−a)⟹a=3x(3−3)=x⋅0=0 Let's substitute 3 for a in the given inequality and see if the solution set is all real numbers. 3x+8+2ax≥3ax−4aa=33x+8+2(3)x≥3(3)x−4(3) Solve for x Multiply3x+8+6x≥9x−12Add terms9x+8≥9x−12LHS−9x≥RHS−9x 8≥-12 ✓ We found that a=3 produced a true statement, where the value of the x-variable is irrelevant. Therefore, the solution set is all real numbers. | |

Exercises 41 To begin, we will determine which of the symbols, <,>,≤, or ≥, correspond with the statement. Less than or equal to is symbolically represented by ≤. Also, we can write 6 times a number y as: 6 times a number y.6 × y Notice that 6×y can be written as 6y. Writing a complete inequality, we have the following. 6 times a number y is less than or equal to 10.6y≤10 | |

Exercises 42 To begin, we will determine which of the symbols, <,>,≤, or ≥ correspond with the statement. Greater than is symbolically represented by >. We can write a number p plus 7 as A number p plus 7. p + 7 Writing a complete inequality, we have the following. A number p plus 7 is greater than 24.p + 7>24 | |

Exercises 43 The expression no more than can be written as the symbol ≤. So far we have: …≤… To find the expression on the left-hand side of the inequality sign, we substitute the information that comes before the phrase no more than. In this case, we are given the quotient of a number r and 7. This indicates that we need to be divide the variable q by the number 7. 7r≤… For the right-hand side of the inequality sign, we should substitute the information given that comes after no more than, which is 18. We get the following inequality. quotient of a number r and 7 no more than 187r ≤ 18 |

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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
- Quiz
- Solving Compound Inequalities
- Solving Absolute Value Inequalities
- Chapter Review
- Chapter Test
- Cumulative Assessment