Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition

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SOLUTIONSMANUALELEMENTARY ANDINTERMEDIATEALGEBRACONCEPTS ANDAPPLICATIONSSEVENTHEDITIONMarvin L. BittingerIndiana University Purdue University IndianapolisDavid J. EllenbogenCommunity College of VermontBarbara L. JohnsonIvy Tech Community College of Indiana

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CONTENTSCHAPTER 1INTRODUCTION TO ALGEBRAIC EXPRESSIONS.....................1CHAPTER 2EQUATIONS, INEQUALITIES, AND PROBLEM SOLVING .....33CHAPTER 3INTRODUCTION TO GRAPHING ................................................90CHAPTER 4POLYNOMIALS ...........................................................................151CHAPTER 5POLYNOMIALS AND FACTORING ..........................................198CHAPTER 6RATIONAL EXPRESSIONS AND EQUATIONS .......................247CHAPTER 7FUNCTIONS AND GRAPHS........................................................311CHAPTER 8SYSTEMS OF LINEAR EQUATIONS ANDPROBLEM SOLVING ...................................................................347CHAPTER 9INEQUALITIES AND PROBLEM SOLVING.............................422CHAPTER 10EXPONENTS AND RADICALS...................................................467CHAPTER 11QUADRATIC FUNCTIONS AND EQUATIONS ........................518CHAPTER 12EXPONENTIAL FUNCTIONS ANDLOGARITHMIC FUNCTIONS .....................................................592CHAPTER 13CONIC SECTIONS ........................................................................645CHAPTER 14SEQUENCES, SERIES, AND THEBINOMIAL THEOREM ................................................................684

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Chapter 1Introduction to Algebraic ExpressionsExercise Set 1.11.In the expression 4 +x, the number 4 is aconstant.2.In the expression 4 +x, the symbol + indicates theoperationof addition.3.Toevaluatean algebraic expression, we substitute anumber for each variable and carry out the operations.4.Anequationcontains an equal sign.5.101ndoes not contain an equals sign, so it is anexpression.6.321xcontains an equals sign, so it is an equation.7.259xcontains an equals sign, so it is anequation.8.5(2)xdoes not contain an equals sign, so it is anexpression.9.451acontains an equals sign, so it is anequation.10.45abdoes not contain an equals sign, so it is anexpression.11.238xycontains an equals sign, so it is anequation.12.(7)5r tdoes not contain an equals sign, so it is anexpression.13.Substitute 9 foraand multiply.55 945a14.11 77715.Substitute 4 forrand subtract.124816.82810t17.4559ab18.14132793319.214164444xy20.546921.5520355777pq22.99 65431818mq23.55 94531515zy24.2081262225.2(6 ft)(4 ft)(6)(4)(ft)(ft)24 ft , or 24 square feetbh26.27,00024 hr112527.22121 (5 cm)(6 cm)21=(5)(6)(cm)(cm)25=6 cm215 cm , or 15 square centimetersAbh28.(a)3(30 sec)90 sec;(b)3(90 sec)270 sec;(c)3(2 min)6 min29.2(67 ft)(12 ft)=(67)(12)(ft)(ft)=804 ft , or 804 square feetAbh30.80.57114ha31.Letrrepresent Ron’s age. Then we have5,ror 5.r32.4 , or4aa33.6 , or6bb34.Letprepresent Patti’s weight. Then we have7,por7.p35.9c36.4d37.6, or6qq38.11, or11zz39.pt

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2Chapter 1:Introduction to Algebraic Expressions40.nm41.yx42.Letarepresent Kurt’s age. Then we have2.a43., orxxww44.Letsandtrepresent the numbers. Then we have,stor.st45.Letlandhrepresent the box’s length and height,respectively. Then we have, or.lhhl46., ordffd47.9 2, or 29mm48.Letarepresent Abby’s speed andwrepresent thewind speed. Then we have2.aw49.Letyrepresent “some number.” Then we have113, or13.44yy50.Letnrepresent the number; 104n51.Letaandbrepresent the two numbers. Then we have5().ab52.Letxandyrepresent the numbers. Then we have1 (), or.33xyxy53.Letwrepresent the number of women attending. Thenwe have 64% ofw, or 0.64w.54.Letyrepresent “a number.” Then we have 38% ofy,or 0.38y.55.Letxrepresent the number.Translating:What numberadded to73is201?73201x73201x56.Letxrepresent the number.71596x57.Letxrepresent the number.Rewording: 42 times what number is 2352?Translating: 422352x422352x58.Letxrepresent the number.345987x59.Letsrepresent the number of unoccupied squares.The numberofRewording:added to 19 is 64.unoccupiedsquaresTranslating:1964s 1964s60.Lethrepresent the number of hours the carpenterworked.$35$3640h61.Letxrepresent the total amount of waste generated, inmillions of tons.the total87 millionRewording:34.5%ofamountistons.of wasteTranslating:34.5%87x34.5%87, or 0.34587xx62.Lettrepresent the length of the average commute inFort Bliss, in minutes.51.259.8t63.We look for a pattern in the data. We try subtracting.8351165945127510551385The amount is the same, 5, for each pair of numbers.Letarepresent the age of the child andfrepresent thenumber of grams of dietary fiber.We reword and translate as follows:dietarychild'sRewording:isadded to5fiberageTranslating:5fa5fa64.Letcrepresent the cost of tuition andhrepresent thehours of classes.100ch65.We look for a pattern in the data. We try subtracting.6.594.172.427.184.762.428.766.342.42The amount is the same, 2.42, for each pair ofnumbers. Letnrepresent the nonmachinable cost andnrepresent the machinable cost.We reword and translate as follows:nonmachinablemachinableisadded to 2.42costcost2.42nm2.42nm

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Exercise Set 1.1366.Letrrepresent the amount received andsrepresent theamount spent.3rs67.We look for a pattern in the data. We try dividing.10,00030,00010,00010,0001320,00040,00010,00010,00024The amount is the same, 10,000, for each pair ofnumbers. Letvrepresent the number of vehicle milestraveled anddrepresent the number drivers.We reword and translate as follows:number ofnumberRewording:is 10,000 timesmiles traveledof driversTranslating:10,000vd10,000vd68.Letwrepresent the depth of water andsrepresent thedepth of snow.10ws69.The sum of two numbersmandnis,mnand twicethe sum is 2().mnChoice (f) is the correct answer.70.Five less than a numberxis5.xIf this expressionis equal to 12, we have the equation512.xChoice (h) is the correct answer.71.Twelve more than a numbertis12.tIf thisexpression is equal to 5, we have the equation125.tChoice (d) is the correct answer.72.The product of two numbersaandbis.a bHalf ofthis product is12.a bChoice (c) is the correctanswer.73.The sum of a numbertand 5 is5,tand 3 times thesum is 3(5).tChoice (g) is the correct answer.74.The sum of two numbersxandyis,xyand twicethis sum is 2().xyIf this expression is equal to 48,we have the equation 2()48.xyChoice (b) is thecorrect answer.75.The product of two numbersaandbisab, and1 less than this product is1.abIf this expression isequal to 48, we have the equation148.abChoice(e) is the correct answer.76.The quotient of two numbersxandyis,xyand6 more than this quotient is6.xyChoice (a) is thecorrect answer.77.Writing Exercise. Avariableis a letter that is used tostand for any number chosen from a set of numbers.Analgebraic expressionis an expression that consistsof variables, constants, operation signs, and/orgrouping symbols. Avariable expressionis analgebraic expression that contains a variable. Anequationis a number sentence with the verb =. Thesymbol = is used to indicate that the algebraicexpressions on either side of the symbol represent thesame number.78.Writing Exercise. To evaluate an algebraic expressionmeans to find the value of the expression when itsvariables are given values.79.Writing Exercise. No; for a square with sides, thearea is given by.As sThe area of a square withside 2sis given by (2 )(2 )4sss s42.AA80.Writing Exercise. Answers may vary.Juliet was born in 2006. Find her age in 2014.81.Area of sign:212(3 ft)(2.5 ft)3.75 ftACost of sign: $120(3.75)$45082.The shaded area is the area of a rectangle withdimensions 20 cm by 10 cm less the area of a trianglewith base 20 cm – 4 cm – 5 cm, or 11 cm, and height7.5 cm. We perform the computation2221(20 cm)(10 cm)(11 cm)(7.5)2200 cm41.25 cm158.75 cm , or 158.75 square centimeters83.Whenxis twicey, thenyis one-halfx, so126.2y12662333xy84.6,22 612612189222xyxxy85.Whenais twiceb, thenbis one-halfa, so1628.b168246444ab86.Whenais three timesb, thenbis one-thirda, so1836.b186124333ab87.The next whole number is one more than3 :w314ww88.The preceding odd number is 2 less than2 :d22dd89., or 22lwlwlw90., or 4sssss91.Iftis Molly’s race time, then Dion’s race time is3tand Ellie’s race time is358.tt

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4Chapter 1:Introduction to Algebraic Expressions92.27, or9aa93.Writing Exercise. Yes; the area of a triangle with baseband heighthis given by12.AbhThe area of atriangle with baseband height 2his given by1122(2 )22.bhbhAExercise Set 1.21.Equivalentexpressions represent the same number.2.Changing the order of multiplication does not affectthe answer. This is an example of acommutativelaw.3.The result of addition is called asum.4.The numbers in a product are calledfactors.5.Commutative6.Associative7.Distributive8.Commutative9.Commutative10.Distributive11.11Changing the ordert12.2a13.84x14.cab15.39yx16.73ba17.5(1)a18.9(5)x19.7Changing the orderx20.yx21.ts22.13m23.5ba24.3xy25.(1)5a26.(5)9x27.(8)xy28.5()mr29.()7uv30.(2)xy31.()abcd32.()mnpr33.10()xy34.4()uv35.(2 )a b36.(9 7)r37.(3 2)()ab38.(5 )(2)xy39.(6)()6(6)ststts40.7()()7(7)7()(7)(7)vwvwvwvwvwvwAnswers may vary.41.(17 )17()Using the associative lawa bab(17 )(17 )Using the commutative law( 17)Using the commutative law againa bbab aAnswers may vary.42.(3 )(3 )3()(3 )(3)(3 )xyy xyxxyxyx y43.(1)2(1)2 Commutative law(12) Associative law3Simplifyingxxxx44.(2 )44(2 )Commutative law(4 2)Associative law8Simplifyingaaaa45.(3)7(3 7) Associative law21Simplifying21Commutative lawmmmm46.4(9)(49)Associative law13Simplifying13Commutative lawxxxx47.1xxyzThe term are separated by plus signs. They arex,xyz,and 1.48.9, 17 ,aabc49.253aabbThe terms are separated by plus signs. They are 2a,,3aband 5b.

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Exercise Set 1.2550.43, 20,axyb51.4 , 4xy52.14, 2y53.55nnThe factors are 5 andn.54.uvu vThe factors areuandv.55.3()3 ()xyxyThe factors are 3 and ().xy56.()12() 12ababThe factors are ()aband 12.57.The factors are 7,aandb.58.The factors arem,nand 2.59.()()() ()abxyabxyThe factors are () and ().abxy60.(3)()(3) ()abcabcThe factors are (3) and ().abc61.2(15)22 15230xxx62.315x63.4(1)4 1444aaa64.77y65.10(96)10 910 69060xxx66.5463m67.5(23 )55 25 351015rtrtrt68.203212xp69.()2(2)(2)22ababab70.714x71.(2)5(5)(5)2(5)5510xyxyxy72.1266ab73.22Using the distributive law2()The common factor is 2.ababCheck: 2()2222ababab74.555()yzyzCheck: 5()55yzyz75.777 17The common factor is 7.7(1)Using the distributive lawyyyCheck: 7(1)7 1777yyy76.131313(1)xxCheck: 13(1)1313xx77.3222 162 12(161)xxxCheck: 2(161)2 162 1322xxx78.2055(41)aaCheck: 5(41)205aa79.5101555 25 35(23 )xyxyxyCheck: 5(23 )55 25 351015xyxyxy80.32763(192 )bcbcCheck: 3(192 )3276bcbc81.73577 57(5 )abababCheck: 7(5 )77 5735ababab82.3243(8 )xyxyCheck: 3(8 )324xyxy83.44112211 41111 211(42 )xyzxyzxyzCheck: 11(42 )11 41111 2441122xyzxyzxyz84.145677(281)ababCheck: 7(281)14567abab85.3(2)3(2) Commutative law of addition33 2 Distributive law36Multiplyingxxxx86.(4)55(2) Commutative law of multiplication55 4 Distributive law520Multiplyingyyyy87.7(23 )7(2 )7(3 ) Distributive law(7 2)(7 3)Associative law of multiplication1421Multiplyingxyxyxyxy88.(42)88(42)Commutative law of multiplication8(4 )8(2) Distributive law(8 4)8(2) Associative law of multiplication3216Multiplyingaaaaa89.Writing Exercise. No; in general, when subtracting,the result depends on the order in which the operationis performed.90.Writing Exercise. No; in general, when dividing, theresult depends on the grouping.91.Writing Exercise. Terms are added; factors aremultiplied. Examples will vary.

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6Chapter 1:Introduction to Algebraic Expressions92.Writing Exercise.2(34 )2(3 )2(4 )Distributive law(2 3)(2 4)Associative law68xyxyxyxy93.2(1)322 1)3Distributive law= 223Multiplying223Associative law of addition232Commutative law of addition232Associative law of addition(23)2Distributive lxxxxxxxxxxxxxaw52Adding52xx94.124(5)(4 3)4(5) Writing 12 as a product4(3 )4(5)Associative law of multiplication4(3 )4( )4(5) Distributive law4(35)Distributive lawababababab95.The expressions are equivalent by the distributive law.84()8444(2)ababab96.The expressions are not equivalent.Let2 and3.abThen we have:5(2 3)5 630, but5 2 5 310 5 350 3150.97.The expressions are not equivalent.Let1.mThen we have:7773 11, but3331 3737.798.The expressions are equivalent by the commutativelaw of multiplication and the distributive law.()55()5()5 ()rtstrtstt rst rs99.The expressions are not equivalent.Let1 and0.xyThen we have:30 01 1501515, but5[2(13 0)]5[2(1)]5 210.100.The expressions are equivalent by the commutativeand associative laws of multiplication and thedistributive law.[ (23 )]55[ (23 )]5 (23 )1015cbcbcbcbc101.Writing Exercise.3(2)3(20)3 26x;6606xThe result indicates that 3(2)xand 6xare equivalent when0.x(By the distributive law,we know they are not equivalent for all values of.x)102.Writing Exercise.15405(38)15 44060401005(3 48)5(128)5 20100xxAlthough the expressions 1540xand 5(38)xareequivalent when4x, this result does not guaranteethat the factorization is correct. (See Exercise 101.)103. Answers may vary.a.Letxrepresent the number of overtime hoursworked in one week.Aidan: 10(1.540)xBeth: 10 4010(1.5)xCody: 15400xb.We simplify each expression.Aiden: 10(1.540)10(1.5 )10(40)15400xxxBeth: 10 4010(1.5)40015xxCody: 15400xAll expressions are equivalent to 15400.xExercise Set 1.31.The top number in a fraction is called thenumerator.2.Aprimenumber has exactly two different factors.3.To divide two fractions, multiply by thereciprocalofthe divisor.4.We need a common denominator in order toaddfractions.5.Since 355 7, choice (b) is correct.6.Since 603 20, choice (c) is correct.7.Since 65 is an odd number and has more than twodifferent factors, choice (d) is correct.8.The only even prime number is 2, so choice (a) iscorrect.9.9 is composite because it has more than two differentfactors. They are 1, 3, and 9.10.15 is composite because it has more than two differentfactors. They are 1, 3, 5, and 15.11.41 is prime because it has only two different factors,41 and 1.12.49 is composite because it has more than two differentfactors. They are 1, 7, and 49.13.77 is composite because it has more than two differentfactors. They are 1, 7, 11, and 77.14.37 is prime because it has only two different factors,37 and 1.15.2 is prime because it has only two different factors, 2and 1.

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Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition - Page 10 preview image

Exercise Set 1.3716.1 is not prime because it does not have two differentfactors. It is not composite because it does not havemore than two different factors. Thus 1 is neitherprime nor composite.17.The terms “prime” and “composite” apply only tonatural numbers. Since 0 is not a natural number, it isneither prime nor composite.18.16 is composite because it has more than two factors.They are 1, 2, 4, 8, and 16.19.Factorizations:1 50, 2 25, 5 10List all of the factors of 50:1, 2, 5, 10, 25, 5020.Factorizations: 1 70, 2 35, 5 14, 7 10Factors: 1, 2, 5, 7, 10, 14, 35, 7021.Factorizations:1 42,2 21,3 14,6 7List all of the factors of 42:1, 2, 3, 6, 7, 14, 21, 4222.Factorizations:1 60, 2 30, 3 20, 4 15, 5 12, 6 10Factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 6023.393 1324.2 1725.We begin factoring 30 in any way that we can andcontinue factoring until each factor is prime.302 152 3 526.2 7 727.We begin by factoring 27 in any way that we can andcontinue factoring until each factor is prime.273 93 3 328.2 3 3 329.We begin by factoring 150 in any way that we canand continue factoring until each factor is prime.1502 752 3 252 3 5 530.2 2 2 731.31 has exactly two different factors, 31 and 1. Thus,31 is prime.32.1802 2 3 3 533.2102 1052 3 352 3 5 734.79 is prime.35.1155 2336.11 1337.217 3Factoring numerator and denominator357 573 Rewriting as a product of two fractions753711573Using the identity property of 1538.2 102010262 131339.162 828221567 8787740.897282739341.121 12Factoring and using the identity484 12property of 1 to to write 12 as 1 121124 121114442.63183846 141443.5213 44141313 1144.12 11132111 111245.191 19Factoring and using the identity764 19property of 1 to write 19 as 1 1919141919Removing a factor equal to 1:1191446.1 1717513 171347.1506 25Factoring and using the identity251 25property of 1 to write 25 as 1 2525612525Removing a factor equal to 1:125616Simplifying48.536180361365

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Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition - Page 11 preview image

8Chapter 1:Introduction to Algebraic Expressions49.422 21Factoring the numerator502 25and the denominator2212Removing a factor equal to 1:21251225250.5 157515805 161651.1202 60Factoring822 412602Removing a factor equal to 1:12412604152.5 1575453 155353.2102 7 15Factoring982 7 727 152 7Removing a factor equal to1:17751272 754.225714023502557555.131 3Multiplying numerators and252 5denominators31056.11 2411811 84410510 525 52557.3322949 46232 32358.By the commutative law of multiplication,11 12and12 11are equivalent. Thus, the product of thenumerators will be equivalent to the product of thedenominators, so the result is a number divided byitself, or 1.11 1211 121121112 1159.1313Adding numerators; keeping888the common denominator481 41Simplifying24260.1782 441010102 5561.Using 18 as the common4134213denominator918921881318182118737Simplifying63662.48128204515151515363.Multiplying numerators33and denominators77bbaa64.55yxyxzz65.Adding numerators; keeping4610the common denominatornnn66.954xxx67.Using 30 as the383382common denom91630302inator1015103152530555Simplifying65668.7521103181224242469.1147177770.12210255571.Using 18 as the1341342common denominator1189183818188925172.1311391128154545454573.Using 90 as the1121132 10common denominato33209090139r309309100374.55151051421424242

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Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition - Page 12 preview image

Exercise Set 1.3975.Multiplying by the reciprocal of7375the divisor6385563176.7107321535105077.43 9412912279141 478.9 19911944949479.Note that we have a number divided by itself. Thus,the result is 1. We can also do this exercise as follows:777137 13113131377 1380.11155110510110281.2725232 36573757 535382.3835151818583.91929 291812111 1284.373116761485.Writing Exercise. If the fractions have thesame denominator and the numerators and/ordenominators are very large numbers, it wouldprobably be easier to compute the sum of the fractionsthan their product.86.Writing Exercise. If the fractions have differentdenominators, it would probably be easier to computethe product of the fractions than their sum.87.Writing Exercise. Bryce is canceling incorrectly. Thenumber 2 is not a common factor of both terms in thenumerator, so it cannot be canceled. For example, let1x. Then (21)/83/8but (11)/42/41/2.The expressions are not equivalent.89.Product5663367214096168Factor772361488Factor89182101221Sum1516203824202990.We need to find the smallest number that has both 6and 8 as factors. Starting with 6 we list some numberswith a factor of 6, and starting with 8 we also listsome numbers with a factor of 8. Then we find thefirst number that is on both lists.6, 12, 18, 24, 30, 36, …8, 16, 24, 32, 40, 48, …Since 24 is the smallest number that is on both lists,the carton should be 24 in. long.91.44332216 9 4215 8 12352434592.92419 81236249xyxyxyxy93.5945599pqrspqrsqprstprstt94.13 192473231719131795.1515 496 2515xyxy2233232515xyx625y96.101012 2523020xyzxy223552310xyz52225zxy97.272515271827181525151825272539551518352952abnpmnabbcabbcmnnpmnbcnpabnpabnpmnbcmnbcapcm98.4545243230243032315xyzxyzabacxzabxzac22838215xyzacab2cybxz99.323235239444199424 222232232494 92322322 918rsrsrsrsststststrsrsststrsrstt100.5263 772654147142 552 1356572749mnmnmnnpnpnpmnmnpp

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Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition - Page 13 preview image

10Chapter 1:Introduction to Algebraic Expressions101.247mm5947 (m)(m)592828m , orsquare meters4545Alw102.22211 105mm22741 105 (m)(m)2741 10 5 m2 7 41 25 5 m27 42525m , orsquare meters2828Abh103.53212844 3m4mm,9992or 14m9Ps104.47222m2m59814mm59814m5989145m59957270m45451427m, or 3m4545Plw105.There are 12 edges, each with length3102cm. Wemultiply to find the total length of the edges.32312 2cm12cm101012 23 cm1026 23 cm251383cm, or 27cm55Exercise Set 1.41.Since30.15,20we can write320 as aterminatingdecimal.2.If a number is aninteger, it is either a whole numberor the opposite of a whole number.3.0 is the onlywhole numberthat is not a naturalnumber.4.A number like5, which cannot be written preciselyin fraction notation or decimal notation, is an exampleof anirrational number.5.Theoppositeof 1 is –1.6.When two numbers are opposites, they have the sameabsolute value.7.–nis the opposite ofn.8.x9.10x10.6y11.The real number –9500 corresponds to borrowing$9500. The real number 5000 corresponds to theaward of $5000.12.–150; 6513.The real number 100 corresponds to 100°F. The realnumber –80 corresponds to 80°F below zero.14.–1312; 29,03515.The real number –777.68 corresponds to a 777.68-pointfall. The real number 936.42 corresponds to a 936.42-point gain.16.The real number 10,000 corresponds to a $10,000grant, and the real number –4500 corresponds to$4500 spent.17.The real number 8 corresponds to an 8-yd gain, andthe real number –5 corresponds to a 5-yd loss.18.4, –519.20.21.The graph of –4.3 is310 of a unit to the left of –4.22.23.Since 1013,33its graph is 13 of a unit tothe right of 3.24.17233.455  

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Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition - Page 14 preview image

Exercise Set 1.41125.From lowest to highest: points-per-game26.From lowest to highest: field-goal-percentage27.78means 78,so we divide.0.8758 7.0006 4605640400We have 70.875.828.0.1258 1.0008201640400110.125, so0.125.88 29.We first find decimal notation for 3 .4Since 34 means34,we divide.0.754 3.002 820200Thus,33440.75, so0.75. 30....1.8336 11.00065048201820111.83.631.76 means 76,so we divide....1.1666 7.0006106403640Thus 771.16, so1.16.66 32....0.416612 5.00004 82012807280728550.416, so0.416.1212 33.2 means 23,3so we divide....0.6663 2.00018201820182We have 20.6.334.0.254 1.0082020010.25.435.We first find decimal notation for 1 .2Since 12 means12,we divide.0.52 1.01 00Thus, 110.5, so0.5.22 

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Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition - Page 15 preview image

12Chapter 1:Introduction to Algebraic Expressions36....0.1119 1.00091091091110. 1, so0. 1.99 37.Since the denominator is 100, we know that130.13.100We could also divide 13 by 100to find this result.38.0.4520 9.00801001000990.45, so0.45.2020 39.40.41.42.43.Since 5 is to the right of 0, we have 50.44.88 45.Since –9 is to the left of 9, we have99.46.07 47.Since –8 is to the left of –5, we have85. 48.43 49.Since –5 is to the right of –11, we have511. 50.34 51.Since12.5is to the left of –10.2, we have12.510.2. 52.10.314.5 53.We convert to decimal notation.5110.416 and0.44.1225Thus,511 .122554.27140.82353 and0.77143.1735  Thus,2714.1735 55.2xhas the same meaning as2.x 56.9a57.10yhas the same meaning as10.y58.12t59.583,4.7, 0,, 2.16, 62960.6261.–83, 0, 6262.,1763.All are real numbers.64.0, 6265.|58|58since –58 is 58 units from 0.66.4767.|12.2|12.2since –12.2 is 12.2 units from 0.68.4.369.|2 |2since2 is2units from 0.70.45671.9977since99is77units from 0.72.373.|0|0since 0 is 0 units from itself.74.3475.|||8 |8x76.|||5 |5a77.Writing Exercise. Yes; every integer can be written asn/1, a quotient of the forma/bwhere0.b78.Writing Exercise. No; for instance, –2 is an integerand –2 is not a natural number.79.Writing Exercise. No;|0|0which is neitherpositive nor negative.

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Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition - Page 16 preview image

Mid-Chapter Review1380.Writing Exercise. There are infinitely many rationalnumbers between 0 and 1. Consider only rationalnumbers of the form1,nwherenis an integer greaterthan 1. There are infinitely many integers greater than1, so there are infinitely many numbers1,nallbetween 0 and 1. (These numbers are a subset of therational numbers between 0 and 1.)81.Writing Exercise. No; every positive number isnonnegative, but zero is nonnegative and zero is notpositive.82.List the numbers as they occur on the number line,from left to right: –17, –12, 5, 1383.List the numbers as they occur on the number line,from left to right: –23, –17, 0, 484.353211,,,,,324686can be written in decimalnotation as0.666, 0.5,0.75,0.833, 0.375,0.166, respectively. Listing from least to greatest (infractional form), we have532131,,,,,.64368285.Converting to decimal notation, we can write4444444,,,,,,5386923as 0.8, 1.33, 0.5, 0.66,0.44, 2,1.33,respectively. List the numbers(in fractional form) as they occur on the number line,from left to right:4444444,,,,,,398653286.|5|5 and |2|2, so |5||2|. 87.|4|4 and |7|7, so |4||7|. 88.|8|8 and |8|8, so |8||8|.89.|23|23 and |23|23, so |23||23|. 90.||19xxrepresents a number whose distance from 0is 19. Thus,19 or19.xx 91.xrepresents an integer whose distance from 0 is lessthan 3 units. Thus,2,1, 0, 1, 2.x 92.2||5xxrepresents an integer whose distance from 0 isgreater than 2 and also less than 5. Thus,4,3, 3, 4.x 93.10.3331110.1 13333994.130.993(0.33)3 3395.1505.5550(0.1 1)50 99(See Exercise 93.)96.1707.7770(0.1 1)70 99(See Exercise 94.)97.a< 098.Nonpositive numbers include zero. Thus,x< 0.99.||10x100.Distance from zero can go in the positive or negativedirection. Thus,| |20.t101.Writing Exercise. The number entered by hand is anapproximation of2while the value that is squaredimmediately after being calculated is actuallyregarded by the calculator as2.102.Writing Exercise. The statement2||aafor anyreal numberais true. Ifais nonnegative, then2.aaIfais negative,2aa since2amust be nonnegative. Thus, for a nonnegative number,the result is the number and, for a negative number,the result is the opposite of the number. This describesthe absolute value of a number.Mid-Chapter Review1.2210Substituting3312Subtracting34Dividingxy2.1477 27 17(21)xxx3.312 Substituting15xy4.22 10455a5.d– 106.Lethrepresent the number of hours worked; 8h7.Letsrepresent the number of students originallyenrolled;527s8.No; 1313 810494t9.10x+ 710.(3a)b11.4(2x+ 8) = 8x+ 3212.3(2m+ 5n+ 10) = 6m+ 15n+ 3013.18x+ 15 = 3(6x) + 3(5) = 3(6x+ 5)14.9c+ 12d+ 3 = 3(3c+ 4d+ 1)

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