Solution Manual for College Algebra, 6th Edition

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’SSOLUTIONSMANUALEDGARREYESSoutheastern Louisiana UniversityCOLLEGEALGEBRASIXTHEDITIONMark DugopolskiSoutheastern Louisiana UniversityBostonColumbusIndianapolisNew YorkSan FranciscoUpper Saddle RiverAmsterdamCape TownDubaiLondonMadridMilanMunichParisMontrealTorontoDelhiMexico CitySão PauloSydneyHong KongSeoulSingaporeTaipeiTokyo

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Table of ContentsChapter P...……………………………………………………………………………………1Chapter 1……………………………………………………………………………………..34Chapter 2…………………………………………………………………………………….121Chapter 3…………………………………………………………………………………….179Chapter 4…………………………………………………………………………………….255Chapter 5…………………………………………………………………………………….306Chapter 6…………………………………………………………………………………….388Chapter 7…………………………………………………………………………………….458Chapter 8…………………………………………………………………………………….501

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34Chapter 1Equations, Inequalities, and ModelingFor Thought1.True, since 5(1) = 6−1.2.True, sincex= 3 is the solution to bothequations.3.False,−2 is not a solution of the first equationsince√−2 is not a real number.4.True, sincex−x= 0.5.False,x= 0 is the solution.6.True7.False, since|x|=−8 has no solution.8.False,xx−5 is undefined atx= 5.9.False, since we should multiply by−32 .10.False, 0·x+ 1 = 0 has no solution.1.1 Exercises1.equation2.linear3.equivalent4.solution set5.identity6.inconsistent equation7.conditional equation8.extraneous root9.No, since 2(3)−4 = 26= 9.10.Yes11.Yes, since (−4)2= 16.12.No, since√166=−4.13.Since 3x= 5, the solution set is{53}.14.Since−2x=−3, the solution set is{32}.15.Since−3x= 6, the solution set is{−2}.16.Since 5x=−10, the solution set is{−2}.17.Since 14x= 7, the solution set is{12}.18.Since−2x= 2, the solution set is{−1}.19.Since 7 + 3x= 4x−4, the solution set is{11}.20.Since−3x+ 15 = 4−2x, the solution setis{11}.21.Sincex=−43·18, the solution set is{−24}.22.Sincex= 32·(−9), the solution set is{−272}.23.Multiplying by 6 we get3x−30=−72−4x7x=−42.The solution set is{−6}.24.Multiplying by 4 we obtainx−12=2x+ 12−24=x.The solution set is{−24}.25.Multiply both sides of the equation by 12.18x+ 4=3x−215x=−6x=−25.The solution set is{−25}.26.Multiply both sides of the equation by 30.15x+ 6x=5x−1016x=−10x=−58.The solution set is{−58}.27.Note, 3(x−6) = 3x−18 is true by thedistributive law. It is an identity and thesolution set isR.

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1.1 Linear, Rational, and Absolute Value Equations3528.Subtract 5afrom both sides of 5a= 6atoget 0 =a. The latter equation is conditionalwhose solution set is{0}.29.Note, 5x= 4xis equivalent tox= 0.Thelatter equation is conditional whose solutionset is{0}.30.Note, 4(y−1) = 4y−4 is true by thedistributive law. The equation is an identityand the solution set isR.31.Equivalently, we get 2x+ 6 = 3x−3 or 9 =x.The latter equation is conditional whosesolution set is{9}.32.Equivalently, we obtain 2x+ 2 = 3x+ 2 or0=x.The latter equation is conditionalwhose solution set is{0}.33.Using the distributive property, we find3x−18=3x+ 18−18=18.The equation is inconsistent and the solutionset is∅.34.Since 5x= 5x+ 1 or 0 = 1, the equation isinconsistent and the solution set is∅.35.An identity and the solution set is{x|x6= 0}.36.An identity and the solution set is{x|x6=−2}.37.Multiplying by 2(w−1), we get1w−1−12w−2=12w−22−1=1.An identity and the solution set is{w|w6= 1}38.Multiply byx(x−3).(x−3) +x=92x=12A conditional equation with solution set{6}.39.Multiply by 6x.6−2=3 + 14=4An identity with solution set{x|x6= 0}.40.Multiply by 60x.12−15 + 20=−17x17=−17xA conditional equation with solution set{−1}.41.Multiply by 3(z−3).3(z+ 2)=−5(z−3)3z+ 6=−5z+ 158z=9A conditional equation with solution set{98}.42.Multiply by (x−4).2x−3=52x=8x=4Since division by zero is not allowed,x= 4does not satisfy the original equation. We havean inconsistent equation and so the solutionset is∅.43.Multiplying by (x−3)(x+ 3).(x+ 3)−(x−3)=66=6An identity with solution set{x|x6= 3, x6=−3}.44.Multiply by (x+ 1)(x−1).4(x+ 1)−9(x−1)=34x+ 4−9x+ 9=3−5x=−10A conditional equation with solution set{2}.45.Multiply by (y−3).4(y−3) + 6=2y4y−6=2yy= 3Since division by zero is not allowed,y= 3does not satisfy the original equation. We havean inconsistent equation and so the solutionset is∅.

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36Chapter 1Equations, Inequalities, and Modeling46.Multiply byx+ 6.x−3(x+ 6)=(x+ 6)−6−2x−18=x−6=xSince division by zero is not allowed,x=−6does not satisfy the original equation. We havean inconsistent equation and so the solutionset is∅.47.Multiply byt+ 3.t+ 4t+ 12=25t=−10A conditional equation with solution set{−2}.48.Multiply byx+ 1.3x−5(x+ 1)=x−113x−5x−5=x−116=3xA conditional equation with solution set{2}.49.Since−4.19 = 0.21xand−4.190.21≈ −19.952,the solution set is approximately{−19.952}.50.Since 0.92x= 5.9, we getx=5.90.92≈6.413.The solution set is approximately{6.413}.51.Divide by 0.06.x−3.78=1.950.06x=32.5 + 3.78x=36.28The solution set is{36.28}.52.Divide by 0.86.3.7−2.3x=4.90.86−2.3x=4.90.86−3.7x=4.90.86−3.7−2.3x≈−0.869The solution set is approximately{−0.869}.53.2a=−1− √17a=−1− √172a≈−1−4.12312a≈−2.562The solution set is approximately{−2.562}.54.3c=√38−4c=√38−43x≈6.1644−43x≈0.721The solution set is approximately{0.721}.55.0.001=3(y−0.333)0.001=3y−0.9991=3y13=yThe solution set is{13}.56.Multiply byt−1.(t−1) + 0.001=0t−0.999=0The solution set is{0.999}.57.Factoringx, we getx(10.376 +10.135)=2x(2.6596 + 7.4074)≈210.067x≈2x≈0.199The solution set is approximately{0.199}.

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1.1 Linear, Rational, and Absolute Value Equations3758.1x=10.379−56.721=x(10.379−56.72)110.379−56.72=x0.104≈xThe solution set is approximately{0.104}.59.x2+ 6.5x+ 3.252=x2−8.2x+ 4.1214.7x=4.12−3.25214.7x=16.81−10.562514.7x=6.2475x=0.425The solution set is{0.425}.60.0.25(4x2−6.4x+ 2.56)=x2−1.8x+ 0.81x2−1.6x+ 0.64=x2−1.8x+ 0.810.2x=0.17x=0.85The solution set is{0.85}.61.(2.3×106)x=1.63×104−8.9×105x=1.63×104−8.9×1052.3×106x≈−0.380The solution set is approximately{−0.380}.62.Note, 3.45×10−8≈0 .x=1.63×104−3.45×10−8−3.4×10−9x≈−4.794×1012The solution set is approximately{−4.794×1012}.63.Solution set is{±8}.64.Solution set is{±2.6}.65.Sincex−4 =±8, we getx= 4±8.The solution set is{−4,12}.66.Sincex−5 = 3.6 orx−5 =−3.6, we getx= 8.6 orx= 1.4. The solution setis{1.4,8.6}.67.Sincex−6 = 0, we getx= 6.The solution set is{6}.68.Sincex−7 = 0, we getx= 7.The solution set is{7}.69.Since the absolute value of a real number is nota negative number, the equation|x+ 8|=−3has no solution. The solution set is∅.70.Since the absolute value of a real number is nota negative number, the equation|x+ 9|=−6has no solution. The solution set is∅.71.Since 2x−3 = 7 or 2x−3 =−7, we get2x= 10 or 2x=−4. The solution setis{−2,5}.72.Since 3x+ 4 = 12 or 3x+ 4 =−12, we find3x= 8 or 3x=−16. The solution setis{−16/3,8/3}.73.Multiplying 12|x−9|= 16 by 2 we obtain|x−9|= 32. Thenx−9 = 32 orx−9 =−32.The solution set is{−23,41}.74.Multiplying 23|x+ 4|= 8 by 32 we obtain|x+ 4|= 12. Thenx+ 4 = 12 orx+ 4 =−12.The solution set is{−16,8}.75.Since 2|x+ 5|= 10, we find|x+ 5|= 5.Thenx+ 5 =±5 orx=±5−5.The solution set is{−10,0}.76.Since 8 = 4|x+ 3|, we obtain 2 =|x+ 3|.Thenx+ 3 =±2 orx=±2−3.The solution set is{−5,−1}.77.Dividing 8|3x−2|= 0 by 8, we obtain|3x−2|= 0. Then 3x−2 = 0 and thesolution set is{2/3}.

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38Chapter 1Equations, Inequalities, and Modeling78.Dividing 5|6−3x|= 0 by 5, we obtain|6−3x|= 0. Then 6−3x= 0 and thesolution set is{2}.79.Subtracting 7, we find 2|x|=−1 and|x|=−12 . Since an absolute value is not equalto a negative number, the solution set is∅.80.Subtracting 5, we obtain 3|x−4|=−5 and|x−4|=−53 . Since an absolute value is not anegative number, the solution set is∅.81.Since 0.95x= 190, the solution set is{200}.82.Since 1.1x= 121, the solution set is{110}.83.0.1x−0.05x+ 1=1.20.05x=0.2The solution set is{4}.84.0.03x−0.2=0.2x+ 0.006−0.206=0.17x−0.2060.17=x−0.2060.17·10001000=x−206170=xThe solution set is{−10385}.85.Simplifyingx2+ 4x+ 4 =x2+ 4, we obtain4x= 0. The solution set is{0}.86.Simplifyingx2−6x+9 =x2−9, we get 18 = 6x.The solution set is{3}.87.Since|2x−3|=|2x+ 5|, we get2x−3 = 2x+ 5 or 2x−3 =−2x−5.Solving forx, we find−3 = 5 (an inconsistentequation) or 4x=−2.The solution set is{−1/2}.88.Squaring the terms, we find (9x2−24x+ 16) +(16x2+ 8x+ 1) = 25x2+ 20x+ 4. Setting theleft side to zero, we obtain 0 = 36x−13. Thesolution set is{13/36}.89.Multiply by 4.2x+ 4=x−6x=−10The solution set is{−10}.90.Multiply by 12.−2(x+ 3)=3(3−x)−2x−6=9−3xx=15The solution set is{15}.91.Multiply by 30.15(y−3) + 6y=90−5(y+ 1)15y−45 + 6y=90−5y−526y=130The solution set is{5}.92.Multiply by 10.2(y−3)−5(y−4)=502y−6−5y+ 20=50−36=3yThe solution set is{−12}.93.Since 7|x+ 6|= 14,|x+ 6|= 2.Thenx+ 6 = 2 orx+ 6 =−2.The solution set is{−4,−8}.94.From 3 =|2x−3|, it follows that2x−3 = 3or2x−3 =−32x= 6or2x= 0.The solution set is{3,0}.95.Since−4|2x−3|= 0, we get|2x−3|= 0. Then2x−3 = 0 and the solution set is{3/2}.96.Since−|3x+1|=|3x+1|, we get 0 = 2|3x+1|.Then|3x+ 1|= 0 or 3x+ 1 = 0. The solutionset is{−1/3}.

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1.1 Linear, Rational, and Absolute Value Equations3997.Since−5|5x+ 1|= 4, we find|5x+ 1|=−4/5.Since the absolute value is not a negative num-ber, the solution set is∅.98.Since|7−3x|=−3 and the absolute value isnot a negative number, the solution set is∅.99.Multiply by (x−2)(x+ 2).3(x+ 2) + 4(x−2)=7x−23x+ 6 + 4x−8=7x−27x−2=7x−2An identity with solution set{x|x6= 2, x6=−2}.100.Multiply by (x−1)(x+ 2).2(x+ 2)−3(x−1)=8−x2x+ 4−3x+ 3=8−x7−x=8−xAn inconsistent equation and the solution setis∅.101.Multiply (x+ 3)(x−2) to both sides of4x+ 3 +3x−2 =7x+ 1(x+ 3)(x−2).Then we find4(x−2) + 3(x+ 3)=7x+ 14x−8 + 3x+ 9=7x+ 17x+ 1=7x+ 1.An identity and the solution setis{x|x6= 2 andx6=−3}.102.Multiply byx(x−1) to both sides of3x+4x−1 =7x−3x(x−1).Then we get3(x−1) + 4x=7x−33x−3 + 4x=7x−37x−3=7x−3.An identity and the solution setis{x|x6= 0 andx6= 1}.103.Multiply by (x−3)(x−4).(x−4)(x−2)=(x−3)2x2−6x+ 8=x2−6x+ 98=9An inconsistent equation and so the solutionset is∅.104.Multiply by (y+ 4)(y−2).(y−2)(y−1)=(y+ 4)(y+ 1)y2−3y+ 2=y2+ 5y+ 4−2=8y.A conditional equation and the solution set is{−1/4}.105. a)About 1995b)Increasingc)Lety= 0.95. Solving forx, we find0.95=0.0102x+ 0.6440.95−0.6440.0102=x30=x.In the year 2020 (= 1990 + 30), 95% ofmothers will be in the labor force.106.Lety= 0.644. Solving forx, we find0.644=0.0102x+ 0.6440=0.0102x0=x.In the year 1990 (= 1990 + 0), 64.4% ofmothers were in the labor force.107.SinceB= 21,000−0.15B, we obtain1.15B= 21,000 and the bonus isB= 21,0001.15= $18,260.87.108.Since 0.30(200,000) = 60,000, we findS=0.06(140,000 + 0.3S)S=8400 + 0.018S0.982S=8400.The state tax isS= 84000.982 = $8553.97 andthe federal tax isF= 0.30(200,000−8553.97) = $57,433.81.

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40Chapter 1Equations, Inequalities, and Modeling109.Rewrite the left-hand side as a sum.10,000 + 500,000,000x=12,000500,000,000x=2,000500,000,000=2000x250,000=xThus, 250,000 vehicles must be sold.110. (a)The harmonic mean is5115.1+17.3+15.9+13.6+12.8which is about $4.96 trillion.(b)Letxbe the GDP of Brazil, andA=115.1 +17.3 +15.9 +13.6 +12.8.Applying the harmonic mean formula,we get6A+ 1x=4.266=4.26A+ 4.26x6−4.26A=4.26xx=4.266−4.26Ax=$2.49 trillion111.The third side of the triangle is√3 by thePythagorean Theorem. Then draw radial linesfrom the center of the circle to each of the threesides.Consider the square with siderthatis formed with the 90◦angle of the triangle.Then the side of length√3 is divided into twosegments of lengthrand√3−r.Similarly,the side of length 1 is divided into segments oflengthrand 1−r.Note, the center of the circle lies on the bi-sectors of the angles of the triangle.Usingcongruent triangles, the hypotenuse consists ofline segments of length√3−rand 1−r. Sincethe hypotenuse is 2, we have(√3−r) + (1−r)=2√3−1=2r.Thus, the radius isr=√3−12.112.The hypotenuse is√2 by the PythagoreanTheorem.Then draw radial lines from thecenter of the circle to each of the three sides.Consider the square with siderthat is formedwith the 90◦angle of the triangle. Then eachside of length 1 consists of line segments oflengthrand 1−r.Note, the center of the circle lies on the bi-sectors of the angles of the triangle.Usingcongruent triangles, the hypotenuse consists ofline segments of length 1−rand 1−r. Sincethe hypotenuse is√2, we have(1−r) + (1−r)=√22− √2=2r.Thus, the radius isr= 2− √22.115.−3,0,7116.20−24 =−4117.−5x+ 20−24 + 12x= 7x−4118.14−1315−14=−112−120= 2012 = 53119.√5√3·√3√3 =√153120.Sincex6= 0, domain is (−∞,0)∪(0,∞).121.$9, $99, $999, $9,999, $99,999,$999,999, $9,999,999 $99,999,999,$999,999,999122.Notice, 5! + 6! +· · ·+ 1776! is a multiple of5! = 120. Then the units digit of 1!+· · ·+1776!is the same as the units digit of 1!+2!+3!+4! =33. Thus, the units digit of the sum is 3.

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1.2 Constructing Models to Solve Problems411.1 Pop Quiz1.Since 7x= 6, we getx= 6/7.A conditionalequation and the solution set is{6/7}.2.Since 14x= 13 + 16 = 12 , we getx= 4·12 = 2.A conditional equation, the solution set is{2}.3.Since 3x−27 = 3x−27 is an identity, thesolution set isR.4.Sincew−1 = 6 orw−1 =−6, we getw= 7orw=−5.A conditional equation and thesolution set is{−5,7}.5.Since 2x+ 12 = 2x+ 6, we obtain 12 = 6 whichis an inconsistent equation.The solution setis∅.6.Sincex2+ 2x+ 1 =x2+ 1, we obtain 2x= 0.This is a conditional equation and the solutionset is{0}.1.1 Linking Concepts(a)The power expenditures for runners withmasses 60 kg, 65 kg, and 70 kg are60(a·400−b)≈22.9 kcal/min,65(a·400−b)≈24.8 kcal/min, and70(a·400−b)≈26.7 kcal/min, respectively.(b)Power expenditure increases as the mass of therunner increases (assuming constant velocity).(c)Sincev= 1a(PM+b), the velocities arev= 1a(38.780+b)≈500 m/min,v= 1a(38.784+b)≈477 m/min,andv= 1a(38.790+b)≈447 m/min.(d)The velocity decreases as the mass increases(assuming constant power expenditure).(e)Note,P=M va−M b. Solving forv, we find52(480)a−52b=50va−50b52(480)a−2b=50va52(480)a−2b50a=v498.2≈v.The velocity is approximately 498.2 m/min.(f )With weights removed and constant powerexpenditure, a runner’s velocity increases.(g)The first graph showsPversusM(withv=400)204060M2550pand the second graph showsvversusM(withP= 40)255075M50010001500vFor Thought1.False,P(1 +rt) =SimpliesP=S1 +rt .2.False, since the perimeter is twice the sum ofthe length and width.3.False, sincen+ 1andn+ 3 are even integers ifnis odd.4.True5.True, sincex+ (−3−x) =−3.6.False, sinceP= 2S

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42Chapter 1Equations, Inequalities, and Modeling7.False, for if the house sells forxdollars thenx−0.09x=100,0000.91x=100,000x=$109,890.11.8.True9.False, a correct equation is 4(x−2) = 3x−5.10.False, since 9 andx+ 9 differ byx.1.2 Exercises1.formula2.function3.uniform4.rate, time5.r=IP t6.R=DT7.SinceF−32 = 95C,C= 59 (F−32).8.Since 95C=F−32, 95C+ 32 =ForF= 95C+ 32.9.Since 2A=bh, we getb= 2Ah.10.Since 2A=bh, we haveh= 2Ab.11.SinceBy=C−Ax, we obtainy=C−AxB.12.SinceAx=C−By, we getx=C−ByA.13.Multiplying byRR1R2R3, we findR1R2R3=RR2R3+RR1R3+RR1R2R1R2R3−RR1R3−RR1R2=RR2R3R1(R2R3−RR3−RR2)=RR2R3.ThenR1=RR2R3R2R3−RR3−RR2.14.Multiplying byRR1R2R3, we findR1R2R3=RR2R3+RR1R3+RR1R2R1R2R3−RR2R3−RR1R2=RR1R3R2(R1R3−RR3−RR1)=RR1R3.ThenR2=RR1R3R1R3−RR3−RR1.15.Sincean−a1= (n−1)d, we obtainn−1=an−a1dn=an−a1d+ 1n=an−a1+dd.16.Multiplying by 2, we get2Sn=n(a1+an)2Sn=na1+nan2Sn−nan=na1.Thena1= 2Sn−nann.17.SinceS=a1(1−rn)1−r, we obtaina1(1−rn)=S(1−r)a1=S(1−r)1−rn.18.SinceS= 2LW+H(2L+ 2W), we obtainH(2L+ 2W)=S−2LWH=S−2LW2L+ 2W .19.Multiplying by 2.37, one finds2.4(2.37)=L+ 2D−F√S5.688−L+F√S=2DandD= 5.688−L+F√S2.20.Multiplying by 2.37, one finds2.4(2.37)=L+ 2D−F√SF√S=L+ 2D−5.688F=L+ 2D−5.688√S.

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1.2 Constructing Models to Solve Problems4321.R=D/T22.T=D/R23.SinceLW=A, we haveW=A/L.24.SinceP= 2L+ 2W, we have 2W=P−2LandW=P/2−L.25.r=d/226.d= 2r27.By using the formulaI=P rt, one gets51.30=950r·10.054=r.The simple interest rate is 5.4%.28.Note,I=P rtand the interest is $5.5=100r·11260=100r0.6=rSimple interest rate is 60%.29.SinceD=RT, we find5570=2228·T2.5=T.The surveillance takes 2.5 hours.30.SinceC= 2πr, the radius isr= 72π2π= 36 in.31.Note,C= 59 (F−32). IfF= 23oF, thenC= 59 (23−32) =−5oC.32.SinceF= 95C+ 32 andC= 30oC, we getF= 95·30 + 32 = 54 + 32 = 86oF.33.Ifxis the cost of the car before taxes, then1.08x=40,230x=$37,250.34.Ifxis the minimum selling price, thenx−0.06x−780=128,0000.94x=128,780x=$137,000.35.LetSbe the saddle height and letLbe theinside measurement.S=1.09L37=1.09L371.09=L33.9≈LThe inside leg measurement is 33.9 inches.36.Leth,a, andrdenote the target heart rate,age, and resting heart rate, respectively.Substituting we obtain144=0.6[220−(a+r)] +r144=0.6[220−(30 +r)] +r144=0.6[190−r] +r144=114 + 0.4r30=0.4r.The resting heart rate isr= 300.4 = 75.37.Letxbe the sale price.1.1x=50,600x=50,6001.1x=$46,00038.Letxbe the winning bid.1.05x=2.835x=2.8351.05x=2.7The winning bid is 2.7 million pounds.

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44Chapter 1Equations, Inequalities, and Modeling39.Letxbe the amount of her game-showwinnings.0.14x3 + 0.12x6=40006(0.14x3 + 0.12x6)=24,0000.28x+ 0.12x=24,0000.40x=24,000x=$60,000Her winnings is $60,000.40.Letxand 4.7−xbe the amounts of the highschool and stadium contracts, respectively, inmillions of dollars.0.05x+ 0.04(4.7−x)=0.2230.05x+ 0.188−0.04x=0.2230.01x=0.035x=3.5The high school contract and stadium contractare are $3.5 million and $1.2 million, respec-tively.41.Ifxis the length of the shorter piece in feet,then the length of the longer side is 2x+ 2.Then we obtainx+x+ (2x+ 2)=304x=28x=7.The length of each shorter piece is 7 ft and thelonger piece is (2·7 + 2) or 16 ft.42.Ifxis the width of the old field, then (x+ 3)xis the area of the old field. The larger field hasan area of (x+ 5)(x+ 2) and so(x+ 5)(x+ 2)=(x+ 3)x+ 46x2+ 7x+ 10=x2+ 3x+ 464x=36x=9.Sincex+ 3 = 9 + 3 = 12, the dimension of theold field is 9 m by 12 m.43.Ifxis the length of the side of the larger squarelot then 2xis the amount of fencing neededto divide the square lot into four smaller lots.The solution to 4x+ 2x= 480 isx= 80.Theside of the larger square lot is 80 feet and itsarea is 6400 ft2.44.Ifxis the length of a shorter side, then thelonger side is 3xand the amount of fencing forthe three long sides is 9x. Since there are fourshorter sides and there is 65 feet of fencing,we have 4x+ 9x= 65.The solution to thisequation isx= 5 and the dimension of eachpen is 5 ft by 15 ft.45.Letddenote the distance from Fairbanks toColdfoot,which is the same distance fromColdfoot to Deadhorse.d50 +d40=11.25 hr90d=11.25(2000)d=250 miles46.Letxbe the distance from Dawson City toInuvik.x23 + 3=x20x=−3123−120x=460 miles47.Note, Bobby will complete the remaining 8laps in890 of an hour. If Ricky is to finish atthe same time as Bobby, then Ricky’s averagespeedsover 10 laps must satisfy 10s=890 .(Note:time=distancespeed).The equationis equivalent to 900 = 8s. Thus, Ricky’saverage speed must be 112.5 mph.

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Solution Manual for College Algebra, 6th Edition - Page 15 preview image

1.2 Constructing Models to Solve Problems4548.Letdbe the distance from the camp to thesite of the crab traps. Note,rate=distance÷time.againstd1/22dgoingd1/66ddistancetimerateSince the speed going with the tide is increasedby 2 mph, his normal speed is 6d−2. Similarly,since his speed against the tide is decreasedby 2 mph, his normal speed is 2d+ 2.Then6d−2 = 2d+ 2 for the normal speeds are thesame. Solving ford, we getd= 1 mile whichis the distance between the camp and the siteof the crab traps.49.Letdbe the halfway distance between SanAntonio and El Paso, and letsbe the speedin the last half of the trip. Junior tookd80 hours to get to the halfway point and thelast half tookdshours to drive. Since the totaldistance is 2danddistance=rate×time,2d=60(d80 +ds)160sd=60 (sd+ 80d)160sd=60sd+ 4800d100sd=4800d100d(s−48)=0.Sinced6= 0, the speed for the last half of thetrip wass= 48 mph.50.LetAbe Parker’s average for the remaininggames and letnbe the number of games in theentire season. The number of points she scoredthrough two-thirds of the season is 18(2n3)points.If she must have an average of 22points per game for the entire season, then18(2n3)+A(n3)n=22363 +A3=2236 +A=66A= 30.For the remaining games, she must average 30points per game.51.Ifxis the part of the start-up capital investedat 5% andx+ 10,000 is the part invested at6%, then0.05x+ 0.06(x+ 10,000)=58800.11x+ 600=58800.11x=5280x=48,000.Norma invested $48,000 at 5% and $58,000at 6% for a total start-up capital of $106,000.52.Ifxis the amount in cents Bob borrowed at8% andx2 is the amount borrowed by Betty at16%, then0.08x+ 0.16(x2)=240.08x+ 0.08x=240.16x=24x=150 cents.Bob borrowed $1.50 and Betty borrowed$0.75.53.Letxand 1500−xbe the number ofemployees from the Northside and Southside,respectively. Then(0.05)x+ 0.80(1500−x)=0.5(1500)0.05x+ 1200−0.80x=750450=0.75x600=x.There were 600 and 900 employees atthe Northside and Southside, respectively.54.Letxbe the number of ounces of a30% solution.70% soln40 +x0.7(40 +x)80% soln400.8(40)30% solnx0.3xamt of solnamt of alcohol

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Solution Manual for College Algebra, 6th Edition - Page 16 preview image

46Chapter 1Equations, Inequalities, and ModelingAdding the amounts of pure alcohol we get0.3x+ 0.8(40)=0.7(40 +x)0.3x+ 32=28 + 0.7x4=0.4x10=x.Then10ouncesofthe30%solutionareneeded.55.Letxbe the number of hours it takes bothcombines working together to harvest an entirewheat crop.combined1/xnew1/48old1/72rateThen172 + 148 = 1x. Multiply both sides by144xand get 2x+ 3x= 144. The solution isx= 28.8 hr, which is the time it takes bothcombines to harvest the entire wheat crop.56.Letxbe the number of hours it takes Rita andEduardo, working together, to process a batchof claims.together1/xEduardo1/2Rita1/4rateIt follows that 12 + 14 = 1x. Multiplying bothsides by 4x, we get 2x+x= 4.The solution isx= 43 hr, which is how long it will take bothof them to process a batch of claims.57.Lettbe the number of hours since 8:00 a.m.Robin1/12tt/12Batman1/8t−2(t−2)/8ratetime work completedt−28+t12=124(t−28+t12)=243(t−2) + 2t=245t−6=24t=6At 2 p.m., all the crimes have been cleaned up.58.Lettbe the number of hours since noon.Don1/15tt/15Della1/10t−3(t−3)/10ratetime work completedSince the sum of the works completed is 1,t−310+t15=130(t−310+t15)=303(t−3) + 2t=305t−9=30t=7.8.They will finish the job in 7.8 hrs or at 7:48p.m.59.Since there are 5280 feet to a mile and thecircumference of a circle isC= 2πr,the radiusrof the race track isr= 52802π .Since the length of a side of the square plot istwice the radius, the area of the plot is(2·52802π)2≈2,824,672.5 ft2.Dividing this number by 43,560 results to64.85 acres which is the acreage of the squarelot.60.Since the volume of a circular cylinder isV=πr2h, we have 12(1.8) =π(2.3752)2·h.Solving forh, we geth≈4.876 in., the heightof a can of Coke.

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