Solution Manual for College Algebra, 3rd Edition

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Solution Manual for College Algebra, 3rd Edition - Page 1 preview image

CONTENTSChapter PBasic Concepts of AlgebraP.1The Real Numbers and Their Properties...................................................... 1P.2Integer Exponents and Scientific Notation .................................................. 6P.3Polynomials................................................................................................ 10P.4Factoring Polynomials ............................................................................... 14P.5Rational Expressions.................................................................................. 19P.6Rational Exponents and Radicals............................................................... 27Chapter P Review Exercises ................................................................................... 35Chapter P Practice Test ........................................................................................... 40Chapter 1Equations and Inequalities1.1Linear Equations in One Variable.............................................................. 411.2Quadratic Equations................................................................................... 551.3Complex Numbers: Quadratic Equations with Complex Solutions........... 691.4Solving Other Types of Equations ............................................................. 771.5Inequalities................................................................................................. 971.6Equations and Inequalities Involving Absolute Value............................. 113Chapter 1 Review Exercises.................................................................................. 129Chapter 1 Practice Test A ..................................................................................... 140Chapter 1 Practice Test B...................................................................................... 141Chapter 2Graphs and Functions2.1The Coordinate Plane............................................................................... 1442.2Graphs of Equations................................................................................. 1542.3Lines......................................................................................................... 1672.4Functions.................................................................................................. 1802.5Properties of Functions ............................................................................ 1892.6A Library of Functions............................................................................. 1982.7Transformations of Functions .................................................................. 2092.8Combining Functions; Composite Functions........................................... 2222.9Inverse Functions ..................................................................................... 238Chapter 2 Review Exercises.................................................................................. 250Chapter 2 Practice Test A ..................................................................................... 260Chapter 2 Practice Test B...................................................................................... 261Cumulative Review Exercises (Chapters P−2) ..................................................... 262Chapter 3Polynomial and Rational Functions3.1Quadratic Functions ................................................................................. 2663.2Polynomial Functions .............................................................................. 2853.3Dividing Polynomials .............................................................................. 2983.4The Real Zeros of a Polynomial Function ............................................... 3093.5The Complex Zeros of a Polynomial Function ........................................ 3283.6Rational Functions ................................................................................... 3373.7Variation .................................................................................................. 355Chapter 3 Review Exercises.................................................................................. 360Chapter 3 Practice Test A ..................................................................................... 376Chapter 3 Practice Test B...................................................................................... 378Cumulative Review Exercises (Chapters P−3) ..................................................... 379

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Chapter 4Exponential and Logarithmic Functions4.1Exponential Functions.............................................................................. 3824.2Logarithmic Functions ............................................................................. 3944.3Rules of Logarithms................................................................................. 4044.4Exponential and Logarithmic Equations and Inequalities........................ 4154.5Logarithmic Scales................................................................................... 427Chapter 4 Review Exercises.................................................................................. 438Chapter 4 Practice Test A ..................................................................................... 446Chapter 4 Practice Test B...................................................................................... 447Cumulative Review Exercises (Chapters P−4) ..................................................... 448Chapter 5Systems of Equations and Inequalities5.1Systems of Linear Equations in Two Variables....................................... 4515.2Systems of Linear Equations in Three Variables..................................... 4695.3Partial-Fraction Decomposition ............................................................... 4905.4Systems of Nonlinear Equations .............................................................. 5075.5Systems of Inequalities ............................................................................ 5215.6Linear Programming ................................................................................ 534Chapter 5 Review Exercises.................................................................................. 549Chapter 5 Practice Test A ..................................................................................... 562Chapter 5 Practice Test B...................................................................................... 565Cumulative Review Exercises (Chapters P−5) ..................................................... 566Chapter 6Matrices and Determinants6.1Matrices and Systems of Equations ......................................................... 5696.2Matrix Algebra......................................................................................... 5866.3The Matrix Inverse................................................................................... 6046.4Determinants and Cramer’s Rule............................................................. 625Chapter 6 Review Exercises.................................................................................. 637Chapter 6 Practice Test A ..................................................................................... 647Chapter 6 Practice Test B...................................................................................... 649Cumulative Review Exercises (Chapters P−6) ..................................................... 650Chapter 7The Conic Sections7.2The Parabola ............................................................................................ 6537.3The Ellipse ............................................................................................... 6687.4The Hyperbola.......................................................................................... 685Chapter 7 Review Exercises.................................................................................. 708Chapter 7 Practice Test A ..................................................................................... 717Chapter 7 Practice Test B...................................................................................... 719Cumulative Review Exercises (Chapters P−7) ..................................................... 720Chapter 8Further Topics in Algebra8.1Sequences and Series ............................................................................... 7238.2Arithmetic Sequences; Partial Sums ........................................................ 7318.3Geometric Sequences and Series ............................................................. 7378.4Mathematical Induction ........................................................................... 7448.5The Binomial Theorem ............................................................................ 7528.6Counting Principles.................................................................................. 7598.7Probability................................................................................................ 766Chapter 8 Review Exercises.................................................................................. 770Chapter 8 Practice Test A ..................................................................................... 774Chapter 8 Practice Test B...................................................................................... 775Cumulative Review Exercises (Chapters P−8) ..................................................... 776

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41Chapter 1 Equations and Inequalities1.1Linear Equations in OneVariable1.1 Practice Problems1. a.Both sides of the equation753x−=aredefined for all real numbers, so the domainis (,)−∞ ∞.b.The left side of the equation242x=−isnot defined ifx= 2. The right side of theequation is defined for all real numbers, sothe domain is()(), 22,.−∞∞∪c.The left side of the equation10x−=isnot defined ifx< 1. The right side of theequation is defined for all real numbers, sothe domain is[)1,.∞2.23173263xx−=−To clear the fractions, multiply both sides ofthe equation by the LCD, 6.491144914114144514541453535535xxxxxxxxxxx−=−−+=−++=+−=−= −−== −Solution set:{}35−3.()()()3261713266713467134671767176771761xxxxxxxxxxxxxxxxxxxx− ⎡−+⎤ =−⎣⎦−−−=−− −−=−++=−+=−+−=−−= −Since 6 = −1 is false, no number satisfies thisequation. Thus, the equation is inconsistent,and the solution set is.∅4.()()2 36512196612512196677667676677777700xxxxxxxxxx−+=−−−+=−+−= −+−−= −+−−= −−+= −+=The equation 0 = 0 is always true. Therefore,the original equation is an identity, and thesolution set is(),.−∞ ∞5.93259503259503232325918555918 99510FCCCCCC=+=+−=+−=⋅=⋅=Thus, 50ºF converts to 10ºC.6.22Plw=+Subtract 2lfrom both sides.22Plw−=Now, divide both sides by 2.22Plw−=7.Letw= the width of the rectangle.Then 2w+ 5 = the length of the rectangle.P= 2l +2w, so we have282(25)2284102286101863wwwwwww=++=++=+==The width of the rectangle is 3 m and thelength is 2(3) + 5 = 11 m8.Letx= the amount invested in stocks. Then15,000 −x= the amount invested in bonds.()3 15, 00045, 0003445, 00011, 250xxxxxx=−=−==Tyrick invested $11,250 in stocks and$15,000 − $11,250 = $3,750 in bonds.

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42Chapter 1Equations and Inequalities9.Letx= the amount of capital. Then 5x=theamount invested at 5%, 6x=the amountinvested at 8%, and195630xxxx⎛⎞−+==⎜⎟⎝⎠theamount invested at 10%.PrincipalRateTimeInterest5x0.0510.055x⎛⎞⎜⎟⎝⎠6x0.0810.086x⎛⎞⎜⎟⎝⎠1930x0.110.11930x⎛⎞⎜⎟⎝⎠The total interest is $130, so190.050.080.11305630Multiply by the LCD, 30.0.30.41.939002.639001500xxxxxxxx⎛⎞⎛⎞⎛⎞++=⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠++===The total capital is $1500.10.Letx=the length of the bridge.Thenx+ 130 = the distance the train travels.rt=d, so()25 21130525130395xxx=+⇒=+⇒=The bridge is 395 m long.11.Following the reasoning in example 10, wehavex+ 2x= 3xis the maximum extendedlength (in feet) of the cord.371012031712031717120173103310334.333xxxxxx++=+=+−=−==⇒≈The cord should be no longer than 34.3 feet.1.1 Basic Concepts and Skills1.The domain of the variable in an equation isthe set of all real number for which both sidesof the equation are defined.2.Standard form for a linear equation inxis ofthe formax+b= 0.3.Two equations with the same solution sets arecalled equivalent.4.A conditional equation is one that is not truefor some values of the variables.5.False. The interestI= (100)(0.05)(3).6.False. Since the rate is given in feet persecond, the time must also be converted toseconds. 15 minutes = 15(60) = 900 secondsTherefore,d= 60(900) feet.7. a.Substitute 0 forxin the equation256xx−=+:025(0)626−=+⇒ −≠So, 0 is not a solution of the equation.b.Substitute –2 forxin the equation256xx−=+:225( 2)6410644−−=−+⇒ −= −+⇒−= −So, –2 is a solution of the equation.8. a.Substitute –1 forxin the equation83141xx+=−:8( 1)314( 1)183141515−+=−−⇒ −+= −−⇒−≠ −So, –1 is not a solution of the equation.b.Substitute 2 3 forxin the equation83141xx+=−:22162883141313333252533⎛⎞⎛⎞+=−⇒+=−⇒⎜⎟⎜⎟⎝⎠⎝⎠=So, 2 3 is a solution of the equation.9. a.Substitute 4 forxin the equation21132xx=++:21111111434223622=+⇒=+⇒=+So, 4 is a solution of the equation.b.Substitute 1 forxin the equation21132xx=++:211112221312333=+⇒=+⇒≠+So, 1 is not a solution of the equation.10. a.Substitute 1 2 forxin the equation(3)(21)0xx−+=:1153210(2)022250⎛⎞ ⎛⎞⎛⎞−⋅+=⇒−=⇒⎜⎟ ⎜⎟⎜⎟⎝⎠ ⎝⎠⎝⎠−≠So, 1 2 is not a solution of the equation.

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Solution Manual for College Algebra, 3rd Edition - Page 6 preview image

Section 1.1Linear Equations in One Variable43b.Substitute 3 for x in the equation(3)(21)0xx−+=:()()332 310(0)(7)000−⋅+=⇒=⇒=So, 3 is a solution of the equation.11. a.The equation 235xxx+=is an identity,so every real number is a solution of theequation. Thus 157 is a solution of theequation. This can be checked bysubstituting 157 forxin the equation:2(157)3(157)5(157)314471785785785+=⇒+=⇒=b.The equation 235xxx+=is an identity,so every real number is a solution of theequation. Thus2046−is a solution of theequation. This can be checked bysubstituting2046−forxin the equation:2( 2046)3( 2046)5( 2046)4092613810, 23010, 23010, 230−+−=−−−= −−= −12.Both sides of the equation(2)473(4)xxx−−=−+are defined for allreal numbers, so the domain is (,)−∞ ∞.13.The left side of the equation312yyy=−+isnot defined if1y=, and the right side of theequation is not defined if2y= −. The domainis (,2)( 2, 1)(1,)−∞ −−∞∪∪.14.The left side of the equation 12yy=+isnot defined if0y=. The right side of theequation is not defined if0y<, so thedomain is (0,)∞.15.The left side of the equation329(3)(4)xxxx=+−−is not defined if3x=or4x=. The right side is defined for all realnumbers. So, the domain is(, 3)(3, 4)(4,)−∞∞∪∪.16.The left side of the equation211xx=−isnot defined if0x≤. The right side of theequation is defined for all real numbers. Sothe domain is (0,)∞.17.Substitute 0 forxin 2351xx+=+. Because31≠, the equation is not an identity.18.When the like terms on the right side of theequation 3462(32)xxx+=+−−arecollected, the equation becomes3434xx+=+, which is an identity.19.When the terms on the left side of theequation 11222xxx++=are collected, theequation becomes 2222xxxx++=, which is anidentity.20.The right side of the equation11133xx=++isnot defined for0x=, while the left side isdefined for0x=. Therefore, the equation isnot an identity.In exercises 21–46, solve the equations using theprocedures listed on page 79 in your text: eliminatefractions, simplify, isolate the variable term, combineterms, isolate the variable term, and check thesolution.21.35143551453939333xxxxx+=+−=−===Solution set: {3}22.2177217177172242242212xxxxx−=−+=+===Solution set: {12}23.10123210121232121020102010102xxxxx−+=−+−=−−=−=−−= −Solution set: {−2}24.2562556521211222xxxxx−+=−+−=−−=−=⇒= −−−Solution set:12⎧⎫−⎨⎬⎩⎭

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44Chapter 1Equations and Inequalities25.34334377yyyy−= −−−= −−−= −⇒=Solution set: {7}26.2723272232721721377yyyyy−=−−=−−=−=⇒= −−−Solution set: {−3}27.772(1)7722777227725722525555155xxxxxxxxxxxxxxx+=++=++−=+−=−−=−−= −−=⇒= −Solution set: {−1}28.3(2)43643644644664642421442xxxxxxxxxxxxx+=−+=−++=−++=+−=−= −−=⇒= −Solution set:12⎧⎫−⎨⎬⎩⎭29.3(2)536353623622326yyyyyyyyyyyyy−+=−+=+=+−=−⇒=Solution set: {6}30.93(1)693366366365365336353533555yyyyyyyyyyyyyyyyy−−=+−+=++=++−=+−+=+−=−==⇒=Solution set:35⎧⎫⎨⎬⎩⎭31.4372(7)Distribute1 to clear the parentheses.72775755512yyyyyyyy−+−=−−−=−+= −++= −++⇒=Solution set: {12}32.3(1)642433443334433413441411yyyyyyyyyyyyyyyy−=−+−−=−−+=−+=−−=−−−= −⇒=Solution set: {1}33.3(2)2(3)1366211xxxxx−+−=−+−=⇒=Solution set: {1}34.23(31)6Distribute1 to clear the parentheses.2331626226288xxxxxxxx−−−=−−−+=−−=−−+=+−=⇒= −Solution set: {−8}35.23(4)7102312710512710512127101257225772272222221122xxxxxxxxxxxxxxxxxxx+−=++−=+−=+−+=++=+−=+−−=−=⇒= −−−Solution set: {−11}36.3(23 )4310694310613310613631061331613331631616161611616xxxxxxxxxxxxxxxxxxx−−=−−−=−−=−−−=−−−=−−−=−−−= −−−=⇒=−−Solution set: {1}37.4[2(3)]21Distribute 2 to clear the inner parentheses.4[62 ]21Combine like terms within the brackets.4[6]21Distribute 4 to clear the brackets.24421xxxxxxxxxx+−=++−=+−=+−=+2442421244223422362362323666xxxxxxxxx−−=+−−=−−−= −−= −−−=⇒=−−Solution set:236⎧⎫⎨⎬⎩⎭

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Section 1.1Linear Equations in One Variable4538.3[3(2)]4Distribute3 to clear the parentheses.3[36]4Combine like terms in the brackets.3[ 26]4Distribute1 to clear the brackets.32642942994925255222xxxxxxxxxxx−−+=−−−−=− −−=−++=+=+−=−= −−=⇒= −Solution set:52⎧⎫−⎨⎬⎩⎭39.3(43)4[(43)]Distribute 3 on the left side and1 onthe right side to clear parentheses.1294[43]Combine like terms in the brackets.1294[ 33]Distribute 4 to clear the brackets.1291yyyyyyyyy−=−−−−=−+−=−+−= −2121299121291212211212122112242124212172424248yyyyyyyyyyyy+−+= −++= −++= −++==⇒==Solution set:78⎧⎫⎨⎬⎩⎭40.5(69)22(1)Distribute1 on the left and 2 on theright to clear the parentheses.5692224422444224426422626666661yyyyyyyyyyyyyyyyyyy−++=+−−−+=+−−=+−−+=++−=+−−=+−−=−=−−= −Solution set: {−1}41.23(2)(3)21Distribute3 on the left to clearthe parentheses.2633215632xxxxxxxxxx−−=−++−−+=−++−=−566326534533432424222xxxxxxxxxxx−+=−+=+−=+−==⇒=Solution set: {2}42.5(3)6(4)5Distribute 5 to clear the first set ofparentheses. Distribute6 to clearthe second set of parentheses.51562459599591414xxxxxxxx−−−= −−−−+= −−+= −−+−= −−−= −⇒=Solution set: {14}43.214196To clear the fractions, multiply bothsides of the equation by the leastcommon denominator, 36.2143636(1)964(21)6(4)36846243622036220203620256256222xxxxxxxxxxxxx++−=++⎛⎞−=⎜⎟⎝⎠+−+=+−−=−=−+=+==⇒=8Solution set: {28}44.2313737To clear the fractions, multiply both sidesof the equation by the least commondenominator, 21.231321217373(23 )7(1)3(3 )69779129122921111xxxxxxxxxxxxxxxxxxx−−+=−−⎛⎞⎛⎞+=⎜⎟⎜⎟⎝⎠⎝⎠−+−=−+−=−−=−−+=+−=−111111111xx=⇒= −Solution set:111⎧⎫−⎨⎬⎩⎭

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46Chapter 1Equations and Inequalities45.1512(1)3428To clear the fractions, multiply bothsides by the least common denominator, 8.1512(1)88 3428Distribute the 8 on both sides.1512(1)888(3)842xxxxxxxxx−+++=−−++⎛⎞⎛⎞+=−⎜⎟⎜⎟⎝⎠⎝⎠−++⎛⎞⎛⎞+=−⎜⎟⎜⎟⎝⎠⎝⎠82(1)4(51)8(3)2(1)Simplify by collecting like terms andcombining constants.22204242218622218622222206222066226201620162020164205xxxxxxxxxxxxxxxxx⎛⎞⎜⎟⎝⎠−++=−+−++=−−+=−++=−++=+−=−====Solution set:45⎧⎫⎨⎬⎩⎭46.41322326xxx+++−=To clear the fractions, multiply both sides ofthe equation by the least commondenominator, 6.4132626326xxx++⎛⎞⎛⎞+−=⎜⎟⎜⎟⎝⎠⎝⎠Distribute the 6 on both sides.()()413266 2663262412332xxxxxx++⎛⎞⎛⎞⎛⎞+−=⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠++−=+Simplify by collecting like terms andcombining constants.281233214532145332311521155251131133111111xxxxxxxxxxxxxx++−=++=++−=+−+=+−=−= −−=⇒= −Solution set:311⎧⎫−⎨⎬⎩⎭47.To solvedrt=forr, divide both sides of theequation byt.drt=.48.To solveFma=for a, divide both sides ofthe equation by m.Fam=.49.To solve2Crπ=for r, divide both sides ofthe equation by2π.2Crπ=.50.22222222To solve2for, subtractfrom both sides.22Divide both sides by 2.2222ArxrxrArrxrrArrxrArrxrrArxrππππππππππππππππ=+−=+−−=−=−=51.To solvefor, multiply both sides by.EIRRRERIRRIER=⎛⎞=⇒=⎜⎟⎝⎠Divide both sides by.IRIEERIII=⇒=52.To solve(1) for , distribute.Subtractfrom both sides.Divide both sides by.APrttPAPPrtPAPPPrtPAPPrtPrAPPrtPrPrAPtPr=+=+−=+−−=−=−=53.()To solvefor, multiply both2sides by 2.2()Divide both sides by ().2()2ab hAhAab habAab hAhababab+==+++=⇒=+++

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Section 1.1Linear Equations in One Variable4754.To solve(1)for, subtractfrom both sides.(1)(1)Divide both sides by (1).(1)111TanddaTaandaTandnTandTadnnn=+−−=+−−−=−−−−−=⇒=−−−55.111To solvefor, clear the fractionsby multiplying both sides by the least commondenominator,.111ufuvfuvfuvfuvfuv=+⎛⎞⎛⎞=+⎜⎟⎜⎟⎝⎠⎝⎠111Simplify.Subtractfrom both sides.Factor the left side.()Divide both sides by.()fuvfuvfuvfuvuvfvfufuuvfufvfufuuvfufvu vffvvfu vffvfvuvfvfvf⎛⎞⎛⎞⎛⎞=+⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠=+−=+−−=−=−−=⇒=−−−56.2121212121212121122111To solvefor,clear the fractions by multiplyingboth sides by the least commondenominator,.111111SiRRRRRRRR RRR RRRRRR RRR RRRRRRRR=+⎛⎞⎛⎞⋅⋅=⋅⋅+⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞⎛⎞⋅⋅=⋅⋅⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞+⎜⎟⎝⎠iiii1221mplify.R RRRRR=+21222121221Subtractfrom both sides.RRR RRRRRRRRRR RRRRR−=+−−=211121112111Factor the left side.()Divide both sides by ().()RRRRRRRRRRRRRRRRRRRRR−=−−=⇒=−−−57.To solvefor, subtractfromboth sides.Divide both sides by.ymxbmbybmxbbybmxxybmxybmxxx=+−=+−⇒−=−−=⇒=58.To solvefor, subtractfromboth sides.Divide both sides by.axbycyaxaxbyaxcaxbycaxbbycaxcaxybbb+=+−=−⇒=−−−=⇒=59.0.065x60.56x61.$22, 000x−62.229.5072x−1.1 Applying the Concepts63.The formula for volume isVlwh=.Substitute 2808 forV, 18 forl, and 12 forh.Solve forw.280818 122808216280821621621613wwww=⋅⋅===The width of thepool is 13 ft.64.The formula for volume isVlwh=.Substitute 168 forV, 7 forl, and 3 forw.Solve forh.1687 3168211682121218hhhh=⋅⋅===The hole must be 8 ftdeep.65.Letw= the width of the rectangle.Then 2w− 5 = the length of the rectangle.()22 25802410806108069015, 2525wwwwwwww+−=+−=−===−=The width of the rectangle is 15 ft and itslength is 25 feet.

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48Chapter 1Equations and Inequalities66.Letl= the length of the rectangle.Then132l+=the width of the rectangle.122 336226363636330110, 382llllllll⎛⎞++=⎜⎟⎝⎠++=+===+=The length of the rectangle is 10 ft and itswidth is 8 ft.67.The formula for circumference of a circle is2Crπ=. Substitute 114πforC. Solve forr.114211425722rrrππππππ=⇒=⇒=The radius is 57 cm.68.The formula for perimeter of a rectangle is22Plw=+. Substitute 28 forPand 5 forw.Solve forl.2822(5)28210281021010182182229llllll=+=+−=+−===The length is 9 m.69.The formula for surface area of a cylinder is222Srhrππ=+. Substitute 6πforSand 1forr. Solve forh.262(1)2(1 )622622224242222hhhhhhπππππππππππππππππ=+=+−=+−==⇒=The height is 2 m.70.The formula for volume of a cylinder is2Vr hπ=. Substitute 148πforVand 2 forr.Solve forh.21482148414844437hhhhππππππππ=⋅⋅===The height of the can is37 cm.71.The formula for area of a trapezoid is()1212Ah bb=+. Substitute 66 forA, 6 forh,and 3 for1b. Solve for2b.()()22222221666 32663 366936699395733573319bbbbbbb=⋅+=+=+−=+−===The length of the second base is 19 ft.72.The formula for area of a trapezoid is()1212Ah bb=+. Substitute 35 forA, 9 for1b, and 11 for2b .Solve forh.()()1359112135202351035103.51010hhhhh=+===⇒=The height of the trapezoid is 3.5 cm.73.Letx= the cost of the less expensive land.Thenx+ 23,000 = the cost of the moreexpensive land. Together they cost $147,000,so(23, 000)147, 000223, 000147, 0002124, 00062, 000xxxxx++=+==⇒=The less expensive piece of land costs $62,000and the more expensive piece of land costs$62,000 + $23,000=$85,000.74.Letx=the amount the assistant managerearns. Thenx+ 450=the amount themanager earns. Together they earn $3700, so(450)370024503700232501625xxxxx++=+==⇒=The assistant manager earns $1625, and themanager earns $1625 + $450=$2075.

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Solution Manual for College Algebra, 3rd Edition - Page 12 preview image

Section 1.1Linear Equations in One Variable4975.Letx= the lottery ticket sales in July. Then1.10x= the lottery ticket sales in August.A total of 1113 tickets were sold, so1.1011132.101113530xxxx+==⇒=530 tickets were sold in July, and1.10(530) = 583 tickets were sold in August.76.Letx= Jan’s commission in March. Then15 + 0.5x= Jan’s commission in February.She earned a total of $633, so(150.5 )6331.5156331.5618412xxxxx++=+=⇒=⇒=Jan’s commission was $412 in March and15 + 0.5(412)=$221 in February.77.Letx= the amount the younger son receives.Then 4x= the amount the older son receives.Together they receive $225,000, so4225, 0005225, 00045, 000xxxx+=⇒=⇒=The younger son will received $45,000, andthe older son will receive4($45,000)= $180,000.78.Letx= the amount Kevin kept for himself.Then2x=the amount he gave his daughter,and4x=the amount he gave his dad.He won $735,000, so735, 0002444(735, 000)24422, 940, 00072, 940, 000420, 000xxxxxxxxxxx++=⎛⎞++=⎜⎟⎝⎠++==⇒=Kevin kept $420,000 for himself. He gave$420, 000 2$210, 000=to his daughter and$420, 000 4$105, 000=to his dad.79. a.Letx= the number of points needed toaverage 75.87597375421930081xxx+++=+==You need to score 81 in order to average75.b.87597327552192375215678xxxx+++=+===You need to score 78 in order to average 75if the final carries double weight.80.Letx= the amount invested in real estate.Then 4200 –x= the amount invested in asavings and loan.InvestmentPrincipalRateTimeInterestReal estatex0.1510.15xSavings4200 –x0.0810.08(4200 –x)The total income was $448, so0.150.08(4200)4480.153360.084480.073364480.071121600xxxxxxx+−=+−=+==⇒=So, the real estate agent invested $1600 in realestate and 4200 – 1600 = $2600 in a savingsand loan.81.Letx= the amount invested in a tax shelter.Then 7000 –x= the amount invested in abank.InvestmentPrincipalRateTimeInterestTax shelterx0.0910.09xBank7000 –x0.0610.06(7000 –x)The total interest was $540, so0.090.06(7000)5400.094200.065400.034205400.031204000xxxxxxx+−=+−=+==⇒=Mr. Mostafa invested $4000 in a tax shelterand 7000 – 4000 = $3000 in a bank.82.Letx= the amount invested at 6%. Then4900 –x= the amount invested at 8%PrincipalRateTimeInterestx0.0610.06x4900 –x0.0810.08(4900 –x)The amount of interest for each investment isequal, so0.060.08(4900)0.063920.080.143922800xxxxxx=−=−=⇒=Ms. Jordan invested $2800 at 6% and $2100at 8%. The amount of interest she earned oneach investment is $168, so she earned $336in all.

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Solution Manual for College Algebra, 3rd Edition - Page 13 preview image

50Chapter 1Equations and Inequalities83.Letx= the amount to be invested at 8%.PrincipalRateTimeInterest50000.051250x0.0810.08x5000 +x0.0610.06(5000 +x)The amount of interest for the total investmentis the sum of the interest earned on theindividual investments, so0.06(5000)2500.083000.062500.08500.060.08500.022500xxxxxxxx+=++=++==⇒=So, $2500 must be invested at 8%.84.Letx=the selling price. Thenx– 480=theprofit. So4800.2xx−=⇒4800.8x−= −⇒600x=. The selling price is $600.85.There is a profit of $2 on each shaving set.They want to earn $40,000 + $30,000 =$70,000. Letx= the number of shaving sets tobe sold. Then 2x= the amount of profit forxshaving sets. So,270, 00035, 000xx=⇒=They must sell 35,000 shaving sets.86.Lett=the time each traveled.Then 100t=Angelina’s rate and 150t=Harry’s rate.RateTimeDistanceAngelina100tt100Harry150tt150Harry’s rate is 15 meters per minute fasterthan Angelina’s, so we have1001501510015150155010 min3ttttt+=+===So, Angelina jogged at1003010 3=meters perminute. Harry biked at 1504510 3=meters perminute.87.Letx= the time the second car travels.Then 1 +x= the time the first car travels. So,RateTimeDistanceFirstcar501 +x50(1)x+Secondcar70x70xThe distances are equal, so50(1)7050507050202.5xxxxxx+=+==⇒=So, it will take the second car 2.5 hours toovertake the first car.88.Letx=the time the planes travel. So,RateTimeDistanceFirstplane470x470xSecondplane430x430xThe planes are 2250 km apart, so470430225090022502.5xxxx+=⇒=⇒=So, the planes will be 2250 km apart at 2.5hours.89.At 20 miles per hour, it will take Lucas twominutes to bike the remaining 2/3 of a mile.20 mi20 mi1mi1 hr60 min3 min⎛⎞==⎜⎟⎝⎠So his brotherwill have to bike 1 mile in 2 minutes:1mi30 mi30 mi2 min60 min1 hr==90.Driving at 40 miles per hour, it will takeKaren’s husband 45 40 hours or 1 hour and7.5 minutes to get to the airport. Driving at 60miles per hour, it will take Karen 45 minutesto get to the airport. Her husband has alreadydriven for 15 minutes, so it will take him anadditional 52.5 minutes to get to the airport.Karen will get there before he does.91.Letx=the rate the slower car travels. Thenx+ 5=the rate the faster car travels. So,RateTimeDistanceFirst carx33xSecond carx+ 53()35x+(continued on next page)

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Solution Manual for College Algebra, 3rd Edition - Page 14 preview image

Section 1.1Linear Equations in One Variable51(continued)The cars are 405 miles apart, so()3354053315405615405639065xxxxxxx++=++=+=⇒=⇒=One car is traveling at 65 miles per hour, andthe other car is traveling at 70 miles per hour.92.Letx=the distance to Aya’s friend’s house.RateDistanceTimego16x16xreturn80x80xShe traveled for a total of 3 hours, so316808080(3)16805240624040xxxxxxxx+=⎛⎞+=⎜⎟⎝⎠+=⇒=⇒=So, her friend lives 40 km away.93.Substitute 170 forPinto the formula2000.02Pq=−. Solve forq.1702000.021702002000.02200300.02300.0215000.020.02qqqqq=−−=−−−= −−−=⇒=−−Note that the solution must fall between 100and 2000 cameras. 1500 cameras must beordered.94.Substitute 37,000 for Q, 1500 forL, and 3200for I into the formulaAIQL−=. Solve forA.320037, 000150032001500(37, 000)1500150055, 500, 000320055,500, 00032003200320055,503, 200AAAAA−=−⎛⎞=⎜⎟⎝⎠=−+=−+=The current assets are $55,503,200.95.Letx= one number. Then 3x= the other number.3284287, 33(7)21xxxxx+=⇒=⇒===The numbers are 7 and 21.96.Letx= the first even integer. Thenx+ 2 = thesecond even integer, andx+ 4 = the third eveninteger.()()2442364233612214,416xxxxxxxx++++=+==⇒=+=+=The numbers are 12, 14, and 16.97.Letx= one number. Then 2xis the secondnumber.21414xxx−==The numbers are 14 and 28.98.Letx= one number. Thenx+ 5 = the secondnumber. Note that the second number is thelarger number.()2549210493104933913xxxxxxx++=++=+===The numbers are 13 and 18.1.1 Beyond the Basics99. a.The solution set of2xx=is {0,1} , whilethe solution set of1x=is {1}. Therefore,the equations are not equivalent.b.The solution set of29x=is { 3, 3}−, whilethe solution set of3x=is {3}. Therefore,the equations are not equivalent.c.The solution set of211xx−=−is {0, 1},while the solution set of0x=is {0}.Therefore, the equations are not equivalent.d.The equation222xxx=−−is aninconsistent equation, so is solution set is∅. The solution set of2x=is {2}.Therefore, the equations are not equivalent.100.First, solve 7216x+=. Subtracting 2 fromboth sides, we have 714x=. Then divideboth sides by 7; we obtain2x=. Nowsubstitute 2 forxin 31xk−=. This becomes3(2)1k−=, so5k=.

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Solution Manual for College Algebra, 3rd Edition - Page 15 preview image

52Chapter 1Equations and Inequalities101.Letx= the average speed for the second halfof the trip.RateDistanceTime1sthalf75D75D2ndhalfxDDxEntire trip602D260D275602300300756043005 (2)430010300650DDDxDDDxxxDxDxDDxDDxDDxx+=⎛⎞⎛⎞+=⎜⎟⎜⎟⎝⎠⎝⎠+=+==⇒=The average speed for the second half of thedrive is 50 mph.102.First we need to compute how much time itwill take for Davinder and Mikhael to meet.Lett= the time it will take for them to meet.So,RateTimeDistanceeDavinder3.7t3.7tMikhail4.3t4.3t3.74.32820.25tttt+=⇒=⇒=They will be walking for 0.25 hour until theymeet.The dog starts with Davinder. Let1dt=theamount of time it takes for the dog to meetMikhail. So,RateTimeDistancedog61dt16dtMikhail4.31dt14.3dt11164.3210.320.19dddtttt+===The dog meets Mikhail for the first time whenthey have walked for 0.19 hour. The dog willhave traveled 1.14 mi.While the dog has been running towardsMikhail, Davinder has continued to walk.During the 0.19 hour, he walked 0.70 mi, sonow he and the dog are 1.14 – 0.70 = 0.44 miapart. Let2dt=the time it takes the dog tomeet Davinder.RateTimeDistancedog62dt26dtDavinder3.72dt23.7dt22263.70.449.70.440.05dddtttt+===So, the dog meets Davinder when they havewalked for another 0.05 hour. The dog willhave traveled 0.3 mi. They have now walkedfor 0.19 + 0.05 = 0.24 hr. Since Davinder andMikhail don’t meet until they have walked for0.25 hours, the dog must walk for 0.01 hrmore. In that time, the dog will travel 0.06 mi.So in total the dog will travel1.14 + 0.3 + 0.06 = 1.5 mi.103.Letx= the number of liters of water in theoriginal mixture. Then 5x= the number ofliters of alcohol in the original mixture, and6x= the total number of liters in the originalmixture.x+ 5 = the number of liters of waterin the new mixture. Then 6x+ 5 = the totalnumber of liters in the new mixture. Since theratio of alcohol to water in the new mixture is5:2, then the amount of alcohol in the newmixture is 5/7 of the total mixture or5 (65)7x+.There was no alcohol added, so the amount ofalcohol in the original mixture equals theamount of alcohol in the new mixture. Thisgives5 (65)575(65)353025352555xxxxxxxx+=+=+=⇒=⇒=So, there were 5 liters of water in the originalmixture and 25 liters of alcohol.104.Letx= the amount of each alloy. There are 13parts in the first alloy and 8 parts in the secondalloy. We can use the following table toorganize the information:TotalZincCopperAlloy 1x513x813xAlloy 2x58x38xTotal2x55138xx+83138xx+(continued on next page)

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Solution Manual for College Algebra, 3rd Edition - Page 16 preview image

Section 1.1Linear Equations in One Variable53(continued)The amount of zinc in the new mixture is55105138104xxx+=, and the amount of copper inthe new mixture is 83103138104xxx+=.So, the ratio of zinc to copper in the newmixture is 105103:104104xxor 105:103.105.Letx= Democratus’ age now. Then6x=the number of years as a boy,8x=thenumber of years as a youth, and2x=thenumber of years as a man. He has spent 15years as a mature adult. So,156822415246824312360241936024360572xxxxxxxxxxxxxxxx+++=⎛⎞+++=⎜⎟⎝⎠+++=+==⇒=Democratus is 72 years old.106.Letx= the man’s age now. When the womanisxyears old, the man will be 119 –xyearsold. So the difference in their ages is(119)1192xxx−−=−years. So thewoman’s age now is(1192 )3119xxx−−=−. When the man was3119x−years old, she was 31192x−yearsold. Since the difference in their ages is1192x−, we have()3119311911922623831192384311923847119238735751xxxxxxxxxxx−−−=−−−+=−−=−−==⇒=So the man is now 51 years old.Check by verifying the facts in the problem.When she is 51 years old, he will be119 – 51 = 68 years old. The difference intheir ages is 68 – 51 = 17 years. So she is51 – 17 = 34 years old now. When he was 34years old, she was 17 years old, which is 1/2of 34.107.There are 180 minutes from 3 p.m. to 6 p.m.So, the number of minutes before 6 p.m. plus50 minutes plus 4 × the number of minutesbefore 6 p.m. equals 180 minutes. Letx= thenumber of minutes before 6 p.m. So,504180xx++=⇒550180x+=⇒513026xx=⇒=So it is 26 minutes before 6 p.m. or 5:34 p.m.Check this by verifying that 26 + 50 = 76minutes before 6 p.m. is the same time as4(26) = 104 minutes after 3 p.m. Seventy-sixminutes before 6 p.m. is 4:44 p.m., while 104minutes after 3 p.m. is also 4:44 p.m.108.Letx= the number of minutes pipeBis open.PipeAis open for 18 minutes, so it fills 18/24or 3/4 of the tank. PipeBfillsx/32 of the tank.So, 3313232(1)432432xx⎛⎞+=⇒+=⇒⎜⎟⎝⎠24328xx+=⇒=PipeBshould be turned off after 8 minutes.109. a.Because of the head wind, the plane flies at140 mph from Atlanta to Washington and160 mph from Washington to Atlanta. Letx= the distance the plane flew beforeturning back. So,RateDistanceTimeto140x140xfrom160x160x1.51401601601401.5(140)(160)30033, 600112xxxxxx+=+==⇒=The plane flew 112 miles before turningback.b.The plane traveled 224 miles in 1.5 hours,so the average speed is 224149.331.5=mph.

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