Solving Equations with Variables on Both Sides

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Exercises 1 If the equation is an identity, it means that no matter what value we substitute for x, the equation will hold true. By trying to solve the equation, we can determine if the equation is an identity. -2(4−x)=2x+8Distribute -2-8+2x=2x+8Rearrange terms2x−8=2x+8LHS−2x=RHS−2x-8≠8 When trying to solve the equation we reached an inconsistency. Therefore, the given equation will never be true. It not only can't be an identity, it has no solution at all.
Exercises 2 To solve an equation we have to isolate the variable, x. Since we have a product of a factor and parentheses, the first thing we have to do is to distribute the 3 to eliminate the parentheses. 3(3x−8)=4x+6Distribute 33⋅3x−3⋅8=4x+6Multiply9x−24=4x+6 Having distributed 3 throughout the parentheses, we can continue solving the equation by isolating x on one side by conducting a series of inverse operations. 9x−24=4x+6LHS−4x=RHS−4x5x−24=6LHS+24=RHS+245x=30LHS/5=RHS/5x=6
Exercises 3 To solve an equation, we need to isolate the variable. In the given equation, notice that there are two variable terms. We will need to move them to one side to solve it. 15−2x=3xLHS+2x=RHS+2x15−2x+2x=3x+2xAdd terms15=5xLHS/5=RHS/5515​=55x​Calculate quotient3=xRearrange equationx=3 The solution to the equation is x=3. We can check our solution by substituting it into the original equation and simplifying. 15−2x=3xx=315−2⋅3=?3⋅3Multiply15−6=?9Subtract term9=9 The solution is correct.
Exercises 4 To solve an equation, we need to isolate the variable. In the given equation, notice that there are two variable terms. We will need to move them to one side to solve it. 26−4s=9sLHS+4s=RHS+4s26−4s+4s=9s+4sAdd terms26=13sLHS/13=RHS/131326​=1313s​Calculate quotient2=sRearrange equations=2 The solution to the equation is s=2. We can check our solution by substituting it into the original equation and simplifying. 26−4s=9ss=226−4⋅2=?9⋅2Multiply26−8=?18Subtract term18=18 The solution is correct.
Exercises 5 To solve the equation, we have to isolate p on one side. To begin, we will move the variable terms to one side. 5p−9=2p+12LHS−2p=RHS−2p5p−9−2p=2p+12−2pSubtract terms3p−9=12LHS+9=RHS+93p−9+9=12+9Add terms3p=21LHS/3=RHS/333p​=321​Calculate quotientp=7 The solution to the equation is p=7. By substituting p=7 into the equation and evaluating, we can check if the solution is correct. 5p−9=2p+12p=75⋅7−9=?2⋅7+12Multiply35−9=?14+12Add and subtract terms26=26 Because the left-hand and right-hands are equal, we know that the solution is correct.
Exercises 6 To solve the equation, we have to isolate g 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. 8g+10=35+3gLHS−3g=RHS−3g8g+10−3g=35+3g−3gSubtract terms5g+10=35LHS−10=RHS−105g+10−10=35−10Subtract terms5g=25 Finally, we need to divide both sides by 5 to fully isolate g. 5g=25LHS/5=RHS/555g​=525​Calculate quotientg=5 Therefore, g=5 is the solution to the equation. We can check our answer by substituting it back into the original equation. 8g+10=35+3gg=58⋅5+10=?35+3⋅5Multiply40+10=?35+15Add terms50=50 Because both sides are equal, we know that our answer is correct.
Exercises 7 To solve the equation, we have to isolate t 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. 5t+16=6−5tLHS+5t=RHS+5t5t+16+5t=6−5t+5tAdd terms10t+16=6LHS−16=RHS−1610t+16−16=6−16Subtract terms10t=-10 Finally, we need to divide both sides by 10 to fully isolate t. 10t=-10LHS/10=RHS/101010t​=10-10​Put minus sign in front of fraction1010t​=-1010​Calculate quotientt=-1 Therefore, t=-1 is the solution to the equation. We can check our answer by substituting it back into the original equation. 5t+16=6−5tt=-15(-1)+16=?6−5(-1)a(-b)=-a⋅b-5+16=?6−5(-1)-a(-b)=a⋅b-5+16=?6+5Add terms11=11 Because both sides are equal, we know that our answer is correct.
Exercises 8 To solve the equation, we have to isolate r on one side. We accomplish this by moving the variable terms to the right-hand side and the constant terms to the left-hand side. -3r+10=15r−8LHS+3r=RHS+3r-3r+10+3r=15r−8+3rAdd terms10=18r−8LHS+8=RHS+810+8=18r−8+8Add terms18=18r Finally, we need to divide both sides by 18 to fully isolate r. 18=18rLHS/18=RHS/181818​=1818r​Calculate quotient1=rRearrange equationr=1 Therefore, r=1 is the solution to the equation. We can check our answer by substituting it back into the original equation. -3r+10=15r−8r=1-3⋅1+10=?15⋅1−8Identity Property of Multiplication-3+10=?15−8Add and subtract terms7=7 Because both sides are equal, we know that our answer is correct.
Exercises 9 To solve the equation, we have to isolate x on one side. We accomplish this by moving the variable terms to the right-hand side and the constant terms to the left-hand side. 7+3x−12x=3x+1Subtract term7−9x=3x+1LHS+9x=RHS+9x7−9x+9x=3x+1+9xAdd terms7=12x+1LHS−1=RHS−17−1=12x+1−1Subtract terms6=12x Finally, we need to divide both sides by 12 to fully isolate x. 6=12xLHS/12=RHS/12126​=1212x​Calculate quotient21​=xRearrange equationx=21​ Therefore, x=21​ is the solution to the equation. We can check our answer by substituting it back into the original equation. 7+3x−12x=3x+1x=21​7+3⋅21​−12⋅21​=?3⋅21​+1a⋅b1​=ba​7+23​−212​=?23​+1Subtract fractions7−29​=?23​+1a=22⋅a​214​−29​=?23​+22​Add and subtract fractions25​=25​ Because both sides are equal, we know that our answer is correct.
Exercises 10 To solve the equation, we have to isolate w on one side. We accomplish this by moving the variable terms to the right-hand side and the constant terms to the left-hand side. w−2+2w=6+5wAdd terms3w−2=6+5wLHS−3w=RHS−3w3w−2−3w=6+5w−3wSubtract terms-2=6+2wLHS−6=RHS−6-2−6=6+2w−6Subtract terms-8=2w Finally, we need to divide both sides by 2 to fully isolate w. -8=2wLHS/2=RHS/22-8​=22w​Put minus sign in front of fraction-28​=22w​Calculate quotient-4=wRearrange equationw=-4 Therefore, w=-4 is the solution to the equation. We can check our answer by substituting it back into the original equation. w−2+2w=6+5ww=-4-4−2+2(-4)=?6+5(-4)a(-b)=-a⋅b-4−2−8=?6−20Subtract terms-14=-14 Because both sides are equal, we know that our answer is correct.
Exercises 11 On both sides of the equation, we have products that can be simplified using the Distributive Property. 10(g+5)=2(g+9)Distribute 10 & 2g⋅10+5⋅10=g⋅2+9⋅2Multiply10g+50=2g+18 Now, we can continue to solve using the Properties of Equality. 10g+50=2g+18LHS−2g=RHS−2g10g+50−2g=2g+18−2gSubtract term8g+50=18LHS−50=RHS−508g+50−50=18−50Subtract terms8g=-32LHS/8=RHS/888g​=8-32​Put minus sign in front of fraction88g​=-832​Calculate quotientg=-4 The solution to the equation is g=-4. We can check our solution by substituting it into the original equation. 10(g+5)=2(g+9)g=-410(-4+5)=?2(-4+9)Add terms10(1)=?2(5)Multiply10=10 Since the left-hand side is equal to the right-hand side, our solution is correct.
Exercises 12 On both sides of the equation, we have products that can be simplified using the Distributive Property. -9(t−2)=4(t−15)Distribute -9 & 4-9t+18=4t−60 Now, we can continue to solve using the Properties of Equality. -9t+18=4t−60LHS+9t=RHS+9t-9t+18+9t=4t−60+9tAdd terms18=13t−60LHS+60=RHS+6018+60=13t−60+60Add terms78=13tLHS/13=RHS/131378​=1313t​Calculate quotient6=tRearrange equationt=6 The solution to the equation is t=6. We can check our solution by substituting it into the original equation. -9(t−2)=4(t−15)t=6-9(6−2)=?4(6−15)Subtract terms-9(4)=?4(-9)Multiply-36=-36 Since the left-hand side is equal to the right-hand side, our solution is correct.
Exercises 13 There are a few ways to solve this exercise. We will show how to solve it by first clearing the fraction from left-hand side of the equation. 32​(3x+9)=-2(2x+6)LHS⋅3=RHS⋅33⋅32​(3x+9)=3(-2)(2x+6)3⋅3a​=a2(3x+9)=3(-2)(2x+6)a(-b)=-a⋅b2(3x+9)=-6(2x+6) Now that we have removed the fraction, the exercise becomes much easier. Let's continue by distributing the 2 and -6 to the terms inside the parentheses. 2(3x+9)=-6(2x+6)Distribute 2 & -63x⋅2+9⋅2=2x(-6)+6(-6)Multiply6x+18=2x(-6)+6(-6)a(-b)=-a⋅b6x+18=-12x−36 Now we can continue solving using the Properties of Equality. 6x+18=-12x−36LHS+12x=RHS+12x6x+18+12x=-12x−36+12xAdd terms18x+18=-36LHS−18=RHS−1818x+18−18=-36−18Subtract term18x=-54LHS/18=RHS/181818x​=18-54​Put minus sign in front of fraction1818x​=-1854​Calculate quotientx=-3 We can check our solution by substituting it into the original equation and simplifying. 32​(3x+9)=-2(2x+6)x=-332​(3(-3)+9)=?-2(2(-3)+6) Simplify both sides a(-b)=-a⋅b32​(-9+9)=?-2(-6+6)Add terms32​(0)=?-2(0)Zero Property of Multiplication 0=0 Because the left and right-hand sides are the same, we know that our solution is correct.
Exercises 14 There are a few ways to solve this exercise. We will show how to solve it by first clearing the fraction from right-hand side of the equation. 2(2t+4)=43​(24−8t)LHS⋅4=RHS⋅44⋅2(2t+4)=4⋅43​(24−8t)4⋅4a​=a4⋅2(2t+4)=3(24−8t)Multiply8(2t+4)=3(24−8t) Now that we have removed the fraction, the exercise becomes much easier. Let's continue by distributing the 8 and 3 to the terms inside the parentheses. 8(2t+4)=3(24−8t)Distribute 8 & 32t⋅8+4⋅8=24⋅3−8t⋅3Multiply16t+32=72−24t Now we can continue solving using the Properties of Equality. 16t+32=72−24tLHS+24t=RHS+24t16t+32+24t=72−24t+24tAdd terms40t+32=72LHS−32=RHS−3240t+32−32=72−32Subtract terms40t=40LHS/40=RHS/404040t​=4040​Calculate quotientt=1 We can check our solution by substituting it into the original equation and simplifying. 2(2t+4)=43​(24−8t)t=12(2⋅1+4)=?43​(24−8⋅1) Simplify both sides Identity Property of Multiplication2(2+4)=?43​(24−8)Add and subtract terms2(6)=?43​(16)ca​⋅b=ca⋅b​2(6)=?43⋅16​ba​=b/4a/4​2(6)=?13⋅4​Multiply12=?112​1a​=a 12=12 Because the left and right-hand sides are the same, we know that our solution is correct.
Exercises 15 On both sides of the equation, we have products that can be simplified using the Distributive Property. Let's do that first. 10(2y+2)−y=2(8y−8)Distribute 10 & 22y⋅10+2⋅10−y=8y⋅2−8⋅2Multiply20y+20−y=16y−16Subtract term19y+20=16y−16 Now, we can continue solving for y using inverse operations. We need to move all the constants to the right and the variable terms to the left. 19y+20=16y−16LHS−16y=RHS−16y19y+20−16y=16y−16−16ySubtract terms3y+20=-16LHS−20=RHS−203y+20−20=-16−20Subtract term3y=-36LHS/3=RHS/333y​=3-36​Put minus sign in front of fraction33y​=-336​Calculate quotienty=-12 The solution to the equation is y=-12. We can check this solution by substituting it into the original equation. 10(2y+2)−y=2(8y−8)y=-1210(2(-12)+2)−(-12)=?2(8(-12)−8) Simplify both sides a(-b)=-a⋅b10(-24+2)−(-12)=?2(-96−8)Add and subtract terms10(-22)−(-12)=?2(-104)a−(-b)=a+b10(-22)+12=?2(-104)a(-b)=-a⋅b-220+12=?-208Add terms -208=-208 Since both sides are equal, we know that the solution is correct.
Exercises 16 On both sides of the equation, we have products that can be simplified using the distributive property. Let's do that first. 2(4x+2)=4x−12(x−1)Distribute 2 & -124x⋅2+2⋅2=4x+x(-12)−1(-12)Multiply8x+4=4x+x(-12)−1(-12)a(-b)=-a⋅b8x+4=4x−12x+12Subtract term8x+4=-8x+12 Now, we can continue solving for a using inverse operations. We need to move all the constants to the right and the variable terms to the left. 8x+4=-8x+12LHS+8x=RHS+8x8x+4+8x=-8x+12+8xAdd terms16x+4=12LHS−4=RHS−416x+4−4=12−4Subtract term16x=8LHS/16=RHS/161616x​=168​Calculate quotientx=21​ The solution to the equation is x=21​. We can check this solution by substituting it into the original equation. 2(4x+2)=4x−12(x−1)x=21​2(4⋅21​+2)=?4⋅21​−12(21​−1) Simplify both sides a⋅b1​=ba​2(24​+2)=?24​−12(21​−1)Calculate quotient2(2+2)=?2−12(21​−1)a=22⋅a​2(2+2)=?2−12(21​−22​)Subtract fractions2(2+2)=?2−12(2-1​)a⋅b1​=ba​2(2+2)=?2−(2-12​)Calculate quotient2(2+2)=?2−(-6)a−(-b)=a+b2(2+2)=?2+6Add terms2(4)=?8Multiply 8=8 Since both sides are equal, we know that the solution is correct.
Exercises 17 We need to solve the given equation for h in order to determine when we will meet our friend. 50h=190−45hLHS+45h=RHS+45h95h=190LHS/95=RHS/95h=2 We will meet in 2 hours.
Exercises 18 We need to solve the given equation for r in order to determine how many movies must we rent to spend the same amount at each movie store. 1.5r+15=2.25rLHS−1.5r=RHS−1.5r1.5r+15−1.5r=2.25r−1.5rSubtract term15=0.75rLHS/0.75=RHS/0.750.7515​=0.750.75r​Use a calculator20=rRearrange equationr=20 We must rent 20 movies.
Exercises 19 To solve an equation, we need to isolate the variable. First we will move the variable terms to one side of the equation. 3t+4=12+3tLHS−3t=RHS−3t3t+4−3t=12+3t−3tSubtract term4≠12 The equation simplified to a false statement. This means that there are no solutions to this equation.
Exercises 20 To solve an equation, we need to isolate the variable. In the given equation, notice that there are two variable terms. We will need to move them to one side to solve it. 6d+8=14+3dLHS−3d=RHS−3d6d+8−3d=14+3d−3dSubtract term3d+8=14LHS−8=RHS−83d+8−8=14−8Subtract terms3d=6LHS/3=RHS/333d​=36​Calculate quotientd=2 The equation has only one solution, d=2.
Exercises 21 To solve an equation, we need to isolate the variable. Before we solve using the Properties of Equality, we will distribute the 2 on the left-hand side. 2(h+1)=5h−7Distribute 22h+2=5h−7 Now we can continue by moving the variable terms to one side of the equation and the constants to the other. 2h+2=5h−7LHS−2h=RHS−2h2h+2−2h=5h−7−2hSubtract terms2=3h−7LHS+7=RHS+72+7=3h−7+7Add terms9=3hLHS/3=RHS/339​=33h​Calculate quotient3=hRearrange equationh=3 There is one solution to the equation, h=3.
Exercises 22 To solve an equation, we need to isolate the variable. Before we can do that, we must distribute on the right-hand side. 12y+6=6(2y+1)Distribute 612y+6=12y+6LHS−12y=RHS−12y12y+6−12y=12y+6−12ySubtract terms6=6 The equation reduced to a statement that will always be true. This is called an identity. This means that the solution is all real numbers.
Exercises 23 To solve an equation, we need to isolate the variable. Before we can do that, we must distribute on both sides of the equation. 3(4g+6)=2(6g+9)Distribute 3 & 24g⋅3+6⋅3=6g⋅2+9⋅2Multiply12g+18=12g+18LHS−12g=RHS−12g12g+18−12g=12g+18−12gSubtract terms18=18 The equation reduced to a statement that will always be true. This is called an identity. This means that the solution is all real numbers.
Exercises 24 To solve an equation, we need to isolate the variable. Before we can do that, we must distribute on both sides of the equation. 5(1+2m)=21​(8+20m)Distribute 5 & 21​1⋅5+2m⋅5=8⋅21​+20m⋅21​Multiply5+10m=8⋅21​+20m⋅21​a⋅b1​=ba​5+10m=28​+220m​Calculate quotient5+10m=4+10m Now that we have finished with the distribution, we can move the variable terms to one side of the equation. 5+10m=4+10mLHS−10m=RHS−10m5+10m−10m=4+10m−10mSubtract term5≠4 The equation simplified to a false statement. This means that there are no solutions to this equation.
Exercises 25 To solve an equation, we first have to isolate the variable on one side. This involves using inverse operations and making sure that you do the same operation on both sides. First, let's eliminate the 3c from the right-hand side. To get rid of subtraction, we add. 5c−6=4−3cLHS+3c=RHS+3c8c−6=4 Already, we can see the error. Instead of adding 3c to both sides of the equation, the exercise shows that it is added to the right-hand side but subtracted from the left-hand side. You need to perform the same operation on both sides in an equation if it is to be solved correctly. Let's continue correctly solving for c. 8c−6=4LHS+6=RHS+68c=10LHS/8=RHS/8c=810​ba​=b/2a/2​c=45​
Exercises 26 Let's solve the given equation for y to spot the mistake. 6(2y+6)=4(9+3y)Distribute 612y+36=4(9+3y)Distribute 412y+36=36+12yLHS−36=RHS−3612y=12y From here, rather than dividing both sides of the equation by 12y, let's try only dividing by 12. 12y=12yLHS/12=RHS/12y=y This equation, y=y, is an identity and is true for all values of y. Therefore, it has infinitely many solutions. In the given solution, the algebraic manipulations are correct but the interpretation of the result is wrong.
Exercises 27 The total amount we will pay for each internet service has two components:The installation fee which is flat and one-time. The reoccurring monthly fee which remains constant. We can model these costs with algebraic expressions. Let's write the expressions for the total cost for each company separately.Company A The installation fee is given to be $60 and the price per month is $42.95. If we let the number of months be x, then we can write the total cost expression as: 42.95x+60.Company B The installation fee is given to be $25 and the price per month is $49.95. If we let the number of months be x, then we can write the total cost expression as: 49.95x+25When are costs the same? To determine the month when we will have paid the same total amount, we need to know when these two expressions will be equal to one another. We can do that by equating them and solving for x. 42.95x+60=49.95x+25LHS−49.95x=RHS−49.95x-7x+60=25LHS−60=RHS−60-7x=-35LHS/(-7)=RHS/(-7)x=5 After 5 months the total costs will be identical.
Exercises 28 To solve this problem, we can use a percent proportion: ba​=100p​ where a is the part of the whole, b is the whole, and p is the percent. After we have stuffed our face with a granola bar, we still need 48 more grams of protein to cover our daily recommended requirements. Also, if a granola bar covers 4% of your daily needs, then these remaining 48 grams represent 100%−4%=96% of your needs. In other words, we have a=48 and p=96. Let's substitute this into the percent proportion and solve for b. ba​=100p​a=48, p=96b48​=10096​ Solve for b Write as a decimalb48​=0.96LHS⋅b=RHS⋅b48=0.96bRearrange equation0.96b=48LHS/0.96=RHS/0.96 b=50 We need 50 gram of protein daily.
Exercises 29 Let's remove the parentheses through distribution and then isolate r on the left-hand side. 8(x+6)−10+r=3(x+12)+5xDistribute 88x+48−10+r=3(x+12)+5xDistribute 38x+48−10+r=3x+36+5xAdd and subtract terms8x+38+r=8x+36LHS−8x=RHS−8x38+r=36LHS−38=RHS−38r=-2 We have found that r=-2.
Exercises 30 To solve for r, let's begin by removing the parentheses through distribution. 4(x−3)−r+2x=5(3x−7)−9xDistribute 4 & 5x⋅4−3⋅4−r+2x=3x⋅5−7⋅5−9xMultiply4x−12−r+2x=15x−35−9x Now we can continue solving using the Properties of Equality. 4x−12−r+2x=15x−35−9xAdd and subtract terms6x−12−r=6x−35LHS−6x=RHS−6x-12−r=-35LHS+12=RHS+12-r=-23Change signsr=23 We have found that r=23.
Exercises 31 Let's start by writing equations that describe the surface area and volume for this particular cylinder.Surface area (A) The surface area of a cylinder can be calculated as: A=2πrh+2πr2, where r is the radius of each circular end and h is the height of the cylinder. In our case the radius is given as 2.5 cm and the height is given as x. Let's substitute these values into the formula and simplify. A=2πrh+2πr2r=2.5, h=xA=2π⋅2.5⋅x+2π⋅2.52 Simplify RHS Calculate powerA=2π⋅2.5⋅x+2π⋅6.25Multiply A=5πx+12.5π The cylinder's surface area is: A=5πx+12.5π.Volume (V) The volume of a cylinder can be found using the formula V=πr2h Let's substitute our values for radius and height into this formula. V=πr2hr=2.5, h=xV=π⋅2.52⋅x Simplify RHS Calculate powerV=π⋅6.25⋅xMultiply V=6.25πx The cylinder's volume is: V=6.25πx.Calculating area and volume To determine what value of x gives the same numerical value for surface area and volume, we simply equate their formulas and solve for x. V=AV=6.25πx, A=5πx+12.5π6.25πx=5πx+12.5πLHS−5πx=RHS−5πx1.25πx=12.5πLHS/1.25π=RHS/1.25πx=10 When the height x is equal to 10 cm, the numerical value is the same for the surface area and volume. Knowing x, we can find the surface area and volume by substituting x=10 into the two formulas: AV​=5π⋅10+12.5π=62.5π cm2=6.25π⋅10=62.5π cm3​ Notice that we could also type this into the calculator and get an answer of 62.5π≈196.5 as the numerical value for each of the measurements.
Exercises 32 Let's start by writing equations that describe the surface area and volume for this particular cylinder.Surface area (A) The surface area of a cylinder can be calculated as: A=2πrh+2πr2, where r is the radius of each circular end and h is the height of the cylinder. However, we have been given the diameter, which is 751​ ft. We can first rewrite this number as a decimal. 751​=7.2 Now we can find the radius using the fact that the radius is half of the diameter. 2d​=r⇒27.2​=3.6 The height is given as x. Let's substitute these values into the formula and simplify. A=2πrh+2πr2r=3.6, h=xA=2π⋅3.6⋅x+2π⋅3.62 Simplify RHS Calculate powerA=2π⋅3.6⋅x+2π⋅12.96Multiply A=7.2πx+25.92π The cylinder's surface area is: A=7.2πx+25.92π.Volume (V) The volume of a cylinder can be found using the formula V=πr2h. Let's substitute our values for radius and height into this formula. V=πr2hr=3.6, h=xV=π⋅3.62⋅x Simplify RHS Calculate powerV=π⋅12.96⋅xMultiply V=12.96πx The cylinder's volume is: V=12.96πx.Calculating area and volume To determine what value of x gives the same numerical value for surface area and volume, we simply equate their formulas and solve for x. V=AV=12.96πx, A=7.2πx+25.92π12.96πx=7.2πx+25.92πLHS−7.2πx=RHS−7.2πx5.76πx=25.92πLHS/5.76π=RHS/5.76πx=4.5 When the height x is equal to 4.5 ft, the numerical value is the same for the surface area and volume. Knowing x, we can find the surface area and volume by substituting x=4.5 into the two formulas. AV​=7.2π⋅4.5+25.92π=58.32π ft2=12.96π⋅4.5=58.32π cm3​ Notice that we could also type this into the calculator and get an answer of 58.32π≈183.2 as the numerical value for each of the measurements.
Exercises 33 To find the time at which the cheetah will catch the antelope, we need to use the distance formula d=rt, where r is the rate and t is the time. The cheetah will catch up to the antelope when they both have run the same distance. Thus, we can write the following equation. DistanceCheetah​=DistanceAntelope​​ The cheetah runs at a rate r of 90 feet per second. DistanceCheetah​=90t​ The antelope runs at a rate r of 60 feet per second and it has a 120 feet head start. Its distance can be represented by the following expression. DistanceAntelope​=120+60t​ Now, we can bring all of the information together and form an equation that can be solved for the time t. DistanceCheetah​=DistanceAntelope​Substitute expressions90t=120+60tLHS−60t=RHS−60t30t=120LHS/30=RHS/30t=4 Therefore, the cheetah needs 4 seconds to catch up to the antelope.
Exercises 34 You friend thinks that the antelope will still not be safe, even if he starts out 650 feet ahead of the cheetah. Let's find out if they are correct! If the antelope is running at 60 feet per second and has a 650 feet head start, rather than a 120 feet head start, on the cheetah running at 90 feet per second, our equation would slightly change from the previous exercise. Before we had: distancecheetah​90t​=distanceantelope​=120+60t​ Now we will have: distancecheetah​90t​=distanceantelope​=650+60t​ According to the exercise, the antelope will survive if he can keep running for longer than 20 seconds, because then the cheetah will burn out all of her energy and give up the chase. Let's solve for time t in the new equation to see if it is greater than 20 seconds. 90t=650+60tLHS−60t=RHS−60t30t=650LHS/30=RHS/30t=21.66666… In this situation, the cheetah would not catch the antelope until 21.6ˉ seconds had passed. Since 21.6ˉ>20 the antelope will narrowly escape his vicious hungry and furry predator. Also... your friend is wrong.
Exercises 35 To find the value for a that makes the equation an identity, we need to find a such that our solution is x=x. Let's first use the Distributive Property on the left-hand side and combine like terms on the right-hand side so that it is easier to see. a(2x+3)=9x+15+xDistribute a2ax+3a=9x+15+xAdd terms2ax+3a=10x+15 If the equation is an identity, the variable-term on the left-hand side has to be identical to the variable-term on the right-hand side. Same thing goes for the constants. We have: 3a2ax​=152⇔a=5=10x⇔a=5​ Therefore, a=5 gives us an identity. Let's check our solution. a(2x+3)=9x+15+xa=55(2x+3)=9x+15+x Simplify LHS & RHS Distribute 510x+15=9x+15+xAdd terms10x+15=10x+15LHS−15=RHS−1510x=10xLHS/10=RHS/10 x=x
Exercises 36 To find the value for a that makes the equation an identity, we need to find a such that our solution is x=x. This is easier to see if we first combine like terms and then use the Distributive Property in reverse on both sides of the equation. 8x−8+3ax=5ax−2aLHS−3ax=RHS−3ax8x−8=2ax−2aFactor out 88(x−1)=2ax−2aFactor out 2a8(x−1)=2a(x−1) Now we can find a such that 8=2a. 8=2a⇔a=4 Therefore, a=4 gives us an identity. Let's check our solution. 8x−8+3ax=5ax−2aa=48x−8+(3⋅4)x=(5⋅4)x−2⋅4 Simplify LHS & RHS Multiply8x−8+12x=20x−8LHS+8=RHS+88x+12x=20xAdd terms20x=20xLHS/20=RHS/20 x=x
Exercises 37 Let's call the smaller of the integers n. The second integer, which is consecutive to the first, must be n+1. The phrase "two times the greater" can be written as 2(n+1). This expression is equal to "9 less than three times the lesser." Since the lesser integer is n, this phrase can be written as 3n−9. We can equate these two expressions and solve for n. 2(n+1)=3n−9Distribute 22n+2=3n−9LHS−2n=RHS−2n2n+2−2n=3n−9−2nSubtract term2=n−9LHS+9=RHS+92+9=n−9+9Add terms11=nRearrange equationn=11 If n=11, then n+1 must be 12. These are our two consecutive integers.
Exercises 38 aEnrollment in Spanish and French classes will be equal when the two lines on the graph intersect. The y-axis shows the predicted enrollment numbers for the future of the classes and the x-axis shows the change in those numbers over time. The enrollment is predicted to be equal 6 years from now with approximately 300 students in both courses.bIf we look at the information about Spanish classes in the table, we know that there are 355 students this year and the average rate of change per year is a decrease of 9 students. If we use the variables from the graph, total enrollment is y and years from now is x, then we can express this algebraically as: ys​=355−9x. Similarly, if we look at the French classes, we know that there are 229 students this year and the average rate of change per year is an increase of 12 students. With the same variables, we can express this algebraically as: yf​=229+12x. Now, if we want to find the point at which these two functions intersect, when ys​=yf​, we can set the equations equal to one another: ys​355−9x​=yf​=229+12x.​ This is the equation shown in the exercise. It represents the point at which the enrollments for both courses are predicted to be the same. We can solve this for x to verify our answer in the previous exercise. 355−9x=229+12xLHS−229=RHS−229126−9x=12xLHS+9x=RHS+9x126=21xLHS/21=RHS/216=xRearrange equationx=6 The enrollments will be equal after 6 years time. This matches our previous result.
Exercises 39 aHere we can start with a result that would be considered to have "no solution" and then work backwards to create an equation. Remember to follow the Properties of Equality. Let's begin with 2≠8. 2≠8LHS+3x=RHS+3x2+3x≠8+3xLHS−2=RHS−23x≠6−3xLHS⋅2=RHS⋅26x≠2(6−3x) So, if we were to write the equation 6x=2(6−3x), we know it would have no solutions.bHere we can start with a result that would be considered to have "infinitely many solutions" and then work backwards to create an equation. Remember to follow the Properties of Equality. Let's begin with 8=8. 8=8LHS⋅x=RHS⋅x8x=8xLHS+4=RHS+48x+4=8x+4Factor out 48x+4=4(2x+1)LHS−4=RHS−48x=4(2x+1)−4 So, if we were to write the equation 8x=4(2x+1)−4 we know it would infinitely many solutions.
Exercises 40 Before we can draw a shape that has the same perimeter, we need to find the most simplified expression for the perimeter of the triangle using the general formula: PT​=s1​+s2​+s3​. Let's substitute the triangles sides into the formula and simplify the right-hand side. PT​=s1​+s2​+s3​Substitute expressionsPT​=(x+3)+(2x+1)+(3x)Remove parenthesesPT​=x+3+2x+1+3xAdd termsPT​=6x+4 The most simplified expression for the perimeter of the given triangle is 6x+4. Now we can choose any shape we like as long as we follow the rules for its proper proportions and the sides add to be 6x+4. Here we will look at one possible solution using a rectangle. Remember, for a rectangle we need to have the opposite sides with equal lengths.The perimeter for this rectangle is then: PR​=2(2x+2)+2x⇒PR​=6x+4, which is the same as our triangle.
Exercises 41 So that we may order the given list of numbers from least to greatest, we need to remember that ∣-a∣=a. Currently our list of numbers is as follows. 9,∣-4∣,-4,5,∣2∣ If we simplify the absolute values from the list, ∣-4∣=4and ∣2∣​=2,​ then our list becomes: 9,4,-4,5, and 2. We can use a number line to write these in order from least to greatest.
Exercises 42 So that we may order the given list of numbers from least to greatest, we need to remember that ∣-a∣=a. Currently our list of numbers is as follows. ∣-32∣,22,-16,-∣21∣,∣-10∣ If we simplify the absolute values from the list, ∣-32∣=32,-∣21∣=-21,and ∣-10∣​=10,​ then our list becomes: 32,22,-16,-21, and 10. We can use a number line to write these in order from least to greatest.
Exercises 43 So that we may order the given list of numbers from least to greatest, we need to remember that ∣-a∣=a. Currently our list of numbers is as follows. -18,∣-24∣,-19,∣-18∣,∣22∣ If we simplify the absolute values from the list, ∣-24∣=24,∣-18∣=18,and ∣22∣​=22​ then our list becomes: -18,24,-19,18, and 22. We can write these in order from least to greatest. -19,-18,18,22,24⇕-19,-18,∣-18∣,∣22∣,∣-24∣​
Exercises 44 So that we may order the given list of numbers from least to greatest, we need to remember that ∣-a∣=a. Currently our list of numbers is as follows. -∣-3∣,∣0∣,-1,∣2∣,-2 If we simplify the absolute values from the list, -∣-3∣=-3,∣0∣=0and ∣2∣​=2​ then our list becomes: -3,0,-1,2,-2. We can use a number line to write these in order from least to greatest.
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