Solution Manual for Beginning Algebra, 7th Edition

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INSTRUCTOR’SSOLUTIONSMANUALCTRIMBLE&ASSOCIATESBEGINNINGALGEBRASEVENTHEDITIONElayn Martin-GayUniversity of New Orleans

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ContentsChapter 11Chapter 234Chapter 3107Chapter 4163Chapter 5215Chapter 6255Chapter 7297Chapter 8369Chapter 9411Appendix A467Appendix B469Appendix C471Appendix D473Practice Final Exam474

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1Chapter 1Section 1.2 PracticeExercises1.a.5 < 8 since 5 is to the left of 8 on the numberline.b.6 > 4 since 6 is to the right of 4 on thenumber line.c.16 < 82 since 16 is to the left of 82 on thenumber line.2.a.93 is true, since 9 > 3 is true.b.38 is false, since neither 3 > 8 nor 3 = 8 istrue.c.2525 is true, since 25 = 25 is true.d.414 is true, since 4 < 14 is true.3.a.3 < 8b.159c.674.The integer10represents10 metersbelow sealevel.5.a.The natural number is 25.b.The whole number is 25.c.The integers are 25,15,99.d.The rational numbers are 25,7 ,315,3 ,43.7, 8.8,99.e.The irrational number is5.f.The real numbers are 25,7 ,315,3 ,45,3.7, 8.8,99.6.a.0 < 3 since 0 is to the left of 3 onanumberline.b.15 >5 since 15 is to the right of5 onanumber line.c.1234since124simplifies to 3.7.a.|8| = 8 since8 is 8 units from 0 on anumber line.b.|9| = 9 since 9 is 9 units from 0 on a numberline.c.|2.5| = 2.5 since2.5 is 2.5 units from 0 ona number line.d.551111since511is511unit from 0 on anumber line.e.33since3is3units from 0 ona number line.8.a.|8| = |8| since 8 = 8.b.|3| > 0 since 3 > 0.c.|7| < |11| since 7 < 11.d.|3| > |2| since 3 > 2.e.|0| < |4| since 0 < 4.Vocabulary,Readiness& VideoCheck1.21.Thewholenumbers are {0, 1, 2, 3, 4, ...}.2.Thenaturalnumbers are {1, 2, 3, 4, 5, ...}.3.The symbols,, and > are calledinequalitysymbols.4.Theintegersare {...,3,2,1, 0, 1, 2, 3, ...}.5.Therealnumbers are {all numbers thatcorrespond to points onanumber line}.6.Therationalnumbers areandare integers,0.a abbb7.Theirrationalnumbers are {nonrational numbersthat correspond to points onanumber line}.8.The distance between a numberband 0 on anumber line is|b|.

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Chapter 1:Review of Real NumbersISM:Beginning Algebra29.To form a true statement: 0 < 7.10.Five is greater than or equal to four; 5411.0 belongs to the whole numbers, the integers, therational numbers, and the real numbers; since 0is a rational number, it cannot also be anirrational number.12.Theabsolute valueof a real numbera, denotedby |a|, is the distance betweenaand 0 on anumber line.Exercise Set 1.22.9 < 15 since 9 is to the left of 15 onanumberline.4.2.13 > 1.13 since 2.13 is to the right of 1.13 onanumber line.6.20 > 0 since 20 is to the right of 0 onanumberline.8.4 >6 since4 is to the right of6 onanumber line.10.0 < 100 since 0 is to the left of 100 onanumberline.12.360180since 360 is to the right of 180 on anumber line.14.47 is false, since 4 is to the left of 7 onanumber line.16.17 > 16 is true, since 17 is to the right of 16 onanumber line.18.8887 is false, since 56 is to the left of 64onanumber line.20.4 < 7 is true, since 4 is to the left of 7 onanumber line.22.0 <15 is false, since 0 is to the right of15 onanumber line.24.Fifteen is greater than five is written as 15 > 5.26.Negative ten is less than or equal to thirty-sevenis written as1037.28.Negative seven is not equal to seven is written as77.30.The integer 535 represents 535 feet above sealevel. The integer8 represents 8 feet below sealevel.32.The integer80,784represents80,784fewerstudents.34.30 represents an ascent of feet.50 represents adescent of 50 feet.36.The number14belongs to the sets of: rationalnumbers and real numbers.38.The number12belongs to the sets of: rationalnumbers and real numbers.40.The number 5 belongs to the sets of: naturalnumbers, whole numbers, integers, rationalnumbers, and real numbers.42.The number3belongs to the sets of: irrationalnumbers and real numbers.44.The number51 9belongs to the sets of: rationalnumbers and real numbers.46.False; negative numbers may be irrational.48.True50.False; irrational numbers are real.52.False;12is not an integer.54.False; 0 is a whole number that is not positive.56.200 <20 since200 is to the left of20 onanumber line.58.7.1 >7 since 7.1 is to the right of7 onanumber line.60.81223since 4 = 4.62.|20| >200 since 20 > 200.64.0 = |0| since 0 = 0.

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers366.2255 since22 .5568.500 < |50| since500 < 50.70.24122since2412.272.The 2014cranberry production in Massachusettswas206million pounds, while the 2014cranberry production in Wisconsin was539million pounds.206million <539million74.The bars for Oregon and New Jersey areclosestinlength, so Oregon and New Jersey hadtheclosestcranberry crops in 2014.76.The tallest bars represent 49 players admitted.78.Look for the bar with a height less than 30;19561965.80.answers may vary82.0.96 < 0.98 since 0.96 is to the left of 0.98 on thenumber line.84.Spica is dimmer since 0.98 > 0.96.86.Regulus is dimmest since 1.35 is to the right ofall other numbers listed.88.1313 has the same meaning as1313.90.73 <75 has the same meaning as75 >73.92.2 >4 has the same meaning as4 <2.94.answers may varySection 1.3PracticeExercises1.a.36 = 49 = 2233b.2002 1002 4 252 2 2 5 52.a.633 3 777722 2 2 3 32 2 28b.642 2 2 2 2 22 2 2 216122 2 333c.77255 5There are no common factors other than 1,so725is already in lowest terms.3.373 73 777898 92 2 2 3 32 2 2 3244.a.34393 92749444 416b.5515 151151212 1512 1512 3 536c.777 157 15153 55615676 762 325.a.8383515555b.828265555c.313145555d.5151611212121226.2272 7143373 7217.a.515 71 1111711 77 11351177773511774677b.510214217642131442

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Chapter 1:Review of Real NumbersISM:Beginning Algebra4c.129410294 6330530305 610292430303924301530128.1231225 656531 226 531 2 112 3 534115112215 9.1113767675121212133353535412125104040612Vocabulary,Readiness& VideoCheck1.31.A quotient of two numbers, such as5 ,8is calledafraction.2.In the fraction3 ,11the number 3 is called thenumeratorand the number 11 is called thedenominator.3.To factor a number means to write it as aproduct.4.A fraction is said to besimplifiedwhen thenumerator and the denominator have no commonfactors other than 1.5.In 73 = 21, the numbers 7 and 3 are calledfactorsand the number 21 is called theproduct.6.The fractions29and92are calledreciprocals.7.Fractions that represent the same quantity arecalledequivalentfractions.8.5, Fundamental Principle of Fractions9.The division operation changes to multiplicationand the second fraction120changes to itsreciprocal201.10.Find the LCD; two fractions must havethe sameor common denominatorbefore you can subtract(or add).11.The number74 6is not in proper mixed numberform as the fraction part,76,should not be animproper fraction.Exercise Set 1.32.1 of the 4 equal parts are shaded;144.2 of the 5 equal parts are shaded;256.60 = 415 = 22358.27 = 39 = 33310.56 = 414 = 222712.32 = 48 = 2222214.24 = 46 = 222316.31 3162 3218.153 53204 5420.5599

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers522.422 3 714453 3 51524.3602 2 2 3 3 52 3 3187002 2 5 5 75 73526.737 32111 511 55528.737 37 318218 218 3 7830.71727 27 2712212112 12 6 1632.393 103 103 2 52510595 95 3 3334.3103 103 2 52235 6335 635 7 3 3 73 7 714736.1 191 191 1919464 62 2 2 32438.115155Areasq m22422 4 216bh40.232 36Areasquare meters5 115 1155l w42.61615777744.616171777746.18111811771353535355 7548.13351335132132132481322 2 2 2 32 2 3 112 21141150.22 3633 3952.88 86477 85654.44 52055 52556.313 31 29211464 36 212121258.1114434434112164848484860.787 38 2101510 315 2211630302116305301662.31631631328888864.32113211326111111111111111The unknown part is6 .1166.131013 210163110510105 21010The unknown part is3 .1068.115121 21 4516312126 23 4121224512112The unknown part is1 .12

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Chapter 1:Review of Real NumbersISM:Beginning Algebra670.3711 1516552154848323272.7815351593 5 9271389898358 5 75674.117111426402640201020202566 20567 20167 476.71444816332216161121678.113113142 72353535355 7580.11311 53553527357 53535353582.373 73 7774 124 122 2 2 2 32 2 2 21684.373 123 123 3 49412474 74 7786.1701701717101010101088.5159159659 6599÷÷596666616 1190.777 57 33521352114152515 523 37575757592.57135713444 1111321321321323 4 11394.3713472451051013 2475 21026471010264710731037 1096.8118 41 111 22114211 44 112 2232112244444432112244434498.132332992 162 1228121833991433 4991 4413299441329942314357feet4100.answers may vary

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers7102.11137111773521228422847 411 4177 22 42 484 22844171488882844171481038712 8The distance for all 4 trails is712mi.8104.The sector for digital but not 3D is labeled1120.106.3331518940840404020920of U.S. movie screens are analog or digital3D.108.The work is correct.110.The work is incorrect.162 2 2 24282 2 77Section 1.4PracticeExercises1.a.311 1 11b.255 525c.21111101010100d.199e.32222855551252.a.6 + 39 = 6 + 27 = 33b.348364838311c.22432588or 33999 d.9(146)9(8)7236222e.7117434441616163.2653653131365831 838114.4[253(53)]4[253(8)]4[2524]4[1]45.2369545925322536.a.2x+y= 2(2) + 5 = 4 + 5 = 9b.44(2)833(5)15xyc.3321541925101010xxyd.32322582533xy7.9679(4)67(4)366283028Falsexx004 is not a solution of 9x6 = 7x.8.a.Six times a number is 6x, since 6xdenotesthe product of 6 andx.b.A number decreased by 8 isx8 because“decreased by” means subtract.c.The product of a number and 9 isx9 or 9x.d.Two times a number is 2x, plus 3 is 2x+ 3.e.The sum of 7 and a numberxis 7 +x.

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Chapter 1:Review of Real NumbersISM:Beginning Algebra89.a.A numberxincreased by 7 isx+ 7, sox+ 7 = 13.b.Two less than a numberxisx2, sox2 = 11.c.Double a numberxis 2x, added to 9 is2x+ 9, so 2x+ 925.d.Five times 11 is 5(11), so 5(11)x, wherexis an unknown number.Calculator Explorations1.456252.4724013.5959, 0494.68262,1445.2(205) = 306.3(147) + 21 = 3(7) + 21 = 21 + 21 = 427.24(862455) + 89 = 98578.99 + (401 + 962) = 14629.46231292376363410.9564521688986Vocabulary,Readiness& VideoCheck1.41.In the expression25 ,the 5 is called thebaseandthe 2 is called theexponent.2.The symbols ( ), [ ], and { } are examples ofgroupingsymbols.3.A symbol that is used to represent a number iscalled avariable.4.A collection of numbers, variables, operationsymbols, and grouping symbols is called anexpression.5.A mathematical statement that two expressionsare equal is called anequation.6.A value for the variable that makes an equation atrue statement is called asolution.7.Deciding what values of a variable make anequation a true statement is calledsolvingtheequation.8.The order in whichwe perform operations doesmatter! We came upwithan order of operationsto avoid getting more than one answer whenevaluating an expression.9.The replacement value forzis not used becauseit’s not neededthere is no variablezin thegiven algebraic expression.10.No; the variable was replaced with 0 in theequation to see if a true statement occurred, andit did not.11.We translate phrases to mathematicalexpressions and sentences to mathematicalequations.Exercise Set 1.42.522 2 2 2 2324.444 4 4 42566.811 1 1 1 1 1 1 11      8.18810.299 98112.26666 63611111111 1112114.51111112222221 1 1 1 12 2 2 2 213216.21 51 5 1 52 25( . )( . )( . ).18.30 030 030 030 030 000027( .)( .)( .)( .).20.8 + 53 = 8 + 15 = 23

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers922.12536 = 6018 = 4224.5(62) = 5(4) = 2026.562 2262 23264323428.22 52 255030.31232916254238324242432.3[43(64)]3[43(2)]3[46]3[10]3034.4 3212214743 24610536.153115215213123 21266638.2236(85)36(3)424236(3)16231816221187640.222161354168417517516841751681617540124 104 310342.34[8(5 520)39]34[8(2520)39]34[8(5)39]34[4039]34[1]34744.2311391138482644829136441691612646464376446.a.(1 + 4)63 = 563 = 303 = 27b.1 + 4(63) = 1 + 43 = 1 + 12 = 13c.1 + 463 = 1 + 243 = 22d.(1 + 4)(63) = 53 = 1548.Letx= 1.4x= 4(1) = 450.Lety= 3 andz= 5.3322(5)10yz52.Lety= 3.6y8 = 6(3)8 = 188 = 1054.Lety= 3 andz= 5.525(5)2(3)2561919zy56.Letx= 1,y= 3,andz= 5.3 5115114yzx58.Letz= 5.2222(5)2(25)50z60.Letx= 12,y= 8,andz= 4.888(12)296984yxz62.Letx= 12 andy= 8.223(8)3(12)86436836yxy

Solution Manual for Beginning Algebra, 7th Edition - Page 13 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra1064.Letx= 12 andy= 8.2222(8)126412761914424168423(12)3(8)yxxy66.No;answers may vary.68.Letx= 6.2732(6)73(6)127181918, falsexx006 is not a solution of the equation.70.Letx= 2.4284(2)22882101010, truexx002 is a solution of the equation.72.Letx= 6.31083(6)1081810888, truex006 is a solution of the equation.74.Letx= 10.661061061616, truexx010is a solution of the equation.76.Letx= 1.4141140, falsex01 is not a solution of the equation.78.A number increased by 9 is written asx+ 9.80.Five decreased by a number is written as 5x.82.The quotient of a number and 9 is written as.9x84.Twice a number, decreased by 72 is written as 2x72.86.Four subtracted from eight is equal to two squared is written as2842 .88.The difference of sixteen and four is greater than ten is written as 164 > 10.90.Seven subtracted from a number is 0 is written asx7 = 0.

Solution Manual for Beginning Algebra, 7th Edition - Page 14 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers1192.9.1 times a number equals 4 is written as 9.1x= 4.94.Eight added to twice a number is 42 is written as 2x+ 8 = 42.96.To simplify the expression (1 + 3)6, firstadd.98.To simplify the expression 2042, firstdivide.100.Yes;answers may vary.Length,lWidth,wPerimeter of Rectangle:2l+ 2wArea of Rectangle:lw102.6 in.1 in.2l+ 2w= 2(6 in.) + 2(1 in.)= 12 in. + 2 in.= 14 in.lw= (6 in.)(1 in.)= 6 sq in.104.4.6 in.2.4 in.2l+ 2w= 2(4.6 in.) + 2(2.4 in.)= 9.2 in. + 4.8 in.= 14 in.lw= (4.6 in.)(2.4 in.)= 11.04 sq in.106.answers may vary108.22 (53 )2 (59)2 1428110.a.2326xis an expression since it does not containanequalsign.b.23261xis an equation since it contains anequalsign.c.2x5 = 7x5 is an equation since it containsan equal sign.d.9y+x8 is an expression since it does not containan equal sign.e.234(53)is an expression since it does not containan equal sign.112.answers may vary114.answers may vary, for example, 2(107) + 1:2(107) + 1 = 2(3) + 1 = 6 + 1 = 7116.Leth= 5,B= 15, andb= 7.111()(5)(157)(5)(22)55222Ah BbThe area is 55 sq in.118.LetP= 650,T= 3, andI= 126.75.126.75126.750.0656.5%(650)(3)1950IPT

Solution Manual for Beginning Algebra, 7th Edition - Page 15 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra12Section 1.5PracticeExercises1.2 + 4 = 62.2 + (3) =53.a.5 + (8)Add the absolute values.5 + 8 = 13The common sign is negative, so5 + (8) =13.b.31 + (1)Add the absolute values.31 + 1 = 32The common sign is negative, so31 + (1) =32.4.3 + 8 = 55.a.15 + (18)Subtract the absolute values.1815 = 3Use the sign of the number with the largestabsolute value.15 + (18) =3b.19 + 20 = 2019 = 1c.0.6 + 0.4 =(0.60.4) =(0.2) =0.26.a.3251555  b.3 + (9) =6c.2.2 + (1.7) = 0.5d.2320211710707070 7.a.8 + (5) + (9) = 3 + (9) =6b.[ 85]52[ 3][ 52]3[ 3]6      8.7 + 4 + 7 =3 + 7 = 4The temperature at 8A.M. was 4F.9.a.The opposite of59is5 .9b.The opposite of 8 is8.c.The opposite of 6.2 is6.2.d.The opposite of3 is 3.10.a.Since |15| = 15,|15| =15.b.3355c.(5y) = 5yd.(8) = 8Vocabulary,Readiness& VideoCheck1.51.Two numbers that are the same distance from 0but lie on opposite sides of 0 are calledopposites.2.Ifnis a number, thenn+ (n) =0.3.Ifnis a number, then(n) =n.4.The sum of two negative numbers is alwaysanegative number.5.To add two numbers with the same sign, addtheirabsolute valuesand use their common signas the sign of the sum.6.Negative; when you add two numbers withdifferent signs, the sign of the sum is the same asthe sign of the number with the larger absolutevalue and8.4 has a larger absolute value than6.3.7.Negative temperatures; the high temperature forthe day was6F.8.Example13 is an example of the opposite of theabsolute valueofa, not the opposite ofa. Theabsolute value ofais positive, so its opposite isnegative, therefore the answers toExamples12and 13 have different signs.

Solution Manual for Beginning Algebra, 7th Edition - Page 16 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers13Exercise Set 1.52.9 + (12) =34.6 + (14) =206.6 + (4) = 28.10 + 5 =510.7 + (4) =1112.7 + (5) = 214.5 + 9 = 416.8 + (6) = 218.3 + (6) =320.23 + (23) = 022.53 + (37) = 1624.26 + 14 =1226.18 + (26) =4428.9.2 + (11.4) =2.230.|6| + (61) = 6 + (61) =5532.144 + (88) = 5634.6.7 + (7.6) =14.336.5721121212638.5151 35329393 3999   40.5252 26363 254669632    42.9 + 15 + (5) = 6 + (5) = 144.18 + (6) + (40) =24 + (40) =6446.14 + (3) + 11 =17 + 11 =648.|7 + (17)| = |10| = 1050.8 + (2) + 7 = 6 + 7 = 1352.[2 + (7)] + [11 + 22] = [9] + [11] = 254.|43 + (73)| + |20| = |30| + 20 = 30 + 20 = 5056.3.7[0.1( 0.6)8.1]3.7[ 0.58.1]3.7[7.6]3.9    58.4 + 49 = 45It rose to a temperature of 45F.60.52 + 439 = 387Your elevation is 387 feet.62.3 03 226 463 96 226 463 920 263 984 1.(. )........  The total net income for fiscal year 2013 was$84.1 million.64.6 + (4) + (5) + (6) =21His score was21 or 21 under par.66.The opposite of 4 is4.68.The opposite of8 is 8.70.The opposite of14is1 .472.Since |11| is 11, the opposite of |11| is11.74.(3) = 376.223378.(7) = 780.Letx= 10.737(10)377, falsex  010 is not a solution of the equation.

Solution Manual for Beginning Algebra, 7th Edition - Page 17 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra1482.Lety=6.1716711, truey06 is a solution of the equation.84.Look for thebar that extends farthest in thenegative direction. The month for which thetemperature is the lowest is February.86.Look for the bar whose length has a negativevalue closest to 0; November88.[( 19.3)27.08.8]3[7.78.8]3[16.5]35.5The average was 5.5F.90.4 + 14 = 1092.15 + (17) =3294.Sincebis a negative number,bis apositivenumber.96.Sincebis a negative number,b+bis anegativenumber.98.True100.True102.answers may vary104.answers may varySection 1.6PracticeExercises1.a.76 =7 + (6) =13b.8(1) =8 + 1 =7c.9(3) = 9 + 3 = 12d.57 = 5 + (7) =22.a.8.4(2.5) = 8.4 + 2.5 = 10.9b.515141888882   c.3131454515420201920   3.25 =2 + (5) =74.a.152( 4)715( 2)476    b.3.5( 4.1)( 6.7)3.5( 4.1)6.76.1   5.a.4[( 83)5]4[( 8( 3))5]4[( 11)5]4[ 11( 5)]4[ 16]20         b.2133[2( 7)]139[27]139913 6.a.77( 3)7310222(4)( 3)8( 3)5xyx   b.22(4)( 3)16( 3)13yx  7.282(75) = 282 + 75 =357The overall change was $357.8.a.x= 9062= 28b.y= 18043= 137Vocabulary,Readiness& VideoCheck1.61.7 minus a number7x2.7 subtracted from a numberx7.3.A number decreased by 7x74.7 less a number7x5.A number less than 77x6.A number subtracted from 77x

Solution Manual for Beginning Algebra, 7th Edition - Page 18 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers157.To evaluatexyforx=10 andy=14, wereplacexwith10 andywith14 and evaluate10(14).d8.The expression510 equals5 + (10).c9.To subtract two real numbers, change theoperation toadditionand take theoppositeof thesecond number.10.10 + (8) + (4) + (20); it’s rewritten to changethe subtraction operations to addition and turnthe problem into an addition of real numbersproblem.11.There’s a minus sign in the numerator and thereplacement value is negative (notice parenthesesare used around the replacement value), and it’salways good to be careful when working withnegative signs.12.This means that the overall vertical altitudechange of the jet is actually a decrease in altitudefrom when theExamplestarted.13.InExample9, you have two supplementaryangles and know the measure of one of them.From the definition, you know that the twosupplementary angles must sum to 180.Therefore you can subtract the known anglemeasure from 180to get the measure of theother angle.Exercise Set 1.62.128 =12 + (8) =204.811 = 8 + (11) =36.12(5) = 12 + 5 = 178.373748483 274 28678818 10.20(48) =20 + 48 = 2812.84 =8 + (4) =1214.3(6) = 3 + 6 = 916.4(16) =4 + 16 = 1218.15(33) = 15 + 33 = 4820.8.311.2 = 8.3 + (11.2) =2.922.3651 =36 + (51) =8724.17(17) =17 + 17 = 026.6.1(5.3) =6.1 + 5.3 =0.828.4141377777  30.17171081081 47 510 48 543540403940    32.4.3(0.87) = 4.3 + 0.87 = 5.1734.23 =2 + (3) =536.117 = 1 + (17) =1638.49 =4 + (9) =1340.11(14) = 11 + 14 = 2542.16( 3)( 11)14163( 11)( 14)13( 11)( 14)24( 14)38           44.712( 5)2( 2)7( 12)( 5)( 2)( 2)5( 5)( 2)( 2)10( 2)( 2)12( 2)14                46.9(38) =9(5) =9 + 5 =448.326 386 38188( 18)10  50.46(73)46(4)4244( 24)20  

Solution Manual for Beginning Algebra, 7th Edition - Page 19 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra1652.222(23)5[2( 3)]5[ 1]5[ 1]2524   54.5[(415)( 6)8]5[(4( 15))6( 8)]5[( 11)6( 8)]5[( 5)( 8)]5[ 13]18           56.22226( 38)26[ 3( 8)]26[11]236[ 11]38[ 11]27      58.Letx=5 andy= 4.yx= 4(5) = 4 + 5 = 960.Letx=5,y= 4, andt= 10.75107(4)51028510( 28)5( 28)2323xty      62.Letx=5 any= 4.1515( 5)155202 10102424262 33xy 64.Letx=5 andt= 10.2210( 5)1005105tx 66.Letx=5,y= 4 andt= 10.55(4)( 5)66(10)2056(10)256(10)25605 512 5512yxt 68.5 + (12) =7The new elevation is7 feet or7 feet below sealevel.70.134(80) = 134 + 80 = 214134F is 214F warmer than80F.72.93182693( 18)( 26)75( 26)49   She owes $49 on her account.74.1524 = 15 + (24) =9Her new score is9.76.The difference in elevation isthe higher elevationthe lower elevation.4101(226) = 4101 + 226 = 4327The Bishop airport is4327 feet higher.78.x= 9050 = 90 + (50) = 40The complementary angle is 40.80.y= 180105 = 180 + (105) = 75The supplementary angle is 75.82.Letx= 3.107310777, truex  03 is a solution of the equation.84.Letx=10.61( 10)6( 10)110( 6)10( 1)49, falsexx     0010 is not a solution of the equation.

Solution Manual for Beginning Algebra, 7th Edition - Page 20 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers1786.Letx= 5.4141541( 5)44, falsex  005 is not a solution of the equation.88.The difference of3 and a number is3x.90.Add a number and36 isx+ (36).92.MonthMonthly Increase or DecreaseJuly33.629.7=33.6 + (29.7) = 3.9August33.333.6 = 33.3 + (33.6) =0.3September27.033.3 = 27.0 + (33.3) =6.3October8.827.0 = 8.8 + (27.0) =18.2November6.98.8 =6.9 + (8.8) =15.7December17.2(6.9) =17.2 + 6.9 =10.394.The largest negative number corresponds to October.96.answers may vary98.48 =4 + (8) =12100.3(10) =3 + 10 = 7102.true; answers may vary104.true; answers may vary106.Since 4.362 is less than 7.0086, the answer is negative.4.3627.0086 =2.6466Integrated Review1.The opposite of a positive number is anegativenumber.2.The sum of two negative numbers is anegativenumber.3.The absolute value of a negative number is apositivenumber.4.The absolute value of zero is0.5.The reciprocal of a positive number is apositivenumber.6.The sum of a number and its opposite is0.7.The absolute value of a positive number is apositivenumber.8.The opposite of a negative number is apositivenumber.

Solution Manual for Beginning Algebra, 7th Edition - Page 21 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra18NumberOppositeAbsolute Value9.17171710.12512512511.33312.91191191113.19 + (23) =4214.7(3) = 7 + 3 = 1015.15 + 17 = 216.810 =8 + (10) =1817.18 + (25) =718.2 + (37) =3919.14(12) =14 + 12 =220.514 = 5 + (14) =921.4.57.9 = 4.5 + (7.9) =3.422.8.61.2 =8.6 + (1.2) =9.823.3121421425472828282828   24.27162116215382424242424 25.9( 7)469746974( 6)4     26.1120( 3)1211( 20)( 3)( 12)9( 3)( 12)12( 12)24          27.246(1411)246[14( 11)]246(3)241824( 18)6  28.305(108)305[10( 8)]305(2)301030( 10)20  29.22(717)4[7( 17)]4( 10)166  30.229(1030)9[10( 30)]81( 20)61  31.293( 420)99[ 4( 20)]99( 24)18( 24)6      32.22455( 50)4( 5)5( 50)925( 50)925( 50)34( 50)16         33.7[(12)( 29)]7[( 1)( 11)]7[ 12]19      34.6[( 37)(415)]6[(4)( 11)]6( 7)13     35.15 = 1 + (5) =436.3(2) =3 + 2 =137.12125813454520202038.51254254298104040404040   39.3232322(1917)3( 79)2[19( 17)]3( 79)2(2)3(2)2(8)3(4)161216( 12)4  

Solution Manual for Beginning Algebra, 7th Edition - Page 22 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers1940.2323233(109)6(2019)3[10( 9)]6[20( 19)]3(1)6(1)369  41.xy=2(1) =2 + 1 =142.x+y=2 + (1) =343.y+z=1 + 9 = 844.zy= 9(1) = 9 + 1 = 1045.55(9)( 2)45247471( 2)121zxyx    46.( 2)( 1)922(9)219181218121823xyzz  Section 1.7PracticeExercises1.a.8(5) =40b.(3)(4) = 12c.(6)(9) =542.a.(1)(5)(6) = 5(6) =30b.(3)(2)(4) = 6(4) = 24c.(4)(0)(5) = 0(5) = 0d.( 2)( 3)( 4)(5)6( 20)62026  3.a.(0.23)(0.2) =[(0.23)(0.2)] =0.046b.343 4124595 94515      c.77 24( 24)7 2141212 14.a.2( 6)( 6)( 6)36 b.26(6 6)(36)36   c.3( 4)( 4)( 4)( 4)16( 4)64  d.34(4 4 4)[16(4)]64   5.a.The reciprocal of83is38since8 31.3 8b.The reciprocal of 15 is115since1151.15c.The reciprocal of27is72since271.72d.The reciprocal of5 is15since1( 5)1.56.a.16116822 b.124( 6)2446  c.35355 757777.a.18183 63666b.48483 1616333   c.3136( 2)5255  d.44 1411899 89 4 29 218    

Solution Manual for Beginning Algebra, 7th Edition - Page 23 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra208.a.002b.40is undefined.c.556(0)0is undefined.9.a.( 8)( 11)488484849( 4)9455  b.33( 2)93( 8)9633249333311 10.a.7yx= 7(2)(5) =14 + 5 =9b.2323( 5)( 2)25( 8)25833xy   c.22( 5)10533( 2)63xy11.total score = 4(13) =52The card player’s total score was52.Calculator Explorations1.38(2627) = 382.59(8) + 1726 = 21983.134 + 25(6891) =4414.45(32)8(218) =3045.50(294)163.31752656.444444.81.761813247.59455054, 4998.856259384,3669.2( 125)15,62510.212515, 625 Vocabulary,Readiness& VideoCheck1.71.Ifnis a real number, thenn0 =0and 0n=0.2.Ifnis a real number, but not 0, then00nandwe say0nisundefined.3.The product of two negative numbers is apositivenumber.4.The quotient of two negative numbers is apositivenumber.5.The quotient of a positive number and a negativenumber is anegativenumber.6.The product of a positive number and a negativenumber is anegativenumber.7.The reciprocal of a positive number is apositivenumber.8.The opposite of a positive number is anegativenumber.9.The parentheses, or lack of them, determine thebase of the expression. InExample6,42() ,thebase is2 and all of2 is raised tothefourthpower. InExample7,42,the base is 2 andonly 2 is raised tothefourth power.10.Remember, the product of a number and itsreciprocal is 1,not1.23132,as needed.11.Yes; because division of real numbers is definedin terms of multiplication.12.The replacement values are negative and bothwill be squared. Therefore they must be placed inparentheses so the entire value, including thenegative, is squared.

Solution Manual for Beginning Algebra, 7th Edition - Page 24 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers2113.The football team lost 4yardson each play and aloss of yardage is represented by a negativenumber.Exercise Set 1.72.8(5) =404.7(4) =286.6(11) = 668.28 =1610.60 = 012.3(5) =1514.111 11838 32416.535 3151516106 106015 4418.6(2.5) =1520.0.5(0.3) = 0.1522.20(60) =120024.5(9) =4526.(7)(7) = 4928.222 247117 1177  30.12(12) =14432.25625 61505361536 1548018   34.( 2)( 3)( 4)( 2)6( 4)( 2)24( 2)48 36.8( 3)4( 5)24( 20)24204    38.20( 4)(3)( 2)20(24)20( 24)4   40.True; example: 222 = 842.True; example: 2222 = 1644.42(2 2 2 2)16  46.5( 1)( 1)( 1)( 1)( 1)( 1)1  48.25(5 5)25  50.2( 7)( 7)( 7)49 52.Reciprocal of 100 is1100since11001.10054.Reciprocal of17is 7 since1 71.756.Reciprocal of8 is18since181.8  58.Reciprocal of613is136since6131.136 60.Reciprocal of 1.5 is11.5since11.51.1.562.Reciprocal of18.9is8.9 since18.91.8.9 64.2012021010  66.18118366  68.601601255  70.010099 72.24124388  74.30is undefined.76.45145599  78.14114722  

Solution Manual for Beginning Algebra, 7th Edition - Page 25 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra2280.41424 2852515 15  82.181111 1111101110810 880 84.555125 121121212512 5   86.6( 3)189442 88.1515514390.234941355592.26( 2)6410249495  94.20( 4)(3)20( 12)32815154  96.2242162( 16)1813( 6)3( 6)1818 98.83( 2)8( 6)8614725( 4)2( 20)2202211  100.48( 2)4( 16)41612492( 3)9( 6)963   102.36273535822 2444 104.Letx=5 andy=3.4x+ 5y= 4(5) + 5(3) =20 + (15) =35106.Letx=5 andy=3.22222( 5)2( 3)252(9)25( 18)7xy  108.Letx=5 andy=3.333( 3)3( 5)27( 15)42yx    110.Letx=5 andy=3.2122( 3)12612182454549yx112.Letx=5 andy=3.4242( 5)410 is undefined.3330xy114.The quotient of8 and a number is8xor8x.116.The sum of a number and12 isx+ (12).118.The difference of a number and10 isx(10).120.Multiply a number by17 isx(17) or17x.122.A loss of $400 is represented by400.7(400) =2800His total loss was $2800.124.A drop of 5 degrees is represented by5.6(5) =30The total drop in temperature was 30 degrees.126.Letx=4.212( 4)4185, falsexx 04 is not a solution of the equation.128.Letx=3.4515451531515, truex  03 is a solution of the equation.130.Letx=4.2482( 4)44884444, falsexx004 is not a solution of the equation.132.3303301 The surface temperature of Neptune is330F.134.answers may vary

Solution Manual for Beginning Algebra, 7th Edition - Page 26 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers23136.answers may vary138.Sinceqis negative,ris negative, andtispositive, then2qr tis negative.140.Not possible to determine whethert+rispositive or negative.142.Sinceqis negative,ris negative, andtispositive, thenr(qt) is positive.144.1 + (8)(5) = 1 + 40 = 41146.070775 Section 1.8PracticeExercises1.a.x8 =8xb.x+ 17 =17 +x2.a.(2 + 9) + 7 =2 + (9 + 7)b.4(27) =(42)73.a.(5)9(5)9(59)14xxxxb.5(6x) = [5(6)]x=30x4.a.5(xy) = 5(x)5(y) = 5x5yb.6(4 + 2t) =6(4) + (6)(2t) =2412tc.2(34)2(3 )2( 4 )2()682xyzxyzxyzd.(3y)(1) = 3(1) + (y)(1) =3 +ye.(72 )( 1)(72 )( 1)( 1)( 7)( 1)(2 )72xsxsxsxs     f.1112492492222911xxxx()()( )5.a.5w+ 53 = 5(w+ 3)b.9w+ 9z= 9w+ 9z= 9(w+z)6.a.(73x)4 = (3x7)4; commutativeproperty of multiplicationb.6 + (3 +y) = (6 + 3) +y; associativeproperty of additionc.8 + (t+ 0) = 8 +t; identity element foradditiond.341;43multiplicative inversepropertye.(2 +x) + 5 = 5 + (2 +x); commutativeproperty of additionf.3 + (3) = 0; additive inverse propertyg.(3b)7 = (37)b; commutative andassociative properties of multiplicationVocabulary,Readiness& VideoCheck1.81.x+ 5 = 5 +xis a true statement by thecommutative property of addition.2.x5 = 5xis a true statement by thecommutative property of multiplication.3.3(y+ 6) = 3y+ 36 is true by thedistributiveproperty.4.2(xy) = (2x)yis a true statement by theassociative property of multiplication.5.x+ (7 +y) = (x+ 7) +yis a true statement by theassociative property of addition.6.The numbers23and32are calledreciprocalsor multiplicative inverses.7.The numbers23and23are calledopposites oradditive inverses.8.order; grouping9.2 is outside the parentheses, so the point is madethat you should only distribute the9 to theterms within the parentheses and not also to the2.

Solution Manual for Beginning Algebra, 7th Edition - Page 27 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra2410.The identity element for addition is0because ifweadd0to any real number, the result is thatreal number.The identity element for multiplication is1because any real number times1gives a result ofthat original real number.Exercise Set 1.82.4 +y=y+ 44.2x=x(2)6.ab=ba8.19 + 3y =3y+ 1910.3(xy) = (3x)y12.(y+ 4) +z=y+ (4 +z)14.(3y)z=3(yz)16.6 + (r+s) = (6 +r) +s18.(r+ 3) + 11 =r+ (3 + 11) =r+ 1420.2(42x) = (242)x= 84x22.11(8 )8188zzzz24.7 + (x+ 4) = 7 + (4 +x) = (7 + 4) +x= 11 +x26.3(12y) = (312)y=36y28.272717272rrrr30.72729999991yyyy32.7(a+b) = 7a+ 7b34.11(y4) = 11y114 = 11y4436.5(7 + 8y) = 5(7) + 5(8y) = 35 + 40y38.3(8x1) = 3(8x)3(1) = 24x340.2(x+ 5) = 2(x) + 2(5) = 2x+ 1042.3(zy) =3z(3)y=3z+ 3y44.5(2r+ 11) =5(2r) + (5)(11) =10r5546.8(3y + z6) = 8(3y) + 8z8(6) = 24y+ 8z4848.4(425 )4(4)( 4)(2)( 4)(5 )16820    pqpqpq50.(95)1(95)1(9 )( 1)(5)95rrrr    52.(26 )1(26 )1( 1)(2)( 1)(6 )26qrqrqrqr     54.111(42)(4 )(2)44411424412xxxx56.111(1025 )(10 )(25 )555111025552525ababababab   58.10(46)4010(4 )10(6)404060( 40)4020ssss 60.11(53)1011(5 )( 11)(3)105533105523xxxx    62.6(21)16(2 )( 6)(1)112( 6)1127xxxx     64.14z+ 145 = 14(z+ 5)66.9a+ 9b= 9(a+b)68.(3)a+ (3)b=3(a+b)70.25x+ 25y= 25(x+y)72.4(3 + 8) = 43 + 48; distributive property

Solution Manual for Beginning Algebra, 7th Edition - Page 28 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers2574.(x+ 9) + 3 = (9 +x) + 3; commutative propertyof addition76.19 = 9; identity elementformultiplication78.161;6multiplicative inverse property80.(a+ 9) + 6 =a+ (9 + 6); associative property ofaddition82.(11 +r) + 8 = (r+ 11) + 8; commutativeproperty of addition84.r+ 0 =r; identity elementforadditionExpressionOppositeReciprocal86.23233288.4y4y14y90.7x7x17x92.False; the reciprocal of2ais2 .aThe opposite of2ais.2a94.“Putting on your shoes” and “putting on yoursocks” are not commutative, since the order inwhich they are performed affects the outcome.96.“Reading the sports section” and “reading thecomics section” are commutative, since the orderin which they are performed does not affect theoutcome.98.“Baking a cake” and “eating the cake” are notcommutative, since the order in which they areperformed affects the outcome.100.“Feeding the dog” and “feeding the cat” arecommutative, since the order in which they areperformed does not affect the outcome.102.a.The property illustrated is the associativeproperty of addition since the grouping ofaddition changed.b.The property illustrated is the commutativeproperty of addition since the order in whichthey are added changed.c.The property illustrated is the commutativeproperty of addition since the order in whichthey are added changed.104.answers may vary106.answers may varyChapter 1 Vocabulary Check1.The symbols, <, and > are calledinequalitysymbols.2.A mathematical statement that two expressionsare equal is called anequation.3.Theabsolute valueof a number is the distancebetween that number and 0 onanumber line.4.A symbol used to represent a number is called avariable.5.Two numbers that are the same distance from 0but lie on opposite sides of 0 are calledopposites.6.The number in a fraction above the fraction baris called thenumerator.7.Asolutionof an equation is a value for thevariable that makes the equation a true statement.8.Two numbers whose product is 1 are calledreciprocals.9.In32 ,the 2 is called thebaseand the 3 is calledtheexponent.10.The number in a fraction below the fraction baris called thedenominator.11.Parentheses and brackets are examples ofgrouping symbols.12.Asetis a collection of objects.Chapter 1 Review1.8 < 10 since 8 is to the left of 10 onanumberline.2.7 > 2 since 7 is to the right of 2 onanumber line.3.4 >5 since4 is to the right of5 onanumber line.

Solution Manual for Beginning Algebra, 7th Edition - Page 29 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra264.1282 since 6 >8.5.|7| < |8| since 7 < 8.6.|9| >9 since 9 >9.7.|1| =1 since1 =1.8.|14| =(14) since 14 = 14.9.1.2 > 1.02 since 1.2 is to the right of 1.02 onanumber line.10.3324 since32is to the left of34onanumber line.11.Four is greater than or equal to negative three iswritten as 43.12.Six is not equal to five is written as 65.13.0.03 is less than 0.3 is written as 0.03 < 0.3.14.155 < 400 or 400 > 15515.a.The natural numbers are 1 and 3.b.The whole numbers are 0, 1, and 3.c.The integers are6, 0, 1, and 3.d.The rational numbers are6, 0, 1,11,23,and 9.62.e.The irrational number is.f.The real numbers are all numbers in thegiven set.16.a.The natural numbers are 2 and 5.b.The whole numbers are 2 and 5.c.The integers are3, 2, and 5.d.The rational numbers are3,1.6, 2, 5,11 ,2and 15.1.e.The irrational numbers are5and 2.f.The real numbers are all numbers in thegiven set.17.Look for the negative number with the greatestabsolute value. The greatest loss was on Friday.18.Look for the largest positive number. Thegreatest gain was on Wednesday.19.36 = 49 = 223320.120 = 815 = 2223521.8278 272 4 3 3 31215 3015 303 5 2 3 52522.7217 327 327 8 448328218 218 3 7323.757 25 515615 26 51425303014253039303 133 10131024.333 534204 52015320201532012203 45 435

Solution Manual for Beginning Algebra, 7th Edition - Page 30 preview image

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ISM:Beginning AlgebraChapter 1:Review of Real Numbers2725.35115326484811 2534 282253882253875839 826.12438726363438 263 2431666431662769 32 39214 227.1355153128.3352 353512 8217442 22229.11121 21 3164126 24 31223121212122312712The unknown part is7 .1230.111052105231251010101010The unknown part is310.31.22Plw172 123824271318814388 814 33 88 36442242464422410624104 2454meters12PAlw171 3847384 73 2 47611sq meters6A32.the sum of the lengths of the sidesP5833252642in.1111111111111111Pthe sum of the two areas, each given byAlw553325934sq in.11 1111 11121121121A

Solution Manual for Beginning Algebra, 7th Edition - Page 31 preview image

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Chapter 1:Review of Real NumbersISM:Beginning Algebra2833.11391372312122164161616549741293921641616164049284129391616161616162261621416114lb834.1317353435695224lb8168161616161635.Total weightweight of boysweight of girls22669161629516718lb1636.Look for the largest number.Baby Cweighedthe most.37.Look for the smallest number.Baby Eweighedthe least.38.134974928215311lb1641641616161639.26 32 86 92 8541670The answer is c.40.3685 2685 8684068( 40)28 The answer is b.41.222247774942.333332744446443.3(12 5)43(110)43(11)43343744.83(2 61)83(121)83(11)8334145.24628446446 4424446442472284 184 718746.5[3(25)5]5[3(7)5]5[215]5[16]8047.The difference of twenty and twelve is equal tothe product of two and four is written as2012 = 24.48.The quotient of nine and two is greater thannegative five is written as95.2 49.Letx= 6 andy= 2.2x+ 3y= 2(6) + 3(2) = 12 + 6 = 1850.Letx= 6,y= 2, andz= 8.(2 )6[22(8)]6[216]6[18]108x yz51.Letx= 6,y= 2, andz= 8.6868325222(2)24xzyy52.Letx= 6 andy= 2.22223(6)3(2)363(4)361236( 12)24xy 53.Replaceawith 37 andbwith 80.1801803780180( 37)( 80)143( 80)63ab   The measure of the unknown angle is 63.

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