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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Document preview page 1

Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 1

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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition

Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition is the ultimate guide for understanding and solving textbook problems.

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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 1 preview imageINSTRUCTOR’SSOLUTIONSMANUALELEMENTARY ANDINTERMEDIATEALGEBRAGRAPHS ANDMODELSFIFTHEDITIONMarvin L. BittingerIndiana University Purdue University IndianapolisDavid J. EllenbogenCommunity College of VermontBarbara L. JohnsonIvy Tech Community College of Indiana
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 2 preview image
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 3 preview imageCONTENTSChapter1Introduction to Algebraic Expressions...................................................................1Chapter2Equations, Inequalities, and Problem Solving..................................................51Chapter3Introduction to Graphing and Functions...........................................................157Chapters1–3Cumulative Review...........................................................................................281Chapter4Systems of Equations in TwoVariables...........................................................289Chapter5Polynomials...................................................................................................................377Chapter6Polynomial Factorizations and Equations........................................................465Chapters1–6Cumulative Review...........................................................................................551Chapter7Rational Expressions, Equations, and Functions..........................................559Chapter8Inequalities.....................................................................................................................683Chapter9More on Systems.........................................................................................................751Chapters1–9Cumulative Review...........................................................................................829Chapter10Exponents and Radical Functions.....................................................................837Chapter11Quadratic Functions and Equations.................................................................921Chapter12Exponential Functionsand Logarithmic Functions...............................1015Chapters1–12Cumulative Review.....................................................................................1109Chapter13ConicSections........................................................................................................1119Chapter14Sequences, Series, and the Binomial Theorem........................................1179Chapters1–14Cumulative Review.....................................................................................1223ChapterRElementary Algebra Review..............................................................................1233
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 4 preview imageChapter 1Introduction to Algebraic ExpressionsExercise Set 1.11.Expression2.Equation3.Equation4.Expression5.Equation6.Equation7.Expression8.Equation9.Equation10.Expression11.Expression12.Expression13.Substitute 9 foraand multiply.33 927a==14.Substitute 7 forxand multiply.88 756x==15.Substitute 2 fortand add.6268t+=+=16.Substitute 9 forrand subtract.131394r−=−=17.Substitute 2 forx, 14 fory. Add and thendivide the sum by 4.214164444xy++===18.Substitute 15 forc, 20 ford, add and thendivide the sum by 7.1520355777cd++===19.Substitute 20 form, 6 forn, subtract and thendivide the difference by 2.206221472mn−−===20.Substitute 23 forx, 5 fory, subtract anddivide the difference by 6.235183666xy−−===21.Substitute 6 formand 18 forq.99 65431818mq===22.Substitute 9 forzand 15 fory.55 94531515zy×===23.Enter the expression in the graphingcalculator, replacingawith 136 andbwith13. We see that27183438ab−=for136 and13.ab==24.Enter the expression in the graphingcalculator, replacingxwith 87 andywith 29.We see that1991347,531xyxy−+=for87 and29.xy==
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 5 preview image2Chapter 1:Elementary and Intermediate Algebra:Graphs and Models25.21212(6ft)(3ft)(6)(3 )(ft)(ft)21ft , or 21 square feetAbh====26.1400tv=1400 mi3.5hr400 mph==27.12Abh=21 (5cm)(6 cm)21 (5)(6)(cm)(cm)215cm , or 15 square centimeters===28.a.30 sec. By substitution,t=22 30 sec60 sect=×=b.35min. By substitution,t=22 35 min70 mint=×=c.122hr. By substitution,t=1222 2hr5 hrt=×=29.Abh=2(67 ft)(12 ft)(67)(12)(ft)(ft)804 ft , or 804 square feet===30.Averageha=80.36422=31.Letrrepresent Ron’s age. 5 more than Ron’sage, or Ron’s age plus 5 can be expressed as5, or5rr++32.Letsrepresent Luke’s speed. Timesmeansmultiplication, so we have8s×, or8s33.Product is the result of multiplication, so wehave4, or 4aa34.Letlrepresent Lou’s weight;7l+or7l+35.9 less thanc;9c−36.4 less thand;4d-37.6 increased byq; to increase is to add;6, or6qq++38.11 increased byz; to increase is to add;11,z+or11z+39.The difference ofmandn;mn-40.tsubtracted fromp;pt-41.xless thany, or decreaseybyx;y–x42.Letarepresent Lorrie’s age;a–243.xdivided byw(Note:xis the dividend, ornumerator,andwis the divisor, ordenominator);, orxwxw44.Letsandtrepresent the two numbers;quotient is the result of division;, orstst45.Letlandhrepresent the box’s length andheight, respectively. We have the sumlh+or.hl+46.Sum ofdandf;, ordffd++47.Letprepresent Panya’s speed and letwrepresent the speed of the wind.2pw−48.The product of 9 and twicem. Twicemis2,mor2.mWe have9 2, or 29mm49.Letyrepresent“somenumber”;14412, or12yy−−50.Letnrepresent the number;620n-51.Letaandbrepresent the numbers;8()ab-52.One third of the sum is13×sum; letxandyrepresent the two numbers. We have1133. sum()xy×=+
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 6 preview imageExercise Set 1.1353.64%0.64,=so 64% ofw, wherew=thenumber of women, is 0.64w. (Note: replace“of ” withmultiplication)54.Lety=the number;38%0.38,=so we have0.38y.55.|17321517 ? 323232x+=+Writing the equationSubstituting 15 forx32 = 32 is TRUESince the left-hand and right-hand sides arethe same, 15 is a solution.56.|28937528 ? 9310393y+=+Writing the equationSubstituting 75 fory103 = 93 is FALSENo;since the left-hand and right-hand sidesare not the same, 75 is not a solution.57.|287593 28 ?756575a−=−Writing the equationSubstituting 93 fora65 = 75 is FALSESince the left-hand and right-hand sides arenot the same, 93 is not a solution.58.|8968 12 ?969696t=Writing the equationSubstituting 12 fort96 = 96 is TRUESince the left-hand and right-hand sides arethe same, 12 is a solution.59.76379|? 999t=9 = 9 is TRUESince the left-hand and right-hand sides arethe same, 63 is a solution.60.85286|? 66.56x=6.5 = 6 is FALSESince the left-hand and right-hand sides arenot the same, 52 is not a solution.61.108108336|? 363636x=36 = 36 is TRUESince the left-hand and right-hand sides arethe same, 3 is a solution.62.9494712|? 12313127y=31312 is FALSE.7=Since the left-hand and right-hand sides arenot the same, 7 is not a solution.63.Letxrepresent the number.7 times what number is 1596?Translating: 7159671596xx==64.Letxrepresent the number.What number added to 73 is 201?Translating:7320173201xx¯¯¯ ¯¯+=+=65.Letxrepresent the number.Rewording: 42 times what number is 2352?Translating: 422352422352xx¯¯¯¯¯×==66.Letxrepresent the number.Rewording: what number plus 345 is 987?Translating:345987345987xx+=+=67.Letsrepresent the number of unoccupiedsquares.NumbernumberRewording:plusof occupiedof unoccupiedTranslating:19641964is 64.ss¯¯¯¯¯+=+=
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 7 preview image4Chapter 1:Elementary and Intermediate Algebra:Graphs and Models68.Lethrepresent the numberofhours.Rewording: $35 times the number of hours is $3150Translating: $35= $3150$35$3150hh=69.Letwrepresent the total amount of wastegenerated, in millions of tons.the totalamount ofmillion87Rewording: 34.5% ofwasteistons.Translating:34.5%=34.5%87, or 0.3458787www==70.Lettrepresent the average commute inFort Blissin minutes.The averageThe averagecommutingcommutingtime intime inRewording:ismin.Fort Blissminus 51.2Indian WellsTranslating:59.851.2t¯¯¯¯¯=-71.f72.h73.d74.c75.g76.b77.e78.a.79.Look for a pattern in the data. Observe thatthe daily recommended number of grams ofdietary fiber for each age is 5 more than thatage. We reword and translate.Grams of fiber is 5 more than the age5fa =+We have5.fa=+This equation could alsobe written as5.fa=+80.Observe that the tuition is 100 times thenumber of class hours. Reword and translate.Tuition is 100 times number of class hours100ch=We have100 .ch=81.Look for a pattern in the data. Observe thatitcosts $2.42more to processnonmachinablepackagesthanto processmachinablepackages. We reword and translate.Cost ofCost ofnonmachinablemachinableispackages$2.42packagesmore than$2.42nm=+We have$2.42.nm=+This equation couldalso be written as2.42.nm=+82.Observe that in each case theamount receivedis $3.00lessthan theamountremovedfromher account. Reword and translate.Amountamount$3.00minusis removedreceived$3.00rs=−We have$3.00,rs=−orr=s–3.83.Observe that the number of vehicle milestraveled is 10,000 times the number ofdrivers. Reword and translate.Number ofNumber ofisVehicle milesdrivers10,000 times10, 000vd=We have10,000 .vd=
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 8 preview imageExercise Set 1.1584.Observe that the depth of the water is one-tenth the depth of the snow. Reword andtranslate.110DepthDepthisone-tenthofof Waterof snowws=110,ws=or, we could write this as10.ws=85.Thinking and Writing Exercise.Avariableisa letter that is used to stand for any numberchosen from a set of numbers. Analgebraicexpressionis an expression that consists ofvariables, constants, operation signs, and/orgrouping symbols. Avariableexpression isan algebraic expression that contains avariable. Anequationis a number sentencewith the verb=The symbol‘=’is used toindicate that the algebraic expressions oneither side of the symbol represent the samequantity.86.Thinking and Writing Exercise.To evaluatean algebraic expression means to find thevalue of the expression when its variables aregiven values.87.Thinking and Writing Exercise.No; for asquare with sides, the area is given byA=ss. The area of a square with side 2sisgiven by(2 )(2 )442 .sss sAA==88.Thinking and Writing Exercise.Answer mayvary.Juliet was born in 1998. How old will she bein 2006?89.The sign is a triangle whose area is211, or(3ft)(2.5ft)3.75ft22b h× ×=The cost is $120for each square foot.Thus,$120(3.75)$450C==90.The shaded area is the area of a rectangle withdimensions 20 cm by 10 cm less the area of atrianglewith base 20 cm4−cm–5 cm, or11 cm, and height 7.5 cm. We perform thecomputation:1(20 cm)(10 cm)(11 cm)(7.5)2−222200 cm41.25 cmsquare158.75 cm , or 158.75 centimeters=−=91.Whenyis twicexand6,x=2 612612221892yxy==++===92.Whenais three timesb, thenbis one-thirda,so1863186331243bab==−−===93.The next whole number is one more than3:w+314ww++=+94.The preceding odd number is 2 less than2:d+22dd+−=95.Adding the length of the sides, we have:,or22lwlwlw++++96.Adding the length of the sides, we have:,or4sssss+++97.Ella’s time is 5 more than Kyle’s time,or5.EK=+Kyle’s time is 3 more thanAva’s time (t), or3.Kt=+Substituting3forinto5We have:5(3),or8,tKEKtt+=++++or8.t+Also: Kyle is 3 more than Ava, or3t+andElla is 5 more than Kyle, or(3)58tt++=+98.Ray is currently 2 years older than Monique,or2.a+In 7 years, he will be27, or9.aa+++
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 9 preview image6Chapter 1:Elementary and Intermediate Algebra:Graphs and Models99.Thinking and Writing Exercise.Yes; the areaof a triangle with base b and heighthis givenby12.Abh=The area of a triangle with baseband height 2his given by11(2 )22 .22bhbhA==Exercise Set 1.21.commutative2.associative3.associative4.commutative5.distributive6.associative7.associative8.commutative9.commutative10.distributiveFor exercises 11–18, change the order of theaddends.11.7x+12.2a+13.3yx+14.39yx+15.cab+16.xyuv+17.5(1)a+18.9(5)x+For exercises 19–26, change the order of thefactors.19.2a20.yx21.ts22.4x23.5ba+24.3xy+25.(1)5a+26.(5)9x+For exercises 27–32, regroup theaddends.27.(5)ab++28.5()mr++29.()7rt++30.(2)xy++31.()abcd++32.()mnpr++For exercises 33–38, regroup thefactors.33.7()mn34.13()xy35.(2 )a b36.(9 )r p37.(3 2)()ab+38.(5 )(2)xy+
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 10 preview imageExercise Set 1.2739.2(6)2(6)Commutative lawof addition(26)Associative lawof addition8Simplificationtttt++=++=++=+Answersmay vary.40.(11)4(11)4Commutative lawof addition(114)Associative lawof addition15Simplificationvvvv++=++=++=+Answers may vary.41.(3 ) 7(3) 7Commutative lawof multiplication(3 7)Associative lawof multiplication21 or 21Simplificationaaaaa===Answers may vary.42.5(8)5(8 )Commutative lawof multiplication(5 8)Associative lawof multiplication40Simplificationxxxx===Answers may vary.43.(5)2x++(5)2Commutative law(52)Associative law7Simplifyingxxx=++=++=+44.(2 )44(2 )Commutative lawaa=(4 2)Associative law8Simplifyingaa==45.(3)7(3 7)Associative law(3 7)Commutative law21Simplificationmmmm×=×=×=46.4(9)x++(49)Associative law(49)Commutative law13Simplificationxxx=++=++=+For exercises 47–62, use the distributive law.47.4(3)44 3412aaa+=+=+48.3(5)33 5315xxx+=+=+49.6(1)6 1666xxx+=+=+50.6(4)66 4624vvv+=+=+51.(5)225 2210nnn+=+=+52.(1)31 3333ttt+=+=+53.8(35 )8 38 52440xyxyxy+=+=+54.7(45 )7 47 52835xyxyxy+=×+×=+55.9(26)9 29 6xx+=+1854x=+56.9(67)9 69 7mm+=+5463m=+57.5(23 )55 25 3rtrt++=++51015rt=++58.4(583 )4 54 84 3xpxp++=++203212xp=++59.()22222ababab+=+=+60.(2)772 7xx+=+714x=+61.(2) 5552 5xyxy++=++5510xy=++
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 11 preview image8Chapter 1:Elementary and Intermediate Algebra:Graphs and Models62.(2)62 666abab++=++1266ab=++For exercises 63–66, the terms are separatedby the addition/plus signs.63.,, 19x xyz64.9, 17 ,a abc65.2 ,, 5aabb66.43, 20,axyb67.2222The commonfactor is 2abab+=+2()Using thedistributive lawab=+Check:2()2222ababab+=+=+68.5555The commonfactor is 5.yzyz+=+5()Using thedistributive lawyz=+Check:5()5555yzyzyz+=+=+69.777 17The commonfactor is 7yy+= +7(1)Using thedistributive lawy=+Check:7(1)7 1777yyy+= +=+70.131313 113The commonfactor is 13.xx+=+13(1)Using thedistributive lawx=+Check:13(1)13 1131313xxx+= +=+71.1833 63 1The commonfactor is 3.xx+=+3(61)Using thedistributive lawx=+Check:3(61)3 63 1183xxx+=+=+72.2055 45 1The commonfactor is 5.aa+=+5(41)Using thedistributive lawa=+Check:5(41)5 45 1205aaa+=×+× =+73.5101555 25 35(23 )xyxyxy++=++=++Check:5(23 )55 25 351015xyxyxy++=++=++74.32763 13 93 23(192 )bcbcbc++=× +×+×=++Check:3(192 )3 13 93 23276bcbcbc++=× +×+×=++75.1293 43 33(43)xxx+=+=+Check:3(43)3 43 3129xxx+=+=+76.25305 55 65(56)yyy+=+=+Check:5(56)5 55 6.2530yyy+=×+×=+77.3933 33(3 )ababab+=+=+Check:3(3 )33 339ababab+=+×=+78.51555 35(3 )ababab+=× +× ×=+Check:5(3 )55 3515ababab+=+=+
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 12 preview imageExercise Set 1.2979.44886622 222 422 322(243 )xyzxyzxyz++=×+×+×=++Check:22(243 )22 222 422 3448866xyzxyzxyz++=×+×+×=++80.24486012 212 412 512(245)ababab++=×+×+×=++Check:12(245)12 212 412 5244860ababab++=×+×+×=++81.andst82.5 andx83.3 and (x+y)84.() and 6ab+85.7,a,andb86.m, n,and287.() and ()abxy−−88.(3) and ()abc−+89.Terms are separated by addition andsubtraction signs. A single term may includeplus or minus signs only if they are insideparentheses, brackets, or other groupingsymbols.Factors are numbers, variables, or expressionsthat are multiplied together.90.2(34 )2 32 4Distributive law68Simplificationxyxyxy+=+=+91.Letkrepresent Kylie’s salary. Then we have122, orkk.92.To multiply a sum by a number, put the sumin parentheses;2(7)m+93.Thinking and Writing Exercise.No; ingeneral, when subtracting, the result dependson the order in which the operation isperformed.94.Thinking and Writing Exercise.No; ingeneral, when dividing, the result depends onthe grouping.95.The expressions are equivalent by thedistributive law.84()8444 2444(2)abababab++=++=++=++96.The expressions are not equal. Distributionapplies only to sums and differences.5(a)5, but 555 525.bababa bab===97.The expressions are not equal. For example,let1.m=Then we have7773 11, but3333 1737.7¸× =× =× ¸=¸=98.The expressions are equivalent bythecommutative law of multiplicationandthedistributive law.()55()5()5 ()rtstt rt st rst rs+=+=+=+99.The expressions are not equivalent.5[2(3 )]5 2(3 )10(3 )1010 310303010xyxyxyxyxyyx+=+=+=+=+=+100.The expressions are equivalent by thecommutative law of multiplicationand thedistributive law.[ (23 )]55[ (23 )]5(23 )5 (23 )52531015.cbcbcbcbccbcbc+=+=+=+=+=+
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 13 preview image10Chapter 1:Elementary and Intermediate Algebra:Graphs and Models101.Thinking and Writing Exercise.3(2)3(20)x+=+3 26;6606x==+=+=The result indicates that3(2)x+and6x+are equivalent when0.x=(By thedistributive law, we know they are notequivalent for all values ofx.)102.Writing Exercise.15405(38)15 44060401005(3 48)5(128)5 20100xx+=++=+=+=+==Although the expressions1540x+and5(38)x+are equivalentfor4,x=this resultdoes not guarantee that the factorization iscorrectfor all values ofx. (See Exercise 101.)Exercise Set 1.31.b2.c3.d4.a5.The paired factors of 50 are:1 50, 2 25, and 5 10The factors of 50are:1, 2, 5, 10, 25, 50.6.The paired factors of 70 are:1 70, 2 35, 5 14, and 7 10The factors of 70 are:1, 2, 5, 7, 10, 14, 35, 70.7.The paired factors of 42 are:1 42, 2 21, 3 14, and 6 7The factors of 42 are:1, 2, 3, 6, 7, 14, 21, 428.The paired factors of 60 are:1 60, 2 30, 3 20, 4 15, 5 12, and 6 10The factors of 60 are:1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 609.Factors of 21are 1, 3,7, 21. Composite.10.Factors of 15 are 1, 3, 5, 15. Composite.11.The factors of 31 are:1 and 31. Prime.12.The factors of 35 are:1, 5, 7, 35. Composite.13.The factors of 25 are:1, 5, 25. Composite.14.The factors of 37 are:1 and 37. Prime.15.The factors of 2 are:l and 2. Prime.16.1 is the only factor of1. Itis neither primenor composite.(It is aunit.)17.0 is neither prime nor composite.(Note: The number must be a countingnumbergreaterthan 1).18.The factors of 4 are:1, 2, 4. Composite.19.The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40.Composite.20.The factors of 75 are: 1, 3, 5, 15, 25, 75.Composite.21.262 13=22.153 5=23.302 152 3 5==24.555 11=25.273 93 3 3==26.982 492 7 7==27.402 202 2 102 2 2 5===28.542 272 3 92 3 3 3===29.43; Prime30.1202 602 2 302 2 2 152 2 2 3 5====31.2102 1052 3 352 3 5 7===32.79;Prime33.1155 23=
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 14 preview imageExercise Set 1.31134.14311 13=35.Factoring numerator and142 7denominator213 7=Rewriting as a product27of two fractions372711372Using the identity property of 13=×=×==36.Factoring numerator and202 10denominator262 13=Rewriting as a product2 10of two fractions2 131021113210Using the identity property of 113====37.162 2 2 2562 2 2 7=2222222277221 1 1 7727===  ==38.722 2 2 3 3273 3 3=2 2 2 333 3383==39.661486 88==40.186 33846 1414==41.5213 441313 1×==×42.1321111=121112=43.191976=4 1914=44.171751=3 1713=45.15062525=256=46.17053434=345=47.42221215022525==48.755 1515805 1616==49.1202 6060822 4141==50.755 1545=3 1553=51.210298=3 5 721577 7=52.140270350=57025=53.131 332 72 714==54.9 39 3274 84 832==55.12 1012 103 4 5 24 28595 953 333====56.11 121112 11=1212111=
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 15 preview image12Chapter 1:Elementary and Intermediate Algebra:Graphs and Models57.1313448888++===4122=58.Use 8 as the common114 11denominator28428+=+414188858+=+==59.Use 18 as the common4132413denominator9182918+=×+813213 71818183 676=+===60.483 485153 515+=×+128201515154 543 53=+===61.333777bbbaaa==62.555xyx yxyzzz==63.43437aaaa++==64.75752aaaa−−==65.Use 30 as thecommon denominator38332810153 102 1591691625303030305 555 66+=++=+====66.753 725Use 24 as thecommon denominator8123 82 12+=+211031242424=+=67.9292717777−−===68.122122102 5255555−−====69.1341324Use 18 as the commondenominator1891829138181813851818-=-×=--==70.1383 138Use 45 as thecommon denominator15453 15453984545398314545−=−=−−==71.2022 10303−=3 1023−22033=−=72.553 557213 721−=−155211021−==73.Multiply by the reciprocal737 5of the divisor656 3=7 56 33518==
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 16 preview imageExercise Set 1.31374.Multiply by the reciprocal7374of the divisor5453=7 45 32815==75.848 1591594=8 159 42 43 53 3 4103===76.1114848 14 114 22===77.37121273=3 4 7283==78.1010111099109==79.7771371313137¸=×=13×137×1=80.175176172 35186852 4 520===81.2753252 36737535===82.3815313 515858 18===83.129199 2182=¸=× =84.3733 13 116677 67 3 214====85.Thinking and Writing Exercise.There is aninfinite number of even numbers. There is aninfinite number ofwhole numbers, and eachof these can be multiplied by 2, producing aneven number.86.Thinking and Writing Exercise.If thefractions havethe samedenominators, itwould probably be easier to compute theirsum than their product.87.5(3)(3)5xx+=+Commutative Law of Multiplication5(3)5(3)xx+=+Commutative Law of Addition88.7()7() orabba++=++7()()7abab++=++89.Thinking and Writing Exercise.Bryce iscanceling incorrectly. The number 2 is not acommon factor of both terms in thenumerator, so it cannot be canceled. Forexample, let1.x=Then(21) / 83 / 8+=but(11) / 42 / 41/ 2.+==The expressions arenot equivalent.90.0 is not a natural number, and we consideronly natural numbersgreaterthan 1.91.Product56 63 3672 140 96 168Factor77236 1488Factor89182101221Sum15 1620 3824202992.We need to find the least number that hasboth 6 and 8 as factors. Starting with 6 we listsome numbers with a factor of 6, and startingwith 8 we also list some numbers with afactor of 8. Then we find the first number thatis on both lists.6,12,18, 24,30,36,...8,16, 24,32, 40, 48,...Since 24 is the least number that is on bothlists, thecartonshould be 24 in. long.
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 17 preview image14Chapter 1:Elementary and Intermediate Algebra:Graphs and Models93.16 9 444332 2215 8 123 5 24345==94.92419 81236249xyxyxyxy==95.3 927399p q rspqrsqprstprs tt==96.24713 19323=17 191317=97.1515 496 2515xyxy=2 23 323 2515xyx625y=98.101012 2523020xyzxy××=××223x××××5 52310yz×× ×× ××22 552xyz××× ××=99.27151825271827252725152515181518395 55352 92abmnbcnpabbcabnpabnpmnnpmnbcmnbca bnpapmnb ccm=====100.4524303245322430xyzabxzacxyzacabxz=3=15×228xyz×× ××× ×38ac×××215ab×× ××2xzcyb××=101.341253154442rsrsstst=2392324249232232 2 918rsstrsstrsrs tt==== 334 53Note:55444 142032344=+=++==1241819Similarly: 4 22 1222+=+==102.574535432275mnmnnpnp=2614752657142 1357 2 76549mnnpmnnpm nnpmp=¸=×/ ×× ×/=×/ ×/=103.47mm59Alw ==   247(m)(m)592828m , orsquare meters4545 =   =104.11105mm2274Abh ==  2221105(m)(m)2741 10 5 m2 7 41 2 5 5 m2 7 42525m , orsquare meters2828 =  ===105.544 3m9324m91282m, or 14m99Ps====
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 18 preview imageExercise Set 1.415106.47222m2m59Plw=+=+814mm59814m5989145m59957270m45451427m, or 3m4545=+æö=+ç÷èøæö=×+×ç÷èøæö=+ç÷èø=107.The cube has 12 edges, all with length of32cm.10312 210232 6 2312 102 5138327cm55TL=====Exercise Set 1.41.repeating2.terminating3.integer4.whole number5.rational number6.irrational number7.natural number8.absolute value9.100corresponds tothe highest temperaturerecorded in Alaska.–80corresponds tothelowest.10.–250 corresponds to burning250 calories, and65 corresponds to consuming 65 calories.11.–777.68corresponds tofalling 777.68 points,and936.42corresponds togaining 936.42points.12.–1340 corresponds to 1340 feet below sealevel, and 29,035 corresponds to 29,035 feetabove sea level.13.–12,500corresponds totaking on a debt of$12,500, and 5000 corresponds to receiving$5000 as a scholarship.14.18.7corresponds to the birth rateof18.7births per 1000 people, and–7.89correspondsto7.89deathsper 1000 people.15.8 corresponds to the 8 yard gain, and–5corresponds to the 5 yard loss.16.6corresponds tobeing 6over par, and–17corresponds tobeing 17underpar.17.18.19.20.21.22.23.24.
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 19 preview image16Chapter 1:Elementary and Intermediate Algebra:Graphs and Models25.78means78,so we divide.0.8 7 58 7.0006 46 05 64 04 00780.875.=26.18means(18), so we divide.−−0.1258 1.00 082 0164 04 00180.125.−= −27.We first find decimal notation for3344. Sincemeans 34, we divide.0.7 54 3.0 02 82 02 00340.75.−= −28.116means 116,so we divide.1.833...6 11.00065048201821161.83.=29.76means76, so we divide.--¸1.16 66 7.0 0 061064 0364 0364761.16.-= -30.512means512, so we divide.−−0.416612 5.000 04 82 012807 2807 285120.416.-= -31.23means 23,so we divide.0.6 6 63 2.0 0 01 82 0182 0182230.6.=32.14means 14,so we divide.0.2 54 1.0 082 02 00140.25.=
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 20 preview imageExercise Set 1.41733.12means(12), so−−we divide0.52 1.01 0011220.5, so0.5.=−= −34.29−means29,−so we divide.0.29 2.0 0182290.2.-= -35.Since the denominator is 100, we know that131000.13.=We could also divide 13 by 100 tofind this result.36.720means(720),−−so we divide.0.3520 7.0 06 01 0 01 0 007200.35.−= −37.38.39.40.41.7 > 0, since 7 is to the right of 0.42.8 >–8, since 8 is to the right of–8.43.–6 < 6, since–6 is to the left of 6.44.0 >–7, since 0 is to the right of–7.45.–8 <–5, since–8 is to the left of–5.46.–4 <–3, since–4 is to the left of–3.47.–5 >–11, since–5 is to the right of–11.48.–3 >–4, since–3 is to the right of–4.49.–12.5 <–9.4, since–12.5 is to the left of–9.4.50.–10.3 >–14.5, since–10.3 is to the right of–14.5.51.5111225,− −since512−is to therightof1125.−52.27141735,−−since1417−is to the left of2735.-53.–7 >xhas the same meaning asx<–7.54.a> 9 has the same meaning as 9 <a.55.–10yhas the same meaning as10.y −56.12thas the same meaning as12.t57.311− −is true.58.55 −is false.59.08is false.60.57−is true.61.88− −is true.62.88is true.63.5858-=64.4747-=65.5.65.6=66.2255−=67.22=68.456456-=69.9977−=
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 21 preview image18Chapter 1:Elementary and Intermediate Algebra:Graphs and Models70.33-=71.00=72.4.34.3=73.88x= -=74.55a= -=75.518,4.7, 0,, 2.16,379−−−76.1877.18, 0,37−78.,1779.All of them:518,4.7, 0,,,17, 2.16,379−−−80.18, 081.Thinking and Writing Exercise.Yes; everyinteger can bewrittenasn/1, a quotient of theforma/bwhere0.b82.Thinking and Writing Exercise.No; forinstance,–2 is an integer and–2 is not anatural number.83.33 2 742xy==84.55, orabab+=+55.abba+=+85.Thinking and Writing Exercise.According tothe three given measures, the Spurs are thebest team overall, with the highest number ofpoints per game, the greatest points per gamedifferential, and thethird-greatest field goalpercentage differential.86.Thinking and Writing Exercise.There areinfinitely many rational numbers between 0and 1. Consider rational numbers of the form1,nwherenis an integer greater than 1. Thereare infinitely many integers greater than 1, sothere are infinitely many numbers1,nallbetween 0 and 1. (These numbers are a subsetof the rational numbers between 0 and 1.)87.Thinking and Writing Exercise.No; everypositive number is nonnegative, but zero isnonnegative and zero is not positive.88.List the numbers as they occur on the numberline, from left to right:–17,–12, 5, 1389.List the numbers as they occur on the numberline, from left to right:–23,–17, 0, 490.213531,,,,,324686−−−can be written indecimal notation as0.666,0.5,0.75,0.833,0.375,0.166,−−−respectively. Listing from least to greatest (infractional form), we have532131,,,,,.643682−−−91.Converting to decimal notation, we can write4444444,,,,,,as5386923−0.8, 1.33, 0.5, 0.66, 0.44, 2,1.33,−respectively. List the numbers (in fractionform) as they occur on the number line, fromleft to right:4444444,,,,,,3986532−92.55-=and22,-=so52 .-> -93.44=and77,-=so 47 .< -94.88-=and88,=so88 .-=95.2323 and2323, so 2323 .=-== -96.7x=xrepresents a number whose distance from 0is 7. Thus,7x=or7.x= −97.3x<xrepresents an integer whose distance from 0is less than 3 units. Thus,2,x= −–1, 0, 1, 2.
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 22 preview imageMid-Chapter Review1998.25x<<xrepresents an integer whose distance from 0is greater than 2 and also less than 5. Thus,= −4,x–3, 3, 499.130.31 110.1 1333 39====100.()130.993 0.33 33===101.()1505.5550 0. 150 99===(See Exercise 99).102.()1707.7770 0. 170 99===(See Exercise 99).103.0a104.0x105.10x106.20t107.Thinking and Writing Exercise.The numberentered by hand is an approximation of2while the value that is squared immediatelyafter being calculated is actually regarded bythe calculator as2.108.Thinking and Writing Exercise.Yes. We havethree possibilities:ais negative,0,a=oraispositive. We will show an example of eachpossibility.()2222;242; |2|20;000; | 0 |07;7497; |7|7.aaa= --==-=========Mid-Chapter ReviewGuided Solutions1.2210124333xy−−===2.4088 58 18(51)xxx+=+=+Mixed Review1.x+y= 3 + 12 = 152.22 10204555a===3.10d−4.Lethbe the number of hours worked. Theneight times the number of hours work is: 8h.5.Letnbe the number of students that originallyenrolled in Janine’s class. Then reword “fivefewer than the number that originally enrolled”as “the number that originally enrolled minusfive” and write theequation:The numberthat originallyTwenty-sevenisminusfiveenrolled275n=−275n=−6.13 810494,=so8t=is not a solution to1394.t=7.710107xx+=+8.3()(3 )aba b=9.4(28)4 24 8832xxx+=+=+10.3(2510)3 23 53 1061530mnmnmn++=×+×+×=++11.1899 29 19(21)xxx+=×+× =+12.824204 24 64 54(265)ayayay++=×+×+×=++13.842 422 2 212 2 3 7=×=× ×=× × ×14.8 6486408 55×==×15.1353 453 3 153 3 3 53153 1053 3 353 3 5 733 3 51353315335 77=×=× ×=× × ×=×=× ×=× × ××× ×==×××
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 23 preview image20Chapter 1:Elementary and Intermediate Algebra:Graphs and Models16.122 62 2 3 and 82 2 2, sothe LCM of 12 and 8 is 2 2 2 324.111122233 39and, so1212224883241132292291312824242424=× =× ×=× ×× × × ==×==×=--=-==17.868112 2 2114415111563 52 345× ×¸=×=×=××18.2.5 is 2.5 units to the left of zero on thenumber line:-19.3(320) and200.1532020 3.0020101001000-= -¸¸=So,30.1520−= −20.1624, because16 is to the rightof24 on the number line.->---21.222 11 and 153 5, so the LCM of 22and 15 is 2 3 5 11330.33 15452222 15330222244151522330345442Therefore2233033015since45 is to the left of44 on thenumber line.=×=×× × ×=---=×=---=×=----=<=--22.9 is equivalent to 9.xx³£23.6 is to the left of5 on the number line,so the statement65 is true.---£ -24.|5.6 |5.6-=25.| 0 |0=Exercise Set 1.51.f2.d3.e4.a5.b6.c7.Start at 5. Move 8 units to the left.5( 8)3+ −= −8.Start at 2. Move 5 units to the left.2( 5)3+ −= −9.Start at–5. Move 9 units to the right.594−+=10.Start at–3. Move 8 units to the right.385− +=11.Start at4−. Move 0 units.404−+= −12.Start at6−. Move 0 units.606−+= −
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 24 preview imageExercise Set 1.52113.Start at–3. Move 5 units to the left.3( 5)8− + −= −14.Start at–4. Move 6 units to the left.4( 6)10−+ −= −15.6( 5).−+ −Two negatives. Add their absolutevalues . Make the answer negative.6( 5)(65)11−+ −= −+= −16.8( 12).−+ −Two negatives. Add theirabsolute values. Make the answer negative.8( 12)(812)20− + −= −+= −17.10( 15).+ −The absolute values are 10 and15. Their difference is15105.−=Thenegative number has the larger absolutevalue, so the answer is negative.10( 15)(1510)5+ −= −−= −18.12( 22).+ −The absolute values are 12 and22. Their difference is221210.−=Thenegative number has the larger absolutevalue, so the answer is negative.12( 22)(2212)10+ −= −−= −19.12( 12).+ −The numbers have the sameabsolute value and their difference is 0.12( 12)0+ −=20.17( 17).+ −The numbers have the sameabsolute value and their difference is 0.17( 17)0+ -=21.24( 17).−+ −Two negatives. Add theirabsolute values. Make the answer negative.24( 17)(2417)41−+ −= −+= −22.17( 25).−+ −Two negatives. Add theirabsolute values. Make the answer negative.17( 25)(1725)42−+ −= −+= −23.1313.−+The numbers have the sameabsolute value and their difference is 0.13130−+=24.1818.−+The numbers have the sameabsolute value and their difference is 0.18180−+=25.18( 11).+ −The absolute values are 18 and11. Their difference is18117.−=Thepositive number has the greater absolutevalue, so the answer is positive.18( 11)18117+ −=−=26.8( 5).+ −The absolute values are 8 and 5.Their difference is853.−=The positivenumber has the greater absolute value, so theanswer is positive.8( 5)853+ −=−=27.360.−+0 is the additive identity.36036−+= −28.0( 74).+ −0 is the additive identity.0( 74)74+ −= −29.314.− +The absolute values are 3 and 14.Their difference is14311.−=The positivenumber has the larger absolute value, so theanswer is positive.31414311−+=−=30.13( 6).+ −The absolute values are 13 and 6.Their difference is1367.−=The positivenumber has the larger absolute value, so theanswer is positive.13( 6)1367+ −=−=31.14( 19).−+ −Two negatives. Add theirabsolute values. Make the answer negative.14( 19)(1419)33−+ −= −+= −32.11( 9).+ −The absolute values are 11 and 9.Their difference is1192.−=The positivenumber has the larger absolute value, so theanswer is positive.11( 9)1192+ −=−=
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 25 preview image22Chapter 1:Elementary and Intermediate Algebra:Graphs and Models33.19( 19).+ −The numbers have the sameabsolute value and their difference is 0.19( 19)0+ −=34.20( 6).−+ −Two negatives. Add theirabsolute values. Make the answer negative.20( 6)(206)26−+ −= −+= −35.23( 5).+ −The absolute values are 23 and 5.Their difference is235−or 18. The positivenumber has the larger absolute value, so theanswer is positive.23( 5)23518+ −=−=36.15( 7).−+ −Two negatives. Add theirabsolute values. Make the answer negative.15( 7)(517)22−+ −= −+= −37.31( 14).−+ −Two negatives. Add theirabsolute values. Make the answer negative.31( 14)(3114)45−+ −= −+= −38.40( 8).+ −The absolute values are 40 and 8.Their difference is40832.−=. The positivenumber has the larger absolute value, so theanswer is positive.40( 8)40832+ −=−=39.40( 40).+ −The numbers have the sameabsolute value and their difference is 0.40( 40)0+ −=40.2525.−+The numbers have the sameabsolute value and their difference is 0.25250−+=41.85( 65).+ −The absolute values are 85 and65. Their difference is856520.−=Thepositive number has the larger absolute value,so the answer is positive.85( 65)856520+ −=−=42.63( 18).+ −The absolute values are 63 and18. Their difference is631845.−=Thepositive number has the larger absolute value,so the answer is positive.63( 18)631845+ −=−=43.3.61.9.−+The absolute values are 3.6 and1.9. Their difference is3.61.91.7.−=Thenegative number has the larger absolutevalue, so the answer is negative.3.61.9(3.61.9)1.7−+= −−= −44.6.54.7.−+The absolute values are 6.5 and4.7. Their difference is6.54.71.8.−=Thenegative number has the larger absolutevalue, so the answer is negative.6.54.7(6.54.7)1.8−+= −−= −45.5.4( 3.7).−+ −Two negatives. Add theirabsolute values. Make the answer negative.5.4( 3.7)(5.43.7)9.1−+ −= −+= −46.3.8( 9.4).−+ −Two negatives. Add theirabsolute values. Make the answer negative.3.8( 9.4)(3.89.4)13.2−+ −= −+= −47.3455.-+The absolute values are35and45.Their difference is341555.−=The positivenumber has the larger absolute value, so theanswer is positive.3443155555−+=−=48.3277.−+The absolute values are27and37.Their difference is321777.−=The positivenumber has the largerabsolute value, so theanswer is positive.2332177777−+=−=49.4277.−−+Two negatives. Add their absolutevalues. Make the answer negative.4242677777−−+= −+= −50.5299.−−+Two negatives. Add their absolutevalues. Make the answer negative.5252799999−−+= −+= −
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 26 preview imageExercise Set 1.52351.2153.−+The absolute values are25and13.Their difference is651151515.−=The negativenumber has the larger absolute value, so theanswer is negative.2165651531515151515--æö+=+= --= -ç÷èø52.41132.−+The absolute values are413and12.Their difference is1385262626.−=The positivenumber has the larger absolute value, so theanswer is positive.4181313851322626262626−−+=+=−=53.4293.−+The absolute values are49and23.Their difference is642999.−=The positivenumber has the larger absolute value, so theanswer is positive.42466429399999−−+=+=−=54.1163.−+The absolute values are16and13.The difference is211666.−=The positivenumber has the larger absolute values, so theanswer is positive.11122116366666−−+=+=−=55.35( 14)( 19)( 5)+ −+ −+ −Using theassociativelaw of additionAdding thenegativesAdding apositiveand a negative35[( 14)( 19)( 5)]35( 38)3=+−+ −+ −=+ −= −56.28( 44)1731( 94)+ −+++ −Using the commutativeand associativelaws of additionAdding the positivesand adding the negativesAdding a positiveand a negative[281731][( 44)( 94)]76( 138)62=+++−+ −=+ −= −57.4.98.54.9( 8.5)−+++ −Note that we havetwo pairs of numbers with different signs andthe same absolute value:4.9−and 4.9, 8.5and8.5.−The sum of each pair is 0, so theresult is00,+or 0.58.243.1( 44)( 8.2)63++ −+ −+Using the commutativeand associativelaws of additionAdding the positivesand adding the negativesAdding a positiveand a negative[243.163][( 44)]( 8.2)]90.1( 52.2)37.9=+++−+ −=+ −=59.Rewording:FirstsecondpluspluschangechangeTranslating:0.05+( 0.03)+thirdtotalischangechangetotal0.07change−−=Since0.05( 0.03)0.070.01,−+ −+= −theprice dropped 1¢, or the cost changed$0.01.−60.Rewording:FirstsecondchangechangeplusplusTranslating:( 0.06)+0.12+thirdtotalchangechangeistotal( 0.04)change−−=Since0.06(0.12)( 0.04)0.02,−++ −=theprice rose 2¢, or the cost changed $0.02.
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 27 preview image24Chapter 1:Elementary and Intermediate Algebra:Graphs and Models61.()()()715881971010Rewording: Firstsecondchange plus change plus+( )thirdfourthlevelchangeplusischangechange=totalchange+−+−Since157197881010753576284040404075357628881140405−++−+=−++−+−+−+== −= −the lake level dropped by115feet.62.Rewording:20142015losspluslossplusTranslating:26,500+( 10, 200)2016total profitisprofitor loss32,400total profitor loss−−+=Since26,500( 10, 200)32, 40036, 70032, 4004300,−+ −+= −+= −the loss was $4300, or the profit was–$4300.63.Rewording: FirstsecondthirdTotaltryplustryplustry is gainTranslating: 13+( 12)21Totalgain¯¯¯¯¯¯¯-+=Since13( 12)2122,+ −+=the total gain was22 yd.64.Rewording: Originalchange frombalanceplus writing first check plusTranslating:350+( 530)newchange fromdepositplusisbalancewriting second check75=new(90)balan−++−ceSince350( 530)(75)( 90)+ −++ −(35075)[ 530( 90)]425( 620)195=++ −+ −=+ −= −The balance is$195.−65.Rewording:Originalchangebalanceplusfrom checkplusTranslating:82+( 50)Change fromisNew balanceAugust chargesNew balance63−+=Since82( 50)6395,+ −+=hisnew balanceis $95.66.Rewording:Originalchangebalanceplusfrom checkplusTranslating:470+( 45)newchangetotalplusischargesfrom checkchange160+( 500)new balance−+−=Since470( 45)160( 500)85,+ −++ −=Ianstill owes $85.67.Rewording:baseplusriseis elevationTranslating:( 19,684)33, 480elevation−+=Since( 19,684)33, 48013,796,−+=theelevation of the peak is 13,796 ft above sealevel.
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 28 preview imageExercise Set 1.52568.Rewording:Change fromchange fromwithdrawalsplusadditionsTranslating:5+8change fromchange inoriginal sizeisplus"no shows"=change in( 4)original size¯¯¯-¯¯¯¯+-Since58( 4)−++ −3( 4)1,=+ −= −the class lost 1 student,or the class sizechanged by–1.69.710(710)17aaaa+=+=70.38(38)11xxxx+=+=71.312(123)9xxxx−+=−=72.2( 7)(72)5mmmm+ −= −−= −73.5( 8 )(85)3aaaa+ −= −−= −74.38(83)5xxxx−+=−=75.52(52)7aaaa−+ −= −+= −76.10( 17 )(1710)7nnnn+ −= −−= −77.384( 10 )xx− +++ −348( 10 )Using thecommutativelaw of addition( 34)[8( 10)]Using thedistributive law12Addingxxxx= −+++ −=−+++ −=−78.85()( 3)aa++ −+ −8()5( 3)Using thecommutativelaw of addition[8( 1)][5( 3)]Using thedistributive law72Addingaaaa=+ −++ −=+ −++ −=+79.69( 9 )( 10)(106)(99)4mnnmmnm++ −+ −= −−+−= −80.11( 8 )( 3 )8[ 11( 3)]( 88)14stststs−+ −+ −+= −+ −+ −+= −81.46.3()( 10.2)[( 4)( 1)](10.26.3)53.9xxxx−++ −+ −=−+ −−−= −−82.710.513( 11.5 )(11.510.5)(137)166, or 6yyyyyy−+++ −= −−+−= −+= −+−83.Perimeter8597xx=+++8957(89)(57)1217xxxx=+++=+++=+84.Perimeter8456aa=+++(85)(46 )(85)(46)1013aaaa=+++=+++=+85.Perimeter337594trtr=+++++(35 )(34 )(79)(35)(34)(79)8716ttrrtrtr=+++++=+++++=++86.Perimeter275384xzxz=+++++(23 )(54 )(78)(23)(54)(78)5915xxzzxzxz=+++++=+++++=++87.Perimeter96784nnn=++++97684(97)(684)1816nnnnn=++++=++++=+88.Perimeter265737nnn=+++++(577 )(263)(577)(263)1911nnnnn=+++++=+++++=+
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 29 preview image26Chapter 1:Elementary and Intermediate Algebra:Graphs and Models89.Thinking and Writing Exercise.Answer mayvary. One possible explanation follows.Consider performing the addition on anumber line. We start to the left of 0 and thenmove farther left, so the result must be anegative number.90.Thinking and Writing Exercise.Each nonzerointeger from–10 to 10 can be added to itsopposite with the sum of each pair being 0.When this is added to the remaining integersums, the totalis 0.91.7(32)7 377 2zyzy++=++21714zy=++92.737 8282 37 82 372 42 3283====93.Thinking and Writing Exercise.The sum willbe positive when the positive number isgreater than the sum of the absolute values ofthe negative numbers.94.Thinking and Writing Exercise.No; when weadd real numbers with different signs, wemust know how to subtract absolute values.That is, we must be able to calculateab−whereaandbare both nonnegative with.ab95.We’re looking for the difference between theamount in Travis’s account after the depositand his eventual overdrawn amount.Rewording:OriginaldepositplusminusamountTranslating:257.33+(152)overdrawncheckisamountamountcheck42.37amount¯¯¯¯-¯¯¯-=Since257.33152( 42.37)451.70,+− −=thecheck’s amount$451.70.96.Rewording:Finaloppositevalueof dropplusplusTranslating:61+12+oppositeoriginalof riseisvalueoriginal( 17.50)value−=Since6112( 17.50)55.50,++ −=the originalvalue was $55.50.97.4__( 9 )( 2 )57xxyxy++ −+ −= −−Simplify the left side of the equation.4___( 9 )( 2 )xxy++ −+ −[4( 9 )]___( 2 )[4( 9)]___( 2 )5___( 2 )xxyxyxy=+ −++ −=+ −++ −= −++ −We now have:5__( 2 )5755 , so__+( 2 )7xyxyxxyy−++ −= −−−= −−= −The missing term is–5y, since527yyy−+ −= −98.39__526abaab−+++=−Simplify the left side of the equation.39__5( 35 )9__( 35)9__29__abaaababab−+++=−+++=−+++=++We now have:29__2622 ,so 9__6ababaabb++=−=+= −The missing term is–15b, since9( 15 )6bbb+ −= −
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 30 preview imageExercise Set 1.62799.32__( 2)2( 6)mnmnm+++ −=+ −Simplify the left side of the equation.32__( 2)[3( 2)]2__[3( 2)]2__2__2__mnmmmnmnmnnm+++ −=+ −++=+ −++=++=++We now have:2__2( 6)22 ,so__6nmnmnnmm++=+ −=+= −The missing term is–7m, since( 7)6mmm+ −= −100.__9( 4 )107xyxxy++ −+=−(9)( 4 )__10710( 4 )__1071010 ,so ( 4 )__7xxyxyxyxyxxyy++ −+=−+ −+=−=−+= −The missing term is–3y, since4( 3 )7yyy−+ −= −101.Note that, in order for the sum to be 0, thetwo missing terms must be the opposites ofthe given terms. Thus, the missing terms are–7tand–23.102.22710plwx=+=+We know22 510,l==so 2wis 7x. Thenthe width is a number which yields 7xwhenadded to itself. Since3.53.57 ,xxx+=thewidth is 3.5x, or72.x103.3( 3)2( 2)15− + −++ −+= −Since the total is 5 under par after the fiverounds and51( 1)( 1)( 1)( 1),−= − + −+ −+ −+ −the golferwas 1 under par on average.Exercise Set 1.61.d2.g3.f4.h5.a6.c7.b8.e9.four minus ten10.five minus thirteen11.two minus negative nine12.four minus negative one13.the negative/opposite ofxminusy14.the negative/opposite ofaminusb15.negative three minus the negative/oppositeofn16.negative seven minus the negative/oppositeofm17.The additive inverse of 39 is–39 because39( 39)0+ −=18.The additive inverse of–17 is 17 because17170−+=19.The additive inverse of112−is112,because1111022−+=20.The additive inverse of72is72−because()77220+ −=21.The additive inverse of3.14−is 3.14 because3.143.140−+=22.The additive inverse of 48.2 is48.2−because48.248.20+ −=23.If45,x= −then( 45)45.x−= − −=24.If13,x=then13.x−= −25.If143,x= −then()141433.x−= − −=
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Solution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition - Page 31 preview image28Chapter 1:Elementary and Intermediate Algebra:Graphs and Models26.If1328,x=then1328.x−= −27.If0.101,x=then0.101.x−= −28.If0,x=then00.x−= −=29.If72,x=then()72.xx− −==30.If29,x=then()29.xx− −==31.If25,x= −then25().xx− −== −32.If9.1,x= −then()9.1.xx− −== −33.( 1)1− −=34.( 7)7− −=35.(7)7−= −36.(10)10−= −37.68(86)2−= −−= −38.413(134)9−= −−= −39.050( 5)5−=+ −= −40.080( 8)8−=+ −= −41.43(43)7−−= −+= −42.56(56)11−−= −+= −43.9( 3)936−− −= −+= −44.9( 5)954−− −= −+= −45.Note that we are subtracting a number fromitself. The result is 0. We could also do thisexercise as follows:8( 8)880−− −= − +=46.10( 10)10100−− −= −+=See exercise 45.47.304030( 40)10−=+ −= −48.202720( 27)7−=+ −= −49.7( 9)792−− −= −+=50.8( 3)835−− −= − += −51.9( 9)990−− −= − +=52.40( 40)40400−− −= −+=53.555( 5)0−=+ −=54.777( 7)0−=+ −=55.4( 4)448− −=+=56.6( 6)6612− −=+=57.747( 4)11−−= −+ −= −58.686( 8)14−−= −+ −= −59.6( 10)61016− −=+=60.3( 12)31215− −=+=61.6( 5)651−− −= −+= −62.4( 7)473−− −= − +=63.5( 12)51217− −=+=64.5( 6)5611− −=+=65.0( 10)01010− −=+=66.0( 1)011− −=+=67.5( 2)523−− −= − += −68.3( 1)312− − −= − += −69.7147( 14)21−−= −+ −= −70.9169( 16)25−−= −+ −= −71.80808− −= − += −
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