' +JJJJ ?\>m0M='+l> /+l   d]@ŵLҦ]]L L}BBL] 鷎귭෍ᷩ췩緈JJJJx Lȿ L8ᷭ緍췩 緍i 8 `巬 췌`x (`(8`I`B` ``>J>J>VU)?`8'x0|&HhHh VY)'&Y)xꪽ)' `Hh`V0^*^*>&` aI꽌ɪVɭ&Y&&Y& 꽌ɪ\8`&&꽌ɪɖ'*&%&,E'зЮ꽌ɪФ`+*xS&x'8*3Ixix&& 8  '  & x)*++`FG8`0($ p,&"_]` L/浍굺L  !"#$%&'()*+,-./0123456789:;<=>?  1#"""  (9"1 ( ,.(0# 2  /#0/#0 *?'#07#00/0/'#07#0:"4<*55/**5/*%5/)1/)1/)1/)'#0/#0*5/*75/**5/*:5//#0/#0'#07#0:::*::'#07#0EB H  @H !D)"E` @ $ C ` DQ &J80^݌Hh ü ü݌ ռ ռ ռA ļD ļ? ļAEDE?HJ>h Լ ռ ռ ռ`HJ>݌h Hh݌`HIHHHHhHH݌hHhHh݌H6 ˆʎõĵL õ ĵµ aµ`` L̦µ_bJLuLz`  ȟ QlXJ̥KlV  ȟ QlV eօ3L e3L &RL &QL d L4 Ne)n `@-eff L f`L . tQLѤ LҦL` OPu d L Ne)noon 8ɍ` ^f\õL ^NR  RΩLҦ)\Z ʽ LHv 3h`0h8` [L NС õ`A@` ŵL^L iõ`  \ 濭0 \  ȟ Q ^\lZl^?cqH şch`fhjõĵ@OAP`u@`@&`QR`E Ls  @DAE@u`8` %@ @A@`@`@A`Mµ ) LЦ`8@AWc@8@-@HAȑ@hHȑ@ȑ@hHȑ@Ȋ@ch8&ȑ@Hȑ@Ah@LHȑ@ȑ@ htphso`hMhL`9V8U897T6S67`NONFORLOASAVRUCHAIDELETLOCUNLOCCLOSREAEXEWRITPOSITIOOPEAPPENRENAMCAMONOMOPRINMAXFILEFINBSAVBLOABRUVERIF!pppp p p p p`" t""#x"p0p@p@@@p@!y q q p@  LANGUAGE NOT AVAILABLRANGE ERROWRITE PROTECTEEND OF DATFILE NOT FOUNVOLUME MISMATCI/O ERRODISK FULFILE LOCKESYNTAX ERRONO BUFFERS AVAILABLFILE TYPE MISMATCPROGRAM TOO LARGNOT DIRECT COMMANč$3>L[dmx ( Ϡ@跻~!Wo*9~~~~ɬƬ~_ j ʪHɪH`Lc (L ܫ㵮赎 ɱ^_ J QL_Ls贩紎 DǴҵԵƴѵӵµȴ 7 ַ :ŵƴѵǴҵȴµ納贍﵎ٵ്ᵭⳍڵL^ѵ-I `  4 ò-յ!  8صٵ紭ﵝ 7L (0+BC  7L HH`LgL{0 HH` õL H hBL BH [ h`Lo õ ڬL B ڬ LʬH hB@ յյ [L (ȴ) ȴ 7L L ( L (ȴL{ƴѵ洩ƴǴҵ 7 ^* B0 HȱBh ӵԵ 8 L8 ݲ` ܫ  / / ED B / / ]ƴS0Jȴ ȴ)  紅D贅E B ƴ  / 0L Ν `HD٤DEEhiHLGh ` ŵBѵ-` ѵB-` ܫ XI볩쳢8 DH E𳈈췍Ȍ X0 · JLǵBȵC`,յp` 䯩 R-յյ`յ0` K R-յյ`ɵʵӵԵ` 4 K ( ѵҵLBȱBL8` DBHBH : ַ޵BȭߵBhhӵԵ RBܵmڵ޵ȱBݵm۵ߵ` 䯩LR˵̵ֵ׵`êĪLR E( 8` R` ELRŪƪ`췌 յյI뷭鷭귭ⵍ㵍跬ª 뷰` Lf ݵܵߵ޵ ^`8ܵ i B8` 4L ֵȱB׵ ܯ䵍൭嵍 ` DȑB׵Bֵ  ַ յյ`굎뵎쵬 뵎쵌``õĵBCõĵ`µµ`L õBĵCصص Qƴ0"Bƴ 󮜳` 0۰ϬBƴ8`i#`ЗLw!0>ﵭ` m ﳐ 7i볍 8 ЉLw`H h ݲL~ `浍국䵍뵩嵠Jm赍嵊mjnnn浈m浍浭m䵍䵐`"L ŵ8ŵH ~(` d ֠Ġz# uSSS-`X$P$"+"S"@30H@, "P$"+"S;X$U:37:U$&q"@I@"V$"@I23HU@-"SU;X$P$"@30H@, "P$X$;US:37:V$"@I@"W$"@I23H@";:X$"<"ĺ"-"SU;X$P$X$US:115&r"-"SU;X$P$"@30H@OR "US;Y$P$'s37:W$"@I@"H1$"@I18V2H@<@37H@>":19:C5355:C1:"!";:C:"@20H@0":17,147F1$:H1$"@16V2H@<6> Graph "B1$:Q$"@9V2H@<2> Rewrite as a @D2H@ "G1$" ":"@5V23H@!"P$"+"S"!"X$U:I1$" or ":X$"<"I1$" and"#&p"@I@"O$"@I23H@!"P$"+"S"!"X$U:37:O$"@I@"Q$"@IU23H@-"U;X$P$"+"S;I1$"@D23H@"P$"+"S;X$U:37:Q$"@I@"U$"@IU23H@-"U;X$"<":Y$">":(A(2))X$">":Y$"<"$nSA(5)3:UA(5)3:O$"@8V2H@<1> Write "N$:U$"@11V2H@<3> Separate its @D2H@ components"F1$:V$"@13V2H@<4> Solve each"F1$"@D2H@ independently ":G1$Z$:X$">"G1$C1$%oW$"@15V2H@<5> Recombine "D2H@to the "C1$" A=-B or A=B.@2D2H@<2> The in"E$" !A! The in"E$" !A!>B is@D2H@"I$" to the "C1$" A<-B@D2H@or A>B.":43%$mP14:39:Q9:L262:H120:R36:131:H16:131:"@5V2H@Solve the in"E$:P$(B(1)):solved by@D2H@breaking them into their two parts,";"j"@D2H@transforming each individually, and@D2H@then turning them back into a@D2H@"Z$" or "C1$".":43X#kH104:131:"@5V2H@For every "N$" A and positive@D2H@number B,@2D2H@<1> The "E$" !A!=B is "I$"@E1$"-10<3X+1<10"D$"!3X+1!>10"E1$"3X+1<-10 or":"@19V2H@3X+1>10.":43V!hP105,107,513"iQ9:L262:R36:H80:131:"@5V2H@C"(Z$,10)"s are "D1$"s formed@D2H@by the word 'and.' D"(C1$,10)"s are@D2H@those formed by the word 'or.'@2D2H@Those "D1$"s may be tten as a"D$C1$". For example,@2D11H@!3X+1! = 10@2D12H@is "I$" to@2D8H@3X+1=10 or 3X+1=-10":43D!g"@5V1H@I"(K$,11)" with "A9$"s and"D$"< signs may be rewritten as"D$Z$"s, while those with"D$A9$"s and > signs may be"D$"rewritten as "C1$"s:"D$"!3X+1!<10":96-a"@16V15HR@"C;:37:"@B@ ":1C100mb"@16V19HL@"C;:37:"@B@ @R16V24H@"C;:37:"@B@ ":C6100cT$(C)T$:"@L16V30H@"T$dC:43e138_ f"@5V1H@An equation with an "A9$" on"D$"on one side and an "N$" on the"D$"other may be rewriand X<6 are true.@10K14V11H@";:C13:"@D11H@"7)"@4F@"7);:C_"@0KI15V12H@";:C12:"@D12H@"5)"@6F@"5);:C:"@D1HL@"19)"@R16V13H@1< @24H@ <6@18V5H@X@32H@1":Y$"<":A(1)0X$"<":Y$">"iUSA(8)2:UA(8)2:VA(8)2:WA(8)2:YSV:Y085VXWU:X0XY(XY)85WP$(B(1)):"@14V2H@IF "S;P$"+"U" "X$" "V;P$"+"W:Y0Y$X$EX"@16V2H@Then "P$" ? ?":23:"@16V9H@"A$:C%2:D1:D1:14,39265,39:"@I5V2H@Find the "L$" of the"8)"@D2H@in"E$" below by transforming the@D2H@in"E$" into a simpler "I$"@D2H@one. Replace the first question mark"/T"@9V2H@with either a > or <, and the second@D2H@with a number to make the "D1$" @D2H@true.6H@"SU;P$"-"SV"-"W;P$"@36H@<"Y:37:Q$"@I@"U$"@I26H@"X;P$"-"SV"@36H@<"Y:37QU$"@I@"V$"@I26H@"X;P$"@36H@<"YSV:37:V$"@I@"W$"@I26H@"P$"@36H@<"(YSV)X:37:W$:19:31051:P:51RP19:39:83:P:51SQ9:L262:H64:R36:131:R108:H48:13om "OW$"@16V1H@<5> Divide using @D1H@ multiplication axiom":"@5V1H@Find the "L$" of the@D1H@in"E$" "S"("U;P$"-"V")-"W;P$"<"Y"@8V1H@HOW TO PROCEED@26H@SOLUTION"NP"@I@"O$"@26HI@"S"("U;P$"-"V")-"W;P$"<"Y:37:O$,"@I@"Q$"@IU2SA(7)2:UA(7)2:VA(7)2:WA(8)1:P$(B(1)):XSUW:X0X176mMYA(9)XSV:YSV99Y9ī77NQ$"@11V1H@<2> Use "R$" @D1H@ "S$" ":U$"@13V1H@<3> Collect like terms ":V$"@14V1H@<4> Add using addition @D1H@ axime negative number.@2D7H@If A>B and C>0@D7H@then AC > BC"J"@14V2H@<4> Multiplying or dividing each@D2H@member by a negative number and@D2H@reversing the order.@D7H@If A>B and C<0@D7H@then AC < BC":43KP14:39:O$"@10V1H@<1> Write the in"E$HL2H@<2> Adding to (or subtracting from)@D2H@each member the same real number.@2D6H@If A>B@D6H@then A+C > B+C@D6H@and A-C > B-C":439I131:H24:131:"@5V2H@"M$"s that produce an@D2H@"I$" "E$".@2D2H@<3> Multiplying (or dividing) each@D2H@member by the sa a simpler "I$D$E$" whose "L$" is"D$"obvious.":43BFP71,73,51GQ9:L262:R36:H24:131:H120:131:"@5V2H@"M$"s that produce an@D2H@"I$" in"E$".@2D2H@<1> Substituting for either member@D2H@of the in"E$" an "N$"@D2H@"I$" to that member."H"@12V our example"D$"the domain has been the whole numbers";DD$"1, 2, 3, 4, and 5, but usually it is"D$"assumed to be the set of real numbers.":432E"@5V1H@To find the "L$" of an"D$"in"E$", we can use the axioms of"D$K$" to transform the"D$"in"E$" intos also a member of the "L$D$" 2X>4. So, we call 2X>4 and X>2@2D2H@";:C$I$" "K$:48:".":61:437CC$"conditional in"E$:"@5V1H@The in"E$" 2X>4 is called a@2D2H@";:48:"because it is"D$"true for at least one but not all"D$"elements of the domain. Inwill be true. The "L$D$"can be described by the sentence:"A"@13V16H@"(123)"X:X>2"(125)"@15V1H@which is read 'The set of all X such@2D1H@that X is greater than 2.'":43bB132:"@L5V1H@5 4 3 2 1@R22H@ @I@>2@I10V2H@Every member of the "L$" X>2"D$" i:G$"":H$" 5 4 3 2 1":C15:L(H$)((2C)):L0L0>"@L5V"C43"H@"(H$,(L))"@5V22HR@"C:37:C2G$(C)" "G$:"@L5V27H@"G$?37:C:"@R5V22H@ "::@"@5V1H@Notice that if X is replaced by any"D$"number greater than 2, the sentence"D$"2X>4 ,91:M59,70,75,82/;P60,64,66,67,69,51<132:"@L5V1H@5 4 3 2 1@20HIR@2*@I@ @I@>4@I10V1H@To find the "L$" of the"D$"in"E$" 2X>4, we must find the set"D$"of all real numbers which can replace"D$"X to make the sentence 2X>4 true.":61:437=19ION@2D4H@<2> RULE@2D4H@<3> EXAMPLE@2D4H@<4> SAMPLE PROBLEM@2D4H@<0> RETURN TO CONTENTS@6D5HI@WHICH ONE (0-4) ??@I@":1364(31153)41295P14:119:P:4:5:(4)"RUNALGEBRA 2"6P1:MA%:(36320M),(36320M)1:407P51839:B2000,58:PP1:55 /44Y 0(36)2):Q(36)73:R(37)82:C$" ";:L(C$)76:H11:131: 1B(31153):133:35339:A(Q)((1)Q):B(Q)((1)684):51 24:5:(4)"RUNAM2.4"x3P0:M0:39:B4(36309)ĺ"@12H5V@MODE SELECTION@2D4H@<1> DISCUSSP"@D1H@M"M: (31051:42& )30976L *"@21V1HI@"38)"@D1H@"38)"@I@": +"@5H21VI@PRESS "(1)" TO VIEW THE NEXT PAGE.@3H22V@PRESS "(2)" TO VIEW THE PREVIOUS PAGE.@I@" ,A(16384):10:A12844 -16368,0:A136İ41:PP1:55 .A149İ40C%F%F%Do !AA128:((A47)(A58))((A39)(A91)(E1))(A45)ĖE%F%:(A);:256F%,A:F%F%D:F%C%35w "27 #B$"":C0F%:B$B$((256C)):C:F(B$):E0: $G1:I1J:K:"@D@"N):I: %O800:38 &I1O:I: '"@1H1V@C"B" P"368,0:A%A128:A%60A%6223/ A$(A%):e D%:E%:C%):F%0:C0C%:256C,32:C:16368,0 E%F%:"@I@"((256F%))"@I@"; 10:A12828 16368,0:E%F%:((256F%));:A155G%1: A14135 A136F%F%F%D A149F%Ѱ5:(4)"RUNALGEBRA 2"' A205Bı0 17: :16D 135] "@R15C0K@"::41:51 "@22V1HI@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 10:A12820 16368,9:A16020 "@22V1H@"36)"@I@": A(16384):A12823 16'24576:30719:63900#49)80:1002:B::S10:A1286m16368,0:A155A%0A%A176:A%B%A%C%6  A(16384):A155(36309)ı 16368,0:"@40X40YN@" A(16384):A12812 16368,0:A155İ4:            22)255Ħ\,0::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": AM2.4":Cnd C > 0 then AC < BC;";+w"@D6H@If A > B and C > 0 then AC > BC.@12V2H@<2> If A < B and C < 0 then AC > BC;@D6H@If A > B and C < 0 then AC < BC.":42+x4:5:30976:F3629936315:F,0:F:(4)"RUN ALGEBRA 2"+yN4Č31051+zN:54 ,1002:(2"AM2.PROGRESS":D$"@2D1H@":I$"Equality":E$"statement":J$"Multiplication":L$"divided":K$"multiplied":G$"Addition":T$"inequalities":Z$"inequality":+vH80:112:"@5V6H@"J$" axiom of order@2D2H@For all real numbers A, B, and C:@2D2H@<1> If A < B aQ1:"<";:H1S:">":H73,Q83(HS)73,Q83:F120:OA(F):O0ēO73,Q83O73,Q88O74,Q88O74,Q83)rF:F120:A(F)0:F:)s AXIOMS OF ORDER,TRANSFORMING INEQUALITIES,COMBINING INEQUALITIES,INEQUALITY TEST*tH$"Property":F$V18H@Correct":36299,(36299)1:[(k"@16V18H@Wrong":K1100:K:"@L10C@":R108,109,109h(l15:110}(m19:(S$)0Ė21(n"@13V@>@R15C@":B4ı(o"@18V2H@By the "N$(R):(pO,PHO,POL,POL,PHO1,PHO1,P:OL1,POL1,PH:x)qH1:"@15C@IF @10C@"Q$O$R$"@D2H15C@THEN @10C@"Q$"+"S$"?"R$"+"S$:106`'hS$(B(19)9):S$"0"104'i"@15C@IF @10C@"Q$O$R$"@D2H15C@THEN @10C@"S$"*"Q$"?"S$"*"R$:(S$)0O$P$!(j"@R15C16V2HI@Enter < or >@I@":14,12797,127:23:"@16V15H@"A$:O$A$ĺ"@16cation"M$:TB(2):O$"<":P$">":T1O$">":P$"<"q&cQ$(C(1)):R$(C(1)):S$(C(1)):S$R$S$Q$R$Q$99&dRB(3)1:"@L10C11V2H@";:R101,102,104&e"@15C@IF @10C@"Q$O$R$O$S$"@D2H15C@THEN @10C@"Q$"?"S$:106&fS$(B(19)9):S$"0"102A'g"@I5V2H@In each of the problems below, @D2H@replace the question mark with the @D2H@symbol < or > so that the resulting @D2H@sentence is true."19)"@I@"3&b14,39265,39:M$" axiom of order":N$(1)"transitive"M$:N$(2)"addition"M$:N$(3)"multipli0O,126170O,131163O,126:O:67,13077,130$_"@17V6H@-12@24H@4@D17H@-12<4@L9V1H@"18):35:"@R9V1H@The order of the "Z$" is@D1H@reversed if the number is negative.":42$`N19:37:97:19:31051:N:55%aO9:L262:P36:H40:112:P84:H72:112:r of the "Z$" is@D1H@preserved if the number is positive.":Q13:J12:S36:H2:F824:A(F7)F:F-$^113:Q16:J:"@12V18H@-6<2@2D13H@-6@22H@2":19:"@15V19H@*2":O12:97O,11965O,131:160O,117170O,131:69O,12465O,131:165O,131170O,131:17der"n"["@138C13V8H@>@12H@>@26H@<@30H@<@15C@":P107:O45:D1:5:92:O74:92:D1:O199:92:O228:92:3:42"\F116:O,P:OOD:K1100:K,F:c#]"@5V1H@When each member of the "Z$"@D1H@-6<2 is "K$" by the same nonzero@D1H@real number,"D$"the ordend 4 units in the@D1H@same direction from the graph of 2,""Z"@10V1H@you will arrive at points on the same@D1H@order on the line as the graphs of -6@D1H@and 2.@17V4H@-6+4<2+4@24H@-6+(-4)<2+(-4)@D6H@-2<6@28H@-10<-6@D1H@This shows the addition axiom of orelow, the graph@D1H@of -6 is left to the graph of 2.":H1:Q14:S14:F612:A(F5)F:F:113:H23:F2632:A(F25)F:F:113:!Y"@15V5H@-6@3F@2@27H@-6@3F@2@D6H@-6<2@29H@-6<2":19:"@6V34H@In@D1H@each figure, if you move 4 units from@D1H@the graph of -6 aD1H@left of Z (XY)@D1H@and Y lies to the right of X (Y>X),@D1H@then Z lies to the right of X (Z>X).@17V1H@This is an example of the transitive@D1H@axiom of order.":42| X"@5V1H@On the number lines b18:A(1)19:113:"@12V6H@X@5F@Y@27H@Y@5F@X@17V18H@X,Y":42VH3:Q14:S34:A(1)9:A(2)19:A(3)30:113:"@15V9H@X@19H@Y@30H@Z@5V1H@From the graph below we see that if X@D1H@is to the left of Y (XY@14V4H@X and Y are at the same point@D18H@X=Y":UH2:Q11:S15:A(1)6:A(2)12:113:H22:A(1)27:A(2)33:113:H10:Q16:SD2H@<2> If A If A If A>B then A+C > B+C.":42R118SN84,86,88,93,55T"@5V1H@If two real numbers X and@Axiom of comparison@7V2H@For all real numbers A and B, one@D2H@and only one of the following@D2H@"E$"s is true.@2D7H@A < B@5F@A = B@6F@A > B"&P"@13V8H@Transitive axiom of order@2D2H@For all real numbers A, B, and C:@2D2H@<1> If A>B and B>C, then A>C;@7)" is"D$" unchanged. But when they are "L$" "D$" or "K$" by a negative number,";MD$" the order is reversed. This is the"D$" ";:C$"multiplication property of order":47:".":42NN79,81,82,55OO9:L262:H120:P36:112:9,100270,100:"@5V11H"K19:"@17V5H@"33)"@11V34H@But"D$"if we were to reduce the order by"D$"changing > to <, it would become true.@17V15H@-24 < -12":42QL"@5V2H@When both members of an "I$" are@12B@e"D$" "K$" or "L$" by a positive"D$" number, the order of the e"(I$, > 2":19:"@5V34H@If we"D$"multiply it by 6, it is still true.@17V11H@4*6 > 2*6 or 24 > 12":19J"@9V1H@Look at it again.@17V11H@"20)"@17H@4 > 2":19:"@9V19H@If we multiply it by"D$"-6, it becomes a false sentence.@17V5H@4*(-6) > 2*(-6) or -24 > -12s"D$"of an "Z$", the order of the"D$Z$" remains unchanged."D$"For example,"H"@8H17V@8 > 1@28H@1 < 8":19:"@4H17V@8+(2) > 1+(2)@23H@1+(-4) < 8+(-4)":19:"@17V1H@"38)"@17V7H@10 > 3@27H@-3 < 4":42`I"@5V1H@Look at the true sentence below.@17V17H@4same way as the"D$" second and third, then the first and"D$" third numbers are ordered in that"D$" same way also.":42KGC$"addition property of equality":"@5V1H@The ";:47:D$"says that when the same number is"D$"added to or subtracted from both sidet"D$"number is larger than the third.":42EC$"transitive axiom of order":"@5V2H@The axiom for using >, <, and =, the"D$" three relational operators, is the"D$" ";:47:". It says"D$" that if the first and second numbers";vFD$" are ordered in the 14H@>@24H@>@R15C@"C"@5V1H@If the first number is larger than the"D$"second, but the second is larger than"D$"the third, then the third must be the"D$"smallest of the three.":19:"@L15V14H@"6)"@D14H@"6)"@10C16V19H@>@R15C@"*D"@11V24H@So, the firsr is less than the third."D$"The third number must then be the"D$"largest of the three.";A" So, the first"D$"number is smaller than the third.":42B"@14V10K@";:F13:"@D8H@ ";:F:"@15V@";:F12:"@D18H@ ";:F:"@17V28H@ @0K10CL16Ver than the"r?"@14V1H@second."D$"They are both equal to each other."D$"The first is less than the second.":42Y@"@5V1H@Now let's compare three numbers."D$"Suppose that the first number is less"D$"than the second number, and that the"D$"second numbe55:37:B60,3000,4000,5000%;8<M61,78,83,96W=N62,64,66,69,71,73,76,55 >"@5V1H@In the first module of the ALGEBRA"D$"series, you were told that only one of"D$"these "E$"s may be true for any"D$"two numbers:"D$"The first number is greatAM2.4-2"7N0:M0:37:"@12H5V@MODE SELECTION@2D4H@<1> DISCUSSION@2D4H@<2> RULE@2D4H@<3> EXAMPLE@2D4H@<4> SAMPLE PROBLEM@2D4H@<0> RETURN TO CONTENTS@6D5HI@WHICH ONE (0-4) ??@I@":B%0:C%4:6:A%İ40:498N1:MA%:(36320M),(36320M)1:399N(1)"@D10H@<2> "A$(2)"@D10H@<3> "A$(3)"@D10H@<4> "A$(4)"@2D10H@<0> RETURN TO ALGEBRA MENU@9DI11H@WHICH ONE (0-4) ??@I@":B%0:C%4:6:BA%:B1204C$A$(B):38:39:B1555B4āF15:36298F,0:F:N14:37:97:19:12164:5:31153,B:(4)"RUN3O13,P16:O14,P12O14,P16O17,P132P:21,15414,14821,14242,14249,14842,15420,15413,14820,142:43,14250,14843,154:31,10631,14228,139:32,10632,14235,139:4:19:"<0>":59,3259,15960,15960,323"@20H5V@CONTENTS@2D10H@<1> "A$(37)82:C$" ";:L(C$)76:H11:112:~ 0116:35339:A$(4):A(20):B(O)((1)O):C(O)((1)684):F14:A$(F):F*1B0:N0:M0:37:C$"":38:O18:P469416:H12:L27:112:(P10)8:"@3H@<"(P30)16">":P94ēO13,P12O13,P16:O10,P11V1HI@"38)"@D1H@"38)"@I@": *"@5H21VI@PRESS "(1)" TO VIEW THE NEXT PAGE.@3H22V@PRESS "(2)" TO VIEW THE PREVIOUS PAGE.@I@" +A(16384):10:A12843 ,16368,0:A136İ40:NN1:57 -A149İ39:NN1:57 .43* /(36)2):O(36)73:P(E1))(A45)ĖD%E%:(A);:256E%,A:E%E%C:E%C%34@ !26u "B$"":F0E%:B$B$((256F)):F:G(B$):E0: #I400:36 $K1I:K: %"@1H1V@C"B" P"N"@D1H@M"M: &"@I2V7H@"31):24(C$)2:C$"@I@": '31051:41 (30976 )"@268,0:A%A128:A%60A%6223. A$(A%):R D%E%:"@I@"((256E%))"@I@";d 10:A12827 16368,0:D%E%:((256E%));:A155F%1: A14134 A136E%E%E%C A149E%C%E%E%C8 AA128:((A47)(A58))((A39)(A91)Ͱ5:(4)"RUNALGEBRA 2"' A205Bı0 17: :16D 117\ "@R15C0K@":40:55 "@22V1HI@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 10:A12820 16368,9:A16020 "@22V1H@"36)"@I@": A(16384):A12823 163^$24576:30719:63900#48)80:1002:B::S10:A1286m16368,0:A155A%0A%A176:A%B%A%C%6  A(16384):A155(36309)ı 16368,0:"@40X40YN@" A(16384):A12812 16368,0:A155İ4:              ĦW10::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": AM2.3-2":`in.":390Q134136:Q4,100Q,104Q,144Q4,148:Q,122Q3,124Q,126:Q:"@13V20H@-1 If Y<0@2D21H@0 If Y=0@2D21H@1 If Y>0":390L1:125:" "T;N1$"+"U"@8V7H@"VC$"@26H@"EC$"@D10H@"N1$"@28H@"T;N1$"+"U"@D8H@DOMAIN@27H@RANGE":10711002:(222)255nd"P$"see if both sides are equal.":39r/44:B$"range":" of the "M1$", and R is"P$"called the ";:44:".":390L1:125:" 2X+1"P$"is read ' The "M1$" F that pairs"P$"X with 2X+1.' But to name a "M1$P$"completely, you must also identify"P$"its domaE":EC$"EQUATION":S.35339:A(Q)((1)Q):C16:D$(C):C:B(31153):B650[.46|.B%1:C%9:5:"@15V33H@"A%:.44:" of the other. So"P$" are "Y$" and "Z$".":39'/P$"To check the value, you can "B1$P$"it back into the original "H$" asion":W$"addition":X$"subtraction":M$"transformation".H$"equation":P$"@2D1H@":T$"operation":V$"opposite":B1$"substitute":M1$"function":E$"element":A1$"equivalent":K1$"Perform inverse":G$"RUNAM2.3":J$"Incorrect":SC$"SOLUTION":VC$"VARIABL& SUB.,TRANSFORMATION BY MULT. & DIV.,USING SEVERAL TRANSFORMATIONS,FUNCTIONS,EQUATIONS TEST8-O1$"property":F$"AM2.PROGRESS":P1$"expression":C1$"distributive":E1$C1$" "O1$:I$"solution":U$"subtracting":D1$"variable":Y$"multiplication":Z$"divi:TA(8)2:UA(8)2:N1$(A(6)84):F1$N1$1267++C$(1,2)Y$:C$(2,2)Z$" ":C$(3,2)"*":C$(4,2)"/":C$(1,1)W$" ":C$(2,1)X$" ":C$(3,1)" +":C$(4,1)"-":+C$(1,1)W$:C$(2,1)X$:C$(2,2)Z$:\,EQUATIONS,TRANSFORMATION BY ADD. SQL1,SH:*zA1A(9)1:B1A(9)1:C1A(9)1::TA(9)1:UA(9)1:O$((1)684):D1A1B1:E1D1C1:F1(TU):C1(A1B1)122*{F1E1(F1E1)122*|G1F1E1:*}Q(36):QL1:">";:QQ7:S(37)83:LL73:Q,SQL,S:1+~F1$(A(6)84)8H@= "TJ1"@14H@+ "U:33:"@17V8H@= "TJ1U:)qC15:D(C)0:C:51:Y2:57:Y3:57:P14:35:86:31051:P:P14:35:109:18:31051:P)rI15:36298I,D(I):I)s4:36309,1:(4)"RUNALGEBRA 2" *yQ,SHQ,SQL,SQL,SHQ1,SHQ1,S:QL1,9:126:"@11V3H@"F1$":"N1$;:L1:125:" "T;N1$"+"U"@U21H@What is the value@D21H@of "F1$"("J1")?":C%3:D%12:E%31:E0:D1:22:FTJ1Uĺ"@13V25H@Correct":D(5)D(5)1:(o"@13V24H@"J$:B6ĺ"@17V8H@= "TJ1U:18:-)p"@13V8H@="T"("J1") + "U:33:"@15V%'lP19:35:109:18:31051:P:46'mS36:Q9:L262:H32:121:S76:H80:L126:121:Q143:121:"@5V2HI@With the "M1$" as defined below, @D2H@find the element of the range for @D2H@the given element of the domain. @I@":14,39265,39(nJ1A(19) domain element@D2H@and solve the "H$".":39&jP14:35:Q9:L262:S36:H104:121:9,60269,60:9,92269,92:126:"@5V15H@FUNCTION "F1$"@D15H@"F1$":"N1$;:146&k11:H113:I1H12:"@2D10H@"I1"@23H@"T"("I1")+"U"@31H@= "TI1U;:H1:18:P:31051:46itten in the form:@2D2H@"M1$" name(members of the domain)@D7H@= member of the range.",&i"@13V2H@To find the element of the range@D2H@that the "M1$" assigns to an@D2H@element of the domain, assign the@D2H@"D1$" in the "M1$"'s "H$"@D2H@to the value of theember of the domain set."$g"@11V2H@Its definition is written in the@D2H@form:@2D2H@"M1$" name: "D1$" ";:L1:125:" "H$:39U%hQ9:L262:S36:H56:121:S100:121:"@5V2H@The assignment of a member of the@D2H@range of the "M1$" to the domain@D2H@is wrs not matter, and several"P$H$"s may be used to name a"P$M1$".@15V12H@S(Y) =":145d#eP102,104,46$fQ9:L262:H40:S36:121:S84:121:"@5V2H@A "M1$" is an "H$" or set of@D2H@"H$"s in which every member of@D2H@the range set is paired with every@D2H@mumber 5 to the number 3, you use the"P$H$" F(3)=5, which is read as";"cP$"'F of 3 equals 5.' The number 5, one"P$"of the "E$"s of the range, is also"P$"called a ";:B$"value":44:" of the "M1$".":39R#d"@5V1H@The "D1$" used in defining a"P$M1$" doe the range is the"P$"set <3,5,7>. The "H$P$"that relates them is R = 2D+1.":39!a"@5V1H@F"(M1$,7)"s are usually named by letters"P$"such as F and G. The arrow notation"P$;13)"F:X ";:144E"bL1:"@5V1H@To state that F:X";:125:"X+2 assigns the"P$"n"one "E$" of R. The set D is called"P$"the ";:143 _S44:H48:L18:Q218:"@6V32H@1 > 3@2D32H@2 > 5@2D32H@3 > 7@2D32H@D R":121:Q253:121:S518316:236,S251,S:SO!`"@5V1H@In this example the domain of"P$"the "M1$" is the set"P$"<1,2,3> andE1;O$"@11H@= "F1;I1$:33:I1$"@17V3H@"O$"@11H@= "G1;J1$:33:J1$:18:\\M93,101,106,108v]P94,95,97,98,100,462 ^B$M1$:"@5V1H@A ";:44:B$"domain":"consists of two sets C and"P$"R together with an "H$" that"P$"assigns each "E$" of D exactly"P$@I@"J1$"@I@":18:ZG1$"@14V18HI@Combine like terms ":H1$"@15V18HI@Put vars. on left ":I1$"@16V18HI@"K1$"(+,-)":J1$"@17V18HI@"K1$"(*,/)"F["@14V3H@"D1;O$"+"T"@11H@= "C1;O$"-"U;G1$:33:G1$"@15V3H@"E1;O$"+"T"@11H@= -"U;H1$:33:H1$"@16V3H@"ing inverse @D2H@"T$"s."25)"@I@":14,39265,39:122X"@13V3H@"A1;O$"+"B1;O$"+"T" = "C1;O$"-"U"@18V2H@What is the value of "O$"?":C%3:D%19:E%26:E0:D1:22:FG1ĺ"@18V29H@Correct":D(4)D(4)1:18:Y"@18V29H@"J$:B6ĺ"@17V3H@"O$"@11H@= "G1"19:35:86:31051:P:46VQ9:L262:S36:H120:121:S921008:9,S269,S:S:"@5V2HI@Solve each of the "H$"s below by@D2H@using the "E1$" to @D2H@combine like terms, bringing all ";7W"@D2H@"D1$"s to one side of the"8)"@D2H@"H$", and then us"A1;O$"+"B1;O$"+"T" = "C1;O$"-"U:33:"@I@"G1$,H1$"@I24H@"D1;O$"+"T"@32H@= "C1;O$"-"U:33:"@I@"H1$:I1$"@I24H@"E1;O$"+"T"@32H@= -"U:33:"@I@"I1$,J1$"@I24H@"E1;O$"@32H@= "F1:33:"@I@"J1$,L1$"@I24H@"O$"@32H@= "G1:33:"@I@"L1$T18:31051:P:46UP$"s on "RI1$I1$"@D1H@ one side. ":J1$"@I12V1H@4) "K1$" @D1H@ "T$"s on @D1H@ "W$"s and @D1H@ "X$"s ":L1$"@I16V1H@5) "K1$" @D1H@ "T$"s on @D1H@ "Y$"s @D1H@ and "Z$"s "SG1$"@I24H@I$" of the "H$"@D6H@is now obvious, check its root@D6H@in the original "H$".":39 QP14:35:"@5V1H@HOW TO PROCEED@24H@SOLUTION":122:G1$"@I7V1H@1) Write "H$" ":H1$"@8V1HI@2) Combine like terms":I1$"@I9V1H@3) Subtract terms to @D1H@ get "D1e@D6H@terms from both sides of the@D6H@"H$".":39O121:"@5V2H@<3> If there are "W$"s or@D6H@"X$"s, use their inverse@D6H@"T$"s to undo them.@9V2H@<4> If there are "Y$"s or@D6H@"Z$"s, use their inverse@D6H@"T$"s to undo them."QP"@13V2H@<5> If the "2H@"A1$" one with an obvious@D2H@"I$":@2D2H@<1> Simplify each member of the@D6H@"H$", combining all terms@D6H@with the same "D1$"s by the@D6H@"E1$"."1N"@14V2H@<2> If the terms on both sides of@D6H@the "H$" contain the same@D6H@"D1$", subtract one of thin their";KP$" numerical coefficients. Using the"P$" "C1$" axiom, you can replace a"P$" sum of similar terms with a single"P$" "D1$". For example, 6X + 3X = 9X.":39LP77,79,46MQ9:L262:S36:H120:121:"@5V2H@To transform an "H$" into an@D$"undoing each "T$" until you get"P$"an "H$" which shows the "I$".";:142 JB$"similar terms":"@5V2H@Terms such as 7Y and 5Y are called"P$" ";:44:B$"like terms":" or ";:44:". Two"P$" terms are similar if they are exactly"P$" alike or differ only H@Since adding and "U$" a number"P$" have an "V$" effect, you can undo"P$" one of these "T$"s by doing its"P$" "V$" "T$" on the result."P$" A"(W$,7)" and "X$" are called"P$" ";:141II"@5V1H@To solve an "H$" you transform it"P$"into an "A1$" "T$" by"PXA(20)2:66D@TA(9)1:UA(9)1:XA(9)1:ZTU:N$"-"ZTUPAVXZ:qBZA(X1)2:XZ(XZ)ī66CUA(Z1)1:N$"*"TZU:T(T)ī67DN$"/"TZUEVXZ:FM71,76,81,85GP72,73,74,46HB$"inverse "T$"s":"@5V22)L$=O$((1)684):63:"@9V2H@("T;K$(3);O$")"N$;U"@12H@= "X:C%3:D%18:E%29:E1:D1:"@17V2H@The "I$" set for "O$" is:":22:FVĺ"@18V28H@Correct":D(Y)D(Y)1:18:>Q$(T):R$((U)):S$(X):"@18V28H@"J$"@12V2H@"O$"@12H@= "V:18:?Y35:58:31051:P:8:Y3L$K$(1):K$(1)K$(2):K$(2)L$;Q9:L262:S68:H88:121:S36:H24:121:14,39265,39:"@5V2HI@Use "M$" by "K$(1)" or @D2H@"36)"@2H@"K$(2)" to solve the "H$"@I@":RA(2):N$K$(4):RN$K$(3)<Y3L$K$(1):K$(1)K$(2):K$(D23H@the domain is@D23H@the root of the@D23H@"H$"?":140:A%B(R)ĺ"@17V25H@Correct":D(1)D(1)1:567"@17V25H@"J$"@D24H@Answer is "B(R)818:"@15V33H@ @2D25H@"9)"@D24H@"12)"@12V20H@ ":P:31051:9128:129:C14:K$(C)C$(C,Y1):C:P14:3DOMAIN@11H@EQUATION@23H@SOLUTION SET":P14:35:TA(8)1:UA(8)1:B(1)A(8)15B(2)A(8)1:B(3)A(8)1:B(3)B(1)B(3)B(2)B(2)B(1)53d6RA(3)1:XTB(R)U:"@12V4H@"B(1)"@2DB@"B(2)"@2DB@"B(3)"@12V10H@"T"X + "U" = "X"@12V23H@Which number in@6 1P46*235:B47,47,47,70,92,1133Q9:L262:S36:H32:121:H80:S76:L49:121:Q65:L84:121:Q157:L112:121:"@I5V2H@Choose the one number from the set @D2H@at the left which makes the "H$"@D2H@below true."25)Q414,39265,39:"@I10V2H@ -137 .P0:M0:35:"@12H5V@MODE SELECTION@2D4H@<1> DISCUSSION@2D4H@<2> RULE@2D4H@<3> EXAMPLE@2D4H@<4> SAMPLE PROBLEM@2D4H@<0> RETURN TO CONTENTS@6D5HI@WHICH ONE (0-4) ??@I@" /B%0:C%4:5:A%İ37:4:(4)G$0P1:MA%:(36320M),(36320M)1:3@PRESS "(1)" TO VIEW THE NEXT PAGE.@3H22V@PRESS "(2)" TO VIEW THE PREVIOUS PAGE.@I@"s (A(16384):9:A12840 )16368,0:A136İ37:PP1:49 *A149İ36:PP1:49 +40 ,(36)2):Q(36)73:S(37)82:B$" ";:L(B$)76:H11:121:,A:F%F%D:F%C%31 23R A$"":C0F%:A$A$((256C)):C:F(A$):E0:t G1:I1J:K:"@D@"N):I: !O800:34 "I1O:I: #"@1H1V@C"B" P"P"@D1H@M"M: $31051:38 %30976 &"@21V1HI@"38)"@D1H@"38)"@I@":V '"@5H21VIC:16368,00 E%F%:"@I@"((256F%))"@I@";A 9:A12824s 16368,0:E%F%:((256F%));:A155G%1: A14131 A136F%F%F%D A149F%C%F%F%D AA128:((A47)(A58))((A39)(A91)(E1))(A45)ĖE%F%:(A);:256F%(4)"RUNALGEBRA 2"" A205Bı+ 165 :15? 136W "@R15C0K@":37:46 "@22V1HI@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 9:A12819 16368,9:A16019 "@22V1H@"36)"@I@": D%:E%:C%):F%0:C0C%:256C,32:Y)24576:30719:63900(132:45.A0:1002:::Q9:A1285k16368,0:A155A%0A%A176:A%B%A%C%5 A(16384):A155(36309)ı 16368,0:"@40X40YN@" A(16384):A12811 16368,0:A155İ4:                         ION":m=R$(4)"@I13V18H@"B$"axiom of equal.":(36309)2āC17:S$(C),R$(C):35:R$(C):C:19:31051:=S$(7)"@I@"R$(7):19:31051:=1002:(222)255Ħ=0::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": AM2.3":M1$"dividing":T1$"multiplied":V1$"divided":SS$"SOLUTION":OO$"EQUATION":HH$"Transformation":J2$"Substitution":159<X9:L262:Y36:H56:142:Y100:H32:142:=HC$"TRANSFORMATION":JC$"ADDITION":LC$"SUBTRACTION":MC$"MULTIPLICATION":NC$"DIVISwhose "S$P$"set is obvious.":43?;D$J$" "F1$" of "G1$:;S$(2)"@11V2H@"V$K$(3)"("Y$O1$Z$")@12H@="A1$:P1$K$(4):O1$K$(3):F10P1$K$(3):O1$K$(4);115w<H$"transformation":O$"equation":G1$"equality":P$"@2D1H@":W1$U1$"ing":Xtement":M$"multiplication":N$"division":J$"addition":L$"subtraction":J5$"Addition":157:"@7V2H@PROCEDURE@28H@"SS$:P14:38:"@5V2H@"36)"@15V2H@"35)"@D2H@"20):G10:K3:J5:O36:33:C1A(8)1:D1A(8)1:77 ;48:", it into an"P$E1$" "Q$" e":Q$"expression":A2$Z1$" "F1$:U1$"multiply":Y1$"multiplicative":B2$"Mult. prop. of ":M5$"Multiplication":N5$"Division":L5$"Subtraction"\:R1$"Commutative":S$"solution":E1$"equivalent":H1$"subtracted":I1$"subtracting":U$"variable":X$"sta(4)"CLOSE":35339:8"EQUATIONS","TRANSFORMATION BY ADD. & SUB.","TRANSFORMATION BY MULT. & DIV." ,"USING SEVERAL TRANSFORMATIONS","FUNCTIONS","EQUATIONS TEST" 9F1$"property":F$"AM2.PROGRESS":J1$"substitution":Z1$"distributive":S1$"Associativ6,2)"Div. ":C$(1,1)"A"(J$,7):C$(2,1)"S"(L$,10):C$(2,2)"D"(N$,7):C$(7,1)"+0":C$(7,2)"*1":C$(8,1)J5$" axiom of 0 ":C$(8,2)C$(5,2)"identity of 1 ":74:5:(4)"OPEN"F$:(4)"READ"F$:C15:C(C):C:C(1)5:W3C(1)1:C(2)3C(2)68"+":C$(4,1)"-":C$(5,1)"Additive "6C$(6,1)"opposites ":C$(6,2)"reciprocals ":C$(7,1)"zero ":C$(7,2)"one ":C$(8,1)"+0":C$(8,2)"*1":Z$(1,1)J$:Z$(2,1)L$:Z$(1,2)N$:Z$(2,2)M$:7C$(5,1)"Add. ":C$(5,2)"Mult. ":C$(6,1)"Sub. ":C$(,@2D17H@A@5F@1@D17H@$ = A*$@D17H@B@5F@B":4355103>5109~5X,YHX,YXL,YXL,YHX1,YHX1,Y:XL1,YXL1,YH:)6C$(1,2)M5$:C$(2,2)N5$" ":C$(3,2)"*":C$(4,2)"/":C$(5,2)"M"(Y1$,13):C$(1,1)J5$" ":C$(2,1)L5$" ":C$(3,1)= C*B@2D2H@This "F1$" guarantees that for";4"@D2H@all real numbers A and all non-zero@D2H@numbers B,@16V13H@1@DB@$*(B*A) = A@D13H@B":43,5X9:L262:Y36:H72:142:"@5V2H@Rule for "N$"@7V2H@For all real numbers A and all@2HD@non-zero real numbers Bl real numbers A and all non-zero@D2H@numbers B,@16V13H@1@DB@$*(B*A) = A@D13H@B":43f3P137,139,55+4X9:L262:Y36:H120:142:"@5V2H@M"(M$,13)" "F1$" of "G1$"@2D2H@If A, B, and C are any real numbers@D2H@such that A=B, then@2D7H@A*C = B*C and C*A "P$" is the same as "W1$" by a "P$" reciprocal.":432X9:L262:Y36:H120:142:"@5V2H@"M$" "F1$" of "G1$"@2D2H@If A, B, and C are any real numbers@D2H@such that A=B, then@2D7H@A*C = B*C and C*A = C*B@2D2H@This "F1$" guarantees that for";T3"@D2H@al$" is saved.":431D$H$" by "M$:"@5V2H@E"(E1$,9)" "O$"s are produced by"P$" ";:48:","P$" "W1$" each side of a given"P$" "O$" by the same non-zero number."P$;:D$"T"(H$,13)" by "N$:48:" is just a";42P$" special case of this, since "X1$" 430D$N$" "F1$" of "G1$:"@5V2H@If two equal numbers are "V1$" by"P$" the same non-zero number, the"P$" quotients are equal. This is the"P$" ";:48:". So,"P$" if both sides of an "G1$" are";1P$" "V1$" by the same non-zero number,"P$" the "G1"@5V1H@The ";:48:D$G1$:"@7V2H@";:48:"says that when equal numbers"P$"are "T1$" by the same number, the"P$"products are equal. So, to undo the";0P$M$" of a number by a given"P$"number, "U1$" the product by the"P$"reciprocal of the given number.":(9)1:E1A(9)1:F1C1D1:O1$"-"F1C1D19.yVE1F1:a.zF1A(E11)2:E1F1(E1F1)ī122.{D1A(F11)1:O1$"*"C1F1D1:C1(C1)ī123.|O1$"/"C1F1D1.}VE1F1:.~M127,136,140,141.P128,130,132,55/D$M$" "F1$" of": ":R$(2)"@I11V18H@"S1$" ":R$(3)"@I12V18H@"J2$" ":R$(5)"@I14V18H@"J2$" "-uR$(7)"@I16V18H@"J2$" ":R$(6)"@I15V18H@"K$(8):B$K$(5):O1$K$(3)B$K$(6)-v160-wB3E1A(20)2:122+.xC1A(9)1:D1AA1$:155,sS$(3)"@12V2H@"V$O1$B1$"@12H@="A1$:S$(4)"@13V2H@"V$O1$B1$P1$B1$"@12H@="A1$P1$B1$:S$(5)"@14V2H@"V$K$(7)"@12H@="A1$P1$B1$:S$(6)"@15V2H@"V$"@12H@="A1$P1$B1$n-tS$(7)"@16V2H@"V$"@12H@="(V):R$(1)"@I10V18H@"R1$"")"O1$D1"@12H@="E1:C%3:D%18:E%29:E1:D1:"@17V2H@The "S$" set for "V$" is:":23:FVĺ"@18V28H@Correct":36301,(36301)1:19:31051: ,rY$(C1):Z$((D1)):A1$(E1):"@18V28H@Incorrect":B1$((F1)):S$(1)"@10V2H@("V$K$(3)Y$")"O1$Z$"@12H@="9:L262:Y68:H88:142:Y36:H24:142:14,39265,39:"@5V2HI@Use "H$" by "Z$(1,B1)" or @D2H@"36)"@2H@"Z$(2,B1)" to solve the "O$"@I@":RA(2):O1$K$(4):RO1$K$(3)*pB3T$K$(1):K$(1)K$(2):K$(2)T$+qV$((1)684):119:"@9V2H@("C1;K$(3);V$34:Q1$"@11V6H@"V$K$(8)" ="D1;P1$N;L1$:34:L1$"@14V6H@"V$"@16H@="D1;P1$N;M1$:34:M1$"@17V6H@"V$"@16H@= "E1;N1$:34:N1$"@R15C@":19:31051:P:55)mP19:38:110:P:55)n143:145:C18:K$(C)C$(C,B1):C:B3T$K$(1):K$(1)K$(2):K$(2)T$*oX(2):E1D1N((iB3E1D1N:RE1D1NY(jE1(E1)D1A(9)1:NA(9)1:E1D1N:106(kQ1$"@I8V25HR15C@"K1$"@25HD@"F1$" of @D25H@"G1$" @L10C@":"@4V1HL15C@IF@6H10C@"V$O1$N"@16H@= "D1"@D1H15C@THEN@D6H10C@"V$O1$N;P1$N"="D1;P1$N;Q1$)lK$(6)"@L10C@":M1$"@I14V25HR15C@"K$(5)"@D25H@axiom of "K$(7)"@L10C@":N1$"@I17V25HR15C@"J2$" @D25H@"F1$" @L10C@" (hP14:K1$K$(1):38:V$((1)2665):NA(9)1:D1A(9)1:E1D1N:RA(2):O1$K$(4):P1$K$(3):RO1$K$(3):P1$K$(4):K1$K$@D2H@member of a given "Q$".@11V2H@"HH$" by "L$&f"@13V2H@Adding the additive inverse of a@D2H@real number contained in an@D2H@"Q$" to both sides of the@D2H@"Q$".":43'g143:C18:K$(C)C$(C,B1):C:K1$K$(2):L1$"@I11V25HR15C@Axiom of @D25H@"s@D1H@always gives an "E1$" "Q$":@9V2H@"HH$" by "J1$"@11V2H@Substituting for either member of a@D2H@given "O$" an "Q$"@D2H@"E1$" to that member.":43/&eX9:L262:Y36:H40:142:Y84:H56:142:"@5V2H@"HH$" by "J$"@2D2H@Adding the same real number to each" of "G1$"@7V2H@For all real numbers A and B,@2D14H@A-B = A+(-B)"$c"@12V2H@When "I1$", you can instead@D2H@replace the number that you are@D2H@"I1$" by its opposite and then@D2H@add.":43%dX9:L262:Y68:H48:142:"@5V1H@Each of the following "H$"@5V2H@A"(J$,7)" "F1$" of "G1$"@7V2H@If A, B, and C are real numbers such@D2H@that A=B then@2D8H@A+C = B+C and C+A = C+B@13V2H@For all real numbers A and B,@2D15H@A-B = A+(-B)":43B$bX9:L262:Y36:H48:142:Y92:H40:142:"@5V2H@S"(L$,10)" "F1$P$"two equal numbers, the differences are"P$"equal. This is the ";:48:D$F1$" of "G1$:P$;:48:". So, we can say";"_P$"that if the same number is "H1$P$"from both numbers of an "G1$", the"P$G1$" is retained.":43"`P97,98,100,101,55#a158:"to the sum.":43!]"@5V1H@For example, in the "O$" X+3=5 we"P$"can add the opposite of 3 to both"P$"sides of the "O$" and get"P$"X+3+(-3)=5+(-3). After doing the"P$J$", we find that X+0=2, or X=2.":43v"^D$L$:"@5V1H@If the same number is "H1$" from" "O$", you usually try to"P$"change, or ";:153!\154:P$"The ";:48:P$"says that if the same number is added"P$"to equal numbers, the sums are equal."P$"To undo the "J$" of a given number"P$"to another number, add the opposite of"P$"the given number DX19:"@15V33H@ @2D25H@"9)"@D24H@"12)"@12V20H@ ":P:31051:55YYM90,96,103,109oZP91,92,93,94,55/ [D$E1$:"@5V1H@E"(O$,7)"s that have the same "S$P$"set and domain are called ";:48:D$"transform":P$O$"s over that domain. To solve"P$"an1C1B(R)D1:"@12V4H@"B(1)"@2DB@"B(2)"@2DB@"B(3)"@12V10H@"C1"X + "D1" = "E1"@12V23H@Which number in@D23H@the domain is@D23H@the root of the@D23H@"O$"?":B%1:C%9:6:"@15V33H@"A%:A%B(R)ĺ"@17V25H@Correct":88W"@17V25H@Incorrect@D24H@Answer is "B(R)m the set @D2H@at the left which makes the "O$"@D2H@below true."25)T14,39265,39:"@I10V2H@DOMAIN@11H@"OO$"@23H@"SS$" SET":P19:38:C1A(8)1:D1A(8)1:B(1)A(8)1UB(2)A(8)1:B(3)A(8)1:B(3)B(1)B(3)B(2)B(2)B(1)ī85VRA(3)1:E1$" true "X$:D1$" yes.":TF1C1$" false "X$:D1$" no."R"@15V2H@The result is a"C1$".@D2H@The answer is"D1$:19:P:31051:55ESX9:L262:Y36:H32:142:H80:Y76:X9:L49:142:X65:L84:142:X157:L112:142:"@I5V2H@Choose the one number fro"A1$"="B1$:E$(2)Y$"*"Z$"-"A1$"="B1$"@2D@":E$(3)(C1D1)"-"A1$"="B1$:"@5V2H@Is "D1" a root of "E$(1)" ?":34OE$(4)(C1D1E1)"="B1$:9:C14:"@ID2H@"S$(C)"@28HI@"E$(C);:I1900:I:C2ĺ"@2U2H@"S$(C):81P"@2H@"S$(C);=QC:C:142:H16:142:S$(1)"<1> Write the "Q$:S$(3)"<3> Do the "M$:S$(4)"<4> Do the "L$:152MS$(2)"<2> Replace the "U$"@D6H@X by "(D1)"@U@":E1A(98)1:F1C1D1E1:TF1:RA(2):RF1A(18)1|NY$(C1):Z$(D1):A1$(E1):B1$(F1):E$(1)Y$"X-o check whether a number is a root@D2H@of an "O$":@2D2H@<1> Replace the "U$" in the@D6H@"O$" by the number.@2D2H@<2> Perform the indicated operations"K"@12V6H@to see whether the "X$" is@D6H@then true.":19:31051:55\LY36:X9:L262:H104:142:H801":W$(2)" 3 2":W$(3)" 3":W$(4)" ":".@L138C5V3H@"W$(1)I34:I$"":C24:"@5V3H@"W$(C):34:"@R15C7V15H@"C1"@21H@"C1:35:I$(C1)" "I$:"@138CL5V27H@"I$:C:I1500:I:"@R15C7V15H@ @21H@ ":43J38:X9:L262:H80:Y36:142:"@5V2H@TH@ @DFI@ @I@":43G"@I@":U13:S13:37:"@0KR15C8V5H@DOMAIN@28H@"SS$"@D30H@SET@7V16H@+5 = +5@I7V15H@ @21H@ @11V15H@X+5 = X+5"AHD$"identity":P$"If every element of the domain makes"P$"the "O$" true, the "O$" is"P$"called an ";:48:W$(1)"3 2 V21H@ ":W$"4 3 2 1":Z9:A10:B13174:ZZ2:"@2H10C5VL@"(B13)2)(W$,Z):B13İ34:B1DA1A11:"@15CR7V21H@"A1:A12ĺ"@I8V23H@True @I5V32HL10C@2@R15C@":70E"@I8V23H@False@I@"F35:B1:"@5V17H@ @L5V13H@ @R7V21H@4@DFI@False":34:"@I7V21embers may be used to replace"P$" a "U$".":43BD$"conditional "O$:"@13V1H@If only some elements of the domain"P$"make an "O$" true, the "O$" is"P$"called a ";:48:".@I@":U11:S19:37:"@8V8H@DOMAIN@30H@"SS$"@D32H@SET@22H7V@+5 = 7@I@":34[C"@743@D$"solution set":"@5V2H@The set consisting of all the roots"P$" of an "O$" is its ";:48:D$"truth set":P$" or ";:48:". ";:D$"To solve an "O$:48:D$"domain".AP$" means to find its "S$" set. The"P$;:48:" of a "U$" is the set"P$" whose mside and a right side.":"@L15V8H@LEFT@17H@=@20H@RIGHT@R17V14H@"O$:43?D$"root":"@5V1H@An "O$" may either be true (2+3=5)"P$"or false (2+3=4). When a "U$P$"replaces a number in an "O$P$"(X+3=5), the number it replaces is the"P$;:48:"of the "O$".":M),(36320M)1:409P55B:38:B60,89,126,5000,6000,7000H;[<M61,74,76,83t=P62,63,64,66,71,55F>D$O$:"@5V1H@An ";:48:" is a sentence which uses"P$"the symbol = to say that two algebraic"P$Q$"s are equal. Each "O$P$"has a left B3İ4:5:31153,B:(4)"RUNAM2.3-2"7P0:M0:38:"@12H5V@MODE SELECTION@2D4H@<1> DISCUSSION@2D4H@<2> RULE@2D4H@<3> EXAMPLE@2D4H@<4> SAMPLE PROBLEM@2D4H@<0> RETURN TO CONTENTS@6D5HI@WHICH ONE (0-4) ??@I@":B%0:C%4:6:A%İ41:508P1:MA%:(36320"NC$"@D10H@<4> USING SEVERAL@D14H@"HC$"S@D10H@<5> FUNCTIONS@D10H@<6> EQUATIONS TEST"5"@2D10H@<0> RETURN TO ALGEBRA MENU@3DI11H@WHICH ONE (0-6) ??@I@":B%0:C%6:6:BA%:BāC3629936315:C,0:C:4:5:30976:(4)"RUN ALGEBRA 2"$6D$G$(B):39:40:42,15420,15413,14820,142:43,14250,14843,154:31,13031,14228,139:32,13032,14235,139:4:19:"<0>":59,3259,15960,15960,32T4"@20H5V@CONTENTS@2D10H@<1> EQUATIONS@D10H@<2> "HC$" BY@D14H@"JC$" AND "LC$"@D10H@<3> "HC$" BY@D14H@"MC$" AND @D14H@38:B3:110:P:4:5:(4)"RUNAM2.4-2"2B0:P0:M0:38:D$"":39:X18:Y3811816:H12:L27:142:(Y10)8:"@3H@<"(Y22)16">":Y118ēX13,Y12X13,Y16:X10,Y13X13,Y16:X14,Y12X14,Y16X17,Y133Y:21,15414,14821,14242,14249,148@" ,A(16384):10:A12844C -16368,0:A136İ41:PP1:57\ .A149İ40:PP1:57d /44 0(36)2):X(36)73:Y(37)82:D$" ";:L(D$)76:H11:142:'135339:G$(6):A(X)((1)X):C16:G$(C):C:(36309)2Ĺ36301,0:P14:B0:G1:KK1:J3:33:4 &"@1H1V@C"B" P"P"@D1H@M"M:^ '"@I2V7H@"31):24(D$)2:D$"@I@":m (31051:42x )30976 *"@21V1HI@"38)"@D1H@"38)"@I@": +"@5H21VI@PRESS "(1)" TO VIEW THE NEXT PAGE.@3H22V@PRESS "(2)" TO VIEW THE PREVIOUS PAGE.@I(A);:256F%,A:F%F%D:F%C%32* 24_ A$"":C0F%:A$A$((256C)):C:F(A$):E0: !G1:I1J:K:"@D@"O):I: "Q400:36 #Q900 $I1Q:I: %K3:G9:O36:J2:33:14,63265,63:KS1:OU:G6:J5:"@10C@":33:"@15C@":OO2:G%:256C,32:C:16368,0< E%F%:"@I@"((256F%))"@I@";N 10:A12825 16368,0:E%F%:((256F%));:A155G%1: A14132 A136F%F%F%D A149F%C%F%F%D" AA128:((A47)(A58))((A39)(A91)(E1))(A45)ĖE%F%:5İ4:5:(4)"RUNALGEBRA 2", A205Bı5 17? :16I 151b "@R15C0K@"::41:55 "@22V1HI@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 10:A12820 16368,9:A16020 "@22V1H@"36)"@I@": D%:E%:C%):F%0:C0C524576:30719:63900(149:49.=0:1002:G::X10:A1286r16368,0:A155A%0A%A176:A%B%A%C%6  A(16384):A155(36309)ı 16368,0:"@40X40YN@" A(16384):A12812 16368,0:A15                            :"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": PT2.2":"reciprocal":23C1İ24:CC1:C$(1)(C):106;31053E$R$" inverse":"@9V1H@or ";:82:", of the@11V1H@other. Since there is no number which,@2D1H@when multiplied by 0, gives 1, zero@15V1H@has no "S$".":7731002:(222)255ĦF40:::RULES", "RECIPROCAL OF REAL NUMBERS","MULTIPLICATION TEST" 3T$"property":F$"AM2.PROGRESS":Q$"substitution":O$"distributive":P$"associative":I$"expression":H$O$" "T$:U$"multiply":R$"multiplicative":V$"Mult. ""prop. of ":W$"commutative":S$XL,YXL,YHX1,YHX1,Y:XL1,YXL1,YH:15:6:(4)"OPEN"F$:(4)"READ"F$:E15:B(E):E:B(2)5:W3B(2)1:B(3)3B(3)61(4)"CLOSE":35339:<2"CLOSURE; EQUALITY","COMMUTATIVE; ASSOCIATIVE","DISTRIBUTIVE PROPERTY","MULTIPLICATION C1A%ĺ"@18V7H@Correct":A(2)A(2)1:154H0"@18V3H@No. Answer is ";C1P020{0J13:R17:35:J13:Q3:35:P:31051:0E14:K11:135:0E14:K11:140:0E14:K11:93:31051:0E14:K11:103:31051:0E14:K11:127:11X,YHX,YC@"&/C13ĺ"@2H15VL133C@"T"@R15C@"/20:J13:Q22:O7:R17:35:"@12V21H@Which axiom is@D21H@represented?@D21H@<1> Substitution@D21H@<2> Commutative@D21H@<3> Associative@I18V21H@ WHICH (1-3)? @I@":147,143265,143(0B%1:C%3:7:"@35H18V@"A%:D21H@the question mark@D21H@makes the@21HD@sentence true? @I@ @I@":B%0:C%9:7:"@I36H15V@"A%"@I@":A%Tĺ"@26H17V@Correct":149."@24H17V@Incorrect.@D23H@Answer was "T:20:"@15V@":K12:"@D22H@"15):K:O7:155/C11C12ĺ"@8H15VL133C@"T"@R15-49:TU142`-"@12V2HL15C@(@133C@"T"*"U"@15C@)@133C@*"V"=":C1((1)3)1:C1146,144,145-"@15V2H15C@(@133C@"U"*?@15C@)@133C@*"V:147-"@15V2H@?*@15C@(@133C@"U"*"V"@15C@)@133C@":147-TTU:"@15V8H@?*"V."@R15C21H12V@What value for@tence below @7V2H@true. Then select from the list to @8V2H@the right the axiom represented when@9V2H@the replacement is made."12):"@I@",14,7914,39266,39266,79:X10:L260:H46:Y36:161:X10:Y92:L127:H64:161:X143:Y92:L127:161:P14:72B%1:C%6:7:"@36H17V@"A%:A%C1ĺ"@18V22H@That is correct":A(1)A(1)1:139m+"@18V22H@No. Answer is "C1+20:"@36H17V@ @20H18V@"18):J11:O9:Q3:R16:35:P:31051:,"@I5V2H@Give a replacement for the question @6V2H@mark which makes the sentement to the left."14)"@I@"*"@20H10V@<1> Reflexive@D20H@<2> Symmetric@D20H@<3> Transitive@D20H@<4> Closure@D20H@<5> Substitution@D20H@<6> False-none@2D20HI@ WHICH (1-6) ? @I@ @I@ @I@":140,135266,135266,143150,143:P1E1:72:D$" sum ":55M+:159:160:E15:36298E,A(E):E:31051:5:6:(4)"RUNALGEBRA 2"*X10:Y76:H80:L120:161:X136:L134:161:X10:Y36:H30:L260:161:14,39266,39266,63:"@5V2HI@Choose from the list on the right @D2H@the axiom which best describes the @D2H@staof 1@DB@$ *"G1"? @D24H@"F1:G0:F1:C%2:D%17:E%33:25:ICĺ"@18V29H@Correct":A(5)A(5)1:133("@18V28H@Incorrect":L11000:O5:R36:Q3:J9:"@I@":35:"@7V25H@"10)"@I@":128(20:"@18V28H@"9):P:C)E15:A(E)0:E:36309,1:156:157:15818H@"F1:E1L1:E:"@8V28H@1@9V25H@= $*"F1" *"C"@D28H@"F1^'B5ĺ"@9V18H@="G1F1"@I@":'0:X26:Y10:S0:67:X32:Y10:S1:67:3:E1L1:E:"@12V16H@=1*"C" ="C"@I@":a(P1E1:72:D8:24:F1C2:24:CC2:G1F1C:"@15V2H@What is the value 116:H40:161:14,39265,39:"@I@":R36:O8:Q3:J6:35:"@5V2H@Following the example below,@D2H@simplify the "I$" in the lower@D2H@box.@I@":L12:F19:G163:C7:128:131<'"@I8V11H@1@D11H@$*"G1"@D11H@"F1:B5āE1L1:E:"@8V18H@1@D16H@= $("F1"*"C")@D"s of 5 and 3 are $ and $@U32H@1@38H@1@2D32H@5@38H@3"%}"@14V1H@The products of the "S$"s are@2D2H@1 1@6F@1@15F@1@4F@1 1@D2H@$*$ or $$. We see that $$$ = $*$.@D2H@5 3@5F@15@26H@5*3 5 3":77%~E19:K10:127:31051:87&X9:L262:Y36:H72:161:Y${"@15V34H@1@16V1H@'The "S$" of three-fourths$$$ 4@D1H@equals four thirds' is written 3 =$@D1H@as:@34H@$ 3@4BD@4":775%|"@5V1H@The product of 5 and 3 is 5*3, or 15.@D1H@The "S$" of the product is@D6H@1@6F@1@D5H@$$$ or $$@D5H@5*3@4F@15@11V1H@The "S$24:S1:67:Y17:X18:S0:67:77#z"@5V1H@0.1 is the "S$" of 10 and 10 is@D1H@the "S$" of 0.1 because@D1H@0.1*10 = 1.@2D1H@-1 is its own "S$" because@D1H@-1*-1 = 1.@2D1H@0 has no "S$" because the@D1H@product of 0 and any real number is 0,@D1H@not 1."z:77"wP120,122,124,87"xD7:24:CC2:"@6V1H@Since "C"*$ = 1, $ is the "S$" of@9V2H@"C".@5V9H@1@6F@1@7V9H@"C"@16H@"C:24:CC2:"@12V1H@-"C" and -$ are "S$"s of each@16V1H@other because -"C"* -$ = 1"!#y"@11V9H@1@2DB@"C"@15V21H@1@2DB@"C:Y17:Xuct@8V2H@The "S$" of a product of real@D2H@numbers, each different from zero,@D2H@is the product of the "S$"s of@D2H@the numbers; that is, for all real""v"@12V2H@numbers A and B such that A<>0 and@D2H@B<>0,@15V3H@1 1 1@D2H@$$$ = $*$@17V2H@A*B A B"2:161:"@5V2H@Axiom of "S$"s@7V2H@For every real number A,except zero,@D2H@there is a unique real number@10V2H@1@11V2H@$ such that@12V2H@A@16V2H@A*$ = 1 and $*A = 1":"@15V4H@1@14H@1@17V4H@A@14H@A":77!u161:"@5V2H@P"(T$,7)" of the "S$" of a@D2H@prod:77r"@5V1H@For non-zero numbers, the "S$"@7V1H@of the product of two numbers is equal@9V1H@to the product of their "S$".@11V1H@For example, 1 1 1 1@12V16H@$$$ = $*$ = $$@13V16H@3*5 3 5 15":77sP116,117,87 tX9:L262:Y36:H11238:Y120:H16:L7:161:X245:161:238,128252,128:3:77q"@5V1H@NOTE: If a number is positive, its@7V1H@"S$" is positive. If a number is@9V1H@negative, its "S$" is negative.@11V1H@Also, the "S$" of the "S$"@13V1H@of the number is the number itself."",";:169o"@5V1H@Every real number except zero has a@7V1H@"S$". The symbol 1@DB@$@DB@A":H32:X170:Y52:L16:5:161:3:"@11V1H@represents the "S$" for A@16V10H@4 is the "S$" of $@UB@1@2DB@4":p"@L14V3H138K@ @F@ @17V3H@ @F@ @R16V34H@ @0K@":5:X:D%18:E%16:25:L$"":E110:((B$,E,1)" ")L$L$(B$,E,1)pkE:B$L$:36:20:R36:O3:Q3:J12:35:P:lM109,115,119,126mP110,111,113,114,87 nE$S$:"@5V1H@When the product of two numbers is 1,@7V1H@one number is called the ";:82:R$"@D2H@properties for 1, 0, -1, and@D2H@opposites.@I16V2H@What is the value of the above@D2H@expression?":14,39265,39:P1E1:72:D2:24:D3:167iC$(1)(((1)26)65)@jZ23:24:C$(Z)(C1):Z:"@11V3H@"C$(1)"("C$(2)"+"C$(3)")":G1:F1:C%10"@I21H@"N$(E);:K11000:K:"@I21H@"N$(E):E:102We"@28H18V@ Correct":A(3)A(3)1df20:P:gX9:L262:H40:Y36:161:H32:Y84:161:Y124:161:R36:O4:Q3:J6:"@I@":35h"@5V2H@Simplify the expression below by@D2H@using the appropriate "K$(H1)J$:M$(2)"="(F1)J$"+"(G1)J$"+"(H1)J$"+"(F1)K$"+"(G1)K$:M$(3)"=("(F1)"+"(G1)"+"(H1)")"J$"+("(F1)"+"(G1)")"K$:M$(4)"="(I1)J$"+"(J1)K$2dN$(1)O$:N$(3)O$:N$(2)P$:N$(4)Q$:"@10V@";:E14:"@3H@"M$(E)D%19:E%18:F1:G1:25:L$"":E19:((B$,E,1)" ")L$L$(B$,E,1)aE:(L$(I1)J$"+"(J1)K$)(L$(J1)K$"+"(I1)J$)ī101bK11ĺ"@28H18V@Incorrect@20H9V@="I1;J$"+";J1;K$:102cM$(1)"="(F1)J$"+"(F1)K$"+"(G1)J$"+"(G1)D2H@the one above? "k^14,39265,39:P1E1:72:R31:O5:Q3:J10:35:"@28H18V@"10):J$(((1)6)84)_K$(((1)6)84):K$J$ī95F`D8:24:F1C1:24:G1C1:24:H1C1:I1F1G1H1:J1F1G1:"@9V3H@"F1"("J$"+"K$")+"G1"("J$"+"K$")+"H1;J$:C%9:z!#     Ͳ7 Ͳ+Ͳ&Ͳ) ͲӠ!ŠͲϠ ӳͲ4Ͳ5Ͳ)"Բ.:39:83:P:P14:119:P:4:5:36309,1:(4)"RUNALGEBRA 2"},Q,RHQ,RQL,RQL,RHQ1,RHQ1,R:QL1,RQL1,RH:,"@4V19HL10K@ @5V19H@ @15K5V20H@ @0K7VI1H@"19)"@R7V7H@DOMAIN@29H@SOLUTION@D31H@SET@I@":-F$"AM2.PROGRESS":D$"@2r is "P$" less than?@I26H18V@"P$"<":E%29:D%19:26:FB1ĺ"@18V2H@Incorrect."+}FB1ĺ"@18V2H@Correct.":(36309)1Ĺ36301,(36301)1:128+~36302,(36302)1:128+"@13H18V@"P$">"A1;J1$P$"<"B110)+19:31051:+31051:36309,1=,P14t":128*z"@18V19H@Correct":19:J5:N23:G15:K4:36:"@I14V2H@What number is "P$" greater than?@I18V15H@ <"P$:C%3:D%19:E%14:D1:26:A1(SU)(1):B1US:Z1TB1:B1A1:A1T*{FA1ĺ"@16V3H@Incorrect.":127O+|"@18V21H@"J1$"@I16V2H@What numbe@set."32)"@I@":P$(B(1)):SA(8)2:UA(8)2:X$">":Z1:J1$" or ":A(2)0X$"<":Z0:J1$" and "*y"@11V3H@!"P$"+"S"!"X$;U"@3D3H@Is this "I$" to@D3H@<0> "Z$"@D3H@<1> "C1$"@2D3HI@WHICH (0-1) ?@I@":B%0:C%1:6:"@18V17H@"A%:A%Zĺ"@18V19H@IncorrecUS))4,R:3/(u3:37:H1$:19:31051:P:51F(vP19:119:P:51(w39:Q9:L262:H40:R36:131:H72:R84:131:14,39265,39:"@5V2HI@Determine if the in"E$" below is@D2H@a "Z$" or "C1$", and @D2H@then find the range of its solution ";e)x"@D2H263,147:R139:X$"<"ĺ"@17V"20(SU)"H@)@"20US"H@(@"17(SU)"H@-"SU"@"21US"H@"US:1:3(7(20(SU))),R1(7(20US)),R:117 (t"@17V2H@<@37H@>@"20(SU)"H@(@"20US"H@)@"21(SU)"H@-"SU"@"18US"H@"US:1:17,R1(7(20(SU))),R:261,R(7(20={ÇH3%,J0%,H4%,K1%,H5%,K2%,H6%,K3%,H7%,K4%,H8%,K5%,H9%,K6%C7DZ[1002:(222)255Ħ0::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": EDU-WARE":%,C8%,F9%,C9%,F0%,C0%,G1%&lÇD1%,G2%rnà 90,104,81 ,96,77 ,88,79 ,67,82 ,60, 90,53,106,49,138,45,156,45,162,51oà 168,59,170,68,169,76,161,82,134,88,130,96,123,100xÇG3%,I0%,G4%,J1%,G5%,J2%,G6%,J3%,G7%,J4%,G8%,J5%,G9%,J6%,G0%,J7%,H1%,J8%,H2%,J9%,179,145,156,150,159,168 odÇA1%,D2%,A2%,D3%,A3%,D4%,A4%,D5%,A5%,D6%,A6%,D7%,A7%,D8%,A8%,D9%,A9%,D0%,A0%,E1%gÇB1%,E2%,B2%,E3%,B3%,E4%,B4%,E5%,B5%,E6%,B6%,E7%,B7%,E8%,B8%,E9%,B9%,E0%,B0%,F1%iÇC1%,F2%,C2%,F3%,C3%,F4%,C4%,F5%,C5%,F6%,C6%,F7%,C7%,F83,0,6,1,0,2 SÁI06:C%(I):I{ Uà 68,160, 75,147, 77,134, 75,125,68,108,64,97,64,74, 72,53, 80,46, 96,38,137,37,166,42 Xà 175,45,181,52,185,56,187,64,186,72,184,82,182,88,191,104,193,107,190,112Zà 183,114,184,120,186,124,184,128,185,132,183,141R";:I:"E":I223:I:1:1;:40:1:I:"Z";:I239:"R";:I:"C" '"@0V2H133C0KL@EDU-WARE@R15C1V@TM@I@":I35:I:28:11):I:"@3V28H@ALGEBRA 2@I@": Pà 91,24,64,104,126,64,105,40,77,88,56 QÇN1%,N2%,N3%,N4%,N5%,N6%,N7%,N8%,N9%,N0%,M1% RÃ2,N1%,N3%I3N5%,N3%:I3> /N7%,N8%N9%,N0%:N9%,N8%N7%,N0%:W 35339:10000:50000` 200k K12 I05:C%(I):300:I,K:C%(6):300 I3629936351:I,0:I:I12500:I:26:80:(4)"RUNALGEBRA 2" C ':"@0V0HG10C0K@Q";:I239:"G3%,I0%G4%,J1%G5%,J2%G6%,J3%G7%,J4%G8%,J5%G9%,J6%G0%,J7%H1%,J8%H2%,J9% H3%,J0%H4%,K1%H5%,K2%H6%,K3%H7%,K4%H8%,K5%H9%,K6%G3%,I0% + ,I302:I3%I3:N1%,N2%N1%I3%,N3%N1%,N4%N1%I3%,N3%N1%,N2%:N5%,N3%N1%,N3%I3%N6%,N3%%A3%,D4%A4%,D5%A5%,D6%A6%,D7%A7%,D8%A8%,D9%A9%,D0%A0%,E1% B1%,E2%B2%,E3%B3%,E4%B4%,E5%B5%,E6%B6%,E7%B7%,E8%B8%,E9%B9%,E0%B0%,F1% C1%,F2%C2%,F3%C3%,F4%C4%,F5%C5%,F6%C6%,F7%C7%,F8%C8%,F9%C9%,F0%C0%,G1%D1%,G2%A1%,D2%R 3205,255:1012,0::::255:ZZ(0)::63900j24576:27903:10:8:"CRACKED BY THE NUT CRACKER"(4)"BLOADEWS3"3 COPYRIGHT 1981 EDU-WARE SERVICES, INC. 10000:1002:P:1002:51,0::A A1%,D2%A2%,D3     @@``@|@@@`x6@|`0@ @0 0|| @L << @ 0 0~ff@@@@@@ |@@@`@xp@@@?p6@0 ~@x@|p||||0|p0@| |||| | p| | | | |@|@ @|||@@@@@@@|@ |@@`@x@@@`pF@00@@ @@ < @@ < @ < @ 0 0||||||||||@@@@@@@@@p@@~pp@X@?pF0 p`@@<@ 0 @@@| @ @ 0< @ ``````````@@@@@@ 0<00@@@@p@@?``6@00p`@@ p @ 0 @@ 0 @ @ xxxxxxxxxx@@@@@@@@@x`x@@f|p||@p6000@@<@ 0 @@@@| @ @ 0<0@ xxxxxxxxxx@@@@@@ 0<00@@@@`@?``@00x@x|p|||p~||0||p|||||| p |||||| || |0``````````@@@@@@@0```|@`|||xpp@@|00@l@<| 0 @0|@| @L LL@ L0@0 ||||||||||@@@@@@ L@p@@@@@@@`~@00`p@|||@~pp0p|@|p|||p p p|p|p| || |@@@@@@@@@0`x@@`~|x`ppp6@|00`@l@0| 0 0|@|| @L L L@ L0@0 ~f>@@@@@@ L@|@@@@`~@@@@@@@@~~@@|`~ppppp|F@0@`||~L@@@ |@pp@ ||| || |@< LL | |p@ @@@ @~ffff>>fff@@@@ 0 0`0???6???330>>0<0~ <1<??333>>?3?~6 6?0<33080<0 1 1 6~6<~< <?6?0<<3>8<<0x? <63???16???330>>33?`?y$%GΩϩ  %%%GΩϩ  %%$$'к`&$%GΩϩ  %%%GΩϩ  %%$$м`  p 4114440 4004438 3883339 399322 2222222 2222217 1771117 1771117 177136 3663336 36          D1H@":K$"inequalities":I$"equivalent":E$"equality":L$"solution set":M$"Transformation":R$"distributive":S$"property":J$"junction":Z$"con"J$:C1$"dis"J$:P$"variable"+.D1$"statement":A1$"sentence":N$"expression":E1$" is "I$" to ":F1$" " "  c6`<`@x   p"p>1, >0>"><"">"","">> >"<,<"*" ">>*1,<Acx>``~x?> ><,<"*" "> >>"82 "" ""  >""""""2" "-"0""&2" *"", "*> 62*>~Ac>|>" ~>`0|xw>0"&2" *"", "*0>">"<>, >& > "" >>""2" "2"* ""-0"<&2>"*""&2""**(,~Ac>|" ~>x` xxxx0<&2>"*""&2""*?">"8 >* <0>"">**""""!0r &2"& *&"&2&""*?6 ><Ac>|<<>x` pxxx0 &2"& *&"&2&""*0>">p"p>> 2 "" >"""" *&""""""!"0>,,<>"""">*  6Ac>|>"" >`0`x>0",,<>""""> >>" & " " ""  >"""6&""""""!"" 0  6<& * Acx"" >``@x0   p""6 >>8>0>>><""""<>""!"">>> 8 ( >6 `>`@x>> 8 D "LU :F`F`$L"%e%`$e($`%80%`$80$`'$L:{|0L_`F)׭F F)L i)`) qp`<) Ji L? Ji L? d d Z P F < 2 (  2222P2(2F2 2<(LI GIG`RH`GH`LH`E yIy`OF) ELSSzI zU`TR`P EI E`NLUX F{LY F|LTIF eiLȱ|ȱ{ U`FFB DLF .L`% ImE8Ie$e΅ϩeeυ GIy) e$) =}м`0:)F FmFeF`@D`C r)׍} ~  `K r)׍ `H F($LV F%@ JTL J UEPWRz6F GFy 7 ?` !!""## ( !(!"("#(# P !P!"P"#P# ((ˍ) J ?`DLk@D` LH5Ω " .Ωz % $L.jjΩ .@0~~@@|@>f@@@@ 0 0@pp @@@@@@@p@0`x@@@0@@@pp|@|x0`@ @0| @0|| @L << @ 00 0~fff>>fff@@@@@@ |@@@@@@@pp@0 x@pp|||p0pp@0@ |p|| p pp | p p |@p@ @|||@@@@@@@p@ ||ENTS":3-C%,D%:C%0520.F%,G%:F%045</50:460C%,D%C%25,D%C%25,D%12C%,D%12C%,D%:C%1,D%C%1,D%12:C%26,D%C%26,D%12:1C%,D%C%4,D%4:C%,D%C%4,D%4:2C%,D%F%,G%:D%G%ēC%1,D%F%1,G%3C%F%:D%G%:4:"@20H@"19)"":Q:"@1V26H@ALGEBRA@D25H@MODULE #2@D20H@VER 1.0 01 Apr 82@I@","@2V8H@START@4D8H@MENU@15H@<0>@2H@<9>":"@20H17V@<9> RESET MENU@D20H@<0> STOP":"@I20V@";:G13:"@20H@"19)"":G:"@21V21H@WHICH (0-9) ?? @I@":"@36H21V@ ":"@26H6V@CONT)"WRITEAM2.PROGRESS":I15:B(I):I:(4)"CLOSE":(4)"LOCKAM2.PROGRESS":35339:O+I15:A$(I):I:275,4275,188:133,4133,188:134,4134,188:41,5637,5241,48:94,5698,5294,48:3,4276,4276,1883,1883,4:4,44,188:140,7272,7:"@I1V@";:Q13P1,R:PL1,RPL1,RH:(4:5:P(36310):(4)"OPENAM2.PROGRESS":(4)"READAM2.PROGRESS":I15:B(I):I:(4)"CLOSE":35339:)P(36310):B(P)5:T3B(P)1:P5B(P1)3B(P1)6R*4:5:(4)"UNLOCKAM2.PROGRESS":(4)"OPENAM2.PROGRESS":(4have passed@D3H@";:O5ī61]""the post test. ";:T2ĺ"ARROW SHOWS AREA@D3H@OF WEAKNESS."$"@I@":10:31051:41:4:5:(36309)37,38%36309,0:(4)"RUN AM2."O&I3629936315:I,0:I:(4)"RUN ALGEBRA 2"'P,RHP,RPL,RPL,RHP1,RH:"@I8V@";:I15:D(I)(36298I):C$(I)B$(((36310)),I):E(I)4D(I):C$(I)""32"@1D3H@"C$(I)"@22H@"D(I)"@5F@"E(I);:E(I)1ĺ"@I4F138K15C@<--@0KI@";:TT1 I:"@I@":TT1:T33,33,63,63,63,63,63!21,127258,127:"@16V3H@Congratulations. You )B$(3,1):B$(5,4)"INEQUALITIES":B$(5,5)""P17:R44:L246:H112:39:L203:H64:R60:39:S137:S13717841:S,60S,124S1,124S1,60:S:"@I6V3H@"34)"@6V8H@CONCEPT@20H@RIGHT WRONG ":21,47258,47:J5:E13:F33:B7:14:F4:B17:E3:J32:14QT0TIONS":B$(3,2)"TRANS. BY +, -":B$(3,3)"TRANS. BY *, /":B$(3,4)"SEVERAL TRANS.":B$(3,5)"FUNCTIONS"+B$(4,1)"AXIOMS OF ORDER":B$(4,2)"TRANS. INEQUAL.":B$(4,3)"COMBINING INEQ.":B$(4,4)"":B$(4,5)"":B$(5,1)"ADDITION":B$(5,2)"MULTIPLICATION":B$(5,3O(36310):35339:B$(1,1)"CLOSURE; EQUAL.":B$(1,2)"COMM. AND ASSOC.":B$(1,3)"IDENTITY, DISP.":B$(1,4)"OPPOSITE OF REAL":B$(1,5)""eB$(2,1)B$(1,1):B$(2,2)B$(1,2):B$(2,3)"DISTRIBUTIVE":B$(2,4)"RULES FOR MULT.":B$(2,5)"RECIPROCALS":B$(3,1)"EQUA3,164277,164:3,28277,28:1:"@I@":G12:"@7H@"32)"":G:22:G12:"@1H@"38)"":G:49,7272,7:7,167272,167 64:24(A$(A%))2:"@1V@"A$(A%)"@I@":36310,A%:4:5:A%5Ĺ36309,2:ZZ15:36298ZZ,0:ZZ:(4)"RUNAM2.1" (4)"RUN AM2."A%N:A$(I);:28:(C(I)100M);:NN2:31:"%":I:24: A%9B(1)6:I25:B(I)3:I:I15:C%A(I,0):D%A(I,1):B(I):48:I:4:5:42:18 A%518t "@I@"A$(A%)"@I@"::3:3,1883,4277,4277,1882,1882,4:278,4278,188:45,445,2846,2846,4:V@ ":E%0:B%9:6:"@36H21V@"A%:A%21b ::4:5:M0:I14:C(I)(36320I):MMC(I):I:M0Ā6 :L124:L:40):L::N521:N:5:32):N:A$(1)"DISCUSSION":A$(2)"RULE":A$(3)"EXAMPLE":A$(4)"SAMPLE PROBLEM":16:6:"MODE USAGE":N9:I14:9:24H@OF REAL NUMBERS":A$(3)"@9H14V@<3>@20H2U @<3> SOLVING @D24H@EQUATIONS " A$(4)"@9H17V@<4>@20H3U@<4> SOLVING @D24H@INEQUALITIES":A$(5)"@9H20V@<5>@20H4U@<5> POST TEST ":43:K05:C%A(K,0):D%A(K,1):B(K):48:K& "@36H2116011! "@22V1H@"36)"@I@":C B1:D1E:F:"@D@"J):D:` 40:A(5,1):(36309)26W :K05:C01:A(K,C):C,K:B(0)3:31153,0:A$(1)"@5H10V@<1>@20H2U@<1> ADDITION OF @D24H@REAL NUMBERS ":A$(2)"@12H10V@<2>@20H@<2> MULTIPLICATION @D24576:30719:63900&3:15,;0:1002:E::^A(16384):A1286x16368,0:A155A%0A%A176:A%0A%B%6  "@22V1HI@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 A(16384):A12811 16368,0:A       ."F/"@15V5H@'1>0 or 1<0' is true.@2D5H@'1>2 or 1<0' is false.":43^/1002:(222)255Ħ/0::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": AM2.4-2": ":B1$" "F1$:A9$"absolute value":I.B%0:C%4:6:A%İ41:50Q.54/"@5V1H@A "A1$" formed by joining two"D$A1$"s by the word 'or' is called"D$"a ";:C$C1$:48:". For a "C1$" to"D$"be true, at least one of the joined"D$D1$"s must be trueMBERS":A$(2)"MULTIPLICATION OF REAL NUMBERS":A$(3)"SOLVING EQUATIONS":A$(4)"SOLVING INEQUALITIES":A$(5)"POST TEST":1002:(222)255Ħ0::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": ALGEBRA 2":95,100,-4,-4,-4,-4 , 70,40,45,78,95,78,74 ,110,74,134,74,158,-4,-4 ="unit "O" and may now go on to unit@D3H@"O1".":T2ĺ"@6H18V@ARROW SHOWS AREA OF WEAKNESS.">36?"@5H18V@ARROWS SHOW AREAS OF WEAKNESS.":36x@A$(1)"ADDITION OF REAL NU6,133,52,126,58,105,58,98,52,-4,-4,70,26,70,40,-4,-4,74,100,74,110,-4,-4,74,122,74,134,-4,-4,74,146,74,158,-4,-4,45,68,45,78F; -4,-4,95,90,95,100,-4,-4, -4,-4 , 37,52,49,52,91,52,98 ,52,70,64,70,68,45,68,95,68,95,68,95,78,45,90,45,100,45, 100,C%,D%,F%,G%:C%055?5C%,D%F%,G%:D%G%ēC%1,D%F%1,G%G652Z7C%,D%:C%0ıf849:55911,46,33,78,82,78,61,110,61,134,61,158|: 49,20,56,14,91,14,98,20,91,26,56,26,49,20,-4,-4,70,40,91,52,70,64,49,52,70,40,-4,-4,98,52,105,46,126,4ASSOCIATIVE AXIOMS@D10H@<3> THE IDENTITY ELEMENT@D14H@AND DISPLACEMENT@D10H@<4> THE OPPOSITE OF A REAL@D14H@NUMBER" P"@10H15V@<5> ADDITION TEST@2D10H@<0> RETURN TO ALGEBRA MENU@4DI11H@WHICH ONE (0-5) ??@I@":B%0:C%5:6:BA%:BāI112:36298I,0:I:21,12642,12649,13242,13820,13813,13220,126:43,12650,13243,138:31,11431,12628,123:32,11432,12635,123:4:17:"<0>":59,3259,15960,15960,32sO"@20H5V@CONTENTS@2D10H@<1> AXIOMS OF CLOSURE AND@D12H@ EQUALITY@D10H@<2> COMMUTATIVE AND@D14H@(3)A(4)):4:5:(4)"RUNPT2.2";LI$(5):I15:I$(I):IMB0:P0:M0:60:D$"":61:X18:J3810216:H12:E27:169:(J10)8:"@3H@<"(J22)16">":J102ēX13,J12X13,J16:X10,J13X13,J16:X14,J12X14,J16X17,J13NJ:21,13814,132:X(36)73:J(37)82:D$" ";:E(D$)76:H11:169:yG"@7H14VL@";:NNQ:72:"+";:NNW:72:"=";:NNQW:72:" ":HNN0ĺ"@10C@"NN"@15C@";:I"@15C@(@138C@"NN"@15C@)";:K35339:I14:A(I)0:I:(36309)2D14:168:36299,(A(1)A(2)A30976+@"@21V1HI@"38)"@D1H@"38)"@I@":A"@5H21VI@PRESS "(1)" TO VIEW THE NEXT PAGE.@3H22V@PRESS "(2)" TO VIEW THE PREVIOUS PAGE.@I@"BA(16384):10:A12866C16368,0:A136İ63:PP1:85DA149İ62:PP1:85E668F(36)2)(QW1)1):C1QWA1:(A1Q)((C1)C1)58;"@2H10V@IF@2H13V@AND@2H16V@THEN@5H11VL138C@"QW"="Q"*"W"@3H14V@"Q"*"W"="A1"*"C1"@5H17V@"QW"="A1"*"C1"@R15C@":<"@1H1V@C"B" P"P"@D1H@M"M:="@I2V7H@"31):24(D$)2:D$"@I@":>31051:64?1"@5H17V@"QW"="A1"+"C11"@R15C@":x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e -A1((1)8):A1QW45k . /A1((1)(QW)):A1Q47 0 1B1((1)8)1:43:B150,51,52,53,54,55,56,57 2Q((1)9)1:"@13V7HL138C@"Q"="Q"@R15C@":D3"@2H11V;:A155G%1: $A141413 %A136F%F%F%FL &A149F%C%F%F%F 'AA128:(A47)(A58)(A45)ĖE%F%:(A);:256F%,A:F%F%F:F%C%41 (33 )B$"":I0F%:B$B$((256I)):I:R(B$): *S1:L1T:U:"@D@"V):L:% +N14:8C@<@15C@"q EE7:KQ7135Q7135E2O:K,J13:L125:L,K:19Q(E14):JJ4:(J12)71:"("E7")":7: D%:E%:C%):F%0:I0C%:256I,32:I:16368,0 !E%F%:"@I@"((256F%))"@I@"; "10:A12834 #16368,0:E%F%:((256F%))))6:# 1(E0)4:G1:E0G1J E20:J8:E0ĺ"@10C@>@15C@":27^ "@138C@<@15C@" EE7:K135135E2G:K,J9:L125:L,K:7:19(E14):(J4)71:"("E7")": O1:1:E0O1:5 E20Q:J81:E0ĺ"@10C@>@15C@":31 "@1316368,0:A155İ4:5:(4)"RUNALGEBRA 2"9 A205BıB 17L :16V 173n "@R15C0K@":63:83 "@22V1HI@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 10:A12820 16368,9:A16020 "@22V1H@"36)"@I@": C((1)(D5224576:30719:639005A$"AM2.PROGRESS":75;J0:1002:T::e10:A128616368,0:A155A%0A%A176:A%B%A%C%6  A(16384):A155(36309)ı 16368,0:"@40X40YN@" A(16384):A12812(                  roperty of the "ZZ$" of a sum":"@5V1H@The ";:70:"@7V1H@says that the "ZZ$" of the sum of@9V1H@two numbers is equal to the sums of@11V1H@their "ZZ$"s. For example,":410+'zF11419:F1:"@30H10K@ ":F1:F11518:F1:"@26H138K@ ":F1:"@133K13V25:119,152153,152:3:"@138K16V17H@ @3F@ @D5B@ @3F@ @D5B@ @3F@ @0K@":"@17V19H@"C1:"@133K13V18H@ @D3B@ @D3B@ @U2B0K@"C:19:F11316:F1:"@18H0K@ @D3B133K@ @D3B@ @D3B@ @U2B0K@"C:L1100:L,F1:"@15C17V19H@0":65&yZZ$"opposite":D$"pegative":" or ";:70:" of the other@15V1H@number.":65$wD$"axiom of opposites":"@5V1H@The ";:70:" says that@7V1H@every real number has an opposite and@9V1H@the sum of a number and its opposite@11V1H@is zero.":C1(((1)9))%x98,128175,128:3,127,135#tP117,119,121,83#uD$"additive inverse":"@5V1H@You may pair points that have the same@7V1H@displacements: -6 and 6, -2 and 2,@9V1H@0 and 0. Each number in such a pair is@11V1H@called the ";8$v70:D$"opposite":" or @13V2H@";:70:D$"n + X = "XE"@16V3HI@ What is the value of X? @I18V3H@X = ":21,127195,127:D%19:E%8:C%2:F1:32:400"oD19ĺ"@18V10H@That is incorrect- see above":QX:J104:28:113"p"@18V10H@That is incorrect ( X= "E" )""q19:P:"r62:83 #sM116,124V3H@"15)"@7V12HI@"13)"@I18V3H@"35)"@4H11V@"33)!mD24:23:EC:D30E:23:XC15:"@5V2HI@For the equation written below, find@D2H@the displacement X required to":14,39265,39g"n"@6V32H@"6)"@7V18H@"20)"@7V2H@arrive at "XE" from "X".@I14V3H@"X"number.":J12 j170:D15:23:J100:EC:24:"@16V1H@A displacement in the negative@D1H@direction describes a negative number.":D15:23:J100:E(C):24:65 kD19:A(3)0:108:1143!lX9:J36:H32:E260:169:J76:H80:169:J12:170:P1D1:60:"@110)6:J76:24:EE7:"@12V6H@"E" + 0 = "E:65=hP105,83 i"@1H5V@In the first module of the ALGEBRA@D1H@series you were told about the number@D1H@line. A change in direction, or@D1H@displacement, in the positive@D1H@direction describes a positive :5f71:135,83135((QW)7),83:"@R15C@":19:"@8V2HL@"18):11:K12:"@D2H@"18):K:"@R15C@":7:135,83135((QW)7),83:E1/g"@14V2H@When 0 is added to a number, there is@D2H@no displacement. This is the additive@D2H@axiom of zero.":E((1) A = A":65a"@5V1H@Together, two displacements represent@D1H@a sum of two numbers.":J10:170:E114:D9:23:QC:D15Q:23:WC:E1101,98,99,100bWW1:101cQQ1:101dQQ1:WW1eEQ:24:EW:JJ4:28:1:"@10C@":QW0ĺ"@138C@"equal to its@D2H@displacement from 0, the identity@D2H@element for addition." `"@11V2H@The additive axiom of zero@13V2H@The set of real numbers contains a@D2H@unique element 0 having the property@D2H@that for every real number A,@2D8H@A + 0 = A and 0 +xiom of zero":"@5V1H@When 0 is added to a number, the sum@2D1H@is equal to that number. This rule is@2D1H@called the ";:70:".":65^P95,97,83M_X9:E261:J36:H40:169:J84:H64:169:"@5V2H@For any real number A, the absolute@D2H@value of A is @2D2H@"D$".":65\D$"identity element":"@5V1H@We call 0 the ";:70:"for@2D1H@addition. Adding zero to a number@2D1H@means to take no displacement from the@2D1H@original number. For example,@2D4H@(-4) + 0 = -4 and 0 + (-4) = -4":65]D$"additive ale of the ALGEBRA@2D2H@series, you were told about positive@2D2H@and negative numbers, and absolute@2D2H@value. The absolute value gives the@2D2H@";:70["of the number from zero,@2D2H@and the sign (positive or negative)@2D2H@gives the directiom of theM@2D4H@<0> RETURN TO CONTENTS@6D5HI@WHICH ONE (0-4) ??@I@":B%0:C%4:6:A%İ63:77}TP1:MA%:(36320M),(36320M)1:62UP83V60:B1,1,88,115,144WXM89,94,104,107YP90,92,93,83ZD$"displacement":"@5V2H@In the first modu4:5:30976:(4)"RUN ALGEBRA 2"eQD$I$(B):61:62:B3İ4:5:31152,1:31153,B:(4)"RUN AM2.1-2"RB5P0:M0:ZZ15:A(ZZ)0:ZZ:86TSP0:M0:60:"@12H5V@MODE SELECTION@2D4H@<1> DISCUSSION@2D4H@<2> RULE@2D4H@<3> EXAMPLE@2D4H@<4> SAMPLE PROBLE24H17V@Incorrect-@D23H@answer was "Q:19:"@15V@":L12:"@D22H@"15):L:T7:166z6B11B12ĺ"@8H15VL133C@"Q"@R15C@"6B13ĺ"@2H15VL133C@"Q"@R15C@"V719:S13:U22:T7:V17:42:"@12V21H@Which axiom is@D21H@represented?@D21H@<1> Substitution@D2+"Y:158=5"@15V2H@?+@15C@(@133C@"W"+"Y"@15C@)@133C@":158U5QQW:"@15V8H@?+"Y5"@R15C21H12V@What value for@D21H@the question mark@D21H@makes the@21HD@sentence true? @I@ @I@":B%0:C%9:6:"@I36H15V@"A%"@I@":A%Qĺ"@26H17V@Correct":160R6"@"u414,7914,39266,39266,79:X10:E260:H46:J36:169:X10:J92:E127:H64:169:X143:J92:E127:169:P14:60443:QW1534"@12V2HL15C@(@133C@"Q"+"W"@15C@)@133C@+"Y"=":B1((1)3)1:B1157,155,1565"@15V2H15C@(@133C@"W"+?@15C@)@133C@):S11:T9:U3:V16:42'3P:31051:4"@I5V2H@Give a replacement for the question @6V2H@mark which makes the sentence below @7V2H@true. Then select from the list to @8V2H@the right the axiom represented when@9V2H@the replacement is made."12):"@I@@<6> False-none@2D20HI@ WHICH (1-6) ? @I@ @I@ @I@":140,135266,135266,143150,143:P14:60:C$" SUM ":492B%1:C%6:6:"@36H17V@"A%:A%B1ĺ"@18V22H@That is correct":A(1)A(1)1:1492"@18V22H@No. Answer is "B1319:"@36H17V@ @20H18V@"18169:14,39266,39266,63:"@5V2HI@Choose from the list on the right @D2H@the axiom which best describes the @D2H@statement to the left."14)n2"@I20H10V@<1> Reflexive@D20H@<2> Symmetric@D20H@<3> Transitive@D20H@<4> Closure@D20H@<5> Substitution@D20H1H1))"@R15C@":142O0"@14V3H138C@THE ANSWER IS@17V23H@"(1(G1H1))"@R15C@"\019:P:h062:83036309,1:145:151:167:168:I14:36298I,A(I):I:4:5:(4)"RUNALGEBRA 2"1X10:J76:H80:E120:169:X136:E134:169:X10:J36:H30:E260:1G11$/23:(C2)C2H1H11/"@11V2HL10C@-@15C@(@10C@"G1"+"H1"@15C@)@19H11V10C@=":D%12:E%25:F2:C%6:32:R((G1H1))ĺ"@14V4H@THAT IS CORRECT@R15C@":A(4)A(4)1:1420D19ĺ"@10CL11V24H@-("G1H1")@14V3H138C@THE ANSWER IS@17V23H@"(1(G8:T4:S6:U31:42:"@5V2H@Find the opposite or additive@D2H@inverse of the expression below by@D2H@using the property of the opposite";/"@8V2H@of a sum and enter its value.@I@":P1D1:60:S12:U3:T8:V36:42:D9:23:G1C:23:H1C:23:(C2)C2G Property of@D1H@Equality, -(3+5)=-3+(-5). So, you can@D1H@see that the opposite of the sum of@D1H@two real numbers is equal to the sum@D1H@of the opposites.":65-D19:A(4)0:136:143.X9:J36:E260:H40:169:J84:H72:169:14,39265,39:"@I@":V4H@-8@4V21H@3 + 5":19:"@14V1H@When the opposite of 3 and the@D1H@opposite of 5 are added, the sum is@D1H@-3 + -5 = -8.":19,"@7V11H138C@<@7V16H@<@8V12H15C@-5@8V17H@-3@15C@":5:77,59135,59:3:19:V37:T6:U2:S12:42-"@11V1H@By the Transitive real number, then the@D1H@opposite of A is written -A.@D1H@NOTE: -(-A) = A.":65+J6:170:"@11V1H@When 3 and 5 are added, the opposite@D1H@of the sum -(3+5)=-8.":19:"@138C 5V11H@<@5V22H10C@>@5V27H@>":5:78,43138,43},1:138,43192,43:"@15C4V1 of 0.":J11:170*"@13V19H@^@13V14H@^@13V24H@^":101,116101,108100,108100,116171,116171,108170,108170,116:"@13V11H@^@13V27H@^":80,12480,10879,10879,124192,124192,108191,108191,124:135,106135,112136,112136,106Q+"@16V1H@If A is a opposite of a sum@2D2H@For all real numbers A and B,@2D10H@-(A + B) = (-A) + (-B)":65i)P128,131,83*"@5V1H@On a real number line, any number can@D1H@be paired with another number which is@D1H@the same distance from 0 and on the@D1H@opposite sideive number, then"(}"@12V6H@-A is a negative number.@D6H@If A is a negative number, then@D6H@-A is a positive number.@D6H@If A is zero, then -A is zero.@D2H@<2> The opposite of -A is A;@D6H@ that is, -(-A) = A.":65W)~H48:169:"@5V2H@Property of the6H@ @18V26H@ @0K@":119,127245,127:65='{P124,126,83(|X10:J36:E260:H112:169:"@5V11H@Axiom of opposites@7V2H@For every real number A there is a@D2H@unique real number -A such that@D2H@A + (-A) = 0 and (-A) + A = 0.@2D2H@<1> If A is a posit 3V1H@-9 = -(3+6) = -3 + -6@15V17H@POSITIVE@16V17H@NEGATIVE":122X:1002:(222)255Ħ:0::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": AM2.1":0 11,J4:(B130)35((B130)35)ēB1,J4B1,J5B11,J5B11,J4L9B1:9 "CLOSURE; EQUALITY","COMMUTATIVE; ASSOCIATIVE","IDENTITY; DISPLACEMENT","OPPOSITE","ADDITION TEST" 9REĺ"@18V10H@That is correct":A(3)A(3)1:1139111@:"@1:31051: 8D14:136:31051:`8X,JHX,JXE,JXE,JHX1,JHX1,J:XE1,JXE1,JH:B9J1:4:"<":J1:36:">":J2:20:"0":X21:I1250:JJ83:X,JI1,J:JJ1:XX1:I1I11:X,JI1,J:JJ2:X,JI1,J:B1302407:B1,J4B1,J2B11,J2B1H@<2> Commutative@D21H@<3> Associative@I18V21H@ WHICH (1-3)? @I@":147,143265,1437B%1:C%3:6:"@35H18V@"A%:B1A%ĺ"@18V7H@Correct":A(2)A(2)1:1657"@18V3H@No. answer is ";B17197S13:V17:42:S13:U3:42:P:31051: 8D14:1089:4:5:(4)"RUNAM2."D@HP1:MA%:(36320M),(36320M)1:48KIP71]J47:B76,122cKxLM77,97,103,115MP78,80,82,84,89,91,93,95,71KN"@5V1HR@An assumption is a statement which is@7V1H@taken to be true without proof. You@1H9V@willD14:D2Ģ15E?F35339:J$(5):D(31152):B(31153)GP0:M0:47:"@12H5V@MODE SELECTION@2D4H@<1> DISCUSSION@2D4H@<2> RULE@2D4H@<3> EXAMPLE@2D4H@<4> SAMPLE PROBLEM@2D4H@<0> RETURN TO CONTENTS@6D5HI@WHICH ONE (0-4) ??@I@":B%0:C%4:6:A%İ4@3@F@3@FI24H@=@I30H@6@I@"ABT9:U38:H20:L129:164:T142:164:"@2H5V@The statement@2H6V@below is true@21H5V@Due to this@21H6V@axiom":CT9:U76:H24:L128:164:T142:U76:164:"@2H10V@This statement@11V2H@is true@21H10V@Due to this axiom":D1A$"1+3=2+2":B$"2+2=4":C$"1+3=4":D$"@16H19V10K@ @F@ @F@ @2UB133K@ @0KDF@=":E$"@FU133K@ @F@ @2D3B10K@ @F@ @FU0K@=":F$"@F10K@ @F@ @F@ @F@ @0K@"@D2A$"2*3=3*2":B$"3*2=1*6":C$"2*3=1*6":D$"@18V16HI@3@F@3@FI@=@FI@2@F@2@F@2@FI@=@FI@6@I@":I$"@I:B$"THEN 3+1=2+2":C$"@133K@ @F@ @2D3B10K@ @F@ @U0K@":D$"@133K@ @F@ @F@ @2D5B10K@ @0K4F2U@":D2A$"IF 2*2=4*1":B$"THEN 4*1=2*2":D$"@133K@ @F@ @F@ @F@ @0K@":C$"@133K@ @F@ @2D3B10K@ @F@ @U0K@";<D1Ė22=D2Ė31:15:8):20>?HE NEXT PAGE.@1H22V@ PRESS "(2)" TO VIEW THE PREVIOUS PAGE.@I@"_4A(16384):10:A12852516368,0:A136İ49:PP1:736A149İ48:PP1:737528(36)2):T(36)73:U(37)82:H$" ";:L(H$)76:H10:164:9A$"IF 2+2=3+1"10V@IF@2H13V@THEN@5H11VL138C@"EF"="E"*"F"@14V6H@"G"*"EF1"=@17V4H@"G"*("E"*"F")@R15C@":K6:{.O1Q:R:"@D@"S):O:/"@1H1V@C"B" P"P"@D1H@M"M:031051:501309762"@21V1HI@"38)"@D1H@"38)"@I@":A3"@5H21VI@PRESS "(1)" TO VIEW TF"@3H14V@"E"*"F"="J"*"N"@5H17V@"EF"="J"*"N"@R15C@":+23:"@2H10V@IF@2H13V@THEN@5H11VL138C@"EF"="E"*"F"@14V6H@"G"*"EF"=@17V4H@"G"*("E"*"F")@R15C@":,23:"@2H11V@IF@2H15V@THEN@L138C5H12V@"E"*"F"="EF"@5H16V@"EF"="1E"*"F"@R15C@":^-23:"@2Het.":K6:7'K((1)8)1:K31,40,41,34,43,44,45,38(23:"@2H11V@IF@2H15V@THEN@L138C5H12V@"E"*"F"="EF"@5H16V@"EF"="E"*"F"@R15C@":)23:J((1)(EF1)1):NEFJ:(JE)((N)N)415*"@2H10V@IF@2H13V@AND@2H16V@THEN@5H11VL138C@"EF"="E"*""="E"+"F"@3H14V@"E"+"F"="J"+"N"@5H17V@"EF"="J"+"N1"@R15C@": %23:"@2H11V@IF@2H15V@THEN@5H12VL138C@"E"+"F"="EF"@16V5H@"J"="E"+"F"@R15C@":K6: &"@11V5H@(0,1,2,3,4)@13V3H@The"G$"of@D3H@any two members@D3H@of this set is@D3H@a member of@D3H@this sH@The"G$"of@D3H@any two members@D3H@of this set is@D3H@a member of@D3H@this set.": #23:"@2H10V@IF@2H13V@THEN@5H11VL138C@"EF"="E"+"F"@14V6H@"G"+"EF"=@17V4H@"G"+("E"+"F")@R15C@":> $23:28:N(EF)J:"@2H10V@IF@2H13V@AND@2H16V@THEN@5H11VL138C@"EF"@R15C@":_ 23:"@2H11V@IF@2H15V@THEN@L138C5H12V@"E"+"F"="EF"@5H16V@"EF"="E"+"F"@R15C@": !23:28:N(EF)J:"@2H10V@IF@2H13V@AND@2H16V@THEN@5H11VL138C@"EF"="E"+"F"@3H14V@"E"+"F"="J"+"N"@5H17V@"EF"="J"+"N"@R15C@":S ""@11V5H@(0,1,2,...)@13V3:D%(C)(1)62:C:D1(D%(1)D%(2))923F D2(D%(1)D%(2))923j ED%(1):FD%(2):GD%(3):ID%(4) J((1)9):JEF26  J((1)(EF)):JE28  K((1)8)1:K31,32,33,34,35,36,37,38 E((1)9)1:"@13V7HL138C@"E"="E68,0:A155İ4:5:(4)"RUN ALGEBRA 2"7 A205Bı@ 17J :16T 165l "@R15C0K@":49:71 23:2:"@I@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 A(16384):A12820 16368,9:A16020 23:2:36)"@I@":* C14324576:30719:63900&3:70,;0:1002:E::bA(16384):10:A1286|16368,0:A155A%0A%A176:A%B%A%C%6  A(16384):A155(36309)ı 16368,0:"@40X40YN@" A(16384):A12812& 163                  ty of@D21H@equality@9V2HL@IF@11V2H@AND@15V2H@THEN@5C@":G$"+":D2G$"*":WD%(1)D%(2)~&kXD%(1):YD%(2):D1WD%(1)D%(2)&l"@8H9V@"W"@9V10H@="D%(1);G$;D%(2)"@12H11V@"D%(1);G$;D%(2):D2ĺ"@8H9V@"W&mE%(1)(W1)1:ZE%:ZX109'nA1XYZ:DD%(1)D%(2)%i"@13V2H15C@IF@5C7H@"D%(1);G$D%(2)"="V"@15V2H15C@THEN@17V4H5C@"V"="D%(1);G$D%(2)"@R15C21H9V@The reflexive@D21H@property of@D21H@equality@2D21H@The symmetric@D21H@property of@D21H@equality":51W&j66:23:"@9V21H@The transitive@D21H@proper14V@For every number A and every number@2H15V@B, if A=B then B may be substituted@2H16V@for A, and A may be substituted for@2H17V@B in any expression.":51$gP104,106,112,71 %h66:23:"@2H9VL5C@"D%(1)"@6H@= "D%(1):G$"+":VD%(1)D%(2):D2G$"*":V(1)"addition":B$(1)"sum A+B":A$(2)"multiplication":B$(2)"product A*B":"@3H5V@Axiom of closure for "A$(D)"@2H7V@For all real numbers A and B, the @2H8V@"B$(D)" is a unique (one and@2H9V@only one) real number."$f"@7H12V@The substitution principle@2HFor all real numbers A,B, and C:@2H15V@Reflexive property A=A@2H16V@Symmetric property If A=B,then B=A@2H17V@Transitive property If A=B and B=C@23H18V@then A=C":T9:U82:H76:L262:164:51"dT9:H48:L260:U34:164:T10:U90:L260:H56:164#eA$pression.":51!aP98,100,71!b"@2H5V@The following rules are called@2H6V@axioms. That is, they have been@2H7V@accepted by mathematicians as being@2H8V@true without the need of proofs.":T9:U34:H40:L262:164"c"@11H11V@Axioms of equality@2H13V@:"@30H5V@says that@7V1H@changing the numeral naming a number@9V1H@in an expression does not change the@11V1H@value of the expression. Since "A$(D)!`"@36H11V@is@13V1H@equal to 8, you may always substitute@15V1H@8 whenever "A$(D)" appears in an@17V1H@ex9V2H@of two real numbers gives a real@11V2H@number.@14V2H@A "L$(D)" of numbers does not need to@16V2H@be written with the numerals used to@18V2H@represent the numbers, however.":51 _A$(1)"4+4":A$(2)"2*4":"@5V1H@The ";:H$"substitution principle":56pends upon two things:@15V2H@the operation and the set of numbers@17V2H@being used.":51]K$(1)"addition":L$(1)"sum":K$(2)"multiplication":L$(2)"product":"@5V2H@The ";:H$"axiom of closure for":56:H$K$(D):3:8:56:" says that the "L$(D)^"@:56:".":51["@5V2H@Any set S is said to be ";:H$"closed under":56:H$"an operation":3:8:56:"@16H7V@performed on its"X\"@9V2H@elements, given that each result of@2H11V@the operation is an element of S.@13V2H@";:H$"Closure":56:"@11H13V@den":"@5V1H@You have probably found that if two@7V1H@elements of the set of natural numbers@9V1H@are "K$(D)", the result is always a" Z"@1H11V@natural number. We say that the set of@13V1H@natural numbers is ";:H$"closed under":56:"@15V2H@";:H$L$(D):"@2H17V@If "A$"@2H18V@and "B$"@2H19V@then "C$:D2ĺD$:87MVD$;E$;F$W19:D1ĺ"@18V1H@"28)"@17V23H@"4)"@19V23H@"4)"@18V25H@=":51X"@18V1H@"28)"@18V16H@"I$:51YK$(1)"added":L$(1)"addition":K$(2)"multiplied":L$(2)"multiplicatio ";:H$"transitive property of equality":56:"@7V1H@says that if one number is equal to a@9V1H@second number, and the second number@11V1H@is equal to a third number, then the"?U"@13V1H@first number is equal to the third@15V1H@number. For example:":63H$"symmetric property of equality":56:"@1H7V@says that an equality may be reversed.@1H9V@For example,":57:7:13:A$"@D5F@"C$:28:15:"@L138C@=@R15C@":68:31:D$S19:7:17:B$:14:21:17):16:24:3):60:68:D$"@13V4F@"C$:51T"@1H5V@Theequality to these axioms:@1H9V@The ";:H$"reflexive property of equality":56Q"@1H11V@says that any number is equal to@13V1H@itself. For example, 4=4, or 8=8.@L10K@":17:"@10H@ @13H@ @16H@ @22H@ @25H@ @28H@ @0K138C19H@=@R15C@":51R"@1H5V@The ";: now be told about some of the@11V1H@basic assumptions on which Algebra is"O"@13V1H@built. In mathematics, assumptions are@15V1H@called postulates or ";:H$"axioms":56:".":51MP"@5V1H@The way to use the equals symbol @L138C34HU@=@15CR7V1H@gives P:48:716"@I5V2H@Give a replacement for the question @6V2H@mark which makes the sentence below @7V2H@true. Then select from the list to @8V2H@the right the axiom represented when@9V2H@the replacement is made."12):"@I@"k714,7914,39266,39266,143,144,1455E2:G5:146.5E4:G25:146@5E1:G10:146R5F((1)9)15G$"+":IEG:JFI:D2G$"*":IEG:JFI 6"@13V3H@("E;G$;F")"G$;G"=("F;G$;E")"G$;G"@11H15V@="F;G$"("E;G$;G")@17V11H@="F;G$;I"@19V11H@="J:19:R4:13:Q7:S18:46:ve axiom@2D1H@2.@21H@Associative axiom@2D1H@3.@21H@Substitution@2D1H@4.@21H@Substitution@5V2H@The commutative and associative"4"@6V2H@axioms may be used to find a desired@7V2H@"S$(D)"quickly and easily.":P15:47:D1İ23:147 5K(((1)3)1):Ktive axiom of"T$(D)"@7V2H@For all real numbers A and B,@9V14H@"U$(D)3"@12V2H@Associative axiom of"T$(D)"@14V2H@For all real numbers A,B, and C,@16V10H@"V$(D):51~4S$(1)"sum ":S$(2)"product ":T10:U36:H32:L260:164:67:"@13V1H@1.@21H@Commutati$"8)"G$"2 is not equal to@14V2H@10"G$"(8"G$"2).":51B2P137,712T10:U36:L260:H48:164:U92:164:T$(1)" addition":T$(2)" multiplication"D3U$(1)"A+B = B+A":U$(2)"A*B = B*A":V$(1)"(A+B)+C = A+(B+C)":V$(2)"(A*B)*C = A*(B*C)":"@5V2H@Commuta511O$(1)"addition":O$(2)"multiplication":P$(1)" Subtraction ":P$(2)" Division ":H$"associative axiom of "O$(D):"@5V2H@This assumption is called the":8:2:56:".":G$"-":D2G$"/"42"@10V2H@NOTE:"P$(D)"is not associative@12V2H@since (10"G@numbers at a time. When"Q$(D)"@11V1H@three or more numbers, we assume that@13V1H@we may group the numbers in different@15V1H@ways without changing the"P$(D)".@17V1H@For example, ":G$"+":D2G$"*"1"@17V1H@For example, (1"G$"2)"G$"3=1"G$"(2"G$"3).":perty because, for@2D1H@example, 3"G$"4 does not equal 4"G$"3.":51/P$(1)" sum":P$(2)" product":T$(1)"an ":T$(2)"a ":Q$(1)" adding":Q$(2)" multiplying":"@5V1H@When doing "T$(D)O$(D)", you find@2D1H@a"P$(D)" by working with only two"0"@9V1HO$(1)"addition":P$(1)"Subtraction ":O$(2)"multiplication":P$(2)"Division ":H$"commutative axiom of "O$(D):"@5V1H@This assumption is called the":2:8:56:".@12V1H@NOTE: "P$(D)"does not have the":G$"-":D2G$"/"C/"@14V1H@same commutative pro6V13H@+@15CR@":19:"@L16V13H@ @F138C@+@R15C@":51-D2T122:U116:H32:L42:164:122,132164,132:"@15V18H10K@ @F@ @F@ @2D5B@ @F@ @F@ @0K@"-D2İ19:0:124,132162,132:3:136,117136,147137,147137,117:150,117150,147151,147151,117:51.*2":"@5V1H@In arithmetic, we assume that when two",}"@1D1H@numbers are"O$(D)", the"P$(D)"@2D1H@remains the same no matter what order@2D1H@is used in"Q$(D)"them. For@2D1H@example,"R$(D)2-~D1ĺ"@16V3HL10K10C@ @F@ @F@ @F@ @F@ @F@ @F@ @F0K@=@F@7@138C17V@ @20H18V@"18):10:Q9:R3:S16:46:P10İ47?+yP:48:71V+zM123,136,140,149p+{P124,129,131,134,713,|O$(1)" added":P$(1)" sum ":Q$(1)" adding ":R$(1)" 3+4 = 4+3":O$(2)" multiplied":P$(2)" product":Q$(2)" multiplying ":R$(2)" 2*3 = 3@<6> False-none@2D20HI@ WHICH (1-6) ? @I@ @I@ @I@":140,135266,135266,143150,143:P210:G$" sum ":D1İ30*uD2G$" product ":39*vB%1:C%6:6:"@36H17V@"A%:A%Kĺ"@18V22H@That is correct":120*w"@18V22H@No. Answer is "K0+x19:"@36H1:14,39266,39266,63:"@5V2HI@Choose from the list on the right @D2H@the axiom which best describes the @D2H@statement to the left."14)"@I@"p*t"@20H10V@<1> Reflexive@D20H@<2> Symmetric@D20H@<3> Transitive@D20H@<4> Closure@D20H@<5> Substitution@D20Htion@21H14V@principle@9H9V@is a real@9H10V@number":D2WXY(r"@2H9VL5C@"W;N$X"@2H13V15C@IF @5C@"W"="X;N$Y"@2H15V15C@THEN @5C@"Z;N$W"@2H17V@="Z;N$"@15C@(@5C@"X;N$Y"@15C@)@R@":51)sT10:U76:H80:L120:164:T136:L134:164:T10:U36:H30:L260:1642A1XYZ:A1(A1)109Z'o"@13V10H@="Z;G$;A1"@12H15V@"W"@10H17V@="Z;G$;A1"@R15C@":51'p66:23:XD%(1):YD%(2):WXY:E%(1)72:ZE%:M$"addition":N$"+":D2M$"multiplication":N$"*"=(q"@21H9V@Axiom of closure@21H10V@for "M$"@21H13V@SubstituKA%ĺ"@18V7H@Correct":163$;-162<;1002:(222)255Ħ;0::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": AM1-2":Commutative@D21H@<3> Associative@I18V21H@ WHICH (1-3)? @I@":147,143265,143:B%1:C%3:6:"@35H18V@"A%:300:"@18V3H@No. Answer is ";K:19:12:S17:46:12:R3:46:P:48:71:T,UHT,UTL,UTL,UHT1,UHT1,U:TL1,UTL1,UH:;,"@I@":A%Eĺ"@26H17V@Correct":159O9"@24H17V@Incorrect.@D23H@Answer was "Eu9K1K2ĺ"@8H15VL133C@"E"@R15C@"9K3ĺ"@2H15VL133C@"E"@R15C@"o:19:12:R22:Q7:S17:46:"@12V21H@Which axiom is@D21H@represented?@D21H@<1> Substitution@D21H@<2> 3C@"G$(D);G:157O8"@15V2H@?"G$(D)"@15C@(@133C@"F;G$(D);G"@15C@)@133C@":157h8EEF:D2E(EF)F8"@15V8H@?";G$(D);G#9"@R15C21H12V@What value for@D21H@the question mark@D21H@makes the@21HD@sentence true? @I@ @I@":B%0:C%9:6:"@I36H15V@"A%79:T10:L260:H46:U36:164:T10:U92:L127:H64:164:T143:U92:L127:164:G$(1)"+":G$(2)"*":P19:47|723:EF1517"@12V2HL15C@(@133C@"E;G$(D);F"@15C@)@133C@"G$(D);G"=":K((1)3)1:K155,153,1548"@15V2H15C@(@133C@"F;G$(D);"?@15C@)@13 ";:59:"@5V30H@, or@2D1H@change the form of an algebraic@2D1H@expression from a sum to a product.@2D1H@For example, 2X+3X+4X = (2+3+4)X = 9X.@2D1H@The "J$" is assumed"N"@J15V1H@to be true for subtraction and for@2D1H@more than three numbers.":54ys that@2D1H@the product of one number times the@2D1H@sum of a second and third number is@2D1H@equal to the product of the first and@2D1H@second number plus the product of the@2D1H@first and third numbers.":54MH$"transform":"@5V1H@It can be used to IM74,81,84,89'JP75,76,77,79,68KH$J$:"@5V2H@Since 2(20+3) = 2(23) = 46 and@2D2H@2*20+2*3 =40 + 6 = 46, we see that@2D2H@2(20+3) = 2*20+2*3 by the transitive@2D2H@property. This is an example of the@2D2H@";:59:".":54L"@5V1H@The "J$" saDE SELECTION@2D4H@<1> DISCUSSION@2D4H@<2> RULE@2D4H@<3> EXAMPLE@2D4H@<4> SAMPLE PROBLEM@2D4H@<0> RETURN TO CONTENTS@6D5HI@WHICH ONE (0-4) ??@I@":B%0:C%4:6:A%İ52:61EP1:MA%:(36320M),(36320M)1:51FP68G49:C1,1,73,99,5000HH ONE (0-6) ??@I@":B%0:C%6:6:CA%:CāG3629936315:G,0:G:4:5:30976:(4)"RUN ALGEBRA 2"AH$I$(C):50:51:C3İ4:5:31152,2:31153,C:(4)"RUN AM2.1-2"BC4İ4:5:31153,C:(4)"RUN PT2.2"CC6P0:M0:71DP0:M0:49:"@12H5V@MOEQUALITY@D10H@<2> COMMUTATIVE AND@D14H@ASSOCIATIVE AXIOMS@D10H@<3> THE DISTRIBUTIVE PROPERTY@D10H@<4> RULES FOR MULTIPLICATION@D10H@<5> THE RECIPROCAL OF A REAL"a@"@14V14H@NUMBER@D10H@<6> MULTIPLICATION TEST@2D10H@<0> RETURN TO ALGEBRA MENU@4DI11H@WHIC4,T16S17,T13>T:21,15414,14821,14242,14249,14842,15420,15413,14820,142:43,14250,14843,154:31,13031,14228,139:32,13032,14235,139:4:19:"<0>":59,3259,15960,15960,32?"@20H5V@CONTENTS@2D10H@<1> AXIOMS OF CLOSURE AND@D12H@ (36)2):S(36)73:T(37)82:H$" ";:L(H$)76:H11:126:b<35339:I$(6):G16:I$(G):G=C0:P0:M0:49:H$"":50:S18:T3811816:H12:L27:126:(T10)8:"@3H@<"(T22)16">":T118ēS13,T12S13,T16:S10,T13S13,T16:S14,T12S13 43097625"@21V1HI@"38)"@D1H@"38)"@I@":6"@5H21VI@PRESS "(1)" TO VIEW THE NEXT PAGE.@3H22V@PRESS "(2)" TO VIEW THE PREVIOUS PAGE.@I@"7B(16384):10:B12855816368,0:B136İ52:PP1:709B149İ51:PP1:70:55?;rect":H ,"@17V15H@Incorrect@12V2H@="C$(1)"("C$(4)")@15H@S"(D$,11):b -"@13V15H@"E$C$(1): /(C$(4))1(C$(4))1ĺ"@13V15H@"G$: 0"@13V15H@"E$C$(4): 1"@1H1V@C"C" P"P"@D1H@M"M: 2"@I2V7H@"31):24(H$)2:H$"@I@":331051:5@13V2H@="A$:45? $B$A$(C$(1)"1"C$(1)"0"C$(1)"-1")ī43[ %C$(4)"1"A$C$(1):41| &C$(4)"-1"A$"-"C$(1):41 'C$(4)"0"A$"0":41 (A$C$(4)C$(1):C$(1)"2"A$((C$(4))2) )B$A$ī43 *44:"@13V2H@="A$:47 +"@17V15H@CorI1))(B45)ĖE%F%:(B);:256F%,B:F%F%F:F%C%33? 25t !B$"":G0F%:B$B$((256G)):G:J(B$):I0: "K1:N1O:Q:"@D@"R):N: #C$(4)((C$(2))(C$(3))):C$(1)"-1"C$(1)"1"C$(1)"0"A(C$(1))(C$(4)):A$(A):B$A$İ44:"E%:C%):F%0:G0C%:256G,32:G:16368,0Q E%F%:"@I@"((256F%))"@I@";c 10:B12826 16368,0:E%F%:((256F%));:B155G%1: B14133 B136F%F%F%F B149F%C%F%F%F7 BB128:((B47)(B58))((B39)(B91)(5İ4:5:(4)"RUNALGEBRA 2", B205Cı5 17? :16I 129a "@R15C0K@":52:68 "@22V1HI@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 10:B12820 16368,9:B16020 "@22V1H@"36)"@I@": D1(((1)E)):- D%:'24576:30719:63900(128:60.=0:1002:G::X10:B1286r16368,0:B155A%0A%B176:A%B%A%C%6  B(16384):B155(36309)ı 16368,0:"@40X40YN@" B(16384):B12812 16368,0:B15                !!! ! ! ! ! !!sitive and@D6H@negative number is a negative@D6H@number.@11V2H@<2> The product of two positive@D6H@numbers or of two negative"0'q"@13V6H@numbers is a positive number.@15V2H@<3> The absolute value of the@D6H@product of two real numbers is@D6H@the productA(-1) = -A and (-1)A = -A@3D2H@P"(T$,7)" of opposites in products"%o"@14V2H@For all real numbers A and B,@2D2H@(-A)B = -AB, A(-B) = -AB,@D2H@(-A)(-B) = AB":54~&pS9:L262:T36:H120:126:"@5V2H@Rules for multiplication@7V2H@<1> The product of a poproperty that@2HD@for every real number A,@2D3H@A*1 = A and 1*A = A"$m"@14V2H@"U$" "T$" of zero@2D2H@For each real number A,@2D2H@A*0 = 0 and 0*A = 0.":54C%nT36:H48:126:T92:H56:126:"@5V2H@"U$" "T$" of -1@2D2H@For each real number A,@2D2H@e product of@2D1H@real numbers is the product of the@2D1H@absolute values of the numbers.":54t#kP108,110,112,68D$lS9:L262:T36:H64:126:T108:H48:126:"@2H5V@"U$" "T$" of one@2D2H@The set of real numbers has a unique@2HD@element 1 having the A@2D1H@product of nonzero numbers of which an@2D1H@even number are negative is a positive@2D1H@number. A product of nonzero real""i"@15V1H@numbers of which an odd number are@2D1H@negative is a negative number.":54^#j"@5V1H@The absolute value of thoduct is 0."!g"@13V2H@Multiplying any real number by -1@15V2H@gives the opposite of the number.@17V2H@When both factors are -1, you would@19V2H@have (-1)(-1) = 1.":54"h"@5V1H@By multiplying numbers, you will@2D1H@discover the following properties. 1, the@11V2H@product is that same real number. For@13V2H@example, 5*1=5 and 1*5=5.":54 !fH$"multiplicative":"@5V2H@The sentences 5*0=0 and 0*5=0@7V2H@represent the ";:59:H$T$" of zero":"@9V2H@";:59:". When one of the@11V2H@factors is 0, the prR$(G);:N11000:N:"@I21H@"R$(G):G:98Ba"@28H18V@ Correct"Ob19:P:fcM100,107,114,121dP101,102,104,106,68X eH$"identity element":"@5V2H@'1' is the ";:59:" for@7V2H@multiplication because whenever you@9V2H@"S$" a real number by+"(A1)N$)ī97|_133:Q$(3)"=("(W)"+"(X)"+"(Y)")"N$"+("(W)"+"(X)")"O$:Q$(4)"="(A1)N$"+"(B1)O$)`R$(1)"D"(K$,11):R$(3)R$(1):R$(2)"A"(L$,10)" ":R$(4)"S"(D$,11):"@28H18V@Incorrect@10V@";:G14:"@3H@"Q$(G)"@I21H@"\O$(((1)6)84):O$N$ī92]E8:23:WD1:23:XD1:23:YD1:A1WXY:B1WX:"@9V3H@"W"("N$"+"O$")+"X"("N$"+"O$")+"Y;N$:C%9:D%19:E%18:F1:I1:24:P$"":G19:((B$,G,1)" ")P$P$(B$,G,1)^G:(P$(A1)N$"+"(B1)O$)(P$(B1)O$"9:L260:H24:T36:126:T132:126:T68:H56:126:"@I5V2H@Apply the "J$" and @D2H@simplify the "M$" below"7)"@I17V2H@What is the simplest "M$" for@D2H@the one above? "[14,39265,39:P19:49:R31:O5:Q3:K10:34:"@28H18V@"10):N$(((1)6)84) @"W"("X")=@9H9V@"W"("Y"+"A1")@D7H@= "W"*"Y"+"W"*"A1"@D7H@= "B1"+"C1"@D7H@= "D1:19:R16:O4:K10:Q3:34:P:132X"@5V11H@"W"Y+"X"+"Y"Y@9V2H@"W"Y+"X"+"Y"Y="W"Y+"Y"Y+"X"@D9H@=("W"+"Y")Y+"X"@D9H@= "A1"Y+"X:19:P:31051:68YE19:90:31051:68ZS7V2H@Step@14F@Property":U52:V116:G13:131G,U131G,V:GV"@9V20H@S"(D$,11)"@11V20H@S"(D$,11)"@12V20H@S"(D$,11)"@10V20H@D"(K$,11):P14:49:E7:23:WD2:E88:23:XD11:Y((X10))10:A1XY:B1WY:C1WA1pWD1B1C1:"@5V11H@"W"("X")@9V2H AB+AC and (B+C)A = BA+CA."S"@14V3H@Also:@16V2H@AB+AC = A(B+C) and BA+CA = (B+C)A@D2H@A(B+C+D+...+Z) = AB+AC+AD+...+AZ@D2H@A(B-C) = AB-AC and AB-AC = A (B-C)":54TP85,68,68=US9:T36:H80:L260:126:9,52269,52:9,68269,68:"@5V2H@Simplify@ession with as few terms as@2D1H@possible, you have ";:59:"@33H@it.":54XQP82,68RL262:S9:T44:H64:126:T124:H32:126:"@6V5H@The "J$" of@D4H@multiplication with respect to@D15H@addition@10V2H@For all real numbers A, B, and C,@2D2H@A(B+C) =OH$"equivalent expressions":"@5V1H@(2+3)X = 5X. Since the expressions on@2D1H@either side of the equals sign signify@2D1H@the same number, they are called@2D2H@";:59KPH$"simplified":". When you@2D1H@replace an expression by an equivalent@2D1H@expr^0 "OPENAM2.PROGRESS":"WRITEAM2.PROGRESS"C5:3:3:3:3]"CLOSEAM2.PROGRESS"3!"(X)O$"+"(Y)N$:Q$(2)"="(W)N$"+"(X)N$"+"(Y)N$"+"(W)O$"+"(X)O$:q/1002:(222)255Ħ/0::::"++ ERROR ++";::" "(222):" AT LINE "(218)(219)256": AM2.2": prop. of ":G$"Commutative":f.14,39265,39:P19:49:E2:23:E3:D1İ23:DD1:C$(1)(D):124o.123.49:"@9V20H@A"(L$,10)" @3D20H@"13):P59:34:49:E9:23:WD:23:XD:23:YD:A1WY:88Y/Q$(1)"="(W)N$"+"(W)O$"+"(X)N$"+RIBUTIVE PROPERTY","MULTIPLICATION RULES", "RECIPROCAL OF REAL NUMBERS","MULTIPLICATION TEST" .T$"property":F$"AM2.PROGRESS":D$"substitution":K$"distributive":L$"associative":M$"expression":J$K$" "T$:S$"multiply":U$"Multiplicative":E$"Mult.I1:F1:C%10:D%18:E%16:24:P$"":G110:((B$,G,1)" ")P$P$(B$,G,1),}G:B$P$:35:19:R36:O3:Q3:K12:34:P:31051:68,~S,THS,TSL,TSL,THS1,THS1,T:SL1,TSL1,TH:^-"CLOSURE; EQUALITY","COMMUTATIVE; ASSOCIATIVE","DISTpression below by@D2H@using the appropriate m"(U$,13)"@D2H@properties for 1, 0, -1, and@D2H@opposites.@I16V2H@What is the value of the above@D2H@expression?":130+{C$(1)(((1)6)84)M,|Z23:23:C$(Z)(D1):Z:"@11V3H@"C$(1)"("C$(2)"+"C$(3)")":000:Z,G*x"@11V2H@The absolute@D2H@value of the@D2H@product of two@D2H@numbers":G12:"@I21H11V@!"W"*"X"! = !"W"!*!"X"!":Z11000:Z,G:54*yS9:L262:H40:T36:126:H32:T84:126:T124:126:R36:O4:Q3:K6:"@I@":34+z"@5V2H@Simplify the exH88:126:9,52272,52:S137139:S,TS,TH:S:"@5V2H@Rule@20H@Example":E7:23:WD2:23:XD2*w"@7V2H@P"(T$,7)" of@D2H@opposites in@D2H@products":G12:"@I21H7V@(-"W")"X" = -("W"*"X")@D21H@"W"(-"X") = -("W"*"X")@D21H@(-"W")(-"X") = "W"*"X:Z111000:Z,G:23:DD2:"@10V2H@"U$"@11V2H@"T$" of zero":G12:"@10V21HI@0*"D" = 0 "D"*0 = 0":Z11000:Z(uG:23:DD2:G12:"@13V2H@"U$"@14V2H@"T$" of -1":G12:"@13V21HI@-1*"D" = -"D"@D21H@"D"*-1 = -"D:Z11000:Z,G:54a)vS9:L263:T36: of the absolute@D6H@values of the numbers.":54B'rP115,118,68'sS9:L263:T36:H88:126:9,52272,52:S137139:S,TS,TH:S:"@5V2H@Rule@20H@Example":E7:"@7V2H@"U$"@2HD@"T$" of one":23:DD2l(tG12:"@7V21HI@1*"D" = "D" "D"*1 = "D:Z1A%İ75:855YP1:MA%:(36320M),(36320M)1:74AZP187_[72:B2,2,2,4000,108,134e\]X9:L260:H24:Y36:161:Y132:161:Y68:H56:161:"@I5V2H@Apply the "H$" and @D2H@simplify the "I$" below"7)"@I17V2H@What is the simplest "I$" for@):73:B5P0:M0:31051:72:1080VB6ī134FWB6P0:M0:134 XP0:M0:72:"@12H5V@MODE SELECTION@2D4H@<1> DISCUSSION@2D4H@<2> RULE@2D4H@<3> EXAMPLE@2D4H@<4> SAMPLE PROBLEM@2D4H@<0> RETURN TO CONTENTS@6D5HI@WHICH ONE (0-4) ??@I@":B%0:C%4:7::PP1:90#PA149İ74:PP1:90+Q78rR(36)2):X(36)73:Y(37)82:E$" ";:L(E$)76:H11:161:S35339:(36309)2E14:158:36300,A(3):5:6:(4)"RUNAM2.3"TB(31153):E16:G$(E):E:87"U162:5:6:(4)"RUNAM2.2":E$G$(B"@I2V7H@"31):24(E$)2:E$"@I@":4J31051:76?K30976eL"@21V1HI@"38)"@D1H@"38)"@I@":M"@5H21VI@PRESS "(1)" TO VIEW THE NEXT PAGE.@3H22V@PRESS "(2)" TO VIEW THE PREVIOUS PAGE.@I@"NA(16384):11:A12878 O16368,0:A136İ75"="B1"*"D1"@5H17V@"TU"="B1"*"D1"@R15C@":OCXX7:YY84:H0:S0XX4:70DHH1:X7,Y16X3,Y9X,YX3,Y9X7,Y16:H1XX1:68EFHH1:X7,Y16X3,Y9X,YX3,Y9X7,Y16:H1XX1:70GH"@1H1V@C"B" P"P"@D1H@M"M:%Iw@"@11V5H@(0,1,2,3,4)@13V3H@The"D$"of@D3H@any two members@D3H@of this set is@D3H@a member of@D3H@this set.":C16:A49:B1((1)(TU1)1):D1TUB1:(B1T)((D1)D1)65*B"@2H10V@IF@2H13V@AND@2H16V@THEN@5H11VL138C@"TU"="T"*"U"@3H14V@"T"*"U"*"TU"=@17V4H@"V"*("T"*"U")@R15C@":>53:D1(TU)B1:"@2H10V@IF@2H13V@AND@2H16V@THEN@5H11VL138C@"TU"="T"*"U"@3H14V@"T"*"U"="B1"*"D1"@5H17V@"TU"="B1"*"D11"@R15C@":?"@2H11V@IF@2H15V@THEN@5H12VL138C@"T"*"U"="TU"@16V5H@"B1"="T"*"U"@R15C@":C16:EN@5H11VL138C@"TU"="T"*"U"@3H14V@"T"*"U"="B1"*"D1"@5H17V@"TU"="B1"*"D1"@R15C@":<"@11V5H@(0,1,2,...)@13V3H@The"D$"of@D3H@any two members@D3H@of this set is@D3H@a member of@D3H@this set.":%="@2H10V@IF@2H13V@THEN@5H11VL138C@"TU"="T"*"U"@14V6H@"V697C1((1)8)1:49:C156,57,58,60,61,62,63,64j8T((1)9)1:"@13V7HL138C@"T"="T"@R15C@":9"@2H11V@IF@2H15V@THEN@L138C5H12V@"T"*"U"="TU"@5H16V@"TU"="T"*"U"@R15C@"::53:D1(TU)B1:D1(D1)58R;"@2H10V@IF@2H13V@AND@2H16V@TH$(4):43. .C$(1)"-1"C$(4)"2"A$"-2":43T /C$(1)"-1"C$(4)"-2"A$"2":43k 0A$C$(4)C$(1):43 1N14:H%(N)(1)62:N:H%(1)H%(2)949 2TH%(1):UH%(2):VH%(3):A1H%(4) 3D8:24:B1C:CTU51 45DTU1:24:B1C1:B1T531"A$C$(1):43+ 'C$(1)"1"A$C$(4):43Q (C$(4)"-1"C$(1)"-1"A$"1":43y )C$(1)"-1"A$((C$(4))(1)):43 *C$(4)"-1"A$"-"C$(1):43 +B$A$ĺ"@17V15H@Correct":A(4)A(4)1: ,"@17V15H@Incorrect@11V12H@="A$: -C$(1)"1"A$CG1))(A45)ĖE%F%:(A);:256F%,A:F%F%F:F%C%34? !26t "B$"":E0F%:B$B$((256E)):E:I(B$):G0: #J1:K1O:Q:"@D@"R):K: $C$(4)((C$(2))(C$(3))):C$(1)"0"C$(4)"0"A$"0":43 %C$(4)"2"C$(4)"-2"ī45 &C$(4)"E%:C%):F%0:E0C%:256E,32:E:16368,0Q E%F%:"@I@"((256F%))"@I@";c 11:A12827 16368,0:E%F%:((256F%));:A155G%1: A14134 A136F%F%F%F A149F%C%F%F%F7 AA128:((A47)(A58))((A39)(A91)(A155İ5:6:(4)"RUNALGEBRA 2"0 A205Bı9 18C :17M 166e "@R15C0K@":75:87 "@22V1HI@..PRESS (SPACE BAR) TO CONTINUE...":16368,0 11:A12821 16368,9:A16021 "@22V1H@"36)"@I@": C((1)D):- D%:H,24576:3071963900,165:832A0:1002:K::\11:A1287v16368,0:A155A%0 A%A176:A%B%A%C%7  A(16384):A155(36309)ı 16368,0:"@40X40YN@" A(16384):A12813 16368,0:"" " " " " """""""""