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Exercises 1 We are able to combine terms when they are like terms. (-7)+1=-62x+3x=5x5m−4m=m Like terms can be combined because they have the same exact variable, or because they are constants and have no variable. | |

Exercises 2 There are two ways to approach this problem, you can use the Distributive Property or you can treat (4x−11) as a single quantity. Let's try both and compare our answers.Distributive Property We can solve this problem by first distributing the 2 to the terms inside the parentheses and simplifying. 2(4x−11)=10Distribute 22⋅4x−2⋅11=10Multiply8x−22=10 Now we can continue solving by using the Addition Property of Equality to isolate the 8x. 8x−22=10LHS+22=RHS+228x−22+22=10+22Add terms8x=32 Finally, we can find our answer by using the Division Property of Equality. 8x=32LHS/8=RHS/888x=832Calculate quotientx=4 Using this method, we found that x=4.Treating the parentheses as a single quantity This time, let's try dividing both sides of the equation by 2 first, instead of distributing. We are allowed to do this by the Division Property of Equality. 2(4x−11)=10LHS/2=RHS/222(4x−11)=210Calculate quotient4x−11=5 Next, we can use the Addition Property of Equality to isolate the 4x. 4x−11=5LHS+11=RHS+114x−11+11=5+11Add terms4x=16 And finally, we will once again use the Division Property of Equality to solve for x. 4x=16LHS/4=RHS/444x=416Calculate quotientx=4 Using this method, we also found that x=4.Conclusion Both methods came to the conclusion that x=4. Both methods are equally valid and will find the correct answer when used properly! | |

Exercises 3 To solve the equation, we first have to isolate 3w by subtracting 7 from both sides. 3w+7=19LHS−7=RHS−73w+7−7=19−7Subtract term3w=12 The next step is to isolate w on the left-hand side by dividing both sides of the equation by 3. 3w=12LHS/3=RHS/333w=312Calculate quotientw=4 Therefore, w=4 is the solution to the equation. We can check our solution by substituting it into the original equation. 3w+7=19w=43(4)+7=?19Multiply12+7=?19Add terms19=19 Since the left-hand side and right-hand side are equal, we have the correct solution. | |

Exercises 4 To solve the equation, we first isolate 2g by adding 13 to both sides. 2g−13=3LHS+13=RHS+132g−13+13=3+13Add terms2g=16 The next step is to isolate g on the left-hand side. By dividing both sides of the equation by 2, we can isolate g. 2g=16LHS/2=RHS/222g=216Calculate quotientg=8 Therefore, g=8 is the solution to the equation. We can check our solution by substituting it into the original equation and simplifying. 2g−13=3g=82(8)−13=?3Multiply16−13=?3Subtract term13=13 Since the left-hand side and right-hand side are equal, g=8 is the correct solution. | |

Exercises 5 To solve the equation, we first have to isolate -q. We can do this by subtracting 12 from both sides. 11=12−qLHS−12=RHS−1211−12=12−q−12Subtract terms-1=-q Now, we can continue isolating q by multiplying both sides of the equation by -1. -1=-qLHS⋅-1=RHS⋅-11=qRearrange equationq=1 Therefore, q=1 is the solution to the equation. We can check our solution by substituting it into the original equation. 11=12−qq=111=?12−1Subtract term11=11 Since the left-hand side and right-hand side are equal, we know the solution is correct. | |

Exercises 6 To solve the equation, we first have to isolate -m. We can do this by subtracting 7 from both sides. 10=7−mLHS−7=RHS−710−7=7−m−7Subtract terms3=-m Now, we can continue isolating m by multiplying both sides of the equation by -1. 3=-mChange signs-3=mRearrange equationm=-3 Therefore, m=-3 is the solution to the equation. We can check our solution by substituting it into the original equation. 10=7−mm=-310=?7−(-3)a−(-b)=a+b10=?7+3Add terms10=10 Since the left-hand side and right-hand side are equal, we know the solution is correct. | |

Exercises 7 To solve the equation, we first have to isolate the fraction by removing -3 from the right-hand side. Adding 3 to both sides of the equation will cancel out -3. 5=-4z−3LHS+3=RHS+35+3=-4z−3+3Add terms8=-4z The next step is to isolate z on the right-hand side. Division can be canceled out by multiplying both sides of the equation by the fraction's denominator. 8=-4zLHS⋅(-4)=RHS⋅(-4)8(-4)=-4z(-4)a(-b)=-a⋅b-32=-4z(-4)-4a⋅-4=a-32=zRearrange equationz=-32 Therefore, z=-32 is the solution to the equation. Finally, let's check our solution by substituting it into the original equation. 5=-4z−3z=-325=?-4-32−3-b-a=ba5=?432−3Calculate quotient5=?8−3Subtract term5=5 Because the left-hand side and right-hand side are equal, we know that our solution is correct. | |

Exercises 8 To solve the equation, we first have to isolate 3a by subtracting 4 from both sides. 3a+4=6LHS−4=RHS−43a+4−4=6−4Subtract term3a=2 The next step is to isolate a by multiplying both sides by 3. This will allow us to cancel out the fraction. 3a=2LHS⋅3=RHS⋅33a⋅3=2⋅33a⋅3=aa=2⋅3Multiplya=6 Therefore, a=6 is the solution to the equation. We can check our solution by substituting it into the original equation. 3a+4=6a=636+4=?6Calculate quotient2+4=?6Add terms6=6 Since the left-hand side and right-hand side are equal, the solution is correct. | |

Exercises 9 To solve the equation, we first have to isolate h+6. We can do this by multiplying both sides by 5. 5h+6=2LHS⋅5=RHS⋅55h+6⋅5=2⋅55a⋅5=ah+6=2⋅5Multiplyh+6=10 Now we continue isolating h by subtracting 6 from both sides. h+6=10LHS−6=RHS−6h+6−6=10−6Subtract termh=4 Therefore, h=4 is the solution to the equation. We can check our solution by substituting it into the original equation. 5h+6=2h=454+6=?2Add terms510=?2Calculate quotient2=2 Since the left-hand side and right-hand side are equal, we know the solution is correct. | |

Exercises 10 To begin solving this equation, let's first isolate d−8 by multiplying both sides by -2. -2d−8=12LHS⋅(-2)=RHS⋅(-2)-2d−8(-2)=12(-2)-2a⋅-2=ad−8=12(-2)a(-b)=-a⋅bd−8=-24LHS+8=RHS+8d−8+8=-24+8Add termsd=-16 Therefore, d=-16 is the solution to the equation. We can check if this solution is correct by substituting it into the original equation. -2d−8=12d=-16-2(-16)−8=?12Subtract term-2-24=?12-b-a=ba224=?12Calculate quotient12=12 Since both sides are equal, the solution is correct. | |

Exercises 11 We can solve the equation by combining like terms and then isolating y. 8y+3y=44Add terms11y=44LHS/11=RHS/111111y=1144Calculate quotienty=4 Let's check our answer by substituting y=4 into the original equation. 8y+3y=44y=48(4)+3(4)=?44 Simplify LHS Multiply32+12=?44Add terms 44=44 Since the left-hand side and right-hand side are equal, we know our solution is correct. | |

Exercises 12 We can solve the equation by combining like terms and then isolating n. 36=13n−4nSubtract terms36=9nLHS/9=RHS/9936=99nCalculate quotient4=nRearrange equationn=4 Let's check our answer by substituting n=4 into the original equation. 36=13n−4nn=436=?13(4)−4(4) Simplify LHS Multiply36=?52−16Subtract term 36=36 Since the left-hand side and right-hand side are equal, we know our solution is correct. | |

Exercises 13 We can solve the equation by combining like terms and then isolating v. 12v+10v+14=80Add terms22v+14=80LHS−14=RHS−1422v+14−14=80−14Subtract terms22v=66LHS/22=RHS/222222v=2266Calculate quotientv=3 Let's check our answer by substituting v=3 into the original equation. 12v+10v+14=80v=312(3)+10(3)+14=?80 Simplify LHS Multiply36+30+14=?80Add terms 80=80 Since the left-hand side and right-hand side are equal, we know our solution is correct. | |

Exercises 14 We can solve the equation by combining like terms and then isolating c. 6c−8−2c=-16Subtract terms4c−8=-16LHS+8=RHS+84c−8+8=-16+8Add terms4c=-8LHS/4=RHS/444c=4-8Put minus sign in front of fraction44c=-48Calculate quotientc=-2 Let's check our answer by substituting c=-2 into the original equation. 6c−8−2c=-16c=-26(-2)−8−2(-2)=?-16 Simplify LHS a(-b)=-a⋅b-12−8−2(-2)=?-16(-a)(-b)=a⋅b-12−8+4=?-16Subtract term-20+4=?-16Add terms -16=-16 Since the left-hand side and right-hand side are equal, we know our solution is correct. | |

Exercises 15 We are given a formula for the altitude (in feet) of a plane with two variables, a and t. In this formula, a is the altitude of the plane and t is the time after liftoff in minutes. To find how many minutes after liftoff the plane is at an altitude of 21000 feet, we substitute this altitude into the given formula and solve for t. a=3400t+600a=2100021000=3400t+600LHS−600=RHS−60021000−600=3400t+600−600Subtract terms20400=3400tLHS/3400=RHS/3400340020400=34003400tCalculate quotient6=tRearrange equationt=6 The plane reaches an altitude of 21,000 feet 6 minutes after liftoff. | |

Exercises 16 To write an equation for the number of hours of labor spent repairing the car, let's break down the given information into individual expressions and then combine these parts to form the equation. First, we know that the total bill came to $533. …=533 This cost represents the sum of the cost of parts and the cost of labor. We are given that the parts cost $265. labor cost+265=533 We are also told that the labor cost is $48 per hour, but not the number of hours spent on the car. If we call this unknown variable t, the expression for the labor cost is the product of t and the hourly cost. Finally, our equation becomes: 48t+265=553. Now we can solve for t using the Properties of Equality to isolate the variable. 48t+265=553LHS−265=RHS−26548t+265−265=553−265Subtract terms48t=288LHS/48=RHS/484848t=48288Use a calculatort=6 6 hours of labor were spent repairing the car. | |

Exercises 17 To begin solving this equation, we distribute the 4 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 4(z+5)=32Distribute 44z+20=32LHS−20=RHS−204z+20−20=32−20Subtract terms4z=12LHS/4=RHS/444z=412Calculate quotientz=3 Therefore, z=3 is the solution to the equation. We can check our answer by substituting it back into the original equation. 4(z+5)=32z=34(3+5)=?32Add terms4(8)=?32Multiply32=32 Because both sides are equal, we know that our answer is correct. | |

Exercises 18 To begin solving this equation, we distribute the -2 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. -2(4g−3)=30Distribute -2-2(4g)−(-2)(3)=30Multiply-8g+6=30LHS−6=RHS−6-8g+6−6=30−6Subtract term-8g=24LHS/-8=RHS/-8-8-8g=-824Put minus sign in front of fraction-8-8g=-824Calculate quotientg=-3 Therefore, g=-3 is the solution to the equation. We can check our answer by substituting it back into the original equation. -2(4g−3)=30g=-3-2(4(-3)−3)=?30a(-b)=-a⋅b-2(-12−3)=?30Subtract terms-2(-15)=?30-a(-b)=a⋅b30=30 Because both sides are equal, we know that our answer is correct. | |

Exercises 19 To begin solving this equation, we can distribute the 5 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 6+5(m+1)=26Distribute 56+5m+5=26Add terms11+5m=26LHS−11=RHS−1111+5m−11=26−11Subtract terms5m=15LHS/5=RHS/555m=515Calculate quotientm=3 Therefore, m=3 is the solution to the equation. We can check our answer by substituting it back into the original equation. 6+5(m+1)=26m=36+5(3+1)=?26Add terms6+5(4)=?26Multiply6+20=?26Add terms26=26 Because both sides are equal, we know that our answer is correct. | |

Exercises 20 To begin solving this equation, we can distribute the 2 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 5h+2(11−h)=-5Distribute 25h+11(2)−h(2)=-5Multiply5h+22−2h=-5Subtract term3h+22=-5LHS−22=RHS−223h+22−22=-5−22Subtract term3h=-27LHS/3=RHS/333h=3-27Calculate quotienth=-9 Therefore, h=-9 is the solution to the equation. We can check our answer by substituting it back into the original equation. 5h+2(11−h)=-5h=-95(-9)+2(11−(-9))=?-5a(-b)=-a⋅b-45+2(11−(-9))=?-5a−(-b)=a+b-45+2(11+9)=?-5Add terms-45+2(20)=?-5Multiply-45+40=?-5Add terms-5=-5 Because both sides are equal, we know that our answer is correct. | |

Exercises 21 To begin solving this equation, we can distribute the -3 to each term in the parentheses on the right-hand side. Then we can continue solving by using the Properties of Equality. 27=3c−3(6−2c)Distribute -327=3c+6(-3)−2c(-3)a(-b)=-a⋅b27=3c−18−2c(-3)-a(-b)=a⋅b27=3c−18+6cAdd terms27=9c−18LHS+18=RHS+1827+18=9c−18+18Add terms45=9cLHS/9=RHS/9945=99cCalculate quotient5=cRearrange equationc=5 Therefore, c=5 is the solution to the equation. We can check our answer by substituting it back into the original equation. 27=3c−3(6−2c)c=527=?3(5)−3(6−2(5))Multiply27=?15−3(6−10)Subtract term27=?15−3(-4)-a(-b)=a⋅b27=?15+12Add terms27=27 Because both sides are equal, we know that our answer is correct. | |

Exercises 22 To begin solving this equation, we begin by distributing the -5 to each term in the parentheses on the right-hand side and simplifying. -3=12y−5(2y−7)Distribute -5-3=12y+2y(-5)−7(-5)a(-b)=-a⋅b-3=12y−10y−7(-5)-a(-b)=a⋅b-3=12y−10y+35Subtract term-3=2y+35 We continue solving by using the Properties of Equality to isolate y. -3=2y+35LHS−35=RHS−35-3−35=2y+35−35Subtract term-38=2yLHS/2=RHS/22-38=22yCalculate quotient-19=yRearrange equationy=-19 Therefore, y=-19 is the solution to the equation. We can check our answer by substituting it back into the original equation. -3=12y−5(2y−7)y=-19-3=?12(-19)−5(2(-19)−7)a(-b)=-a⋅b-3=?-228−5(-38−7)Subtract terms-3=?-228−5(-45)-a(-b)=a⋅b-3=?-228+225Add terms-3=-3 Because both sides are equal, we know that our answer is correct. | |

Exercises 23 On the left-hand side of the equation, we have products that can be simplified using the Distributive Property. -3(3+x)+4(x−6)=-4Distribute -3 & 43(-3)+x(-3)+x(4)−6(4)=-4Multiply-9−3x+4x−24=-4Add and subtract terms-33+x=-4 Now, we can continue to solve using the Properties of Equality. -33+x=-4LHS+33=RHS+33-33+x+33=-4+33Add termsx=29 The solution to the equation is x=29. We can check our solution by substituting it into the original equation. -3(3+x)+4(x−6)=-4x=29-3(3+29)+4(29−6)=?-4Add and subtract terms-3(32)+4(23)=?-4Multiply-96+92=?-4Add terms-4=-4 Since the left-hand side is equal to the right-hand side, our solution is correct. | |

Exercises 24 On the left-hand side of the equation, we have products that can be simplified using the Distributive Property. 5(r+9)−2(1−r)=1Distribute 5 & -2r(5)+9(5)+1(-2)−r(-2)=1Multiply5r+45−2+2r=1Add and subtract terms7r+43=1 Now, we can continue to solve using the Properties of Equality. 7r+43=1LHS−43=RHS−437r+43−43=1−43Subtract term7r=-42LHS/7=RHS/777r=7-42Put minus sign in front of fraction77r=-742Calculate quotientr=-6 The solution to the equation is r=-6. We can check our solution by substituting it into the original equation. 5(r+9)−2(1−r)=1r=-65(-6+9)−2(1−(-6))=?1a−(-b)=a+b5(-6+9)−2(1+6)=?1Add terms5(3)−2(7)=?1Multiply15−14=?1Subtract term1=1 Since the left-hand side is equal to the right-hand side, our solution is correct. | |

Exercises 25 It is given that the sum of all interior angles of the triangle is 180∘. We can thus write an expression for the sum of the interior angles and set it equal to 180∘ to form an equation. 45+2k+k=180 Solving this equation for k will give us the measure of the unknown angle. 45+2k+k=180Add terms45+3k=180LHS−45=RHS−4545+3k−45=180−45Subtract terms3k=135LHS/3=RHS/333k=3135Calculate quotientk=45 The unknown variable k is equal to 45∘. The angle measures are therefore: k=45∘,2⋅k=90∘ and 45∘. | |

Exercises 26 It is given that the sum of all interior angles in the given quadrilateral is 360∘. We can thus write an equation that adds all of the angles together and is equal to 360∘: 2a+a+2a+a=360 Solving this equation for a will give us the measure of the unknown angle. 2a+a+2a+a=360Add terms6a=360LHS/6=RHS/666a=6360Calculate quotienta=60 The unknown variable is 60∘ so the angle measures are a=60∘,2⋅a=120∘,60∘ and 120∘. | |

Exercises 27 It is given that the sum of all interior angles in the given pentagon is 540∘. Therefore, we can write an equation that adds all of the angles together and is equal to 540∘. (2b−90)+23b+b+(b+45)+90=540 Solving this equation for b will give us the measure of the unknown angle. (2b−90)+23b+b+(b+45)+90=540Remove parentheses2b−90+23b+b+b+45+90=540 Simplify left-hand side Add terms4b+23b+45=540a=22⋅a28b+23b+45=540ca⋅b=ca⋅b28b+23b+45=540Add fractions28b+3b+45=540Add terms211b+45=540ca⋅b=ca⋅b 211b+45=540 Isolate b LHS−45=RHS−45211b+45−45=540−45Subtract terms211b=495LHS⋅2=RHS⋅2211b⋅2=495⋅2Multiply11b=990LHS/11=RHS/111111b=11990Calculate quotient b=90 The unknown variable is 90∘, so we can calculate the angle measures.Given AngleSubstitute b=90Simplify b9090∘ 23b23(90)135∘ b+4590+45135∘ 2b−902(90)−9090∘ 90—90∘ | |

Exercises 28 It is given that the sum of all interior angles in the given hexagon is 720∘. We can thus write an equation that adds all of the angles together and is equal to 720∘: x+120+100+120+(x+10)+120=720 Solving this equation for x will give us the measure of the unknown angle. x+120+100+120+(x+10)+120=720Remove parenthesesx+120+100+120+x+10+120=720Add terms2x+470=720LHS−470=RHS−4702x+470−470=720−470Subtract terms2x=250LHS/2=RHS/222x=2250Calculate quotientx=125 The unknown variable is 125∘ so the angle measures are x=125∘,120∘,100∘,120∘,x+10=135∘ and 120∘. | |

Exercises 29 The part of the sentence that says "is 75" tells us that something should be equal to 75. …=75 On the other side of the equation, the key phrase is "the sum of," indicating that we are adding two terms. The two terms being added are "twice a number" and "13." The phrase "a number" indicates a variable, which we will call n. 2n +13=… Finally, we can bring both sides of the equation together. The sum of twice a number and 13 is 75.2n +13=75 The next step is to find the number n. 2n+13=75LHS−13=RHS−132n+13−13=75−13Subtract terms2n=62LHS/2=RHS/222n=262Calculate quotientn=31 | |

Exercises 30 We are asked to write an equation and solve it.Writing the equation The last part of the sentence says "is -19", which tells us that something should be equal to -19. The equation so far is shown below. …=-19 The first part of the sentence tells us what should be written on the left-hand side of the equation. The phrasing "the difference of three times a number and 4" means that we should subtract 4 from three multiplied by a number, which can be called x. We can now write the full equation. The difference of three times a number and 4 is -193x − 4=-19Finding the number To find the number, we need to solve the equation we've just written. 3x−4=-19LHS+4=RHS+43x−4+4=-19+4Add terms3x=-15LHS/3=RHS/333x=3-15ca⋅b=ca⋅b33x=3-15Calculate quotientx=-5 | |

Exercises 31 Before we can solve for anything, we need to write an algebraic equation from the given sentence. The word "is" indicates where the equal sign is placed. In this case, we have "is -2." …=-2 The left-hand side of the equation can be broken down into two key phrases, "eight plus" and "the quotient of a number and 3." We can add the first of these to our equation. 8+…=-2 If we let the mystery number be n, then we have n÷3 which can also be written as 3n. We can now form our final equation. Eight plus the quotient of a number and 3 is -28+3n=-2 Finally, let's solve for n. 8+3n=-2LHS−8=RHS−88+3n−8=-2−8Subtract terms3n=-10LHS⋅3=RHS⋅33n⋅3=-10⋅33a⋅3=an=-10⋅3Multiplyn=-30 | |

Exercises 32 The part of the sentence that says "is ten," tells us that something should be equal to 10. So far we have …=10. The first part of the sentence tells us what should be on the left-hand side of the equation. The phrasing "the sum of twice a number and half the number" means that we should add a number multiplied by 2 and the same number multiplied by 21. Let's name the number n. We can now write the full equation. The sum of twice n and half of n is ten 2n + 21n=10 Now, let's solve the equation for n! 2n+21n=10LHS⋅2=RHS⋅22(2n+21n)=2⋅10Distribute 22⋅2n+2⋅21n=2⋅102a⋅2=a2⋅2n+n=2⋅10Multiply4n+n=20Add terms5n=20LHS/5=RHS/5n=4 The solution to the equation is n=4. | |

Exercises 33 The last part of the sentence, "...is -42," tells us that something should be equal to -42. …=-42 The part that comes before "is -42" tells us what should be on the left-hand side of the equation. "Six times the sum" indicates that two numbers should be added together before being multiplied by 6. 6(the sum of two numbers)=-42 The two numbers being added together are "a number and 15." If we call this unknown number n, we can form our final expression: Six times the sum of a number and 15 is -426(n+15)=-42 We can solve the equation by isolating n on the left-hand side. 6(n+15)=-42LHS/6=RHS/666(n+15)=6-42 Calculate Quotient 6⋅6a=an+15=6-42Put minus sign in front of fractionn+15=-642Calculate quotient n+15=-7LHS−15=RHS−15n+15−15=-7−15Subtract termsn=-22 The solution to the equation is n=-22. | |

Exercises 34 The last part of the sentence, "...is 12," tells us that something should be equal to 12. …=12 Before the "is 12," we are given "four times the difference." This indicates that a subtraction should be calculated before being multiplied by 4. 4(the difference of two numbers)=12 The numbers being subtracted are "a number and 7." If we call this unknown number n, we can form our final expression. Four times the difference of a number and 7 is 124(n−7)=12 We can solve the equation by isolating n on the left-hand side. 4(n−7)=12LHS/4=RHS/444(n−7)=412 Calculate Quotient 4⋅4a=an−7=412Calculate quotient n−7=3LHS+7=RHS+7n−7+7=3+7Add termsn=10 The solution to the equation is n=10. | |

Exercises 35 To calculate the total amount earned in a week at one job, we need to calculate the product of earnings per hour and the number of hours worked. At the gas station job, we earn $8.75 per hour and work 30 hours per week. 30⋅8.75 ⇔ 8.75(30)=gas station earnings We will need to apply this operation to the wage and number of hours worked at the landscaping job as well. We make $11 per hour at this job, but, we don't yet know the number of hours we have to work to accomplish our goal. Let's call this unknown value t. t⋅11 ⇔ 11t=landscaping earnings By adding these two expressions, we can form one master expression for calculating the total amount earned at our summer jobs each week. Our given goal is $400, so the only remaining unknown is the number of hours we need to work landscaping t. 11t+8.75(30)=400 Using the Properties of Equality, we will isolate t to find our solution. 11t+8.75(30)=400Multiply11t+262.5=400LHS−262.5=RHS−262.511t+262.5−262.5=400−262.5Subtract terms11t=137.5LHS/11=RHS/111111t=11137.5Use a calculatort=12.5 We must work 12.5 hours at the landscaping job each week to meet our weekly goal of $400. | |

Exercises 36 Using the formula for the area of rectangle, we can write an equation to solve for our missing length d. A=ℓ⋅w We are given that the area is A=210 square feet and the width is w=10 feet. The length is found by adding the length of the deep and shallow ends together, ℓ=d+9 feet. A=ℓ⋅w⇒210=(d+9)⋅10 Now we can solve for d. 210=(d+9)⋅10LHS/10=RHS/1021=d+9LHS−9=RHS−912=dRearrange equationd=12 A solution of d=12, in the context of this word problem, means that the deep end of the pool is 12 feet long. To check that our units are correct, we can think back to the formula for area of rectangle. 10 feet ×(12+9) feet =210 square feet | |

Exercises 37 To write an equation for the given situation, we really need to think about each factor that goes into the calculation of your total cost.Total Cost and Tip The total cost is $13.80 The $3 tip is a constant added after the tax and ordered items are already taken into consideration. …+3=13.80Food Purchased Let the unknown cost of a taco be t. We can write an expression for the ordered items, two tacos and a salad, using the fact that the salad is known to be $2.50. 2t+2.50Sales Tax To account for sales tax, we need to use the percent change in the total cost. The sales tax is 8% so the total cost of the meal is going to be 8% higher. 100%+8%=108% We can multiply the cost of the meal by this percent change. However, for simplicity, let's think about this in the decimal format rather than as a percentage, 108%=1.08.Final Equation Now, we can combine all of the components of the bill to have an equation that we can solve for the cost of tacos. sales tax ×( food cost )+ tip = total cost1.08(2t+2.50)+3=13.80 Now we can solve for t. 1.08(2t+2.50)+3=13.80LHS−3=RHS−31.08(2t+2.50)=10.80LHS/1.08=RHS/1.082t+2.50=10LHS−2.50=RHS−2.502t=7.50LHS/2=RHS/2t=3.75 Each taco costs $3.75. | |

Exercises 38 We are shown the solving of an equation and asked to justify each step. To do this, we need to think about what is changing on both sides of the equation with each step.Line 1 to Line 2 Between the first and second lines, 1 is added to both sides so that the …−1 can be eliminated from the left-hand side. This is allowed by the Subtraction Property of Equality. -21(5x−8)−1=6LHS+1=RHS+1-21(5x−8)−1+1=6+1Add terms-21(5x−8)=7Line 2 to Line 3 Between the second and third lines, the -21 is eliminated from the left-hand side. To cancel out a fraction, we multiply both sides by its reciprocal. The reciprocal of -21 is -2. This is allowed by the Multiplication Property of Equality. -21(5x−8)=7LHS⋅(-2)=RHS⋅(-2)-2⋅(-21)(5x−8)=-2⋅7Multiply5x−8=-14Line 3 to Line 4 The following step shows the −8 being removed from the left-hand side of the equation. The inverse operation of subtraction is addition. Therefore, 8 is being added to both sides of the equation. This step is using the Addition Property of Equality. 5x−8=-14LHS+8=RHS+85x−8+8=-14+8Add terms5x=-6Line 4 to Line 5 The final step to fully solving for x is to remove the multiplication by 5. The inverse operation for multiplication is division. This step is using the Division Property of Equality. 5x=-6LHS/5=RHS/555x=5-6Calculate quotientx=5-6Put minus sign in front of fractionx=-56Conclusion The steps that were taken to solve the equation were:Add 1 to both sides. Multiply both sides by -2. Add 8 to both sides. Divide both sides by 5. | |

Exercises 39 We are shown the solving of an equation and asked to justify each step. To do this, we need to think about what is changing on both sides of the equation with each step.Line 1 to Line 2 Between the first and second lines, 2 is distributed according to the Distributive Property to each term inside the parentheses. 2(x+3)+x=-9Distribute 22(x)+2(3)+x=-9Line 2 to Line 3 Between the second and third lines, the multiplication in 2(x) and 2(3) is simplified. 2(x)+2(3)+x=-9Multiply2x+6+x=-9Line 3 to Line 4 Between the third and fourth lines, the x-terms are combined. 2x+6+x=-9Add terms3x+6=-9Line 4 to Line 5 In the next step, 6 is subtracted from both sides of the equation. 3x+6=-9LHS−6=RHS−63x=-15Line 5 to Line 6 Finally, both sides of the equation are divided by 3. 3x=-15 LHS/3=RHS/3 LHS/3=RHS/333x=3-15Calculate quotientx=3-15Put minus sign in front of fractionx=-315Calculate quotient x=-5Conclusion The steps that were taken to solve the equation are:Distributive Property Simplify. Combine like terms. Subtract 6 from each side. Divide each side by 3. | |

Exercises 40 When solving an equation, we need to remember to watch our positives and negatives. The error in the given solution can be found in the very first step. When using the Distributive Property, the solution shows that a negative times a negative is a negative, but it should be positive. Shown: Correct: -2(-y)=-2y×-2(-y)=2y✓ With that mistake corrected, let's solve for y. -2(7−y)+4=-4Distribute -2-14+2y+4=-4Add terms-10+2y=-4LHS+10=RHS+102y=6LHS/2=RHS/2y=3 | |

Exercises 41 When solving an equation, we need to remember which operations are inverse. The error in the given solution can be found between the second and third lines. When removing the 41 from the left-hand side, the solution shows a division by 4. To remove the fraction, we should instead be multiplying by its reciprocal, 4. With that mistake corrected, let's solve for x. 41(x−2)+4=12LHS−4=RHS−441(x−2)=8LHS⋅4=RHS⋅4x−2=32LHS+2=RHS+2x=34 | |

Exercises 42 The perimeter P of a rectangle is found by adding the lengths of all sides of the figure. P=ℓ+ℓ+w+w⇒P=2ℓ+2w In the above formula, ℓ is the length and w is the width. In this exercise, we are told that ℓ=2w+6 and that P=228 feet. We can substitute these values into our formula and solve for w. Remember, we want to keep our units in mind as well. P=2ℓ+2wP=228, ℓ=2w+6228=2(2w+6)+2wDistribute 2228=4w+12+2w Solve for w Add terms228=6w+12LHS−12=RHS−12216=6wLHS/6=RHS/636=wRearrange equation w=36 A solution of w=36, in the context of this word problem, tells us that the width of the tennis court is 36 feet. Now we can solve for the length of the tennis court using the expression provided in the diagram. ℓ=2w+6w=36ℓ=2(36)+6Multiplyℓ=72+6Add termsℓ=78 The length of the tennis court is 78 feet. | |

Exercises 43 The perimeter P of a rectangle is found by adding together the lengths of all of its sides: P=ℓ+ℓ+w+w⇒P=2ℓ+2w, where ℓ is the length and w is the width. In this exercise, we are told that w=y, ℓ=811y, and P=190 inches. We can substitute these values into our formula and solve for y. P=2ℓ+2wSubstitute P=190,ℓ=811y,w=y190=2(811y)+2y Simplify terms a⋅cb=ca⋅b190=82⋅11y+2yba=b/2a/2 190=411y+2y Combine Like Terms Factor out y190=(411+2)ya=44⋅a190=(411+48)yAdd fractions 190=419yLHS⋅194=RHS⋅194190⋅194=y Multiply a⋅cb=ca⋅b19190⋅4=yba=b/19a/19110⋅4=y1a=a10⋅4=yMultiply 40=yRearrange equationy=40 A solution of y=40 tells us that the width of the flag is 40 inches. Now we can solve for the length of the flag using the expression provided in the diagram. ℓ=811yy=40ℓ=811⋅40 Multiply Multiply fractionsℓ=811⋅40ba=b/8a/8ℓ=111⋅5Multiplyℓ=1551a=a ℓ=55 The length of the flag is 55 inches. | |

Exercises 44 The perimeter P of a pentagon is found by adding together the lengths of all sides of the figure. This crossing sign has two sides equal to s, two equal to s+6 and one side 2s. P=s+s+(s+6)+(s+6)+2s In this exercise, we are told that P=102 inches. By solving for s we will be able to calculate all of the side lengths of the sign. P=s+s+(s+6)+(s+6)+2sP=102102=s+s+(s+6)+(s+6)+2s Simplify Terms Remove parentheses102=s+s+s+6+s+6+2sAdd terms 102=6s+12 Subtract terms LHS−12=RHS−12102−12=6s+12−12Subtract terms 90=6s LHS/6=RHS/6 LHS/6=RHS/6690=66sCalculate quotient 15=sRearrange equations=15 We found that s=15. Let's use this information to find the other side lengths.Side LengthSubstituteSimplify s 15 15 s 15 15 s+6 15+6 21 s+6 15+6 21 2s 2⋅15 21 | |

Exercises 45 There is often more than one way to solve an equation. Here we will look at two methods to solving the same equation and then look at the benefits of each one.Method 1 For the first method, we will use the Distributive Property to solve the equation. 2(4−8x)+6=-1LHS−6=RHS−62(4−8x)=-7Distribute 28−16x=-7 Solve for x LHS−8=RHS−8-16x=-15Change signs16x=15LHS/16=RHS/16 x=1615Method 2 For the second method, we will treat the parentheses as a single quantity. 2(4−8x)+6=-1LHS−6=RHS−62(4−8x)=-7LHS/2=RHS/24−8x=2-7 Solve for x LHS−4=RHS−4-8x=2-7−4Multiply by 22-8x=2-7−4⋅22a⋅cb=ca⋅b-8x=2-7−28Subtract fractions-8x=-215Change signs8x=215LHS/8=RHS/8x=215÷8ba/c=b⋅ca x=1615Benefits of each method In general, there are benefits to both methods. If the numbers are very large, or if the number to be distributed is a fraction, it may be easier to use the second method. In this exercise, the second method proved to be very difficult because 2 does not divide evenly into 7 which caused us to have to work with fractions the entire time. For this exercise, the easier method was definitely Method 1. We were able to avoid fractions until the very end and the numbers remained simple. | |

Exercises 46 To solve for the number of tickets purchased, which we can call t, we should first write an equation to represent the situation. We know that the total cost of the order is $220.70 and that the order is charged a flat fee of $5.90. Let's use these values to begin writing our equation. 220.70=5.90+… Next, let's think about the cost of the tickets themselves. For each ticket purchased, we need to pay the price of the ticket, $32.50, and the convenience charge $3.30. (32.50+3.30)t We can combine the total cost, flat fee, and ticket prices to have our equation. 220.70=5.90+(32.50+3.30)t Finally, let's solve for t. 220.70=5.90+(32.50+3.30)tAdd terms220.70=5.90+35.80tLHS−5.90=RHS−5.90214.80=35.80tLHS/35.80=RHS/35.806=tRearrange equationt=6 6 tickets were purchased through the online ticketing agency. | |

Exercises 47 We must have our units matching throughout, so before we can start solving this exercise, we need to rewrite the total amount of change into cents. $2.80⇒280 cents Since dimes d are worth 10 cents and quarters q are worth 25 cents, we can multiply the number of dimes by 10 and the number of quarters by 25 to know the total value of having that many of each coin. 280=10d+25q Now, if we have 280 cents made up of quarters and dimes, is it possible to have 8 more quarters than dimes? We can substitute q=d+8 into our equation to find out! 280=10d+25qq=d+8280=10d+25(d+8)Distribute 25280=10d+25d+200 Solve for d Add terms280=35d+200LHS−200=RHS−20080=35dLHS/35=RHS/352.2857=dRearrange equation d=2.2857 A solution of d=2.2857 means that, if we have 8 more quarters than dimes and a total of 280 cents, we would need to have approximately 2.3 dimes. It is not possible to have a partial number of dimes so our friend cannot be correct. Alternative solution Alternate Equation Another way to write the equation would be to keep the total in dollars, $2.80, and write the values of the coins in decimal format. Dimes are worth 0.10 dollars and quarters are worth 0.25 dollars. This is how the book chooses to solve the exercise. However, decimals are not as easy to work with so we chose to convert the dollars into cents instead. | |

Exercises 48 Before we begin, please know that there are many possible solutions to this exercise. The first thing we need to do is assign weights to the three remaining columns.Homework weight =0.20 Midterm Exam weight =0.30 Final Exam weight =0.30 Any combination would be fine, as long as the total adds to 1. 0.20+0.20+0.30+0.30=1 Our table now looks as shown below.ComponentStudent's scoreWeightScore × Weight Class Participation92%0.2092%×0.20=18.4% Homework95%0.20Midterm Exam88%0.30Final Exam0.30Total1The next step is to calculate the values in the "Score × Weight" column for the known exam scores. To find these values, we need to multiply the scores by the weight. This column tells us the portion of the final grade awarded by each component of grading.ComponentStudent's scoreWeightScore × Weight Class Participation92%0.2092%×0.20=18.4% Homework95%0.2095%×0.20=19% Midterm Exam88%0.3088%×0.30=26.4% Final Exam0.30Total1If the student wants to earn a 90% in the class, we can use this to solve for the portion of the final grade required from the Final Exam score. Right now the student has an overall 63.8% in the class. 18.4+19+26.4=63.8% To earn a 90%, they need 90−63.8=26.2% in the "Score × Weight" column.ComponentStudent's scoreWeightScore × Weight Class Participation92%0.2092%×0.20=18.4% Homework95%0.2095%×0.20=19% Midterm Exam88%0.3088%×0.30=26.4% Final Examp0.30p%×0.30=26.2% Total190%Finally, we can solve for the score required on the Final Exam to earn the 90% in the course. p%×0.30=26.2%LHS/0.30=RHS/0.30p%=87.3333% The students needs to earn at least 87.3ˉ% on the Final Exam.ComponentStudent's scoreWeightScore × Weight Class Participation92%0.2092%×0.20=18.4% Homework95%0.2095%×0.20=19% Midterm Exam88%0.3088%×0.30=26.4% Final Exam87.3ˉ%0.3087.3ˉ%×0.30=26.2% Total190% | |

Exercises 49 Even and odd integers alternate. If we begin at 0 and label the next few integers as even or odd, we have the following. 0even,1odd,2even,3odd,4even,5odd,6even Therefore, any integer can be represented by the expression 2n. To get to the next even integer we have to add 2. If the first of the three consecutive even integers is 2n, the other integers must be: First Integer: Second Integer: Third Integer: 2n2n+22n+4 To find three consecutive even integers whose sum is 54, we need to add together these three expressions and isolate n. 2n+(2n+2)+(2n+4)=54Remove parentheses2n+2n+2+2n+4=54Add terms6n+6=64 LHS−6=RHS−6 LHS−6=RHS−66n+6−6=54−6Subtract term 6n=48 LHS/6=RHS/6 LHS/6=RHS/666n=648Calculate quotient n=8 Since the first integer in our set of consecutive even integers is 2n, the first number is 2⋅8=16. Therefore, the second number is 18 and the third number is 20. | |

Exercises 50 aA "mean" is just one way of measuring the center of a set of numbers. The number of students who attended the first three meetings were 18, 21, and 17. We can tell that the center of those numbers is below 20 because 17 and 18 are both "more below" 20 than 21 is above it. 17+4=2118+3=2121−1=21 If the average number of students attending meetings is 20, then the fourth meeting must have had greater than 20 students.bTo estimate the number of students at the fourth meeting, knowing that the mean attendance is 20, we need to think about the number of students that would make 20 the central value. Let's look at the differences we have so far. ∣20−17∣=3 students∣20−18∣=2 students In total, between two of the meetings, 5 students less than 20 attended. ∣20−21∣=1 student At one meeting, 1 student more than 20 attended. If we want these values to be equal, the fourth meeting should have had 4 students over 20 attend, or 24 students. This will make the values below and above the average the same. 3+2=4+1cTo check our estimate, we can use formula for calculating the mean. Let the attendance during the four meetings be represented by y1, y2, y3, and y4. 4y1+y2+y3+y4=20 We can substitute in our known values and solve for the number of students at the final meeting. 4y1+y2+y3+y4=20Substitute values418+21+17+y4=20Add terms456+y4=20LHS⋅4=RHS⋅456+y4=80LHS−56=RHS−56y4=24 Our estimation was correct, 24 students attended the fourth meeting. | |

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