Solution Manual For Intermediate Algebra, 13th Edition

Solution Manual For Intermediate Algebra, 13th Edition is your ultimate textbook solutions guide, providing answers to the most difficult questions.

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SOLUTIONSMANUALINTERMEDIATEALGEBRATHIRTEENTHEDITIONMarvin L. BittingerIndiana University Purdue University IndianapolisJudith A. BeecherBarbara L. JohnsonIvy Tech Community College of Indiana

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ContentsJust-In-Time Review...................... 1Chapter1..........................11Chapter2..........................39Chapter3.........................73Chapter4......................... 113Chapter5.......................... 143Chapter6......................... 183Chapter7......................... 211Chapter8......................... 259Chapter9......................... 295

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Just-In-Time Review1. The Set of Real Numbers1.We specify conditions by which we know whether a numberis in the set.{x|xis a whole number less than or equal to 5}, or{x|xis a whole number less than 6}2.We specify conditions by which we know whether a numberis in the set.{x|xis an integer greater than 3 and less than 11}3.Consider the following numbers:20,10,5.34, 18.999,0.583, 1145 ,7,2,16, 0,23 , 9.34334333433334....a) The natural numbers are numbers used for counting.The natural numbers in the list above are 20 and16 (16 = 4).b) The whole numbers are the set of natural numberswith 0 also included. The whole numbers in the listabove are 20,16, and 0.c) The integers are the set of whole numbers and theiropposites.The integers in the list above are 20,10,16, and 0.d) The irrational numbers are the numbers that can-notberepresentedasaquotientoftwointe-gers.Their decimal notation neither terminatesnor has a repeating block of digits.The irra-tional numbers in the list above are7,2, and9.34334333433334....e) Rational numbers can be named as quotients ofintegers with nonzero denominators. Decimal no-tation for rational numbers either terminates orhas a block of repeating digits. The rational num-bers in the list above are 20,10,5.34, 18.999,0.583, 1145 ,16, 0, and23 .f) The set of real numbers is composed of the ratio-nal numbers and the irrational numbers. All of thenumbers in the list are real numbers.4.Consider the following numbers:6, 0, 1,12 ,4, 79 , 12,65 , 3.45, 5 12 ,3,25,123 , 0.131331333133331....a) The natural numbers are numbers used for counting.The natural numbers in the list above are 1, 12, and25 (25 = 5).b) The whole numbers are the set of natural numberswith 0 also included. The whole numbers in the listabove are 0, 1, 12, and25.c) The integers are the set of whole numbers and theiropposites. The integers in the list above are6, 0,1,4, 12,25, and123(123=4).d) The irrational numbers are the numbers that cannotbe represented as a quotient of two integers. Theirdecimal notation neither terminates nor has a re-peating block of digits.The irrational numbers inthe list above are3, and 0.131331333133331....e) Rational numbers can be named as quotients of in-tegers with nonzero denominators.Decimal nota-tion for rational numbers either terminates or hasa block of repeating digits.The rational numbersin the list above are6, 0, 1,12 ,4, 79 , 12,65 ,3.45, 5 12(5 12 = 112),25, and123 .f) The set of real numbers is composed of the ratio-nal numbers and the irrational numbers. All of thenumbers in the list are real numbers.2. Order for the Real Numbers1.Since5 is to the left of4 on the number line, we have5<4.2.Since14 is to the right of12 on the number line, we have14>12 .3.Since 87 is to the right of 67, we have 87>67.4.Since9.8 is to the left of4 23 , we have9.8<4 23 .5.Since 6.78 is to the right of6.77, we have 6.78>6.77.6.45 =0.8; since0.8 is to the right of0.86, we have0.8>0.86, or45>0.86.7.We convert to decimal notation:1429 = 0.48275. . .and1732 = 0.53125.We see that1429is to the left of1732 , sowe have 1429<1732 .8.We convert to decimal notation:1213 =0.923. . .and1415 =0.93. We see that1213 is to the right of1415 , sowe have1213>1415 .9.Since 1.8 is to the right of 1.08, we have 1.8>1.08.10.Since 0 is to the right of4, we have 0>4.11.x >6The inequality 6< xhas the same meaning.12.4<7The inequality 7>4 has the same meaning.

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2Just-in-Time Review13.6≥ −9.4 is true because 6>9.4 is true.14.18≤ −18 is true because18 =18 is true.15.7.6≤ −10 45 is false because neither7.6<10 45 nor7.6 =10 45 is true.16.2427 =0.8 and2528 =0.892. . .so2427>2528 . Thus,2427≥ −2528 is true.3. Graphing Inqualities on the Number Line1.x >1We shade all the numbers to the right of1 and use a leftparenthesis at1 to indicate that it is not a solution.2.x5We shade all the numbers to the left of 5 and use a rightbracket at 5 to indicate that it is also a solution.3.x >0We shade all the numbers to the right of 0 and use a leftparenthesis at 0 to indicate that it is not a solution.4.x≤ −52We shade all the numbers to the left of52 and use a rightbracket at52 to indicate that it is also a solution.4. Absolute Value1.The distance of4 from 0 is 4, so| −4|= 4.2.The distance of 0 from 0 is 0, so|0|= 0.3.The distance of 252.7 from 0 is 252.7, so|252.7|= 252.7.4.The distance of 31 from 0 is 31, so|31|= 31.5.The distance of12 from 0 is 12 , so12= 12 .6.The distance of0.03 from 0 is 0.03, so| −0.03|= 0.03.5. Add Real Numbers1.8 + (3)We find the difference of the absolute values, 83 = 5.Since the positive number has the larger absolute value,the answer is positive.8 + (3) = 52.8 + 3We find the difference of the absolute values, 83 = 5.Since the negative number has the larger absolute value,the answer is negative.8 + 3 =53.8 + (3)The sum of two negative numbers is negative. We add theabsolute values, 8 + 3 = 11, and make the answer negative.8 + (3) =114.8 + 3We add the numbers. The result is positive.8 + 3 = 115.7 + (2)We find the difference of the absolute values, 72 = 5.Since the positive number has the larger absolute value,the answer is positive.7 + (2) = 56.8 + (8)The sum of two negative numbers is negative. We add theabsolute values, 8 + 8 = 16, and make the answer negative.8 + (8) =167.13 + 13We have a positive number and a negative number withthe same absolute value, so the sum is 0.13 + 13 = 08.26 + 0One number is 0. The sum is the other number.26 + 0 =269.6 + (15)The sum of two negative numbers is negative.We addthe absolute values, 6 + 15 = 21, and make the answernegative.6 + (15) =2110.9 + (2)We find the difference of the absolute values, 92 = 7.Since the positive number has the larger absolute value,the answer is positive.9 + (2) = 7

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Just-in-Time Review311.8.4 + 9.6We find the difference of the absolute values, 9.68.4 =1.2.Since the positive number has the larger absolutevalue, the answer is positive.8.4 + 9.6 = 1.212.2.62 + (6.24)The sum of two negative numbers is negative. We add theabsolute values, 2.62 + 6.24 = 8.86, and make the answernegative.2.62 + (6.24) =8.8613.25 + 34We find the difference of the absolute values.3425 = 34·5525·44 = 1520820 =720Since the positive number has the larger absolute value,the answer is positive.25 + 34 =72014.56 +(78)The sum of two negative numbers is negative. We add theabsolute values.56 + 78 = 56·44 + 78·33·2024 + 2124 = 4124The answer negative.56 +(78)=41246. Opposites, or Additive Inverses1.The opposite, or additive inverse, of 10 is10, because10 + (10) = 0.2.The opposite, or additive inverse, of23is23 , because23 + 23 = 0.3.The opposite, or additive inverse, of0.007 is 0.007, be-cause0.007 + 0.007 = 0.4.The opposite, or additive inverse, of 0 is 0, because 0 + 0 =0.5.Whenx= 9, thenx=9.Whenx= 9, then(x) =(9) = 9.6.Whenx=23, thenx=(23) = 23.Whenx=23, then(x) =((23)) =(23) =23.7. Subtract Real Numbers1.511 = 5 + (11) =62.5(11) = 5 + 11 = 163.511 =5 + (11) =164.5(11) =5 + 11 = 65.17(17) = 17 + 17 = 346.614 =6 + (14) =207.9(9) =9 + 9 = 08.813 = 8 + (13) =59.31(16) = 31 + 16 = 4710.15.827.4 = 15.8 + (27.4) =11.611.18.0111.24 =18.01 + (11.24) =29.2512.56(112)= 56 + 112 = 1012 + 112 = 111213.13(112)=13 + 112 =412 + 112 =312 =1414.1345 = 13 +(45)=515 +(1215)=7158. Multiply Real Numbers1.3(7)The product of a positive number and a negative numberis negative.We multiply the absolute values, 3·7 = 21,and make the answer negative.3(7) =212.(4.2)(6.3)The product of two negative numbers is positive. We mul-tiply the absolute values, 4.2(6.3) = 26.46, and make theanswer positive.(4.2)(6.3) = 26.463.5·9The product of a negative number and a positive numberis negative.We multiply the absolute values, 5·9 = 45,and make the answer negative.5·9 =454.137·(52)The product of two negative numbers is positive. We mul-tiply the absolute values, 137·52 = 6514 , and make the an-swer positive.137·(52)= 65145.0·(11) = 0

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4Just-in-Time Review6.35·47The product of a negative number and a positive numberis negative. We multiply the absolute values, 35·47 = 1235 ,and make the answer negative.35·47 =12357.4(13)The product of two negative numbers is positive. We mul-tiply the absolute values, 4·13 = 52, and make the answerpositive.4(13) = 528.3(23)The product of a negative number and a positive number isnegative. We multiply the absolute values, 3·23 = 63 = 2,and make the answer negative.3(23)=29.911·(119)The product of two negative numbers is positive. We mul-tiply the absolute values,911·119= 9999 = 1, and make theanswer positive.911·(119)= 110.3(4)(5)= 12(5)The product of two negative numbersis positive.= 60The product of two positive numbersis positive.9. Divide Real Numbers1.When a negative number is divided by a positive number,the answer is negative.84=22.When a negative number is divided by a negative number,the answer is positive.7711 = 73.50Not defined:Division by 0.4.032 = 0 because 0·32 = 0.5.The reciprocal of 34 is 43 , because 34·43 = 1.6.The reciprocal of78 is87 , because78·(87)= 1.7.The reciprocal of 25 is125 , because 25·125 = 1.8.The reciprocal of 0.2 is10.2 .This can also be expressed as follows:10.2 =10.2·1010 = 102= 5.9.27÷(113)= 27·(311)=67710.103÷ −215 =103·(152)= 1506, or 2511.48÷0.4 =480.4=480.4 =12012.819÷(2) =819·(12)=838 =2·42·19 =41913.The opposite of38 is 38 , because38 + 38 = 0.The reciprocal of38 is83 , because38(83)= 1.14.The opposite of 6 is6, because 6 + (6) = 0.The reciprocal of 6 is 16 , because 6·16 = 1.10. Exponential Notation (Part 1)1.y·y·y·y·y·y︷︷=y66 factors2.8·8·8·8︷︷= 844 factors3.(3.8)(3.8)(3.8)(3.8)(3.8)︷︷= (3.8)55 factors4.(45)(45)(45)︷︷=(45)33 factors5.(34)1= 34(For any numbera,a1=a.)6.170= 1(For any nonzero numbera,a0= 1.)7.(2)6= (2)·(2)·(2)·(2)·(2)·(2)︷︷= 646 factors8.(7)3= (7)·(7)·(7)︷︷=3433 factors9.(13)3= 13·13·13︷︷=1273 factors10.(34) =(3·3·3·3) =(81) =81

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Just-in-Time Review511.(0.1)6= (0.1)(0.1)(0.1)(0.1)(0.1)(0.1) = 0.00000112.(57)2=(57)(57)= 254913.y5=1y514.(4)3=1(4)3=164 =16415.(34)2=1(34)2=1916= 1·169= 16916.1a2=a217.(14)2=1(14)2=1116= 1·161= 1618.134= 3419.1b3=b320.1(16)2= (16)211. Order of Operations1.9[87(52)] = 9[87·3]= 9[821]= 9[13]=1172.(912)2= (3)2= 93.4÷(810)2+ 1 = 4÷(2)2+ 1= 4÷4 + 1= 1 + 1= 24.[64÷(4)]÷(2) =16÷(2)= 85.92122= 81144 =636.[24÷(3)]÷(12)=8÷(12)=8·(2)= 167.9÷(3) + 16÷8 =3 + 2=18.20 + 43÷(8) = 20 + 64÷(8)= 20 + (8)= 129.256÷(32)÷(4) =8÷(4)= 210.9[(811)13] = 9[313]= 9[16]=14411.(182·3)9 = (186)9= 129= 312.(13·28·4)2= (2632)2= (6)2= 3613.124(51) = 124(4)= 1216=414.[2·(53)]2= [2·2]2= 42= 1615.[5(86) + 12][24(84)] = [5·2 + 12][244]= [10 + 12][244]= 2220= 216.3218÷[4(2)] =3218÷[4 + 2]=3218÷6=323=3517.25[4·32(1 + 4·2)] = 25[4·32(1 + 8)]= 25[4·32·9]= 25[1218]= 25[6]= 25+ 6= 32 + 6= 3818.103{8÷[4(116)]}= 103{8÷[45]}= 103{8÷[1]}= 103{−8}= 10 + 24= 3419.4|67| −5·46·78|41|= 4| −1| −5·46·78|3|= 4·15·46·78·3 =4204224 =1618=89

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6Just-in-Time Review20.(83)2+ (710)23223= 52+ (3)23223= 25 + 998= 341= 3412. Translate to an Algebraic Expression1.xdivided by 6We havex÷6, orx6 , orx/6.2.Seven more than four times a numberLetyrepresent the number. We have 7 + 4y, or 4y+ 7.3.25 less thantWe havet25.4.One third of a numberLetbrepresent the number. We have 13b, orb3 .5.The sum ofwand twiceqWe havew+ 2q, or 2q+w.6.18 multiplied bymWe have18·m, or18m, orm(18).7.msubtracted fromsWe havesm.8.Nineteen percent of some numberLetxrepresent the number. We have 19%x, or 0.19x, or19100x.13. Evaluate Algebraic Expressions1.Substitute8 foryand carry out the calculation.57y= 57(8) =4562.Substitute 30 forxand6 foryand carry out the calcu-lation.xy= 306 =53.Substitute 20 forpand 30 forqand carry out the calcula-tions.5p+q=520 + 30 =550 =1104.Substitute 7 formand 18 fornand carry out the calcula-tions.18mn= 18·718= 1818·71 = 75.Substitute 3 forxand2 foryand carry out the calcula-tions.4xy= 4·3(2) = 12(2) = 12 + 2 = 146.Substitute 4 forband 6 forcand carry out the calculations.2c÷3b= 2·6÷3·4= 12÷3·4= 4·4= 167.Substitute 3 forrand 27 forsand carry out the calcula-tions.25r2+s÷r2= 2532+ 27÷32= 259 + 27÷9= 259 + 3= 16 + 3= 198.Substitute 15 formand 3 fornand carry out the calcula-tions.m+n(5 +n2) = 15 + 3(5 + 32)= 15 + 3(5 + 9)= 15 + 3·14= 15 + 42= 5714. Equivalent Fraction Expressions1.Since 8x= 8·x, we multiply by 1 usingx/xas a name for1.78 = 78·1 = 78·xx= 7x8x2.Since 50y= 10·5y, we multiply by 1 using 5y5yas a namefor 1.310 =310·5y5y= 15y50y3.Since 51b= 17·3bwe multiply by 1 using 3b3bas a namefor 1.117 =117·3b3b=3b51b4.Since 98w= 49·2w, we multiply by 1 using 2w2was a namefor 1.30049= 30049·2w2w= 600w98w5.25x15x= 5·5x3·5xWe look for the largest commonfactor of the numerator and de-nominator and factor each.= 53·5x5xFactoring the expression= 53·1(5x5x= 1)= 53

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Just-in-Time Review76.36y18y= 18y·218y·1We look for the largest commonfactor of the numerator and de-nominator and factor each.= 18y18y·21Factoring the expression= 1·21(18y18y= 1)= 27.100a25a=25a·425a·1Factoring numerator and denominator=25a25a·41Factoring the expression=1·41(25a25a= 1)=41=48.625t15t=125·5t3·5tFactoring numerator anddenominator=1253·5t5tFactoring the expression=1253·1(5t5t= 1)=1253,or125315. The Commutative Laws and the Associative Laws1.w+ 3 = 3 +wCommutative law of addition2.rt=trCommutative law of multiplication3.pq+ 14 = 14 +pqCommutative law ofadditionpq+ 14 =qp+ 14Commutative law ofmultiplication4.m+ (n+ 2) = (m+n) + 2Associative law ofaddition5.(7·x)·y= 7·(x·y)Associative law ofmultiplication6.(a+b) + 8 =a+ (b+ 8)Associative law=a+ (8 +b)Commutative law(a+b) + 8 =a+ (b+ 8)Associative law=a+ (8 +b)Commutative law= (a+ 8) +bAssociative law(a+b) + 8 = (b+a) + 8Commutative law=b+ (a+ 8)Associative lawOther answers are possible.7.9·(x·y) = (9·x)·yAssociative law= (x·9)·yCommutative law=x·(9·y)Associative law9·(x·y) = (9·x)·yAssociative law=y·(9·x)Commutative law=y·(x·9)Commutative law9·(x·y) = (x·y)·9Commutative law= (y·x)·9Commutative law16. The Distributive Laws1.3(c+ 1) = 3·c+ 3·1= 3c+ 32.3(xy) = 3·x3·y= 3x3y3.2(3c+ 5d) =2·3c2·5d=6c10d4.5x(yz+w) = 5x·y5x·z+ 5x·w= 5xy5xz+ 5xw5.12h(a+b) = 12h·a+ 12h·b= 12ha+ 12hb6.2πr(h+ 1) = 2πr·h+ 2πr·1= 2πrh+ 2πr7.4a5b+ 6 = 4a+ (5b) + 6The terms are 4a,5b, and 6.8.2xy1 = 2x+ (y) + (1)The terms are 2x,y, and1.9.9a+ 9b= 9·a+ 9·b= 9(a+b)10.7x21 = 7·x7·3= 7(x3)11.ab+a=a·b+a·1=a(b+ 1)12.18a24b48 = 6·3a6·4b6·8= 6(3a4b8)13.8m+ 4n24 = 4·2m+ 4·n4·6= 4(2m+n6)14.xyxz+xw=x·yx·z+x·w=x(yz+w)

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8Just-in-Time Review17. Collecting Like Terms1.7x+ 5x= (7 + 5)x= 12x2.8b11b= (811)b=3b3.14y+y= 14y+ 1y= (14 + 1)y= 15y4.12aa= 12a1a= (121)a= 11a5.t9t= 1t9t= (19)t=8t6.5x3x+ 8x= (53 + 8)x= 10x7.3c+ 8d7c+ 4d= (37)c+ (8 + 4)d=4c+ 12d8.4x7 + 18x+ 25 = (4 + 18)x+ (7 + 25)= 22x+ 189.1.3x+ 1.4y0.11x0.47y= (1.30.11)x+ (1.40.47)y= 1.19x+ 0.93y10.23a+ 56b2745a76b=(2345)a+(5676)b27=(10151215)a+(26)b27=215a13b2718. Removing Parentheses and Collecting Like Terms1.(2c) =1(2c)= [1(2)]c= 2c2.(b+ 4) =1(b+ 4)=1·b+ (1)4=b43.(x8) =1(x8)=1·x+ (1)(8)=x+ 8,or 8x4.(ty) =1(ty)=1·t(1)·y=t+ [(1)y]=t+y,oryt5.(r+s+t) =1(r+s+t)=1·r+ (1)·s+ (1)·t=rst6.(8x6y+ 13)=8x+ 6y13Changing the sign of everyterm inside parentheses7.4m(3m1) = 4m3m+ 1=m+ 18.7a[93(5a2)] = 7a[915a+ 6]= 7a[1515a]= 7a15 + 15a= 22a159.2(x+ 3)5(x4)=2(x+ 3) + [5(x4)]=2x6 + [5x+ 20]=2x65x+ 20=7x+ 1410.8x(3y+ 7) + (9x11)= 8x+ 3y7 + 9x11= 17x+ 3y1811.[7(x+ 5)19][4(x6) + 10]= [7x+ 3519][4x24 + 10]= [7x+ 16][4x14]= 7x+ 164x+ 14= 3x+ 3012.5{−2 + 3[42(3 + 5)]}= 5{−2 + 3[42(8)]}= 5{−2 + 3[416]}= 5{−2 + 3[12]}= 5{−236}= 5{−38}=19019. Exponential Notation1.82·84= 82+(4)= 86=1862.a3a2=a3(2)=a3+2=a53.103106= 1036= 103+(6)= 109=1109

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Just-in-Time Review94.b2·b5=b2+(5)=b3= 1b35.24x6y718x3y9=2418x6(3)y79=2418x6+3y2=43x9y2=4x93y26.(14m2n3)(2m3n2) = 14·(2)·m2·m3·n3·n2=28m2+3n3+2=28m5n57.(3t4a)(5ta) =3(5)·t4a·ta= 15·t4a+(a)= 15t5a,or 15t5a8.18x2y312x5y5=1812x2(5)y35= 1812x2+5y2= 32x3y2= 3x32y29.(r9)5=r9·5=r4510.(4334)3= (43)3(34)3= 43·334·3= 49312=149·31211.(5a2b2)3= 53(a2)3(b2)3= 125a2·3b2·3= 125a6b612.(a4c7)2= (a4)2(c7)2=a8c1413.(43)2= 43·2= 4614.(2x3y23y3)3= (2x3y2)3(3y3)3= 23(x3)3(y2)333(y3)3= 8x9y627y9= 8x9y6(9)27= 8x9y32715.(8x)4y= 8x·4y= 84xy16.(4x4y25x1y4)4= (4x4y2)4(5x1y4)4= (4)4(x4)4(y2)454(x1)4(y4)4= (1)4·44x16y854x4y16= 44x164y8(16)54= 44x20y2454= 54y2444x2020. Scientific Notation1.4 . 7,000,000,000.10 placesLarge number, so the exponent is positive.47,000,000,000 = 4.7×10102.0.00000001.68 placesSmall number, so the exponent is negative.0.000000016 = 1.6×1083.0.0000002.637 placesSmall number, so the exponent is negative.0.000000263 = 2.63×1074.2 . 600,000,000,000.12 placesLarge number, so the exponent is positive.2,600,000,000,000 = 2.6×10125.200 billionths = 200×0.000000001 = 0.00000020.0000002.7 placesSmall number, so the exponent is negative.0.0000002 = 2×1076.6.73000000.8 placesPositive exponent so the number is large.6.73×108= 673,000,000

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10Just-in-Time Review7.We move the decimal point 28 places to the left.9.11×1028g = 0.000000000000000000000000000911 g8.We move the decimal point 9 places to the right.1.86×109= 1,860,000,000 active Facebook users9.(6.5×103)(5.2×108)= (6.5×5.2)(103×108)= 33.8×105= (3.38×10)×105= 3.38×10410.3.6×1037.2×1012= 3.67.2×1031012= 0.5×109= (5×101)×109= 5×108

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Chapter 1Solving Linear Equations and InequalitiesExercise Set 1.1RC2.(f)RC4.(c)CC2.423x= 52a) The LCM of the denominators is 6, so wemultiplyby6.b)6(423x)= 6·52244x= 15CC4.0.067x= 1.2xa) The greatest number of decimal places is2, so we multiplyby100.b) 100(0.067x) = 100(1.2x)6700x= 120x2.47x= 234724 ? 2323TRUE24 is a solution of the equation.4.3x+ 14 =273(10) + 14 ?2730 + 1416FALSE10 is not a solution of the equation.6.x8=3328?34FALSE32 is not a solution of the equation.8.45x= 5945(11) ? 594 + 5559TRUE11 is a solution of the equation.10.9y+ 5 = 869·9 + 5 ? 8681 + 586TRUE9 is a solution of the equation.12.x+ 5 = 5 +x13 + 5 ? 5 + (13)88TRUE13 is a solution of the equation.14.x+ 7 = 14x+ 77 = 147x+ 0 = 7x= 716.27 =y1727 + 17 =y17 + 1710 =y+ 010 =y18.8 +r= 1788 +r= 8 + 170 +r= 25r= 2520.37 +x=893737 +x= 37890 +x=52x=5222.z14.9 =5.73z14.9 + 14.9 =5.73 + 14.9z+ 0 = 9.17z= 9.1724.x+ 112 =56x+ 112112 =56112x+ 0 =1012112x=1112

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12Chapter 1:Solving Linear Equations and Inequalities26.5x= 305x5= 3051·x= 305x= 628.4x= 1244x4= 12441·x= 1244x=3130.x3 =2513x=253(13)x=3(25)x= 7532.120 =8y1208=8y81208= 1·y15 =y34.0.39t=2.730.39t0.39=2.730.391·t=2.730.39t=736.76y=7867(76)(y) =67(78)1·y= 4256y= 3438.4x7 = 814x= 88x= 2240.6z7 = 116z= 18z= 342.5x+ 7 =1085x=115x=2344.92y+ 4 =9129y+ 8 =919y=99y= 1146.2x+x=13x=1x=1348.2x+ 6x2 = 114x= 13x= 13450.95y+ 410y= 661018y+ 4y= 66Multiplying by 1022y= 66y= 352.0.8t0.3t= 6.50.5t= 6.5t= 1354.15x+ 40 = 8x915x= 8x497x=49x=756.3x15 = 15 + 3x15 = 15False equationNo solution58.9t4 = 14 + 15t9t18 = 15t18 = 6t3 =t60.67x=x14207x=x20 = 8x208=x52 =x62.5x8 =8 + 3xx5x8 =8 + 2x3x= 0x= 064.6y+ 20 = 10 + 3y+y6y+ 20 = 10 + 4y2y=10y=5

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Solution Manual For Intermediate Algebra, 13th Edition - Page 16 preview image

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Exercise Set 1.11366.3t+ 4 = 53t4 = 5False equationNo solution68.52y=2y+ 55 = 5True equationAll real numbers are solutions.70.3458m= 12m3865m= 4m3Multiplying by 89 = 9m1 =m72.0.2t+ 1.7 = 5.8 + 0.3t2t+ 17 = 58 + 3tMultiplying by 1041 =t74.20.01x= 0.4x+ 2.5200x= 40x+ 250Multiplying by 10050 = 41x5041 =x76.23x+ 116 = 12x56x4x+ 61 = 3x5x4x+ 5 =2x5 =6x56 =x78.3(y+ 6) = 9y3y+ 18 = 9y18 = 6y3 =y80.27 = 9(5y2)27 = 45y1845 = 45y1 =y82.210(x3) = 840210x630 = 840210x= 1470x= 784.8x(3x5) = 408x3x+ 5 = 405x= 35x= 786.3(42x) = 4(6x8)126x= 46x+ 8126x= 126x12 = 12True equationAll real numbers are solutions.88.40x+ 45 = 3[72(7x4)]40x+ 45 = 3[714x+ 8]40x+ 45 = 3[14x+ 15]40x+ 45 =42x+ 452x= 0x= 090.16 (12t+ 48)20 =18 (24t144)2t+ 820 =3t+ 185t= 30t= 692.6[4(8y)5(9 + 3y)]21 =7[3(7 + 4y)4]6[324y4515y]21 =7[21 + 12y4]6[1319y]21 =7[17 + 12y]78114y21 =11984y20 = 30y23 =y94.34(3x12)+ 23 = 1354x9 + 16 = 8Multiplying by 2454x= 1x=15496.9(4x+ 7)3(5x8) = 6(23x)5(35 + 2x)36x+ 6315x+ 24 = 46x310x21x+ 87 =16x+ 137x=86x=863798.a9a23=a32=1a32100.2x8y3102.5 + 6x104.10x+ 35y20106.4(x6y), or4(x+ 6y)108.5(2x+ 7y4), or5(2x7y+ 4)110.{−8,7,6,5,4,3,2,1};{x|xis a negative integer greater than9}112.0.00458y+ 1.7787 = 13.002y1.00513.00658y=2.7837y0.214
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