#### Solving Inequalities Using Multiplication or Division

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###### Exercises
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Exercises 1 Inequalities follow various Properties of Equality, just as with equations, that allow for using inverse operations to systematically isolate a variable. However, when you multiply or divide an inequality by a negative number, the inequality sign changes direction.First Inequality To isolate x, we need to divide both sides of the inequality by 2. 2x<-8LHS/2<RHS/2x<-4Second Inequality To isolate x, we need to divide both sides of the inequality by -2 and reverse the inequality sign. -2x<8Divide by -2 and flip inequality signx>-4
Exercises 2 The inequality sign needs to be reversed whenever we multiply or divide the inequality by a negative number. If we want to write an inequality that would be solved using the Division Property of Inequality and also requires a reversal of the symbol, we need a negative coefficient for our variable. An example of this would be: -3m≥9, which we would solve by dividing both sides of the inequality by -3. -3m≥9⇒m≤-3
Exercises 3 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. 4x<8LHS/4<RHS/4x<2 This inequality tells us that all values less than 2 will satisfy the inequality. Notice that x cannot equal 2, which we show with an open circle on the number line.
Exercises 4 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. 3y≤-9LHS/3≤RHS/3y≤-3 This inequality tells us that all values less than or equal to -3 will satisfy the inequality. Notice that y can equal -3, which we show with a closed circle on the number line.
Exercises 5 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 −20≤10nLHS/10≤RHS/10-2≤nRearrange inequalityn≥-2 This inequality tells us that all values greater than or equal to -2 will satisfy the inequality. Notice that n can equal -2, which we show with a closed circle on the number line.
Exercises 6 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that, when you divide or multiply by a negative number, you must reverse the inequality sign. 35<7tLHS/7<RHS/75<tRearrange inequalityt>5 This inequality tells us that all values greater than 5 will satisfy the inequality. Notice that t cannot equal 5, which we show with an open circle on the number line.
Exercises 7 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 2x​>-2LHS⋅2>RHS⋅2x>-4 This inequality tells us that all values greater than -4 will satisfy the inequality. Notice that x cannot equal -4, 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. 4a​<10.2LHS⋅4<RHS⋅4a<40.8 This inequality tells us that all values less than 40.8 will satisfy the inequality. Notice that a cannot equal 40.8, which we show with an open 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. 20≥54​wLHS⋅45​≥RHS⋅45​420⋅5​≥45​⋅54​wba​⋅ab​=125≥wRearrange inequalityw≤25 This inequality tells us that all values less than or equal to 25 will satisfy the inequality. Notice that w can equal 25, 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. -16≤38​tLHS⋅83​≤RHS⋅83​8-16⋅3​≤83​⋅38​tba​⋅ab​=1-6≤tRearrange inequalityt≥-6 This inequality tells us that all values greater than or equal to -6 will satisfy the inequality. Notice that t can equal -6, which we show with a closed circle on the number line.
Exercises 11 Inequalities can be solved in the same way as equations. By dividing both sides by -6, we can isolate t. Because we are dividing the inequality by a negative number, we have to reverse the inequality sign. -6t<12Divide by -6 and flip inequality signt>-2 This inequality tells us that all values greater than -2 will satisfy the inequality. Notice that t 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 dividing both sides by -9, we can isolate y. Because we are dividing the inequality by a negative number, we have to reverse the inequality sign. -9y>9Divide by -9 and flip inequality signy<-1 This inequality tells us that all values greater than -1 will satisfy the inequality. Notice that y cannot equal -1, which we show with an open circle on the number line.
Exercises 13 Inequalities can be solved in the same way as equations. By dividing both sides by -2, we can isolate z. Because we are dividing the inequality by a negative number, we have to reverse the inequality sign. -10≥-2zDivide by -2 and flip inequality sign5≤zRearrange inequalityz≥5 This inequality tells us that all values greater than or equal to 5 will satisfy the inequality. Notice that z can equal 5, which we show with a closed circle on the number line.
Exercises 14 Inequalities can be solved in the same way as equations. By dividing both sides by -3, we can isolate c. Because we are dividing the inequality by a negative number, we have to reverse the inequality sign. -15≤-3cDivide by -3 and flip inequality sign5≥cRearrange inequalityc≤5 This inequality tells us that all values less than or equal to 5 will satisfy the inequality. Notice that c can equal 5, 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. -3n​≥1Multiply by -3 and flip inequality signn≤-3 This inequality tells us that all values less than or equal to -3 will satisfy the inequality. Notice that n 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. -5w​≤16Multiply by -5 and flip inequality signw≥-90 This inequality tells us that all values greater than or equal to -90 will satisfy the inequality. Notice that w can equal -90, which we show with a closed circle on the number line.
Exercises 17 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. -8<-41​mb1​⋅a=ba​-8<-4m​Multiply by -4 and flip inequality sign32>mRearrange inequalitym<32 This inequality tells us that all values less than 32 will satisfy the inequality. Notice that m cannot equal 32, which we show with an opened circle on the number line.
Exercises 18 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>-32​yca​⋅b=ca⋅b​-6>-32⋅y​Multiply by -3 and flip inequality sign18<2yLHS/2<RHS/29<yRearrange inequalityy>9 This inequality tells us that all values greater than 9 will satisfy the inequality. Notice that y cannot equal 9, which we show with an open circle on the number line.
Exercises 19 To find the solution for this exercise, we will need to translate the given information into algebraic symbols and operations. There are three key pieces of information we will use to find our solution."\$12" "five fish" "price per fish" Our fish budget is \$12. This means, we can spend no more than \$12 on fish. The phrase "no more than" can be expressed as ≤. This symbol will be at the center of our inequality. …≤… The amount we spend on fish must be less than or equal to \$12. amount spent on fish≤12 We are also given that we will buy 5 fish. If we call the price per fish p, and multiply this number by the number of fish we will buy, we can form an expression for the total amount spent on fish. 5p≤12 To find the number of fish we can buy, we will solve the inequality by isolating the variable. 5p≤12LHS/5≤RHS/5p≤512​Use a calculatorp≤2.4 Since p is the price per fish, if we want to buy 5 fish on a \$12 budget, we can spend no more than \$2.40 per fish.
Exercises 20 If the temperature is decreasing by 8∘F each hour h, we can express the change in temperature algebraically as -8h. We want to know when the temperature has dropped by at least 36∘F, the other side of an inequality would then be -36. So far we have: -8h … -36. To decide the direction of the inequality sign, let's look at a number line.We want to know when the drop in temperature is greater than 36. This means that we need to go more negative which, in an inequality and on a number line, means we need to be less than -36. So our inequality becomes: -8h≤-36. We can solve this for h. -8h≤-36Divide by -8 and flip inequality signh≥4.5 Any time after 4.5 hours, the temperature will have dropped greater than 36∘F.
Exercises 21 Let's solve the inequality 36<3yLHS/3<RHS/312<yRearrange inequalityy>12 To solve this inequality using our calculator, we press Y=, and then proceed with writing it on the first row. Notice that we will use x as our variable as this is the only variable we are allowed to write using the calculator.To draw the inequality, we press GRAPH.We can't see the graph as the windows settings don't allow it. We have to extend it in the positive direction of the x-axis. We can do that by pressing WINDOW, and then make the necessary changes. Let's change Xmax from 10 to 20.By pressing GRAPH again, we draw the inequality using the new window settings.We have the correct answer.
Exercises 22 Let's solve the inequality 17v≥51LHS/17≥RHS/17v≥3 To solve this inequality using our calculator, we press Y=, and then proceed with writing it on the first row. Notice that we will use x as our variable as this is the only variable we are allowed to write using the calculator.To draw the inequality, we press GRAPH.
Exercises 23 Let's solve the inequality 2≤-92​xMultiply by (-1) and flip inequality sign-2≥92​xLHS⋅9≥RHS⋅9-18≥2xLHS/2≥RHS/2-9≥xRearrange inequalityx≤-9 To solve this inequality using our calculator, we press Y=, and then proceed with writing it on the first row. Notice that we will use x as our variable as this is the only variable we are allowed to write using the calculator.To draw the inequality, we press GRAPH.
Exercises 24 Let's solve the inequality 4>-4n​Put minus sign in front of fraction4>-4n​Multiply by (-1) and flip inequality sign-4<4n​LHS⋅4<RHS⋅4-16<nRearrange inequalityn>-16 To solve this inequality using our calculator, we press Y=, and then proceed with writing it on the first row. Notice that we will use x as our variable as this is the only variable we are allowed to write using the calculator.To draw the inequality, we press GRAPH.We can't see the graph as the windows settings don't allow it. We have to extended it in the negative direction of the x-axis. We can do that by pressing WINDOW, and then make the necessary changes. Let's change Xmin from -10 to -20.By pressing GRAPH again, we draw the inequality but using the new window settings.
Exercises 25 Let's solve the inequality 2x>43​LHS/2>RHS/2x>43​/2ba​/c=b⋅ca​x>83​ To solve this inequality using our calculator, we press Y=, and then proceed with writing it on the first row. Notice that we will use x as our variable as this is the only variable we are allowed to write using the calculator.To draw the inequality, we press GRAPH.
Exercises 26 Let's solve the inequality 1.1y<4.4LHS/1.1<RHS/1.1y<4 To solve this inequality using our calculator, we press Y=, and then proceed with writing it on the first row. Notice that we will use x as our variable as this is the only variable we are allowed to write using the calculator.To draw the inequality, we press GRAPH.
Exercises 27 The error in the shown solution occurs between the first and second lines. The person solving the inequality correctly multiplied by the reciprocal of the fraction to isolate x. However, the reciprocal is not negative so there was no need to change the direction of the symbol. If we fix this error, we can find the correct solution. -6>32​xLHS⋅23​>RHS⋅23​23​⋅(-6)>23​⋅32​xba​⋅ab​=123​⋅(-6)>xMultiply-218​>xCalculate quotient-9>xRearrange inequalityx<-9
Exercises 28 The error in the shown solution occurs between the first and second lines. The person solving the inequality correctly divided by -4 to isolate x. However, dividing by a negative number requires us to change the direction of the symbol. If we fix this error, we can find the correct solution. -4y≤-32Divide by -4 and flip inequality sign-4-4y​≥-4-32​-b-a​=ba​44y​≥432​Calculate quotienty≥8
Exercises 29 Your bedroom floor is a square that measures 14 feet by 14 feet. We can find the area A of the room using the length of the sides s and the formula for the area of a square. A=s2s=14 ftA=(14 ft)2(ab)m=ambmA=196 ft2 We have a maximum budget of \$700 to buy carpet for the room. If we let c represent the cost per square foot of carpet, we can write an inequality to represent this situation. 196c≤700​ We can now isolate c to find the maximum allowed cost per square foot of carpet by dividing both sides of the inequality by 196. 196c≤700LHS/196≤RHS/196c≤3.57142…c≤3.57 The cost per square foot is less than or equal to \$3.57. c≤3.57;dollars/ft2​
Exercises 30 Before we can match the inequalities with their corresponding graphs, we should solve them for x. It is not easy to visualize the graphs with the fractions.OptionInequalityIsolated x amx​<-1x<-m bmx​>1x>m cmx​<1x<m d-mx​<1x>-m Now we can easily match the graphs with the inequalities.Graph A Graph A shows a positive value for m with the shading to the right. This would correspond with the inequality x>m, option b.Graph B Graph B shows a positive value for m with the shading to the left. This would correspond with the inequality x<m, option c.Graph C Graph C shows a negative value for m with the shading to the left. This would correspond with the inequality x<-m, option a.Graph D Graph D shows a negative value for m with the shading to the right. This would correspond with the inequality x>-m, option d.
Exercises 31
Exercises 32 Let's begin by solving the given inequality. 4x​≤5LHS⋅4≤RHS⋅44x​⋅4≤4⋅5Multiplyx≤20 We are told that this solution set, x≤20, contains x=p. Therefore, we know that: p≤20. To create another inequality that also contains the solution x=p, we can start from the solution set we already found and do any arbitrary operation to both sides. For example, we can multiply by 4. x≤20LHS⋅4≤RHS⋅4x⋅4≤20⋅4Multiply4x≤80 This inequality has the same solution set which means it contains x=p.
Exercises 33 Let's call the number of pennies produced p. To produce 1 penny, it costs \$0.02. When producing p pennies the U.S. Mint pays: 0.02p. Since the U.S. Mint pays more than \$6 million in production, we can write the following inequality: 0.02p≥6 Let's solve this inequality. 0.02p≥6LHS/0.02=RHS/0.02p≥0.026​Use a calculatorp≥300 More than 300 million pennies are produced.
Exercises 34 If they are equivalent, we should be able to rewrite one as the other. The first inequality is solved for x, let's solve the second one for x as well. -3x≤-2Divide by -3 and flip inequality signx≥-3-2​-b-a​=ba​x≥32​ Let's compare the two inequalities: First inequality:Second inequality:​x≤32​x≥32​​ Since the inequality sign of the first inequality is ≤ and in the second one it is ≥, we know the inequalities aren't equivalent.
Exercises 35
Exercises 36 To begin, we will solve the given inequality. x4​≥2LHS⋅x≥RHS⋅x4≥2xLHS/2≥RHS/22≥xRearrange inequalityx≤2 The solution set x≤2 means that x can take on any value less than or equal to 2. However, notice that if x=0 the original inequality becomes 04​≥2. This possible value of x makes it so we divide by 0 which is not allowed. Thus, solving the given inequality leads to an error. center
Exercises 37 We want to determine the possible values for the circumference of a circle whose radius is greater than 5. To do so, we will consider the formula for the radius r in terms of the circumference C. r=2πC​​ Now, we will substitute 2πC​ for r in the inequality given in the diagram, r>5. r>5⇒2πC​>5​ Finally, we will use Inverse Operations to isolate C. 2πC​>5LHS⋅2π>RHS⋅2πC>10π A circle with a radius greater than 5 will have a circumference that's greater than 10π.
Exercises 38 To find the solution for this exercise, we will need to translate the given information into algebraic symbols and operations. There are three key pieces of information we will use to find our solution."less than" "18 mph" "45 minutes" The phrase "less than" can be expressed as <. This symbol will be at the center of our inequality. …<… Coming immediately after the "less than" is speed limit , 18 mph (notice that mph= h mi​). This should be expressed on the right-hand side of the inequality. …<h18 mi​ We are also informed that practice time lasts 45 minutes. It is necessary that we rewrite either the speed limit or the practice time so that they are both in the same unit of time. We will arbitrarily choose to convert 45 minutes to hours. 45mina=60 min60 min⋅a​60 min60 min⋅45 min​ba​=a⋅b1​60 min⋅60 min45 min​Calculate quotient60 min⋅43​1 h=60 min1 h⋅43​Multiply43​ h If we divide the distance by the practise time in hours , the product should be less than the right-hand side of the inequality.To find the distance that beginer can travel in the given time, we will solve the inequality by isolating the variable. LHS⋅43​ h<RHS⋅43​ hd<h18 mi​⋅43​ hMultiply fractionsd<4 h54 mi⋅ h​Calculate quotientd<13.5 mi The distance a beginner can travel in one hour going less than 18mph is 13.5 miles.
Exercises 39 The key phrase for writing the inequality is that "at most 24" of the 36 employees are full time. This means that at least 12 employees are part time, which gives us: …≥12. The left side of the inequality is going to tell us the portion p of the employees, out of the total 36, which are part time. We can write this as 36p and our inequality becomes: 36p≥12. We can solve this using the Division Property of Inequality. 36p≥12LHS/36≥RHS/36p≥3612​ba​=b/12a/12​p≥31​ We have that at least 31​ of the employees are part time. We can graph this solution set on a number line.Alternative solution info Alternative way of thinking We can skip a step if we think about the fact the number of part time employees p is at least 12 out of 36. Then we can skip straight to the inequality: p≥3612​.
Exercises 40 To solve the equation, we first have to isolate 5x by subtracting 3 from both sides. 5x+3=13LHS−3=RHS−35x+3−3=13−3Subtract term5x=10 The next step is to isolate x on the left-hand side by dividing both sides of the equation by 5. 5x=10LHS/5=RHS/555x​=510​Calculate quotientx=2 Therefore, x=2 is the solution to the equation. We can check our solution by substituting it into the original equation. 5x+3=13x=25(2)+3=?13Multiply10+3=?13Add terms13=13 Since the left-hand side and right-hand side are equal, we have the correct solution.
Exercises 41 To solve the equation, we have to isolate y on one side. We accomplish this by moving the variable terms to the left-hand side and the constant terms to the right-hand side. 21​y−8=-10LHS+8=RHS+821​y−8+8=-10+8Add terms21​y=-2 Finally, we need to multiply both sides by 2 to fully isolate y. 21​y=-2LHS⋅2=RHS⋅221​y⋅2=-2⋅2b1​⋅a=ba​22​y=-2⋅2Calculate quotient and producty=-4 Therefore, y=-4 is the solution to the equation. We can check our answer by substituting it back into the original equation. 21​y−8=-10y=-421​(-4)−8=-10Multiply-2−8=?-10Subtract terms-10=-10 Because both sides are equal, we know that our answer is correct.
Exercises 42 To solve the equation, we have to isolate n on one side. We accomplish this by moving the variable terms to the left-hand side and the constant terms to the right-hand side. -3n+2=2n−3LHS+3n=RHS+3n-3n+2+3n=2n−3+3nAdd terms2=5n−3LHS+3=RHS+32+3=5n−3+3Add terms5=5nRearrange equation5n=5 Finally, we need to divide both sides by 5 to fully isolate n. 5n=5LHS/5=RHS/555n​=55​Calculate quotientn=1 Therefore, n=1 is the solution to the equation. We can check our answer by substituting it back into the original equation. -3n+2=2n−3n=1-3⋅1+2=2⋅1−3Multiply-3+2=2−3Add and subtract terms-1=-1 Because both sides are equal, we know that our answer is correct.
Exercises 43 To solve the equation, we have to isolate z on one side. We accomplish this by moving the variable terms to the left-hand side and the constant terms to the right-hand side. 21​z+4=25​z−8LHS−21​z=RHS−21​z21​z+4−21​z=25​z−8−21​zSubtract terms4=24​z−8LHS+8=RHS+84+8=24​z−8+8Add terms12=24​zba​=b/2a/2​12=2zRearrange equation2z=12 Finally, we need to divide both sides by 2 to fully isolate z. 2z=12LHS/2=RHS/2z2z​=212​Calculate quotientz=6 Therefore, z=6 is the solution to the equation. We can check our answer by substituting it back into the original equation. 21​z+4=25​z−8z=621​⋅6+4=?25​⋅6−8Multiply3+4=?15−8Add and subtract terms7=7 Because both sides are equal, we know that our answer is correct.
Exercises 44 To begin, it is necessary that we rewrite one of the numbers so that they are both in the same form. We can write 0.8 as a percent or we can write 85% as a decimal. It does not matter which. We will arbitrarily choose to write 85% as a decimal. 85%a%=100a​10085​Calculate quotient0.85 In decimal form, 85% is equivalent to 0.85. Also 0.8 can be written as 0.80. Now its easier to compare them: 0.80<0.85.
Exercises 45 To begin, it is necessary that we rewrite the numbers so that they are both in the same form. We will write both in decimal form. This will make comparing them easier. To write 3016​ as a decimal, we must carry out the division: 3016​=0.53. To write 50% as a decimal, we must first write it as a fraction with a denominator of 100. Then, calculating the division gives us our number in decimal form. 50%a%=100a​10050​Calculate quotient0.50 Now that the values are in the same form comparing them is easier. We have 0.53>0.50.
Exercises 46 To begin, it is necessary that we rewrite one of the numbers so that they are both in the same form. We can write 0.12 as a percent or we can write 120% as a decimal. It does not matter which. We will arbitrarily choose to write 120% as a decimal. 120%a%=100a​100120​Calculate quotient1.2 In decimal form, 120% is equivalent to 1.2. Now we can compare both numbers: 1.2>0.12
Exercises 47 To begin, it is necessary that we rewrite the numbers so that they are both in the same form. We will write both in decimal form. This will make comparing them easier. To write 32​ as a decimal, we must carry out the division. 32​=0.67 To write 60% as a decimal, we must first write it as a fraction with a denominator of 100. Then, calculating the division gives us our number in decimal form. 60%a%=100a​10060​Calculate quotient0.60 Now that the values are in the same form, comparing them is easier. 0.60<0.67
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