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The FublisJieys fuvnisli Descrtptire Ch-eiiinr.s of the ahovc ^lathe- tnatical Text-BooliS, tuith J^ficis and otJier htforntatioti conecrnint/ them. Entcrt'il according to Art of Congress, in tho year Eighteen Httndred and Fifty-Two, by M'jNTiiROP B. Smith, in the Clerk's OIBce of the District Court of tlio United States for the District of Oliio, KEY KAY'S ALGEERA, PART FIRST. IE7" The numbers in parentheses, as seen in the margin of this Key, refer to the corresponding number of example, under the same article iu the Algebra. INTELLECTUAL EXERCISES. To Teachers. As these exercises, except, perhaps, a few in Lesson XIV, can be readily solved without the aid of Algebra, by pupils having a good knowledge of Mental Arithmetic, it is unnecessary to occupy space with their solution. Some instructors who use the Algebra, pay no attention to the intellectual exercises, but permit their pupils to begin with the preliminary definitions and principles. This course is proper with pupils of considerable maturity of mind, and who possess a good knowledge of arithmetic ; but in the case of learners generally, and especially those who are young, the intellectual exercises should be thoroughly studied. Lesson 14. (33) Let x= the 1st, then g= the 2d, and g+2a;= (he 3d. Then, a;+|+|+2x=44, 2x adding, ^^+"3 =44, 11a; -^=44, X 3=4, ie=ia. the numbers are 12, 4, and 28. KEY TO PART FIRST. X (34) Let c= the distance from A -.o B; then ^= (list, from fl tH C; and ar-f-r dist. from A to C 2^ Whence, 2x-\--r= dist. from C to D; and, a;-|--+2a;4-g-=72, 3x or. 3ic+-g-=72, 18a; -^=72, o ' 5=4, a;=20, the distance from A to jB. a; r=4, the distance from i? to C. 2a; 2x-l-"r=48, the distance from C to D. rsS) Lot a:=the number. Then, x+|-|-j+26=5x, 3a; or, a;+Y4-26=5a;, 7x 20a; 13a: 26=—, a; 2=4' (36) Let x=len(Tth of the body, then g-r6= length of tail Then, as=|-f6+6. a; -=12 ' 2 a;=24, the length of the body ; 2+6=18, the length of the tail ; and, 6+24+18=48, the lengtli of the whole fish. MULTIPI, TnATTOnj (37) Let x= his age. then, a;+-4-g+28=3a;, or, a;+-g4-28=3a;, 11a; 18a; 7a; 28=-^, a; 4:=^, and a;=24, his age ADDITION AND SUBTRACTION. Remark. Pupils sometimes experience difficulty from not attending to tlie definition of similar quantities ; for e.\ample, by regarding sucii terms as %L^h and 3a62 as similar. By attending to tliis point, and beijig careful to write sirnilar terms under each other, no difficulty need b^. experienced in solving all the questions, in either addition or subtraction MULTIPLICATION. Remarks. In algebraic multiplication, there are a few things which, although they affect no principle, are of sufficient importance to claim the pupil's attention. 1st. In multiplying two monomials together, it is customary to write the sign first, then the numeral coefficient, and then the letters of the product from the left toward the right. Thus, in finding the product of — 2a2 by -|-3ac, we first write the sign of the product — , then 6, then o', and then c, making the whole — &a^c. This is more convenient thau writing the letters in the reverse order, because it corresponds to the manner of writing words. 2d. In multiplying by a polynomial, it is customary to multiply, /5rat by the left term of the multiplier, next by the second term from the left, and so on. Although the result would evidently be the same if the operation were performed in a reverse order, yet this method is now so well established, that a different one would be regarded as unscholarly. When a pupil understands addition of idgebraio quantities, ami how to multiply one monomial by another, he can encounter no real difiiculty ill performing any of the operations in multiplication. It is not, tlisrs- fore, deemed necessary to Insert the work of any of the exair.ples. KEi* TO PART FIRST DIVISION, Remarks. For reasons similar to tliose given under mu liplicatjon, il is customary, in dividing one monomial by another, to write, iirst tlie s.gn of the quotient, then the numeral coeiBcient, if any, and then the literal part from left to right. lu dividing one polynomial by another, in order to conform to the general method of proceeding from the left toward the right, it is cus- tomary to divide the first term of the dividend by the first term of thfa divisor ; this, however, affects no principle, as the division may be com menced at the right hand, by dividing the last term of the dividend b) the last term of the divisor. The divisor may be written either on the left or the right of the divi- dend ; the latter is the French method, and is more convenient, because the quotient being written beneath, the quantities to be multiplied together before making each subtraction, are the most conveniently situated with regard to each other. We here present the operation of a few of the more difficult pjKimples, with which, and similar ones, pupils sometimes find difficulty. Article 79. Note. The terms in examples 12 and 17, require lu uo arrange ' which, the operations present no difficulty. (14) 6a^x — laV Ans. 6a^x — 9a-x--|-3ax' (15) (18) ix*- 4x*- -64 |2a; (2i —4 34-4a;2_j_8a;+16 Ans. Sx'—l Ir' 16j;2- -33j; 3;i.r— 64 a;4 — x'(/ {x^-\-x^y-\-\))^-\-}f* -\-x^y Ans. F A T O R I iV a . (24) 3(fi~3xY+ZxY—y' \x^—3x^y+3m/^—y^ -\-3x^y~6xY+ x'y' +3a;y +3xY—Sxy+S^Y _(_a;3y3_3j,y_j.3^5_^« FACTORING. Kemakks, In solving the examples in factoring at the blackboard, the pupil should always explain why the given quantity can be separated Into factors. Thus, ia^ri — 9bh/e=(2ax2-[-3bjf){2ax''—3by'), because it is the difference of the squares of two monomials, 2aar^ and 3by^. Again, x^-^-l can be separated into two factors, because it is the sum of the odd powers of two quantities x and 1, (Art. 94. 5th) ; and one of the factors is x-[-l. It is shown in Art. 215, that the direct method of resolving a qaadratic trinomial into its factors, is to place it equal to zero, and then find the roots of the equation ; yet as the indirect method explained in Art. 95, presents no difficulty to an intelligent pupil, and is much shorter than the direct method, it should always be taught. Let it be kept distinctly before the mind of the pupil, that the whole difficulty consists in finding two numbers whose sum is equal to the coefficient of the second term, and whose product is equal to the third term. Thus, in example 1, " What two numbers are those whose sum is 5, and product 6 ? " Any Intelligent pupil will soon discover that 2 and 3 are the numbers required. We here present the solution of the examples in Article 95. (2) a2+7a-f-12=(a+3)(a+4) ; because +3+4=7, and 3X4=12. (3> x^—5x+6=(x—2){x—3) ; because — 2 and — 3=— 5, and— 2X— 3=+6. (4) a;2 — 9x-\-20=(x — i){x — 5) ; because — 4 and —5= — 9, and— 4X— 5=+20. (5) x^+x—6={x-{-3){x—2); because — 2+3=+l, and —2X3=— 6. (6) x' — X — 6=(x — 3)(j;+2) ; because — 3+2= - 1, and —3X2=— 6. t; KEYIOPARTFIRST. (7) a;'-t-x-2=(a;+2)(a;— 1); because +2— 1=+1, anrf —1X2=— 2. (8) i2— I3x+40=(a;— 5)(a^— 8); because — 5 and — 8= — 13, and— 5X— 8=40. (9) x^—1x—8={x—8)(x-{-l); because —84-1=— 7, and —8X1=— 8. (10) a;'+7a:— 18=(a;— 8)(a;— 2) ; because — 2+9=+7, and —2X9=— 18. (U) a:'-^— 30=(.i;— 6)(a;-l-5) ; because — 6+5=— 1, and —6X5=- 3U. (12) 3x^+12x—16=3{!c''+4x—6)=3(x+5)(x—l.) (13) aV—9a^x-\-Ua^=a\x^—9x-{-U)=aX:t^^)ix—2). (14) 2abx^—lialx—60ah=2ab(x^—'7x—30)=2ah(x—10X3:+3). (15) 2x^—4x''—30x=2xix^—2x—16')=2xix—5)(x+3). Article 96. Note. In performing the operations on the slate or black-board, a line fuould be drawn across each canceled factor. We have not the means, s.icept in the case of figures, of representing this by type. Thus, in ' *].tample 4, following, a line should be drawn across each of the c's, anr) •jIso across (a — b) in the numerator and denominator .o^ (■^-3)(^^-l ) {x~3){x+l)(x -l) (2) J=l = (^=1) =(a^3)(x+l) —X- — 2a;— 3. ,„ (^'+1)(^'-1) i'J+ l){z-l )(,z+l) C3) -^^1)— = ^ =(^3+l)(^_l^ =2"— 2^+2— 1. ea^c— 12aic-t-6&=c ^c(a— Zi)(a— t) ^^-^^ 2^^=267 =~^(^^)~=^'^"~*)- (eax-\-9ay){ix^—iy '') 3g (2j:4-3j/)(23:-f3y )(3j— 3y) ^^^ 4x^+\2xy-\-9y^' ~ {2x-{-3i/X2x-f3y) =3a(2a^-3y). (j'— 5 j+ 6)(a^— 7.r4- 12) (x—2)(ix—3){x—3){x—i) (^) a;2_Gx+9 ■" (x— 3)(x--3)~' =(a;— 2)(.r— 4V GREAT LST COMMON DIVISOR. GREATEST COMMON DIVISOR. Note. All the examples for exorcise, Art. 11)6, may be solved by merely separating the quantities into tlieir factors, by tlie rules for fac- toring; Arts. 94 ; 95. But as the application of the direct rule for finding the great*:«t common divisor of two polynomials, is generally regarded by pupils as a difficult operation, we here present the solutions of all flia examples. Article 106. (5) 5a''--\-5ax^5a{a-\-x) By omitting the factor 5a, (see Note 2), and dividing a' — x^ by the other factor a-\-x, we find there is no remainder ; tliere- fore a-\-x is the ff. c. d. (6) a:'— a'.r=x(x2— 0=) a' — a^x (x a'x — a^ a\x — a) x^ — a' \ X — a g. c. d. x^ — ax {x-\-a ax — a' ax — a^ After dividing we find the first remainder contains a factor a'' not contained in x^ — a^, hence it is not a factor of the greatest common divisor, and should be omitted. See Note 3. (7] a:' — c'^x=x{x'^ — c^) x^-^-icx-^-c- I a;'— 2ca;+2c' 2c(a;-|-c) (1 I ir+o g. c. d. x'^-\-cx ( X — c — ex — c' — CX — c^ (8) a;2+5x+6 j;^+2j— 3 3x+9 3(x+3") a;2+2x— 3 a;2-l-3x [xy-2x— 3 — X — 3 — X — 3 10 MCYTOPARTPIRST. (9) 6a^+llax-\-3x^ ]^6a^-1ax—3x^ ea^+lax—Sx^ (. 1 4ox+6x' ; by factoring this wc got 2a;(2a+3x) Ga^-\-lax — 3x= | 2a-\-Sx By completing this division, we find there is no remainder hence, 2a-l-3x is the greatest common divisor. (10) a<— x" I n'+a'x— ax=— x5_ a^-\-a^x — a^x- — ax' ( a — x — (z'x-|-«'x'-j-ax-' — x'' — a'x — a^x^-\-ax^-{-x'' 2a^x- — 2x'; then by factoring 2xHa^—x^) a^-\-a-x — ax- — .r^ | a- — x' By completing the division, we find there is no remainder , hence, a^ — x^ is the greatest common divisor. (11) a'— a'x+Sax-'— 3x' \ a^—5ax-\-ix' o' — 5a^x-\-4ax- ( a-j-4x -|-4a^x — ax^ — 3x^ 4a^x — 2 0gx°+ 16x3 -\-19ax'' — 19x' ; by factoring this we get 19x2(a— x) By dividing a' — 5ax-|-4x= by a — x, we find there is no remain- der ; hence, a — x is the greatest common divisor. (12) a^x^—aY=a\x''—y'>) : x'-\-xY=xKx^+f)- By the principle of Note 3, neither of the factors a' or x', can form factors of the greatest common divisor ; then by dividing x^ — y* by x--\-y^, we find there is no remainder ; hence, the latter quantity is the required greatest common divisor. (13: «" — a'x* ( o^-j-a'x* 4-aV— x'3 a'- -x* 1 a^—x^ —aV ( a- aV — x* x'(a2— x2) +a'x'»-xi5 x'»(a5— x') a'— x' 1 a-— a? — ax^ ( a -\-ax- — x' x^(a — ^.r) By dividing a' — x' by a — x, we find there is no remainder; hence, a — x is the greatest conunon divisor sought. ALGEliKAIC FRACTIONS, 1) LEAST COMMON MULTIPLE. Nora. The pupil should be reminded, lliat the operation of finding the least common multiple in algebra, involves precisely the same priii ciples as in arithmetic. Ans. a — X 4a-(a — x) 6ax\a^—x') 2a 4a^ 6ax^(a+x) 2a 3x\a+x) (7) 2 &x\x—y) Sa^x'' ISam/' 2 4j:\x—y) 3a''x^ Qaxy'^ 3a 1x\x—y) Za^x^ Zaxy'^ X 2x\x—y) a'x^ xy- X '2x{x — y) a^x f 2{x-y) a? f 2X2X3aX^XJ:X2(j;— i/)Xa^X»/'=24a'"x'(a;— ^). Ans. (8; x—y \ X a? — x' a x(a-\-x')(a — x)a a a-\-x x^ a — X {a-\-x')x'^(a — x) x' ,,„, a^H-y'^— y= a {x-^+y-'){x^-y){x-y) a (12) l^y^l^jX\= ix-y)i.x-^) =aW+y-).A ex c' c^ c^ — x' ^^^-{-xYc — x) c'CcA-x) 06') cA- = X =-^ — ■ — -=-- — ■ — - ^ ' '^c — X c — x' c — 3! x-\-l (c — a;)(a;-|-l) a;-|-l Article 141. ~2b+ikr ■ ^"*' a;' — y^ x^ — y'^ 1 [jic-\-y'){x — i/) avf-j ^2^> ^3^~ ■=-(^ ~^'^=~Yar ^x^—xy ""' A^{ii^^^^~3i^ Article 142. a—b. a^—V _a.—h o^+aai+i^ (a-i)(a-f;,)( a+t) 2x2 J. 2x2 a-\-x 2x2(a-|-a;) * ' /v3_J_T'3 * flA—V /»3_t Tr.3'^ V a3_|_j.3 • a-|-x a'-|-x^ a; {a-\-x){a? — flx-fx').!" _ 2x a' — ox-j-.x-' 14 KEY TO PART FIRST Article 144. EESOLUTION OF FRACTIONS INTO SERIES. (2) 1 \l+x l-\-x (1 — x-\-x' &.C. X — X — x^ +x' 4) l+x I 1 — X 1—x ~ {l-\-2x-\-2x^+&c. +2.r -if-2x—2x^ +2x^ -]--2x^ —2x^ +2x^ (3) (6) ax 1 a — X X2 X^ a;3 x^ x^2 1 x+] 1+; 1 j; 1 1 X X-i EQUATIONS OF THE FIRST DEGREE. OR SIMPLE EQUATIONS. Remark. The only difficulty pupils will be likely to experience in wiving the examples in Articles 154 and 155, will be where a fraction whose numerator contains two or more terms, is preceded by the sign minus, as in example 10, Art 154, or in examples IG, 17, &c., Art. 155. This may be obviated by the pupil writing the numerator of the fractiou In a vinculum when the equation is cleared of fractions, and then pro- ceeding to pt^rform the operations indicated. It will thus be seen, that the efleot of the minus sign before a fraction, is, to change the sign of each SIMPLEEQUATIONS. 15 lerm of the Jiuinerator. (See Art. 132). For illustration, take example .0, Art. 155. a:-\--i X — 3 X — 1 Multiplying both sides by 12, to remove the denominators. 4(x+2)— 3(a;— 3)=12x— 24— 6(x— 1), or, 4a;-[-8— 3x+9=12x— 24— 6j;+6, by transposing, 4r — 3a; — 12x-\-Gx= — 24-)-6 — 8 — 9 ; by reducing, — 5x= — 35 ; x=T. QUESTIONS PRODUCING EQUATIONS OF THE FIKST DEGREE. Article 156. (16) Let a;=A's share, then 2j:=T3's, and x4-2a:=42. Whence x=14. (17) Let x= the first part, then 2x^ the second, and 3x=tne third, and x-|-2x-j-3x=48 ; from which x=:8. (18) Let x:=the first part, then 3x= the second, and 3xX2 =6x= the third part. Therefore, x+3x4-6x=60 ; from which x=6. (19) Let x= the number of each, then Ix or x= cost of the apples, 2x= " " lemons, and 5x= " " oranges. Therefore, x+2x+5x=56 ; from which x=:7. (20) Let x= cost of an apple, then 2x cost of a lemon. 5x= cost of all the apples, and 3X2x= cost of all the lemons. Therefore, 6x+6x=22 ; from which x=2. (21) Letx^ C's age, then 2x= B's age, and 4.x= A's age. Therefore, x+2x-|-4x=98 ; from which x=14. 16 KKY TO PART FIRST, 1,22) Lee x= A's cents, then 3x= If's, x-\-—=1x= C's, and %x-\-^x=hx= D's. Therefore, a;+3x+2ar+5x=44 ; from which x=4. (23) Let x= age of youngest, then 2a;== common difference of their ages, and 3x= age of second, 5a;= age of third, and 1x=i age of fourth. Therefore, x-|-3a;-|-5x+7x=48 ; from which a;=3. (25) Let bx and Ix represent the numbers, since 6a; is to 7* as 5 to 7. Then 6a;+7a^=60 ; from which j;=5. Hence, 5a;=25, and 7a:=35. (26) Let 2a;, 3x, and ox represent the parts ; then 2x+3a:4-5a;=60 ; .from which a;=6. Hence, 2x=12, 3j:=18, and 5x=30. (27) Let 3a;, 5x, 7a;, and 8a; represent the parts ; then 3a;-f 5a;+7a:+8a'=92 ; from wliich x=\. Hence, 3x=12, 5x^20, 7a;^28, and 8x=32. (28) Let 2a;, 'ix, and 5a; represent the parts ; these will evi- dently fulfill the second condition, since i of the first, \ of the second, and \ of the third, are each equal to :r Then 2a;+3a;+5x=60 : from which x=6. Hence, 2a;^12, 3a;=18, and 5a;=30. (29) Let x= the number. Then 1+1+^=65 ; from which x=fiO. Or, let 12x= the number; then fia;-|-4x+3a;=66 ; from which a;=5 ; and 12x==60, To avoid fractions, we choose 12x, because it is a multi- ple of 2, 3, and 4. SIMPLE EQUATIONS. 17 ; 30) Let a;= the number. X X Then 5-;^=4 ; from which x=10. By putting 35a; for the number, we may avcid tractions. (31) Let x= A'a age, then ^x= B's. Therefore, x-{-—^^=1& ; 5x 14x 19x by clearing of fractions, 19a;=76X5, by dividing by 19, x=4X5=20, A's age. 2x 3x (32) Let x= A's part, then ~= B's, and -=-=C'b. 2x 3x Therefore, a;+y+Y=88 ; from which a;=42. 3x 3 3x 9x (33) Let x= C's share, then — -=:B's, and ^ of — -:=— = A's. 9x 3x 20~'"T" from which a;=200. Therefore, — +— +a;=440 ; ' 20 ' 4 ' (34) Let 3x= distance from A to B, then 5x= distance from B to C; 3x 6x also, — -}-5a;=6x, and --=2a;= distance from C to D. o o Therefore, 3x-\-5x-\-2x=\2Q ; from which a;=12. Hence, 3x=36, and Sa;^=60. (35) Let 3x= capital 3x Then 3x — —=2a;= capital at close of 1st year; 2 ix 14ar 2x-l-r of 2a;=2a7-|--r-=-r-== cap. close 2nd year ; 14x 1 14a; 14a; 2a; 12x -fi 7 °f "^-=-7— T=^'= cap- 3d year. 12a; from which 3a;=1545. Therefore, -^-=1236 ; 18 KEYTOPARTFIRST. (36) Let a;= rent last year. 5x Then x+Y^g=l68 ; from which as=]60. (37) Let 1= the less part, then x-{-6= the greater. Therefore, x+a;-)-6=32 ; from which x=13. (38) Let x= votes of unsuccessful candidate, Then x-\-50= votes of successful candidate. Therefore, a;+a;+50=256 ; from which a;=103. (39) Let x= A's, then a;+100=B's, and a;4-100+270=x+370= C's. Therefore, x+a;+100+a;+370=l520 ; from which a;=350. (40) Let x=: number of women, then x-\-i= " men, and 2x-l-44-10=2a;-l-14= number of children. Therefore, a;+a:4-44-2a;-l-14=90 ; from which a;=18. (41) Let 1= number of yards cut oif, then X — 9= number of yards remaining. Therefore, x-\-x — 9=45 ; from which 1=27. (42) Let x= the number. Then 7x— 20=20— a; ; from which x^b. (43) Let^=each daughter's share, then 2x^ each son's snaie, 3x^ what all the daughters will receive ; ' ix= " both the sons will receive ; then 7i — 500= what the widow will receive. Therefore, 3x+4x+7j.' — 500=6500 ; from which a;^500. (44) Let x= the number of days, then 20a:= distance 1st travels, and 30a::= " 2nd travels. Therefore, 20x+30x=400 ; from which x=8. SIMPLE EQUATIONS. (45) Let »;:=: the number of hours. Then Sx-j-SO^ miles B travels, and bx= miles A travels. Therefore, 5a:=3x-|-30. from which x=\5. (46) Let x=: the number. Then 1+1-44=^-6; from which a;=60. (47) Let x= time past noon. Then 12 — x time to midnight. X X 2x 12 — X Therefore, ^+2+3+y=^6~~ ' clearing of fractions and reducing, we find 731=60 j 5 whence a=s hr. =50 min. o (48) Let x= one part, then 120 — x= the other. 120— a; 3 Therefore, — - — =1 J or ^ ; from which x=48. (49) Let x= the number. 7X-P3 Then -^—4=16; from which, x=5. (5b) Let x= tlie number. 5j^24 Then — g — +13=a; from which x=54. (51) Let 3x== A's capital, then 2x=:B'b. Then 3x — 100= A's after losing $100; 2x4-100= B's after gaining $100. 5 Therofore, 2x+100— ^(3x— 100)=134 ; from which x=262. Hence, 3x=786, and 2x=524. (52) Let x= his money. / 2x \ X Then X— (y+3j=g+7; from which x==75. ',i! KEY TO P A RT FI EST. (53) Let 5x= annual income of each. Then x= what A saves, and 4x= what he spends yearly i also, 4a;-|-25= what B spends yearly ; and 5x — (_4x-\-2b)=x — 25= what B saves yearlv Therefore, 5(a;— 25)=200 ; from which a:=65. Hence, 5x:=325. (54) Let x= the number of pounds. 2x Then ---flO= lbs. of nitre ; 2x 5^+1= lbs. of sulphur ; r — 17=:lbs. of charcoal ; Therefore, — +io+^+l+-— 17=*;. 2x X By omitting y+o on the right, and its equivalent x on the left, and reducing, 2x we find — — 6=0 ; whence 2x=6X23, and 1=3X23=69. (55) Let x= cost of harness, then 3x=cost of horse ; , 32j; , 32a; 4xX2f=-g-, therefore —-—19= cost of chaise. Z2x Hence, a;+3a;+—-— 19=245 ; 44.r transposing and reducing, — ~=264 ; whence 44x=264X3, and a;=6x3=18. (56) Let Zx and ix represent the number. Then 3i+4 : ix-\-i ■ : 5 ; 6, whence 6(3a;4-4)=5(4x4-4) ; from which x=2. Hence, 3a:=6, and 4x=8. (67) Let 2x and hx represent the numbers. Then 2a;— 2 : 5x— 2 : : 3 : 8, whence 8(2x — 2)=3(5x — 2) ; from which a-^lO. Hence, 2r=20, and 5as=50. SIMPLE F.QU » TI ON S. '21 (58) Let x= the number of years. Then 25+a; : 30+a; : : 8 : 9 ; whence 9(25+a;)=8(30-|-x) ; whence, by reducing, x=15. Again, let x= the number of years, since their ages were as 1 to 2. Then 25— x : 30— a; : : 1 : 2 ; whence 2(25 — a;)=30 — it, and, by reducing, x=20. (69) Let x= the number of hours. Then, since the first fills the cistern in IJ hours, it fills 1 3x — =1 of it in 1 hour, and in x hours it will fill — part of it. In like manner, the second pipe fills — =^j of the cis- 3x tern in 1 hour, and in x hours it will fill — part of it. Also, the third pipe fills i- in 1 hour, and in x hours will X fill r part of it. Sx 3x X Therefore, 'T-\-T7)-\-fr=^> or the whole of the cistern ; (60) Let x= the number of days. Then, since the first does it in seven days, he does If of X it in 1 day, and in x days, ^• X In like manner, the second does ^ in 1 day, and in x days, ^ X The third does ^ in 1 day, and in x days r. Therefore, 7+c+q=^) or the whole ; from which x=2 f J. (61) Let 3x= money, 3x 3x — Y=2a;- Then 2j;+50— ^(2a:+50)+37=I00 ; firom which a;==10. Hence, 3a;=30. 22 KEYTOPARTFIBSTP (62) Let 5x=: yearly salary ; 5x — f of 6x=3x ; 3x — I of 3a;=2a;. 2x Then 2x—-r=i20 ; from which x='15, and 5a;^375. (63) Let a;= value of suit of clothes. Then 80-l-a;= yearly wages ; 80-f-x _ / 80-j-a; ■ and —TV,- = monthly wages. / 80+a; \ Therefore, 7 I ~- I =a;+35, from v/hich a;=28. (64) Let x= days it will last the woman. Then, -= part the woman can drink in 1 day ; since both can drink it in 6 days, they can drink J o' it in 1 day ; since the man can drink it in 10 days, he can diink jij of it in 1 day. 1 Therefore, i—j'^=^ ; from which a;=15. (65) Let x= the distance in miles. X Then tt= hours in going from C to L ; X and rn= " "i going from L to C. Therefore, -^+^=25 ; from which a;=150. (66) Let x= what B lost ; then 2a;= what A lost. 240— 2a; Therefore, — =96— j; ; from wh'^na;=48. (67) Let x= the whole number of gallons. X Then r-j-SS^ gallons of wine ; SIMPLEEQUATIONS 23 X and - — 5= gallons of water. X X Therefore, a;=--l-25-|-r — 5. Whence x=\10, and -+25=85= gallons of wine ; and -z — 5=35= gallons of water. (C8J Let a:= less part, then 91 — a;= the greater part, and 91 — 2x= the difference of the parts. 91— X Therefore, §YZ.2x='^ ' from which a;=42. (69) By representing the four parts by x — 2, a;+2, {x, and 2jC| we at once fulfill the last four conditions. Therefore, a;— 2+a;+2-f^a;+2a;=72 ; by adding, 4^x^72, whence oc=\Q. Then x— 2=14 ; a;-|-2=18 ; \x=S ; and 2a;=3a. (70) Let x^ length of each piece. Then 3(a;—19)-}-a;— 17=142 ; from which x=54. (71) Let x^ the number of sheep. X Then — = acres ploughed ; X and -= acres of pasture. Therefore, —+-=161 ; from which x=46,0. (72) Let x= greater part, then 34 — x= less part, 18— (84— a;)=a^l6; Therefore, a;— 18 : a;— 16 : : 2 : 3. Whence 3(a;— 18)=2(a;— 16) ; from which a;==22. (73) Let x= the number of beggars. Then 3a; — 8=: his money ; also, 2a^+3= his money ; Therefore, 3a;— 8=2a:+3 ; from which a;=ll. iil KEYTOPARTFIRST. (74) To avoid fractions, let 16j;= the number of apples. Sx — 8= number distributed to the first ; 8x+8= number left ; 4x-(-4r— 8= number " " " second; 4a;-|-12= number left ; 2x4-6— 8= number " " " third; 2x-\-lA= number left ; a;_[-7— 8= number " " " fourth; a;-j-l5= number left. Therefore, a;+15=20 ; from which x=5, and 16x=80. The question may be solved in the same manner by let- ting x= the number of apples. (,75) Let x= number of days, in which B alone could reap it. 1 6 Then -^ part B could reap in 1 day, and -^= the part X X he could reap in 6 days. Since A can reap it in 20 days, he can reap ^^ in 1 day, and in 16 days, ^^. 6 Therefore, ^§-|--=l, the whole ; from_ which a;=30. (76) Let |x and fx represent the numbers. Then ^+6 : fx+o : : | : i- Whence Ki^+6)=f(|^+5) ; from which x=60 ; hence, ^x=30, and |a;=40. (77) Let x=: price of a bushel of barley. 4X+90 Then — r — =: price of a bushel of oats ; 4x-j-9o therefore, x-\-3 : — 5 — : : 8 : 5. Whence 5(x4-3)=8 ( ^^±^ ) ; from which as=45. (78) Let 2x== distance from A to B, then 3x= distance from C to D ; 2x 3x -^-\--^=2x:=3 times the distance from B to C ; S I 1\I P L H E Q U A T I O N S . * j therefore, y= distance from B to C. Hence, 2.r-|--5-+3^=34 ; from which x=6 ; hence, 2x=12 ; 3i=18 ; and -^=4. (79) Let x= the lbs. of rice, x+5 then —^z=^ the weight of the flour, since | of | is J, 2x-\-l5 and — r- — = the weight of the flour ; ' 3x4-15 \ a;-|-a 1 „ I — u.. -<■ the water. / 3X+15 \ x+3, i [ — 2 — +^ / ="2~= weight of 3a;+15 a;+3 Therefore, x+^ — 4—^ = 15 ; from which x=2. Article 161. fcwlfSTIONS PRODUCING EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. \o) Lot x= the price of a lb. of coffee, and y=the price of a lb. of sugar. Then 5a;+3^=79, (I), and 3r+5y=73, (2) ; from which .r=:l 1, and y=8. » By subtracting Eq. (2) from (i) we have 1 I_ Z 2=^i'i — r'5=6^(i) (4) ; by adding equations (4) and (3) together 2 or y=30, by clearing of fractions. By subtracting equation (4) from (3) 2 2 = z'ff— 6'ff=3'iT' (6). or 2=60, by clearing of fractions. The value of x may be found by substituting the value of y m fiq. (1), or by subtracting Eq. (3) from (2), and adding the result, ing equation and Eq. (1) together. Note. For another method of finding the values of the unknown ouantities after the equations are formed, see Art. 169, Example 1st. (18) Let ic= digit in hundred's place, y= digit in ten's place, and z= digit in unit's place. Then 100.j;+10!/+z= the number, lOOx-i-lQy-i-z also j— ^-,-^—=19, (1), x-\-y-\-z-\-9 ^ ■" x4-z y=^' (2). and 100x+10!/-|-z+198=1002-|-10y+a;, (3) , from wV ich x=:4, !/=5, and 2=6. (19) Let a;= bushels of barley, y^ bushels of rye, and 2= bushels of wheat. Then x+y-\-z=100, (1), 28a;+36!/+482=4000, (2), 28a;+36X2y+4S(24-10)==40(100-f-y+10), by reducing, 28j;+32s/-i-482=3920, (3) ; from which a;=28, y=20, and 2^62. (20) Let X, y, and 2 represent the birds respectively which A, B, and C killed. X y z y+^ y z X — y — z 2y 22, after Isl division ; x—f — z X — y — z-1-22 22 2x — 2y — 22 3y—x — 2 42, after 2d division ; 2x — 2y — 22 3y — x — z x-\-y — 32 Ax — iy — iz, Qy — 2x — 2z, Tz — x — y, after 3d division. 36 KE Y TO PA RT FI RST. Tlierefore a;-|-y-t-2=96, (1), 4x—4y— 42=32, (2), 6y— 2a;— 22=32, (3) ; from which x=52, y=28, and 2=16 By reversing the operation, as in the solution of example 41 page 31, this qutstion is easily solved by arithmetic thus : at close of 3d division ; at close of 2d division ; at close of 1st division ; previous to the 1st division. Since each had an equal number after the third division, there- fore, each must have had J, of 96, which is 32. And since, in making the third division C gave to A and B as many as they had, it is evident that before the third division, that is, after the second division, A and B must each have had ^ of 32, which is 16, and C 32, and what A and B received at the third division, making in all 64. By reasoning in a similar manner we find what each had previous to the other divisions. A B C. 32 32 32, 16 16 32 16 16 64 8 40 3Q 8 56 32 44 28 16 52 28 16 GENERALIZATION. Article 170. GENERAL PROBLEMS. (1) Let a;= one of the parts, then a — a: will be the other. Therefore, x=n(a — x)=na — nx, transposing, nx-\-x=na, factoring, (n-\-l)x=7ia, dividing, ^=^i ; na na-\-a — na a a — x=^a — — r~r= — "-. =— . — . n+l n+1 n-\-l (2) Let x= one of the parts, then a — x will be the other. Therefore, mx=n{a — x)=na — nx, transposing, mx-\-n-x=na, GENERALIZATION. 37 factoring, (m-\-7i)x=zna, dividing, x=: — i — ; ''' m-\-n ' ma-\-na na ma m-\-n m-\-n m-\-n {¥) Let a;= one part, then a — x will be the other. Then mx-\-n{a — x)^b . trms. and fact'ng, (m — n)x=b — na, b — na dividing, x-. m — n ma — 710 b — na ma — b a — x= — n m- -ra m — n (.4) Let x=: the number. X X Then -+-=a, m ' n nx-\-mx=mna, by clearing of fractions, mna whence x= 'm-\-n' (5) Let jr=the first part; then ma;= the second; and nx= the third part. Therefore, x-]-mx-\-nx=a, factoring, il-\-m-\-n)x=a, a "•""l+m+ra • ma l-Jf-m-^n ' na "■^-l+m+re (6^ Let x=^ one part, then a — x= the other. X , a — X c cx-\-ab — bx=bcd, by clearing of fractions , ex — bx=bcd — ab, by transposing, or, bx — cx=ab — bed, by changing the sijrns of all the terms on both sides, (i — c)x=h{a — cdj, b{a—cd) b — c a{b — c) b(a — cd) c(bd — a) b—c b — c b — c 35 K E Y T O P A R T F - R S r . (7) Let xz= the number. Tlien a-\-x : b-\-x : : m : n, whence n{a-\-x)=m{b-\-x) ; transposing, nx — mx^mh — na, ■Ml — na n — m (8) Let a?= the number. Then a — x : h — x : : m : n, whence n{a — x)=m{b — x). transposing, mx — nx=mb — na, mb — na na — mb x= , or . £.fleAtt. 133 m — n n — m (9) Let x= the number. Tlien a-\-x b — x : : m : n, whence n(a-{-x)^=m(b — x), transposing, mx-\-nx=rab — na, 'Itch — na m-\-n (10) Let 1:= the number of dollars he had at first. 1 1 Then x — —x — -a;=a, m n mnx — nx — mx=mna, by clearing of fractions ; {mn — m — n)x=mna, mna mn — m — n (11) Let a;^ the quantity. m p Then —x — - a:=a, n q mqx — npx=anq, by clearing of fractions ; anq mq — np (12) Le' x= the number of persons. Then ax= the number of cents paid ; also (x — b)c= the number of cents paid ; therefore, {x — b)c=ax ; ex — ax=bc, _hc_ G E N E in A li I Z A T 1 O N 39 (13) Let a:= the number of persons. Then ax-\-b= the number of cents the person had , also cx-\-d= the number of cents the perso.ti had ; Therefore ax-\-h=cx-\-d ; (a — c)x=d — b, by transposing ; d—b x= . (14) Let a;=the number of bushels of oats, then n — a:=tn» number of busliels of rye. Then ax= cost of x bushels at a cents per bushel ; (ra — x)b= cost of n — x bushels at b cents per bushel; therefore ax-\-(n — x)b=nc, (a — b)x=nc — nh^n(c — b) , n{c — J) ''=~ci:—b~' n{a — i) n(c — b) n{a — c) a — h a — h a — b (15) Let x= the money he had in his purse. a;-(-a:=:2a;,then2j; — a= money he had after 1st spending ; 2x — a-\-(2x — a) — a^4x — 3a= money after 2d spnd'g ; 4x — 3a-|-(4x — 3a) — a=8x — 1a= money after 3d spnd'g ; Sx — 7a-l-(8x — 7a) — a=16x — 15a= money after 4th " Therefore I6.1— I5a=0, 16.r=I5a, x={la. (16) Let x=: number of pieces of 1st kind, then c — x= num- ber of pieces of .second kind. Since o pieces of the first kind make 1 dollar, or 100 cts. 100 ^ , , therefore = value in cents of a piece of the first kind. a '^ 100 In like manner -j—= value in cents of a piece of *h9 second. 100 x= value in cents of x pieces of first kind. — T-(f — j;)= value in cents of (c — x) pieces of the second kind. 40 K K Y T O PAR '1' FIRST 100 100 Therefore x'+-r-(c— x)=100 ; -_|--^ — :=1, by dividing both sidea by 100; bx-\-ac — ax=ah, by clearing of fractions. (J — a)x=za(_h — c) ; x=—, . I) — a c(b — a) a{b — c) i(c — a) i— a b — a b — a To illustrate this question by numbers, take the rollowing : How many 5 and 25 cent pieces must be taken, so that 8 shall make a dollar ! Arts. 5 five cent pieces, and 3 twenty-five cent pieces. Article 171. (11) Let x= the less number ; then x-\-2= the greater. Therefore x{£+-2)=x^+8, whence x=4, and j;-|-2=6. (12) Let x^the greater part, then a — x:= the less. Therefore x' — (a — xy=c a^-\-c a^ — r. whence x=—z — , and a — x^- 2a ' 2a ' (13) Let x=: number of pages, and y=: number of lines on a page ; then xy^ number of lines in the book. Therefore ('a;+5)(j/+10)=X2/+450, (1), and (x— 10)(y— 5)=.'n/— 450, (2) ; from which we find x=20, and ^^40. Article 172. KEGATIVE SOLUTIOHS. Enunciation of questions 2, 3, 4, and 5, so that the results will be true in an arithmetical sense. What number must be added to 20, that the sum may be 25 ! Ans. 5. RADICALS OF THE SECOND DEGRF, E 41 3. What number must be subtracted from 11, tbat the remain- der being multiplied by 5, the product shall be 40 ! A?is. 3 4. What number is that, of which the | is less than the the J by 3? Ans. 36. 6 A father, whose age is 46 years, has a son \ged 15 ; hmo many years sirice, was the son ^ as old as his father ] Ans. '-). RADICALS OP THE SECOND DEGREE. Note. All the e.xamples in the formation of powers, and extraction OF the square root, being performed by direct, straightforward metliods of operation, can present but few difficulties, if any, to the careful stu- dent. In the examples Art. 196, before commencing the operation the pupil must be careful to arrange the terms of the polynomial with refer- ence to a certain letter. Article 199. KEDUCTION OF RADICALS OF THE SECOND DEGREE. (1) 78a2=^4ffl=X2=V4o=Xs/2=2aV2. (2) J12a'=Jia^XSa-—^4a^Xjia=2aj3o,. (3) ^16a'b=J16a^X a.b= ^l6a^Xj a6=4a J ah. (4) V ISa" J3c3= J9a^b^c^x 2bc= J 9a'bV X-J^bc =3a-bcj2bc. (5) ^20aVc^=^ia^^c'X&abc=J'ia^bVX ^5abc =iabc^oabc. Remark. It is not necessary tbat the second step of the operation pbould always be written down, as in tlie preceding solutions; it should be done, however, by the pupil, on the slate or blackboard, until the princi- ples are well understood (6) 3v'24aV=3.,y4a''c2x6=3X2a=cV6=6a%v'6. 4 42 KEY TO PART FIRST. (7) 4J27aV=4;^9a2c2x3ac=4X3aCv'3a;£ ;:=12ac^3a J18=3J2 J2'7=3JS ^80=4^5 Sum =5^2 Sum=5j3 Sum=6J5 (.6) J24=2J6 (7) V8=2J2 (8) ^40=2^10 v^ 150=5^ 6 ^32=4^2 ^90=3^ 10 Sum =7^6 ^50=5^2 ^250=5^ 10 Sum = llV2 Sum =107)0 (9) j28aV,'=2ahJ7 (10) ^Toa^-c=-^5aj3c (11) JJ=^J3 Sum=Qahjn Bum=VMj3c Sum=j«5V3 (12) J^=i75 (13) Jl=lJ2 (14) 2Vf=73 J t%=\J^ J8=2J2 3JV2 =6J3 Sum=J|V5 Sum=2^V2 Sum=7j3 (15) lJi=iJ2 (16) 3V!=n/6 (17) j48aVx=4acj3x Sum=V2 Sum=?lJ6 Sum=(4ac-l-2i)V3c (18) V(2a'— 4a=c+2ac2)=VC«-— 2«c+c^)X2a=(a— c)V2a J (2a^+4a^c+2ac^)=V(ffl^+2ae+c^)X2g =(a+c) J 2« Sum^2a;«y2a (19) Ja-\-x=Ja^x J ax'^-\-x'^^xJ a-\-x J (a+xy=(a.+x) J a+x Sum =^{l-\-a-\-'2x) J a-\-x Article 201. SUBTRACTION OF RADICALS OF THE SECOND DEGREE. (o) ^/]8=3v'2 (3) j46^=SaJ5 (4) J54h=3j6b V2=V2 J6a^=aj6 J6h= j6b Dif. =2J2 Dif. =2ajb Dif. =2^64 44 KEYiOPART FIRST. (5) Jn2aV=4acJY Dif. =2acJ1 (7) V36^=6aV« Dif. =ia'Ja' Dif. =2bcjah (6) J'21b'c==SbcJ3bc JV2bV=2bcj3bc I)lL=bcj3bc (8) JidabV-^'lbcJab J25a/j'c^=5bcjab (.9) Vl60a'S'c=4ffiJJ10aJc (10) 6aV27=15aV3 J \Oa ^b^c=ab J Wabc. 3a ^48= 120^3 Dif. =3a6VlOaic Dif.=3a^3 (U) 1Jl=JZ (12) Vi=W30 (13) .^12=2^3 3Vi=V3 x/M= W30 x/i =W3 Dif. =0 Dif. =1)5^30 Dif. =|V3 (14) 3V1=|V2 (15) ^/i==W6 (16) jAa-^x=iaJx J2=J2 n/Zi^WS ajx^=axj x Dif.=i^2 Dif. =1^6 Dif. =(2(z— ar)Va: (17) ^3m'i-|-6?nrea:-|-3ra'j;=(m-|-M)\/3a; ^ 3m'^x — 6mnx-{- 3n-x:=(m — n) ^ 3x Jjil.=2nj3x . Article 202. MULTIPLICATION OF RADICALS OF THE SECOND DE G K E E. (3) V8Xj2=v'l6=4. (4) 2jaX3^a=6Ja'-=6a. (5) V27XV3=J81=9. (6) 3^2X2^3=6^6. (71 3^3X2.^3=6^9=18. (8) V6X ^15=^90=3^ 10 (9) 2715X3^35=6^3X5X5X7=30^21. ( 1 0) Ja'b'cX J abc= Jli¥^=^aVc. (11) ^/iX^/3 = ^/i=^ RADICALS OF THE SECOND DEGREE. 45 (12) V|XVf=Vif=VFV5X5=AV5- a a a' I a^ 6a 3a (13) .2^-x3^--6^-=e^--X2=-^2=j^2. (14) 2+V2 fl5) 1+V2 2—72 1— V 2 4+2^2 1+^2 —2^2—2 — V2— 2 4—2=2. 1— 2=— 1. (16) Jx+2X^x—2=^ix+2)ix—2)=^x^—4:. (17) 7 B+a; X V a+^= ^ (a-\-x'){a-\-x) ^a-\-x. (18) ^ab+bxX J ab—bx= J {ab-\-bx)iab—bx)=^a^^—b^x^ (19) Va+2X V^+3= V(^+2)(x+3)=: V-<;=+5a;+6. (20) cja+d^b (21) 7+2^6 clja—dljb 9—5^6 c'^a^cd^ah 63+1876 —cdja b—d^b —3576— 10X6 c^a — d^fi. 3—1776. ^ " (22) Ja-^x-\-Ja—x (23) a;+27oa;+a Ja-\-x — 7*^ — ""^ •'^ — 'iJax-^-a (a-\-x)-\- J {a'' — x'^) x''-\-'2,x^ax-\-ax — 7C''^ — '^^) — ('^ — •'^) — ixjax — iax — 2a^ax {a-\-x') — (a — x)=2.r. -\-ax-\-2a J ax-\-ii' x^ — 2ax -^-a' (24) a;=— a:72+l a;^+a72+l xi—x^^2+x'' +x'j2—2x'+xJ2 +a°— j:7 2+1 «« ' +1. 46 KEY TO PAR T F 1 f ST. Article 203. DIVISION OF RADICALS OF THE SF, COND DEGREE. ^ ^ Jbi ,54 , 6^28 6 ,28 (4) 2-7"7-=2-^/T=V4=«• 6V54_6 54_ JI6O 160 (6) (7) 15^37 8 15 ,378 378 / ^-— =3V63=3^9X7==9V7. ahja^b^ ah a}V a Ja^ a? a^i aba hd a 11 a .^ .d a c ac achd 1 (11) -Ji^'J-=^^X;^=Jq=>J^,j,=^^abcd. /TQi /^ /I /3 3 9 3 I ^IK^ 3 ,1 1 3 3 2 /1~T 6 /i 6 2 (16) 5^/3-2^^5=6XW3X3=5^/9X5=l-6^/5=5V5. rtADIOALS OF THE SECOND DEGREE. 47 1 1 1 h 11, 1 , j2+3/-=^2+3^lx 2=^2+1^2=1^2 4V2-r-2v2— 4X5V2— 20— 10- Article 204. J 1 2— V 3 2—^3 W 2+v'3~2+V3^2— V3- 4—3 -2— V^- ,^, _3 3_ 6+V3 3(6+^3) 1 W 6— V3~6— ^3^6+^3"" 36—3 — u'-^'rV^^ _ 5 . 5 V7— ye 5(^7-^ 6) =5(V7-V6). 8 8___ V5+V3 8(J5+V3) ^^^ 75-73-75-^3^ V5+n/3~ 5-3" =4(75+73). ^') ^5=75X^H^/^=5'^2-^^^°^'^+)=l-'^l'^°'+ 3 3^ 75+ 73 3(75+72) ■*^ 75^72"~75-72^ 75+72 3 =75+72=2.2360679+1. 4142135+=3.650281+. 72 72 ,75+73_7_10±76 (9) 75=-73-75— 73^75+7:J- 5-3 — V^10+76)=t(3.162277+2.449489+)=2.805833+ 48 KEY TO PA RT FIRST. Article 205. SIMPLE EQUATIONS CONTAINING EADICALS OF THE SECOND DEGREE. (1) ^(^+3)+3=7. Transposing, ^(x-{-S)=4: ; squaring, x-\-Z^l6 ; fronn which , a;=13. (2) a:+V(x2+ll)=ll. Transposing, ;y(x^-j-ll)=ll — x ; squaring, a;--|-n = 121 — 22x-\-x' ; from which x=£i. (3) V(6+Vx-1)=3. Squaring, G-\-^x — 1^9 transposing, ,Jx — 1=3 squaring, x — 1=9 from which x^lO (4) iJx{a-\-x)=a — x. Squaring, x{a-\-x)=a'^ — ■2ax-\-x^. reducing Zax^a^ ; a fiom which x=-. (5) Vx— 2=V(^— 8). Squaring, x — i^x-\-i^=x — 8 ; reducing, — 4^x= — 12 ; dividing, ^a;=3 ; squaring, x:=9. (6) x+^x^—^=z^. Transposing, ^x- — 7=7 — x ; squaring, x'' — 7=49 — 14a;-l-ar' J from which x^i. 17) 2+^Sx=^5x+4. Squaring, 4+4^3x-|-3x=5i'-|-4 ; reducing, 4^'3x=2x; dividing, 2^3x=a;; squaring, 4X3a:=a;2; from which .x'^12. SIMl'LE EQUiTl'INS CONTAINING RADICA;3.-i9 (8) ji+7=6— V^^. Squaring, i-(-7=-?6 — \2jx — o+x — 6i transposing and reducing, Jx — 5=2 ; squaring, x — 5^4 ; from wliich x=9. (9) ^x — a=Jx — -^a. Squaring, x — a=^x — ^ax-\--a ; — 5a transposing, and reducing, ^ax=— ; squaring, ar=— - ; 25a whence a;^— -. Id CIO) 7a;-|-235—V'r— 424— 11=0. Transposing, V'^+2i^f' = ll + ^/ic— 424 ; sqiia-:rg, ic+225=121+227x— 424+J.'~ 424 ; reducing, 528=22^1—424; dividing, 'H=^x — 424 ; squaring, 576=1 — 424; from which a;=1000. (11) a;+^2ffi^+a;==a!. Transposing, ^2ax-]-x^=a — x ; squaring, •2ax-\-x^^a^ — 2ax-^x^ ; reducing, 4aa:=a' ; whence x=-:a. 4 (12) ^^-{-a — Jx — a=^Ja. Trt.r.sposing, ^ x-\-a= ^ a-\- ^ x — a; squaring, x-\-a:=a-\-2 ^ ax — a^-\-{x — a); reducing, a=2^ax — a'; squaring, a^=4(ax — a^) ; 5a whence a;=-r. 4 5 50 K K y ]■ O P A R T F J K S 1 . (13) v'^+12=2+Vx. Sauarinsr, x-\-y2—4->r4 ^ x-\-x ; [■•aucing, 2^yjx, squaring, a;=4. (14) j8-\-x=2^1+x—Jx. Squaring, 8-{-x=4{l-lrX)—4 ^ x+x'+3 , transposing, and reducing 4^x-\-x''=4(x — 1) ; dividinir, iJx-\-x-=x — 1 ; squaring, x-\-x^=x^ — 2x-\- 1 ; reducing, 3x=l ; whence a:=x. 12 £15) ^5x+-j=^^=Jbx+6. Multiply by Jdx^, V25a:2+30.i;+]2=5i+6 ; transposing, ;^25x--l-30a:=5a; — 6 ; squaring, '2bx^-\-30x=25x^ — 60a;-(-36 :, reducing, 90.r=:36, whence, ^=k' 237— 10a; (16) ^,_4=-^^^-. Multiplying by 4-\-,Jx,x — 16=237 — lOx; transposing, llj;^253; dividing, a;^23. V^^+V (17) %/a;24-\/4c2-[-a;+^9a;=+12a-=J+a;. Squaring, x^-\-'44x^-\-x+j9x^-{-V2x=l+2x-\-x^ ; omitting x^ on each side, and squaring again, we have 4x^-\-x-{-^9x^^ V2x=l~\-4x+4x^ ; reducing, ^9x'-+]2x= 1 +3x ; squaring, 9j:'-i-12x=l-f6x-(-9x= j reducing, 6a:=;l, and ic=- 6 KQUATIONS OF THE SECOND B r. r. n F. F. 51 1,18) ^a-\-^ax=^a — ^a — ^ax. Squaring, a-\-Jax=a — 2\/ a' — asJoix-\-\a — ^ax) ; transposing and reducing, 2%/a^ — a^ax—a — i^ax ; squaring, 4(o' — a^ax)=a^ — ia Jax-\- icu- ; reducing, 3a'=iax, 3a whence x= — . 4 (19) b(^x+^h)=aUx-Jb). Transposing, a ^ x—b J x=a ^ h-\-b ^ b ; factoring, (a — h) ^x={a-\-b)Jb ; squaring, (a — i)2x=(a+'^)'A ; whence x^-. .^^ . (o— i)2 (20) ^x-\- Jax=a—\ . Factoring, ;,y.r(I4-^a)=a— l=(^a-|-l)(^o— 1; , dividing both side? by \-\-^a, and observing tliat \-\-iJa is the same as ^a-\-\, squaring, x^^a — 1)^. Article 211. ;inESTIONS PRODUCING INCOMPLETF EQUA- TIONS OF THE SECOND DEGEKK (2) Let a;= the number. Then |x|=108 ; x^ whence t^==108 ; a;==1296 ; a;=36. (3) Let ,r. — the number. x^ Then a;'— 16=^+16 ; a;= whence — 32, and x — S. 52 K K Y T O 1' A II T FIRST (4) Let ar=llie niiiiiber. Tlier a;5~5-l=(|) +54; a;2— 54=j+54; transposing, —-=108 ; from whic (8) Let x= the number. Then g=- ; multiply by 9i to remove the denominators, a:==9X16 ; x=3X4=I2. (7) Let 3a; and Ax represent the numbers. Then 16^;'— 9^2=63 ; from which 7x-^G3, and x=3. Hence, 3x=9, and 4x=12. (8) Let 3x and 4-r represent the numbers. Then 9x2+16x==100 ; from which a;^^4, and x^2. Hence, 3,r=6, and 4x-=8. (9) Let a;=the number. Then (a;+3)(a:— 3)=40 ; a;2— 9=40 ; from which x'^49, and a:=7. (10) Let 5x= the breadth and 9x=: the length Then 5x'X9x=i5x^= the number of square I'eut ; therefore, 45.t-=1620 ; from which x-=36, and ■r=6. Hence, 5a;=30, and 9x^54. (11) Let 10j:= the number of acres in the farm ; then r= thu cost per acre in dollars. Therefore, 10a,X-'c=10x==1000, the cost of the farm ; trom which x^=100, and x=10, the cost per acre. Hence, 10.r=100, the number of acres. ^12) By placing 10x= their sum, wc have the greater=7j:, since their sum is to the greater as 10 to 7. And if FCiUAT<0,\S OF THE SKCOND D K (; 11 E B. S3 10j;= Uieir sum, and 7j;= the greater, the lcss=10a>- Therefore, 10a:X3j;=30.r^=270 ; from which x-'^9, and x=3. Hence, 7x=21, and 3x=9. \13) Let 2j:=: their difference, then 9j: will be the greater, tad 9>r — 2x=7x, will be the less. Therefore, (9x)2— (7a;)-'=128 ; 81x2—49x2=128 ; from which x-=-A, and x^2. Hence, 9x=]8, and 7x=14. - t.14) Let x= the cost of an orange, then 3x^ the number ol oranges. Then ' 3xX^=3x==48 ; from which x=4, the cost of an orange ; and 3x=12, the numbei- of oranges. (15) Let 4x= the cost of I yard in cents, then 9x= the nam bor of yards. Then 9xX4x=36x2=324 ; from which x^=9, and x^3. Hence, 4x=12, the cost per yard ; and 9x=27, the number of yards. (16) Let ^x and |x represent the numbers. Then ix-+lx'^=225 ; multiplying by 4X9i to remove fractions, we have 9x2-|-16x-'=225X4X9 ; 25x==226X4X9 ; dividing x2=9X4X9; extracting the sq. root, x=3X2X 3=18. Hence, ix=9, and |x^l2. We may avoid fractions by representing the numbers by 3x and >x, as recommended in the book. (17) By reducing i, f, and J-, to a common denominator, vg find they are to each otlier as 6, 8, and 9 ; therefore, let the three numbers be represented by Gx, 3x, and 9x. Then 36x=-l-64x=+81x2=724 ; adding, 181x2=724; from which ,r-'=4, and x=2. Hence, 6x=12, 8x=16, and 9x=18. B4 KEYTOPARTFIRST. (18) Let 4a=the price of a yard, then 5a;= the number ol yards. Then 20x^= whole cost ; and — ^ cost of a yard if he had received 45 cents more for the same piece. ^, L>0,c=-f45 Tlicrcforo, r : 5a; : : 5 : 4 ox 80:c=+I80 whence, =25x ; multijjlyin;? by 5x, 80a:--|-I80=]2oa;' ; from which x'^4, and x=2. Hence, 4x=8, and 5x=10. Article 212. COMPLETE EQUATIONS OF THE SECOND DEGREE. (50) 2ax—x^=~2ab—h^. x^ — 2ax=2ali-{-b^, by changing the signs , X- — 2ax-\-a^=a^-\-2ab-\-b', by completing the square ; X — a=±(a-|-t), by extracting the sq. root ; transposing, x=^.aziz{a-\-b)^2a-\-b, or — b. (51) x'—2ax=b^—a\ x' — 2ax-|-a-=i-, by completing the square ; X — 0=^=/', by extracting the square root; x==:a±b:=a-\-b, or a — b. (52) x'+3fcx— 4/)==0. x'-'-\-Sbx=Ab- , by transposing ; 9/,2 9/i- 25i^ x=-|-36x+— -=4^=4— ^-=— -, by completing the sq. ; 3b 5i ^+■^=±^7, by extracting tlie sq\iare root ; 3b bb •»=— "^i^=- \-b, or — 4i. (53) X- — ax — /)x= — ab. x^ — (a-|-i)x= — ab, by factoring ; * — (a+i)x+ , — = — —ab= — ; 4 4 4 EQUATIONS OF THE SECOND DEGREE 55 a-\-b a — 6 a-\-b a — b a;=-^±-Y-=+a. or+b. X b * ^ x-\-a X — b x^ — bx=bx-\-ab, by clearing of fractions ; x^ — 2bx=ab ; x—b=dzjab+b' ; x=bzizjab+b^. (66) 2bx^+(a—2b)x=a. a — 26 a a— 2b {a~2by (a—2by a a'+4a.')— 4f' ^'+■"2*""^+ \m "^ 1662 "+^= f^ij ; a— 26_ a+26 a— 2b a+2b a ''=-^b-^-^-=^'°'-2b- \ 2b ) ' x^ X 2a' (56) :j-T=Ti- x^ — -j-x^^-j-^-, by multiplying by a 8a'> Da" "462~4J2"T"462'~4i2 a' 3a- ''~2b""^W' ■ _a^ Zf 2a? a^ '^2b'^'2b^T' °'' ~T- (5") x'—(a—l)x—a=0. x^ — (a — l)x=a ; a:=-( (19) Let a;= the nuin icr in the company at first, 175 then = what each ought to have paid, 175 and -=: what tliose paid who remained. „. 175 175 Therefore =10. X — 2 X 175a; — 175x+350=10x' — 20a;, by clearing of frct'ns ; x' — 2a;=35, by reducing ; from which x=-\-T, or — 5. 100 (20) Leta;= the larger number ; then = the smaller. / 100 \ Therefore, (x— 1) ( — +1 j =120 ; ^^-=120 ; X ' 1 00x+a;2— 1 00— «= 1 20a; ; a:2— 21a;=100; from which a;=-l-25, or — 4 ; 100 and =+•*• or — 25. a;2— 4 (.21) Let a;2= the father's age ; then — ^ — = the son's age. 1 / a;2 — 4 \ Then-(- 1) = a;=— 4 — - — — l^2x, by multiplying by 2, x'^ — 4 — 3=6a;, by multiplying by 3, a;^ — Sx=7, by transposing ; from which a;=+7, or — 1 a;^— 4 Hence, a;^=49, and — ^ — =15. (22) Let x^= her age. Then •g-+a:=10 ; 20 from which a;=-l-4, or — ~r-. Hence, a;^=16, or 44|, the former of which satisfies the conditions of the question in its arithmetical sense. 60 KEY TO PART FIRST. (23) Let a:'= the number. 3a; Then a'—- t-=22 ; 22 from which x=-]-5, or — -7- ; Hence, ar'=25, or 19/;, the furmer of whicli satistiet the conditions of the questio.T in its arithmetics! senso. (24) Let Then and a;= the number of yards. 600 = cost per yard in cents, selling price per yard in cents. X 540 Therefore from which Hence, X— 15 600 540 a; ' x — 15 ' x=+15, or— 120. 600 —-=+8, or —5. X The negative values are the answer, in an arithmetical sense to the following question. A merchant bought a piece of muslin for 6 dollars : after adding to it 15 yards, he sold the whole for 5 dollars and 40 cents ; at which rate he received 1 cent a yard less than the piece cost him ; liow many yards did he buy, and at what price t (25) Let : cost, then T7j^= per cent of loss, x^ Therefore j: — r--=:24. whence a;=+60, or +40. Article 219. EQUATIONS OF THE SECOND DEGREE, COa- TAININO TWO UNKNOWN QUANTITIES. (6) x'+f=34, (1); a;2— !/2=i6, (2). By adding these equations together, and dividing by 2, wf fiad i(^=--25 ; from which x=+5, or — 6. EHiiATiONS OF THE SECOND de(;ri:f. 61 The value of y may be found either by substituting 25 instead '»'■ i^, or by subtracting the second equation from the first. (7) rJ-y=16, (1); a?y=63, (2). Th3 values of x and y are readily obtained by finding the value of either in terms of the other from equation (2), and substituting It in equation (1) ; or thus : x^-}-2j^-)-!/'=256, by squaring (1) ; ixy =252 by multiplying (2) by 4 ; X- — 2r(/-l-2/^=4, by subtracting ; X — y=2, by extracting the square root ; From these equations, by adding and subtracting, the values of ,r and y are readily found. (8) x—y=b, (1) ; a,T/=36, (2). From Eq. (1) y=x — 5 ; this being substituted instead of y in Eq. (2) gives x'^ — 5x=36, from which a: is readily found, and then y. Or, by squaring Eq. (!), then adding it to 4 times Eq. (2), and extracting the square root, we find x^y=--\\ ; from which, and Eq. (1), by adding and subtracting, the valups of a: and y are readily found. (9) x+y=9, (1) ; a:2+y==53, (2). From Eq. (1), y^S — x ; this being substituted instead of y ir Eq. (2) gives, after reducing, a;' — 9ar= — 14 ; from which we find ar=7, or 2 ; consequently 2/^=2, or 7. (10) x—y=b, (1) ; a;2+y2=73, (2). From Eq. (1) y=^x — 5 ; -this being substituted instead of y in Eq. (2), gives, after reducing, x'' — 5a;=24 ; from v/hich x is ibund =8, or — 3 ; hence y=3, or -t-8. (U) a;'+2/'=152, (1); x+y=B, (2). Dividing Eq. (1) by Eq. (2), we find x''—xy-\-jf=\^, (3), From Eq. (2) y=8 — x; substituting this value of y in Eq. (3; and reducing, we have x'' — 8x= — 15 ; from vvhicii we find ,r=^5) «r 3 ; hence !/=3, or 5. 62 K E Y TO PA RT FIRST. (12) x'— 1/=208, (]) ; x—y=i, (2). Dividing Eq. (1) by Eq. (2), we find x''-\-xy-\-y^=5'i, (3). From Eq. {T),y=x — 4; substituting this value of y in Eq. (3; and reducing, we find x' — 4j;=12 ; fi-om which x=Q, or — 2 : hencs v^2, or — 6. (13) a;3+y'=19(a;+y), (1); a;— !/=3, (2). By dividing both sides of Eq. (1) by x-\-yj a;2— x!/+y==l9, (3). From Eq. (2) y=x — 3 ; substituting this value of y in Eq. C3) and reducing, we find x- — 3x=10 ; fi-om which x^b, or — 2; hence 2/^=2, or — 5. (14) x+y=U, (1); x'^—y-—U, (2). From Eq. (1) 2/=ll — x ; this being substituted in Eq. (2) in- stead of y, and the equation reduced, gives 22i:=132, from which aj=6 ; hence !/=5. (15) (a;— 3)(y+2)=12, (1); xy =12, (2). Performing the operations indicated in Eq. (1) and then sub- tracting Eq. (2) from it, we find 2.r — Zy—Q, (3). From Eq. (3) 6+3y we find x= — r--, and this being substituted in Eq. (2), gives, after reducing, y'+2^=8; from which 2/^2, or — 4; hence x=Q, or — 3 (16) y-x=2, (1); 3x!/=10x+y, (2). From Eq. (1) y=x-\-1 ; this value of y being substituted in Eq. (2), gives, after reducing, Zx'' — 5,1=2 ; from which a-:=2, or — ^ ; lience, y=4, or 1|. <17) 3a;2-l-2.ry=24, (1) ; bx—iy=\, (2). 5a;— I , from Eq. (2) y= — - — ; this being substituted in Eq. (1), gives, 36 after reducing, 19x' — 20=72 ; from wliich x^i, or — r^ ; lieiico, 199 y=3,or— ^y. EQUATIONS OF THE SECOND DEGREE. C3 1 1 5 1 1 13 Let -=t), and -=2 ; then ti+2=-, fS) ; X y '6 ^ ' and v^+z^=^, (4). 5 - 5— 6v From Eq. (3) i^j — v^ — ^ — ; substituting this value instead of z in Eq. (4) and reducing, we find 1 1 6d^ — 5v= — 1 ; from which v=t, or- ; and substituting in the equation, 2=- — v, we find 2=5, or -. 1 1 1 Hence, 1)=-=-, or x ; from which x=2, or 3. 1 1 1 z=-=r, or p ; from which !/=3, or 2. (IQ) x-^j=2, (1). a;Y-'=21— 4ry, (2). In Eq. (2) let xy=:z, the equation then becomes 2^=21 — 42 ; from which 2 or xy^3, or — 7. We then have a; — y^=2 xy=3, to find x and y. These equations are similar to those in example 8, and we readily find x=3, or — 1 ; hence y^l, or — 3. From the equations x — y=2 and xy= — 7, we may also finfl two other values of x and y, but they are imaginary. Article 219. PEOBLEMS PKDDUCING EQUATIONS OF THJ SECOND DEGREE, CONTAINING TWO UNKNOWN QUANTITIES. (1) Let X and y represent the numbers^ Then x+y=lO, (1); x'-+y^=52, (2). Solved like question 9, preceding. 04 KEYTOPARTFIRST. (2) Let X and y represent the numbers. Then x—\j—-i, (1); a;2— ^/'=39, (2). Divide Eq. (2) by Eq. (1) and we get a:-|-y=13, (3) ; thea from this and Eq. (1), we readily find x^=.%, and i/=3. (3) Let x' and y' represent the parts. Then x'4-y'=25, (1) ; .r+y=7, (2). The values of x and y may now be found in the same manner US in question 9, preceding. (4) Let x= the digit in ten's place, and y= the digit in unit's place. Then 1 Ox-|-3/= the number ; (10j;+2/)(:r+y)=160, (1); Dividing Eq. (1) by Eq. (2) we get 10— u^ 4!/(a;-|-i/)=40 ; from which x= --. Substituting this instead of x^ in equation 2, we get / 10—2/2 \ clearing of fractions, 100 — \'iy^-\-y'^=\%y'^ ; from which y=2 ; hence x^3. (5) Let x= the greater number, and y= their difTerence, then X — y^= the less. Then a;y=16, (1) ; and xy— 2/2=12, (2). Subtracting the 2d equation from the 1st, we get y2=4 ; hence y=2 ; IG a;= — =8, and a; — y=6. (6) Let X and y represent the numbers. Then x+2/=io, (1); m/— (a:— ^)=22, (2). Find the value of either x or y from Eq. (1), and substitute il instead of Iho same unknown quantity, in Eq. (2). K Q U A T I O N S OF THE S 1" C O N D D F. O .{ E ti . 65 This may also be easily solved by means of one ui:knovvn ijuaii- (ity ; thus, let x= one of the parts, then 10 — x= tiie other, and 10— 2a;= their diiference. Then a:(10— x)— (10— 2x)=22 ; from which x:=8, or 4 ; hence 10 — j;=3, or 6. The numbers 4 and 6, satisfy the conditions of the qiiestion in an arithmetical sense. The numbers 8 and 2 satisfy the following problem. Divide 10 into two such parts, tliat tlieir product plus their difference, may be 22. (7) Let X and y represent the numbers. Then a;-l-!/=10, (1); 3;'4-y'=370, (2). For the method of solution, see question 11, preceding. (8) Let X and y represent the numbers. Then x—y=2, (1) ; a;'— 1/'=98, (2). For the method of solution, see question 12, preceding. (9) Let x^ the greater, and ?/=: the less of the two numbers. Then 6x+5(/=50, (1) ; xy=20, (2). 20 The value of y, from Eq. (2), is — ; this being substi tuted in Eq. (1) and reduced, gives 6x'^ — 50x^ — 100 ; from which , x^5, or 3^ ; hence, y=4, or 6. The first values of x and y satisfy the question ; the other two -■atisfy a question precisely similar, except that the words greater and less are transposed. (10) Let x= the digit in ten's place, and y= the digit in unit's place. Then 10x-\-y= the number. 10a;4-?/+27=10y+a;, (2). From the 2d equation y=x-\-3 ; this being substituted in tlia Ist equation, we get, after reducing, 2a;---5a;=3 ; from which we find 2=3 ; lience y=6, and the number is 36. 6 ^(i K E Y T O P A R T F I R S T . dl) Iiet X, y, and z represent the numbers. Then ^=a, (1) ; xz -=6. (2); f=c, (3). Multiply the three equations together, and we have xyz=:abc, (4). Divide this successively by each of the equations (l), (2)i and (3), and we obtain z-=bc, y'^=zac, and x^=ab. Hence, x=±Jdb, y=^riz,Jac, and z=^^hc. (12^ Let X and y represent the numbers. Then a:+y=9, (1) ; a;'+y'=2](a;4-y), (2). Divide both sides of Eq. (2) by x-\-y, and we get a;2— X2/+y2=21, (3). The value of y, from Eq. (1), is 9 — x ; this being substi- tuted in Eq. (3), we have, after reducing, x'^ — 9x=: — 20, from which x^b, or 4 ; hence y=4, or 6. (13) Let x-\-y, and x — y represent the numbers. Then the sum of their squares =2x'-|-2j/' ; the difference of their squares =4i^; and their product =a:' — y''. Therefore 2a;2-f-2y2— 2(a;=— j/2)=4, (1) , ixy—kW^')=i, (2). Reducing Eq. (1) we readily find y=l ; this value being sub- Eti*uted in Eq. (2), we have, after reduction, x^ — &x= — 7 , from which x^7. Hence, a;-l-y=8 ; and x — ^!/=6. (14) Let x= the circumference of the less wheel, and j/= tlm circumference of greater. 120 120 Then --=--+6, (1) ; 120 120 120y=120a;-|-6m/, by clearing Eq. (1) of fractions ; .+ I=^-.H^+4' (2) E C[ L A TI O N S OF THE SECOND D E O R E B . 07 V20y-{-'l20='[2Qu:-]-120-\-4xy-\-Ax-\-4T/-\-i, by clearing E(]. (3) of fractions. 116i/:=12^-\-4mj-{-4, by reducing. From Eq. (1) after clearing, we find y=^{i ; this being sub- Btituted in the last equation, gives, after clearing of fractions and reducing, 1 Ix^ — 39j;=20 ; from which a;=4 ; hence y^&. (15) Let X and y represent the rates of travel of A and B ; then 2a= distance A travels in 2 hours at j; miles per hour, and 2a;-|-l== distance A travels in 2 hours at x-\-^ miles per hour. 30 — '2,x= distance A travels after B starts, in 1st case ; 42 — (2a;+l)=41^2a;= distance A travels after B starts, in 2d case. 30— 2:c 30 Therefore —~=~, (1) ; 41— 2 j; 42 Clearing Eq. (1) of fractions, and reducing, we find _ \bx y^lE'—x' Clearing Eq. (2) of fractions, and reducing, we get 4l2/— 2jn/=43x+J. Substituting the value of y before found, in this equation clearing of fractions, and reducing, we get 26a;2— 59:r=15 ; from which x=2^ ; hence y^3. {\0) Let. x^ the number of miles B traveled ; then a;-f-30=: the number of miles A traveled. Then since the distance traveled, divided by the number of days spent in traveling, gives the number of milea traveled per day, X -= A's rate of travel ; a;+30 — ^ — = B's rate of travel. 6S KEYTOPARTFIRST. Then dividing tho distance traveled by each man's rate of travelj X 4(x+30) , . {x-\-Zff)-^-—. = days A traveled. x+SO 9.r , ^ But they both traveled the same number of days, tlier*. 4(a;+30) Qx fore, ^7"~=^-f30 ' 4(x4-30)^=9j;', by clearing of fractions ; 2(a;-j-30)=3j;, by extracting the square root ; from which a;=60 ; hence :c-4-30=90, and 60-1-90=150 miles, the distance from A to B. ARITHMETICAL PROGRESSION. Article 222. (11) Here re=20, n=16yV. f?=48i— lRJj=32J. i=a-|-(ra— iy=16i'j-|-(20— l)32i = lGi'j-|-611J=627i Article 223. (5) Here/=a-l-(n— iy=IO— 3X9=— 17. n 10 *-=y+«)^= (-17+ 1 0)-^ =-35. Article 225. EXAMPLES. n 1000 (1) x=(Z+a)-=(l-f 1000)--^=500500, (2) Z=fl-l-("—iy=l-Kl01 — 1)2=201. n 101 s=(;-l-aX^=(201-[-l)-,-=10201. A RITll M KTIC A I, 1' R O O R K SS I O N. 09 (.3; First find liow many limes a clock strikes in 12 hours Here a=l, ;=12, n=12, «=(12+l)-^=78. 78X2=156= strokes per day ; 156X7=1092= strokes in a week. (,4) Since the second term is 2, the 3d term 3, and so on, the nth term is evidently n. Or thus, l=a+{n—l)d=l+(n—l)l=l-\-n—l=n. (5) Here d=2 ; Z=I+(7i— I)2=l+2ra— 2=2ra— 1. s=(l+a)^={2n-l+l)~nK (6) Substituting the values of I, a, and d, in the formula, l:=a-\-(n — l)d, we have 29=2-i-(ra — 1)3 ; from which «=10. n 10 s=(Z+a)-=(29+2)y=155. V.7) Substituting the values of 1, a, and n, in the formula, l^a-\-(n — l)d, we have 10=6-|-(9— l)d; from which d=J. s=(Z+a)|=(10+6;^=72. fS) Substituting the values of s, a, and n, in the formula, n s={l-\-a)-, we have 10 85=(Z+I0)7r ; from which 1=1. Substituting the values of I, a, and n, in the formula, Z=a — (ra — l)d, we have 7=10 — (10 — V)d; from which £Z=J^. (9) Substituting the values of a, h, and ra, in the formula, b—a , 16—1 d= — r— -, we have a= . . , -=3. m-\-l 4+1 Hence the series is 1,4, 7, 10, 13, 16, &c. 70 K E Y T O P A R T F I R S T . ClO) Substituting the values of a, and d, in the formula, 2=a+(7i— ly, we have /=24 — 4(n — 1)=28 — 4re; substituting tliis value of I, and tiiose of s and a in the formula, n n s=(;+a)--, we have 72=(28— 4ra+24)- ; from which, by reducing, n^ — 13?!.= — 36. From this equation m=:-|-9, or -|-4, (.1 1) Let ra=: the number of acres ; then the nth acre evidently cost n dollars. n Substituting n for I in the formula, s=(Z-|-a)r, and for » and a their values, we get n 12880=(7i-f-l)r ; by reducing B2+n=25760 ; from which re=:160. Having the number of acres, the average price per acre is easily found. (12) Let a;= the number of days ; then on the a;th day, A will travel xa, or ax miles. Hence, to find the sum of the series, we have the first term a=a, d=a, and l=ax; sub- n Btituting these in the formula, s=^Q-\-d)-, we have X 1 s=:{ax-\-a)-=-ax(x+ 1 ). Then in (x — 4) days, B travels 9a(x — 4) miles ; 1 Therefore -ax{x-\-l)=9a(x — 4); reducing, x'^ — 17x= — 72 ; from which x=8, or 9. (13) Let n=:the number of hours. Then Z=o+(ra— l)d=5-|-(n— l)l=4+7i; s=CZ+<')f=(4+«+5)2=|(9+n). Therefore o(^+'*)=6(3i+n). By reducing n^ — 3n=40 ; from which n=8. (14) Let x= the number of hours ; then the formula, 1 ~2 l=a—{n—l)d becomes Z=4— -(x— 1)=4J— J^; GEOMETRICAL PROGRESSION. [\ i=Q-\-a)-, becomes •s=(4i—ia;+4)~(8^—i,i:)| : but in X days, A travels 3x miles ; X therefore 3r=:(8^ — \x)^ ; dividing both sides by x and reducing, we find as=5. GEOMETRICAL PROGRESSION. Article 229. (J) In this example the ratio is — -. a 113 '-l-^-l-(-^)-J-4' (5) Here the ratio is -;. x' a 1 x' ^~l--r~ 1 ~x'—l- ' x^ (.«) b Here the ratio is — -. a a a a ~1 — r / b \ b a+b 1— — - 1+-- ^ ^7) Here the ratio is r. a 10 10 S=, =:; r=— =20. l—r I— I i AMBIGUOUS AND ERRONEOUS EXPRESSIONS. As this work is intended especially for the assistance of young teachem It is thought proper, in conclusion, to call attention to some loose and Inaccurate expressions, that are occasionally used in the school room, and which are also to be found in some of the works intended for text books 72 KEYTOPART FIRST. Too much importance cannot be attached to clearness and propriety of expression. Accuracy of style has a tendency to produce accuracy of thought. Every definition should be expressed in language the most precise, brief, and clear, of which it is susceptible ; while all explanations and directions, whether contained in the text book or delivered by the teacher, should be given in such a manner, that the pupil cannot possibly mistake the meaning. Lost some should regard matters of this kind as unworthy of notice* it is proper to add that great attention is paid to them by the French Mathematicians ; hence, many of their works exhibit a perspicuity and simplicity, and a logical clearness of arrangement, which add greatly to their value. " Place the two quantities under each other." This is not possible, one of the quantities may be under the other, but each cannot be under the other at the same time. It should be specified which is to be placed below the other. " Subtract the numerators from each other." It should be specified which is to be taken from the other. " Find the diiFerence of x'' — a^ and a^." When numbers are referred to, this expression is correct, but in the case here presented, the differ- ence is either x' — 2a.^, or 2a^ — x^. It should always be specified which of the two quantities is to be subtracted. " Divide tlie numerators by each other, if thejj will exactly divide." This expression has no clear meaning, and the word divide at the close of the sentence is used improperly. "The two first numbers." " The three first numbers," &c. These expressions are frequently used. When two or more things are consid- ered in regard to order, only one can proj>erly be called first ; hence, there is no such thing as the two first. However, we can with propriety say " the first two," because there may be a second two, a third two, and so on. " Neither the first nor the last terms are squares." This should be, " neither the first nor the last term is a square." "This value is the greatest of all others." Here others ought to be omitted, or it might be, " this value is greater than any other." " An equation of the second degree or power." Equations are of different degrees, but no equation is of the second or any other power. These examples might be greatly extended. The preceding are given merely as specimens. Such expressions confuse the mind of the pupil and often prevent a clear and accurate understanding of the subject under examination. Thoy cannot, therefore, be regarded with indiifor- enc« by aaf one who aspires to the character of an accomplished teacher. KEY TO RAY'S ALGEBRA, PART SECOND, 0:5" The numbers in parentheses, as seen in the margin, refer to the corresponding number of example, under the same article in the Algebra. GREATEST COMMON DIVISOR. Article 108. Note. — This article contains the first examples in the Algebra which the attenlive student will find any real difficulty in solving. (5) a'' — x^ \a'-\-a^x — 5a;' — a;' — a'x-\-a-x''-\-ax' — x^ =-x(ia^-a^x-ax''+x^) ^fter dividing we find the a'—a^x—ax^+x' ^"^^^ remainder contains a fac- a'+a'x—ax^—x^ ^°^' — ^' "°^ ^°^^^ '" ^^ '^'" ■ visor, hence it should be can- -Ha^-x +2x^ ^.g,gj_ See Note 3. — 2x(o'— x2) By dividing a^-\-a^x — ax^ — x^ by o' — x^ we find there is no re- mainder, hence the latter is the greatest common divisor required. (6) x'—5x'+nx—9 x^—2x^+ix—3 |a;s— 2x=4-4x— |1 — 3x2+9x— 6=- k5— 2x2+4x— 3 x3_3x=+2x -3(x'— 3x+2) Ix=— 3i+2 1^+1 x2+2x— 3 x=— 3x+2 X — 1 74 K E Y T O P A E T S B C O N D . X — 1 will be found to divide x' — 3j;-1-2 without a remainder; it is, therefore, the greatest common divisor. Note. — In the solution of the remaining questions in this article, wa shall merely exhibit so much of the operation as is necessary to show how the greatest common divisor is obtained. . The reasons for the dif- ferent steps of the operation will be found in the rule, or in the rotes following it. (7) a^— 5a;'+16a:— 12 |a;'— Si^— I5a;-|-16 jy *— 23;'— 15j4-16 |1 — ax'^i+Sla;— 28 —3x3+ 6i=-|-45ar— 48 j-Sa^^+S 12—28 — 3x5+3 lx=— 28a; |x+25 —25x2+ 73;c— 48 . . . Mult, by 3. — 75x=+2l9^— 144 The 4th line is obtained — 75x +775x — 700 by multiplying the divisor ' — 556X+656 by —3. — 556(x — 1) Ans. x — 1. (8) Multiplying the first polynomial by 2 to render it divisible by the second, and dividing by x (Note 3), we have 42x2— 52X+16 | 6x2— X— 2 42x2— 'jx.—li ~]7 ^45x +30 ■ — I5(3x— 2). Ans. 3x— 2. (9) 2x^+1 Ix'— 13x2— 99x— 45 j2x'— 7x2— 46x— 21 2x''— 7x^—46 x2- 2]x |x-|-9 18x='+33x2— 78x— 45 1 Sx^ — 63x2— 41 4x— 1 89 96x2+336x+144 48(2x2+7x+3) Am. 2x2+7x+3. (10) Multiplying the first polynomial by 7 to render it divisible by the second, we have 7x''+ 14x2+63 |7x3— 11x^+15j+ 9 7x''— l]x-'+15x2+9x |x+n ll'^'— x'— 9x+63 . . .. . Multiply by 7. 77x3— 7x2— 63X+441 77x3— I21x2+l 65x+ 99 1 1 4.1--'— 228X+342 114(x=— 2.(;+3) Alls. x'—2x-\-3. i; R G A T E S T COMMON DIVISOR. / 1) ill) Jlultipiying the second polynomial by 2, and dividing by the first, we have 48;r=i— 44j:=+34a;— 10 |48a:^+16j— 15 48a;=+16x'— lox |x— 5 ' — 60x^4^ 49x— 10 ' — ;240,r2-l-196.-<;— 40 — 240x=— 80X+75 276x— 1 15=23(12x— 5) Ans. 12x — 5. (12) This example presents no dilSculty whatever (13) x^+aV+a" Ix^+ax^ —a^x—a* x''-{-ax^ — a'x — a^ 1 1 — ax'-\-a^x''-\-a^x-\-2a'* — a(x' — ax'' — a^x — 2 a^ ) x''-\-ax' — a'x — a^ \x' — ax^ — a^x — 2a' x^ — ox' — a''x^-^2a^x |x+2 a) +2ax'+ a^x^-\- a'x — a'' -|- 2x^— 2ax^— 2a^x— 4g-^ +3g) + 3ax'+3a^x+3a3 x^-j-ffx-j-a^ Ans. (14) In this example 2i is a factor of the first polynon-.ial, and 33 of the second. Canceling these factors, arranging the terma in both, and multiplying the second by 4, to render it divisible b? the first, we have 12a3— I2a26+4ai=— 4&3 | 4a°— 5a&+6^ 12a3— 15g^&+3a&^ |3a+3A 4 I2a^6+ 4ab'—l6b^ 12g'&— 15a6^+ 36^ 19i=(a— 6) Ans. a—b. (15) x*^px^-\-(,q — l)x''-\-px — q |xi — qx^-\-(j) — l)x^-\-qJil—p x^ — 9x'+(p — l)x^-\-qx—p |1 (See next page). 70 K rj Y T O P A R T S E C O N I) . '^q—p)r'-\-iq—p )a:'— (?— p)^— (7— P)' or x'^x^ — X — 1. by dividing by 9 — p x'^-{- x' — x^^x \x — (?+0 —(7+ 1 )x'+px''+{q+\)x—p -(q+\)x'-{q+\)x'+iq + \)x+(q+i) LEAST COMMON MULTIPLE. Article 113. (4) From Arts. 85 and 86 it is obvious that a-\-x is the onl^ divisor of both the quantities. Hence, (Art. 113) (a'-|-a;') (a^ — x^)-i-(a-\-x)={a^-\-x^){a — x)=a'' — a^x-]-ax^ — x*. Atis. (5) The quantities separated into their prime factors are 2X2a(a+a:), axaxSjJ'Ca — a:), and 3x3X2(a+a:)(a — x) ; from which we readily see that the least common multiple is 2X2X3 X 3ax\a+x){a—x)=zeax\a^~x-). (6) The first quantity divides the second, but not the third, and the second and third have no common factor ; therefore, the least common multiple of the three quantities is the product of the second and third. (7) By examining these quantities we see that the second quantity is divisible by the first, and the fourth by the third, and that these are the only cases of divisibility among the four quan- tities ; hence, their least common multiple will be the product of the second and fourth quantities. (8) By factoring the several quantities, we find the first is =(a;-|-l)(a;— 1), the 2"^ z=x^+\ ; 3"' =(x—l){x—l); 4"* =(x-\-l)(x+l); 5'* =:(x~l)(x'+x+iy, &" =(x+l){x^—x+i , It will now be seen that if we omit the third and fourth quantities the remaining quantities vill contain the factors of these, and no other factor not necessary to be found in the last common multi- ple. Hence, the 1. u. m. will be {x' — I)(a;=+l)(i3_i)(x3-f i) =(a;<— 1)(j:8— l)=a;'»— a;6— a;''+l. i9) It is easily seen that 8 is the least common n ultiple of LEtSTCOMMONMUT.TIPLE. 77 the numerical factors, and that of the literal factors, 1 — x, is the only one common to two of them. Hence, the least common multiple is 8(1— j;)(l— x)(l-l-j;)(H-a:')=8(l— a;)(l— x')(l+a;2) =8(1— x)(l— x^). (,10) We first find the greatest commcn divisor of the 1" and 2"°* polynomials, to be x — 3 ; then of the 1" and 3"' to be 3x— a. Hence, „ , , , , „ , „, ,„ „. "1 It is evident the least W — lla;+6= (a; — 3)(33: — 2) \ , . , . 2i2— 7x4-3- (v—3Yix—l^ I common multiple is TO REDUCE A FRACTION TO ITS LOWEST TEKMS. Article 119. The only difficulty in solving any of the examples in this arti- cle, consists in finding the greatest common divisor of t!ie two terms. In general it may be easily fo\jnd by the rule (Art. 108) and in most cases by mere inspection. Thus : (^13) From Art. 86 we know that x-\-l is a divisor of the de- nominator ; and, by trial, it will be found to divide the numerator. (14) The numerator is the square of 2j; — 3a (Art. 79), and from Art. 83 this will also divide the denominator, since 8x^ — 27a' =(2x)'— (3a)'. (15) Canceling x in the denominator and multiplying the other factor by 5, we have 135x'-|-315x2— 60x— 140 |i5x'-|-35x^+3x-|-7 135x'+315x='+27x+63 |9__ — S7x— 203 — 29(3x-j-7). ■ g. c. d. =3x-[-7. (16) Setting aside the factor 2, which is common to all lb* terms of the numerator and denominator, as a part of the coin mon divisor, and then multiplying the numerator by 4 to render it divisible by the denominator, the reniaindei of the operation to 6iid the g. c d. is, 78 KEYTOrAETSECOND. 4x^+16x^y+32xjf+32y^ \4x^+2xy—12y^ 4x3+ 2 x'y—\2 xi / \x 2 28x=7/+88a;^2+64y |7y aSj^y+ltoy^— 84i/3 ' 74j;(/2+ 1487/3 74i/'2(a;+2)/}. T+ay will be found to divide 4x^+2xy—l2y', -herefore 2{x-\-2y) is the greatest common divisor of both terms. (18) ac+ by+ ay+bc=(a-\-b)c-]-(,a+b)y =ia+b)(r,+y) ; af+2hx+2ax+hf={a+b)f+{a+.b)2x={a+b)(f-Jr2x). Hence a-\-b= greatest c. d. of both terms. ( 1 9) 6fflc+ 1 0hr.+9ax-[- 1 5ix=i2c-\-Sx) 3u+(2c+3x)56 =(2c+3x)(33+5t). 6c2+9ca;— 2c— 3j:=(2c+3j;)3c— (2c+3x) =(2c+3x)(3c— 1). Hence 2c-j-3x= greatest c. d. of both terms. (20) x8+x62^'+x22/4-y5=(x^4-!/2)x6+(x'4-y=)!/ =(x'+3/=)(x'i+2/) ; x*-^'={x^+y'){x'^'). Hence x^-{-y^=: g. c. d. of both terms. (21) a'+{a-\-b)ctx+hx-'=(a^+bx)a+(a'^-{-bx)x=(a'+hx)(^a-\-x), a< — b^x^={a^-\-bx){a' — bx). Hence a'-\-bx= g. c. d. (22) ax" — ix'"+'=(ai — bx'^)x"'-'^x{a — ix)x'"-' ; a^bx — I'l' =bx{a^ — bV)=bx(a-\-bx)(a — bx). Hence x(a — tx)= greatest c. d. of both terms. C23) acx''-[-{ad-\-bc)x-\-bd=ax{cx-\-d)-\-b{cx-\-d) ={ax-\-b){cx-\-d) ; a'x- — b^=(ax-\-b)(ax — b). Hence ax-\-b= g. c. d. (24) a^+ab^—aVi—P=a{a'+b^-)—b{a'+b')=(ia—b)(a'+r-). 4a^—2a'^b^—ia^h-\-2ab^=2a\2a''—b'')—2abi^2a^- — b') =(2a2_2a6)(2a=— /;')=2a(a— i)(2a=— i2). Hence a — b is the greatest c. d. of both terms. (25) 2a''+ab—b^=a'+ab-\-a'—b^=a{a+b)+(a+b)(a—b) =(a+i)(a+a— /0=(a+6)(2o— '0- a^+rt^i— a— i=a=(a+i)— (aH-;;)=(a=— l)(<7+l;|. Hence a-\-b is the greatest c. d. of liotb terms. ALGEBRAIC FRACTIONS. 79 Art. 120. Ex. 7. — The following is the operation of finding Jie greatest common divisor of the two terms : 3a;'— 3x2— 63X+135 lSjc^^2x—2\ —x^— 42a;-)- 1 35 " 3 -3x2— 126X+405 —3x2-)- 2x4- 21 — 128x-f384 — 128(x— 3) X— 3= g. c. d. ADDITION AND SUBTRACTION OF FRACTIONS. Article 130. EXAMPLES IN ADDITION OF FRACTIONS. (12) Reduce the first fraction to the same denominator as tho second, by multiplying both terms by x-\-a ; the sum of the first two fractions will then be found to be "^ The sum of this {x+ay and the third fraction is then readily found by multiplying each numerator by the denominator of the other and taking the sum of the products for the numerator of the result, and the product of the two denominators for the denominator of the result. (13) The least common multiple of the denominators is readily found to be 4a^ia+x)ia—xXa^-{-x')=4,aKa*—x^). We then find for the numerators of the respective fractions, the following quantities : l*" (a — ^x)(a'-)-x') . . . . = a^-\-ax^ — a'x — x* 2"^ (a-)-x)(a'-)-x2) . . . . = a^-lrax'-\-a^x—x' 3"' 2a(_a-\-x)(_a—x) . . . =2a? — 2gx' Sum =-^— -; r = -T-^4- A^- 4aXa'^—x'^) u'—x (14) It is most convenient to make the common denominator o( the fractions, ahc(_a — b)(a — c)(h — c). In doing this we must change the signs of the factor, b — a, in the denominator of the second fraction, which may be done if at the same time we change 'Jie sign of the numerator (Art. 124). The value of the third 80 K E Y T O P A R T S E C O N I). fraction will not be altered if we change the signs of loth the factors c — a and c — b, so as to have a — c and b — c. Hence ] _ hcQ)—c) . a{a — b){a — c) abcia — b){a — c)(6 — c) 1 — ac{a — c) b{b — a)(b — c) abc{a — b){a — c)(i — c) " ] ab{a—h') :{r. — a)(c — b) abc(a — A)(a — c)(6 — c)' The sum of the numerators is bc(J) — c) — ac(a — c)-\-ab(a — A), v.liich, by performing the multiplications indicated, and reducing gives the same result as (a—b)(a — c)(i — c). Hence, this product may be canceled in both terms, and the sum of the three frac- tions is found = — . abc EXAMPLES IN SUETKACTION OF FKACTIONS. (14) By reducing the third fraction to its lowest terms it be- X comes The second fraction subtracted from the first leaves — "r y~ry_ Subtracting the preceding from this leaves ^+y' = l. Ans. 1 1 _2{x-\-l)—(x—l)_ x+3 . (15) x—1 2(x+l) 2(1=— 1) 2(0:2—1) a+3 _ x+3 _ {x+3){x^+\)~ix+S){x^—i )_ x-\-3 2(x=— 1) 2(ap2+l) 2(a;''— 1) a;'— 1" U6) 1+1_1^1+--' Ant X^ X' X X' x—1 _ I _ x^—x''-\-x—\—l _^ x^—x''+x— 2 x2+l (x2+l)2 (x2+l)2 (a:24-i)r-- \J^x—x^_\-\-x—x^' W^-'^y ^ —x^-\-x^—x '+'ix^ +x^-\-x^l ? x^ (^2+1)2 a''(j;2-t-l)* x^—x^-\-x—'ix^ _!fi—x^+x*—-2x^ ■ X - (x'+i)' x' x'(x-'+^y ~ x--l-a'-l- 1 . Sum ?= - -i - ' — Ans x'ix-'+iy ALGEBRAIC FRACTIONS. 81 MDLTirHCATION AND DIVISION OF FRACTIONS. Article 131 . Remark. — In the solution of all questions in multiplication or division of fractions, it is iniportaiit to separate the quantities into factors, before performing any actual multiplications, as this might so involve the fac- tors that they coukl not be readily discovered. By attention to factoring nearly all tile examples are easily solved. (U) X^-\-X+l 1 +'-+\ ~x—\ 4-x'+j:+1 x=+l+. (12) ia.Zx 3i 2i 26, Sx Zx 4a Sab 8ab 8aA , 2 I 9i^ 9x^ M» (U, pr+{pq-\-(ir)x-{-q^x^=(p-\-qx)r+(j>-[-qx)(iqx=(j)-]-qx) (r-lrqx) ; ps-{-{pt — qs)x-^qtx''=(j) — qx)s-{-(p — qx)tx=(p — qx'){s-\-tx) The factors in tlie denominator of the product will cancel the factors p — qx and p-\-qx in the numerator, leaving for the result (r-\-qx){s-\-tx)=zrs-\-{rl-\-qs)x-\-qlx^. ii-ticle 133. (8) In solving this and tho next two examples, first perform the operation indicated in the parentheses. A similar remark applies to example 12. f,.. ^, 1 _x>-\_ {x'~\){x>+l) _ {x-'^\)ix^-\){x''+l) . ^^^' ^ J^ i^ X' ' 1 X- 1 a^_|_l)(x2_i)(a;<+l X _(x2+l)(r<+l)_Ky-Ky-^H-l 3? ^^,34-^+1+1 or a;3+L+a;+i. This example may also be readily solved by ordinary division,. 82 KEYTOrARTSECOND, Article 133. Remark. — In the solution of the exannples in this article the first step Is to perform tlie operations indicated in the respective terms. By doing this they are all easily solved, e.xcept the 5th, of which the solution i< here given. i+i_='y-J=(*±lX^^:^y±\ (Art. 83); a aJy' a¥ a¥ ^h b ^b+iW-b+i) _b J,+i ^„^ abi '^J3_j_l_i ab" ■ KISCELLANEOUS EXEKCISES IK FKACTIONS. (2) To reduce these fractions to a common denominator, it will be most convenient to change the, signs of the factors as in the solution to Ex. 14, Addition of Fractions, so that the common denominator may be (a — b){a — c)(i — c). a2-|-a-|-l _ a2_|_a_|_i ij—c_a^l^-a^c-lrab—ac+h—c_ (a—b)(a—c) (a—h)(a—c) b—c (a— ?;)(«— c) (6— c) —V—b—i _ —b^—b—l a—c_—ah'^-\-b^c—ab+bc—a-\-c . — - X - — ' {a—b)(h-~c) {a—b){b~c) a—c {a—b){a — c){h—c) c'^-\-c-\-\ c^+c-l- 1 a — h ac~ — hc^-^-ac — bc-\-a — h ■ — X - — (a— c)(i— c) (a— c)(i— c) a—h {a—b){a—c){b—c) ' d a^b — ab' — a'^c-\-b^c-\-ac' — bc^ , Sum =-- — '-- J =. Ans. a^b — ab- — a'c-j-w^c-j-ac^ — tc' (S) Perform the operations indicated before substituting the ralue of X. 3 4 12 ' 'W ^IIl~"6(^— 1) ' -— \ ~^~^ l =3_x(x—l)-f^-j-9_3x'— 2x4-9 2 /6(x— l)i 6 (x— 1) ~6(i::^V' SX'fj"— 8;+9 SB; 170 „., , 6X31 20 60 ALOERRAIO FUACTIONS. 83 (6) x+2a= l"" +2a: a-\-b a-\-b a-\-b a-{-b 1" fraction _eab+2a^_3b+a_ -a-3b 2ab — 2a^ b — a a — b x+2b=^+2b=^.^!!±^; ■ a-\-b a-\-b x-2b=^-2b=^J±^; a-{-b a-\-b 2"^ fraction =e_±t^'^3a±b —a—Sb _^3a+b_ 2a—2b ^^_ j^^ a — b a — b a — 6 (^7) Substituting the value of x we have a" , i» 2na" — na" — rib" 2nb" — na" — nh" _ a" , b" "■" 4_ — ^" na" — nb" nb" — na" na" — nb" na" — nh" a"—b" a" — h" 1 na" — rei" n(a" — b") n Ans. X V (8) ^4i/_^;2!^_l 1 1 xy 1 'ip'x Similarly ^—y _ ^y ay _l_l xy \ y x' By dividing bntb terms of each frac- tion by xy, and re- ducing. (9) Let X represent the value of one fraction, and y that of the other ; then x-\-y:=\. Multiply each side of this equality by x — y (see Note, page 61), and we have x^ — y^=x — y, which proves the proposition. f]0). Let X and y represent the fractions, then -^=E; m ? Qc—y)=p. X — y=- ; multiply bot?i sides by q, then ? 84 KEy rOTARTSECOND. Multiply both sides of this equality by ^-(-y ; then or p{x-{-y)=q(x^ — y'). (11) First. — Let (a — h)(a — c)(b — c) be the common denomina- tor of the three fractions, then we must change the signs of the numerator of the second fraction, and the signs of the first fac- tor of the denominator. We must also change the signs of both factors of ihe denominator of the second fraction. Tlie numera- tors of the respective fractions when reduced to a common denominator will be (a=+A2)(i— c)= a'b—a^c-\-hV—cV ; (_i2_7j2-)(a_c)=_ai2-l_J2c_aA2_|-cA2; (c=+A=)((z— i)= ac2— fe2+aA2— Wi2. Sum of the numerators =a^(b — c) — h''(a — c)-\-c-(a — i) ; (a— i)(a— c)(6— c)=a'(i— c)— i=(o— O+c^a— i); hence the value of the fraction is 1. Seco7ld. (ffl2+A2)(J_c)(J+c)= a4'—(/c^+hW—c^h^; —!iO^+h^)ia—c)(a+c)=—a''h''+hV—aVi^-jl-c^h^; (.c''+h^)(a—hXa+b)= a'J^—b^c^+a'-h^—b'A". The sum of the numerators is 0, hence the sum of the fractions is 0. TJiird. (flS-f A2)(J_c)ic= a^'c—a^bc^+J'^ch^—^cVi^ ; —(b'^+h^){a—c)ac=—a^^c-\-ab^c^—ahh'+acVi^ (fi'-\-V){a—b)ab= a^c'—ab^c^-\-a^bh^—ab'^hK The sum of the numerators is V^a-(b — c) — i=(a — c)-\-c'^(a — I) ( and since the denominator is tlie quantity within the brackets, the value of the fraction is h^. EQUATIONS OF THE FIRST DEGREE. Remarks. — The attentive student will find no difBcully with the ex- amples in Articles 151, and 153, provided he attends carefully to the rules. (See Remark on page 14, of tlie Key to Part First.) The ease and facility with which several of the examples may be solved, will depend on the particular method of solution The shortest inelliods, however, are not always the best for learners, [t is important EQUATIONS OF THE FIRST. DEGREE H5 even at the risk of being tedious, that the pupil understand every step of the operation. Let the aim be first to perform the operations cor- rectly and understaudingly, and after this with facility. In some cases it is better to perform tlie operations indicated before clearing tlie equation of fractions. To illustrate this we will take example 27, Art. 153. Blultiplying the terms in the parentheses in the second member fcr — and removing it, we have 6x , 1 — .Sx i ( X— ^ ) —1 {\—3x)=x- '' \ 26/13 39 15.6 tiovf 156 is evidently the least common multiple of the denom- inators. Multiplying both members by this, we have 78a;— 153— 24+72a;=156x— 20.-C+10— 3a; ; reducing, 17x=187, whence a;=ll. QUESTIONS PRODUCING EQUATIONS Of THE FIRST DEGREE. Artie 1 e 154. (.9) Let 0!= the first, then 2x= the second, and 3x= the third: and a;4-2a;-(-3a;=133. Whence a;=19, 2a;=38, and 3a;=57. (10) Let x= the first, then 3a:= the second, and 4ia;= the third ; and a:+3a;+4^a;=187, Whence x=22, 3a;=66, and 4^a;=99. (11) Let x= the second, then 3.'x= the first, and Six — a;:=100. Whence a;=40 and 3^x=140. (12) Let x= the first, then 3ja:= the second, and 100— (330; — a;)=100 — 2^x= the third. Then a;+3ia;+100— 2ia;=156. Whence a;=28, 3ia;=98, and 100— 2la;=30. 13^ Let x-= the number, then ^-l-^_-\-^—62. 2 3 4 Whence 1^.^=52, and x=48. 12 86 KRT TO PART SECOND. (14) Let x= the number, thnn x+^— 20=45. Whence x^35. 15) Let x= the number, then 3 4 6 Whence a;=36. (16) Let 3c= the number, then 4a;— 40=40— a;. Whence x=l6. 17) Let x= the number, then 4(a:+16)=10(a;+l). Whence a;^9. (18) Let x^ the less number, then SO — x^ the grcci'er, and -j(30 — x — x)=3. Whence x=9, and 30 — x=21. (19) Let 0,-= the number of days he worked, then 28 — r- < id'e days; then 75a;— 25(28— a;)=1200. or fa; — 1(28— a;)=12. Whence fa; — 7-l-^a;=12, or a;=19. 1,20) Let a;^B's money, then 3a;=A's, and 3a;~l-50=4(a;— 50). Whence a;=250, and 3a;=750. (21 xjBt a;= sum, then x — _a; — 20=- — 20; 2 2 ?_20-i ( ?-20 ) -30=?-?-50+??=^-ii*' 2 3\2 / 26 333 ^_110_l('^-l=L^)-40=0 3 3 4 V 3 3 / Whence a;=290. ^22) Obsei-ve that 20 per cent, is 5 and 25 oer cent, j EQUATIONS OF THE FIRST DEGREE. 87 1 5 -r 1 1 fia: Let x= capital, then x-\- — =— 1-= cap. close 1" yr. 100 100 115a; I 1 f 115x 115a;, 23a; 138x „„j . .+- of = + — = = cap. 2"''yr. 100 6 100 100 100 100 138x , 1 e 138^; 138a: , 69a; 345a; ..^ 4-_ of = +. — = = cap. 3'^'' yr. 100 4 100 100 200 200 .-. il^— a;=1000.50. 200 "Whence a;=1380. ,23) Let x= B's age, then 2a;= A's, and 3(a;— 22)=2a;— 22. Whence a;=44, and 2a;=88. (24) To avoid fractions we may take some multiple of x that is divisible twice by 2. Thus, I^et 4x= cost of 1" house, then 4a;-l-2a;=6af= coat of 2"'', and 6a;-l-3x=9x= cost of 3'''', also 4x-\-9cc=l'ix= cost of 4'*. Hence 4a;4-6a:4-9a:+l 3x=8000. Whence 4a;=1000, 6a;=1500, 9a:=2250, and 13a;=3250. C25) Let a;= gallons third conveys in 1 minute, then 8a:=: " " " and 3x+8= galls first " ajso 3a; — 7= " second " 9a;-l-I= " a-U convey 72a:4-8= " " .'. 72a;+8=i050. Whence a;=14|g, U3x+8)=nis ; -J(3a;— 7)=12/g, (27) Let x= the number of days in which B can do it, then -= part B does in one day : but A does — , ar J A X 10 and B together do _ in one day ; .-. 1— J_=l Whence x=23- 7 10 X 3' (28; Let x-= the number of days in which A can do it, " 3 minutes, " 3 (( " 3 C( « 3 (( " 24 c 88 KEYTOPARTSBCOND. 1 12 1 then ^= part A does in une day : but A does _ of _= — f X 4 7 14 in one day ; .-. -=— . Whence x=14. ' a: 14 If A and B finish | of the work in 6 days, they do \ of 1=4^. m one day ; and since A does yj in one day, B does jj — ^'4=2'^ io one day, or the whole in 21 days. The solution of this question mainly depends on arithmetical analysis, and the employment of algebraic symbols can scarcely be said to be of any advantage. (29) Leta;= number of each, then %x=- cost of sheep, 12x=^ cost of cows, and 18x= cost of oxen. .-. 3J:+12x+18a;=330. Whence 1=10. (30) Let x= sum A rec'd, then x — 10= what B rec'd ; a:— 10+16=x+fi= wliat C rec'd ; x+6— 5=^4-1= what D rec'd ; x4-l+15=x+16= what E rec'd. .-. a;+I+a:+16=x+x— lO-fx+e. Whence a;=21, what A rec'd, from which, what the others rec'd, is readily found. (-il) Let x= the number of eggs, then — = number of dozen, and — X18= — = cost. 12 2 "' = number of dozen if he had bought 5 more, and since the whole cost divided by the number of dozen, must give the cost of one dozen, therefore 3x X 1 5 — -i— ^!^= cost of one dozen under second supposition Zx_^-\-h_^x^ 12 _ 18x . Y ' ~12~ ¥ x+5 x+5 ' .-. 18^=18-21=151 a;+5 2 2" Whence z=31. y 91 (32; Let x= the number bought, then _= cost of caco ■ X and \{x — 7)=: one-fourth of the remainder. EaUATIONS OF TITE FfRST DEGREE. 89 20-r-Ka>— 7)=20X --=-^= what each soli) for. X — 7 X — 7 . 94 80 ,„, • • — = — . Whence x=Al. X X — 7 133) Let x= the number ui" hours each traveled, tlien -X3 2 33? X hx = — = miles A traveled, and -X5^ — = miles B traveled; 2 '44 .-. !^-|-5^=154. 2 4 Whence x=56, ^=84, and ^=70. 2 4 (34) Let x= the number, then 5^=?i+i3=^. 6 ^ Whence x=54. (35) Let a;= number of dollars, then 3x=r. number of eagieg. .-. 5(x— 8)=3a;— 8. Whence a;=16, and 3x=48. (36) Let x= number of apples, then 100 — x= number of \aeara then -^XI= — = cost of apples ; 10 10 , 100— a:, ^„ 200— 2a; . , and X2= ; — = cost of pears; 25 25 X , 200— 2x_g , To 2~5 -' Whence a:=75, and 100— j:=25 (37) Let x= number of sheep, then _= acres ploughed, and _= acres of pasture; 8 6 .-. ?+?=325. Whence a=1000. 8 5 (38) Let x= miles he can ride, then — =: time of riding and -= time of vralking; 12 4 .-. -^+f=2. Whence x~6. 12 4 90 KEYTOPARTSECOXD. Q (30) Let a;= number of ]bs, then = lbs of salt in 1 Ibi 65-1-a; and 25 | _1_ ) = lbs of salt in 25 lbs. V 65+x / .-. 25 I ^ ] =1. Whence x=U5. \ 654-x / ■* (40) In every 10 lbs of the mass there are 7 lbs of copper and 3 lbs of tin ; hence in 80 lbs there are fgx'7=56 lbs of copper, and foX3=24 lbs of tin. Let 3;= lbs of copper to be added, then b6-\-x= lbs of copper in the new mass, and 24= lbs of tin ; and since there are 11 lbs of copper for every 4 lbs of tin, one-eleventli of the copoer must be equal to one-fourth of the tin. 56+a; 24 ■.,., .-. — '—= — Whence a;=10. 11 4" (41) Let x= stock, then a;— 250+i(a;— 250)=f^ -1^=: O do Stock at the close of the 1" year. 4a;_1000_25o='*^_''SO ;andl5-i!^+i(t-i^) 3 3 3\ 3 3 f = — — = stock at close of 2"^ year. 9 9 •' 16.r_7000_^gQ_16j_ 9250 . ^^^ 16^_9250 "9~ IT " "^ 9 ' ~9~ 9 , 1 / 16x 9250 \ 64r 37000 . , » , , ,, +- I — — = — = stock at close o*^ 3"^ vr 3\ 9 9/27 27 64x_ 37000 _^^ "27" 27 '" Whence j;=3700 SIMULTANEOUS EQUATIONS OP THE FIRBI DEGREE, CONTAINING TWO UN KNOWN QUANTITIES Article 158. f 17) Divide the second equation by a — b and we have (a+i)(x+y)=-^; a — EaUATIONS OF THE FIRST DEfiREH. 01 or (a+6)x+(a+%= " ; (3). a — (a — b)x-\-(a-l-b)i^=c ; equation (P 2J>r= — c ; by subtracting (1) from (3; a — b Again, dividing the second equation by a-\-b, we have a-\-b or (a— J)a;+(a— %= -^ ; (4) (3— 6>+(«+%=c ; (1) 2by=c — , by subtracting (4) from (1). a-\-b ^■'~2b \ '^~^b I ■ (.8). By multiplying eq. (1) by m, and (2) by re, and subtract- ing, we find the value of x. Again, by multiplying (1) by n and (2) by m, and subtracting, we find the value of y. '19) Multiplying both equations by abc, transposing and factor ing, we have {bc-\-ab)x-\-acy=abc ; (1) acx-\-{bc — ab)y=abc ; (2) Multiplying the first equation by be — ah, and the second by ac, we have (J)V—a-b'^)x-\-(bc—ab)acy=abc{bc—ab) ; (3) a'c'^x -\-(bc — aV)acy=a?bc''. (4) Subtracting equation (3) from (4), and factoring, we have {aW-\-a^c''—b^c'')x=abc{ab-\-ac—bcy „., abc(ab-\-ac — be) Whence x= — i — ! -' . a'^b-'^aV—Vc^ Similarly, we may find the value of y by multiplying equation (1) by ac, and (2) by bc-\-db, and subtracting. (20) Transposing Vy in equation (2), multiplying by 3, ai;d factoring, we have (a2_i2)3y+Ca+J-f c)35x=(a4-2i)3(Z?)+?-''i!£ ; (4) a-\-b 02 KEyTOrARTSECOND Separating equation (1) into its parts, we have (a2— i2)5a;+(a2_J2)3^=(4a_/j)2aJ ; (l^ Subtracting equation (4) from (1) we have (,5a^—5h'—3ab—3b'—3hc)a:=8a^b—2ab^—3a^-b—Gah''—^'^ Retucing and factoring, ^5a^-~8b^—3ab—3bc)x=—(5a^—8V—3ab—3bc). a-\-b Whence x= a-\-b' Substituting the value of x in equation (1) we have i^(a2— i=')+3y(a2— J2)=8a=i— 2ai' ; He^lucing, a-\-b 3y(a2— i2)=3 a^b+3ab^=3ab(^a+b), or, ^(a — b)=ab ; ab ••■ y= r. a — b QUESTIONS PRODXrclNG SIMULTANEOUS Eqli- TIONS CONTAINING TWO UNffVr'IVN QUANTITIES. Article 159. (4) Let x= number of slieep, and y= number o cow* then 5x+7!/=lll, 7a;+5?/= 93. Whence x=:4 and y=13. (5) Let x= cost of 1 lb tea, and y= cost cf 1 ih ^•^ft^' then 1x-\- 9y=520, 4x4-1 lj/=385. Whence x^55, and y=15 cts. (C) Lot .r= A's money, and y= B's, then x-]-^0=!/—20, 3x-{-5y=2350. Wlience a;=250, and y=320. (7) Let Gx= A's money, and 5y= B's, then 6j;-1-.5_v=9800 E U IJ A T 1 () N S OF THE FIRST DEGREE. 93 aUo fix — x=by — y, or bx — %=0. Whence x=800 and y=1000 ; .-. 6j;=4800 and 6j/=5000. (8) Let a;= the numerator and y the denominator of the fraction, then ^±1=1, and ^=1=1 y+1 2 2,-1 3- Whence a7=3, and y=^T. (9) Let x= the first number and y= the second, then -=?^+3, 3 4 Whence a;=24 and y=20. (10) Let a;=: number of lbs, and y= cost per lb, then xy= cost . . 30x — xy=\OQ, (I) xy — 22x=300. (2) Adding equations (1) and (2) together we have 8j;=400, whence a;=50. By substitution, the value of y is found =28. (11) Let x= number of bushels of wheat, and y= bushels of corn, then 65a;=33y ; and b5x-\-Z3y= rent ; also Gbx-\-i\y — 140= rent. .-. 65a;-[-41y— 140=55a;+33y, or 10x+8y=140. Whence a;=:6 and y=10. (12) Let X and y= the cubic feet which each discharges, then X -.y : : 5X8 : 13X7; .-. 40y=91a;; (1) also y— 07=561. (2) Whence a;=440 and y=1001. From (1) it is evident that y is greater than x, therefore in (2) we write y — x. (13) Let bx and 1x represent the first two numbers, and 'iy and by the other two, then 191 KEYTOPARTSECOND. bx^Zy : nx-\-by : : 9 : 13; .-. 65:r+39i/=63j;+452/, or 2a?=6!/; (1) also 1x-\-by — {5x-\-'iy^=\&, or 2x-f2!/=16. (2) Whence x=Q and 2/=2, ••. 5x=30, 7x=42 ; and 2y=i and 5j/=I0. (14) Let x= number of apples, and y= number of pears, then ^+^=30, 4 5 and 1 of ?+i oil, or ^+y =13. 2 4 3 5 8 ' 15 Whence x='Ti and y=60. (15) Let x=: acres of tillable land, and y= acres of pasture, then 200x-|-140y==24500 ; (1) also X : ^-^ : : 28 : 9 ; 2 .•. 9:c^l4a: — 14?/, or 5x=14j/. (2) Whence a;=98, and y=35. (16) Let x= digit in ten's place, and !/= digit in unit's place, then 10x+y= the number, and \f)y-{-x= the number when the digits are inverted, then 10a;+y4-10i/+a;=121, or lla;4-lli/=121 ; (1) and lOx+y— (10^+x)=9, or %x—9y=%. (2) Dividing (1) by 11, and (2) by 9, and adding and subtracting we find x=Q, and y=5. ReiMark. — It may be asked why, in ohtaining equation (2), do we Biibtract 10?/-|-:c instead of 10jr-[-?/, since we do not linow whicli is the greater: The answer is, we can not tell wliich to subtract till we pro- ceed to verify tlie resnlt, but if we had subtracted the wrong quantity, the error would bo made linown in verifying the resnlt, by some quau- Lty being' negative that ought to be positive. (See Art. 164.) (17) Let 2x — 6, Zx — 6, and y be the numbers, which fulfilU the first condition., The second condition gives 2j;— 1 :y+5 : : 7 : 11, or 22a;— 1 1 =7!/+35 ; (1) also 3a;— 42 : y— 36 ; 6 : 7, EQUATIONS OF THE FtRST DEGREE. JlS or 21x— 294=6^— 216. (2) Whence a:=18, and y=50 ; .•. the numbers are 30, 48, and 50 (18) Let X and z represent the days respectively in which a and B can do it, then - and _= parts wliich each can do in a day. X z Then i+i=l. (1) X z \Q Also in 4 days A and B do -+-, and in 36 days B does ■ X z — parts of the work ; z . . _-f---f- — =1 (the whole work) ; X z z or f+15=l. (2) X 2 » Multiplying equation (1) by 4, and subtracting it froni (2) we have — =- ; whence 0=48, and by substitution is z 4 readily found =:24. (19) First, 2 hrs 48 min. =2| hrs, and 4 hrs, 40 min. =4j h's. Let X and z represent the hours respectively in which A and B ccn drink it, then - and _= parts which each can drink in a/- X z 2 2 hour ; and _-|— ^^ parts drank by both in 2 hours X z 2f_14_ _._.. j_._,. u.. r, ._ .4 i. ... parts drank by B in 2- hours ; z bz 5 4-14 2 _i= = parts drank by A in 4- hours ; X Zx ^ 3 .-. E+?+li=i (the whole) ; (1) X z bz ?+l'*+?=l " (2). X Zx z By adding together the terms containing x, and those con- taining z, 96 KEY TO PART SECOND, ?+^'=i; (3) ?2+?=--l. (4) 3x z Bv multiplying (3) by y and subtracting (4) from it, we have _=1 whence 2=6, and by substitution x is easily found =10. z 3 (20) Let x^= numerator and y= denominator of 1'' fraction, then f= 1" fraction, and ?— ?=?ZZ:^= g'"' fraction. y ^ y i>y By adding the numerators together, and the denominators together, we have x-\-Sy — 5x=y-\-5y, or 2y=4x, or 2x=y ; X 1 whence _=_= the first fraction, y 2 and I — 5=:j|,= the second fraction. (21) In solving questions of this kind, it is convenient to de- note the capacity by 1 ; it may, however, be denoted by c, the object of the question being not to find either the size of a crown or guinea, or the size (capacity) of the purse, but the ratio of the size of a crown or guinea to the size of the purse. Let x= number of crowns and z= number of guineas, then -:= part filled by 1 crown, and _= part filled by i X z guinea. Also, i^+?=l, (1) X z and -+-=—■ (2) X 2 63 Multiplying equation (1) by 5 and (2) by 6, and subtract- ing, we find a;=21 ; then by substitution 2=63. 22) Let x= number of bushels of wheat, and y= number of bushels of rye. .*. 5x-\-2y= his money. Observe tliat 7 bushels of rye will cost 21 shillings, and 6 EtlUATJONS OF THE FIRST DEGREE 97 bushels of wheat 30 shillings. Then from the nature of the question, we have the following equations • 30+3(a;+y— 6)=5a;+3]/— 6. (2) Whence a:=9 and y=12. SIMULTANEOUS EQUATIONS OP THE FIRST DKSKEE, INVOLVING THREE OR MORE UNKNOWN QUANTITIES. Article 160. C9) (3) from (2) gives l—l=h—c ; U) X y n Sum of (1) and (4) gives -=a-\-h — c. X Whence x= a-\-h — c (2) from (1) gives \—\=a—b ; (6) y 2 o Sum of (3) and (5) gives -^a — 6+c. y Q Whence y^:^ a — b-\-c' (1) from (3) gives 1— i=c— a, (6) z X Q Sum of (2) and (6) gives -=b-\-c — a. z Whence z= b-\-c — a ( 10) Multiplying equation (1) by 3, and (2) by 2, to render tne coefficients of _ alike, and subtracting the former from X the latter, we have _LV-?=|. (4) y 2 3 Multiplying equation (1) by 2 and adding the result to (3) we have — — -=- (5) y z Z- 9 W8 KEY TO PARTS ECOND, Mulr'clying equation (5) by 11, and adding the result t« (4"), we have — == — ; whence y=12 y 3 (11) Multiplying equation (1) by 2, and subtracting (2j from the result, we have — — __== — (4) ix 3y 216 Multiplying equation (2) by 2, and subtracting (3) from the result, we find — _-t-_=_ (5) 3x^2/ 18 Multiplying equation (4) by 3, and (6) by '3', and adding and reducing, we find at=6 ; then by going back and sub- stituting, we readily find the values of j/ and z. (12) Adding the four equations together, and dividing by 2, we find the value of x-^y-\-z-\-v. Then subtracting from this each of the equations successively, and dividing by 2, we get the values of x, y, .i, and 1;. QUESTIONS PRODUCING SIMULTANEOUS EQUA- TIONS CONTAINING THREE OB MOKE UNKNOWN QUANTITIES. irticle 161. (1) Let X, y, and z represent the respective shares, then a;+y+j=760, (1) 1+^—2=240, (2) I/+Z— a;=360. (8) Whence a;=200, 2/=300, and z=260. (2) Let X, y, and z represent the numbers respectively, then ^+y+2=20 ; (1) x-^y : y-\-z : : 4 : 5, or 5a;+5^=4y4-42 ; (2) y—x : z — X : : 2 : 3, or 3ij—3x=2:—2x. (3) Whence a;=5, y=7, and 2=8. (3) Let X, y, z, and v represent the numbers respectively, then ^+y+2=13, (1) a:+y+«=15, (:2) a;+2+i;=18, (3) V+2+t)=23. (4^ E a U A T 1 O N S OF T H F, FIRST D E O E E E . OS Adding the four equations together and dividing by 3, we have x-\-y-\-z-\-v=22, from which, by subtracting equations (4), (3;, (2), and (1) respectively, we find x==2,y=i, z=1, and v^9. (4^ Let x= digit in hundred's place, i/= digit in ten's place, and z^ digit in unit's place, then 100x-\-10i/-\-z= the number, and x-{-y-\-z=l6 ; (1) also x-\-y : !/+2 ^ : 3 : 33, or 35a;+3'3!/=3j/+3z ; (2) and 100x-\-}Oy-\-z-]-198=l(!Oz-li-10y-\-x; or 99x4-198=992. (3) From these equations we readily find x=5, y=4, and z^7. (5) Let x= number of votes for A and B, y= do. for A and C, and z= do. for B and C ; then a;-f-y-|-26=158, (1) a;-l-2+30=132, (2) y+z+28= 58. (3) Whence ar=:102, y=30, and 2=0. (6) If X, y, and z represent the three numbers, then 2^+ly+52=46, (1) i^4-?i/+5^=3S, (2) ?^+Jy+B^=28-J. (3) By clearing these equations effractions, the values of x, y, and t are readily found by elimination by addition and subtraction. (7) Let X, y, and 2 represent the three numbers, tuen x->ry=a, (1) x-\-z=b, (2) y-\-z=c. (3) Whence x, y, and 2 are readily found. (8) Let X, y, z, and v represent the capacity of the respective casks, then x— y=— , (I) y—z=y, (2) 100 KEY TO PART SECOND. .=- x==z-^v+15. (4) Whence a;=140, ^=60, z=45, and t)=80. (9) Let a;, y, and z, represent tlie number of guns, Boldieri and sailors, respectively, then fX22+10=2, (1) y+z=5(a;4-y), (2) Since the number slain in the engagement was one-fourth of the survivors ; therefore, i(y-\-z) represents the slain, and i(y-\-z) the survivors. ••• ^(y+^)+5=|xi3. (3) Frcm these equations we readily iinda:=90,y^55 andz=670. GENERALIZATION, Article 163. (8) Representing the parts by x — m, a;+jn, _, and mx, we m have X — m-\-x-{-w,-\-—-\-mx=a ; m 2x-\-—-\-mx=a ; m 2mx-\-x-\-m^x=ma ; xim^+2m+l)=x(m-\-iy=ma. Whence x=. , from which the parts are easiW found. (9) Let x=^ distance he may ride, then -= time employed in riding, and -= time " " walking. . ■" I ^ - u "be • ' rn -=' i whence x^ c b+c ISaTTATIONS OF THE FrUST DEnREE. 10 (10) Let x= the less number, then /)a;= the greater, since tha quotient of the greater divided by the less is h. .'. bx-\-x=:a, or (i-|-l)=a. Whence x= = less, and bx=: = greater. 6+1 i+1 ^ (11) Let x= the number of beggars that received l> cts. each, then n — x= the number that received c cts each. . . bx-\-c{n — x)=a. Whence a:=?=:^^ and n-x^'^^ b — c b — c (12) Let x= the greater part, and n — x= the less, then n — X n — X Whence x== ''' , and n — x= 1+? 1+9 ,13) Letar, t/, and 2 represent the days respectively in which A, B, and C can perform the work. Then, if A can do it in x days, he can do - part in one X day ; in like manner B can do - part, and C _ part in V one day. W-- _i . CI) X y a W-- _1 . 7 J (2) X z b W-- _1 (3) y 2 c' Fjr the method of solution see examule 9, Art. ; 60. fl4) Let a;= A'a share, then -= expense of one ox for m a months, and _-j-m= — = expense of one ox fo 1 month a 7na .-. ^xnb=~= B's share, and ma ma .^Xpc=^= C's share, 1 02 KEYTO PARTSECOND Wlience x= , from whicli the shares of B and ma-\-7ib-\-pc C are easily found. (15) IjCt x= cost of 1 lb of the mixture, then (a-{-h-\-^;)x=: cost of the whole mixture. But ma= cost of a lbs at m shillings per lb, ni= " b " n " « pc= " c " c " " .•. (a-\-li-\-c)x^ma-\-Tib-\-pc. , , , , 7na-\-nh-\-pc W hence x^ ! L£__ a-{-b-\-c (IG Instead of representing either of the quantities to be found by a separate symbol, the simplest solution is ob- tained by taking x to represent the number of miles per hour the waterman goes when he rows with the current ; then since he can row c miles with the current for d miles against it, we have c : d : : X : _= rate of sailing up stream. And since the number of hours employed in sailing any given distance, is equal to the whole number of miles sailed, divided by the number of miles sailed in 1 hour therefore, _= number of hours in sailing down stream, and X u-——=—= number of hours in sailing up stream. c dx " ^ : "LJr-=h, whence a:=2£±f!?, X dx bd and dx ac-\-ad c he . ac-\-ad hd ^. , n-i- — ;-; — = — = time down ; bd (•+(/ . ac-l-ad he a-r — r = = time up. he c\d ' EaUAT ONS OF THE FIRST DEGREE. 103 It ia evident that the rate of the current will be half the difference of the rates of sailing down and up ; that is 1 ( ac-\-ad ac-\-ad} __a(c^ — d') 2 1 Id be S ~2bcd ' Lastly, the rate of rowing will be the difference between the rate of sailing and the rate of the current ; that is, ac-}-ad a(c'' — d'') a(c-\-d)'' bd 2bcd ' 2bcd ' NEGATIVE SOLUTIONS. Article 164. Enunciations of questions 2, 3, 4, 5, and 6, so that the results shall be true in an arithmetical sense. 2. What number must be svhtracted from the number 30, that the remainder shall be 19 ? Ans. 11. 3. The difference of two numbers is 9, and their sum ,26 ; re- quired the numbers. Ans. 17 and 8. 4. What number is that whose third subtracted from its lialf leaves a remainder 15 ] Atis, 90. 5. A father's age is 40 years ; his son's age is 13 years ; how many years since was the age of the father 4 times that of the son ] A.ns. 4. 6. The triple of a certain number iticreased by 100, is equal to 4 times the number diminished by 200. Required the number. Ans. 300. Article 169. (1) Here we find x= — -i. m — n l"". There will be a negajve solution when n is greater than m. 2"^. The value of x will be infinite when m ia equal to n (Art. 136). 3"'. When q is 0, and m is equal to n, there will be an indeterminate solution ; that is, z may have any value whatever. (2) 1". The boats will meet half way between C and L, when m is equal to n. 2"''. They will meet at C when m is 0. 3"*. They will meet at L when n is 0. 4". 'Ihev 1 04 KEYTOrARTSECOND. will meet atove C when m is less than n, and the boat A runs in an opposite direction from C to L. 5". They will meet below L when m is greater than n, and the boat B runs in an opposite direction from L to C. 6"". They will never meet if m and n have different signs and are equal to each other. 7"". They will sail together when a is zero, and — m=n, or 7n= — n. (.3) Let a;== the number, the 4 whence 2x-\-i=2x — 7, or 11=0. This result is absurd, therefore the question is absurd or im- [cesible. (4) Let x= A's age, then x — 6= B's, and x — 10= C's ; .-. -+'^~'"=l(a— 6)+l. 3 4 12 or 4a;-l-3a; — 30=7a;— 42+12. 0=0. Hence x may have any value whatever, thus if A is 80 years of age, B will be 24, and C 20. (5) We shall find the same values for x and y from any two of the equations, for example, from the 1'' and 2"'', 1 and 3"', 1" and 4", 2"'' and 3"', 2"'' and 4'*, or 3"' and 4'*. Hence we may take either two of the equations and the other two will be redundant. (6) Prom the 1'' and 2"'' equations we readily find x=5 and y=3. From the 1" and 3'-'',x=6|, and i/=2|. From the 1'' and 4"' x= — 6 and y=8. Hence the equations can not all be true at the same time. EXAMPLES INVOLVIHG THE SECOND POWER Of THE UNKHOWN QUANTITY. Article 171. (9) First divide both members by x". (11) Let x= the number, then 4('?X-)=-X-X-, or x'='^- ^ 2 2 / 3 3 3 27- FORMATION OF POWERS. 105 Multiplying both members by 27 and dividing by x', we find r=27. (12) Let x= the length and y= the breadth, then an/= the number of square feet. From 1" supposition (x+4)(y+5)=a:y+ll«. (1) From the second supposition (r+5)(y+4)=^i/+113. (2) Performing the operations indicated, omitting xy on each side, and reducing, we have 5a;+42/=96 ; 4a;+6y=93. Whence x=12 and y=:9. INVOLUTION OR THE FORMATION OF POWERS. Note. — Most of the examplef? in the Formation of powers, and the Extraction of roots, being performed by direct methods of operation, which the attentive student will readily understand, it is not deemed necessary to give these solutions liere. Those only will bo given which present some peculiarity. Article 172. There is a theorem by means of which the cube of any bino- mial may ha written directly, which the pupil will sometimes find useful, viz. : Theorem. — The cube of any hinomial is equal to the sum of the r.vhes of the two terms, plus three times their product multiplied by the binomial, if the second term is positive, or minus three times their product multiplied by the binomial, if the second term is negative. Thus, (a— i)'=a'— 3a=J+3ai2— i'=a'+(— i)'— 3ai(a— J); which proves the theorem. This theorem gives at once the results in examples 25, 27, and 28. Observe that xy}-=l, e'^Xc"'=e'X ^- = 1. X e* 106 KEY TO PART SECOND. f29) Let a and a-l-l be two oonseoutive numbers then (a+l)==«2-U2a4-] : cliff."^2a+l=a+(a+n. ( 30) Let a — 1, «, and a-^-lfhe any three consecutive number* ; then (a — l)-|-a-j-(a+l)=3a= their sum. (a— 1)3= o'— 3u2-|-3a_l, {ay = a?, (a+l)3= fflS-l-Sa^+Sa+l, Sum =3a' -l"6'2=3'2(a'+2) which is evidently divisi- ble by 3a. The theorem may be proved in a similar manner by assuming i, a+1, and a-|-2 for the numbers. EXTEACTION OF THE SQUARE ROOT OF POLYNOMIALS. Article 183. (13) The terms arranged with reference to x, give 5 25 5 ri7) 1 513 2 8 16 128 2-^- 2 4 ^—x^—- 4 ~4 8" 64 2— a;2— _ x' 4 16 8 64 , &c. 8 ' 16 64 2,56 gT 64 256" EXTRACTION OF ROOTS, 107 A more elegant metliod of extracting the square root of 1 — x'', 18 by means of Indeterminate coefficients, Art. 317; or, by the Binomial theorem, Art. 321. (18) The operations in this example are similar to those in the preceding. EXTKACTION OP THE CUBE EOOT OF POLYNOMIALS. Article 191. (6) In solving this example let a-\-l be considered a single quantity. It may, for example, be represented by a single letter as h. ] 1 3 .9 , &c 3-.+I' — X 3 3 27 3 27 EXTRACTION OF THE FOUETH BOOT, SIXTH ROOT, &C. Article 192. a;"— 2x'+3— -+— = sq. root. x^ X* (11 a;»— 4c«+10a;i 3» ->-+"-S+2->i 2x'— 2a;2| — 4r«+10a;'' — 4a;«4- 4a;'' 2a;4— 4x= +3 6x<— 16x2-1-19 ea;"— 12x2-i- 9 2x<— 4a;2+6— 2 x^ - -+-s+^ 2x<— 4a;'+6— 1+i x' X --+»-"+l a!^ X* x^ x' 108 KEY TO PART SECOND. The square root of x* — 2x'-{-3 — — +— is now readily found to x' if be a:^— 1+i x^' (12) The terms arranged with reference to the powers ot a, give a8— 6a<-|-15a=— 20+— — ^+1 The square rout a' a* a' of this, found as in the preceding example, is a' — 3a 3 1 +- — — ; and the cube root of this, found by the rule in a a^ Art. 191, is a — _. a It is proper to remark that both the preceding examples may be solved without using fractions in the operation, by multiplying all the terms of the polynomial in example 11, by afl, and writing i' beneath it, and after extracting the fourth root of both terms, dividing by x''. We should thus find x^ — 2x^-\-Zx* — 2x'-|-l for the first square root of the numerator, and x* — x'-{-l for the sec- ond. Similarly, in example 12, we must multiply all the terms by a'. It is recommended to the pupil to solve these examples by both methods. RADICALS. Note. — As most of the examples in Radicals are performed by direct methods of operation which the careful student can scarcely fail to ap- ply properly, it is not deemed necessary to present all their solutious. BEDDCTIOJI OF KADIOALS. In the reduction of fractional radicals of the second degree there is a principle with which it is well pupils should be ac- quainted, as it both facilitates and simplifies the operations. This principle is, that if a numler contains a factor thai is a perfect square, the numher may be made a perfect square ly multiplying il by the other factor. Thus, if the denominator of a fraction is a'J it may be made a square by multiplying it by b. For example, \72 M36X2 M36X2= M36X4 6X2 12^ RADICALS. 109 If the denominator contains no factor that is a perfect square, 't can only be rendered a perfect square by multiplying both terms by itself. Thus, Article 199. (3) s/(.x^^y''){x+y)=^(,x-\-y){x—i/){x+ij}={x+y)^x—y. Article 200. In order to separate a quantity into two factors, one of which Is a perfect power of any given degree, it is necessary to ascer- tain if the quantity contains a numerical factor that is a perfect power of that degree. To do this we must see if the quantity ia divisible by any of the perfect. powers of that degree. Thus, if the radical is of the third degree the perfect powers to be tried as divisors are 8, 27, 64, 125, 216, 343, 512, 729, &c. If the rad- cal is of the fourth degree, the divisors are 16, 81, 256, 625, &c. If the radical is of the fifth degree, the divisors are 32=2'. 243=3', 1024=4', and so on. (41 Vi=l'AX|==V|X4=|V4; 3/1=5/75^=15/6; V1=V3|bX 36=^^36; Vi^=Vi^X|=VTkX45=iV45. ^'^ fHi'^i- ^^^X2X33=y2X34V54. (7) s/64=i/32X2=2V2; V'729a«=V243o'X3a=3aV3a; « /!='' /ix-=' /ix2'=-V2==-V32 ; Aj2 N2 2' N2« 2^ 2^ 110 KEY TO PART SECOND, or thus » /?=! /?X -=\/\x3X4^=-i'768. The first method has the advantage of giving the reEuJt in the moat simple form. ADEITIOIl AND SUBTRACTION OF RADICALS. Article 204. (14) 2^1=2V|X2=V2; 8V^',=8^g'4X2=2V2. .-. Sum =3^2. (17) 3 /?= /9X?=V6; 7 /?2=7 /^><^X?=7 /Ax6 ' Ms \l 3 ^ MSO V25X2 2 N 100 =f JV6 ; — V54=— V9X6=— 3V6 ; ••• 's/6+fW6-3V6=(l+f-J-3)V6=JoV6- (18) -Wl2=— iV4X3=— V3; 4V27=4j9X3=12V3; -V/B=-2JTVX3=-iV3. .. 2^3— v'34-12Vl—iV3=(2— 1+12— i)V3 (20) Vl6=2, V81=l^27X3=3s/3, — «/— S 12=— V— 8' =8, Vl92=5/64X3=4V3, — 7V9=— 7V3; . . 2+32/5+8+4^3—7,5/3=10. (21) 8(3)^=8 JpO=4j"3;^X 12^=1^4X3=73, -4 X27^=— 4 V9X3=-4V3, -2(-^%)^=-2jJ^i ••• 4V"3+V3-4v^— 1J3=(4+1— 4-pV3=W3 23) ^J=^^^xai=^J^; RADICALS. ill ^c ^c ^c -Jab4-.^^Jab=—Jab. c 2c ^ 2c MULTIPLICATION AND DIVISION OF RADICAL* Article 205. (9) V3=Va'. 2/2=5/2=; V3'XV2'=V3'X2^=V108. (10) lfb='ljh\ i/a='y'^; 3'5yFx4>V"«'=12'V^<. (11) V2='V2"S %/'5='yr, V5"='V5'; ^ 'V2'X'V3'X'V5'='V2'X3^X6'='V64X81X126 ='V648000. (12) V2Xi;/3=V23XV3'=^23X3=='5y2=X3^ N2 N3 \/25^V3^ N2'X3'' Or thus, V2X V3X VlX^I=V2=X VjX V3X Vi = V2, Since 22X^=2, and 3X1=1. (13) y^'—^'^x*, 3V^=v^==v^2 (18) V^-^-^/2=V72-^;^F=V9='^^/9=V3^ (19) 45/9-H2V3=4^9=-^28/35=2V3". '20) 20^200^4V2"=20^200-^4«/8=5«/25=5V5. (21) y'f2-^y3=y72-r-x/'^=%/'8=J^8=y2. (22) ?/4-^^5y6=22/|=2=/|X|=23/2'7X18=fVl8. 112 KEY TO PART SECOND \ 3 9 \'21 3 <-' ^H&' •^/S-■^/rC.xH; (24) yy=yix2=y2. V2+3 J-i= V2+3 Vl X2= ^2+1 J'2=^y2 (25) 3+^5 (26) V2+1 2—^1 ^^2—1 6+2 V5 2+V2 — 3V"S— 5 -V2--1 1— V"5. 2—1—1 (27) llV'2— 4jT5 (28) J^+J^. x/6+V5 V2+V3. 11^12—4^90 +11^10—4^75 2+V6. +^/6+3 Ilv'"i2—V 10—4^75 5+^76" =2277-710-20^3" 5-1 a^T =277—710 25+1076" Observe that— 4790=— 12710 -fl076+4X6 49+207'6. 29) 3^4+672X672"=15^4X2+1272" =15^4X2+4X37¥=l5X2j2+372=3cj2^3^ 10) '^12+719X'^12-719=V[(12+7T9)(12— 7l9)( = Vll44—19i = 7l44— 19=7125=5. RADICALS. 118 (33-) 2^8=4^2, ^72=6^2, 5^20=10^2; (2 J8+3V5— 7V2)=3V5— Sv'l ; (V72— 5^20— 2V2)=4V2— 1075" 12710—24 —150+30^10 42710—174. Ans. Article 206. (4) Multiply both terms by 4? or 16. (5, ?=^2^«=^^x!±V2_4+V2-_ - 3—272 3—272 3+272 9—8 7 3+72_73+72 73+72^5+276^g_^3 ,6. 73—72 73— 72 73+72 3—2 „^ ^&_VJ±i^?+7|^J+37C5^3^^ ■' 2—73 2— 73 2+73 4—3 rs^ l=^=l=4v?:r4='=^=2-75: ^ -* 3+75 3+75 3— 75 9-0 ,a^ 37-5-272 27-5+7T8 18+57l^ _,^, ,^, ^ ^ 27"6-7l-8^275+7l8~ 20-18 ^''^ (10) Multiply both terms by 72+73+75", and the fraction 7'2+73+75 becomes = ; then multiply both terms b* 276 _ _ _ 76 and it becomes 27^+372+730 ^ 12 _3+47'3 ^7-6+72+7-5 ^ ^ 76+72-75 76+72+75 Ji+473X7_6±_7E+7a)^ ^ ^ 3_ ^^, j^ 3+473 ^ ^^ ^^ !I11 KEY TO PART SECOND. 1 X — J x'^ 1 X — Jx- — 1 ,-5 -, (121 y ^ — ^ =x — ^x' — ]; x-^^x'—\ x—Jx'^—l a;=— (a;2_l) —4= x-^^^^S=^-±-f,^i-+v^^. x—Jx-'—lx-\-Jx''—\ x-~{x^—\) Sum =2j:. Ans. (13) Multiply both terms by ,J {x-\-a)-\- J {x—a) and the frao tion becomes fa+'^)+V(^^-«^)+(a=-^) {x-{-a) — {x — a) (14) m;l±^l£=l^^^+}M^t:^-x^+j-^^ Sum ^2.r-. , „ l+s/2 2+V2 4+3^2 fl8) V20+n/]2.^ V5+ J 3_ 16+2 V60 =8+2^15=15.745966+. IMAGINARY, OR IMPOSSIBLE QUANTITIES. Article 210. BRACTIONAL EXPONENTS. 115 ^4) (2V'=3)X(3V^)=(2v'"3V— X)X(3v'^>/^) =—676". — l+V— 3 — 1— V— 3 1— V^ 1+V^ — V^3-3 +V"^^(-3) — 2— 2V^^ — 2+2^^ — 1+ V^^ — 1— v^^ 2+27'=3 2—27"^ — 2V^— 2 (— 3) +2^^— 2(— 3) 2 4-6=8 ; 2 +6=8 ; 2'=8, and|=l. Ans. 1=1- ^"^• (6) 6V=3=6V3j"^; 2j=4=2v'4v'^=W^^; 6737-1-^-4^-1=— -5^ =V3. ^ ^ 47—1 2^ „, i+V~i ., 1+7^1 1+27^ -1 27-=! — (^^ i_7=i>/-^- (8) (a;-l-a7^T)X(x— a7"^)=x3— a=(— l)=a;2+a2 . (9) Multiply the quantities together. MULTIPLICATION AND DIVISION OF QUANTITIEI WITH FEACTIONAL EXPONENTS. Article 213. (4) J-\-ahi+bi (5) x^y+y^ a^—b^ x*—y~^ — a^b^—ah^—b — x^y^—y^ a — b Ans. x^y—y^ 116 KEY TO PART SKOOND. (6) (a-t-t)'"X(a—J)'"=[(a+i)(3— '')]'"=(«'— *')■" i (7^ Observe that ?-!=! ; l_?=3^^_2m_ 3 4 12 m ji rare Q8) a^—bi Joi— ji ahi—ah^. a4ji — ji ah'^—bi. (9) a— 52 lai^-a^ji-l-aift+jt fl 1 L S — „ b^—a^b—a^h^—b'' — a*i^ — a'^b—a^b^—b\ POWEKS AND KOOTS OF QUANTITIES W I T J FKACTIONAL EXPONENTS. Article 215. (5) ah^'+a'^x o»=l, x»=l, (Art. 82). a^x '-\-a ^x a^x-'-lf-aOx" +aOx<'+a~^x' Jx~^-\-2 +a"3j;5 a^a;-'+a"^j; ax ^-\-'2aK-c~'-\-a 'x + a^ x '+2a"k + a^x '-\-2a"h+a-^x> EQUATtONS CONTAINING RADICALS. 1)7 (,6) [3(5)^]^=r3^X5^=3^X5^=(33X5)^=(135)i ^ 7a\a)^ I i^ {la^^a^ ^^ 7^a(a)^ _7^a? ^9(3436^)^) (s^VV^)^) 3(7^6^) 36* (.8) 5a;'— 4x(5ca;)^+4c Is^a.^— 2ci Ans. 5a;' 2(6^)— 2c^l — 4a;(5ea:)g+4g — ix{5cxy-\-4c. (9) Separating the third term into its parts, and arranging the terms according to the ascending powers of a, we have 1 3 ^ , 41 3a^ I . 1 — -a +._a — — -\-a- 2 16 2 3ai 1 — — +a.= sq. root 4 2-JA 4 3^^_l41 — -a + — a 2 ^16 2 16 2-3«^+a 2 2a-^l+a^ 2 2a-?|.Va^ (10) iflS — 3a^i-\-6db — 86^1 ^a— 2i^= cube root. |a'— 3ai^+46 — 3a=6^+6a6— 8J^ — fa'i^+eaJ— 85^ EQUATIOHS CONTAINING RADICALS. Article 216. (4) Transpose 3 and then square both sides. (5) Square both sides, transpose 1, and square again. 118 KEY TO PART SECOND. (6) Square both sides, omit x on each side, divide both sides by 2a, transpose ^x, or _ and square. The answer is either ^ or ^ ■' , the two being equal to each 4 4 other. (,7) Square both sides, transpose 2x — 3a-\-2x, divide by 2 and square again. • (S) Square both sides and transpose 13; square again and transpose 7; square again and transpose 3; whence ijx=l and x=\. (9) Multiply both sides by the first term, transpose 2-\-x and square both sides. (10) Multiply both sides by ^x, transpose ^a, then square both sides, and omit x^ on each side. (11) Transpose the second term to the left member and square both sides, omit x on each side, transpose the known quantities to the left side and square again. (12) aJx-\-h^x — c^x=d,or (ffl-j-i — c)Jx=d; d' whence ^x= , and {a-\-b—c) (o+J— c)2' (13) Multiply both sides by x^x to clear the equation of frac- tions, then divide by x and we have (1 — a)x=l. The equation may also be cleared of fractions by multiplying both sides by x. (14) Square both sides, omit a^ on each side, then divide by x and square again. f 15) Since x — 4=(^j:-|-2)(;^a; — 2) the first member becomes ^x — 2 ; then by transposing we have 6=5i.^x, or ll^x=12, whence x=l^i. '16) Since x — a=(^x-\-^a){Jx — ^a) the first member be- comes Jx — ^a ; then by clearing of fractions and '■educing we find J x=iij a, whence a;=16a. EQUATIONS CONTAINING RADICALS. 1]!J (17) Since 3a;— l=(^3i-|-l)(^3!r— 1), the first member be- comes ^3x — 1 ; then by clearing of fractions, reducing and squaring, a; is found =3. (18) J4a+x=2^b+a:—jx, 4a-\-x =46-|-4j; — 4^bx-\-x^-\-x, by squaring, ^bx-\-x^=(b — a)-\-x, by transposing and reducing, hx-\-x'=ib—ay+2{b—a')x-{-x^-, by squaring, (2a — b)x=(b — ay, by transposing, ^^(b-ay 2a— b ' (19) i±^+4 '~=\i-p~,. \ a-\-x \ a — X \ a^ — x^ H=i+N^-^=+^r\/«^=^= by squaring; But 2 /_^= /_i^; .". + =0, whence x= ^ '' \ a-\-x a — X b — c (20) Multiplyingboth terms of the first member by the numerator and then clearing of fractions and transposing, we have 2 J x''-\-ax^a (c — 1) — 2x, 4x^-{-iax=a~(c — 1)^ — 4ax(c — l)-]-4:X^ by squaring, 4acx=a^(c — 1)^ by reducing and transposing ; a(c— 1)2 4c V2 Squaring both sides, and observing that Ljx-{-3 mull* plied by .hjx — 3 produces Jx — 9, we have ^"^+3—2 Jx—9+Jx—3=2 Jx. Reducing, and omitting 2jx on each side, we have. — 2jx — 9=0, and 4(x — 9)=0 by squaring, whence x=9. 120 KEY TO PART SECOND (22) Square both members, omit _ on each side, square again a? and omit _ on both sides ; then multiply both members a;'' by X-, clear the equation of fractions and the value of x is readily found. 23) Square both members, omit equal quantities on each side, place all the terms not under the radical on the right side and divide by 2, and we have V Kl— «')'+2x(l+3a2)+(l— aV}=Ca2_i)_^ ; square both sides, and we have (l_a2)'+2a;(l+3a=)+(l— a>==(a2— 1)2— 2a;(a2— l)+a;2. The square of 1 — a^ is the same as the square of a? — 1, omit- t'ng these and x^ on each side, and dividing by x and transposing we have a^x^'&a? ; whence a;==8. EXAMPLES IN INEQUALITIES. Note. The subject of inequalities, though interesting and highly important in ilself, is not much used in the subsequent parts of Algebra The last eight examples, that is, from the 10th to the close, may be re- garded as so many independent algebraic theorems, the study of w mav be omitted by all except the higher class of students. Article 223. ) Squaring both quantities, subtracting 19 from each, an^ dividing by 2, we have ^7(3>, or <1+3V6 ; 70>, or <^l-|-6^6-(-54, by squaring ; 15^, or <[6;,y6, by subtracting 55 from each member 5]>, or'<^2^6, by dividing by 3 ; 25]>24, by squaring ; . hence ^'5+^1'* ^^ S''^^*®'' '■'^'"^ ^^3+3^2. (8) Multiplying both members of the first comparison by 12, to clear it of fractions, and reducing, wc get x INEaUALITIES. 121 Treating the second comparison in the same manner, wn fina r^4 ; hence if a; is a whole number and is greater than 4 and less than 6, it must be 6. (■9) 2x+7 not >19, 3a;— 5 not <13, or2j:not>I2, 3a; not <18, X not >6. X not <^6. Hence if x is neither less nor greater than 6, it must be 6. (10) In example 6, page 175 of the Algebra, it is shown that Let a=n, and i=l, then by substitution ?i2+l>2n, n^—^-{-l'^n, by subtracting n from each side n^-\-11^n{n-\-l) by multiplying both aides by n-f-1, or re'-j-l^w^+K. (11) Referring again to example 6 we have —-\-—~^2, by dividing by ab. ab ab _+-']>2, by reducing. b a (12) If a;>7/ then ^x'^^y,ani -J ^l>'Jy'K'Jy>or y, •J^y>y' 2V^>22/; .-. (Art. 222) — 2y> —2jxy, add x-\-y to each member then X — y^x — 2jxy-\-y, or x—y'yijx—^y)'. (13) Referring again to example 6, we have a^-|-i^]>2ai, subtract ab from each member ; a^ — ai-j-i^^ ab, multiply each side by a-\-b, we iiave a? -\-¥~^a''b-\-ab- , divide each side by a^i', a' I /;' ^.^ aV} , aV , . — + — > — + — , or reducing, h'^a^^b^a 11 122 KEY TO PAR'i SECOND 04) Fi'om question (j, we have also a2_|_c2^2(zc, una C^-(-'3^^26c, from whir.li, by adding together the corresponding mem- bers, and dividing each member by 2, we have (!5; — II^>, or <^ "^ ; multiply both members by a-{-h, a^-\-li'' a-\-b a^+alA\-a}b+¥^^ or +h''y, or , or <2aJ. But it has already been proved in example 6 that a^-\-b^ is ^2ai, wljen a is not equal to b ; therefore, a'+b^^ a^+l'^ Or thus, a^-{-b^^2ah, multiply both sides by ab, a'6-|-a6^>2a^i^ add a^-{-¥ to each member, a''+a'h-\-a¥+b'ya'+2aVy'+b-' ; dividing each member by a-{-h, and then by a^-|-&', we haie " "^ ]>'' ~r , which establishes the proposition. (16) a:2=a=+i^ and 2/2=c=-fd- ; x'y2=(a2+i2)(c2+d2)=aV+a'(Z-'+i'c=+i2d^ (ac+M)2=. . . . a=c2+2aterf+/P(/2, xy—{ac-\-bdy=aW—2abcd+b^'c^=(ad—bcy, but ai^ji- — (ac;-|-6d)2=Ja;j/-|-(ac-|-6d)J Ji^ — iac-\-bd)\ j divide each member by xy-\-{ac-\-bd), and we have xy—{ac+bd)= *■", ~ /, -. j^^-|-ac-|-«o But tlie second member of this equation is necessarily positive since the numerator is a square and the denominator positive hence the first member is positive ; that is, xi/'^ac-\-bd. flti CATIONS OF THE SECOND DEGREE. 133 (17) a^^pa^ — — c)', since (i — c)^ is neressarily positive, ]>(a4-i — c){a-\-c — li) by factoring ; Xa-[-;,_c)(ft+c— a; ; c'>c2— (.a— i)2, >(o+c— i)(ft+c— a). Multiplying together the corresponding members of these inequalities, a''b^c'^'^{a-\-h — cy{a-{-c — hyQ)-\-c — ay ; extracting the square root of both members we have ahc^{a-\-h — c) {a-\-c — V) {b-\-c — a) . EQUATIONS OF THE SECOND DEGREE. INCOMPLETE EQUATIONS. Article 228. (12) Multiply both members by ,Ja^-{-x^, transpose o'-f-a:' and square again. (13) Multiply both members by Ix, transpose ab and then square each member. (14) Multiply both members oy the product of the denomina- tors and reduce. (15) Multiplying both members by the denominator of ti>» first transposing and factoring, we have a3(l_6)2 =(i-|-l)2(ffi3— j;2), by squaring, (b-\-iyx'=4a^b, by transposing and reducing ; ,-. x'=: , and x=z±: — y-^ (t+l)2 b+1 SUESTIOKS PRODUCING INCOMPLETE IVAi >»» OF THE SECOND DEGKEE Article 229. (2) Let x= the number, then a;'— 17=130— 2x^ Whence x=T. 121 KEYTOPARTSECOND. (3) Let 1=: the number, then (10— x)x=10(a;— 6|). Whence x=8. (4) Let x=: tlie number, then Whence x=6. (5) To avoid fractions let 9x= the greater, then f of 9j[r=2« and 9x — 2x=7x, will represent the less ; .-. (9.r)2— (7x)2, or 81x2—49x2=128. Whence x=2, . . 9x=18, and 7x=14. (7) Let x= the greater number, then 14 — x^ the less ; then -^— : Ll=? : : 16 : 9 ; 14 — X X whence = -: ; clearing of fractions 14 — X X 9x^=16(14 — xy ; extracting the square root 3x =4(14— x) ; Whence x =8, and 14 — x^6. (8) Let x= the number, then (20+x)(20— x)=319, Whence x=9. (9) Let x= the greater, then Jl-= the less, and X _^126_^ X _ x^ _^\ ' ~x 12ti~T2G^ 2' Wlionce x2=441 ; .-. x=21 and i??=6. X 10) Let x= one of the numbers, then "=: the other, and X . f yX x' p Whence x=sjpq\ 1= ^ = P, EaUATIONS OP THE SECOND DEGREE. 125 (11) Let a;= one of the numbers, then its square is x^, and the square of the other is 370 — x^. .-. x^— (370— a:2)=208. Whence x=n, and 370— x==370— 289=81. •. the other =J81=9. (1-) Let a;= one of the numbers, then its square =x', and the square of the other is c — x' ; .-. x^ — (c — x'')=d. 2x'=c-\-d, 4x-—2{c-]-d), 2x= JUic+d) x=l^2{c+d) ^c—x-'=Jc—l(ic-\-d)=J}Xc—d}=y2(,c—d). 6x (13) Let .r= the sum, then — = interest for 1 year, and 100 1 6x 6x 1 _ of — = — = interest for 3 months, or _ of a vear 4 100 400 4 -. rX— -=720, or -1=720; 400 400 5x^=720X400, a;2= 144X400, X =12 X20=240. r]4) Let x= the first, then _= the second, and _^ the third X X (17) Let a;=: number of drawers, tlien xy,x=x^ Ihe number of divisions, and a;^X4x=4x'^5324. Wheice x3=1331, and x=U. 126 KEY TO PART SECOND. (18) The solution of this question involves a knowledge of two elementary principles of Natural Philosophy, with which the student should be rendered familiar by simple illustrations. 1st. In uniform motion, the space divided hy the time is equal to the velocity or rale of moving. 2nd. In uniform motion, tlie space divided hy the velocity is equal to the time. Thus, if a man travels 80 miles in 4 days, his rate of traveling (velocity) is 20 miles per day. Or, if a man travels 100 miles at ihe rate of 20 miles a day, the time of traveling is 5 days. Let x= the distance B traveled, then a;+18= " A " Then since the distance traveled, divided by the number of days, gives the number of miles traveled in one day, or the rate of traveling, we have — , or — ^= A's rate of traveling, and 15] 63 ^±1!= B's rate 28 But tne distance traveled, divided by the rate of traveling, gives the time, therefore (i-t-lS)-^ — = — ^ ~^ — i^ time A traveled, and ^ ^ ^ 63 41 ^_^ x+\S _ 28x _ .. jg ■ 28 a:+18 But since they both traveled the same time we have 63(x+]8) _ 28j; ~4x ^18' Divide each side by 7 to reduce to lower terms, 9(x4-18)_ Ax ix x-\-\S' Multiplying by ix and a;-|-18, and indicating the opera- tions, we have o(a:+18)'=16x' ; Extracting t.ie square root of both members, Z{x-\-\S)=ix. Whence a;=54, and ,r+18=72 ; and 54+72=120, the required distance. EQUATIONS OF THE SECOND DEGEE13. 127 f 19) The solution of this question involves principles analogous to the preceding. Let x= the number of days, then x — 4= days A workedi and X — 7= days B worked. Also = A's daily wages, and = B's daily wages. x—i X — 7 If B had played only 4 days he would have worked x — 4 days, and would have received j 1 (x — 4) shilling If A had played 7 days he would have worked x — 7 days, and would have received I ) {x — 7) shillings. But by the question each \ X — 4 / would have received the same e'_m, therefore, Multiplying each side by x — 4 and x— 7 to clear the equa- tion of fractions, and indicating the mult'plication, we have 75(x— 7)==48(a;— 4)2 ; dividing by 3 to reduce it to lower terms 25(x— 7)2=1 6(a;— 4)2; e.xtraoting the square root of both members, 5(x— 7)=4(x— 4). Whence a;=19. 2C^ Let -^ the part of the wine drawn each time, then _ of X X 1 (the whole) is _= the part drawn at tlie 1'' draught, and X \ X \ 1 — -== = the part remaining after the 1" draught. X X x—\ 1 pj. a;— l_a;(a;— l)_(x— l)__(,r— l)(a:— 1) XXX a;2 a;2 a;' _(^2lll=part left after 2"'' draught. a;2 ( x—\y _\^^ {,x—\y ^ x{x—\y _ {x—\y ji^x-~.)(x-\)' x^ X i^ a;3 ^3 ^ _(^-— !)'_ p^^j jgcj. ^(.(.gj. gj^ draught. 12S KEY TO PART SECOND. (a;_l)3^1 ^(x—TL)^_ x(x—\y _ (x—iy ^iia:—l ){x—l)' Sx-iy_ part left after 4"" draught But by the question there were 81 gallons of wine left after the 4"" draught, or ^Vs of thp quantity at the begin- ning, {x—iy _ 81 x'^) 256' ^'_, by extracting the 4'* root. a; 4 Whence 4x — 4^3a;, or x=4, and _=_= a; 4 the part of the wine drawn at each draught. I of 256=:64 gallons drawn at 1" draught ; 256—64=192, and { of 192=48 gallons at 2"'' draught ; 192—48=144, and | of 144=36 gallons at 3"' draught ; 144-36=108, and | of 108=27 gallons at 4* draught. Another solution. Let x=: the number of gallons of wine drawn at 1" draught, and let 256=a for the sake of simplicity. Then _= the part of the whole wine drawn at 1" draught a and 1 — _= =: part left. a a - of =— r= part drawn at 2"'' draught, a a a^ a — X x(a — x) a{a — x) — x{a — x) (a — x)(a — x) and a _{a—xy_ part left after 2'"' draught. a' By proceeding as in the previous solution, we find ^ i-= part of the whole wine left after the 4" draujhi. a" (a— a;)'"_81 ~a* 256 ' !:=_, wnence x=_a=64 from which the other draunhts a 4 . 4 are easilv found. EQUATIONS OF THE SROOND DEGREE. 129 COMPLETE EQUATIONS OF THE SECOND DEGREE Article 231. (31) Multiplying by x and transposing, wo have 4 a;' — —=x= — 1 ; V3 4 4 4 1 a;' — —=3:-\--^= — l-\-^=i-, by completing the square V3 2 1 2 1 3 _ 11 — (32) Multiplying both terms of the fractions in the, left mem- ber by X, we have x-)-l_|_x-|-l__ — ^ . [miiji plying ijoth terms again by a;' — 1, x^ — 1 X — 1 4 x''-\-l-\-x^-\-2x-\-l=^^x- — ^^ ; transposing and reducing Whence x^3, or — |. (35) Transposing and dividing by c, we have a:^ — _a;= — , completing the square a;2 — _a;-(-"=_ — —=-r ; whence C C C a , b adzb X — _=zfc- ; or x=^i^. (36) Transposing ah and completing the square, we ;iave 4 4 4 Whence x=-J—dz =a or J. 2 2 (37) Dividing by a — b, transposing and completing the square. 130 KEY TO PART SECOND. ^5_ a+/;,^, {a-\-hy- _a'— Sat+Q/;' o^ 4(a— i)2 4(a— i)2~" Whence a; ==-- ^t — ± =1, or "V 2(a— i) 2(ffl— i) a— i' (38) Transposing np, dividing by mq and completing the square we have .2_ "'^-^? ^^("''i— W) '_ {mn+rqy ^ mq 4m''q' 4m'q^ Whence x ^ ^nn-pq ^ mn+pg ^n^ ^^ _p 2mq 2mq q ni (39) Observing that ar' is the same as 1, clearing of fractions X by multiplying by ax, and completing the square, we have , I a'c^ a-c^ — 4ai a:-' — acxA- = 4 4' Whence ^<^^±AS^'^f=^ (40) I'irst, L —1 —1 1 1 1 1 ' (a6=) ^-\-{a?h) 2 _L- +— !— ■— +-- {aVY ia%y ah ah'^ multiplying both terms of this fraction by ab, it becomes ah ; the equation then becomes a^+i 20-71^ — (a* — i^)x= it 1 ' ■ a^+i^ a^+iJ Multiplying ooth members of this equation by a'-j-i'f it becomes a;' — (a — b)x=ab. Whence x^a, or — b (41) Dividing both sides jy --ac, transposing, and completing the square, we have ^^,_ ad—bc ^^{a,l—bcy_(ad+hcy ac Aa^c' 4a-c^ Whence x=''±±±^!i±^=i or -*. ;2(io 2ac c « EaUATIONS OF THE SECOND DEGREE 131 (42) Square both members, and then multiply both sides by a;-f-12. (43') Square both members, omit the terms wliich destroy each other, transpose a, and square again; the equation will then be free from radicals, and the value of j. is easily found. (44 Multiplying both members first by i-\-Jx and then by Jx, we have ^4i;2-l-2^.r=16 — ^x, or, since ^4x^=2x 4x=2o6 — dtix-{-9x''. Whence x=:4, or ^. The first value verifies the equation when ^x is taken plus, and the second when it is taken minus. (45) DividincT each side by the square root of x, and observing that ^x-=x, we have X — i=Jx, or x^ — 4j;-|-4=x, by squaring. Whence x'' — 5x= — 4, and a;=:4 or 1. The first value verifies the equation when ^x is taken posi lively, and the second when it is taken negatively (46) Squaring both sides we have x-\-a-\-x-{-b — 2 J [x''-\-{a-\-b)x-\-ah]=2x ; omitting 2x on each side, transposing a-\-h and squaring again, we have 4x''-\-i{a-\-b)x-\-4ah={a-]-by ; from which by transposing and reducing, we find x^+(a+b)x=^ 4 Whence x^-\-(a-\-b)x-\-i — ! — i-=^ -'-''- ' ^ L 4 4 2 2 132 KEY TO PART SECOND. (47) Transposing the second term, squaring, and omitting — 2alcx on each side, we have (x^->[-c^)ahi=(^a^-\-h^)cx ; transposing and dividing bv ab a/j 4a^0^ 4,a^b^ ' _^ _ (a^-\-b^)c _^ (a-—b')c _ac ^^ be 2ab iab b' a' Another solution : Let Jx=z, tlien x^z'' ; transposing — cjab, dividing by ^ah, and substituting 2' for x, and z for ^x, we have {a—b)^li (a—hyc (a— bye (_a+hyc ^ ~ Jab 'i'^^ "" 4ai '^'^~ 4ab ' (a-b)J^ {a+byji Whence z= =r-± =r- 2jab 2jab ^2nJ~e joe —•2bJ ~i ihc 2j'ab~^b'°' 2V^""^«' ■. Jx=J"f, or /^, and x=l=, or ^1 M w \ a h a' v48) ]\Iultiplying both sides by Ja-{-x, we have 5 J a' — ^-=— — X, by transposing „3 ~3 49a' 14n: , , , a — x'=: — — x-\-x^, by sqiiannsf a;^— yr=— _-_, by reducing. Whence x='^.% or ??- 6 ft' EQUATIONS OF THE SECOND DEGREE. 133 tROBrEMS PRODUCING COMPLETE EQUATIONS OF THE SECOND DEGREE Article 233. (5) Let x= one of tlie numbers, then 20 — x= the other, and x{W—x)=ZQ. Whence x^2 or 1», therefore 20 — x^l8 or 2. (6^ Let a;:= one part, then 15 — x^= the other, and x{\f>—x) : x''-\-{\b—xy : : 2 : 5 ; .•. 4i- — 60x4-460=75a; — 5x^, reducing x- — 15a:= — 50. Whence x=10 or 5, and 15 — :c=5 or 10. (7) Let x= the number, then k(]0— x)=21. Whence x=7 or 3. (,8) Let x= the less part, then 24 — x= the greater, and x(24 — x")=35(34 — x — x), reducing x2— 94x=— 840. Whence x=10 or 84, the'first of which is evidently only admissible ; therefore, the parts are 10 and 24 — 10=14. (9) Denoting the square roots of the parts by x, and 26 — x, we have x=+(26— x)2=346, reducing x' — 26x= — 165. Whence x=lo or 'l, and 26 — x=ll or 15. (10) Let x= the square loot of the number, then x'=: the number, and x'+x=132. Whence x==l], or — 12, and x^=121, or 144. The last number is the answer to the question " What number Jiminished by its square root gives 132 !" (11) Let x= the square root of the numler, then x'=: t'le number, and x^ — x=48f . Whence x=7l, or —6\, and x'=.56j or 42;|-. 134 KEY TO PART SECOND. The last number is the answer to the question " What numbei added to its square root gives 48| '' (12 Let 3c= one of the numbers, then 41 — x=: tlie other, and a;2-|-(4i_a;)2=901. Whence x=lb, or 26 ; and 41 — x=^2B, or 15. 1,13) Let xz= the less number, then a;-[-8= the greater, and a;2+Ca;+8)==544. Whence a;=12, or — 20 ; and a;-|-8=20, or — 12 ; hence the two numbers are 12 and 20. (14) Let x= the first cost, then x= per cent, of gain, and ,, X x' xy, — = = gam. 100 100 .-. a7+-^=2400. 100 Whence a;=20. (15) Let x= the number of miles B traveled per hour, then x-\-\= the number of miles A traveled per hour, thv>n, 39 39 also, and — = the hours respectively which A Hnd x+i X B traveled. -- • 39 . ^_39. K+l X ' Whence x='i, and a;-|-^=3|. (1(!) Let x=: number to whom B gave, then a;-|-40= nifwiiet to whom A gave, then 1200 a;+40 1200 : what A gave to each, and ; « B " " ; . X 1200 I g_ 1200 S+40 IT' Whence a;=80, and a;-}-40=120. (17) Let x= number of miles B traveled per day, then x-\-S= " " A « " ,aad lx= number of days each traveled ; EUUATIONS OF THE SECOND OEGREB. 135 .-. a;X^ar+(x+8)^a:=320, or Whence a;=I6, and a;-l-8=24. (18) Let x= the distance in miles from C to D, then — = number of miles B traveled per day, also — = number of days B traveled, then 32-1-7X — = whole number of miles A traveled, and — X — = = number of miles B traveled. 19 19 361 19 361 Clearing of fractions and transposing x=—228x=— 11552. Whence a;=76, or 152. (19) Let xs= the number bought, then = number of dollars oacji cost, and, since 240-1-E9 X 6=299, -?^= " " " sold for ; X — 3 ... ^_?f2=8. reducing X — 3 X 8a;5— 83x=720. ._ Whence a;^16. (20) Let x= one of the numbers, then 100 — x== the oviier; then a;(100 — x)=x^ — (100 — xf, reducing 100a;-^2-=_iooOO+200x, or a:'+100j;=10000. Whence a:=61.8034-, ana- 100 — a;=38.197, nea.\'.y. Or by subtracting x^ from the square of 100 — x, and leducmg !ve have the equation x^ — 3001=— 10000. Whence x— 38.197 nearly. (21^ Since each received back $450, they both receive^ f 900 and the whole gain was $900 — 500=.'i 13 K E Y T O r A R T b E C O N 1> . Let a;=: A's stock, then 500 — x= B's stoclc. X dollars for 5 months is the same as 5x dollars for 1 month. (500 — x) dollars for 2 months, is the same as 2(50U — xj =(1000— 2a;) dollars for 1 month. Hence the gain, $400, is to be divided into two parts having the same ratio to each other as 5x and 1000 — 2x. But 5x-\-{\000 — 2x) 5x =3a;-|-1000, therefore the parts of the ffain are . ano 3^+1000 , the sum of which is 1, the whole srain. 3j;+1000 " .-. A's gain is ^ of 400^ ^°°°^ - 3j;+1000 Sx+lOOO B's gain is il0.5=^%f 400=1^02^0=80te 3X+1000 3j;-|-1000 ' But A's gain =430 — x. 2000a; .-. =4aO — X. 3^+1000 Whence a;=200, A's stock, and 500— x=300, B's st:>ck. (22) Let x= first part of 11, then 11- a;= the second ; also, Ji= first part of 17, and 11— "11= the second ; then (11— X) ( n—— \ =48, i (11— a;)(17j;— 45)=48j;. Whence x^5, or jl-, 11-^=6, or ??, and 1^=9, or ^1 ; n X \i ... 17-1^=8, or '^ X 11 ■ Hence the numbers are 5, 6, and 9, 8, or fl jf , and jf, 'ff, either of which entirely sa .isfi<>8 the conditions. r23J Let 3,r= the first part of 21, then x= the first part o- 3u and (21— 3x)=+(30— x)==58o; developing apd reducing ^2 9 3_ ST 8 ■»• — 5 * 6 • Whence .r=fi, or 12^', and :;.i = lS, cr 37*. EaUATIONS OF THE SECOND DEGREE. 137 Since the second value of x gives for 3a; a number greater than 31, it is inadmissible. The first value of a; gives for the parts of 21, 18 and 21 — 18=a and for the parts of 30, 6 and 30 — 6=24. (24) Let x= the first part of 19, then 19 — x= the second part ; and since the difference of the squares of the lirst parts of each is 72, therefore, Jx''-\-'Tl must represent the first part of 29, and 29 — Jx'^-\-12 the second part. .-. (29— V^5q:i72)2— (19— a;)2=180, developing and reducing 297x2-f72=19a;+186; squaring each side and reducing 40x2— 589a:=— 2163. Whence x=l, or ^^^ ; this gives 7x^+72=11, or 4^5^. 19— a:=I2, or \%' , 29—7x^+72=18, or ''-^O^K Whence the parts are 7, 12, and 11, 18, O"^ 40 ' 4T) ) anU 4^. 4ij Article 239a. d) In order that the negative answer, — 9, when taken posi- tively, shall be correct, the question should read : Required a number such, that twice its square, diminished by 8 times the number itself, shall be 90. ' (2) From this question we see that the negative values satisfy the question oqually well with the positive, the only dif- ference being that in one case we subtract -\-3 from -[-7 and in the other, — 7 from — 3, (S) Let x= cost, of watch, then x= per cent, of loss and xX = — = actual loss. 100 100 .•. X — =16, or 100 X -100x=— IfiOO. 12 lIvlS KEYTOPARTSECOND. Whence a;=o0±30=20, or 80, either of which fully satisfies the conditions. Thus, 20— fjfo of 20=20—4=16 ; 80— TOO of 80=80—64=16 (4) These values of x show that no j,ositive number can be found which will satisfy the question. But if tlie ques- tion is changed to read thus : Required a number such, that 6 times the number, diminished by the square of tlio number, and the result subtracted from 7, the remainder shall be 2, either of the numbers, 1 and 5, will satisfy the question. (5) There is evidently but one solution, because if 4 is one of the numbers 10 — 4=6 is the other ; or, if 6 is one of the numbers 10 — 6=4 is the other. (6) These results show that the question is impossible in an arithmetical sense. This we also learn from Art. 236, since the greatest product that can be formed by dividing 10 into two parts, is 25. (7) This question is similar to that of the problem of the lights (Art. 239), and the results may be obtained froiu tiie results there given, by making a=a, b^l, and c=re. When a=12 and ?i=4, the parts are 8 and 4, or 24 and —12. When a=10 and n=l, the parts are 5 and 5. (8) Calling X the distance from the earth, a=240000, ^=80 and c=l, we have by the solution to the problem of the V a^b ajb lights, x= — = -, or x^ — - — -■ V'+n/" Jh—Jc To prepare these formula for numerical calculation, mul- tiply both terms of the first by ^b — ^c, and of iba second by ^b-{-Jc ; this gives ^_ a(/>-Vte) _ 240000(80-V80) _.,^,^^^ b—c 19 - ■ -r or .^ — x^=l-]-x^, Squaring x\io — x^)^\-\-^x''-{-x*, transposing and reducing x^ — 2a;':= — 3. Whence x=±J{l±lJ2). Article 243. (2) We find the square root is x^ — x, with the remaindei — 3a:'-|-3x ; hence the equation may be written thus, (a;2— x)'— 3(^2— x)=10S. 144 KEVTOPARTSECOHlJ Putting x'' — x=!/, we find y=^l2. or — 9. .-. x^ — x^l2, or — 9. Whence a;=4, or —3, or ;',(1±V— 35). (3) Tlie square root of the left memher is x^ — x, with the remainder — a;^-|-x ; hence the equation may be written thus, (x2— x)2— (a;2— a;)=30. Putting K^ — x=!/i we find y=6, or — 5. -. x'^ — x=6, or — 5. Whence x=Z, or —2, or J(1±n/— 19)- (4) Blultiplying both sides by x, we then find the square root of the left member is x- — Sx with the remainder ~\-2x- — 6x ; hence, the equation may be written thus, (x2_3x)24-2(a;2— 3a;)=0. Let x' — 3x=!/, then y^-j-2!/=:0, and i/=0, or — 2. .-. x- — 3x=0, or —2. Whence x^O, or 3, or 2, or 1. The value, x=0, does not satisfy the given equation, but is a root of the equation x(x^ — 6x^-|-ll''^ — 6)=U, and was introduced by multiplying the given equation by x. (5) The square root of the left member is x'^ — 3x with the remainder — ix^-\-\2x; hence the equation may be writ- ten thus, (a;2_3x)2— 4(a;2— 3x)=60. Let x^ — Zx=y, then y^ — 4^=60 ; whence 2/=10, or — 6. .-. a;2— 3x=10, or —6. Whence x=b, —2, or ^(3±V— 15)- (6) The square root of the left member is x' — 4x, with the remainder — %x'^-\-'iix ; hence, the equation may be written (x^ — 4x)^ — 6(x^ — 4x)= — 5. Let x' — Ax=y, then y- — 6?/:= — 5 ; from which jf=5 or 1 ; .-. x^ — 4x^5 or 1. Whence x=5, or — 1, or 2±V5. t7) Multiplying both members by 4 to clear the equation of fractions and render the first term a perfect square ; then transposing IBx', wo find the square root of the left mem BaiTATIONS OF THE SECOND DEGREE 145 ber is 4x^ — 2x, with the remainder —ix'^-\-2x; hence ti"i2 equation may be written (4a;2— 2 j;)2— (4x2— 2a;)= j 33 . Let 4x= — 2a;=2/, then y''—y=m, and y=12, or — 11. .-. 4a;' — 2x=12, or — 11. Whence a;=2, — |, or l(l±V— 43). (8) Observe that L+^=_-|-i, then omitting i on each side, 3a; a; 6 6 and multiplying both sides of the equation by 14j;' to clear it of fractions ; after transposing we nave a;^— 1 4x3-[-56a;2— 49a;==60. The square root of the left member is a;' — 7a; with the re- mainder Va:' — 49a; ; hence, the equation may be written (a;'— 7a;)2+7(x^— 7a>)=60. Let x' — 1x=y, then y2+7y=60, from which we find y=5, or — 12. .-. a;= — 7a;=5, or — 12. Whence a;=4, or 3, or i(7±v'^. SI MTTL T A NE O S EQUATIONS OF THE SECOND DEGREE CONTAINING TWO OK MOKE UNKNOWN QUANTITIES. Article 245. Note. — Instead of indicating each step of the solution of the exam- ples in this article, it has only been deemed necessary in most cases to point out the particular step on which the solution depends. (5) Subtract the square of the first equation from the second, then add the remainder to the second, and extract the square root, which will give x-\-y. (6) Add twice the second equation to the first and extract the square root ; also, subtract twice the second equation from the first and extract the square root. (7) Subtract the second equation from the square of the first; then subtract the remainder from the second equation and extracfthe square root. 13 146 KEY TO PART SECOND. (8) Divide the first equation by the second, this will give (9) From the cube of the first equation subtract the second, dividt; the remainder by 3, and we have xy{x-\-y)=3Q8 ; divide by a;-|-!/=ll, and we have !cy^28. Having x-\-y and xy, we can readily find x and y, as in Form 1, Art. 245. Or, thus, Divide the second equation by the first, subtract the quotient from the square of the first, and divide by 3, which will give xy. (10) From the first equation by transposing and extracting the cube root of both members, we have x=2y ; then by substitution in the second we readily find the value of y, and then x. (11) Subtract the second equation from the first; add the re- mainder to the first and extract the square root, which will give x-\-y=zizl2, then divide the second equation by this, and we have x — y=dz2. (12) Divide the first equation by the second, this gives x-\-y^8; from the square of this subtract the -second equation, and divide by 3, this gives ot/=:15 ; subtract this from the second equation, and extract the square root, which will give X — y=-\-2, (13) Subtract the first equation from the square of the second, this gives xy^AS ; subtract three times this equation from the first and extract the square root, this gives X—y=±8. 1 )) Divide the first equation by the second, transpose Slxy, and subtract the resulting equation from the square of the second, this gives ^xy=4, or xy^8 ; then by the method explained in Form 1, Art. 245, we readilv find a;-f-j/==±li. (!6) Dividing the first equation by the second, we find x^ — xry _|-2/-=7 ; subtracting this from the second equation we have 2a:j/=6, or xy=S ; then adding this to the second equation and extracting the square root, we find x-\-y =±4 ; also, subtracting xy=:3 from the equation a'- — xy -|-»/'=;7, and extracting the square root, we find x--y=±.2 EQUATIONS OF THE SECOND DEGREE. 147 (16) Let x*^P and yi=Q„ the equations then become P^— a==P+Q, (1) ps— Ci'=37. (2) Dividing each side of (1) by P+Q, we have P — Q=l, then dividing each side of (2) by P — 0,^1, we have P^+ PQ,+Q^=37, (3) P2— 2PQ+Q,2= 1, by squaring P— Q=l. Subtracting and dividing by 3, we find PQ,^12, then by adding this to (3) and extracting the square root, we find P+Cl=±7, but P— Q=l. Whence P= 4, or — 3, and Q.=3, or — 4, . . x=16, or 9, and y=9, or 16. J. I (17) Let a;4=P and y'^^Q,, then by substitution the equations become P -|-Q, =5, P2-1-C12=13. The values of P and Q, found as in example 1, page 212 are P =2 or 3, Q,=3 or 2. .•. a;4=2 or 3, and a;=16 or 81 ; y3=3 or 2, and y=27 or 8. (18) Let a;5'=P and 2/^=Q, then by substitution the equations become P -|-Q, =5, P3+Q,3=35. The values of P and Q. found as in example 3, page 213, are P=2 or 3, Q,=3 or 2. .-. xi=2 or 3, and x=8 or 27 ; y^^3 or 2, and y=27 or 8. (I9j Let a;^=P and y^=Q„ then by substitution the equations become P +Q =4, P3-1-C13=28. The values of P and Q, found as in the preceding example^ are P=3 or 1, Q,= l or 3. tl8 KEY TO PART SECOND. I ,•. x^=3 or 1, and x=9 or 1. 1 j/J=l or 3, and y=l or 9. (20) Square both members of the first equation, and from tlie result subtract four times the cube of the second, and we have x^—2xY-\-y^=n2225. extracting the sq. root x' — 2/'=±335 ; but a;5+2/'=351. Whence, by adding and subtracting, dividing by 2 and extract- ng the cube root, we ho'e x^l or 2, and y=2 or 7. (21) Raising both sides of (1) to the fourth power, we have x'-\-4x^y+6xY+ixy'+y^=256, but x^ +y=82 ; .-. 2a;<+4a;'j/+6a;2y'+4OT/3_)_22/4=338 ; or x'^+2x^y+3xY+^^y^+ 2/''=169. Extracting the sq. root x^-{-xy-{-y''=13 (3). Squaring eq. (1) x^-\-2xy-\-y^=:l6 ; Subtracting xy=3 ; hence 3a;^=9, and sub- tracting this from (3), and extracting the square root of tlie resulting equation, we get x — y=^dz2 ; from this, and x-\-y=:4, we get x=3, or 1, and y=l, or 3. (22) Adding the three equations together, and dividing bv 2 we have a:y-|-J:z+y^=— ■ "*" , (4). Sibtracting from this successively the three given equa- tions, we have yz=:Jl , (5) xz=^±^, (6) 2 a-\-b — c ,-^ ^=-^=5^ . (7) Multiplying the three equations together, and extracting the square root, ccy^=^ ^( .a+b~c)^a+c-h)(b+c-a-) l _ (8) EQUATIONS OF THE SECOND DEGREE. 140 Dividing eq (8) by equations (5), (6), (7), respectively^ we obtain the values of x, y, and z. Thus, to find x on/z / | (a+/)— e)(a+c-6)(i+c-a) ^ yz fc+e— a 2 = / Ka+i-e)(a+c— 5)(i+e— a) 4 ( ■ ^ < 8 Q,j^c—af\ =-t / (a+i— e)(g+c— t) \ 2(i+c— a) ADFECTED E Q IT A T I H S . Article 250. (,3) Adding the two equations together, and dividing by 2, we ■find x--\-x='iiO. Subtracting the second equation from the first, and dividing by , we find ?/^-|-y=90. (i) Multiplying the first equation by 4, and subtracting the result from the second, and transposing, we find y'— 18,y=45. Whence ^=3, or 15, and x^\i — 4!/=2, or — 46. (5) From the first equation y=_f2li_=l|a;-|-7, tliis substi- tuted in the second gives 3x2+2(1 ^x—4)'=14, developing and reducing, we have x'^—'ix^—'-i. Whence 1=2, or ll, and !/=10, or 8|. (.6) Clearing the second equation of fractions by multiplying by xy, and substituting the value of x=y-\-2, found from the first equation, and reducing, we have V'— 1^='/. and i/=3, or — 1|. (7) Let y=lx, then substituting this instead of y, finding the value of x^ from the resulting equations, placing these values equal to each other, and reducing, we find /=±1^. !..() KEY TO PARTSECONU. Then substituting tliis in the value of x', we find x and thence y. (8) Let y=te, then substituting this instead of y, finding the value of x^ from the resulting equations, placing these values equal to each other, and reducing, wo find i^-j-lii or — f . Having this, the values of x and y are readily found by substitution. (9) Let a:y=t!, then by substituting in the first equation, and transposing, we have ■y^-l-4u=96, from which we find v, or a;!/=-)-8, or — 12. Having the values of x-\-y and xy, we readily find x and y by the method explained in Form 1, Art. 245. (10) Let _=u, then by substituting the values of v and i)' y in the first equation we find ■!>^-l-4i;=%', from which D or _=-(-_, or — . Then, from these equations, and y ^3 3 X — y=2, we readily find x and y, (11) Denoting xy by v, the first equation becomes, by substitu- tion and transposition, 1)^4-8"= 1^0, from which we find v or Xi/=-|-10, or — 18. From the equations a^=-|-10, and a:-|-3i/=ll, we find a;:=5, or B, and i/=2, or |. From the equations xy^= — 18, and a;-l-3y=ll, we find x=:V q=,^v'337, and ^=^±^-7337. (12) Let iJx-\-y=v, then from the I'' equation, we have d2_|-ij^12, from which ■u=3, or — 4; hence, d^, or a:-(-j/=9, or 16. Having the values of x-\-y and x'^-\-y-, we can find the values of x and y by the method e.xpiained in example 1", Art. 245. The equations x+)/=9, and a;'-|"2/^=41, give x=b or 4, and )/=4 or 5. The equationt x-\-y=\^, and x^-|-j/^=41, give J;=8±1V— 174, and y=8=FA V— 174. (13) Adding twice the second equation to the first, we have a:^-|-2x?/+!/=-|-x-|-i/=30, or (x+?/)2_[-(j^+j,)=30. Whence x-\-y=-\-h, or — 6. EaUAlIONS OF THE SECOND DEGREE. 151 From these equatkns and the value of xy=6, we readily find the values of a; and y. (14) Transposing 2xy in the first equation, and then adding both equations together, we have x''-\-2xy+i/-\-4x+Ay=i 17, or ix+yy+4ix+y)=n'7. Whence x-\-y=-\-9, or — 13, and x=9 — y, or — 13 — y. Substituting these values of x in the second of the gwen equa- tions, we have y''-\-2y=36, or y^-{-2y=:5T . From the 1'' equation we find y=5, or — 7 ; whence a;=9 or 16. From the 2"^ equation we find y= — 1±V^8 ; whence «=— 13±s/58. (15) Let -:=« and -=2;, the equations then become X y V +z =a, (1) v^+z^=b. (2) Subtracting the second equation from the square of the tirst. and then subtracting the remainder from the second, and extract- ing the square root, we find v or _= "^ , and z or - X 2 V = ^ ; whence x= , and y=^~ a±j2b—a^ a±^2b—a^' 12 06) From the first equation x=. , and from the second X— ^^ ; whence J^=_iL, or 12(l4-y0=18(y+y'); dividing each member by 6, and then by l-\-y, we have 2(1 — y-\-y^)^3y. From this equation we readily find y=2, or 2 ; whence x=2, or 16. (17) Adding twice the second equation to the first, we have x''-{-2xy-\-y^-\-x-]-y= 156, or (^+y)=+(x+y)=156. Whence x-\-y=-\-12, or — 13. By substituting the value of x-{-y in the second equation, wa find xy=21, or 52. From the equations x-\-y=\2, and xy=21, we find a:=9 or 3 and !/=3, or 9. 152 KEYTOPARTSECIONB. From ihd equations x-\-y= — 13, and xy^b'2, we ftid a:=K— 13±V— 39), and 2/=i(— 13=F J— 39). 18) From the 1" equation, by transposing, we find (x+j,)2_3(a;+2/)=28. Whence x-\-y=T, or — 4, and y=1 — x, or — 4 — a:. Substituting the value- of y instead of y in the second equation, "'6 have 2a;2 — 17a;=35, or 'ix^-\-f)X= — 35. From tlie first of these equations we find x=i, or 3|, whence From the second we find a;= — |±i^/ — 255, and y= — y =F4 ;^— 255. (19) Let f — ) "=t!, then ( ^ ] ^=1, and d+-=2, ot \ a;-|-i/ / \ ^x I V V v^ — 2v= — 1, wlience v=-\-l. Zx .-. =1, whence 2x=y, Substituting 2x instead of y in the second equation we have 2a:2_3a;=54. Whence x^=6, or — 4^ ; hence ^=^12, or — 9. (20) Transposing 3y and adding 5 to each member, we have a;2+3j/+5+4(a;2+33/+5)^=60. Let (,x^+3y+6)^=v, then i.5-}-4u=G0. Whence v=^6, or — 10. . . a;2+32/+5=36, or 100. Finding the vakie of y from the second equation and substitut- ing it in the preceding equations, and reducing, we have x^-{- 'J'x ^an5,or x■'+'ix=^K From the first equation a:=5, or — — , then since ;/= , 7 ■ 7 we find y=2, or — Va"' From the second equation we find x= — §±5^317, and thence E a t ■ A T I O N S OF THE SECOND DEGREE. 1 53 (21) Tlie first equation is {x-\ry)^ '■> ■i-J^+y' JMultiplying botli mombcrs by J x-\-y, we have y -1-5+y^l'^ x^y y 4 ■ Let -l—=v, then v-\--^ — . x-\-y V 4 Wlience v or _Z_r=4, or _. a;+y 4 Fi'om tiie equations ^--=4, or _, we find i= — -y, or +3j/. a;-|-y 4 4 Substituting tlie first value of x in tlie equation x=y''-\-^, wa find y=— 3±iV— 119; lience aizzrijTg^^V— H9. Substituting the second, value of x in the equation a;^y'-}-2, we find y=2 or 1 ; hence x=6 or 3. QUESTIONS PKODUCING S I M U L T A N E O XT S EQUA- TIONS OF THE SECOND DEGREE CONTAINIKC TWO OR HOSE UNKNOWN QUANTITIES. Article 251. Note. — As the first five examples may be solved without completing !h<) square, their solutions will be given in this form. (1) Let X represent the greater number and y the less, then y(a;-|-!/)=4i, (1) and x{x-\-y')='d y . (2) Multiplying the equations together, dividing both members bj ry, and extracting the square root, we find x-\-y=6. Substituting 6 for x-\-y in (1), we have Gyr=ix, or y=f» ^x Then from the eq lation x-\-y^=&, we have j:-|-l_=x ; whence r==3.6, and 2/=2.4. (2) Let x= the digit in ten's place and y= the digit is unit's place, then \Qx-\-y^ the number, and x{Wx-\-y)=:\Qx'^-\-xy=:i&, (1) also a;(*-l-y)= x'^-\-xy=\0 (2) 154 KEY TO PART SECOND. Subtracting (2) from (1), 9a;-=:36, whence x=2, and ;, is readl'y foind =3. (4) Let x= the greater number and y the less, then (ix-y)(x^-,/)= 32, For the method of finding the values of x and y, see the A.ge- bra, example 4, page 213. (5) Let X and y represent the numbers, then xy = 10, a;'-|-3/3=133. For the method of finding the values of x and y, see the soltr- lion to example 20, Art. 245, (page 148, of Key). (6) Let x^ tlie greater number and y th» less, then (^^=24, (1) y (^+2/)y - 6. (2) X Dividing the first equation by tne secoiid, we have y (^+y)y y^ 6 Wlience -^2, and x^2y ; substituting this value of a; in y (1) we find 6y=24, or y=4 ; hence x=8. (6) Let x= the less number, then a;+15= the greater, and j:(a:+15) _^3 2 Dividing by x, and completing the square, we find a;=3, or — 2 1 ; hence a;-}"15=18, or 12-}, therefore the numbers are i md 13, or —2l, and 12 A. (7) Let x=: the greater number and y= the less, tlien xy =24, and x'' — y2=20. From the first equation y= — ; substituting this ior y in the X second equation, and clearing of fractions, we find x'' — 20j,-'=576. Whence x^=5G, and x=6 ; hence y=4. EQUATIONS OF THE SEOOND DEGREE. 155 (8) Let K= the greater number, and y= tlio less, then x'^-\-xy=\2Q, (1) Let y=:vx, then by substitution the equations become 1 on x''-^x'v=120, whence x^= " ; l-\-v anc vx^ — uV= 16, whence x''= V — u'' 120 16 l-\-v V — v'' From this equation wc find t)=|, or i, then x2=i^=72, or 100, 1+v and a:=6^2, or 10, 2/=4VT, or 2. riic answers 10 and 2 are the only ones given in the Algebra, but it may be easily shown that the others are strictly true in an arithmetical sense. (9) Let X and y represent the numbers, then x'+y'+x+y==42, (1) ,r(/=15. (2) If wo add twice the second equation to the first, the resulting equation is (a;+3/)-^+(x+y)=72. Whence x-\-y^8, or — 9. Havinf x-\-y and xy, the values of x and y are to be found as in example 13, Art. 250. (See Key, page 150.) We thus find a;=5 or 3, and y^=Z or 6, or.=rg±VZ^,andy=-^TV^T 2 2 f 10) Let X and y represent the numbers, then x+y+xy:=47, (1) x^+y''—{x+y)=62. (2) For the method of finding the values of x and y see lie soTu- tion to example 17, Art. 250, (Key, page 151). 150 K E Y T O I' A R T B !•, I. O J\ 1^ . (11) Let x= the greater number, and ?/= the lepa, tlien xy =x-{-ij, (1) x-—y''=x-\-y. (2) Dividing each member of (2) by x-\-y, we have X — y=l, or x=y-]-l. Substituting this value of x in (1), and reducing, we find y2— y=l. Whence y=},±:}, jj, and a;=|zbi^/5. In order that the numbers may be positive, we can only use the upper sign ^ this gives a;=2.668, and 1/^1.668 nearly. 12) Let x= the less number, and xy= the greater, then x^y=x^y^ — x^, (1) X'y''-\-x''=x^y^ — a;'. (2) Dividing each member of (1) by x^, we have y^y^ — 1, oi y''z=y-\-\, from which we find y=\-\-}-Jb. Dividing each member of (2) by x-, we have !/'-|-l=:r(y' — 1), but 7/'— y+l, and multiplying both sides by y, we have y'=?/' J^y=:y-\-il-\-l=^y-\-\. Substituting these values of y" and !/', the equation becomes y+2=x(2y) ; y+2 f+j^r (S+iV5) 1-V5 hence x=^-z — = =-= — X = ^y 1+V5 1+V5 1—^/5 4-2 V5 -; -ays i ~ 1—5 ~ —4 — 2^*^- =.1(5+V"5)- 1_13) Let a;= the price in dollars of a pound of mace, and !/:= the price of a pound of cloves, then 80x+ 100^=65 ; (1) 20 — = pounds of cloves for 20 dollars, and y — = " " mace for 10 dollars. X .: ?2-G0=L0, (2) EQUATIONS OF THE SECOND I)i;OREl!. 157 From these equations we readily find x=\ dollar =50 cts, and 'y=l dollar =25 cts. (14) This question may be solved by using only one unknown quantity. Thus, Let Sx= A's gain, then 20x= B's stock, and 100 — Sx= B's gain, and 40x — 200= A's stock. .-. 40a; — 200 : 20x : : 3x ; 100— 3x. Since the product of the means is equal to the product of the extremes, 60x==(40x— 200)(100— 3x) ; reducing x'—^^x=—'-"-^-^. Whence a;=:20, hence 3x^60= A's gain, &c. (16) Let X and y represent the numbers, then by the question xy+x+y=23, (1) x^+y'—5{x+y)=8. (2) Adding twice eq. (1) to eq. (2), we have a;2-f2xy+2/'— 3(j;+i/)=54, or (x-\-yy — 3(a;4-2/)=54. This is a quadratic form and we readily find x-\-y^9, then by substituting the value of x-j-y in eq. (1) we finda;y=:14. Having x-\-y and xy, we can find x and y. (See Form 1, Art. 245.) (16) Let X, y, and z represent the numbers, then x-^y—(ij—z)=x—2y+z=^5, (1) x+ y+z=U, (2) xyz=i950. (3) Subtracting eq. (1) from (2), and dividing by 3, we find y=13; then substituting this value of y in (2) and (3), we have x-\-z—31, 3.2=100. Whence by Form 1'', Art. 245, we readily find x=25, and z=6. (IT) Let X, y, and z represent the parts, then x+y+z=26, (1) a;2— y2=:2/2_z2, (2) ^^+jy'+^'=300. (3^ 158 KEY TO PART SECOND. From eq. (2) by transposing y'^ — 2', we have Subtracting this from (3), dividing by 3, and extracting the square root, we have i/^10; then by substitution, equa- tions (1) and (a) reduce to x -\-y ^16, and x2+y2=200. These equations are similar to those in example 1, Art. 245, and may be solved in a similar manner. (18) Let X and 7/ represent the number of men respectively in the fronts of the columns A and B, when each consisted of as many ranks as it had men in front ; then x^ and y' represent the number of men in the respective columns. .-. _= number of men in rank, when A was drawn up y with the front that B had, and •/-= the number of men X in rank when B was drawn up with the front that A had, hence a7+y==84, (1) iV^=91- (2) y ^ Multiplying both members of (2) by x(/, we have sciJ^y-i=!d\xy (3) Cubing eq. (1) and subtracting (3) from the result, we have 3x^(x+7/)=(84)= — 91x!/, but x-t-y=84, 252x!/=(84)'— 9lxj/, 343xy=(84)3 or, 7'xj/=(7xl2)'=7'Xl-' xy=12==1728. Having x-|-y^=84, and X(/^1728, we find x=48, and t/=3fi, Dy the method of Form 1, Art. 24.'). .-. x-^48'=2304= men in column A ; y2=362=1296= " " " B. F R M U L .S . Article 252. Note. — ExKinples 2 to 5 have pither been solved before, or are so elmple as to require no explanation. 'Wn shall, therefore, merely '.^xjircss the vr.'^pri-livn furimila in the form of Rules. EaUATIONS OF THE SECOND DEGREE, 159 (2) Pkoblem. — To find two numbers, having given the sum of their squares, and the difference of their squares. Rule. — Add the difference of the squares to the sum of the squares, muUipli/ the sum by 2 and extract the square root ; half the result will be the yrealer number. To find t]ie less number, proceed in the same manner, except lliat the difference of the squares must be subtracted from their sum. Ex. The sum of the squares of two numbers is 120', and the difference of their squares 60 ; required the numbers. Ans. 9i, and 5].. (3) Pkoblem. — Having given the difference of two numbers, and their product, to find the numbers. Rule. — To the square of the difference add four times the product, extract the square root of the sum ; add the result to the difference, and also subtract the difference from it, then half the sum will be the greater number, and half the difference the less number. Ex. The difference of two numbers is U, and their product 80 ; required the numbers. Ans. 5 and 16. (4) Problem. — To find a number, having given the sum of the number and its square root. Rule. — To the given sum add |, and extract the square root, sub- tract the result from the given sum increased by }^, and llie remain- der will be the required number. Ex. The sum of a number and of its square root is 8| ; re- quired the number. Ans. 6|. (5) Peoblem. — To find a number having given the difference of the number and its square root. Rule. — To the given difference add \, extract the square root of the sum, and to the result aid 'tlie given difference increased by ^ ; the sum will be the required 7iumber. Ex. The difl^erence of a number and its square root is 8| ; required the number. Ans. 12j. 160 KEY TO PART SECOND. (6) x+y^s. (1) Squaring, a;^-j-2j;y+y^=s', ' but xy=='p, therefore by transposing 'ixy, or 2p, Cubing eq. (1) x^-\-Zx''y-\-ixy'^-\-y^=s^, or x^-\-Zxy(oc-\-y')-\-y^=s^, or x^-\-Zps-\-y^=s^ . .". a;'-)-^^=s' — ^ps. Again, raising x-\-y^=s, to the fourth power, x''-^ix^y-\-exY-\-ixy^-\-y'>=s\ but ix'y-\-6x'^y''-\-Axif=:Axy{x'^-\-y'^)-\-Qp'' =4p(s2— 2;))+6;)2=4jt)s2— 2;;=. . . a;''+yi=s''— 4jt)s2+2p2. As an additional example, let the following problem be proposed: Pkoblem. — To find two numbers having given their product, und the difference of their cubes. Let X and y represent the numbers, then x^ — y^=za, (1) xy =J. (2) Squaring equation (1), adding to the result 4 times the cube of (2) and extracting the square root, we have a,3_|_^3_a2^4J3. (3-) Adding together equations (1) and (3), dividing by 2 and extractinir the cube root, we find Similarly, by subtracting equations (1) from (3), we find y=V\lU^'+'ib'—a)\. These formulae give the following R UI.E. — ■ To the square of the difference of tlie cubes, aid four times the cube of their product, extract the square root of the sum ; add the result to the difference of the cubes, also sutdract the difference Jrom it, then the cube root of one-half the sum will be llie greater number, and the cube root of one-half the difference tlie less number. Ex. The difference of the cubes of two numbers is 604, and their product is 45 ; required the numbers. Ans. 5 and 9. In a similar manner special rules might be formed for the solu- tion of Hourly all the questions on page 214 of the Algebra. EQUATIONS OF THE SECOND DEGREE. 101 SPECIAL SOLUTIONS AND EXAMPLES. Article 253. (.2) By adding 2x to each member, the equation becomes x' — x=2-\-'2x, or x{a^—l)=2(x-{-l), divide both members bv x-\-l, x{x—l)=2. Wlience x= — 1 or 2. 2 (3) Transposing 4 ^.nd — , the equation becomes or (,x+l)(.x-^)=kx+i,), .-. a;-l-|=0, or a;=— |. Also, X— 5=^. Whence a:=I(ld=-y'lO)- (4) Transpose 1 to the left member, then the cqualia .n*-" »e placed under the form 2x\x—l)+x^—l=0, or 2a;=(x— l)+(a;+l)(x— 1)=0 .-. X — 1=0, or x=l. Also, '2x''+x+l=0. Whence x=li—l±J—f). (6) The equation may be placed under the following form xS—2x^—x^-\-2x—x-\-2=0, or x\x—2)—x(x—2)—(,x—2)^0. .-. X — 2=0, or x=2. Also aP — x — 1=0. iJVhence x=l(l±«y5) (6) Multiplying both sides by x, we have :ri=6x2+9x, 14 II 02 KEiTOPARTSBCOND or, .r<-f3x'=9x2+9x, a;^=3x, or a;=:3, or x^+Sx=—3, and a;==I(— 3±V— 3)- Or, thus, x^=6x+9. .-. a;'— 27=6x— 18=6(x— 3), dividing by x — 3, a;^-l-3a;4-9=6, from which the value si X is readily obtained. (7) x+7x^— 22=(x— 8)+7(a;5— 2)=0, 1 dividing by x^ — 2, we have a;3-|-2x^+4+7=0 a;l_|_2x^=_ll a;S-|-2x^4-l=— 10 xi=—ldzj—l0 X =(— 1±V— 10)'=29±7^— 10. From x^ — 2^0, we have a:-'=2 and x-=S (8) This equation may be written under the form x'— 8l4-'3^(x'— 9)x=0, or (:<;2+9)(j:=— 9)+'/(x2— 9)x=0. ... x'^ — 9=0, and x=-\-i, or —3. Also,a;2+9+'/i=:0. Whence x=l{—n±sj —li>t). (IC) Multiplying both members by x, and adding x-\-l to each Bide, we have x^ — 2x-|-l=4-|-4^.r-j-a;, extracting the square root, x — l=±(2-|-;^a:), From the equation x — l=2+;^r, by transposing Jx and — 1 , we have X ;^.C = 3. Whence Jx=l±.\Jvi, am\ .c=.i,('7itVl3^ EQUATIONS OF THE SECOND DEGIIEB. 163 From the equation x — 1= — 2 — J x, similarly we find 4x=—\±\J^, and a;=.i(— iq=V'^). \\ 1) Adding _ to eacli member, we have 4 x'^ X X-' 1x1 I 1 \ extracting the square root, — — -==h ( 3-j-- 1 , 1x 1 1 Prom the equation — — -=3-|-_, by clearing of fractions 2 a; X transposing and reducing, we find x'—lx='{'. Whence x=2, or — ^. 1x1 1 From the equation — — -= — 3 — _, similarly we find 2 a; X a:'4-5.r=V'. Whence a;=:i(— 3zfc J93). — ^ I to each side, we liavo x^-\- -f- I — I ==16-l-34a;-|- 1 , extracting the sq. root, x^+—=± ( 4+lIf ) . 1 7a; 1 7a; From the equation a;'-|- =4-)- , we have 4 4 a;'=4, and a;=rt2. 17a: 1 7j; From the equation I'-J- — = — 4 — , we have 4 4 x'-\- = — 4. 2 Whence a;= — 8, or — '. (13) F'rst — (3a;=+x)=— 3a;= I \-\-L ) . Dividing both members of the equation by x*, and adding to each side — , we have 4t'| IC4 KEYT? TARTSECOND 70 , _9__289 4X'' 4x*' \ ■ 3x / a;2 V ' Sx / 4a;'' a 13 17 extracting the square root, 1-|-— — — ^± — , hence 1+ =+ or — , Sx x' X-' clearing jf fractions, a;'-|-^a;=-|-10, or — 7. Whence x—Z, or — 3J, or '(— IztV— 251), (14) Multiplying by 2, and adding — -j- — to each side, we 36 36 ■ 36 , 18 , 81 x'l 8a; , 16 have — -1- — +— = — +— + — ; a;2 ' a; ' 36 36 36 36 extracting the sq. root, — [--=± ( -+- ) . 16 \ 6 6 / ' Taking the positive sign, we have the equation x^ — 5a:=36, from which x=9, or — 4. From the equation _-l-_= — ( ~+_ ) , we have .r 6 V 6 6 / the equation a;'-l-13x=: — 36, from which x= — 9, or — 4. .•. the values of x are -|-9, — 9, and — 4. (lb) Mult'plying both sides by 3, transposing and _ ap^i .84^1 X- adding 1 to each side, we have 8lx'+18+l=?--*i+^+16, x' X' X 1 / "9 \ extracting the square root, 9x+-=dz — +4 I , X \ X / Taking the plus sign we find x=2, or — 'g'' ; taking the minus sign " " x=l{ — 2^=^ — ii<-i6). (19) Let x-|-2/=s, and xi/=p, then x2-|-j/'=s' — 2p, and x'-|-2/'=s' — 3sp, and by substitution the first equation becomes 2s'-|-l=(s'— 2rtO+s'— 3sp), but s=x4-i'=3, hence by substitution tlie equation EQUATIONS OF THE SECOND DEGREE. 1G5 becomes 55=i9—2p){p-\-27—9p)=2i3—126p-]-\Gp', or 16^2_]26^=_188. Whence p=*g or 2. Taking a;-(-y=3, and xi/=2, we readily find x=2, ani y=l. (20 Dividing both sides of the equation by l-\-x, we have 1 — x-\-x''=a{l-\-xy=a-\-2aa:-\-ax^, transposing and reducing, (a— l)x=+(2a+l)a;=l— a or x^-\- — ! — x= = — = — 1. a — 1 a — 1 a — 1 ^;, 2a+l^,(2a4-l)^_ C2a+l) '_^_ 12a-3 a— 1 4(a— 1)2 4{a—iy 4(ffi— l)^" _2a+\__-±Jl2ar—3 2(a— 1) 2(ffi— 1) ^_ — 2a— l±Vl2a— 3 _ l+2ffl±Vl2a— 3 2(a— 1) 2(1—3) ■ Since both members of the equation are divisible bv l-f-*> therefore l-|-^=0, and x= — 1. (21) «-l /:r-2a-?=l. ^ ^ x' X^ X Transposing, multiplying by x, and arranging the terms under the radical, we have j ? — X ) 24- ( - — X J = — 2a, by squaring and transposing. Putting _ — x= a single unknown quantity, and finding X its value, we have f — a;=_ 1 ± 1 Jl_8a=J, X whence a — x^=hx, or x'-\-hx=a. 2 -^ I GO KEY TO PART SECOND, Substituting the value of h, we find =J \l±JlSa±j2±:'2jl—8a+8al . (22) Let x-]-y=:s, and 3:y=p, then the equations become s+jE,5+P==85, (1) p+s24-ps=97, (2) adding s--\-2ps-{-p^-\-p-\-s=182, or, (s+;))2+(;)+s)=182. Whence s4-^=-l-13, or — 14. Taking s+|'=13, s=13 — ■;>, and substituting this in (1) 13— ;)+ja(13— p)4-p==85, whence p=6. .'. x-\-y=T, and xy=G, from which we find as=6 or and y=l or 6. (23) ^+«^-(a-&)(2c+ad)^=(a+i)^-(a=-6')x'. a' a da (a' — b^^d^x- — 2acdx — aH''x-\-2bcdx-\-ahd^x — acdx- -hcdx = — acd- -2c^, {a? — i^)(?'x' — 3acdx-\-bcdx — a^d'x-\-abd'x= — a,r,d — 2c' 3 3aC(Z — bcd-\-a'd'' — abd^ acd-\-2c^ ^ ~~ (a'— i=)d2 ^ (a=— 6=)d2 ^, _(^a— i)ci+(a— J)a^=^ I (3a— ;')'c'rf='+(a— ?>)°a'd^4-3acd3(3g— &)(a— 6) ^—ia>cd'—8a^cH '^-{-iaVcd'>-\-8bVd'^ (3a— 6)'^c^d=i+(a— t )^a'.£-'+ 2ge(;^(3a— ;))(a- ji) "*" Aia^'—l^d^ ' _ (a— 3t)'c''fl;3+(g— ^)'g°'^^+2act;3(a— 3i) (a— 5) EQUATIONS OF THE SECOND DEGREE. 101 _ {Sa— b)ca-\-ya —!i)a(P (a—ib)cd-[-ia—h)ad^ _(Za—h)cd+(a—b)id^_^{a—3b)cd-\-{a—b)ad^ 2ia^—b^)d^ ~ " "" 2(s=— i^)d' _ 4{a—b)cd-\-'2(a—b)ad^ _ 2c+ad ii(,a''—b')d' Ca+6)d _ 2(a+A)cd _ c *"" ~ 2{a^—b'')d'~'(^b)d' (24) (a;'+l)(x=4-l)(x4-l)=30x% or, (x'+i+^T+i) {x+l)=30, Let x4-l=s; x2+i= ( x+1 ] '— 2=s»— 3 (s2_24-s)s=30, or s^— 2s+i-^=30, s''+s'— 2s==30s, 4 4 / s^+i \ '=?^+30s ( s^+i ) +10 ( s'+i ] +25=l^-+35s+.M s2+l+5='!f+5 2 2 s'=3s, and s=3=x+-, X X Whence a:=i(3±V5). (25) ifi+y'=35, and a»+2/'=13. Let x-|-y=v, xy=z, then u'— 3»2=35 (3) and »)=— 20 =13, (4) 108 KEY TO I' AKT SECOND. 2d'— 6ra=70, (5) by multiplying (3) by 2 3v^—evz=S9v, (6) by multiplying (4) by 3d v^^SQv — 70, by subtracting (5) from [t 1,5— 39d=— 70, t)"— 39u2=— 70d, 25d2=2Sv=, V* — 14d2=25d2— 70u, „4_14u2_l-49=25t>2— 70t>+49, «=— 7=±(5d— 7), v'=6v, and ij^5, or v^-{-5v=14, and t,=:-l-2, or —7 ; but «=— 2z=13, 2o — 2z=13, and z=6, or, 4 — 22==13, and z= — |, or, 49-22=13, and 2=18. JVom iC-j-!/=5,) a;^3,or2, a;j/=6,) y=2, or 3. Froma;4-y=2, ^ a=l±iV'22 From x-f j/=— 7,^ a;=— |d=W— 23-. y=I8, S y=— |=Fi V— 23, (86) Let syz^p, andi a;-f-y-l-2r=s, then the equations becomo s — z (1) s—y (2) S — X (8) hiunce z — s — ?, a (4^ * 6 (6) x=s — ?, «♦> EQUATIONS OF THE SECOND DEGREE. 169 adding, x-\-y-\-z, or s=Zs — p I i-j-1-f-I ] _ Whence s=Sj^±^S^d^, 2abc Substituting this value of s in equations (4), (5), and (6), we ^et ^^ («^+«c-fc)y , . ^ 2abc ^ ^ ^ 2abc ^^^ * 2abc Multiplying equations (7), (8), and (9) together, we find ^^ iac+bc—ab){ab+ic—ac)(ab+ac—bc)p^ wnence p=2abc I \ ^^ l , N ({ac-{-bc — ab){ab-\-bc — flc)(a6-^-ao — bcp Substituting this value of p in equations (7), (8), and (9), we get x=l \ 2afe(ac+fe-«fc) I ^ M ((ab-\-ac — bc){ab-\-bc — acy) I ^ 2aftc(ffi6+fc — ac) } \ l{ac-\-bc — ah){a.b-\-ac — bcp / 5 2abc{ah^ac — be) ) \ i(_ac-\-bc — ab){ab-\-bc — acp (27) Dividing both members of (1) by a;', and both members of (2) by y, we have or, a^+i=2/ +1. (3) a;' y and y»+i=9(a:+l), 15 ^ 170 KEY TO PART SECOND root .■. «-l-i= I x-i-- ) £/3, by extracting the cube y \ xJ And a;5+i= ( x+l ) tj3 by (3), x' \ X I dividing both members by x-\--, we have a;=4-2+i = 5/M-3. by adding +3, a-j-l = /3/3_)_3, by extracting the equE root. Similarly, x— 1 = /^3— 1, by subtracting 1. But jf+l= ( x-\-l. ) V3=n/3^V3+3. Whence y=i |V3.^2y3+3±^3V9— 1^ RATIO, PROPORTION, AND PROGRESSIONS. EXERCISES IN RATIO AND PROPORTION Article 278. Note. — The solutions of these exercises are given, not because tlief are difficult, but because many of them are of a cliaracter not lierelo f jre presented to the notice of Teachers. (1) 3 to 4 =1 ; 3» to 4= = V ; 3 = 'i?. and since V is greata than ■/, the ratio of 3^ to 4^ is greater than the ratio a 3 to 4. R4T10 AND PROPORTION. 171 (2) Duplicate ratio of 2 to 3 is 2^ to 3= = 4 to 9, triplicate ratio of 3 to 4 is 3^ to 4' =27 to 64, Bubduplicate ratio of 64 to 36 is ;^64 to ^36 =8 to 6, 4X27X8 to 9X64X6=864 to 3456=1 to 4 ; Or, by canceling thus, ^X04X6 _^^X0^ 2X2 ^4^.) ^g ^ 4X#X^ ^X^ 1 1 (3) Let 3:= the quantity, then n-\-x __q m-\-x p whence np-{-px=mq-\-qx, and x= — i — £ p—q (4) _=2|=| ; dividing both terms of each frartion by 3 ''=4=11. Multiplying both terms of the fractions -=|, by | we have t*=|X4='7?=3|. 3a 3^3 — V ■ (6) ^=5|=2_i ; dividing both fractions by ">, *=|. Multiplying both terms of the fractions _>~- s, by | have^Xf=|X|.org=3. --Is-^ ' • • ^ 3+5' a+J 5 Since *=| ; .-. "=f , and ^=^=| c * h ^ b—a 5—3 -^ (7) -=4, and -=2, also 4m=7ra, and m=_ . ?7i ^ ra * 4 7re 3re m — re= — — n= — ; 4 4 dividing 6m by each member of this equalitv ll'Z KEYTOFARTSECOND --6m-i—= — =8X3=14 ; m — n 4 n Also, dividing 5n by each member, we have 5» -=5;,X-=^=6|. ^8) m — n 3re 3n ^ clearing of fractions 10m-l-15»=13m, or m^5n. and — =i, or 5 to 1. m (9) -=l, and m=2re, also 12m=?l^; m "• 7 dividing wi-j-ra by both members of this equality, 12m 7 24?i 24n 24?i ^ Also, since m=3^?n, 12ra=42m ; dividing n — 2m by both members of this equality, and substituting the value of re in the second member, we have n — 2m n — 2m Sim — 2m 3m __ , 1271 42m~ 42m 84m -^" (10) '^^~^y=6, or 7a;— 53/=30y— 48j:, 5y — 8x 55x=35?/ lli=7y, y =y, or X : y : : 7 : 11. X (11) ab:=a' — x'', or ay,h={a-\-x){a — a;) ; whence (Art. 268), a : a-\-x : : a — x : b. (12) x^-\-y'^2ax, or y^=^2ax — x^, or yXy=^(2a — i) ; whence (Art. 268), x : y : : y : 2a — x. (13) Let x= the number, then a-\-x : b-\-x : : c-\-x : d-\-x ; .-. (a+x)(d+x)=(i+x)(c+x), or, ad-\-ax-\-dx-{-x'=bc-{-hx-\-cx-\-x^. RATIO AND PROPORTION. 173 or, cue — hx — cx-^dx=hc — ad, or, (a — h — c-\-d)x^bc — ad, be — ad a — b — c-)-rf' The pupil should verify this answer by using numbers. (]4) Let a, b, c, and d, be four quantities in proportion, and if possible, let a; be a number that being added to each wiJ. make the resulting four quantities proportionals ; then a-\-x : b-\-x : : c-\-x : d-\-x. .■. (a-\-x)(d-\-x)=(b-\-x)(c-\-x), or ad-\-ax-{-dx-\-x^=bc-{-bx-\-cx-\-x^ ; , be — ad whence a:= a — b — c-\-d' But since a, b, c, d, are in proportion (Art. 267), ad=le, . be be n /-A 1 •■oe\ •■• ^= — -. f-r= — i 1-,="' ^^'■*- ^^^) '• a — b — c-j-a a — b — c-[-d hence there is no number which being added to each will leave the resulting quantities proportional. ,'15) Cubing each term of the second proportion, we have a' : 6^ : : c-\-x : d-\-t/, but a; : y : : a' : i'. .-. X -.y : : c-\-x : d-\-y, by Art. 272. Placing the product of the means equal to the product of the e^itremes, and omitting xy on each side, we find x=^y. d f 16 ) Let ma and mb be equal multiples of two quantities, i and 6 ; then since — =-, we have (Art. 263), ma a ma : mb : la :b (17) Let - and _ be like parts of two quantities, a and b; then i-;-?=_X-=- is equal to -, anJ we have (Art. 263), n ' n n a a a t b .1 _ : - : : a : 0. n n 174 KEY TO PART SECOND. (18) Let a [h : : c : d, then ma and mc will be equal multiples of the antecedents, and nb and nd equal multiples of the consonants ; then it is required to prove that ma : mc : : a : c, (1) and nb : nd : : a ic. (2) First, . — =- is equal to -, hence (1) is a true proportion, ma a a Second, _ =-, but since a ib : :c :d,we have nb b d=— (Art. 267), hence _= — i-J=_X-= , a b a a b "• which is the ratio of a to c, therefore (2) is a true proper tion. (Art. 263.) (IQ) Since a :b -.-.cvd, .-. -=- (Art. 263) ; a c . ,. mb b , Tul d ., , but — =-, and — =_, therefore ma a iia c ma :mb::nc: nd. Again, if we take the equation -=-, and multiply bcth a c sides by _, we have ^= — , which gives the propor- nh nd m ma mc tion ma : nb : : mc : nd, (Art. 263). (30) Since a:b::c:d, .: -—^, (Art. 263) ; a c but l-^1=^x'^=i and ^H--=-X^-, m m m a a n n n c c . a _ b _ _c ,d mm n n Again, if we take the equation -=-, and multiply botli a c u , m , . mb md members by _, we have =_ ; n na nc . , . mb b , a , md d c ,. ^ . ., . .a but — —-—-, and — =_- ; — ; that is, the ratio of — na n m nc n m m to - is equal to the ratio of — to -, hence n m n a , h , , c . rf m n m' n EATIO AND PROPORTION. 175 (21) Let a lb: -.c -.d, (1) and e if : Iff ih, (2) from (2) by Art. 271,/: e: :ft:^, (3) multiplying together the corresponding terms of fl) and (3) (Art. 277), we have of : be : : ch : dg, , be dg whence _=.£, af ch but %=lx'-=l-^^-, and ^|=^X?=f-*-i ; :2b- -2H 1, (Art. 276), J2b—b', (Art. 267), ±(a +x)^iib—b^= ±.xj2b—b'= a^aj2b~ Thj2b- 1 \^^2l ■H^aJZb- -b^ -b'. V ±j2b. -b^J- (iO) Let x-\-y= the greater number, and x — y= the less, then {x+i/)(x—y)=x^—y^=320, and {x-\-yy — (a; — yy:=6x''y-{-2y^ ; also, ■^+y— (*— J')=22/. and (2yy=8yK .-. 6x^y+2y^ :8y5 : :61 : 1, or, by dividing the first and second terms by 2y, Sx^+y^ ■■ Ay^ : 61 : 1, (Art. 267), 3a;2+^2_244y2 . whence 3:c^=243y^, and x=±%y. Substituting the value of a;^ in the equation ^2 — y2=32(), we have 817/2—2/2=320 ; whence y=±2, and a;:=±18. .-. a;4-v=±20, •c — 2,=±16. Article 280. Ex. 2. Let x= the number, then a:x:: a — b : c — x, (Art. 267), ax — bx=ac — ax 2ax — lx=ac, x(2a — b)=ac, ac 2a— b' RATIO J»ND PROPORTION. 179 VAKIATIOH Article 290 Note. — The solutions to these examples are given for the same reason as those following Article 278, not because thev are diiTicult, but because to many Teachers they will be new. 8"* Since y varies as x, let y=:mx, then since .. x^2, y='ia, we have 4a=2m, or m=2fl, ■. i/=2ax (4) Since y varies as -, let y=— , then since if x=h v=8 X OS we have 8=— or m=4. 1 2 4 ••• y=-. X 5) Let j/'=:m(o' — a'), tlien _=ml/', or m=_ ; 0= a= .•. «2_._^j(3 — a:'), and j(=_»/a2 — a;^. a^ a (6) Here we have y=v-\-z, where vcnx, and zee — Let v=mx, and z=— , then 2/=mK+— . Since y=Q when a;=l, we have 6= m+w, (1) and " y=6 when a;=2, " " 6=2m+-. (2) From equations (1) and (2) we readily find ot=2 ana 4 71=4, hence y=2x-\--, x' (7) Here we have y^a'\-v-\-Zi where a is constant, vdx, and zo^x''. Let v=mx, and z=nx'' then y=:ffl-)- mx-1- «'^^- • 6=a-l- m -\- n, (1 180 KEY TO PART SECOND. ll=a+2m -j-4n, (2) ]8=a-l-3m -\-9n. (3) From the equations (1), (2), (3), by elimination (Art. lo8), we find (2=3, m=2 n=l ■ hence, y=3~{-2x-l[-x'. (8) Since so:t^ when_ is constant; and so; when t is con- stant, therefore, when both Tar' B evident from Ar* 283 (3), that safi' ; then let s^mff ; but since 2s=/, or s=^when /=1, therefore if=mfl\ whence m=\, and s=lfp. (9) Since roc x, let r=nx, and since sx^x, let s^-^n^x. then i/^7nx-\-iijJx ; hejice, by substituting the corresponding vahies of y and X, we have 5=4m-l-2n, (1) 10=9m+3n. (2) From these equations we find 7?j=g and n=|, hence y=^x-{-^Jx=^(x-\-^x). (10) Let x:=±-, and ij- =-i, p and being 2" then 1/"=' .., and 2""' a= ^-.c""", and 5'" p 5"' c"'"" and c'""a;^az'"". It is proper to observe that all the preceding examples adm of proof. Tims, in the answer to example 9, if we substitute AEITIIMETtCAL PROGRESSION. 181 ^t X we ought to find y=5, or if we substitute 9 for t we ouglit •"o find y^lO. Die subject of Variation is of considerable use in Natural Philosophy, and though not quite so easily understood as the other parts of Proportion, is worthy the careful study of the learner. AKITHMETICAL PKOGRESSION. Article 294 . Note. — The learner who wishes to understand tlie subject thoroughly should derive each of the formulce on page 245, by talcing the two tquations at the beginning of this article, and finding from them the '-ralue of the quantity marked " Requi.ftv. We shall illustrat*" the method of doing this by the solution of two of the most difficult cases, Nos. 2 and 14. Formula 2. Talking the equations Z=a+(re— l)d, (1) and S=(a+0-, (2) 2 we have given a, d, and S, and it is required to find /. The first step is to eliminate n. This may be done by finding die value of 7i from each of the equations, and putting these »a ues equal to each other. Eq. (1) gives n= — _2^=_-+l, a a eq. (2) gives ra=— — ; l-\-a clearing, V—a'^-\-dl-\-ad=2dQ, pj^dl-\--=1dS,-{- i a^—ad+- \ =2dS+{a—ldy, l+i=±Jl2dS+{a—ldyi, i=-ld±jms+(.a-ldy^. 182 KEY TO PART SECOND. Formula 14. Here we have the same formula;, and the same ouantities a, a, and S given, to find n. Finding the value of I in equation (2), and substituting it , / ,N 7 2S — na a (\\ we have a-\-{n — l)d= , n , ■ . 1 ■ 2 1 2a— (i„ 2S clearing and reducing, ra=+—y-»=-y> whence ,^^J^^^^z±^. Ex. \\Q) Here d= — \, and we have given a, d, and n. to find S ; we, therefore, use formula 5, whence S=ln\2a+(n—l)dl=lnl26—{n—l)il =ln)18—(n—l)] =ln(19—n). ni) Here d= — g, and a, d, and n are given to find S. S=inJl-|(n-l)j=J2»l6-7(»-l)|=iL(13-7n). '12) Here a= , and we have a, d, and n given to find S, ^ a-\-b Substituting in formula 5, we have =Ai— ■*-'i-.T. i"'-^t (13) Here d= — -, and we have a, d, and n given, to find S. n S=i. jH^^=±)-l(n-l)S =jn f-=l( ■* ( re n > ~ ( n > m— 1 (14) Here 0=16^,, (^=167^X2=32^, and n=30, to find I and S. ARITHMETICAT. PROGRESSION. 183 Formula 1 gives Z=I6J2+(30— 1)32^=948}^. S=(Z+a)-=(948j^.+16^i2)3j^o=14475. (.15) Since there are 200 stones, there are 200 terms, increfore »=200 ; and since the person travels 20+20=40 /ards, or 120 feet for the first stone, therefore a=120. And since the stones are 2 feet apart, he must travel oyer twice this distance to reach each successive stone, there- fore the common difference ^=4. Applying formula 5 to find the sum of the series of which the first term is 120, the common difference d=4, and the number o;' te/ma n=200, we have S=2g5j2(120)+(200— 1)4|=100(1036) =103600 feet =19m. 4 fur., 640 feet. ^^17) Here a=3, 6^18, and m=4, p= -= — ^_:=3, hence the means are '^ m+1 4+1 3+3=6,9, 12, 15. (18) Here a=l, i= — 1, and m=9, J h — a — 1 — I 1 a= = = — ^, m+1 9+1 ^ (19) Here a=:19, d= — 2, and S=9l ; and it is required to find n, which may be done by formula 14, where „=±^A2^=d)H:!^S--2a+d 2d hence .^ ±^(38+2)^-16X91-38-2 —4 _±12— 40__|_3_j_jQ__j_j3^ or +7 Hence, the sum of either 13 terms, or 7 terms, wil. oe equa. to 'il. To explain the reason of this let the first thirteen terms of the series be written thus, No. of term, 1 , 2 , 3 , 4 , 5 , 6, 7, 8, 9, 10, 11, 12, 13, terms 19, 17, 15, 13, 11, 9, 7, 6, 3, 1, —1—3 -5 184 KEY TO PART SECOND. Here we see that the sum of the first 7 terms is 91, and tlip reason that the sum of 13 terms is the same, is owing to the fact that the sum of the last six terms is zero, the sum of the positive and negative quantities b^ing equal to each other. (20) Here a=.034, (Z=.0344— .034=.0004, and S=2.''4l Substituting these values in formula 14, we have ^^ ±^(■068— ■0004)'+. 00 87936— ■068+.0004 ±.1156— .0676 -60. .008 (21) Let a;= the first term, and y= the common difference then x-\-y= second term, and x-{-x-\-i/=2x-\-y=^4: ; fifth term =a-|-(re— l)i=x+(5— l)y=a;-t-4y=9. From these equations we readily find x=l and y=2, hence the series is 1, 3, 5, 7, 9, &c. (22) Let x= the first term, and y^ the common difference, then the series is X, x-iry, x-Jf-2y, x-\-3y, a;+4y, whence x-\-(,x-\-y)=2x-\-y:=lS, and {x+2y)-{-(x+3y)-\-(,x-\-4y)=dx+9y:=12. From these equations we find a;=:10, and y= — 2. It is now required to find n, having given the first term =10, the common difference — 2, and the sum of the series 28. „ , ,. . ±>/(20-l-2)— 448— 20— 2 Formula 14 gives n= ^ ^ 1 — i _=F6— 22 —4 =q=li-|-5-i=4, or 7. The series is 10, 8, 6, 4, 2, 0, —2, &c. Here we readily perceive why the sum of 4 terms is the same as that of 7. (23) In this example we car readily find the. first, second, iSw, terms by making n=l, 2, 3, &c. Let ?t=l, then 1" term =i(3— 1)=J, " n=2, then 2'"' term =^(6—1)=-'*. ARITHMETICAL PROGRESSION. 1*^5 f]— ;^ = |=j= the common difference. Sum of n term =H+^(3?i-l)?|=a«+J)| (.24) Here a=l, and (i=2, to find the sum of r terms, and also of 2r terms. From formula 5, we find the sum of r terms =i?-j2+(r — 1)2|=H, of 2r terms =:Jx2r)24-(2r— ])2^=4r'. .-. 4r' : r' : :x : 1, whence r'^x=ir'', and a;:=4. (25) The sum S of n terms =^ra!2ffi+( nr—\)d\=an-\-\n'^(l—{nd. The sum S' of 2m " =— J2a+(2n— l)d|=2ara+2ray— nd; Sum of 2ra terms — sum of m terms, or the) ^^anA-inH "^nd second half of 2ra terms. ) a a Sum S" of 3ra terms =^]ia-{-(Zn—\)d\=ian^ln''d—^,nd. A Z an^-lnH-\y,A _^ the required ratio. an-\-'in'^d — \7id This is an interesting general theorem, which the pupil should illustrate by numbers ; thus, if we take the series 1, 3, 5, 7, 9; &c., the sum of the second 4 terms is 48, and the sum of the first 12 terms is 144, being 3 times that of the second half of 2n terms n being 4. (■26") 2^-=.d, l4-_ii-=-il^= 1" arithmetical mean. „+(._l)d='H^9 ._ iL=L9gi= .'^ term. ^ n+1 «+l «+l „+(„_2-l)i=-+L9+(._2-l)JJ.-=l^^ ' ' n-j-l 71+1 w-f 1 =(w — 2) term. 16 180 KEY TO PART SEUOND. / 19n+l ^^+19 \ »_10n!+10«_jo»= sum of n terms / 19ra— 35 ,*7J+19 \ (ra— 2) _ 10?t^— 28n+16 \ ra+1 71+1 / 2 »+l = sum of (n — 2) terms. m+1 whence 30»=5_2^!=li5^+LO, ra+1 or 30»2-[-30m=507i2— 14072+80, reducing, 2ra^ — 17?2= — 8, whence 7i=+8. (27) Let x= the number if days the first travels before he is overtaken by the second. It is then required to find tne sum of X terms of the arithmetical series whose first term li, is 1, and common difference d=:l. S=i7i|2ffi+(ra— iy|=-|2+(x— l)}=|a;2+|a;. The second travels (a; — 5) days at the rate of 12 miles a day Fence the whole distance he travels is represented by 12(x — 5). ••■ 2J:'+ia:=12(x— 5), or, x' — 23a;=— 120. Whence a;=8 or 15. and X — 5=3 or 10. .•. the second travels 12X3=36 miles, or 12X10=120 miles. The second traveler overtakes the first at the end of 3 days, when each has traveled 36 miles ; the second then passes the first, but as the first increases his speed each day, at the end of the 10" day he overtakes the second and they are thus twice together, Tli's example furnishes a beautiful illustration of the mannei In which the different roots of an equation correspond to th« ievei&l circumstances of the problem. GEOMETRICAL PROGRESS ON. 181 GEOMETRICAL PROGRESSION. Article 300. NorE.^AU the formuIiE in this Article are deiived (rom tlie two equations Z=a;-"- , (1) and 8^"Jl^=±=^, (2) r — 1 r — 1 by supposing any three of the quantities to be known, and then finding the values of the other two. In general, the formulse are very easily found, but where n is large the resulting numerical equation is hard to solve, and can only be uuderstood by the learner, after he becomes ac- quainted with the numerical solution of equations, as contained in the Algebra, Articles 428, to 444. After the pupil becomes acquainted with e.\j>onential equations, Articles .382, 383, he will find no difficulty in obtaining the last four formulcB, 17 to 20. To illustrate the method of finding these formula3 from the two preceding equations, we shall find I, formula 4. From (1) a= , " (2) a—rl—S(r—l). Placing these values of a equal to each other, we fin 1 y_ S(?-"+?-"-') ^ ,r— 1)S-"-' (.1) r=2, ?'"-'=2'=128 ; ffir''-'=5X 128=640. (2) r=l, r'-'=(|)==5'j ; ar'-^^diXl 7 4 — s a • 5=? ».n-l— c5^5_J-2. (3) r=2.|^3|=f, r"->=(| (4) '= JT 3' (. — 3) —liV 72SX -il 27(3. on— 1 on— 1 on— S (5) ,-=J_-i-l = ?, r"-'=(|)"-'=l- ; ar"-'=iX—=~. ^ «* -^ ' 3 z ^z^ 2"~ 2"~* 2* ^10) r=3, and Z== vf" term =1X3"- '=3-'-', ^_rl—a_Z X 3-1— I^ 1 -3„_-, . r—\ 3—1 - 1*18 KEY TO PART SKCONT I 11) Here r=— 2, and I, or »"' term =1X(— 2)»" '==F2"- according' as n is odd or even. r—1 —2—1 ^ W— a_— 2X(=i=2" ~')— 1 ^1 ^ ' 12) Here r= — 1^, and Z, or »" term =x ( — ^ ) — !/ ?•— 1 -M^. ^(-f)"-^ -^-1 -?^-l \ xl _ \ xl ___^L_/'_J/\"? a;-|-»/ x-{-i/ ^-\-y ^ ^ .-c / ' (13) Comparing the given quantities witli tliose in formula 131 we have a:=4, /=12500, and ?i=6, to find )-. ^_W-a_1250QX5-4_^p^g^ ?•— 1 5—1 (14) Let x= the 1" term, and !/= the ratio, tlien x, xtj, an2 a;^' represent the first three terms, and a;+a;y=9, (1) a;+X!/'==15. (2) From these equations wo readily find y=2, or — |, hence x=i, or 13l ; therefore the series is 3, 6, 9, &c. ; or, 13!, —4!, -\-\\, &.C. (15) Here a=';r, and )■=;; ; S= = _i_= 1 — )• 1 — I GEOMETRICAL PROGRESSION. 189 (16) Here a=9, and r=| ; S=— =-=27. (17) Here b=6 and ?•=» ; S=-A =?=9. ' ^ 1 1 2 3 S (,18) Here o=|, and r=— ^ ; S=-l-=f (19) Here a=100, and r=| ; S=i^=l?2=16t)|. 5 5 (20) Here a=a, and r=- ; S= = a . ^6 a — b a (21) If we begin at the second term, the series is a regular geometric series, of which the first term is 2a, and the ratio r=a, hence the sum of this series is Then, 1 — a adding 1 to this, the sum of the series l-\-2a-\-2a'-\-2a^-\-, &c., is 1+_?^=L=?+^^=1±? 1 — a 1 — a 1 — a 1 — a (22) Let x= the !•' term, and y= the ratio, then x+xy=2^^, and S=3=_^, from the formula S=_?_ 1— y 1 — r' From these equations we find 2/=-l-|, or — j, and a:=2 or 4 ; hence there are two series, the first being 2+f +f +, &c., and the second 4 — 1+| — , &c. 25) Here m=2, and ?•=•»+ i/-=s/^=:=s^!y=|. ■• 37X|=|, and |xi=3, are the means. C26) Here m=7, and ?-=VJ-^|^^=1 /V6561=V81=3. .-. the means are 2X3=6, 6X3=18, &c. »90 KEYTO PARTSECOSn Article 301. OlECULATING DECIMALa .,. „ 63 63 1 1 II) Here a= — = — ,r — = — , 100 10= 100 102 g_ .63 _.63 63_7 T-h -99 99 11 Or, thus, S= .63636363. 1008=63.63636363. 998=63. (2) Here S=.54123123123. . 1000008=54123.123123. . 1008= 64.123123. . 999008=54069. ' Q 54 009 J 8023 •=> — l^'SOD 3 3 3 0- HAHMONICAL PR0SRESSI05. Article 303. (3) Inverting the terms 3 and 12, they become | and ^2- Let us now insert two arithmetic means between | and Yi and the reciprocals of these will be the harmonic means between 3 and 12. 8ee example 16, page 246. 0=73, l=J, and m^=2 ; ^—<^ — i—7\ — 3 ^3=J m+l 2+1 '^ ■ '^' YT5-|-y3=f2=^Bi hence 6 is one of the harmonic means ; g+7o=:T2^.i! hence 4 is the other harmonic mean. (4) 2 and 5 inverted become ^ and 5. Let us now insert two arithmetic means between 2 and 6. /)— a _6— l_jj • m+l 3" ^' ARITHMETIC AND GEOMETRIC PROGRESSION. 191 !,-\-ll=:2, hence I is K,ne of the harmonic means, 2-1-1 '=31, hence -1=3 is the other • J. 2 3',- (6) I and yj inverted become 2 and 12, let us now insert 4 arithmetic means between 2 and 12. = =1 0=2, hence we have for the arithmetic m-\-l 4+1 ** means, 4, 6, 8, 10, and for the harmonic means, 1111 4' B' B' To- (,6) Since li, h, c, are in arithmetical progression, we have a — b=b — c ; and since b, c, d, are in hiirmonical progres- sion, we have _, -, and - in arithmetical progression. he d 1_1_.1_1 c b d c or, by reducing the fractions on eacli side to a common denominator. b — c c — d be cd ,^. , . , b — c c — d multiplying by c, = . b d hence (Art. 263), b : b — c : : d : c — d, but b — c=a — b, .■. b : a — b : : d : c — d, by Inversion (Art. 271), a — b : b : : c — d : d, by Composition (Art. 273), a ib : : c : d, whi'-h W8» required to be proved. PROBLEMS IN ARITHMETICAL AND G 1. O * , T • RICAL PROGRESSION. Article 304. (8) Let X — y, x and x-\-y, be the numbers, then a; — y-\-x-\-x-\-y=ix=Zi), anda;^10, also, (x— y)2-}-ar2-[-(a;— y)==3a;24-2y2=308. By substituting the value of x we find y=:i, hence X — y=S, a;=10, and a;-)-y=12, are the numbers. 19'Z KEY TO PART SECOND. (4) Let X — 3y, X — y, x-\-y, and x-{-Zy, be the numbers, then X — 'iy-\-x^-\-x-{-y-\-x-\-Zy=Ax=2Q, and a;=6j, also, (a;— 32/)(a;+3y)(j;— y)(a:+!/)=880, or, (.j;^— 9i/2)(a;2— !/2)=880, or, a;"— 10a:22/2_[-9y=880 ; substituting the value of a;, and reducing, we find j/^| ' henqe the numbers are 2, 5, 8, 11. (5) Let x= the first term, and y= the ratio, then x, xy, xy^ represent the terms, and a;+a3/+a^==31, (1) x-\-xy : x-\-xy'' : : 3 : 13, or, ^±iy_ =IiX=i! (2) ' x-\-xy l+y 3 From (2) we find y=5, and by substituting this in (1), we find 1=1 ; therefore, the numbers are 1, 5, and 25. (6) Let X — y, x, and x-f-y, represent the numbers, then (a:-]/)2+x=+(x+y)2=3x=+2!/2=83, (1) x^ — (x — y){x-\-y)=x^ — (x^ — y'')=y'i=4. (2) Prom (2) y=2, and by substituting this value in (1), we find x=5 ; hence the numbers are 3, 5, 7. (7) Let X — 3y, x — y, x-\-y, and x-4-3j', represent the numbers then (a>-3y)(x+3!/)=x2— 9y'=27, (1) (x— yXx+ y)=x^— 2/==35. (2) Prom these equations we easily find y=l, and x=6 ; hence the numbers are 3, 5, 7, 9. (8) Let X — y, x, and x-\-y, represent the numbers, then (x — ^?/)-|-x+(x4-y)=3x=:18, and x^6 ; also, 2x — 2y, 3x, and 6x-{-6y are in geometrical progression .-. 2(x— )/)(x+y)6=9x2, or, 12(x'— !/2)=9x2, whence 2y=x, and y=3, therefore the numbers are 6 — 3=3, 6, and 6-(-3=9. (9) Let X — 1, X, and x-|-l, represent the numbers, then (x- l)i+x''-f (x+l)''=3x'+12x2+2=962 ; whence x=4, and the numbers are 3, 4, 5. ARITHMETIC AND GEOMETRIC PR O RES S I ON 193 CIO) Let X — 3;/, X — y, x-\-t/, and x-\-Sy, represent the numbers, then {x—-iy){x—y){x-L-i/) (,x+3y)=(x—3y ) {x-\-3y){x—y) But since the common difference between the iMimbers is 1, therefore 2^=1, and y=:A ; substituting this value of y and reducing, we find x=bh ; hence the numbers are 4, 5, 6, 7. (11) Let 1, — Zy, X — y, x-\-y, and x-\-Zy, represent the three numbers then (a;— 32/)(x— !/)(a:4-2/)(a;4-3^)=a;''— 10a;23/2_j_9y4__280, (1) and(x— 3j/)24-(a;— y)2+(a;+!/)2+(x+3!/)2=166, or, 4a;2+20i/2=166, (2) .-. x''=2i\\—by^. Let 41 ^=a, then x''=a2~10aj/2-|-25y'. Substituting the values of x* and x' in equation (1), and reducing, we have 842/"— 830!/==— '^"V-". Whence J/=l2, and by substitution a; becomes 5^, whence the numbers are 5'— 3(l-i)=], 5i— U=4, 5^+li=7, &c. (12) Let X — iy, x — Sy, x — 2y, x — y, x, x-\-y, x-\-2y, x-\-3y, and x-\-4y, represent the numbers ; then their sum =9a;=45, whence x^=5 ; also, the sum of their squares =9a;2-j-60i/==285, from which, by substituting the value of x, we find 3/=! ; hence the numbers are 1,2, 3, &e., to 9. (13) Let X — 3y, x — 2y, x — y, x, x-\-y, x-\-2y, and x-\-3y, repre- sent the numbers ; then their sum =7x^35, whence x=5 also, the sum of their cubes :='7x^-\-84xy^=i295, from vrhich, by substituting the value of x, we find y=l , hence the numbers are 2, 3, &c., to 8. (14) Let X and y represent the numbers, then 2 ^ or, x-\-y : 2^3cy : : 5 : 4, (Art. 276), x2+2xy+3/= : 4xy : ; 25 . 16, 17 11)4 KEYTOl'ARTSECOND CArt. 274, Note,) x^—ixy-^y'^ : x^+Siy+y^ : : 9 : 25, (Art. 276), a;— y : x-{-y : : 3 : 5, (Art. 275), 2a; • 2y : : 8 : 2, o>, a; y : : 4 : ] . This theorem may also be proved oy multiplying together the BT^anp and extremes of the first proportion and finding the value. it X in terms of y, by which we find x=^y or {y. The converse of the preceding proposition is also true ; that is, if one of two numbers is 4 times the other, then their arithmetic mean is to their geometric mean as 5 to 4. Thus, let a and 4a be two numbers, then 2^a is their arithmetic mean, and 2a their geometric mean, and 2^a : 23 : : 5 : 4. (15) Let x^, xy, and y'^, represent the numbers, then ^'+^y+y'='7, (1) i+ -+-=!• (2) x' xy y' Multiplying both members of equation (2) by x'^y'^, we have x''-\-xy-\-y'^='lx''y'', (3) .-. |a;^)/^=7, whence x^y''=4, and xy=2. Substituting the value of xy in (1), we find x^-\-y^=5 then from this, and xy^=2, we readily find j;=2 and 3/=!; hence the numbers are 4, 2, and 1. (16) Let ^, a;, 2/, and ^, represent the numbers, y ^ then ^l-\-y =10, (1) y a:+^'=30. (2) X Clearing these equations of fractions, by multiplying ^Ij by y, and (2) by x, we have x'^-\-y''=\'iy, and whence 10!/=30a:, and y=3x. Substituting this value of y in either of the equations (1) and f2), wc find a:=3 ; hence y=9, and the numbers are 1, 3, 9, 27. ARITHMETIC AND GEOMETRIC PROGRESSION 19.1 f 17) Let X, xy, xy^, xy^, be the numbers ; then x-\-xy^^Zb, and xj/-)-a;j/'=3C. Dividing one equation by the other ; xy-\-xij^ 30 y+2/^ 6' But 1+2/' is divisible by \-\-y, and y+!/'=y(I+y} ._. i+]/^_ (i+y)(i— y+.v') _ i— y+y' _7 . y+y^ yi^+y) y 6 ' whence 6^' — 13i/= — 6, and^=| or f. And x=-^=8 or 27. y+y" Hence the numbers are 8, 12, 18, 27. 118) Let X, xy, xy'', xy', be the numbers when increased ; .■. X- — 2, xy — 4, xy' — 8, xy^ — 15 are in arithmetical pro grassion ; hence 1" + 3"' = 2"'' X2 ; and 2"^ -\- 4'* =• S'" X2 ; ... (x--2)+(i^'—S)=^2(xy~i) ; or, x—2xy+xy'=2 ; .-. a;(l— 2y+y-)^2, (1) also, (xy—4)+ixy^—\5)=2{xy'—8) ; or, xy—2xy'-\-xy'=3 ; .-. a:j/(l— 2y+^')r-=3. (2) Dividing equation (2) by (1), we have xy(l~2y+y') ^Sory=?, x(l-2y+y') 2 whence a;(l— 3+|)=2. .•. x=8, xy='l2, xy'=18, and a7y'=27 ; and subtracting 2, 4, 8, and 15 from these niinjbej-s, th« remainders 6, 8, 10, 12, are the numbers required. (IS) Let X, xy, xy^, be the numbers, then a;X^i/X^^=^y=6'ii a^=s/64=4 ; also, x^-\-x'y^-\-x'y^=584:, a;3_[_a;3,/6=g84_a.y=520. From the equation xy=4, we have x^- ; V .W6 KEY TO PAET SECOND. substituting this value of x in the last equation, we haye ^-j_64y'=520. dividing by 8, l+83/'=65 ; clearing, 8y^ — 65y'= — 8 ; v?hence (Art, 242), !/'=8 or J, and y=2 or j .•. «=2 or 8, and the numbers are 2, 4, 8. PERMUTATIONS, COMBINATIONS, AND BINOMIAL THEOREM. Articles 305 — 309. (1) (Art. 306), Pj=»(re— 1)=5(5— 1)=20 ; Pj=K(ra_l)(re— 2)=5X4X3=60 ; P^=?i(ra— l)(n— 2)(ra— 3)=5X4X3X2=12'J (2) (Art.308),C.=»g)=g.^=10; Q _ n(?t— l)(m— 2) _ 5X4X3 _^Q . ' 1X2X3 1X2X3 Q ._ w(m— l)(m— 2)Cb— 3) _ 5X4X3X2 _.g . * 1X2X3X4 1X2X3X4 n _n{n—'i){n—2)(n—3){n — 4) 1X2X3X4X5 _ 5X4X3X2X1 _, 1X2X3X4X5 (3) (Art. 306a), P,=P3= 1X2X3=6. Thus, NOT, NTO, ONT, OTN, TNO, TON. Ps=lX2X3X4=24. (4) This is a case of permutations, when all the letters sre taken together (Art. 306a). •■• P6=1X2X3X4X5X6=720. (5) This is similar to the preceding. .-. P, = 1X2X3X4X5X6X7=5040. PERMaTATiONS AND COMBINATIONS. 19/ (6) Tlie whole number of arrangements is evidently equal to the sum of the different permutations of six letters taken 1 together, 2 together, and so on P,=7l= 6 Ps=n(ffl— 1)=6X5= 30 P3=:n(TO— l)(?i— 2)=6X5X4= 120 P,=n(n— l)(n— 2)(n— 3)=6X5X4X3= . . . .360 Pj=n(?i— I)(n— 2)(n— 3)(m— 4)=6X5X4X3X2= . . 720 P , =n{n— 1 )(re— 2)(re— 3)(ra— 4)(n— 5)=6 X5X4X3X2 Xl= 720 Ans 1956. (7) Here the number of different products will evidently be equal to the number of combinations of 4 things taken 2- together. . c _ "("— 1) ^4X3^^ ' 1X2 1X2 Let the learner verify this result by finding the different prod- acts ; they are 12, 15, 18, 20, 24, 30. (8) Here it is merely required to find the number of combina- tions of 5 things, taken 3 together. . Q _ 7!(n— l)(n— 2) _ 5X4X3 _jQ ' 1X2X3 1X2X3 (9) The number of permutations of n things, taken 4 together is P4=n(re— 1)(«— 2)(n— 3) ; taken 3 together, is V ^=n{n — l)(n— 2) ; . . n{n — \){n — 2)(n — 3)=6ra(n — \){n — 2) ; dividing each member by n{n — l)(ra — 2), we have n — 3=6, or n=9. llO) By Art. 306, the number of permutations of 15 things taken r together, and r — 1 together, is P, =15X14X13X12- • . • (15— »•— 2)(16— r^), P,_, = 15X14X13X12 (15— r— 2). Here we see '.hat the two quantities are the same, except the last factor of the first quantity, which, by the terms of tho question must therefore be equal to 10 ; that is. 108 KEY TO PART SECOND 10=15— ?■—!, whence r=6. Thus the permutations of 15 letters, taken 6 together, are 15X14X13X12X11X10, and the permutations of 15 letters, taken 5 together, are 15X14X13X12X11, whence it is readilr geen that the former is equal to 10 times the latter. (11) Ci=n= 4 C — "(^—1) ^4X3^ g ' 1X1 1X2 C — "("— 1) ('»— 2)— ^X3X2 _ 4 ' 1X2X3 1X2X3 ^ _ n(n—\)(n—'r ) {n 3) ^ 4X3X3X] _^ j * 1X2X3 1X2X3X4 Atis. 15. The learner may easily verify this result by taking the coins, or by finding the different sums that can be formed of the numbers 1, 3, 5, 10 ; the sums are 1, 3, 5, 10 ; 4, 6, 11, 8, 13, 15 ; 9, 14, 16, 18 ; 19 (12) Here it will be necessary to find the different combinations of six things laken singly, two together, three together, four together, five together, and six together. C,=ra= 6 Q _ «X(«— 1) _,6X5^ j5 ' 1X2 1X2 (. _ ?i(n—l)(re— 2) ^ 6X5X4 ^ 20 ' 1X2X3 1X2X3 — "("— ^)("— ^)('^— 3) — 6X5X4X3_ jj ''* 1X2X3X4 1X2X3X4 p _ ?i(w— l)(n— 2)(m— 3)(w— 4) _ 6X5X4X3X2 _ _ g ' 1X2X3X4X5 1X2X3X4X5 p ^ n(w—l)(n—2)(TO—3) (??.—4 )(??— 5) ^ 6X5X4X3X2X1 ^ j • 1X2X3X4X5X6 1X2X3X4X5X6 _ Arts. 33. In this solution we notice an illustration of the principle of Art. 309. Thus the number of conihiniitions of 6 things, taken 1 together, is the same as \iben taken (6- — 1), or 5 together ; the PERMUTATIONS AND COMBINATIONS, 199 ^l^mber, when taken 2 together, is the same as when taken (6 — 2), or 4 together. '13 He may vote for 1 candidate only, or for any 2, or for any 3 ; hence the whole number of ways in which he can vote will be equal to the number of combinations of four things taken singly, of four things taken two together and of four things taken three together ; thus, Ci=re= 4 ^^^ TO(ro— 1 )^4X3_ g 1X2 1X2 r;^^ w(w—l)(?i— 2) ^ 4X3X2 ^ 4 1X2X3 1X2X3 Total number of ways = 14. (14) If we reserve a, and take the different combinations of the four remaining letters h, c, d, e, taken two together, we may then unite a to each of them, hence the required number will be obtained by finding the diiTerent combi- nations of foU7- letters taken two together. Cj=_ ■=.2-^^6 ; and the combinations are 1X2 1X2 abc, abd, abe, acd, ace, ade. (15) A different guard may be posted as often as there are dif ferent combinations of 4 men out of 16. n, — "fa— l)(n— 2)(re— 3) ^16X15X 14X13 ^^ggri 1X2X3X4 1X2X3X.4' To find the number of times any particular man will be on guard, it is merely necessary to find the different combinations of (4 — 1)^3 men that can be formed out of (16 — 1)=15 men, since .he reserved man may be combined with each combination of 3 men, giving a combination of 4 men. ^^_ TO(7i— l)(?t— 2) _ 15X 14X '3_^55 1X2X3 1X2X3 (16) 04="-^-'^^"-- ^^^"-^^ 1X2X3X4 n(n — 1) Ue , 1X2 200 KEV TO PART SECOND. w(m— l)(ra-3)(w— 3) . w(w— 1) . j^ . ^^ 1X2X3X4 1X2 I'Art 9fi7v 2w(re— 1)(to— 2)(m— 3) _ 15?i(m— 1^ ' ' ' 1X2X3X4 1X2 Dividino' both members by ^ •', ^ 1X2 2(m— 2)(m— 3)_^g 3X4 reaucing, ?i^ — 5ra=84, and 71^12. (17) To find the number of peals that may be rung with 5 bells out of 8, find the number of different combinations of 5 things out of 8, then each combination will give as many changes as there are permutations of 5 bells, and the whole number of changes will be equal to the number of combinations multiplied by the number of permuta- tions in each combination. C _ w(w— l) ( w— 2)(ra — 3) (m— 4 )_ 8X7X6X5X4 _pfi . ' 1X2X3X4X5 1X2X3X4X5 Pj=lX2X3X4X5=120; 56X120=6720. The number of changes with the whole peal will evidently be' equal to the number of permutations of 8 things taken all together. P, =1X2X3X4X5X6X7X8=40320. (IS) Had the letters been different the number would be P, =1X2X3X4X5X6X7=5040 ; but there are 2 a's, and therefore (Art. 307), we must divide by 1X2 : .5040-^2=25L!0. Am. (19) Since tliere are 3 a's, 4 b's and 2 c's, in all 9 letters ; .•. (Art. 307a), the number of ways is 1X2X1X4X 5X6X7X8X J =3X7X4X9-1260. (1X2X3)(1X2X3X4)(1X2) f20) The ujmber of terms in which a' will stand first, wili nvidontly be equal to the number of permutations lake PERMUTATIONS AND COMBINATrONS. 201 all together of the letters in h^c^, whicli, by Art. 307 since there are 6 letters, is 1X2X3X4X5X6 ^,^ (1X2X3X4X1X^) v21) Reserving two letters, there are 5 letters remaining, then each permutation of 5 letters may be preceded by ah, therefore the whole number of permutations of 7 letters in which ab, or any other combination stands first, will be equal to the whole number of permutations of the remain- ing letters taken all together. 1X2X3X4X6=120. When abc stands first there are 4 letters remaining ; 1X2X3X4=24. When abed stands first there are 3 letters remaining ; 1X2X3=6. Thus, ahcdiefg'), abcdiegf), abci{Jeg), abcd{fge), abcd(gef), abcd{gfe). (22) The number of different combinations of 2 consonants, ^ , ,_ . n{n—1) 17Xlfi ,„„ out of 17, IS -^^ .-'= — i;^ — =136. 1X2 1X2 Each of these combinations may be united with each of the 5 vowels, giving 136X5=680 different combinations of 2 conson- ants and 1 vowel ; now each of these combinations of 3 letters will give 1X2X3=6 permutations, therefore the whole number of words will be 680X6=4080. (23) In the word " Notation " there are 5 different letters ; and the number of difl^erent combinations of 5 letters, taken 3 together, is J^ ;= __=10. But there are 2 n's, 1X2 1X2 2 o's, and 2 t's, each of which pairs may be combined with each of the other 4 letters, and form 4 combinations of 3 letters, making altogether 3X4, or 12 such combination* where the letters are repeated. .". the number required ^1U-|-12=22. The learner may easily write out the several combinations ; thus the first ten formed of the letters "NOTAI" may be ar- -angod as in Art. 309, and the remaining twelve are nno, nni, nnu, nni ; oon, ool, ooa, ooi ; tin, Ito, tta Ui. 202 KEY TO PART SECOND. Remark. — The term " difFereiit " is sometimes used ir. the preceding BOmtions in connection with combinations ; this is not ntended, liow- ever, to change the meaning of tlie word comhinalions, as given in th» Algebra (Art 308), but merely to render it more emphatic. B I N O Jl . A L THEOREM, WHEN THE EXPONENT IS A POSITIVE IKTEGEB (2) By comparing the quantities with those in the formula (Art. 310, Cor. 3), we find re=10, n — ?--|-l=6, a=x and x=y. Since n — 7--j-l=6, we have 10 — r-)-l=6, and ?-=:S; hence n — r-(-2=10 — ,9+2^7, and r — 1=4 ; therefore the coef- ficient of the r" term, that is, the term in which thB ex- ponent of the leading letter is 6, is ,t(„_l)(n-2)(TO— 3) ^10X9X8X7^^^^ ^^^^ 1X2X3X4 1X2X3X4 The coefficient, however, is most readily found by writing out the whole development, thus, (x+2/)"'=x'°+10x«2/+45a;8u2-Ll20xY4-210xy+, &c. (3) If instead of a, x, n, and r, we substitute c', — d^, 12, and 6 in the formula, Cor 3, Art. 310, we have 1X2X3X4 (4) Comparing the quantities with those in the formula, Cor. 3, Art. 310, we have a^a^ x=3ab, n=9, and r^7. . . the T" term is ^X8X7X6X5X4 . 3.3^3^^.6 1X2X3X4X5X6 =84ffl9X729ffiW=61236a"is. (6) Referring to the same formula, a=Za'', x= — 7x', 7i=8 and r=5 ; .-. the 5" term =-^-''>^^^^(3a2)<(— 7.i^V 1X2X3X4 =70X81a'X2401a; 2=1361 3670a«x'2. (6) Here a=ax, x-—hy, ra=10, and r=6 ; . . the 6" tern =252a'6'ay. . the 6" term =1221i^.l^iI^(aK)s(JyV 1X2X3X4X5 '^' BINOMIAL THEOREM. 203 (7) Since the exponent of the binomial is 12, there will be 13 terms (Art. 310, Cor. 4), hence the middle term will be the 7", and a=a''^, x=x:", n—\i, and ?-=7, (Art. 310, Cor. 3) ; •. the middle term = — Ci — Ci — Ci—L^—Oi^(a^y(x^'y 1X2X3X4X5X6 :=924a^'"x*". (.8_) Since the exponent of the binomial is 13, there will be 14 terms, and the two middle terms will be the 7"' and 8'*, the coefficients of which will be the same, (Art. 310, Cor. 5). (Art. 310, Cor. 3), a=a, and x=x, Ji=13, and r^7 ; -,1^ 13X12X11X10X9X8 7 s ,»,c 7 8 .-. 7"" term = — — — — — — — — — i^fl'a;»=17l6a'a;° ; 1X2X3X4X5X6 Since the exponent of the leading letter increases by unity in each term, and the exponent of the other letter decreases by unity, .-. 8'" term =1716aV. (9) (Art. 310, Cor. 3), a=l, x=x, ra=Il, r=8 ; . . 8.^ term =ll><^><^><^><^(l)Kx)'=330:.'. 1X2X3X4X5X6X7 (10) (Art. 310, Cor. 3), a=x, x=— !/, w=30, r=6 ; ... 6- term ^ 30X29X28X27X26 1X2X3X4X5 =— 142506a;2y. ; 1 1) Comparing this with the general expansion of a-\-x Art. 310, we have a=Zac, x= — 2bd, and re=5 ; and we have (3ac—2bdy=i3acy-\-b(3acy(i—2bd) + 10(3ac)3(— 2M)2+10(3ac)2(— 2M)»-f-5(3ac)(— aid)" +(— 2M)'=243a5c''— 810a"'c''M+1080aVA2d2 —n20aWd^-\-240acb''d''~3'2¥dK 02- ia+2l>—cy=)(a+2b)—c\^=ia+2by—3{a+2byc +3(a+2hy—c'=a'-\-6aH -]-12ab^+8¥ —3a'c—12abc—12b^c-{-3ac^+6bc^—cK tl3) Since the coefficients in the expansion of (a-j-a;)" do not contain either a or x, they will be the same when a=l 204 KEY TO PART SECOND. or x=l, or both a and x at the same time =1. (See Art 310, Cor. 6). For the sake of brevity let the coeffi''.ients of the ex pansion of (l+x)" be represented by A,, Aj, Aj, &c. then (l-|-a;)'>=l_l-A,a:+A2a:2+A3a;S+AiX<-l-Aja;5+, &c. Writing — x, instead of x, (l_x)"=l— AjX+AjX^— AjX'+AjX^— AsX«+, &,c. Now if X be made ^1, tlien since (1 — 1)"^0, we have 1— A.+A^— A3+A,— A,+, &c., =0. .-. 1+A2+A,+Ae+, &c., =A,+A3+A^+, &c. Tiiat is, the sum of the coefficients of the odd terms is equal to the sum of tlic coefficients of the even terms. fNDETERMINATE COEFFICIENTS; BINOMIAL THEOREM WHEN THE EXPONENT IS FRAC- TIONAL OR NEGATIVE; SERIES. INDETERMINATE COEFFICIENTS Articles 314 — 318. (1) Let li?-^=A+Bx+Cx=+Dx3-f Ex''+, &c. 1 — 3x Clearing of fractions, we have l-f2x=A-f(B— 3A)x-f(C— 3B)x'+(D— 3C)x2+, &c., from wliich, by equating the coefficients of tJic same powers of x, A=l ; B— 3A=2, whence B=5 ; C — 3B=0, whence C=15 ; D— 3C=0, whence D=45, &c, 1+2j;_ 1— 3x --l+5x-\-i5x-+45x'-\-, &.C. (2) Let - ^'^^■'' =A+Bx+Cx'+l)x'+Ex'+, &.c 1 X — X- Clcaring, H-2x=A+(n— A)x-1-(C— A— B)x' -1-(D— B— C)x'+, &c. INDETERMINATE C () E F 1'' 1 C ( E N T S . 20E • (Art. 314), A=l, B — A=2, whence B=3 C — A — B=0, vvlieiice 0=4 D— B— C=0, whence D=7 ■ ^+2^ =l-i-3x+4j^+7x^+lla''+, &c. 1 — X — x^ Here we easily perceive that the law is, that the coefficient of &ny term is equal to the sum of the coefficients of the two pre- eeding terms. (3^ Let 1— 3^+^^°^a+Bz4-Cx'4-Dj:^+, &c. ' l-\-x-\-x'^ Clearing, 1— 3x+2x3=A+(A+B)a;+(A+B+C)a;2 +(G+C+D)x'+, &c. .-. (Art. 314), A= 1 ; A+B=— 3, whence B=— 3— A=— 4 ; A-4-B+C= 2, whence C=2— B— A=5 ; B-|-C4-D= 0, whence D=— B— C=— 1 ; C-j-D+E= 0, whence E=— C— D= — 4 ; &c. . . the series is 1 — ix-\-bx^ — x^ — ix*-\-, &.C. r4) Let ?±?f=A+Ba;+Ca;2+Dj;3+Ej:'+, &c. b-{-lx Clearing, 34-2x=5A+(7A+5B)x+(7B+5C)a;» -l-(7C+5D)x=+, &c. .-. (Art 314), 5A=3, whence A=f ; 7A+5B=2, whence B=— 11=— li ; ^ 25 52 7B+5C=0, whence C=T!=Z:Ii ,■ 5^ 5' 7C+5D=0, whence D=— IjII ; &c. • • 3 11 , 7.11 , 7'.11 3 , „ .•. the series is - — ^-x4- x' — x^+i &c. 5 5=^ 53 6< ^5^ Let J+f_— =A+Bx+Cx2+Dx3+Ex4 4-, &o ' (1 — xY 5J06 KEY TO PART SEOONI). Clearing, by muLi plying botli sides by (1 — x)', l+a;=:A+(B— 3A)x+(3A— 3B+C)x2 +(3B— A— 3C+D)x3-[-(3C— B— 3D+E)a;''+, &c. .-. (Art 3)4), A=l ; B— 3A=1, whence 6=4=2^ ; 3A—3B-|-C=0, whence C=9=3= ; 3B— A— 3C+D=0, whence D=16=42, 3C— B— 3D+E=0, whence E=25=52 ; &c. .-. the series is P+2^x+Z^x^-ir'iV-\-5'x*+, &c. 6) Let jT^=A+Bx+Cx^+'Dx'+'Ex''-\-Fx'+, &c. Squaring both members, 1— a;=A2+2AB:>;+(2AC+B>^+(2AD4-2BC)a;3 +(2AE+,2BD+C>''+, &c. .-. (Art. 314), A== 1, whence A=l ; 2AB = — I, whence B=— i ; 2AC+B'= 0, whence C=— -=— _i- ; ' 8 2.4 2AD+2BC= 0, whence D— 2AE4-2BDH-C== 0, wlience E= 1 3 16 2.4.6' 5 _ 3.5 128 2.4.6.8 2 2.4 a . 4 ,"6 2.4.6.8 ,1 • t ^ X X oX o • fjX p •. the series is 1 — - — — — — , oic. (7) If we assume (l-j-ar+x') equal to the preceding series, A-|-Bx-(-> &o., and square both members, the coefhcienta of the different powers of x will be the same as in the preceding solution. By equating the corresponding coef- ficients, we have A-=l, whence A=l 2AB =1, whence B=! 2AC+B2=1, whence C=|; 2AD+2BC =0, whence D=— /g ; &c. -. the series is 1-1— +— — "— 4-i &c. 2 8 16 INDETERMINATE COEFFICIENTS. 207 (8) The solution of this example is exactly like the preceding except that in equating the corresponding coefficients the right member of each equation is 1. (9) Since x—x^=x(l—x), let 1+^=:^+^. X — X^ X 1 — X Reducing the fractions to a common denominator, we have 1+^ ^A(l— x)+Bj; . X — x'^ x{\ — x) or, l-j-a;=A+(B— A)a; ; whence A=l, and B — A=l, or B=2. X — X^ X 1 — x' (10) Since a;^— 4=(x+2)(a;— 2), let 8^—4— ^ ' ^ x^ — 4 x-\-2 x—a' 8a:— 4 ^ A(x— 2 )+B(a;+2)^(A+B)x+(2B— 2A) , x^—4: (x-(-2~)(x— 2) (x-[-2)(x— 2) 8x— 4=(A+B)x+(2B— 2A) ; . . A+B=8, and 2B— 2A=— 4 ; Solving these equations, we find A=:6, and B=3 ; 8x— 4_ 5.3 x=— 4 x-1-2 x—2' (11) Since x'— 7x+12=(x— 4)(x— 3), let — ^.il x' — 7x-(-13 _ A , B X — 4 X — 3' x+1 __ A(x— 3)+B(r— 4 ) x'— 7x+12 (x— 4)(x— 3) _ (A+B)x— (3A+4B) . (x— 4)(x— 3) ' a;+l=(A+B)x— (3A+4B) ; .-. A+B=l, and — 3A— 4B=1 ; whence A=5, and B= — 4 ; x+1 __ 5 _ 4 x= — 7X+12 X— 4 X— 3- (13) (x2— IX^— 2)=(x— 2)(x— lj(x+l). 208 KEY TO PART SECOND. ,. x^ A , B , C (x'— l)(j;— 2) X— 2 X— 1 x+1 _ A(x^— 1)+B(x— 2)(x+])+C(x— 2)C;c— IJ (x— 2)(x— l)(x+l) ■. x^=(A+B+C)x2— (B+3C)x+(2C— A— 2B) ; Solving these equations, we find A=|, B^ — i, C=g ; ._ x^ 4__ _ 1 . 1 (x2— l)(x— 2) 3(x— 2) 2(x— l)~'"6(x+l)' (13) x"— a<=(x'— a2)(x2+a2)=(x— a)(x+a)(x2-fa2). 1 A , B , C Let i^ X — a x-\-a x--\-a' ^ A(x+g)(x^+a^)4-B(x— g )( x^+a^)+C(x^— g^) . (x — a){x-\-a){x--\-a^) .-. l=(A+B)x3+(Aa— Ba+C)x2 +(Aa'-\-Ba^)x+Aa^—Ba'—Ca\ .-. (Art. 314), A+B=0, (1) Aa— Ba+C=0, (2) Aa'-j-Ba^ =0, (3) Aa'— Ba'— Ca2=i. (4) Equation (3) is the same as (1) ; then finding the values of A, B, and C from (1), (2), and (4), we obtain A=L, B=— J-, and C=— J- ; 4a^ 4a3 2a^ x"— a'' Aa^x—a) 4a%x-{-a) 23=(x=+a=)' (14) x«— l=(x'— l)(x3+l)=(x— l)(x2+x+l)(x+l) (a;2_x+l)=(x— I)(x+l)(x2— X+l)(x'+X+l). 11- 1 1 A , B , C , D II we place _ — = + -\- + tl , and re- x«— 1 X— 1 x+1 ' x^— x+I r»+x+r duce the fractions to a common denominator, and equate tlie cot'fficients of tlie same powers of x, ue shall find the equations incompatilile, hence we must make a different assumption. A little reflection will show that in reducing the above frac- tions to a common denominator, and comparing the coefficients of the same powers ol' x, we shall have six independent equa- tions, hence wu may ashume the numerators of the fractions so as BINOMIAL THEOREM, SOf) to involve six unknown quantities, and as x may appear in the oumerator of some of the fractions, we may assume it as a factor af one or more of the unknown coefficients. • let —- = ^ I B I Cx+C , Dx+D' x^ — 1 X — 1 x-\-\ x^ — x-\-l x'^-\-x-\-l' Reducing the fractions on the right to a common denom inator, the numerators are A{x^+x'^-Jf-x^^x^-\-x+1)+Bix^—x'+x^—x''+x—l) -j-C{x'-\-x^—x^—x)+CXx'+x'—x—l) +I>{x^—x'^-\-x'—x)-\-'D'{x''—x^^x — 1), which, are =1. Equating the coefficients of the same powers of x on both sides, we have the following equations : A+B+C+D =0, (1) A— B+C — D +C'+D'=0, (2) A+B+C— D' =0, (3) A— B— C +D =0, (4) A+B— C — C— D +D'=0, (5) A— B— C— D' =1. (6) Solving these equations, we find A=J, B= — g, C^(j, C'= 3, D=: — g, I) = 3. Substituting these values, and writing 5 as a factor of tht whole, we find 1 _1 < 1 _ I , x— 2 _ x-\-2 ) a:^ — 1 6 Ix — 1 x-\-l x'' — x-\-l x^-{-x-\-l' BINOMIAL THEOREM, WHEN THE EXPONENT IS FKAOTIONAL OK NEGATIVE. Note. — Instead of finding the general law of the coefficients by the method given in the Algebra, page 277, it is proper to inform the stu- dent that there is another method, which is more simple in theory, but far more difficult in practice. We shall explain the method and show where the difficulty occurs. Second. — -To find the general law of tlie coefficients. Let (l+xy=l+nx+'Bx^+Cx^+Dx''+. &c., where B C, D, &c., depend upon n. IS 210 KEY TO PART SECOND. Squaring both sides, we liave (l4-a;)2"=l+7JX+Bx2+Ca;'+Di''+, &c., -\-nx-\-n^x--{-'Bnx^-{-Cnx^-\-, " But (Art. 201), (l+»;)2«=|(l+a;)2("=5l+(2i+a=)(". Considering (2x-{-x'') as one term, we have { \-{-(2x+x^)l''=l+n(2x+x')-\-B(2x+xy+C(2x+x'y+,. &.C. =l-\-2nx-\- nx^ +ABx^-\-4'Bx^+ Bx* +8Ca^+12Ca;''4-, &c. +16Dr'4- " + " Now since tliis series and the former must be identical, we have, by equating the coefficients of the lilce powers of x, 2n=2n, 2B+n2=4B+re, .-. B='i^-) ; ' ' 1.2 2C+2B»=8C+4B, .-. c=^±^)^M^^)(n-^) ' 3 1.2.3 2D+2C+B==16D-1-12C4-B. To find D from this equation in terms of 'i is a difficult opera- tion, and the finding of E would be still far more difficult. This, renders the demonstration given in the Algebra, on the whole much the easier of the two. Article 320. 12) Here (n+l)±^{^^ = ^Xfj,X\!i==tl ; . r >1; l-.ence the 2"'' term is the greatest. (3) Here (n+1) J-=(8-H)J ^=^9X^X^=if=4,V. The/)-s< whole number, greater than 4yj, is 5 ; therefore the B"' term is the greatest. BINOMIAL THEOREM, 2li Article 321. (1) Herea=l,i= — x,n= — 1. .■ (1— .r)-'=l— IXIX— j:— ^^~'~^V =^\-\-x-\-x'^-\-x^-\-, &c. (2) Here a=l, 1= — x, ra== — 2. .-. (i_a;)-==I— 2X1X— X— ^t±Zy(— i;)J 1.2 = l+2x+3x'+4a;2+, &c. (3) To develope this expression, expand tlie part in toe pa rentliesis, and multiply by a?. Comparing (a-)-x)"^ with (a-J-^)") we have a=a, i=a;, and n= — 2. 1.2 _ 2(-3)(- 4)^^,^3^, &c., 1.2.3 1 3a; I 3x2 4^3 a^ a^ a'' a' «'(a+x)-== l_??+3x2_4^ +, &c. (4) Herea=l,J= — a:', ra=J. .-. (l_x=')i=l— JXlX^H-l^i^C— a:')' 1 .2 _L. i(— i)(— 3 )r_.r3)3_. &c., 1.2.3 " 3" 9 sT ' ■ ' (6) Here a^a?, b^x, n=l. 212 KEY TO PART SECOND. 1 • ^ -a(—i)(—i) ^a2-)-f^3_|_ i(— ■')(—'.)(— I) (a^)-?a^4-, &c., 1.2.3 1.2.3.4 =^a+- — — +— - — ?^+, &c. 2a 8a' lea* 128a' In making these reductions the pupil must notice that 1(3=) ^x= X TTV..T 1)^ 0\~? 9 1-2 8(a^)^ «''' i(^)(r:i)(a^)-t.'=: ^^=J^, &e. 1.2.3 jg^^,^| 16a= 6) Here 0=^0?, h^ — x, n=^. .-. (a'— a:)i=:(a')^+ira')~^X— a:+"^i^t-'(a')"*(— x)' 1 • )i I a(— s)(— s) (^3s-§(_.^y_|_ -3(— i)(— 3)(— 3) c„3y- V/j-y 1.2.3 1.2.3.4 +, &c., =a—JL—^—^l— '0;^-' — , &c. Sa' 9a5 81a» 243a" f7) Here a^l, h=2x, ?i=^l. .: (l+2a;)^=l+.K2i)+i^Ili\2.r)2+^l"i:i^.^rri\2x)« 1.2.3.4 ^ ^ ^' (8) Here a=a2, i=— x', n=\. i . 2 ( x=)^+, &c., BINOMIAL THEOREM 213 X3 rviH ryif ft lyiO =/9(l+i)=37(l+-J); (See Formula, Art. 322. J =]+.055555— .001543+.000085 —.000005+, &c., =1.064092, and 1.054092X3=3.16227+. (2) ^27+3= V27(l+p=3V 1+1 ; Viqri=i+^xi-i xL+3Lx5--Lxl2+, &c., =l+.037037—.001371+.000084— .000006 = 1.035744; and 1.035744X3=3.10723+. BINOMIAL THEOREM. fZ) J/27-3=5y27(l-^)-.--3V(l-i). The development of (1 — ')* is the same as that of (!-(-')', except that all the terms after the first are nega- tive. To get the result accurately requires that we should calculate five terms of the series after the first. These carried to nine places of decimals are —.037037037— .001371742— .000084676 —.000006272— .0000005 U—, &,c. Subtracting these from 1, and multiplying the remainder by 3, we have 3/24=2.8844992+. (4) V256+4=V256(l+gi^)=4Vl+i;V 1 = 1+.003906— .000022+, &.C., =1.003884, and 1.003884X4=4.01553+. (S) V128—20=V 128(1— j5J=2Vl—/2. In calculating the value of each term, the shortest method is to find it from the preceding term. Thus, by consid- ering the formula, Art. 322, we notice that each term in the development after the first, is equal to the preceding term, multiplied by two factor;s, one of which is _, and a" the others successively -, , , , , n 2n 3n An bn and so on ; therefore calling the terms A, B, C, and so on, we have -w ^ 32 ^ 7-32-^ T^^° aT'slZ^ 7^5^ — II./tjE— , &c. =1— .0223214— .0014947— .0001446 —.0000161— .0000019=.9760213, and .9760213X3 = 1.95304+. 216 KEY TO PART SECOND. THE DIFFERENTIAL METHOD OF SERIES. Article 325. ;2) Here «=2. a=l,i=4, c=9. . . I>,= l— 2X4+?^=1— 8+9=2. ' 1.2 (.3) Here n=3, a=l, J=3, c=6, d=10. ... D,=-l+3X3-!^+!Ml^°=-l+9-18+ie =0. '" ' 1.2 1.2.3 (4) Here n=5, a=l, 6=3, c=9, rf=27, e=81, /=243. .-. Dj=— 1+15— 90+270— 405+243=32. (5) Here re=5, a=\, b=l, c=j, d=g, e=j\, f=s\- ,,_5.4 , I 5.4.3 ,_ 5.4.3.2 _, 1.2' 1.2.3"^ 1.2.3.4''' S t 5 1 o*^ 1 O O'^ 1 O Q -t"i5 j_ ^ .4. 3. 2.1 , iJlQi 214-1' 6 _(_ 1 — 1 J 1 2.3.4.5 '^^ ^ 2T^*4 Tg^lU— 4 T _1_ 1 1 + 3 2 32- Article 326. (3) Here a=l, D,=3, Dj=2, and D3=0. .-. 15" term =1+(15— l)3+ll:-i!x2=l+42+182 1 • ^ =225. n« term =l+(re— l)3+^-5z:i)^^=H2x2=n=. 1 • ^ (4) Here a=l, D,=4, Dj=6, ©5=4, Di=l, Dj=0. .-. 12" term =1+11X4+— -li^X6+" ' ^° ' ^X4 1.2 1.2.3 , 11.10.9.8^ 1=1+44+330+660+330=1866. 1.2.3.4 (5) Here a=l, D,=2, D,=l, D,=0 SERIES DIFFERENTIAL METHOD. 217 .•■ n'nerm =l+(«-l)2+ ^"-^)( -"=g.)==g!+!' ' 1.2 2 2 (6) Here a=l, D,=3, Dj=3, D3=I, D4=0. .-. n'* term =1+Cra— l)3+^"^^^^"~^)x3 1 • a8X17s^g_gO_|_j520-|-i3680+29070 1.2.3.4 =44330. (5) Here the terms are 6, 24, 60, 120, 210, and so on ; hence a=6, D, = 18, Dj=18, 03=6, and D^=0. .-. Sum of 20 terms =20X6+?i^X18 1.2 20X19X18 , 20X19X18X17 ^^,,„,,^^„ 1.2.3 1.2.3.4 +20520+29070=53130. (6) Here a=l, D,=7, Dj=12, D3=6, D^=0. ci r * T n(n — 1).,!, , n(n — l)(n — 2) .*. Sum of n terms =n+_^ ^X"+-^ 1.2 ' 1.2.3 X i9_L. ''("— !)(''— 2)("— 3) y_ fi_,4ra I 14a^— 1471 1.2.3.4 4 4 , Sre-''— 24ra='+16CT I TO''— Sm^+l ITO^— 6re "^ 4 4 = "'+^"'+^ =^V^+2TO+l)=['^=0. .-. Sum of 25 terms =25+?illixi5+-^-?il^ix32 1.2 1.2.3 , 25 . 24 . 23 . 22^ig_25_j_4gog_|_73gQQ_^227700 1.2.3 =305825. PILING OF CANHOn BALLS AND SHELLS. Articles 328 — 332. (3) Comparing the number 15 with Formula B in Art. 332, w» have ra=15. .. number ^<>»+l)(3>»+l)^ 15X 16X31 ^ 6 6 (4) See Formula C, Art. 332, 1=52, and 7J=34^ ~ 3 «(»+l)(3Z— ra+l)=VX35X(156— 34+l-i=V'X35 X 123=17X35X41=24395. (5) Number of balls in a complete triangular pile of which each side of the base is 25, is (Art. 332), Jre(»+l)(n+2)=Vx26X27=25X 13X9=2925. Since the number of balls in a side of the top course ia 220 KEY TO PART SECOND. J 3, the number in a side of the pile that is wanting is 12, hence the number in this pile is 'g X-3X14=:364. .-. 2925—364=2561, the number required. (6) Number in the pile considered as complete, (Art. 332J, =-(ra+l)(ra-[-2)=Sg8x 39X40=19X13X40=9880. Since there are 15 courses, and the number of balls is one less in each course than in the next preceding course therefore 38 — 15^23 is the number of balls in a side of the incomplete pile, and the number in this pile is VX24X25=23X4X25=2300. . . 9880—2300=7580, the number required, f7) Number in the pile considered as complete (Art. 332), = -(reH-l)(2»+l)=YX45x89=22X 15X89=29370. Number of balls in a side of the pile that is wanting is 21, and the number in the incomplete pile is "g'X22X43=7XllX43=3311. .-. 29370—3311=20059, the number in the incomplete square pile. (8> ^1521^39= number of balls in a side of the base course, Jl69 =13= " " " " " top " ^S^'X 40X79= 13X20X79=20540, the number of balls in the pile considered as complete. 13 — 1^12, the number of balls in a side of tlie base of the pile that is wanting ; and 'g" X 1 3X25=650. .-. 20540—650=19890, the number of balls in tha in2om. plete pile. (9) Here we have the equation (Art. 332), i'.w(ra+l)(3;— ra+l)=6440, in which ra=20, to lind I. .-. \°X2](3Z— 19)=6440, 70(3;- 19)=6440, 3<— 19=92, and Z=37. 37X20=740, tlie number of balls in the base. INTERPOLATION OF SERIES 22 1 110,. Here we have the proportion ln{n+i){n+2) : in(re+l)(27i+l) : : 6 : 11. Placing the product of the means equal to the product of the extremes, and canceling ln(n-]-l) on each side, we have 1271+6= Ura+22, whence m=16, the number of balls in a side of the base of each. in(7i+l)( n4-2)=VX 17X18=816 = balls in tr. pile, j7i(7i+l)(2re+l)='a''X 17X33=1496= " " sq. pile. (11} Since the number of balls in each side increases by 1 aa we descend, and since there are 7 courses below the upper one, tlierefore 36-|-7=43, and 17-|-7=24, are the num- ber of balls in the longer and shorter sides of the lower course, iXre4-l)(3Z— 7!+l)=VX23(129— 24+l)=10600, the number of balls in the pile considered as complete. It is evident that 35 and 16 are the number of balls in the longer and shorter sides of the pile that is wanting, hence the number of balls in this pile, is V X 17(105— 16+1)=4080. .-. 10600—4080=6520, the number of balls in the incom. plete pile. INTERPOLATION OF SERIES. Article 333—335. 1 1) Since the 4" differences vanish, we have (Art. 325), e — id-{-Sc — 4i-|-a=0, where a=3, c=15 d=30, and e=:55, to find b, .-. 55-4X30+6X15—46+3=0, whence J=7. Having the terms of the series, viz. . 3, 7, 15, 30, 55, we readily find the first terms of the several orders of differ- ences (Art. 325) to be D,=4, 0^=4, D3=3, and 0^=0, therefore by making re=6, 7, and 8 successively, and sub- stituting the values of D,, Dg, and Dj in the formula „+(„_l)D , ^ (n-l)(^-2) p^ („_i)(^_ 2)(^z--3)p ' 1.2 ' 1.3.3 ' 222 KEY TO PART SECOND. we obta'.n the 6'*, 7", and 8'* terms. Thus the 6"' term Is 3+5X4+8X4X2+^^^1^^X3 1X2X3 =3+20+40+30=93. 7'* term is 3+6X4+6X5X2+-^^^^X3=3+24+60 1X2X3 +60=147. ■ 8»term is 3+7X4+7X6X2+'^^^^^X 3=3+28+84 1X2X3 +105=220. '2) Let x= the 5"" term, then writing down the terms, and ■finding the respective orders of differences, we have I , 18 , 30 , 50 X , 132 , 209, , 12 , 20 , 1—50 , 132— x , 77 , 5,8, a;— 70 , 182— 2x , x — 55 , 3 , a;— 78 , 232— 3x, 3a;— 237, 1—81 , 330— 4j;, 62—489, 411— 5x , IOj,'— 819, 15x— 1330. - 15x— 1230=0, and a;=82. Or, thus, since the 6"" differences vanish, or become 0,it is merely necessary to find the first term of the 6"' order, by means of the Formula, Art. 325, by calling n=6, a=ll, i^l8, &c., thus, ,— ^^+ "("— 'V — "(^— ')'^»— ^) ,f+ '^("~')("~^)("— ^) a 1.2 1.2.3 -1.2.3.4 _ TO(m— l)(7t— 2)(re— 3)(ra— 4) ^ 1.2.3.4.5 (n— l)(n_3)(n— 3)(n— 4)(ra— 5) . 1.2.3.4.5.6 . „-6X18+^>^X30-g> &c.,) ad inf.S ■ 1+2+3+4- • • • „ — (s+i+j- • n'^^Il) (4) Since 9=1, and ;5=3, _ 5 i+i-+i+i+i+.&«-'l _, , , , ,_,5 ■ ■ '-a+i+^+^+^+. &c.) s -'+3+3-'^' and _'* of this sura =|i= sum required 71+1 7l-fl' (5) Since 5=1, jo=2, and ?i=l, 2, 3, &c. . 51+2+ i+i+*+. &c., ' -(l+l+i+. &c.) ] i+i='' (6) To find the series let n=l, 2, 3, 4, &c., successively, thct the terms are ..i-+_L+J_+J_+,&c. 1 . 5 2 . 6 3 . 7 ' 4 . 8^ Also, 5=1, and p=i. and -'* of this sum =1 of |^=||. . 5l+Ri+3+ i+s+.- &c., ) _ , , ■• i _a+i+.&c.^^ -1+2+3+3 (7) Dividing each term of this series hy 2, it becomes 1 I 1 1 1 I 1 I ^ „ _4- -4- —4- +, &c. 2 2.33.44.5 Tlie sum of this series has been found (see example 2.) to be 1 therefore llio sum of ihc ffiveii series is lX2=i!. INFINITE SERIES. 225 (8) ]\[ultiplying each term of this series by 3X4, or 12, it becomes _+ — 4- 4-, &c., the sum of which has 2 2.33.4 been found to bo 1 ; therefore the sum of the series is 1^12=t'2. KECURRING SERIES. Articles 339 — 343. \2) Here A=l, B=6x, C=\2x^, D=48a:', E=120x^ &c. Making x=l, and substituting in the formula, (Art. 341), we have^=^^X^^-6X'^»=l. ^- '^X 120-48X48 _, 12X12—6X48 12X12—6X48 (Art. 343), s- A+B-Apx _ l+6a;-:c _ l+5x 1 — px — qx'^ 1 — X — 6x- 1 — X — 6x'^' (3) Here A=l, B=2x, C=3x^ D=4a;', E=5x\ Making x=l, and applying formula (Art. 341), we have 3X4-2X5^2 3X5— 4X 4_ ^ 3X3—2X4 ■" 3X3—2X4 (Art. 343), ci_ A+B-Apx_\+2x-2x _ 1 1 — px — qx'^ 1 — '2x-\-x'' (1 — xy k4) Here A=^, B=-'^^ C="J!^, T)=-'^, &c. c c c^ c* If we make x=\, and apply the formula, (Art. 341), we shall find jt)= — _, and q=Qi, but the scale of relation is c easily seen to be — -, since if any coetEcient is multi- c plied by this quantity it will give the coefficient of the next following term. a ahx I ahx A+B — Apx '„ 72~ 75' „ (Art. 343), S=-^^^^ ^=° 1 - ^ ^—P^ J , ?5 — c+te' c (5) Ht le A=0, B=i, C=j;', &c., and the scale of relation is that is p=l, and 7=0. 22G KEY TO 1-/1 RT SECOND. „_A4-B — h.px 0+a; — 0_ x 1 — px 1 — X 1 — x' (6) Here A=0, B=a;, C=— x-, &e., and the scale of re.aticB is — 1, that is p= — I, and j=0. „ A-|-B — A-px_d-\-x — X I — px \-\-x l-\-x' (T) Here A=l, B=2x, C=8x2, T>=28x\ E=100a;<. Making a;==l, and applying formula, (Art. 341), we have _ 8X28— 2X100 ^3 _ 8X100— 28X28 ^g, ^ 8X8—2X28 '* 8X8—2X28 (Art. 343), g^A+B-Ag_^^l+2.-3x^ 1-x l—px—qx'^ 1— Sx- 2^2 1— 3j;— 2a;2 (8) Here A=l, B=3x, C=52:^ D=7x3, E=9a;''+, &c. Making ar=l, and applying the formula, (Art. 341), we have 5X7-3X9 _o 5X9— 7X 7__, 5X5— 3X'7 " 5X5—3X7 (Art. 343), s=A+B-Ap_x_l+3x-2x^^J+^_ 1^33; — qx^ l—2x-\-x'^ (1 — xY ({)■) Here A=l, B=4x, C=9a;', D=16i', E=25:i:S &c. Making x=l, and applying the formula, (Art. 341.) The values of p and q thus found will not reproduce the series, hence we must apply the equations in Art. 342, and find the values of p, q, and r, when a;=l. These equa tions give 16^ 9p-(- iqAri", 25=16;)+ 9y+4)-; 36=25;!+16g+9r. From these equations we find p=3, 5= — 3, and r^l. We shall now extend the principle of Art. 343 to finding the gum of an infinite recurring series when the scale of relation onsiste of three terms. The 1" term A=A ; the 2"" CC B=B; the 3,i (£ C=0; the 4'» it T)—Cpx-[-Bqx^+Arx' ; the 5'* (( E='Dpx-{-Cqx''+Brx^ ; the 6"- it F=Fjpx-]-'Dqx''+Crx' ; &. c., = &c. REVERSION OP SERIES. 227 Now if S represents the required sum, by adding togetlier the corresponding members of these equalities, and ob serving that C+D+E+, &c.,=S— A— B ; B+C+D+, &c., =S — A, we have S=A+B+C+(S— A— B)px+(S— A)7a;2+Srx', whence g^ A+B+C-(A+B);>x-Ag^^ ^ 1 — jix — qx'^ — rx^ Substituting in this formula the values of A, B, C, and of p, q, and r; we have ^_Tl-\-4x+j0c'—3x—\2x'+Sx^_ 1+x 1— 3j+3x=— x' (1— a;)'' KEVEKSION OF SERIES. Articles 344 — 346. (!■* Comparing the series with the formula, (Art. 344), 'we have a=-|-l, i= — 1, c=+l, d= — 1, &,c., hence by substitution, we have 1 — 1 2 — 1 , — 1+5 — 5 , , . ir=-2/— — y'+ -y-2/'— — ^ y*+, &c., (2) Here a=l, i=l, c=l, d=l, &c. 1 1 , , 2—1 3 1—5-1-5 4 . o (3) Comparing the coefficients with those of the series lo Art. 346, we have a=2, A=3, c^4, d=5, &c. 1 3,1 27—8 5 „ „ .'. x=-ii — — v-\- v — , &c., 2 16 128 (4) Applying the formula, (Art. 345), we have a'=], a= — 2, and i=3. ... ^=_l(y_l)+^(y_l)^_i|z:?(y-l)3+, &c., =— i(2/-i)4 i(y-i)'-/5(y-i)'+. &c. 228 KEY TO PART SEOND. (5) See formula, Art. 345. Here a'=l, a=l, J=— 2, c:=+ ,/ ,^ I o/ ,N5 1 8—1, ,,, 0+10—40 (y-^y, =!/-l+2(y— l)»+7(y— l)'+30(y— l)^+,&c. (6) See formula, Art. 344. Here a=l, i=J, c=|, d=j'4. &c. ill 1 S I 5 .-. a:=ly— %=+i l2/3_54 L2TIIy4_|_^ Slc, (7; Let X =Ay-\-Bf+Cy^+, &c. ; tlien x^=AY+2A.By'+, &c. ; 3.3— A^y'+, &c. Substituting these values for x, x^, a;', . . in the second member of the given equation, and transposing the first member, we have O^ffAly+gH —1 7jA= -i-27iAB + kA' f+- Hence A^— 1=0, A=A+By— fl=0, A3/t+2ABi+^C— i=0 ; whence, A=l, B=:-(a—A'h)=l I a—- ) =L(o^'--»j ff 9 9^ g'^i f C=l(i— A'/c— 2ABA) 9 g i 9^ gg'' » _ bg''—'kg—'ih{ag'^—K) ~ 9' ^y_^{ag'-h)y\ [bg^-l-9-2h{ag-'-h) J,/^ ^ ^ 9 9^ 9 CONTINUED F K A T I O N g . 2U9 CONTINUED FRACTIONS; LOGARITHMS; EXPONENTIAL EQUATIONS; INTER- EST AND ANNUITIES. CONTINUED FEACTIONS. Articles 347 — 356. (.i; Dividing the greater term by the less, the last divisor by the last remainder, and so on, the quotients ai-e 3, 4, 5, and 6; henco the integral fractions are -j, |) j, and j, and thb "onverging fractions are 1 1X4 _4 4X5+1 __21 21X6+4 _130 3' 3X4+1 is' 13X5+3 68' 68X6+13 421" The S""* and S"' examples are worked in a similar manner. (4) Making 3900 the numerator, and 10963 the denominator of a fraction, and proceeding as in the preceding exam- ples, the successive quotients, that is the denominators of the respective integral fractions, are 2, 1, 4, 3, 2, 2, 1, 30; hence the first approximate fraction is I, the second, _-L^L=l ; the third, ^J1±±I= s ; and so on. 2X1+1 3X4+2 '* C5) Making 4900 the numerator, and 11283 the denominator of a fraction, and proceeding as above, we find the suc- cessive quotients to be 2, 3, 3, 3, 2, 7, 1, 1,1,2; hence 1X3 the approximating fractions are -I ; — — — =3 ; ^^ - 2X3+1 ^ 3X3+1 _, n . 10X3+3_33 . 33X2+10_ ^, ^^ 7X3+2 -^'23X3+7 '^'* ' 76X2+23 ' '^^' (6) Making 1 the numerator, and 3.1415926 the denominator of a fraction, or 10000000 and 31415926, and dividing the greater by the less, the less by the remainder, and so on, the quotients are 3, 7, 15, 1, 243, &o. Operating in a similar manner with 1, and 3.1415927, the quotients are 3, 7, 15, 1, 354, &c., then finding the approximating or converging fractions, corresponding to these quotients, wa ^^^^ , . 1X7 _7 . 7X15+1 _,„„. 106X1+7 ^'3X7+1 -^'22X15+3 ^'-^ ' 333X1+22 230 KEY TO PART SKCUND. The ratio of 113 to 355, that is ff § = 3.1415929+ ; and since the true ratio lies between 3.141526, and 3.1415927, and since the diiTerence between 3.14159:29 and 3.1J15926 is .0000003, I I'ere- fo''6 3.-, i expresses the part that the diameter is of the circum ference to within less than .0000003. (.7) 5 hrs, 48 min., 49 sec, =20929 seconds, 24 hrs, = 86400 " Operating with these numbers as before, we find the euc- cessive quotients to be 4, 7, 1, 3, 1, 16, 1, 1, 15 ; and from these the converging fractions are readily found. (8) Dividing the greater term b'- the less, the less by the remainder, and so on, the quotients are 1, 1, 2, 1, ], 1, 3, 2, 1, ], 2, 3 ; and the successive con- verging fractions found from these are {> l, |-i *i -^^t \^, |§, &c., whence -j-c,- is the required fraction. (9) In solving this example it is most convenient to consider 1 as the numerator, and 27.321661 the denominator, and then invert the resulting converging fractions. Dividing 27.321661 by 1, or 2732166] by 1000000, as in the pre- ceding examples, the quotients are 27, 3, 9, 5, 2, &c. ; these give for approximating fractions 2 7' 8 -i ' 7 II a ' 3 27 ^^^f, &c., hence the required ratios are ii H ' I 4 3 > ' (10) Referring to Art. 353, we have a=l, 1 hence J2^\- 2+- ^2+, &c. The integral fractions are ;',, :',, -',> ;,, &c., " converging fractions are 5, §> jri; -ofi, &,c. Adding 1 to each of these we have fi J> }?, -^,^, &o (11) Referring to Art. 353, wo have rt=2, and 2a=4, CONTINURD FRACTIONS. 231 hence ^4+1=24-- ^+i+, &c. The int. fractions are ^, |, {, | |, &c. " conv. fractions are |, j",, fj /tf^. tW^. &c. Adding 2 to each of these, we have -|, f|. Vs ' l§|. fill' &c. Now the fourth fraction being in an even place is less than the true value, and the fifth being in an odd p.ace is greater than the true value, therefore ^5 is greater than |g|, ana less than 2 88| (.12) Since 8'=8, and 8'=64, x lies between 1 and 2 hence let x=l-\--. y .-. 8'+7=32, or 8X8r=32, or, 8v= 3g2=4. or, 8=4", by raising both members to the y power , Now since 4'=4, and 4^=:16, the value of y lies between 1 and 2, hence let j/=l-|-_. z' 1 1 1 .-. 4'+i=8, or 4X4^=8, or 4«=2; raising both members to the z power, we have 2'=4, whence z=2. 1 3 3 ^ 2 ^13) 3'=25 ; 32=9, and 3'=27. .-. ar=2+l. x' 32+fcl5, or 3=X33=16, or 3x'='p==|. Since 3?=6, we have (-p'^'siS ; here x'=2-\---, 232 KEY TO PART SECOND •• (|)='+^=3,or (|)^X(|F=3, 1_ whence (2)z"='^2. or, mr"=l i here a;"=6-|- — . x'"' .-. ar=2+l hence the approximating fraction to be added to 2, i* h or T3 ; -r%=2.46+, which is true to within {j\y=j^-g. This method of finding the value of x is more curious thin useful, as the same thing may be accomplished directly, and wi'h but little labor, by means of logarithms. „, log. 15 1.1760913 „ .„- , Thus a;=— e = =2.465 nearly. log. 3 .4771213 ^ LOGAIilTHMS. Article 366. (1) The result in this example follows directly from Art. 360, the pupil, however, may prove the principle generally in the case of three factors ; thus, < =N (1), a" =N' (2), a'"=N" (3). Multiplying equations (1), (2), and (3) together, we have But, by the definition of logarithms, if we consider a the base of the system, then u:, x', and x" are the logarithms of N, N', and N", and {x-\-x'-\-x") is the logarithm of NN'N", bonce, the sum of the logarithms of three numbers is equal to the logarithm of their product. (2)- By Art. 361, log. ( ^'"^ ] = log. (afc)— log. {de) • but LOGARITHMS. 233 log. i^abc)= log. a-j- log. h-\- log. c ; and log. (&)= log. d -\- log. e ; hence log. ( — 1 = log. a-{- log. h-\- log. c \ de / — (log. d-{- log. e)^ log. a-|- log. b-\- log. c — log. d — log. e. (3) By Art. 360, log. (a™ . J" . (?>.)= log. a'"+ log. i»+ log. cp; but (Art. 362), log. a"'=m log. cf, log. i"=;i log. b, and log. c?'=p log. c. .'. log. (a™ . b" . dP^^m log. a-\-n log. Z)-|-p log. c. — ^ — ) = '°?- (""" • i") — log- C=m log. a-\- n log. J — ■p log. u. (5) a^ — a;^=(a-|-a:)(o — a:), and log. (a' — je^) = log- \{a-\-x){a—x)\= log. (a+a:)+ log. (a— x). (6) Since log. (a? — a;')=: log. (a4-^)+ log- (<2 — ^)) I log. (a' — ^•*^;=:3 log. (a+x)-j-i log. (a — ir), but, (Art. 363), }, log. (a=— a;2)= log. (o='-^i;=')^, or, log. Jo' — x'^ ; .-. log. ^a^ — a;'=.l log. {a-\-x)-{-y log. (a — a;). (7) a^X iJa^=a^Xa^='^ ■ ; and log. (a * ) = 'j log. a, or, 3| log. a. (8) ;^a' — x^ J a'-' — x^ Ka-]-x)(a — x) / a — x 7^T^4~V("^+^XM^~^'(^=Kr'' (a+x)= ^{a+xy = log. ( = log. (a — x) — 3 log. (a+x) ; log- /, ^, = log. (a— x)— log. (a+x)3, (a-j-x)' hence log. ^/ , or log. .1^ _ \/(ffl+x)' (3+a;)= =AJlog. (a— x)— 3 log. (a+x);; or, =1 log. (a — x) — I log (u+x") ; but the first form h the best. 20 234 KEY TO PART SEC ON I), Article 370. (l; Since 14=2X7 ■•■ log. 14= log. 2+ log. 7, = log. 3+ log. 5 ; = (log. 2)X4; = (log. 3)X2+ log. 2 ; = (log. 2)X2+log. 5; log. 3-1- log. 7 ; :(log. 2)X3+log. 3; :(log. 5)X2; :(log. 3)X3; :(log. 2)X2-(-log. 7; log. 3+ log. 10. ''2) The numbers will evidently be those that can be formed by multiplying together any two or more of the factors 2, 3, 5, 7, either of which may be taken more than once if necessary, thus, 2', 5X7, 3=X2^23X5, 2X3X7,3^X5, 2 m or, m. log. a.x-\-n. log. i.x=m. log. c ; , ?n. lotr. c whence a;=: m. log. a-\-n. log.. 6 nx n. log. c m m. log. a-\-n. log. i' 112) First, log. 2000= log. (1000X2)= log. 1000+ log. 2= 3+ log. 2, and 2'.3'=2000 ; log. (2'. 3")= log. 2'000=3+ log. 2 ; log. (2' 3')= log. 2^+ log. Z'=x. log. 2+z. log. 3 j .-. X. log. 2+2. log. 3=3-1- log- 2 ; and 3z==5x, or z= — ; hence x. log. 2-\- — log. 3:=3-(- log. 2 ; or, 3 log. 2.X+6 log. 3.a;=3(3+ log. 3) ; , 3(3+ log. 2) whence x^- ' = ^ ■ 3 log. 2+5 log. 3 end .-5^- 5(3+ log- 2) 3 8 log. 2+5 log. 3' 240 KEY TO PART SECOND. (13) Let a'^z, then a'^=z'^, and the equation becomes 22—22=8, or z^— 2z+l=9 ; whence 2=±3-|-l=4, or — 2. .-. o'=4, or — 2, hence x log. a= log. 4, or log. ( — 2), but the last is in;iu missible (Art, 369) ; also, 4=2^, and log. 4=2 log. 2 ; .-. X. log. a=2 log. 2, and a;=?i£^. log. a (14) Let 2^=z, then 22^=^^ and 224-2=12. Prom the equation z^-}-z=12, we find 2=:-l-3, the nega- tive value being omitted (Art. 369) ; .-. 2='=3, and x log. 2= log. 3 ; , log. 3* .477121 , KOHoe whence x^-P — = ^=1.58496. log. 2 .301030 (15) 2a'''+a2'==a6', divide each side by a'"' ; or, a"'— 2a2^=l, let a-^=z, then 2^—22=1 ; 22—22+1=2, and 2=,y2+l ; 2x log. a= log. (^"2+1), or xX2 log. a= log. (^2+1), whence a=: °" '''^"'i" \ 2 log. a (16) Let a'=z, then z-j— =J, or z^ — bi=z — 1, whence z or a'=-±},,Jb' — 4=i(6±V6' — 4) ; .-. X log. a.= log. i(i± V° — 4) ; ■ log. -},(bdzJb^—i) whence x^= ° -^ ^ '. log. a (17) Here a;!'=y'(l), and a;3=i/2(2). Extracting the j/ root of both members of eq. (I), and the cube root of both members of eq. (2), we have x=yv, and x=y^. EXPONENTIAL EaUATlONS. 241 « 1 J. ■. Vv=v', whence -=S, and x=^y ; •'• f y^^y" ' divide each member by y' ; lyi=h or yi=| ; cubing each side y^(|)'="g''=3g ; fl8J Here {a^—h^y^'-<)=(a—by'. Extracting the square root of both members, we have whence (x — 1) log. (a' — &')=a: log. (a — b) ; but log. («'—&')= log. [(a+i)(a!— 5)]= log. (a+J) + log- (i— fc)- .-. (x— 1)J log. (a+t)+ log. (a— J)f=x log. (a— J) ; or, X log. ((i+Z/)-t-a; log. (a— A) — log. (a-\-b) — log. (o— 6) =x log. (a — b) ; oniitting x log. (a — i) on each side, and transposing, X log. (a+t)= log. (a!+i)+ log. (a — b) ; whence 3;=]-|-- ' — log. (»+&)■ (191 ,(a^—2a^b^+¥y-'=zl(a^—b'>yi^<=(a^—b^)^>^^ _(a^—by . and (a-i)=-(fl+5)-»=^?=^'; (a'— &')^" _ (g— ;.)''' . (^2—^2)2 (a+by ' Extracting the square root of both members, we have (.a^—b''y__(a—'>y^ a?:Z62 a+b ' but (a2._j2)r^ J (^a+b)(a—b) ] ''=(a+hy(^a—by ; (n-\-by(a—by _ (a—by . (a+b)(a—b) a+b ' dividing both members by (a — 6)"^, and multiplying Oy 21 242 KEY TO PART SECOND. a-\-b, we have ^° ' ^ =1, or (a-\-by=Mi — b ; a — b whence x log. (a-|-i)=: log. (a — i), and ;c= '°g- ^"-^^ bg. (a+i)' (.20) Here xy=jf{l), and xf=y^(2). v p From (l)as=jr, and from (2)a;f=v; .•. Xx=xg , and "=?; or, 4=P; a; 5 •. xq =£, or a; ) =i- ; (f)A' (21) Let a;'— 41+5=2;, then 3^=1200, and z log. 3= log ^ whence z=l°g-li?°°=H:^«l=6.4536. log. 3 .477121 .-. a;2— 4a;+5=6.4536 ; I-— 4a;+4=5.4536 ; a;— 2=±2.33, a;=2±2. 33=4.33, oi —0.33. IHTEREST A H » A K N U I T I E S . Articles 384—391. (1) I+r=1.06 and log. 1.06 = .025306 .025306X100=/ log. (1+r) =2.530600 log. P= log. 1 =0.000000 log. A= log. (339.30) =2.530600 (a) This e.xample is similar to the preceding ; if we mult'oh .025306 by 1000, the product is 25.306000, which is th» INTEREST AN V ANNUITIES. 24:? log. of the amount, and as tho index is 25, the corres- ponding natural number will contain 2b-\-l=26 figures. (Art. 358.) (3) See Art. 386, Cor. 3. For 5 per cent. R=1.05 ; for 6 per cent. R=1.06 ; for 7 per cent. R=1.07; for 8 per cent., R=1.08. Tor 5 per cent., i=J^Mll. =:!H^0=14.206 yrs. j log. 1.05 .021189 for 6 per cent, <=_Mil_=:!E010=n.8956 yrs. ; log. 1.06 .025306 for 7 per cent., fc.i5fll_=l3£i^0= 10.2 447 yrs. ; log. 1.07 .029384 for 8 per cent., t= '"g' ^ ==-J£l^"=9.0064 yrs. log. 1.08 .033424 (.4) See Art. 386, Cor. 3. Here m=10, and R=1.05. . , log. 10 1.000000 .„ ,o .• t= — 2 =: =47.19 yrs. log. 1.05 .021189 (5) Let x= the sum, then (Art. 386), M=P.R', and P=a!.R' r whence ^=^^==1, ^ni .■ . ^=^1 P x.R' X M' (6) Let X, y, z, denote the three shares, then we shall have x+y+z=V ; also, X. R'"=y.R''=2:.R'', are the equations of condition ; whence y==&'''''x, and z=K'-~'x ; .-. x-\-R''-''x-\-'R''-'x=P ; P whence x= l+R°-'+R«- p Similarly, y= _ _, and l+R»-"+R'-' 14-R'^-»-(-R«-»" (7) If we take out the logarithm of 1.06, and multiply it bjr 20, and take out the number con-esponding thereto, we shall have 3.20713546. Subtracting 1 from this, and di- viding by .06, the quotient is 36.785591, which is thu amount of an annuity of $1, for 20 years, at 6 per cent, Then multiplying this by 120 the product is $4414.27. 244 KEY TO PART SECONFl. In finding tlie SO" power of 1.06 the tables of logarithms in common use give a result too small. The learner may satisfy himself of this by actually involving 1.06 to the 20"' power. The ■brmula for solving the question is found in Art. 390. (8) See the Formula, Art. 391. R'=(1.05)'°=4.321942, and i.=.23137746 ; 1— 1=.76862254 ; R' R' J_ ( 1—1 l =J_(.76862254)=15. 37245, the present R— iV R'/ .05 value of an annuity of $1, to be paid for 30 years : 15.37245X250=3843.11+. (9) Here jo=— ( 1— i ) (Art. 392) ; R=1.05, n=10, and ?R" V R' ' «=20; R'=(1.04)2»=2. 191123 ; i=.=.45638697 : 1— 1=.54361303 ; R' R' R"=(1.04)"'=1.480244, •■R"=.05920976, and Ji- ( 1—— ) =9.181138 ; J R" \ R' / 9.181138X1 12.50=$1032.87-|-. 'JC) The amount of o$ at compound interest for n years, r being the rate per cent., is a(l+r)". The amount of an annuity of h$, for the same period, at the same rate, is i (^+'")"~^ (Arts. 386 and 390). r .-. ?, ^'+'-)"-^ =«(l+r)"; )• or, i(l+)-)"— i=ra(l4-r)» ; or, i(l+?-)"— ra(l+r)"=t ; or, (J— ra)(l-i-r)"=6 ; or, log. (b — ra)-\-n log. (1-f r)= log. b ; n log. (!+)•)= log- ^ — log- (* — ''"J ! log. b — log. (6 — ra) " log. (H-rt • This formula also solves the following problem : " What sum must be paid annually to sink a given debt in a certain numboi of years, tho interest on said debt being payable annually." aBNERAI, ThEORV OF EQUATIONS. 245 GENERAL THEORY OF EQUATIONS. Article 396. Note. — Ai.houjrh tlie Synthetic Mctliod oj Dimsiiin is not explaiiied lill Arlicio 4U'J, page 35C, yet we shall employ it, instead of the cominoa method, on account of its conciseness. The Teacher who prefers lo use the Syntlietic method, can require liis pupils to study Art. 409, before commencing the theory of equations. (1) 1—11+23+35 1—1, since i+l is the divisor, — 1+ 1 2—35 1—12+3.5+0. Atis. x2— 12a:+35=0. (2) 1 — 9+26—24 1+3, since x—3 is the divisor, +3—18+24 1_6+ 8+0 . . a:'— 6j:+8=0 ; whence (Art. 231), a;=4 or 2. (3) 1±0— 7+6 |-t-2, since x — 2 is the divisor, +2 +4—6 1+2—3+0. .-. a;2+2x— 3=0, and (Art. 231), x=l, or —3. (4) 1+2—41—42+360 |+3, since x—3 is a divisor, +3+15-78-360 1+5 — 26 — 120+0 I — 4, since x-\-4 is a divisor, —4— 4+120 1+1—30+0. .-. a:=+a;— 30=0, and (Art. 231), x=5, or ~6. (5) 1 — 3 — 5+9 — 2 1+1, since x — 1 is a divisor. -2, since x+2 is a divisor. + 1—3—7+2 1—2—7+2+0 —2+8—2 1—4+1+0. .-, k'— 4a;+l=0 and (Art 231), .t=2+7'3, or 2— ^3, 240 KEY TO PART SECOKD. Article 398. (•2) x=2 .■. X— 2=0, x=3 .-. X— 3=0, !=— 5 .-. x-f5=0, . . (x— 2)(x— 3)(a;4-5)=x'— 19r+30=O. (3) x= 3 .-. X — 3=0, x=— 2 .-. x+2=0, x= 7 .-. X— 7=0. ... (x— 3)(x+2)(x— 7)=x'— 8x'+x+42=0. (4) x= .-. X— 0=0, .x=—l .-. x+l=0, x= 2 . . X— 2=0, x=— 5 .-. x+5=0. ... (x_o)(x+l)(x— 2)(x+5)=ar'+4x'— 7x=— 10x=0 (5) a:=— 2 .-. x+2=0, x=+4 .-. X— 4=0, x=+4 .-. X— 4=0. .-. (x+2)(x—4)(x—4)=:x=— 6x2+32=0 (6) x=l+^l ■■■ X— 1— ^"3=0, . x=l— Vl . . X— 1 + J'3=0. .-. (x— 1— V^)(x— l+^/3)=x'— 2x— 2=0. (7) x=l +^"2 .-. X— 1— ^"2=0, x=l— ^2 .-. X— 14-V2'=0, X=2+J3 .-. X— 2— ^¥=0, x=2— V3 .-. X— 2+^3=0. .-. (x-l-j2)(x— l+V2)(x-2— V3)(x-2+V3) =(x2_2x— l)(x'— 4x+l)=x''— fit'+Sx=+2x— 1=0. (8) It has been shown, see Art. 398, that the coefficient of the fourth term is equal to the sum of the products of all the roots talcen three and three with their signs changed. The roots with their signs changed are +l!, +1, — 1, — 3, GENERAL THEORY OF EQUATIONS. 247 — 4, and the sum of their products taken three and three is (2X1X— l)+(2XlX-3)+(2XlX— 4) +(2X— IX— 3)+(2X— IX— 4)+(2X— 3X— 4) +(1X— IX— 3)+(lX— IX— 4)+(lX— 3X— 4) +(— IX— 3X— 4)=— 2— 6— 8+6+8+24+3+4+12 — 12=29 ; and since x^ appears in the P' term, 29a:^ ia the fourth term. (9) It is evidept there will be 7 terms in the required equa- tion, hence, the middle term will be the 4'*, hence it is re- quired to find the sum of the products of the numbers 6, 3, 1, — 1, — 2, — 4, taken three and three with their signs changed. But the shortest method is to find the product of the several binomial factors, thus, (x— 5)(x— 3)(i— l)(a:+l)(a;+2)(a;+4) =(x2— 8i;+15)(x2+6a;+8)(a;2— l)=a;«— 2x5— 26x<+28a;' +145i:' — 26a; — 120, where the middle term is 28a:'. (10) We may take any two numbers as the other two roots, but the equation will be of the simplest form if we sup- pose the roots to be +.^2, and — ^ — 3, since this assumption will cause the middle terms of the binomial factors to cancel each other. Thus, (i -^"2)(x+V'2)(a:+^'=3)(x— ^■=3)=(j;2— 2)(a;'+3) =x''+a;=— 6=0. Article 400. (1) Since x' — 2a; — 24^0 is the same as x^-]-2x — 24=0, ex- cept that the sign of the 2"^ term is changed, (the fourth being wanting,) and since the roots of the latter are +4, . and — 6, therefore (Art. 400), the roots of the former are —4 and +6. (2) Since x^-\-3x^ — lOx — 24=0, is the same as x^ — 3a;' — 10a; +24=0, with the signs of the alternate terms changed, and since the roots of the latter are 2, — 3 and 4. there- fore (Art. 400), the roots of the former are — 2. +3, — 4. Article 401. (1) Dividing the given equation by x-\-S^x — ( — 6)1 the quo- tient is x^ — 6a;+10^0, of which the roots are 3+^ — 1 and 3—J^l. 248 KEV TO PART SE(, OND. (2) Dividing the given equation by x — ( — 4)=a,-|-4, the quo tient is x' — 4a;-l-l=0, of which the roots are 2-\-Ji, and 2— V3 (3) Since x=Z-}-J%, and a;=3 — ^2, therefore {x—^—j2){x—Z+j2)=x^ — 6a;+7; dividing the given equation by this, the quotient is x^-{-2x — 2^0, of which the roots are — lrt^3, (.4) Since one root is 2 — JZ, therefore (Art. 401, Cor. 1,) 2-|-«y3 is another root, (x—2-\-^3)(x—2—^'3)=x^—4x-\-l, and dividing '.s given equation by this, the quotient is x — 3 ; he..''.e X — 3=0, and x=-)-3. (5) Since — ;',(3+V — 31) is one root of the given equation, therefore (Art. 401), — 1,(3— ^/— 31) is another root ; [a;+.J(3+V"=3'l)][a;+K3-V=3T)]=a;=+x(3) + 1[9— (— 31)] = a;2+3a;+10, and dividing the gi"«n equation by this, the quotient K x^ — 3a; — 4 ; hence x'^ — "ix — 4=0, and x=\, or — 1. (6) Since -|-^/2 is one root of the- given equation, therefore (Art. 410, Cor. 2), — ^2 is another root, and three of the binomial factors of the given equation are (a;— J2)(ar+V2)(x— 3)=a:3— ar3— 2a;4-6. Dividing the given equation by this, the quotient is x" —Iv 4-10 ; hence ac' — 7x-|-10=0 ; from which x=2, and 6. Article 403. Ex, 2. If we substitute 6 for x in tha equation x' — 5*' — x ^-1;=0, we have — 4:=0, and if we substitute 6 for x, we have "1-31=0, and since the results have contrary signs, one root lies aa' woen 5 and 6, that is, 5 is the first figure of one of the roots TEANSFORMAT[ON OF EQUATIONS. '249 T B A N S F R M AT I N OF EQUATIONS. Article 405. (1) Here x''-\-'7x'—ix+i=0 ; let x=J-, then .1 +1^."— 1^4-3=0, multiply by 81, to clear ot fra"- 81 9 3 ^^ ^ tions, ),'''-)-63/— 108^+243=0. (2) Here x^+2x^—lx—l=0; ,et a:=?, then 6 ^—-{-J^ — jy. — 1^0, and clearinc; of fractiona 625 126 5 j/<+10j/5—S75y— 625=0. (3) Here x'— 3a;'+4i-+10=0 ; let a;=2y, then Sy— 12j/=+8y+]0=0, or, dividing b^r 2, 4y3_ 6y=i+4y4- 5=0. (4) Here a;3-|-18x2+99x4-81=0 ; let x=Zy, and change tha signs of the alternate terms (Art. 400), then 27j/'— 18x9y2+99X3y— 81=0, or dividing by 27, y_6y2+ily_3=0. (5) Here x^ — ix'^-\-\x—W=Q ; let a;=|y, then i- — ^-|-.^y — 10=0 ; clearing of fractions, j,3__6^2_(_3^_27o=o. Articles 406 — 407. ()) Here a;' — 7a;+7=0 ; let y—x — 1, then x—-y-\-\. •, (i/+l)'— 7(y+l)+7=0, or fJ^Zy^-iyJ^\=0. (2) Here x''— 3x'— I5x2-f-49x— 12=0, let y=x— 3, .hen a;=y+3 ; .-. (y+3)^-3C2/+3)'— 15(y+3)2+490/+3) — 12=0, or, by developing and reducing, y''-|-9j/'+12j/=— 14y=0. (3) Here x^—Gx'^+Sx — 2=0, and x=y-\-2. .-. (y+2)'— 6(y-)-2)2-|-8(j/+2;— 2=0, or, reducing, y'— 4v— 2=0. 350 KEY TO PART SECOND (4) sc'^-{-2px—q=0. Here A=+2p, n=2. .•. ?-= — j> ; hence x=y — p, substituting ; (y — J)y-\-2p(y — p) — 9=0, or, reducing, y^—p^—g=0. Articles 408-410. (4") Let y=x — 3, then it is required to divide the given eqiia tion, and the successive quotients, by x — 3. 1 ±0 —27 —36 (-1-3, since the divisor is a;— 3. -1-3 -)- 9 —54 +3 —18 —90 .-. —90= 1" R +3 +18 -f 6 -f- .-. 0= 2"'' R. +3 -\-0 .-. -1-9= 3"' R. Ans. y'-\-9y^—90=0. \5) 1 ±0 —27 —14 -1-120 (-1-3, since the divisor is a;— 3, -1-3-1-9 —54 —204 +3 —18 —68 — 84 .-. —84= 1" R. +3 -1-18 — +6 +~0 —68 .-. —68= 2"'' R. +3 +27 +9 +27 . . +27= 3"* R. +3 +12 .-. +12= 4'*R. Ans. y'+12y'+27j/2— 68!/— 84=0. (6) ] —18 — 32 + 17 + 9 (+5, since the divisoi + 5 —65 —485 —2340 is i— 5, —2331=: 1" R. —13 — 97 — 468 —2331 + 5 — 40 — 6S5 — 8 —137 —1153 . •. — 11£ + 5 — 15 — 3 —152 .-. -152= 3"'R. ±-1 + 2 .-. +2= 4"' R. Ans. »/''+2^'— 152^5— 1153;/— 2331=0. TRANSFORMATION OP EOUATIONS. 251 (^) We sliall first diminish tin; roots by 1, and tlien by .2, in- dicatiiig the remainders after each transformation by stars. J —6 +7.4 + 7.92 —17.872 — .79232 (+1.2 +1 —5. + 2.4 + 10.32 +10.32 — 7.552 —7.552 —5 +2.4 —8.34432* +1 —4. — 1.6 + 8.72 + 1.168* + .34432 — i —1.6 + 8.72 —8* +1 —3. — 4.6 + .5536 + 1.7216 —3 —4.6 + 4.12* +1 —2 —6.6* — 1.352 + .2784 + 2.0000* —2 + 2.768 +1 — .16 — 1.376 — 1* —6.76 + 1.392 +.2 — .12 — 1.392 —.8 —6.88 + 0.000* +.2 — .08 —.6 —6.96 4-2 — .04 —.4 —7.* . +.2 —.2 +.2 0* Ans. j,5_7j,3+23/ —8=0. t) Here A= — 6, n=3, .: r=2 ; hence x=y-\-2, or y=x — 2. 1 — 6 +7 — 2 (+2, since the divisor is x — 3. +2 —8 —2 '^ ^T ^^^.•. — 4= 1" R. +2 —4 ^^ ^ .: —5= 2"'' R. +2 .•. 0= 3"* R. Ans. y^ — 5y — 4=0. (3) Here A=— 6, n=3, .-. r=2 ; hence x=y-\-2, or y=-x — 2. Z52 KI3Y TO PART SECOND. I —6 ±0 4- 5 (+2, since the divieor is X—-2 +2 —4 —8 —16 _8 —11 .-. —11= 1"R. +2 —2 —4 — 12 . . —12= 2"^ R. +2 .■ . 0=3"'R. Ans. y'— 12w— 11=0. (,10) Here A=— 6, n=3, .-. r=2 ; hence x=y-\-2, or y=x — 2. 1 — 6 +12 +19 (+2, since the divisor is c— 3 +2—8 +8 ^^ +4 +27 .-. +27— 1" R. +2 —4 —2 ~~0". . 0= 2"'' R. +2 .-. 0= 3"* R. Ans. y'+27=0. (11) Dividing each term by 3, the given equation becomes Here A=5, n=3, .: r=— | ; hence x=y — f , or y=a;+|. 1 +5 +V — 1 ( — i> since the divisor is a;+|. 1 2 5 — 1 -%' 5 +¥ 1 52 , 1 S:2 irt 1} +1 .-. 0= a""" R. 5 — 5 ~0^ .-. 0= 3"' R. Ans. f—\^^=0. or 273/'— ]5S!=0 (13^ Here A=— 6, B=+9, 71=3, -in(n— l)r2+(n— l)Ar+B=?(2)r2+(2)X— 6r+9=0, or, r' — 4r+3=0, whence r=:3, or 1. hence a;=y+3, or y+1, and i/==x — 3, or x — 1. —6 +9 —20 +3 —9 +0 —3 " —20 +3 +3 +3 Ans. 2/'+3i/' ~6 +9 —20 +1 —5 —5 +4 -1-4 —16 +1 — 4 —4 + 1 —3 Ares, y' — 3; E Q [J A L Ji II () I S . *•'■ (-j-3, since the divisor is x — 3. or, 1 ~6 -(-9 — 20 (4-1, since thi) divisor is X—. (12) Here A=— 4, B=5, k=3. ^ra(?i— l)7-2-l-(7j— l)Ar-|-B=3r'— 8r-l-5=0, and r=^ or 1 hence x==y-\-%, or y-f-l, and ^/=a; — :J, or x — 1. -2 (-t-'i) since tiie divisor is x — f. — ¥ +M. or 1 —4 +5 —2 (+1. \ +1+3+2 —3 -1-2 +1 +2_ —2 +1 —1 Ans. y' — y'=0. EQUAL KOOTS. Article 4 14i (3) Here x'— 2a;'— 15j;+36=0, 3a;' — 4x — 15= P' derived polynomial, and the greatest common divisor of this and the given equation (Art. 108,) is J— 3; hence x — 3=0, and a:=-[-3, therelbre +3 and -j-? are two roots of the given equation. —4 +5 +1 35 — "g _i 1 + ^ +i 2 — S 1 — 15 +1 + 1 ,. ,f^f- -4-,=^. 254 KEY TO PART SFCOND. Dividing the giveji equation by (a; — 3)(ar — 3) the quotient is ii'-|-4, hence x+4=0, and a= — 4. (3) Here x^— Qx'-{-ix-\-\2=0, 4x^ — lSx+4= 1" derived polynomial, and the great- est common divisor of this and the given equation is x — 2; hence x=-\-2, and -\-2. Dividing the given equation by (x — 2){x — 2) the quotient is i'-|-4x — 3 ; hence a;'-'-|-4x — 3 = ; from which we find x= — ^1, and — 3. (4) Here a;^—6x3-[-12x2— 10x4-3=0, 4x' — 18x'4-24x — 10= 1'' derived polynomial, and the greatest common divisor of this and the given equation is x^ — 2x+l ; but x= — 2x-|-l=(x — 1)^, therefore the given equation has three roots, each equal to 1. Dividing the given equation by (x — l)(x — ^l)(x — 1) the quotient b X — 3, hence x — 3=0, and x^3. The operation of dividing by x — 1 should be performed by syn- thetic division on account of its brevity, thus, 1 —6 +12 —10 +3 (+1 +1 — 5 + 7 —3 —5 + 7 — 3 + 1 —4 — 4 + 3 + 3 +1 —3 — 3 0. Q,uc itient (5) Here x"— 7x'+9x2-|-27x— 54=0, 4x3— 21x=+18x+27= thg flrst derived polynomial, and the greatest common divisor of this and the given equa- tion is x' — 6x+9 ; but x^ — 6x+9=(x — 3)-, therefore the equation has three roots, each equal to 3. Dividing the given equation by (x — 3)(x — 3)(x — 3), the quo- tient is x4-2, hence x-|-2^0, and x= — 2. (6) Here xi+2x'— 3x'— 4x+4=0, 4x'-|-6x' — 6x — 4= 1" derived polynomial, and the greatest common divisor of this and the given equation ia a;^+^^2=(x+2)(x — 1) ; therefore the equation contains two factors of the form x+2, and of the form x — 1, hence the four roots are — 2, — 2, -|-1, -(-1. K Q O A L. ROOTS. 255 If the learner does not readily see that x^-f-*-' — 2=(x+2)(j: — 1), et liim place it equal to zero and find the roots. (7) Here a:<— 12x'+50x2— 84x+49=0, 4x' — 36x2-l-100x— 84= P' derived polynomial, and the greatest common divisof of this and the given equa- tion is x^ — 6x-|-7. Placing tliis equal to zero, vce find its roots are 3-|-^3, and 3—^2", that is, x^—6x-\-'7={x—S—^'2)(x—3+J2), hence the four binomial factors of the given equation are (a>— 3— J2)(x— 3— 72)(x— 3+ J2)(x— 3+^2), and the four roots are 3+^2, 3+j2, 3—^2, i—Ji. (8) Here x^— 2x^+3x3— 7x2-l-8x— 3=0, 5X'' — 8x'-|-9x^ — 14x-|-8= 1'' derived polynmoidl, and the greatest common divisor of this and the given equii- tion is x' — 2x-|-l=(x — 1)^, therefore the eq■^dtion has tliree roots, each equal to -|-1. Dividing the given equation by (x — l)(x — l)(x — 1) the quotient is x'^-\-x-\-Z, thus, 1 —2 +3 —7 +8 —3 [+1 +1 —1 4-2 —5 +3 "" — 1 +2 _5 +3 0= 1" R. +1 +0 +2 +2 —3 —3 0= 2»'' R. +1 +1 +1 +3 +3 0= 3'-'' R. .-. x2+x+3=0, from which x=— '±1.^—11. As learners sometimes experience difficulty in finding the great- est common divisor in this example, we will here give the opera- tion. Multiplying the given equation by 5, the operation is ai follows : 5x=— lOx^+lSx'— 35x2-l-40x— 15 j Sx"— 8x^+9x^—1 4x+8 5x'— Sx'+ 9x'— 14x2+ 8x \x—2 — 2x*-|- 6x3— 21x2+ 32x— 15 — 10x<+30x3— 105x2+l60x— 75 Multiply the first — 10x^ +16x''— 18x2+ 38x— 16 divisor by 14. +14x'— 87x2+132x— 59 256 K K Y TO PART SECOND. 70a;<— 112x'+126j;'— 196X+I12 1 1 4x^—81x^+ 1 32x— 5 9 70x<— 43oj;'+660j;2— 295a; |oj;+32iS +323x'— 53U2+ 99^4- 1 1 2 X by 14, 4522j;3— 7476x^4- 1386a;+ 1568 4522x3— 2810 lx2+42636a:— 19057 20625x2— 412o0x+20625, or, 20625(x2— 2x+l) x2 — 2x+l will be found to divide 14x'—87x'+ 132.x— 59, and it is therefore the gi-eatest common divisor required! (9) Here x«-|-3x=— Bx" — Bx^-f-gx^+Sx- 4=0. 6x*4-15x^— 24x3— i8x2-|-18x+3= p- derived polynomial, 30x1+60x3— 72x2— 36x-f-18 = 2"'' derived polynomial. We find the greatest common divisor of the given equation and the first derived polynomial is x' — x' — x-j-l : if we put this equal to zero, it is easily seen that x=], then dividing by x — 1, the quotient is x^ — 1, of which the factors are x-|-l and x — 1, hence x'- x' — x+l=(x— l)(x— ])(x4-l)=(x — l)=(x+l), therefore the given equation contains x — 1 as a factor three times, and x-\-l as a factor twice ■ hence three roots of the equation are -\-l, -\-l, -|-1, and two roots — 1, — 1. Dividing 'the given equation hy (x — iy{x-\-\y, the quotient is x+4 ; hence x-|-4=0, and x= — 4. Otherwise thus ; After finding the greatest common divisor of the given equa- tion, and its first derived polynomial, we may proceed to find the greatest common divisor of the 1" and 2'"' derived polynomials, which is X — 1 ; hence, since the 2'"' derived polynomial contains X — 1 as a factor once, the 1" derived polynomial must contain it as a factor twice, and the given equation thi-ee times. Also, by dividing x' — x' — x+l by (x — 1)^, the quotient is x-j-l, which is therefore contained twice as a factor in the given equa- tion. The operation of finding the greatest common divisor of the 1'' and S'"* derived polynomials is quite tedious, but it enables us to determine that x — 1 is a factor of x^ — x- — x-|-) without wiving an equation of the third degree, tlie method of doing vhich has not yet been explained. T u E o R !■; nt o F s T i; it m . SSI THEOREM OF STHUM. Articles 420—427. (3) Here X = x'—2x'—a:+2, and (Art. til) Multiplying X by 3, to render the first teim divisible by the first term of X,, and proceeding according to Art. 108, we luive for a remainder — l4.r-{-16. Canceling tlie factor -f2, and changing the signs (Art. 420), we have X2=+7x— 8. Multiply- ing X, by 7 to render the first term divisible by the first term of Xj, and proceeding as before, the remainder is — 81 ; hence X2=-[-81, and the series of functions is X = x^—2x^—x+2 X,=3x^—4x—\ X^=7x—8 X3=+81 X, X, Xj Xg For x^ — 00 the signs are — -|- — -|-, 3 var. .•. k =3 a;=+co the signs are + + + -f-, o var. . . k'=0. ■ ■• k — k'=2 — 1=1, the number of real roots. By substituting the whole numbers from — 2 to +3, we find the roots are — 1, -)-l, and -|-2. (4) Here X = 8j;'— 36.^2+46^;- 15, and (Art. 411), X,=24x=— 72X-I-46, (or ISx^— 36.r+23). Multiplying X by 3, and dividing by Xj, the first remainder is — 36x^+92^—45 ; multiplying this by 2, and continuing the division, the remainder is — 32a;-l-48=16( — 2x-\-3), hence X =2,r — 3. Dividing X, by X^ the remainder is — 8, hence X3= -\-8, and the series of functions is X = 8a;'— 36x2+46a;— 15 X,=24a;2— 72a;+46 X2= 2a;— 3 X3=+8. For x= — 00 the signs are 1 1-, 3 var. .•. k =3, x=:-\- 00 the signs are -|- -]- _|- -|-, var. .•. k'=0 .•. k — k'=3 — 0=3, the number of real roots. 22 '58 KEY TO PART SECOND. By substituting the wliole numbers, from to 3, we find thai one variation is lost in passing from to 1, one from 1 to ii, and one from 2 to 3. (5) Here X = x'—3x^—4x-\-n, and (Art. 411;), X,=3x2_6a;— 4. Multiplying X by 3, and dividing by X,, the remainder is — ]4a;+29, hence Xs, = 14a; — 29. Multiplying X, by 14, and dividing by Xj, the first remainder is -|-3x — 56, multiplying thig by 14, and continuing the division, the remainder is — 697, hence .5Tg=-j-697, and the series of functions is X = x'—3x^—4x-\-U X,= 3x2— 6x— 4 X2=14a;— 29 X3=+697. For x^ — OD the signs are ] \-, 3 var. .•. K =3, ar=-|- OD the signs are -| — | — | — [-, var. .-. k'=0. .-. k — k'=3— 0=3, the number of real roots. By substituting the vs^hole numbers from — 2 to -]-4, we find tl at one variation is lost in passing from — 2 to — 1, one from -f 1, to -\-2, and one from -f-3 to -\-4. (6) Here X=x^—2x—6, and X,=3a;2— 2. Multiplying X by 3, and dividing by X,, the remainder fs ■ — 4x — 15, hence X2=4x-|-15. Multiplying X, by 4, and divid- ing by Xg, the first remainder is — 45a; — 8 ; multiplying this by 4, and continuing the division, the remainder is +643, he.ice ''j= — 643, and the series of functions is X = x'— 2a;— 6 X,=3x'— 2 X2=4x-1-15 X3=— 643. For x^= — OD the signs are 1 , 2 var. .•. k =2, x^+ QD the signs are + + H , 1 var. .-. k'=l. .•. k — k'=2 — 1 = 1, the number of real roots. We also find that one variation is Inst in passing from 2 to S therufore the root lies between 2 and 3. TrrEOREM OF STURM. 259 (7) Here X=a;3— 15a;— 22, and X,=3a;=— 15, or x^—5. , Dividing X by X,^.r' — 5, the remainder is — Ita — 22=2 ( — 5x— U), hence X2=+5x-|-ll. Multiplying X, by 5, and dividing by X^, the first remainder is — 11a; — 25; multiplying this by 5, and continuing the division, the remainder is — 4, hence X^ — | 4, and the series of functions is X — a;'— 15a;— 22 X,= x^—6 Xj=5a;+ll X3=+4. For x= — OD the signs are ] 1-, 3 var. . . k :=3, as=-|-ao the signs are + + + +>0 var. .. k'^0. .'. k~k'^3 — 0=3, the number of real roots. By substituting the whole numbers from — 3 to -|-5, we find that two variations are lost from — 3 to — 2, and one from -|-4 to 4-5 ; we also find that — 2 is a root. For a:= — 2.', there are three variations, and for x= — 2| there are two variations, hence one root lies between' — 2j and — 2^. (8) Here X = x'>-\-x^—x^—2x-]-4, and (Art. 411), X,=4a;3+3a:'— 2:i;— 2. Multiplying X by 4, and dividing by X,, the first remainder is ^;' — 2a;- — 6a;-|-16, multiplying this by 4, and continuing the divis- ion, the remainder is — llx^ — 22a;+66 = ll( — x^ — 2a;-|-6), hence X2=a;'-|-2a; — 6. Dividing X, by Xj the remainder is -(-32a; — 32=32(a;— 1), hence X3=— a;+l. Dividing X^ by Xg, the re'iainder is — 3, hence Xj=-f-3, therefore the series of func- Ijpis is X = x^-{- a;' — x' — 2a;-j-4 X , =4a;'-|-3x=— 2a;— 2 Xj=x2-[-2x— 6 X,=-x+l X,=+3. For x^ — 00 the signs are -| 1--]--|-, 2 var. .•. k =2. a;=-|- OD the signs are -}--|--| \-,2 var. .-. k'=2. .•. k— k'=2 — 2=0 ; hence there are no real roots. (9) Here X=x''— 4a;'— 3x-l-23, and X,=4a'— ISa;'— 3. Multiplying X by 4, and dividing by X,, the remainder is yCO K r, y TO PART SECOND. — 12a;5— 9r+89, iieuce X^=-\--i2x^-\-9x—89. Multiplying X, by 3, and dividing by X^, the first remainder is — 45£^-|-8to — 9, multiplying this by 4, and continuing the division, the remainder is 4-491X— 1371, hence X3= — 491;r+I371. Multiplying X^ by 491, and dividing by Xg, the first remainder is 20871x — 43699, multiplying this by 491, and continuing the division, the remain- der is -|-7157932, hence X^= — 7157932, and the series of funo- lions is X = x" — 4^3— 3;t;+23 Xj=4x'— 12^:2—3 X2 = I2a;'4-9x— 89 X3=— 49la;+1371 X,=— 7157932. For x= — 00 the signs are -] \- -\ , 3 var. .-. k=3, :c=-l- 00 the signs are -\ — | — | , 1 var. .". k'^l, .". k — k'=3 — 1=2, the number of real roots. By substituting the whole numbers from 1 to 4, we find that one variation is lost in passing from 2 to 3, and one from 3 to 4. (10) Here X=a;''— 2a;3— 7x2+10a:+10, and Xj=4a;»— 6x' — 14^-(-10, or 2x^ — 3x^ — 7x+5. Multiplying X by 2, and dividing by X,, the first remainder is — x^ — 7x^-|-15a;+20, multiplying this by 2, and continuing the division, the remainder is — 17x--|-23x-l-45, hence Xj:=17j;-' — 23x — 45. Multiplying X, by 17, and dividing by X^, the first remainder is — 5x^ — 29x+85 ; multiplying this by 17, and continuing the division, the remainder is — 608j:+ 1220=4 (—152x4-305), hence X3=162x— 305. Multiplying Xj by 152, and dividing by Xj, the first remainder is 1689x — 6840, multiply- ing this by 152, and continuing the division, the remainder is — 524535, hence Xj=:+524535, and the series of functions is X = x''—2x'— 7x2-1-10x4-10 X,= 2x5— 3x3— 7X-1-5 X2=17x2— 23a>— 45 X3=I52x— 305 X^=4-524535. For x^ — 00 the signs are -| 1 \-, 4 var. .■. k =4, x=-l-ao the signs are -j- -j- -f- -J- +> var. .-. k'=0. .'. k — k'=4 — 0=4, the number of real roots. We also find that one variation is lost in passing from — 3 to — 2, one in passing from — 1 to 0, and two in passing from 4-2 to 4-3. R A T 1 r) N A I, R O T S . 201 (11; Here X=xs— 10a;'+6x-!-l, and X,=5x''— -SC-c^-l-e. Multiplying X by 5, and dividing by X,, the remainder ia — 20a;'4-2-1^+5, hence X2=20a"'— 24x— 5. Multiplying X, by i, and dividing by X^, the remainder is — 96i'-|-5x-|-24, hejice X^=9Gx'^ — ox — 24. Multiplying Xj by 24, and dividing by X,, the first remainder, is 25a;' — 456a; — 120; multiplying this by 96, and continuing An division, the remainder is — 43tiola; — 10920, hence X,=4365la;+10920. Multiplying Xj by 43651, and di- viding by X^, the first remainder is — 1266575x — 1047624 ; mul- tiplying this by 43651, and continuing the division, the remainder > — 1372624203024, hence Xj=+1372624203024. It is not necessary, however, to obtain any thing more than the sign of the 'ast function. The series of functions is X = x-"^— 10a:'-f6a;4-l X,= Sa;"— 30a;2-f6 X2=20x'— 24a;— 5 X3=96x'-'— 5a;— 24 X5=43651a;+10920 x,=+. For x= — QD the signs are 1 j 1-, 5 var. .•. k =5, " x=-\- OD the signs are + + + + + +>" var. .-. k'=0. .-. k — k'^5 — 0=5, the number of real roots. By substituting the whole numbers from — 4 to -{-4, we find tnat one variation is lost in passing from — 4 to — 3, two in pass- ing from — 1 to 0, one in passing from to 1, and one in passing from 3 to 4. RESOLUTION OF NUMERICAL EQUATIONS. RATIONAL BOOTS. Article 429. '2) Here a;' — 7a;-'-l-36=0. -\-l and — 1 are not roots. Limit of positive roots =1-|-7=S. . Changing the signs of the alternate terms C.Art. 418), the equation becomes a;^-|-7a;-±0a; — 36^0. 262 K E V TO PART SECOND. .■. limit of negative roots = — (l-|-^36), or — 6. Last term -|-36 Divisors . . . +6, +4, + 3, + 2, — 2, — 3, —4 Quotients . . +6, +9, +13, +18, —18, —12, —9 Add . . . . +6, +9, +12, +18, —18, —12, —9 Quotients . . +1, * + 4, + 9, +. 9, + 4, * Add — 7 . . . —6, — 3, + 2, + 2, ~ 3. Quotients . . — 1, — 1, + 1, — 1. + 1 Add +1 . . . 0, 0, + 2, 0, + 2. Hence tlie roots are +6, +3, and — 2. (3) Here a;' — e^^+llo; — 6=0, and +1 is found to be a loot, Limit of positive roots =1+6^7. Limit of negative roots =0, since when the signs of the alter- nate terms are changed, all the terms are positive, therefore, thia equation has no positive root, and therefore the given equation has no negative root (Art. 402). Last term — 6. Divisors . Quotients Add +11 Quotients Add —6 . Quotients Add+1 . + e, +3, +2 — 1, —2, —3 + 10, +9, +8 *, +3, +4 —3, —2 —1, —1 0, 0. Hence the roots are +3, +2, and 1. (4) Here a?'+a;' — 4a; — 4--=0,, and — 1 is found to be a root. Limit of positive roots l+,^4=3. Limit of negative roots — (1+4)^= — 5. Last terra — 4. Divisors +2, —'2, — 4 Tli«refore the root* Quotients —2, +2, +1 are +2, —2, — 1. Add —4 —6, —2, —3 Quotients — 3, +1, * Add+1 —2, +2 Quotients — 1, — 1 Add+1 0, 0. KATIONAL ROOTS, 263 (5) Here x' — 3x' — 46.x'— ''2=0, and -|-1 and —1 are not roots. Limit of positive roots 72, of negative roots — {i-\-iJ i^)i or —8. Last term — 72 Divisors, +72, +36, +2-1, +18, + 12, + 9, + 8, + 6, + 4, + 3, + 3 - 2, — 3, — 4, — 6, — Quotients, 8. — 1. — 2, — 3, — 4, — 6, — 8, — 9, —12, —18, —24, — 3fa, +36, +24, +18, +12,+ 9. Add —46, -47, —48, -49, —50, — 52, —54, — 55, —68, —64, —70, —82, —10, —22, —28, —34, — Quotients, 37. ", *, *, *, *. 6, *, *,— 16, *, —41 + 5. *, + 7. *, Add —3, *, *. + 2, + 4, Quotients, — 9, —19, —44. — 1. — 1, Add+1, — 1, *, —22 0, 0, .•. the roots are — 0, ■2, -4, +9. —21, (6) Here x^ — 5j;' — 18a;+72=0, and +1 and — 1 are no; -oots Limit of positive roots 1+18=19, of negative roots -72 Last term +72. Divisors, +18, +12, + 9, + 8, + 6, + 4, + 3, + 2, — 2, — 3, 4, — 6, — 8, — 9, —12, —18, —24, —36, —72. Quotients, + 4, + 6, + 8, + 9, +12, +18, +24, +36, —36, —24, —18, —12, — 9, — 8, — 6, — 4, — 3, — 2, — 1. Add —18, —14, —12, — 10, — 9, — 6, 0, + 6, +18, —54, —42, —36, —30, —27, —26, —24, —22, —21, —20, —19. Quotients, ', — 1, *, *, — 1. 0, + 2, + 9, +27, +14, + 9, + 6, *, *, + 2, *, *, *, *.. Add — 5, *; — 6, *, *, — 6, — 5, — 3, + 4, +22, + 9, + 4, 0, *, *, — 3. a64 KEV TO PART SECOND. 1,+ 2,-11,— 3, Quotients, • *j *j »j Ij •^1 0, *, *, *. Add+ ], 0,4- 1, 0, 0, -f 3,-10,- 2, .-. -\-(}, -j-3, and — 4 are the roots. (,7) Here x''— lOx'-f 35x=— 50x+24=0, and -fl, is found tc be one of the roots (Art. 429, Cor. ]). The limit of the positive roots is 24, and since when we change the signs of the alternate terms, all the terms are positive, this equation has no positive roots, (Art. 402, Cor. 1), therefore the given equation has no negative roots. Last term -|-24. Divisors. . . . +2-1, -{-12, + 8, + 6, + 4, + 3, + 2'. Quotients . . . -f 1, -|- 2, -|- 3, -j- 4, -f- 6, -(- 8, -\-12. Add —50 . . . —49, —48, —47, —46, —44, —42, —38. Quotients. . . *, — 4, *, *, _ii, _]4, _i9. Add -1-35 —31, Quotients *, Add— 10 Quotients Add -1-1 -1-24,-1-21,-1-16. + 6, -I- 7, 4- 8. — 4,-3,-2. — 1,— 1,— 1. 0, 0, 0. .•. -\-4, -\-S, -\-2, and -|-1, are the roots. (8) Here x^-\-4x^—x^—\6x—12=0, and — 1 is found to be a root. Limit of positive roots 14-^16=5 ; of negative roots — (l4-4)=-5. Last term — 12. Divisors. . Quotients . Add— 16 . Quotients . Add —1 . . Quotients . \dd -|-4 . . Quotients . Add 4-1 . , . . +2, ■ 4- 4, 4- 3, 4- 2, — 2, — 3, — 4. — 3, — 4, — 6, -{- 6, 4- 4, 4- 3. —19, —20, —22, —10, —12, —13. *,— 11,4- 5,4- 4, *. -12,4- 4,4- 3. -6,-2,— 1. -2,4-2,4-3. -1,-1,-1. 0, 0, 0. , — 3, and — ], are the roots. RATIONAL ROOTS. 205 f9) Here x'— 4,t'— I9,r2+46;c+120=0, and -f-1 and —1 are found not to be roots. ]jimit of the positive roots I-|-19=20 ; of the negative roots --(1+^T6), or — 8. Last term -\-V20. DiviscTS 4- 20, +15, +12, +10, -f 8, + 6, + 5, + 4, + 3, + 2, — 2, — 3, — 4, — 5, — 6, — 8. Quotients, + 6, + 8, +10, +12, +15, +20, +24, +30, +40, + 60, —60, —40, —30, —24, —20, —15. Add +46, + 52, +54, +56, +58, +61, +66, +70, +76, +86 +106, —14, + 6, +16, +22, +26, +31. Quotients, *, *, *, *, *, +11, +14, +19, *, + 53, + 7, — 2,-4, *, *, *. Add —19 -8,-5, 0, + 34,-12,-21,-23. Quotients *, — 1, 0, + 17, + 6, + 7, *. Add —4 —5,-4, — 13,+ 2,+ 3. Quotients — 1, — 1, *,— !,— 1. Add +1 0, 0, 0, 0. .•. +5, +4, — 2, and — 3 are the roots. riO) Here a;''+03;'— 27a;2+14a;+120=0, and +1 and —1 are not roots. Limit of positive roots 1+^^27 or 7 ; of negative roots —(1+27) or 28. Last term +120. DiTisors, + 6, + 5, + 4, + 3, + 2, — 2, — 3, — 4, — 6, — 6,-8, —10, —12, —15, —20, —24. Quotients, +20, +24, +30, +40, +60, —60, —40, —30, —24, —20, —15, —12, —10, — 8, — 6, — 5. Add +14, +34, +38, +44, +54, +74, —46, —26, —16, —10, — 6, — 1, + 2, + 4, + 6, + 8, + 9. 23 266 KTi Y TO PART SECOND. Quotients, *, *, +il, +18, +37, +23, *, + 4, + 2. _4_ 1 * * * *,*,*. Add -27 —16,— 9, +10, — 4, —23,-25, -26. QuotJer.ls — 4, — 3, + 5, + 2, *, + 5. *. Add — 4, — 3, + 5, + 2, + B. Quotients — 1, — 1, *, — 1, — 1. Add +1 0, 0, 0, 0. .-. tlie roots are +4, +3, — 2, and — 5. (11) Here a;'+a;'— 29x5— 9x+180=0, and +!, and —1 are not roots. Limit of positive roots, 1+^29, or 7 ; of negative roots, —(1+29)=— 30. Last term +180. Divisors, + 6, + 5, + 4, + 3, + 2, — 2, — 3, — 4, — 6 — 6, — 9, — 10, — 12, — 15, —18, —20, —30. Quotients, +30, +36, +45, +60, +90, —90, —60, —45, —36, — 30, —20, —18, — 15, —12, —10, -9,-6. Add —9, +21, +27, +36, +51, +81,-99, —69, —54, —45, —39, —29, —27, —24, —21, —19, —18, —15. Quotients, *, *, + 9, +17, *, *, +23, *, + 9, *, *, *. + 2, *, *, *, *. Add —29 —20, —12, — 6, —20. —27, Quotients — 5, — 4, +2, +4. *^ Add+1 + 4,-3, +3, +5. Quotients ... . — 1, — 1, — 1, — 1, Add+1 0, 0, 0, 0. . . the roots are +4, +3, — 3, and — 5. (12) Here x^ — 2x'' — 4j;+8=0, and +1 and — 1 are not roots. Limit of positive roots, 1+^4=3 ; of negative roots ■-(1+4)=— 6. Last term +8. Divisors +2, —2, — 4, Quotients +4, — 4, — 2, RATI ON A L ROOTS 267 Add —4 0, —8, — e. Quotients 0, -f-4. * Add — 2 — 2, -j-2. Quotients — 1, — 1. Add +1 0, 0. .•. -1-2 and — 2 are roots, and by dividing the (:-|-4=0. Let j;=i, then the transformed equation (Ait. 405, Cor.) IS !/■* — 2y2 — 18y-|-36^0, and -f-l, and — 1 are not roots. Limit of positive roots, 1-|-18^19 ; of negative roots, ■-(l+jTS), or —6. Last term -(-36. Divisors, ^-18, -fl2, + 9, -f 6, -f 4, -f 3, -f 2, — 2 — 3, — 4, — 6 Quotients, + 2, + 3, 4- 4, + 6, -f 9, +13, +18, —IS, —12, — 9,-6 HATIONAL ROOTS. 269 Add— 18, -16, —15, —14, —12, — 9,— 6, 0, —36, —30, —27, —24 Quotients, *, *, *,— 2, ^— 2, 0, +18,+10, *,+ 4. Add —2 .... — 4, -4,-2, +16, + 8, +2. Ciuotionts ... *, *, — 1,— 8, *, *. 'idd+l 0,-7. .-. -|-2 is a root of the transformed equation, and dividing by y — 2 the quotient is y'^ — 18, hence y' — 18=0, and 2/=±3V2. .-. i/=2, +3^2, —3^2. (17) Here 8a;3—26a;'+l 1x4-10=0. Let a;=:i, then the transformed equation is 8 y' — 26i/--|-88i/+640=0, and -[-1 and — 1 are not roots. Limit of positive roots l-[-26=27 ; of negative roots -(l+>/~640), or— 10. Last term -|-640. Divisors, -|-20, +16, +10. +8, +5, +2, —2, —5, —8, —10. Quotients, +32, +40, +64, +80, +128, +320, —320, -128,-80, —64. Add +88, +120, +128, +152, +168, +216, +408, —232, —40, +8, +24. Quotients, +6, +8, *, +21, *, +204,-116, +8, —1, *. Add— 26. —20, —18, —5, +178,-142, —18, —27, *. Quotients, —1, *, *, +89, +71, * *. Add+1, 0, +90, -{-72. .•. +20 is a root of the transformed equation, and by di- viding by X — 20, the quotient is !/^ — Qy — 32, heuca y2_6y— 32=0, and y=^±Ji\. 870 KEY TO PART SEUONn. •■• i'=+20, and 3±j41 ; x=l=+l, or |(3±V^)- 8 OS) Here 6a;'— 25x3+26a;2+4a>— 8=0. Let x==!i, then the transformed equation is 6 y— 25y'+1562/2_|_i44^_1728=0, and +1 ina —1 are not roots. Limit of positive roots 14-25=26 ; of negative roots — (1+^T44), or —7. Last term — 1728. Divisors, -(-24, +18, +16, +12, +9, +8, +6, +4, +3, +2, —2, —3, —4, —6. Quotients, —72, —96, —108,-144, —192,-216,-288, —432, —576, —864, +864, +576, +432, +288. Add +144, +72, +48, +36, 0, —48, —72, —144, —288, —432, —720, +1008, +720, +576, +432. Quotients, +3, +2, *, 0, *, — 9, —24, — 72,-144,-360,-504, —240, —144, —72. Add +156, +159, +158, +156, +147, +132, +84, +12, —204,-348, —84, +12, +84. Quotients, *, *, +13, *, +S2, +21, +4, —102, +174, +28, —3. —14. Add— 25 — 12, —3, —4, —21, —127, +149, +3, —28, —39. Quotients — 1, *, —1, —7, *, *, —1, +7 • Add +1 0, 0, —6, 0, +8. .■. y=+12, +4, and — 3 ; and by dividing by {y — 12) (y — 4)(y+3), the quotient is y — 12, lience y — 12=0, anJ 2/=12. .-. x=l=-\-'2, +2, +§, and —},. a9) Here x'— 9a^+'/a;2+yx— *','=±:0. RATIONAL ROOTS. 271 Let a;=-i, then the transformed equation is 2/^— 18j/'+452/'-^+ 1 08^^—324=0. Limit of positive roots, l-(-18=19 ; of negative roots, — (1+VT08), or —6. Last term — 324. Divisors, + 18,+ 12,+ 9,+ 6, -f 4,+ 3,+ 2, — 2,— 3,— 4,— 6. Quotients, — 18, — 27, — 36, — 54, — 81, — 108, — 162, + 162, +108, + 81,+ 54. Add +108, + 90, + 81, + 72, + 54, + 27, 0, — 54, +270, +216, +189, +162. Quotients, *, *, + 8, + 9, *, 0, _ 27, —135, — 72, *, *. Add +45 + 53, + 54, + 45, + 18, — 90, — 27. Quotients, *, + 9, + 15, + 9, + 45, + 9. Add— 18 — 9, _ 3, _ 9 + 27, — 9. Quotients *, ], *^ *,+ 3. Add +1 + 4, 0. . . +3 is a root of the transformed equation. The first derived polynomial of the transformed equation is 4y3 — 54^2+90^+108 ; now we shall find that y — 3 is a divisor of this as well as the transformed equation, therefore +3 and +3 are two roots of the transformed equation (Art. 414), and if we divide it by (y— 3)(y— 3) the quotient is y'— 12y— 36, henca yi_]2y — 36=0, and J'=6±6^2^. .-. 2/=+3, +3,+6+6V2',+6— e^i'. ^=+2' +1' +3+3V2, +3— 3V2. Note. — This example may be solved by Art. il4, but tlio above ii Ihe shortest method. 272 KI3Y TO PART SECOND. HORNER'S METHOD OF APPROXIMATIOS. Articles 430 — 434. ,1) x^-\-6x —12.24=0 r 1 -f5 — 12.24 (1.8=a;. It is readily found lliat +1 +6 ''^ is greater tlian 1, and -j-6 — 6.24 less than 2, hence 1 is -\-l + 6-24 the integral part of the 1 4-7* root. (2) .8 +7.8 ,_V'_6.24__g T' 7 ^ '4-12X —35.4025=0. rs +12 —35.4025 |2.45=r. + 2 +28 +14 — 7.4025* + 2 + 6.56 + 16* — .8425* A + .8425 + 16.4 .0 .4 7.4 . , r= — =4+. +16.8* 16 05 .84 „^ S= =:.05. + 16.85 16.8 When the operation gives a remainder zero, we kno\v from Art. 395, Cor., that the exact root is obtained. Note — In the solution of the succeeding problems, we shall merely present the operation, without exhibiting the work by whicii tiie suc- cessive figures of the root are obtained, since they can gonerallj- be ilf'.or- niineJ mentally, as in Long Division. (3) ..x^— 28x —61.25=0. 4 —28 —61.25 |S.75=a;. + 32 +32 ' +4 — 29.25* BORNBR S METHOD OF APrBOXIMATION. '-i^U 32 4-37.16 4 +36* — 2.09 2.8 + 2.09 +38.8 .0 4 +41.6* .2=.05X4 +41.8 (4) 8r'— 120a; +394.875=0. 8 _120 +394.875 |10.125= + 80 —400 — 40 — 6.125* + 80 + 4.08 8 +40* — 1.045 .8 + .83 52 +40.8 — .2098 ,8 + .3098 8 +41.6* .0 ^16_ +41.76 .16 8 +41.92* .0 4=.0 05X8 +41.96 «5) 5x'— 7.4j! —16.08=0. 5 — 7.4 —16.08 |2.68j +10 + 5.3 + 3.6 +10 6 +12.6* 3 +15.6 3 6 +18.6* .4= .08X5 19.0. — : 10.88. + 9.36 — 1.53 + 1.52 874 KEV TO I'ART SECOND, (6) X''+X —1=0. —1 1.618034. 0000^6^.000034 nearly. ] 2.236*. (1) x^—6x +6=0. 1 _6 -f6 |4.73205=J. +4 —8 —2 —2 +4 +1.89 1 +2* .7 +2?7 .T_ 1 +3.4* .jOOT]^ (,0205_ .03 3.46 +3.43 .03 — .11 + .1029 — .0071 .007 1_ 3.46*. (8) i'+4x' — 91 —57.623625=0. 1+4 _ 9 —57.623625 |3.45=T. +3 +21 +36 +r +12 —21.623625* +3 +30 18.944 +10 +42* - 2.679625 r Operation 3 6.36 + 2.679 6251 continued on 1 +13* +47.36 ^0 [ page 275. Horner's method of approximation. 275 +13.4 4 -1-13.8 4_ -1-14.2* .0^ +14.25. '5.52 +52.88* ■7125 +53.5925 (9) 2x'— 50a;+32.994306=0. CO) ±0 +8 -50 32 +8 —18 +8 +64 +16 +46* 8 +15.12 +24* +61.12 + 1.2 15.84 +25.2 +76.96* 1.2 .8298 +26.4 +77.7898 1.2 +27.6* .06 +32.994306 —72 —39.005694* +36.672 — 2.338694* + 2.333694 .0 +27.66. a:'+a;'+x — 1=0. +1 +1 .75 +1.5 .5 2.0 _5_ 2.5* .04 2.54 .04 2^68 |4.63=r. —1 1-543689 .871 +1.75 1 +2.75* .1016 +2.8516 .1032 +2 9548* .007869 +2.962669 — .125* .114064 .010936* 8888007 .002047993* J Operation continued on page 276. 21 R KEY TO PART SECOND .04 .007878 <") 2.62* +2.970547* .003 2.623 .002047993_ 2.97 =.000689. .003 2.626*. +4 —20=0. —5 —20 [2.23608. +2 +6 +2 +8 2 +12 +7 +16 +23* 2.04 +14 —6* +5.008 — .992 + .823167 +10* V .2 +25.04 2.08 +27.12* .3189 — .168833 +10.2 .2 +10.4 .2 +27.4389 .3198 +10.6* 27.7587* .03 +10.63 .03 .168833_^,^^n„ +10.66 27.76 Note. — In the solution of the remaining examples in this arftcle we eliall use tlie abridged method employed in example 3, page 382, of ihe Algebra. The learner, however, who chooses to carry out the -ecimala fully will find no difficulty in so doing '12) 1 a:'— 2x- +2 2 2 4 2 +6* -5=0. —2 4 +2 8 —5 |2.0945515=a;. +4 — 1* +0.949329 ] +10* .5481 + HJ.5481 — .050671* + .04451752 JSe — .00615348* '-''' page BORNE E'S method OF APPROXIMATION. 277 .09 .5562 557875 6.09 4-11.1043* 57473*- .09 .02508 55800 6.18 11.12938 1673* .09 .02508 1116 6.27* +11.15446* 557* .0031 658 11.1575 .003 +11.16* In finding the result true to seven places, it is not necessary C> add any more figures to 6.27, as it will not alter the result. This example is found in the work of M. Fourier, " Analyst del' Equations" where the result is gi-ven to thirty-two places of deiimals : the result to forty places, as given by Gregory is, x=rr2.0945514815423265914823865405793029638576. We appre- hend few students will undertake to verify this result. The other Iwa values of x are imaginary. CSS) a;'+10a;2— 24a— 240=0. + 10 — 24 —240 |4.8989795=a:. + 4 + 56 +1 128 +14 + 32 — ; 112* + 4 + 72 + 97.792 +18 +104* — 14.208* + 4 18.24 + 12.899169 +22* 122.24 — 1.308831* .8 18.88 - 1.165869792 22.8 +141.12* .142961208 .8 2.2041 .131358087 23.6 143.3241 11603121 .8 2.2122 +145.5363* 10218295 +24.4* 1384826 .09 .197424 1313781 24.49 145.733724 71045 .09 .197488 72988 24.58 + 145.931212* [Over.] 27S KEY TO PART SECONT .09 .02222 1 +24.67* 145.95343 .008 .02222 24.678 +145.97665* .008 24.686 .008 1 +24.694* (14) a:'+]2a;2_i8r=216. +12 — 18 —216 (4.2426407: 4 + 64 + 46 184 16 — 32* 4 80 26.168 20 +126* — 5.832 4 4.84 5.468224 +24* J30.84 .363776 .2 4.88 .275484488 24.2 +135.72* 88291512 .2 .9856 82683912 24.4 136.7056 5607600 .2 .9872 5512892 +24.6* +137.6928* 94708 .04 24.64 .049444 137.742244 96475 .04 .049448 137.791692* 24.68 .04 .01483 +24.72* 137.80652 .002 24.722 .0148 137.8213* .002 .0010 24.724 137.8223 .002 1 +24.726* Horner's method of apvrcximation. 379 (15) a;<— 8a;3f20a;2— 15a;+.5=0. -1-20 —15 . -f .5 |1.28.1724= a -1-1 — 7 4-13 —2.0 —1.6* -1-1.0496 — .4504* .4255385 248616* 210536 —7 +13 + 1 — 6 —6 + 7 + 1 — 5 —5 + 2* +1 — .76 _4* 1.24 .2 — .72 —3.8 .52 .2 — .68 —3.6 — .16* .2 — .2496 —3.4 — .4096 .2 — .2432 —3.2* — .6528 .08 — .24 —3.12 — .89* .08 — .01 —3.04 — .90 .08 — .01 —2.96 — .91 .08 — : 15 + ! L3 — 2 + 7 + 6* .248 -t-6.248 .104 5.352* — .032768 6.319232 — .0522 + 5.2670* — .0036 5.2634 — .0036 + 5.2598* .0006 38079* 36820 1259* 1052 207 210 5.26 (16) a:'-|-a:=— 8x— 15=0. 1 -fO + 1 — 8 —15 I2.302775o3= -1-3 + 4 +W + 4 +2 -f 5 +2 —11* 2 _8_ 26 10.8741 +4" 13 -1-28* — .1259* 2 12 8.247 .09066704 -f6 -f25* 36.247 3523296 2 2.49 9.021 3179543 -f8* ^.49 -l-45.2''.8* 343753 [Over 280 KEY TO I'ART SECOND. .D 8.3 2.58 30.07 .06552 45.33352 .3 2.67 32.74* .06556 8.6 45.39908* .3 .02 02296 8.9 32.76 45.42204 .3 .02 32.78 .02296 -1-9.2* 45.4450* .02 23 32.80* 45.447 2 318129 25624 22724 2900 2726 174 136 45.449* (17) x<— 59j;'+840=0. Note. — This is a trinomial eqnntioii nna may be solved as a quadratic .Art. 242), but it is placed here to be solved by Horner's method. 1 —59 +840 (4.8989795 4 16 —43 — 172 _i_ 388 + 4 • — 172 152* 4 32 — 44 — 140.5184 8 — 11 —216* + 11.4816 4 48 +37* 40.3.")2 — i0.5nG;i7359 12 —175.648 + .97462641 4 13.44 51.616 — .868978336 1 +16* 50.44 — 124 032* .105648074 0.8 14.08 7.287849 97080S04 16.8 64.52 -116.744151 8567270 .8 14.72 +79.24* 7.441S27 7544964 17.6 —109.299324* 1022306 .8 1.7361 80.9761 .677032 — 108.622292 970011 18.4 62295 .8 1.7442 82.7203 .678280 — 107.944012* 53885 I +19.2' .09 1.7523 +84.4726* 07645 — 107 80756 Tsi 19.29 ;c page 28 1. J HOKiJIER'S METHOD OF APPROXIMATION. SSI .09 .1564 .07645 19.38 .09 19.47 .09 84.629 .156 —107.79111* .0059 84.785 —107.7852 .156 .0059 1 +19.56* +84.941* —107.7793. (18) 2x«+5a;3+4>;2+3a;=8002. + 3 959 + 5 14 IF 14 33 14 + 4 133 137 231 —8002 (7.33555 +6734 (40314 962 2576 —1268* 1125.7932 368 329 47 14 8 +61* 697* 18.48 715.48 18.66 3538* 214.644 3752.644 220.242 61.6 6 62.2 6 62.8 6 734.14 18.84 3972.886* 22.646514 — 142.2068* 119.86597542 —22.34082458* 20.11015620 "2^23066838* 2.01310385 3995.632514 22.703682 -63.4* .OS 63.46 .oe 63.52 .95 63.58 .06 752.98* 1.9038 754.8838 1.9056 756.7894 1.9074 4018, 3. 236196* 79505 4022, 3, 03124 79665 .21750453 .20133125 402666) 1623328( 1610665 758.6968* .32 759.01 ■ 32 759.33 .32 4025 ,82789* 3798 4026, 4026 4026 2077 3798 5875* 38 625" 38 12663 12080 583 403 180 161 3 +63.94* 759.66* 4026.663* a:=7.3355540314. Remark. — It is sometimes convenient by drawing lines, as in tha preceding solution, to render separate and distinct the operation for find- ing each successive figure. (19) a:«+4a;*— SajS+lOar^i- 2x=962. 1+4 — 3 +10 — 2 3 +21 54 +192 +7 +18 +64 +190 24 —962 1 3. 385777 +570 —392* rover 882 KEY TO PART SECONI. 3 30 48 144 208 624 290.21133 10 + 814* —101.78867 3 39 261 153.3711 + 94.64260 13 87 +469* 967.3711 —7.14607 3 48 42.237 511.237 166.5714 ! 133.9425* 6.18160 16 +135* — .96447 3 5.79 44.001 49.090 86793 +19* 140.79 555.238 a83.0325 —9654 .i 5.88 ^5.792 601,030* 50.096 1233.128* 8682 19.3 146.67 —972 .3 19.6 5.97 152.64 12.6 613.6 3.19 1236.32 868 —104 .3 19.9 6.06 158.70* 12.6 626.2 3.19 1239.51* .3 12.6 638.8* .4 1239.9 20.2 .3 .4 a;=:3. 385777 1 +20.5* 1240.3* (20) a;5+2x'+3j:'+4«^+5^=54321. - 2 + 3 + 4 + 5 -64321 (8 414 8 80 664 5344 42792 (455 10 83 668 5349 -11529* 8 144 227 1816 2484 19872 11088.97344 18 25221* —440.02656 8 208 435 3480 5964* 2501.4336 27722.4336 304.1105122 26 —135.9160478 8 272 289.584 2620.0064 122.0290372 34 +707* 6253.584 30342.4400* — 13.SS70106 8_ 16.96 296.432 68.61122 12.215 0180 1 +43* 723.96 6550.016 30411.05122 — 1.6719926 0.4 17.12 303-344 68.68888 1.5270320 42.4 741.08 6853.360* 50479.74010* .4 17.28 7.762 27.5192 42.8 758.36 6861.122 30507.2593 .4 17.44 7.766 27.5316 43.2 +775.80* 6868.888 30534.7909" — .1449606 .1527032 uornkr's method of approximation. 283 .4 .4 7.770 2.754 43.6 776.2 6876.658* 30537.545 .4 .4 776.6 3.1 2.754 20540.299* I -f44.0* 6879.8 s=8.414455, .4 3.1 .34 777.0 6882.9 30540.64 .4 3.1 777,4* 6886.0*. ADDITIONAL EXAMPLES. Note. — As some instructors may desire additional eEamples to exer- cise tiioir pupils, we subjoin the foIlowing^, selected froii a collection by Olinthus Gregory, Professor of Mathematics in the Royal Military Academy, Woolwich. They may also be employed as exercises in Sturm's Theorem, in finding tlie number and situation of the real roots. (I) x^—-[2x=15. (See Art. 434-6). C 3.971960, X— .^—1.576334, ^—2.395426. (2) a;3— 13a;2-f49z— 45=0. (5, x= .?6.64575, n. 35425. (3) a;3— 6x=2. C 2.6016791318, ^—2.26180224.52, f— .3398768866. (4) x3— 27a;=36. C 5.7657415977, ?— 4.3206356862, ^—1.4451059115. (5) a;'— 13.r24-38.r-|-16=0. x= { 5.3722813, f— .3722813. (6) a;'— 7a;+7=0. ( 1.69202147163009586962781489, x= } 1.35689586789220944389439951, f— 3.04891733952230531352221440. (7) a;3_7035a;='-(-15262754a:— 1000073088(1=0. a;=1234, or 2345, or 3456. 284 KEY TO PART SECONJ) a;4_6x2— 16x+21=0. ( 3, or],o' (8) ^ ' (See Art. 429). (9) a:<— 19a;3_[_i23a;2_302a;+200=0. •=i- -2ztv'— 3. n. 02803, J 4.00000, I 6.57653, I 7.39542. (10) a:"— 4a;'— 3x2— 4a:+l=0. The two real roots are a;=4.7912 and a:=.2087. (11) a;'— 36x2+72a;— 36=0. (12) xi+xs— 24a;=+43x=21. (See Art. 429). (13) x"— 27x'+162a;2+356j:=1200 (14) a;'"- 17x^+20x— 6=0. f 0.872983, 1.267949, 4.732050, —6.872983. C 1, or 3. x= \ 1.1400549, (—6.1400549. r 2.05607, _ I — 3.00000, *~) 13.15306, I 14.79085. -\ 4.6457513, .6457513, 2±V2"- (15) a:<_li2.3x'+1243.53j;'— 2244.341a;+1112.111=0. a;=l, or 1.1, or 10.1, or 100.1. Article 4 35. to extract thekoots of n u m b e r s b y Horner's method. (.2) To find the cube root of 34012224. 3 9 3 9 3 18 6 27* 3 184 9* 2884 34012224 :S24. Alls. 27 , 7012 5768 1244224 [See pajfe 285. HORNER'S METHOD OF APPROXIMATION. 2SS 2 188 92 3072* 2 3856 94 311066 2 96 4 964 (3) To find the cube root of 9. 10 9 | 2.080084. An*. 2_ _4_ 8_ 2 4 1* 2 8 .998912 T 12* .001088 2 .4864 .001038 6*_ 12.4864 50 .08 .4928 51 6.08 12.9792* .08 6.16 .08 6^* (.4) To find the cube root of 30. 10 30 |3.107233 ._ 3 9 27 T T 3* 3 18 2.791 ~6~ 27« .209 3 .91 .20223 I 9» 27.91 .00677 .1 .92 579 gT 28.83* 98 .1 6^ 87^ ^ 28.89 11 .1 ^ » 9.3* 28.95* S8B KEY TO PART SECOND. (5) To find the fifth root of 68641485507. 68641485507 fUl 1 1 1 1 1 • 1 1 1 1 586414 1 2 3 4 437824 2 3 4 5* 14859085507 1 3 6 59456 14859085507 3 6 10* 109456 1 4 10* 4864 82624 4 14864 192080* 1 216 6792 201926501 5* 1216 20666 2122726501 4 232 6784 54 1448 27440* 4 248 1406643 58 1696 28846643 4 264 62 1960* 4 4949 66 200949 4 70* 7 707 A PPE XIM AT 10 N BY DOUBLE POSITIOH. (2) a:'-|-30J:=420. 6 X 7 216 x' 343 180 30x . 210 396 results 563 653 7 420 396 _6 396 157 : 1 : : 24 : 0.\ 6.1 X 6.2 226.981 «» 238.328 APPROXIMATION BY DOUBLE POSITION. 287 183.0 SOx 186.0 409.981 results 424.328 14.347 : .1 :: 10.019:0.163 By trial x is found to be greater than 6.17, therefore let a:=6.17 and 6.18. 6.17 X 6.18 234.885 x' 236.029 185.10 30a; 185.40 419.985 results .... 421.429 1.444 : ,01 : : .015 : 000103. .-. a:=6.17+.000103=6.170103 nearly. (3) 144x5— 973x=3 19. 2 X 3 11.52 144x3 3888 —1946 — 973x —2919 — 794 results +969 3 + 969 2_ 319 1763 : 1 : : 650 : .3 x= 3— .3=2.7. 2.7 X 2.8 2834.352 144x3 3161.088 — 2627.1- — 973x —2724.4 + 207.252 results + 436.688 As 229.436 : .1 :: 111.74^:. 048 .-. x=2.7+.048=2.748, and by trial 2.75 is found to verify the equation exactly, hence x=2.75. In the application of the rule of Double Position to the solu- tion of equations, the first correction is generally too small, as in the two preceding solutions, and as may be seen more particularly in the solution of example 5. To see the reason of this, it must be noticed that the sums, or the differences, of the higher powers of numbers, increase very rapidly as the numbers increase. Hence if two numbers equally distant from the true number, are substituted in any equation containing the second or higher powers of the unknown quantity the result, arising from the substitution of the greater number, will be farther from the true result than that obtained by the sub 288 KEY TO PART SECOND. stitution of the smaller. And hence, by the operation of the rule, the correction will give for the true number a number too small: To illustrate this by an example, suppose we have the equation x' — x=24, of which the root is 3. Let us notice the results obtained by the substitution of 2 and 4 for X. 2 X 4 8 x^ 64 — 2 —X — 4 6 results 60 — 18 errors +3t> Difference of the errors ==36 — ( — ]8)=54, then 54 : 2 : : 18 : |. Now the true correction is 1, but we obtained | because al- though the- suppositions, 2 and 4, are eqtially distant from the true number, yet the corresponding results are unequally distant from it. Now the rule proceeds on the hypothesis that the errors of the results are proportional to the errors of the suppositions. But this is never exactly true, and is only nearly so when each of the suppositions is very near the true number. Attention to •;his principle will often guide the pupil in selecting trial numbers for the second operation. (4) a;'+10x2+5a;=2600. By trial we find that 11 is so near the true number that we may a' once make trial of 11 and 11.1. 11 X 11.1 1331 a* 1367.631 1210 10x= 1232.1 65 bx 55.5 2596 results 2655.231 — 4 errors -[-55.231 59.231 : .1 : : 4 : .006 .-. a;=:ll-f-.006=l 1.006 nearly. 11.006 X 11.007 1333.179188 a;' 1333.542617 1211.32036 10a:' 1211.54049 APPROXIMATION BY DOUBL'l POSITION. 289 55.030 5x . . 3599.529548 results .470452 errors .588559 : .001 : , . . . 55.035 . . . 2600.118107 . . . +.118107 .470452 : .00079 (6) .-. a:=11.0064-.00079==l 1.00679. 2a;'+3a;2— 43,>=10. 2 2x3 Ig 3 +3a:2 12 —4 —ix — 8 +1 results +20 — 9 errors -t"'" 19 : 1 : : 1 : .5 nearly. .•. a;=l+.5=1.5 nearly. By trial, however, we find that 1.6 is too small, and 1.7 too great, let these therefore be the next two assunied numbers. 1.6 X 1.7 +8.192 2x' + 9.826 +7.68 +3a;2 _1_ 8.67 —6.4 —ix — 6.8 +9.472 results +11.696 — .528 errors + 1.696 2.224 : .1 :: .628 : .024 nearly. . . a;=1.6+.024=1.624. By trial we find 1.624 is too small, and 1.625 too great; using these as the next two assumed numbers, we readily find the next two figures of the root. (6) x*—x^+2x''+x=4. It is easily seen by inspection, that a; is a little more than 1 and by trial it is found greater than 1.1, and less than 1.2; le' {hesBj therefore, be the two assumed numbers. 1.1 X 1.2 + 1.4641 +3!^ +2.0736 ~].331 — x' — 1.728 +2.42 +2x^ +2.88 +1.1 + X +1.2 +3.6531 results +4.4256 — .3469 errors ......+ .4256 7725 : .1 : : .3469 : .045 nearly. 25 290 KEY TO PART SECOND By trial x is found greater than 1.146, and less than 1.147. By repeating the operation with these numbers, we readily find the next two figures of the root. (7) x'-\-x^+2x^—x=4. It is easily seen that a; is a little more than 1, and by trial it is found less thai 1.1 ; therefore let 1 and 1.1 be the two assumed mimbers 1 1 2 —1 +3 —1 1.U51 1.1 1.4641 1.331 2.42 — I.l results -{-4.1151 errors -\- .1151 : : 1 : .09 nearly. + x' . -f a;' . -j-2x^ . — X . By ,iial a; is found to be greater than 1.09, therefore let 1.09 and 1.1 be the next two assumed numbers. 1.09 1.1 1.4)1581 x^ 1.4641 1.2950^9 + x' 1.331 2.3762 -\-2x^ 2.42 — 1.09 — X 1.1 -{-3.99281 results -1-4.1151 — .00719 errors -|- .1151 .12229 : .01 •: .00719 : .00059 nearly. .-. a;=1.09-|-.00059 = l. 09059 nearly (8) x^— 12j;-1-7=0. It is easily seen that a; is a little greater than 2. and by trial it is found less than 2.1, therefore let these be the first two assuipwd 1) imbers. 2 X 2.1 -{-16 i" .... —24 —12a; + 7 +7 — 1 results and errors .... 2.2481 : .1 : 1 04 • ■. a;=2-{-.04=2.04 nearly. -fI9.448] —25.2 + 7- -1-1.2481 APPROXIMATION BY DOUBLE POSITION. 29 Bv trial we find 2.04 too small, and 2.05 too great. 2.04 X 2.05 17.3189. —24.48 . 17.661 —24.6 + 7 .... x" . . .... — 12x . .... 7. . . — .1611 . . . results and errors ....-[- .061 As .2221 : .01 ; .1611 : .0072 .-. a^=2.04+.0072=2.0472 nearly. By trial we find that 2.0473 is too small, and 2.0473 too groat; then by using these as the next two assumed numbers we readily find the remaining figure of the root. (9) 2x''— 1312+10^;— 19=0. Here it is readily found that x lies between 2 and 3, let these therefore be tlie two assumed numbers. 32 2a;'' 162 —52 —13x2 _Ij7 +20 +10x +30 —19. —19 — 19 — 19 results and errors +56 75 : 1 : : 19 : .3 nearly, and x=2.3 nearly. JBy trial we find that x is greater than 2.4, and less than 2.5 let these therefore be the two assumed numbers. 2.4 2.5 66.3552 21" 78.125 —74.88 — 13x^ —81.25 +24 +10x +25. —19 —19 —19. — 3.5248. . . results and errors . . . . 2.875 6.3998 .1 : : 3.5248 : .06. . . x=2.4+.05=2.45 nearly. By trial it is found 2.45 is too small, and 2.46 too great, using (hese as the next two assumed numbers, we obtain the next two figures of the root. flO) V7a;'+4a;2+Vl0.<2a;— 1)=28. 202 KEY TO PAET SECOND. By trial we readily find that cc lies between 4 and 5 ; we thero (uie take these as the first two assumed numbers. 4 X 5 + 8 V ^x''-\-4x'' 9.91 21.21 —28 +16.73 +v'10J;(2a;— 1) . . —28 —28 — 3.27 errors + 3.12 6.39 : 1 : : 3.27 : .51. .•. a;=4-(-. 51=4.51 nearly. By trial we find x greater than 4.51, and less than 4.52 ; there- fore let these be the next two assumed numbers. 4.51 X 4.52 4- 8.9773 V 7a;'+4x2 +19.0185 +v'10x(2x— 1) — 28 — 28 . . , — .0042 errors. . . .064 : .01 : : .0042 : .00066 nearly .-. a;=4.51+. 00066=4.51066 nearly. 8.9965 +19.0633 —28. + .0598 Article 437. NEWTON's method of APPEOXIMATION. The learner must observe that A is what the proposed equa- tion becomes when x=a, and that A' is what the first derived polynomial, or first derived function (Art. 411) becomes when x=a. (1) Proposed equation X=a;' — 2x — 5=0 ; First Derived function X'=3x'— 2. When 2 is substituted for x the result is — 1, and when 3 ia 8ub.9tituted the result is +16 ; therefore (Art. 403), one real root of the equation lies between 2 and 3, and is not mucli greater than 2. By trial we find that 2.1 gives a positive result, there- fore the root lies between 2.0 and 2.1. . . let a;=:a+)/=2+?/. then A=(2)3— 2(2)— 5, and A'=3(2)=— 2, A _ 8—4—5 12—2 :=+.l. cardan's solution uf cubic EauATioNS, 293 .-. x=a+y=2 + ( — ^, j =2+. 1=2.1. Next let x=b-\-z=2.1-\-z, then B=(2.])'— 2(2.1)— 5, and B'=3(2.1)=— 2. B' 3(2.1)2—2 11.23 .-. a;=i+2;=2.1-f(— .0054)=2.0946. Next let a;=c+2'=2.0946+a , then C = (2.0946)'— 2(2.0946)— 5=.00O54155O536, C'=3(2.0946)'— 2= 1 1. 16204748, . =„C__-000541550536___00004g5j_ C 11.16204748 .-. a:=C+2'=2.0946+(— .00004851)=2.09455149, which is true to the seventh place of decimals ; and by proceeding in a similar manner the value of x may be found to any required degree of accuracy. Remark. — Tiie great objection to Newton's Method of Approxima- tion is, that we are obliged after each operation to commence with tho entire approximate value of x in tlie same manner as at iirst, and no assistance is derived from the previous calculations except in having found a nearer value of the root. But in Horner's method we approxi- mate continuously to the true value of the root by the evolution of single figures fi.s in Long Division, and the Extraction of the square root in arithmetic, and each previous figure is of use in finding the next. Newton's method is now rarely used, and may be classed among th« scientific curiosities of a past age. Articles 438 — 441 . CAKDAN'S SOLUTION OF CUBIC EQUATIONS. Formula;. x^-\-3qx+2r=0. (2) a;3— 9j;-f28=0. Here 5=— 3, and r=-|-14. ^=^(— 14+V196— 27)+=/(— 1*— ^196— 27) = ^(— 14+13)+ V(— 14— 13)=V^+V^^^7="1 — 3=— 4. o"" K — 5±v'21). (3) k"— |a;3+2a;=>— |a+l=0 x^ — ^x-\-2 — A-j-L=0, by dividing by a' 2x x^ x' V a; / Let a;+-=2, then x2-|-L=2;=— 2, and a; x^ 2^ — 712=0, whence 2=0, or -|-;}. .-. a;+i.=0, or +4 ; X whence a;=±V — 1, or 2, or i.. (4) x"— 3a;'+3a;— 1=0. Ft is proved in Art. 4 12, Prop. Ill, that this equation is d'uisihl* by x^—^l, .'. a;'^ — 1=0, und a:=drl. RECURRING EaUATIONS. 297 Dividing tlie given equation by a;' — 1 the quotient is %^ — Sa-]-!, tnerefore x' — 3x4-1=0, whence .r=i(3=hV5)- (5) a;'— l]ar'+17j:'4-17x'— 11x4-1=0. It follows from Art. 442, Prop. II, that — 1 13 a root of ihia equation, therefore it is divisible by x+l (Art. 395). 1—114-174-17—114-1 (—1 — l-fl2— 294-12— 1 1—124-29—12-1- 1 .-. a;<—12x'4-29x2— 12x4-1=0. x'— 12x 4-29— — 4-i-=0, by dividing by «» X x^ «24-l— 12 { x4-l ) =—29. x' \ X / Let x^-=-z, then x^4-_=2' — 2, and X x^ z' — 122=^27, whence 2=9 or 3. .-. x-l-2=9 or 3 ; X 9±777 „^ SzbVl wnence x= — , or y~ . (6) 4x«— 24x'4-67x''— 73x'-f57x'— 24x-|-4=0. 4x3— 24x24-57x— 734-— — — -|--=0, by dividing by x», X x^ x' 4(x34-L)-24(x'-fiJ-F57(x+l)=73. Let x+l=z, then x24--=2^— 2. X x^ and x'4-— =2' — 32. x' .-. 4(23-32)— 24(2'— 2)4-572=73, 23—6224-^2=4-2/. To solve this equation by Cardan's Rule, Art. 441, let 2=^4-2, then y^- |y4-]=0, y=V(-i+N/g\-g',)+V(-i-Vs'4-BV 298 KEY TO PART SECOND. Dividing y' — ly+j by y+l,tho quotient is y^ — y-\-\ i therefore, y^ — y-\-^=o, and y=-{-!,, and -I-'. .-. j=!/+2=— 1+2=1, or .^+2=:], and j. . . a;+-=l, whence x=- or, x-\--=^, whence x=2, or \. X - .-. the six roots are 2, ', 2, ' lXj?Zz:?, and ^~'^~^ 2 ~ *> o Article 443. BIBOMIAL EQUATIONS. (1) Let x*=l, then x"— ]=0, and (a:^— l)(jc2+l)=0 .•. a;' — 1=0, whence x-=l, and a!=-f-l, or — 1. Also, x--\-l=^0, whence x^=: — 1, and x=-\-J — 1, or, — 7— 1. (2) Let x':=l, then x^ — 1=0, and the equation is divisible by X — 1, .•. X — 1=0, and a:=+l. Dividing x^ — 1 by x — 1, and placing the quotient equal to zero, we have a;<4-a;'-f-a;2+.i:+l=0, x^-\-x-\-\-\- --\-—=Q, by dividing by x'', X x^ ^'+-,+^+-=-1- Let x-{--=z, then x^-\-—=z' — 2, and z^-\-z=l ; whence z= — ^ =a. '. x-\--=a, whence X *=-+i^/a^ — 4, or - — IJa'' — 4, and since a has two J * 2 " values X will have four values. BINOMIAL EQUATIONS. 299 „:.^ (— 1±J"5)'' _6— 2^'5^ ^^ 6+2 V 5 ' 4 4 ' 4 ■ = 4^5— 1— V(— 10— 2V6)i ; =-i!V5+i-V(-io+2v'5)? ; 300 APPENDIX. APPENDIX, INDETERMINATB ANALYSIS. Art. 1. Indeterminate Analysis is the resolution of equa> ticvns where the number of unknown quantities is greater than the number of independent equations, and where the results are required in positive integers. It is shown (Alg. Part II,. Art. 168,) that whenever the num- ber of unknown quantities is greater than the number of inde- pendent equations, an unlimited number of values may be found for each of the unknown quantities. But such conditions may exist as to limit the number of results,- or even render the ques- tion impossible. Thus the equation 3a;-|-5y^4"2, may be satis- fied by an infinite number of values of x and y ; but if it be re- quiied that these values shall be integral and jjosiZiw, then we can only find a;=9 or 4, and y='& or 6. Problems of this kind are called indeterminate, and the results are generally required in positive integers. An indeterminate equation of the ^7'sf degree, cT^taining two unknown quantities, is of the form ax-\-by=c, where a, b, and c are either positive or negative whole numbers. Art. 2. Proposition I. — If an equation of tlie form ax-\-hy =c, is in its lowest terms, it can not he solved unless a and i are prime to each other. For, if possible, let a=md, and h=nd ; then mdx-\-ndy=:^, and .'. mx-\-ny=' \ d but, by hypothesis, a, b, and c contain no common factor, there* fore - is a fraction, and we have the sum of two whole .lumberi d equal to a fraction, which is absurd ; nence the proposition > true. Art. 3. Pkofosition II. — If a aid b are prime tc ".ach other, each term of the series, b, '2b, 3 J, cj-c. . . . (a — l)t, jchen di- vided by .a, will leave a different positive remainder. INDETERMINATE ANALYSIS. 301 For, if possible, let any two of tlie terms, as mb and lib, when Jivided by a, leave the same remainder r, so that mb , r , nb , r . — =p-\--, and ^-^q-{--; a a a a Subtracting the second equation from the first, mb nb h, — — —=p—g, or -{m—n)=p—q; a a a Dut the left liand member of this equality is a fraction, since - ia a fraction in its lowest terms, and m — n less than a, each be- a ing less than a, therefore we have a fraction equal to a whole number, which is absurd ; hence the remainders are all different. Illustration. — Let a=4, and i=7 ; then 7,7x2, and 7x3, when divided by 4, leave the different remainders 3, 2, and 1. Cor. Since the remainders are all different, and are a — 1 in number, each being less than a, therefore they include all num- bers from 1 to a — :1. Art. 4. Proposition III. — The tqualion ax — 6i/=±,l ts always possible in integers, if a and b are prime to each other. By the Corollary to the preceding proposition, if b, or some multiple of b less than ab, be divided by a, the remainder will be 1 ; let^ be that multiple, then " =x-\-- ; X being the integral part of the a a quotient and 1 the remainder. By clearing and transposing this gives ax — by= — 1, which proves part of the proposition. Again, by the same corollary, y may be some coefficient of b less than a, such that yy^b, when divided by a, will leave a remainder a — 1, that is y^ =x-\- , X being the integral part of a a the quotient, and a — 1 the remainder ; Dy clearing this gives by=ax-\-a — 1 ; by transposing and factoring a(a:-|-l) — by=\ ; let x-\-l=x', this gives ax — ?>«/=!, which proves the remaining part ot the proposition. Hence, if a and b are prime to each other, such values of x and y may always be found as will satisfy the equation ax — iy=±l. 302 APPENDIX. Aet. 5. Proposition IV. — If a and h are prime to each oilier, the equation ax — fry=±c, is always possible, and an indefinite number of integral values may he assigned to x and y, which will satisfy the equation. For ax — i!/'=±l, is always possible (Art. 4). . . c{ax' — by')=zizc, or acx' — bcy'=^±c, is always possible. Let cx'=x and cy'^y, then ax — by=:i±c, is always possible.. Let one solution be x=p and y=q, then ax — hy=ap — bp, or ax — ap=:hy — hq. __ a(a:— p) _^ ^ a„j x^p_h_mh ^ Ky—q) ' y—q « ^a' or, X — p-=mb, and y — q=ma ; .'. x=p-\-mb, and y=q-\-ma. and since m may be either positive or negative, and have any value whatever from to infinity, the number of values of x and y are indefinite. Cor. Since p and q are integers, and since m may be either positive or negative, m may be so assumed, that x shall be less than b, or that y shall be less than a ; for making m equal to 0, — 1, — 2, — 3, &c., successively, we shall have x=p,p — b, p — 2h, &c., and y=q, q—a, q — 2a, &,c., where it is obvious that one of the values of x must be less than h, and one of the values of y less than a, whatever be the values of j> and q. Art. 6. Proposition V. — The equation ax-\-by=c, is al- ways possible in positive whole numbers, provided a and b are pritae to each other, and c is greater than ab — a — b. For, if c^{ab — a — b)-\-r, the equation becomes ax-\-by=ab — a — b-{-r ; .-. ^_ gi— a— t— ty+) -.^,,_^_ (y+1 )b—r . a a Bince b — 1 is an integer, the possibility depends on VJ — i =z, being an integer. a Let 3/+l=!/', then we have az — by'= — r, which is alw«y« INDETERMINATE ANALYSIS. 303 possible, ( Art. 4 ); let then y be less than a, or y-\-\<^a (Prop. IV, Cor.), then in the equation ^"~^ ' ==z- a I mu?, be less than h — 1 , therefore a;=&— 1 — ^ithlltl!'=i— 1— z, must be some in- a teger number ; hence the equation ax-\-by=:c is always possible when a and h are prime to each other, and c^{ah — a — h). Remark. — The last two propositions are of great practical utility, in- asmuch as they show the possibility or hnpossibility of eouations of thia kiud. Art. y. Problem I. — To find positive integral values of x and y in the equal io7i ax — by=c, or, ax-\-hy^c, a and b being prime to each other, and c being either plus or minus. ax — by=c, gives x=JtJ^=:py-\-q-\---01-, where py-\-q rep- a a r?sents the integral part of the quotient, and b'y-\-c the remain- der, b' and c being less than u. Now in order that the value of J. shall be integral, the remainder b'y-\-c' must be divisible by y'=z\ , then acx' — hcy'=c, and — abm-{-ahm=iO, whence, a(cx' — im)-|-i(am — cy')=:c; but ax-{-by=c, .■. x=cx'- — Am, and y^am — cy'. Now it is evident that the number of solutions will be the same ail the number of values that can be assigned to ro that will ren- der bm less than ex' , and am greater than cy', bm.=1, then x'=9, and v'^4. Cj/'_ 254X4 _.)(,3, ^_^'=4. a 5 '' b a This result may be verified by actually finding the valdes of eandy. Thus, ar=9, 20, 31, or 42; and 2/=19, 14, 9, or 4. 2. Given 7a;+9y=2342; to find the number of values of « and y in positive whole numbers. Ans. 37. 3. Given ]la;-)-17!/^987, to find the number of values of x and y in positive integers. Atis. 5. 4. Given 9^+13^^2000, to find the number of solutions in positive integers. Ans. 17. 5. In how many ways can j£100 be paid in crowns and guineas, the crown being 5s. and the guinea 21s. Ans. 19. 6. In how many ways can £1053 be paid in guineas and moidores, the guinea being 21s. and the raoidore 27s. ! Ans. Ill ways. Art. 9. Problem III. — To find the integral values of x,y, and 2, in the equation ax-[-6i/-f-c2=d. Let c be the greatest coefficient in this equation, then since the values of X and y can not be less than 1, the value of z can not , d — a — b exceed . But jl, — by — cz a therefore, by operating on this equation, according to the method employed in Art. 6, we shall obtain a result of the form y ^ . ; let this equal w, then a y=aw^^z±r, where z may have any value from 1 to ° , that will give positive integral c jalues to X and y. EXAMPLES. 1. Given 3a;-l-5i/-f72=50. „ . 50— 3-,5 „ Here z am not exceed =0. INDETERMINATE ANAL^ISIS y—2z- ^Qi) 2—2y—z 3 'iy^ay, 2-2y-z y=2w-\-z—2 y-z+2 ■ 50— 5(3«)+K— 2)— 72 „„ . . ^= ^^ g — ^ =20 — ou) — 42, o If ic=l , a:=15 — Iz, let £= 1,2, 3, y=\-\-z, then x=ll, 7, 3, y= 2,3,4. If «;=2, a;=10— 40, let 2=1, 2, y= 4+2, then a:=6, 2, If 2c^3, a;=:5 — 4j, 2/=5,6. let 2=1, then a;=l, y=8. Therefore, the whole number of solutions is 6. When the number of solutions is numerous, the pre cess will become tedious ; but the object of inquiry in such problems is generally not to find the solutions themselves, but to determine the number of which the equation admits, the method of doing which will be explained in the next problem. 2. Given 2a;-|-3y-|-42=:21, to find all the positive integi-al values of x, y, and z. Ans. 2 2=3; _1 . 3^='^. 4, 1 ; I „ . fe=5, 2; -^ ■ )2/=l,3,5. r-^- ^y=l,3. 3. Given 2j;-|-5!/-|-42=27, to find all the positive integral values of X, y, and 2. 2 3. 2 4. Given 17a;-(-19y-|-2l2=400, to find the integral values of *, y, and z. Ans. z= 1, 2, 3, 4, 5, 6, 11, 12, 13, 14, y=ll, 9, 7, 5, 3, 1, 8, 6, 4, 2, 07=10. 11, 12, 13, 14, 15, 1, 2, 3, 4. Akt. 10. Problem I"V'".'=— To determine the number of solutiona of which the equation aa;-(-fr!/+ct=;d 2 3 4 5 1 1 1 1 1 3 7 5 3 1 4 310 APPENDIX. will ailmit, at least two of the coefficients, a, b, c, being prime lo (jach other. By Art. 8 the number of solutions of which the equaticn ax-\-by=c will admit, is expressed by the difference between the integral parts of — , and X, where x' and y' are to be found b a from the equation ax' — by'=\. Now in the equation ax-\-by-\-cz=^d, if we transpose cz, we have aic-\-lry=d — cz ; therefore, if we make 2=1, 2, 3, 4, &c., successively, the number of solutions in the equa- tions ax-{-by=d — c i '—, and -^ ^ b a ax+hy=d-2c e^'" ^^ ^he differ-^ (d-2c)x' ^ ^^^ {d-OcW ■lence of the inte->- b a 17 J Q tsrsX parts of ) (d — 3c)a;" , (d — 3c)v ax-\-by=d — 3c, " "^ ^ L , and ^^ '-2. b a &c. &c. Now the sum of these differences will be the whole number ot solutions of which the equation admits. Therefore, if we take the sum of the integral parts of the arithmetical series (d—c)x'{d—2c)x' ^ ^^ . {d—z c)x' ^ b b ' ■ • • ^ > and also of the arithmetical series {d^c)y'_^ {d—2 c)y'_^^ &c _[_ ('^— ^'g)y ^ a a a the difference of the two will be the whole number of integral solutions. Now in each of these series the first and last terms, and also the number of terms are known ; for the general term in the first series being ^ ~, and in the second, ^ ^'^^^ the b a extreme terms will be found by taking z=l, and z the greatest whole number in _ ; the last value of z being found by making x and y, each equal 1 in the given equation. It is also obvious that the last value of z expresses the number of terms in the series. Therefore, if we find the sums of the terms in each series, and deduct from each the sum of its fractional parts, we shall obtain the sums of the integral parts of each series. IN DETBRMINATE ANALYSIS. 311 In finding the sums of the fractional parts, since the denomina« tor is the same, it is obvious that the fractions must recur in periods, and that the greatest number of fractions in each period can never exceed the denominator, since any divisor, as b, can leave no other remainder than those from 1 to b — 1 ; hence, the shortest method of finding the sum of all the fractions, will be to find the sum of the fractions in one period, and multiply this by the number of periods. If there are not an exact number of periods, the overplus fractions must be summed by themselves, observing that they will recur in the same order as in the first period. Also, in the first series, _ must be considered as a frao *^ b tion. (See Art. 8.) EXAMPLES. 1.' Given 3a;-|-.'ij/-|-72=100, to find the number of solutions of which it admits in positive integers. Here 3x+.57/=10()— 72. If we make z=l, 2, 3, 4. . . 13, in succession, then the number of solutions, of which the equa- fijr fit/ tion Sx-\-dj/^d will admit, is expressed by — — -i , where x' and y' are to be found from the equation Hx — 5j/'=1, {x' being =2, and y'=l). Therefore, in the equation 'ix-\-!iy=H)() — 7z, if we take s=l, 2, 3, to 13, which is the limit to the value of z, the number of solutions in the equations 'ix-{-5y^9, Sx-\-5y:=lii, and to 3a;+5v=93, will be expressed by — — -i-, — — — _.?., ^ 5 3 5 3 „ , 93a;' 93»/ , &c., to — — ^, or by 5 3 ■' Wy , 16^ , 93^V _ ^9y' ley 93y'> ir}"^ 5 ■ ■ ■ ■ "^ 5 S rs "^ 3 ■ ■ ■ "^.3 S • Or, by substituting the values of x' and y' the number of solu lions will be expressed by the difference of the arithmetical seriea, 2.93 5 ' ""' -3-+~3-+-l-+^ -3-- The sum of the first series is 265|,and of the second 221. But as it is only the sum of the integral numbers in each that ia wanted, we must deduct from each the sum of the fractions in ir. 2.9 ,2.16 ,2.23 2.30 5+ 5 ' 1.9 ,1.16 ' 5 I 1.23 ' 5 , 1.30 3+ 3 " ' 3 ' 3 312 ArrENDix. The fractions in the first .series occur in the following order* h f ' 6' t' i ' ^"d as there are five terms in this period there will be (V='^tI) ^^^ ^"i^h periods, and 3 terms besides. i+§+5 +f+l=¥=3. and 3X2=6, |+f+,^=l^, and 6+1^=71, the sum ol the fractions in the first series. In the second series 3-(-|-l-|=3^1; y=^^3> hence there are 4 periods of fractions, and the whole is 1 x4=4. 265i— 7^=258^^ ; 221—4=217. 258j — 217=41 1, hence the number of solutions required is 41. 2. Given the equation 2a;-|-3^+52=41 , to find the number of solutions of which it admits in integers. Ans. 21. 3. Given the equation 5a;-j-7j'-|-llz=224, to find the number of solutions of which it admits in integers. Ans. 59. 4. It is required to determine the number of integral solutions of which the equation 17a;-|-2l2/-|-30z=3000 will admit. Ans. 40o. 5. It is required to determine the number of integral solutions of which the equation 7x+9!/+232=9999 will admit. Ans. 34365. Art. is. In the preceding problem it is required that at least '.wo of the coefficients shall be prime to each other. When this is not the case, the proposed equation may be easily transformed to another possessing the required property, as is shown in the following example. Given 12a;+152/+202=601. Transposing 20», and dividing by 3, we have 4a;+5y=200— Gz+lziHi ; o Z i •_^__2+___ jjgjjgg 2=3^ — 1 . whence 3 ^3 3 by substitution the proposed equation becomes 12a;+15y+20(3M— 1)~601, which, by reduction, oecomes ix-\-5y-\-'20u=20T . Now in this equation x and y have the same values as in the one proposed, and therefore the number of solutions must be the same. i N i« K T E U M I N A T E ANALYSIS. 313 Art. la. Problem V. To find the values of three unlinoiun quanlilies in two equations. !f two equations, containing three unknown quantities be giVeii one of the unknown quantities may be eliminated, and the value of the other unknown quantities found as in Art. 7. E X A M PL E S . 1. Given 3a;+.5j/+2z=40) to find all the integral values of 4x+4y-|- z=:33S x, y, and z. By eliminating 2, we obtain bx-\-?ixj=-'3.Q, then by Art. 7, we find a;=:l-)-3ti), and y=l — bw, and by substituting these values in either of the equations, we find z=l-\^Qw. ' By taking m)=0, we find x^\, y^l, and ;=1. " " w^l, we find a;=4,y=2, and 2=9, which are the only values. 2. Given x — 2y+:=5) to find the values of x, y, and 2. 2x-\- y—z=lS Ans. x=5,' 6, 7, ?/=3, 6, y, 2^6, 11, IG, &c. 3. Given 2x-{-6y-\-iiz= 51) to find all the integral values of 10a;-(-3^+2r=120i x,y,a.ndz. Ans. a;=10, y=2, and 2^7. Art. 13. Problem V. To find the least whole number, which being divided by given numbers, shall leave given remainders. Let X represent the required number ; a, b, c, &c., the given di- visors ; and/, g, h, &.C., the respective remainders. Then, by subtracting each of the remainders from x, and dividing by a,b,c &.C., we have — ^, i, , &c., where it is required to find a b c BUC.1 a vrlue of x, tha", each of these expressions shall be whola numbers. x-—f Let — i=p, then x=ap-\-f; substituting this value of a; in the a lecond expression, we have jLiZ- — ", in which it is required to b find such a value of p as will render the expression a whole num- ber. This may be done as in Prob. 1, Art. 7. Having found a value o( p, substitute it in the expression ap-\- 27 304 A P P E N D 1 X . Now if we divide Ic by a, and call the quotient q, and the re» mainder r, we shall have yj^=q-\-±Jl- , or i. — = — q-\-i. — , a a a a which ars evidently whole numbers when tllL, or i — are a a whole numbers. Now let ill — =w, a whole number ; then a y=zzaw — r, where w may be any whole number that will render y positive. In a similar manner, if r ia negative, we find y=^aw-\-r. It is evident that the same general method may be applied to find the value of y in the equation ax-\-by:=c. Since the subtraction of fractions does not produce any change .n the common denominator, this may be omitted in the operatioji, and we m.ay proceed according to the following Rule. — Reduce the equation to Me form x:=hy-\-c ; perform the d - vision of hy-\-c by a, and call tlie remainder Vy-\-c'. Take the difference between ay and that multiple of b'y-\-c' in which the multiple of b'y is the nearest to ay, and call tlie remainder V'y-\-c". Again, take the difference between Vy-\-c', and that muUiple of b"y-\-c", in which the mulliple of b"y is the nearest to b'y. And so on, till we get a remainder of the form y-\-k, or y — k. Lastly divide k by a and call the remai?ider r ; then y=aw — r, or aw-{-r, accord- ing as k is plus or minus ; and w may be any whole number that will render y positive. Having the value of y, the general value of x is obtained by substituting the value of y in the given equation. When the given equation is of the form ax-\-by=c, the value of y is found on the same principles, except that it may be necessary to add instead of subtracting, to reduce the coefficient of !/. The preceding rule depends on the principle that the sum or difference of two whole numbers is a whole number; and that ant/ multiple of a whole number is also a whole number. EXAMPLES. 1. Given Ix — 12y=15, to find x and y in positive whcle num bers. Here x=L-i^=y+2+^Mi, and a=T . INDETERMINATE ANALYSIS. 305 1y =ay by-\-\=b'y-\-c' , and the multiple =1, 2y — \=b"y-\-c", Ay — 2^pb"y-{-pc", where p='2, 5y+i =yy+c'. 2/+3= diff. of last two quantities, ^Ij^=w, and y==tw — 3, 7 Let jc=l,2, 3,&c. Then a;=9, 21, 33, &c., »nd y=4, 11, 18, &c., where it is obvious the numbei af values of x and y are unlimited. 2. Given 7a;-l-lly=47, to find x and y in positive whole num- bers. Here x=^:^l}y=.G^+^J=^V, and a=7. 5— 4y=i'y+c', 10— Sy=pi'!/+pc', 7y=ay, 10— J, io-y -i-j-3-y. -^^ — '+-^' 7 -^ x=l!=li^?=Z!f)=2+ll.. 7 ^ Let «)=0, then a;^2, and y^S, the only values. 3. How can 78 francs be paid with pieces of 5 francs and of 3 francs, and in how many ways ] Let x^ the number of 5 franc pieces, and y=: the number of 3 franc pieces. Then 5x+3y='78, 78—3!/ ,^ , 3— 3y 5 5 3 — 'ij/^b'y-\-c', ■' '^ 2G 316 APPENDIX. ^ Find two numbers whoso sum and product are equal. 4ns. a and — "— , wliere a may be any number whatever. a — 1 4. Find the number of solutions in the equation 9j;-(-13J r=2000. Ans. 17. 5. What formula gives numbers which, 'vhen divided by 3, 4, 6, respectively, leaves the remainders 2,3,4] Ans. x:=(iOp — 1. 6. Divide 1591 into two such parts that the one may be divisi ble by 23, and the other by 34. Ans. 1081, and 510; or 299, and 1292. 7. Into how many pairs of numbers may 350 be divided, such that one number, wiien divided by 5, shall leave a remainder 3, and if 5 be taken from the otlier number it shall be a multiple of 7 3 Avs. 10, one of Ihem being 23, 327. 8. Required the least number which is divisible by 5 and 7, but leaves 1 when divided by 6. Ans. 175, 9. Divide 100 into three such parts that the first may be divis- ible by 13, the second by 15, and the third by 27. Ap!:. 13,60,27. 10. A man buys oxen and horses for $1000, he gives $19 tor each ox, and $29 for each horse. How many did he buy ! Ans. 4-3 oxen, and 5 horses ; or 16 oxen, and 24 horses. 11. Divide the fraction V/ into two others whose denominators shall be 9 and 11. Ans. f and -j-^j. Suggestion, — Let x and y represent the numerators of the fractions, then ?4-l^=^^.^, or 11j-1-9»=U8, whence x=9;)— 4, and w=18— lip. 9 11 99 12. Find three fractions, Avhose sum is "f-!, and whose denom- inalors are 5, 7, and 11. Ans. |, i, y'j. 13. Find the least number which, being divided by 2?, 19, and 15, shall leave respectively the remainders 19, 15, and 11. Ans. 7691. 14. Divide 200 into two such parts, that if one of them be di. vided by 6, and the other by 11, the respective remainders may bo 5 and 4. DIOPIIANTINK ANALYSIS. 3l7 SucoESTiON. — Li_'t X and y denote liic quotients, then the parts will bo IJx-t-S, and lii/-^4 ; and i;x-|-ll//=191. Ans. 185 and 15, 119 and 81, or 53 and 147. 15. In what year of the Christian era, was the solar cycle 8, the lunar cycle 10, and the Roman Indiction lOi A?is. 15t)7. 16. A shepherd has a flock of sheep less than 200; when he counts them by fours, sixes, or nines, he has 3 over each time ; when he counts them by sevens or thirteens he has I over, and when he reckons them by elevens the remainder is 7. How many sheep has he 1 Ans. 183. 17. A person wishes to purchase 20 animals for 20£. (400 shillings) ; viz. . sheep at 31s., pigs at lis., and rabbits at Is. each ; in how many ways can he do itt Ans. Three ways ; one of which is 12 sheep, 2 pigs, and 6 rabbits. 18. A wheel in 36 revolutions passes over 29 yards; and in x of these revoltitions it describes z yards -\-y feet 4"'' inches ; required the values of a;, t/, and z. Ans. a;==13, y=l, z=:10. DIOPHANTINE ANALYSIS. Art. 14. The object of the Diophantine Analysis is, '.o render algebraic expressions containing one or more unknown quantities, exact powers, such as squares or cubes ; or, what amounts to the same, to find such values of a quantity as shall render a radical expression depending on it rational. Ex. Let it be required to find such values of x as shall render 4a;-|-5 a square ; that is, so that J4cX-f5 can be exactly deter- mined. If we assume 4a;-(-5=m^, we find x= — IT-, where m may be any number whose square is greater than 5. If m=3 x=l ; if m=4, a;=2|. Remark. — The Diopliantine is pioperly a branch of Indeterniinats Analysis ; it derives its name from Diophantus of Alexandria, in Egypt who lived about A. D. 350. In its full extent it is a comprehensive sub- ject, and has occupied the attention of some of the greatest mathema ticians. The following is designed to present merely the elementary principles of the subject. Those who desire a thorough knowledge of h are referred to " Euler's Algebra, with Lagrange's additions," " BarlowN Theory of Numbers," and "Legendre Theorie des Nombres" 318 APPENDIX Art. 15 Problem 1. — To find such values of x as will render rational the expression ^ ax^-\-lix-\-c. The solution of this prohlem assumes different forms, depending on the values ol a, b, and c. Art. 16. Case I. — When a=0, or when tlie expression becomes Jbx-\-c, Let ^bx-\-c=p, where p may be any number whatever then bx-\-c=p^, and x=C !_ EXAMPLES. 1. Five times a certain number, diminished by 4, makes a square, required the number. Here i=5, c= — 4, and x= ^ ' , where p may be any number 5 wliatcvur. If jci^6, then x=;8; if ^=1, then a:=l; if ^=2 then x:=l.Ci, and so on. 2. Find such values of x as will render the following expres- sions square numbers, and verify the result. 5x+3, 5x— 3, 10— 3a;, 3a:+|. Art. II 7. Case II. — When c=0, or when llie expression heeorres Ja.'c^-\-bx. Let Jax^-\-bx=p.v, then ax'-\-bx:=p-X', and x= , where p may be pny num- p^ — a ber whatever. EXAMPLES. 1 . Find such a value of x as will render 5x--]-8x a square. Q Here a=^5, i=:3, and x= p'—y If p-S,x=->,U p=-2 x=—S. 2. Divide tnc number a into two parts, such th.it tbelr product shall be a square. DIOPIIANTINE ANALYSIS. 31 Lot x= one part, then a — x=: the other, and their product is ax — x^ which it is required to make a square. Let ax — x'^=p''x-, whence x= — —^P being any number what- ever. If a—\Q let v='i, then a;=2, and a-^x—H. 3. Find such a value of x as shall render Ix'^ — 15a; a square. 15 7— b2 ; if^=:2,x=5. 4. Required a number such, that if its half be added to double Its square, the result shall be a square. p being any number. 2j!)'— 4 Aet. 18. Case III. — When a is a square, or wlien the expression IS of the form J a'^x''-\-bx-\-c. Let ^ a''x''-\-hx-\-c^ax-\-p, then a''x''-\-bx-\-c=a''x^-\-'2apx-\-p'^, whence a:^--TI^ , or -^ £-, 'Zap — b b — 2ap' EXAMPLES. 1. Find such a value of x as will render ^x--\-Zx — 7 a square. Here a=2, i=3, and c= — 7, whence x^i—±— Let »=X 3 — 4p' then a;=7|. 2. Find a number, such that if it be increased by 2 and 5 Bfiparately, the product of the sums shall be a square. Le'' x= the number, then it is required to make ixA^'Z){x-{-f>) t square. If p=i, oc=^. 3. Fina a number, such that twice the number int piised by ], multiplied by eight times the number diminished by I, shall be a square. Let a;= the number, then it is required to make I ^x-\-\){Bx — ^2) a '•quarc. If ^=|, a;=l jj. 320 APPENDIX. Akt. 19. Case IV. — When c is a square, or when the expression IS of the form J ax''-}-bx-\-c^. Let J ax'^-\-bx-\-c^=px-\-c, then aa;2_|_jj,_j_(.2__^2j,2_j_2pcar-)-c' - , 2pc — b whence x= . a — y' EXAMPLES. 1. Find such a value of a; as shall render Sx'-}-5x-\-Q a square. ][ p=l,x=z\. 2. Divide the number 16 into two parts, such that the sum of their squarrs slial) be a square. Let X and 16 — x represent the parts, then it is required to make 2x- — 32a;-)-256 a square. If 7)=3, then the ports are 9|,and6|. Similarly, we may find two numbers whose difference shall be equal to a given number d, and the sum of whose squares Bhall be a square. Akt. -i'J, Case V. — When neither a nor c are squares, hut ohen b^ — iac is a square. Let r and ?•' be the roots of the equation a;2_|_*a;4-?=0; a a .-. ax'-\-bx-\-c^=a(x — r){x — r ). Let ij ax--\-bx-[-c=p(x — r'), then ax^-\-bx-\-c^p'^{x — r')' ; •- a(x — r)(_x — r')=p''(x — r')^, a{x — r)=p-(x — r') ; 1 ar — bV whence a;= £1 — _ a — p^ Now the values of r and r' are '2a ' -2a which will be rational when J' -4ac is a square. IMOPMANTTNK A N A I, V S I S. 321 Lft h'^—4ar=tr-, theri d—b , , —d—l) r=_ — , and r = by substituti( za -la 'Za{a — /)-) E X A M 1' L K S . 1. Find such a value of x as will render the expressiuB ^x^-\-Vix-\-iy a square. Here 6'— 4ac=lG9— 144=25, and i=5. _ 30— 78+18p= 3»2_8 • ■ ^= ! i-= _i ; let p=2, then a;=l. 72—12;)^ 12—2^2 lr;5=2J,x=7i. 2. Find such a value ofa; as will render 2,r^-|-10x-|-12 a square. 3t|2 4 Ans. x=.i ; ify=4,a;=6; if p=|, a;= 2— p2 -j> "— -? . 3. Find such a value of x as will render Sa;' — Si-j— 5 a square. Aks. x= i_ ; if p^l , x=:2 3-jo2 ■' Art. 21. Case VI. When the proposed expression can be separated into two parts, one of which is a square, and the other t\t product of two factors. If none of the precedin<^ methods be applicable, still the solu- tion can be effected, if the proposed expression is equal to a square increased or diminished by the product of two factors. The dif- ficulty, however, consists in decomposing the e.xpression, which can only be done by trial. If ax^-\-hx-\-c={dx-\-ey-\-ifx+9){hx+k), let the latter = ^dx-{-e-\-p{fx-\-g)l^. Squaring this, omitting equal quantities on each side, and reducing ^_ p('2e+pff)—k h-p(2d+pf)- EXAMPLES. 1. What value of x will render 5x^ — 1 a square? Bv trial we find 5x=— 1=(2x)=+(j:— l)(x+l). *22 APPENDIX. Comparing this witli the formula, we have i=2,e=0,/=l 0=— 1, A=l, and /i=l ; whence x= ^-T" — Ifp=],a:=< ■ if p=]^,x=\. 2. What vahie of x will render 2x^-\-Sx-\-1l a square's Here 2x-'+8x+'7=ix+2y-\-(x+l){x+:i). ] f p=3 , X=i'i , Art. 22. When all the preceding methods fail, we mar often find, by trial, such a value r of x, as shall render ax^-\-bx-{-c a square. Having' done this, substitute y-\-r for x, and the resulting equation will be a(i/-\-ry-\-l)(y-\-r)-\-c=ay'^-\-2ary-\-bi/-\-ar^-\-l- -\-c ; but by hypothesis ar^-\-br-\-c is a square ; calling this n^, the expression becomes ay''-\-2ary-\-n^, which can now be rendered a square by Case IV, Art. 19. EXAMPLES. 1. Find such values of x as will render 6x^ — 10x^3 a square. By trial, we find x=2 renders the expression a square. Let x=y-\-2; then by substitution, and reduction, the expression becomes &y''-\-\^y-\-\. Let this =(^^+1)2, then y='^^~^^, 6 — f'^ and since x=y-\-2, we have x^^£ l£_Xr, from which, by giv- p- — 6 ing various values to j,, we may find as many values of x as we please. 2. Find a general expression for the value of x that will render 10-j-8x — 2x2 ^ square, which is a square when x=L Ans. x= ""' 4-1. Problem IL Tofnd such values of x as will render the expression 11X^-4'''''^"+^''^+'' '^ s^i.w.7-e. There are but two cases in which this problem admits of a direct solution : 1st, when the last two terms are wanting; or 2d, when the last term is a square. Art. 23. Case I. When the expression is of the form bx^-\-hx' Let xx^-\-bx-=(pxy^p-x'', then x=P^ flioi'iiANiiNji ANALvrsis. 323 EXAMPLES. 1 . Kind X such tliat 2x^-\-Qx^ shall be a square, 2. Find a number, such that 5 times its cube, increased by 10 times its square, shall be a square. Ans. z=3. 3 Find a number, such that 3 times its cube, diminished by 10 times its square, shall be a square. Ans. x=5. Art. 24. Case II. When the expression is of the form ax^-{-bx^-\-cx-\-d^. Let ax'+lx^+cx-\-d^= ( ~x+d ] ^=^'>+cx-{-d^ ; whence x^ Aad^ 1. Find such a value of a: as shall render a;' — x'-\-2x-\-l a siiuare. Here — =1. Ans. a:=2. ' 2d 2. What value of x will render 3a;' — 5x'-j-6a;-l-6 a square? Ans. a;=y|. 3. What value of x will render 2a;' — 5x^-\-l2x-\-4: a square! Ans. x=7. Art. 25. If we know one value r of x, that will render *x'-{-bx'-\-cx-\-d a square, we may find others as follows : Let ar'-\-br^-\-cr-^d=m'', and transform the equation ax^-\-bx'' -\-cx-\-d=0 into another whose roots shall be x — r, (Algebra, Art. 400) ; the transformed equation will be of the form ay^-{-b'y^-\-c'y-\-m^=0 . We may then, by Art. 24, find a value q o( y which will render this e.xpression a square, then the general value of x will be x=q-\-r. Ex, Find such a value of x, other than 2, as will render r'. — a;^-|-2a;-|-l a square. By suhstit'tiae y+2 for x, the resulting equation is 2/'+52/=+10y+9. By assuming this equal to (Ij'+S)^ and reducing, we find 24 A p p E N n 1 X pKOELEB III. To find audi values of x as shall render ax^-]-hx'-\-cx^-{-dx-\-e a square. Aet. 26. Cabe I. When the first term only is a square, thai is, to make a^x' -'rbx'-\-cx'-'-\-dx-{-e a square. Let a^x'' + lix^ -\- cx^ -{-dx-\- e=(ax -{- nix-\-ny=a^.x''-\-2amx^-^ {m'-{-2an)x^-\-2mnx-\-n^. In order tliat the first three terms on each side shall be lh» same, we must make m= — h=2a7n } , I 2a , I .-, V whence \ ■> a ■> t> c=m-'-{-2an) \ c — m-' 4a-'c — b-' _ [ " 2a~ Ua^ ' U is gives dx-\-e=2nmx-\-n'', and x^ d — 2mn EXAMPr, ES. 1. What value of x will render x* — 3x-J-2 a square! Ans. 3:='l. 2. Required a value of x, such that the expression 4a;''4-4x' -\-4x^-\-2x — G may be a square. Aits, x^l'i^. Art. fi7. Case II. When the last term only is a square, that ts, to make ax'^-{-bx^-\-cx''-\-dx-\-e^ a square. Let x=-, then the expression becomes y y' The numerator of this expression may be rendered a square by the preceding article, and the denominator is already a square, therefore the whole will be a square. EXAMPLES. J . What value of x will render Si'' — 3,r^+l a square 1 Here it is required to malie J)'' — Sjz+'-i a square. y=l .'. x^^^- When the first iiid last terms are both squares, the problem may be solved by either of the preceding cases. 1. Wliat value of x will render x'' — 6x^-|-'lx' — 24x+lG a square] Aiis. a;=j. DIOPII ANTING ANAI, YSIf5. li'-i.l Art. 29. Wr might now proceed to consider licw an expres aion of the lorm ax''-\-bx''-\-cx'^-]-dx-\-e can be rendered a square The geno-ai principle is, to assume the given expression equal lo such a '.uantity, that, after squaring, all the terms may disap- pear, or be made to do so, except those containing two consecu- tive powers of a; ; as the value of this quantity can then be obtained in a rational form. The terms to bo destroyed may be at the beginning of the given expression, or at its end, or both, according to its nature. Ex. 1. What value of x will render 4.x'<-\-12x^—2x^—2x-{-l a square. Assume this equal to (2x^-\-px-\-qy, then squaring and reduc- ing, we have 12a;' — 'ix^ — '2x-\-l=z4px^-\-(j)''-\-iq)x''-\-'2pqx-\-q'K Equating the coefficients 12 and 4p, and also — 3 a.ndp'^-\-4q,\ve get p^S, and ^= — 3; this reduces the equation to — 2x-[-l =2pqx-\-q-^ — 18a;4-9; whence a:=o. When we know one value of the unknown quantity that satis- fies the conditions other values mnv be found as in Art. 26. Ex. What other value of x, besides l,will render 3x' — 2 a square 1 By substituting y+1 fti" ^> '^^ get 3^''+12ji'+18)/^+12)/-l-l. Since the last term is a square, assume this equal to (py^-\-qy-\-iy, then the last three terms in each member will disappear by taking^=: — 9, and 9=6; whence y=i'i, and there- fore a:=f|. By assuming a;=j/-|-j|, and performing a similar process, we can find another vai'ue of x, and so on. Art. 29. Quantities of the form ax'-\-hx''-\-cx-\-d, can be ren- dered cubes on principles exactly similar to those that have been employed in rendering quantities squares. Thus, if a be a cube, we may destroy ti.c first and second terms ; if (i be a cube, the third and fourth terms can be destroyed : if a and d be both cubes, we can destroy the first and last terms. When neither a nor d is a cube, if we can tind a value r of x, which being substituted for *, will render the expression a cube, we may substitute i/-\-r for X, and then obtain an expression which can be rendered a cube by the principles just explained. As an example let it be required to find a value oT x which will make 2x^-{-d a cube. ' Here we see that x^ — 1 satisfies the conditions; to find another value substitute y — 1 for x, and the expression becomes 2y3 — Qy^-\-Gy-\-l . Assume this equal to (;)^/+l)^ and then by 26 ,1 P P E N D I X. taking p=2, to make the last two terms of each member disap pear, we get y= — 3, therefore x= — i. By substituting y — 4 for X, we might obtain another value, and so on. EXAMPLES. 1. Find a value of x that will render 2ix^-\-2x-\-\ a cube. Ans. x=i,^. 2. Find a value of x, besides x=\, that will render 2x' — \x -j-6x+4 a cube. Ans. x=\7^. 3. Find a value of x, besides x= — 1, that will render x--\-x-\-X a cube. Ans. x= — 19. DOUBLE AND TRIPLE EQUALITIES. Aet. so. Double, triple, or higher equaliiie':, are problems in which two, three, or more functions of a quantity are to be made squares or cubes, for the same value of x. Thus, if it be required to find a value of x that shall render both the expressions, ax-\-b, and cx-{-d squares, the problem presents a double equality. In the following solutions, for the sake of brevity, the symbol n is used to represent a square number. Pkinciple — It is sometimes convenient to use the following principle : If a square he multiplied by a square, the product will be a square , or if a square be divided by a square, the quotient will he a squaie. Art. 31. Case I. — To solve the double equality ax-\-b= □ , cx-^d= D . Let ax-\-b==p-, and cx-\-d=q-, then equating the two values ol X, and reducing, we find c'^p^=caq'^ — cad-\-c-h. Since the left member is a □ , 9 must have such a value as to render the right member a n > which may be ascertained by som" of the preceding methods. E.\. What vaUie of a; will render x~\ and 2x — 1 both squares' .1) j D I O r H ■ N T I N R A N A I, y i3 1 S . - 327 i\r.T 32. Case il.— To solve the douhk equal Itu. Let as=- then we have Hojice, by the principle, Art. 30, it is only necessiiry to render a-\-hij, and c-\-di/ both squares, which belongs to the preceding ■ problem. Art. 33. Case III. — To solve the double equality ax--\-bx-\-e=: D > dx^-\-ex-irf=a. Here we must solve the equality ax''-\-hx-\-c=:0, by methods already e.xplained, and then substitute the value of x so found in the equality dx--\-ex-\'f= O , which will rise to the fourth degree, and which must then be solved by methods explained in Art. 28. Art. 34. Case IV. — To solve the triple equality ax-{-hy= n ) cx-\-dy= D ) ex-\-fy=U. Put ax-\-hy=i^, cx-{-dy=u^, and ex-\-fy^=s''. By eliminating y from the first two equations, and x from the /7/2 lyii, fiu^ ct^ Bamt equations, we find x= , and y= ; substituting ad — be ad — be these for x and y in the third equation ; putting u=tz. and dividing the expression by f^, we have {_af—hcy^-{cf—de) _^ ad — be When it is possible the value of z may be found by Problenj 1, Articles 15 to 22. Having the value of i', we may assume t of any convenient value, this will give the value of u; then b}' substitution, the values of X and y are easily obtained. The preceding are the most general methods hitherto discovered. Tn tlie resolution of most problems, however, much will depend f 8 A P P R N D I X . cn the judgment a/id skill of the operator, and the most irap'^r- tant and ditiicult problems are solved by methods for which no Bpecial rules can be given. MISCELLANEOUS EXERCISES. i. To divide a given square number, a^, into two squares. Let x'^= one part, then a' — x'=^ the other. Assume a' — «' :=(a — vx)^, whence a;= v^-\-l .•. the parts are _i 1_, and , Suppose a^=100, then if v=2, the parts are 36 and 64 , '.f . , , 22500 , 6400 ■, V)==4, the parts are , and , and so on. ^ 289 289 By means of this formula we can divide a given square into any assigned number of squares, by first dividing it into two squares, and then subdividing one or both of these into others. The solution of this problem gives the following equation ; Dividing both members by a'\ and multiplying by {v'^-\-iy we have {v--\-iy=iv'—iy+'i:v'- ; Substituting c for v, and multiplying both members by q', we 9 get (p2_j_(^2)5=(jo2_y2)=4-4;)=9^. Hence, the square of p'-\-(j^ being equal to the sum of the squareti of jo^ — q- and 2^^, it follows (Legendre IV, 11,) that if p^-\-q^ be the hypothenuse of a right-angled plane triangle, y^ — ij^ and :ipq will be its legs ; this gives the following useful Rule. — To find the sides of a right-angled triangle in lohole numbers, tale two unequal whole munbers ; then the sum of their squares, the di_fference of their squares, and twice their product, wilt be tlie three sides. Thus, by taking 1 and 2, we find the sides to be 5, 3, and 4 ; if we tuive 1 and 3 the sides will be 10, 8, and 6. '2. To divide a number which is the sum of two known squares B^ and i', into two other squares. Let .r' be one of the parts, then a--\-b- — a'':=G. But one \'alue of X is i or a ; therefore, substitute y-\-b (Art. 22,) for x. OIOPHANTINE ANALYSIS. 329 and we get a- — if — 2?))/=n ; assume this equal to (a — ui/;^, and we find y=?.^Illt, ,. ..-,,+;,-K''— 1 )+^'»' Example. Let the given number be 185=4=-|-] ;'.-. Here a— 4, and 6=13; let d=2, then x=\\, and a^—i-jj^ anj 185—121=84=8=. If y=4 the parts are (-7^)^ and (■l:^)^. 3. To find three square numbers in arithmetical progression. Assume {x — yf, x--\-'f, and {x^yY for the three numbers, Wiiose common difference is 2xy, and of which the first and third are already squares. It only remains then to make x'^-\-y'^ a square, which may be done in the manner explained in the latter pan of the solution to Ex. 1. Thus, let the two unequal numbers be 2 and 1, then x, (the diff. of their squares,) =3, and y, (twice their product,) =4. Hence (a; — y)^, a;''-|-2/'i and (jc-|-y)= are 1, 2.J, and 49. 4. To find any assigned number {n) of squares whose sum shall be a square. By assuming as the required squares a?, h^, c' . . . and x^ where a-, b^, &c., are numbers assumed at pleasure, it only remains to find such a value of x as shall make a-+/<'+c-. . -]-x^ a square, which may be done by assuming it equal to (x-\-py, and resolving the equation so found for x. 5. Find two whole numbers, such that their difference shall be a square, and the sum of their squares a cube. Assume ix^ and Zx^ for the numbers ; then 4x^ — 2x-=x^^n, and it remains to make ('lx-y-{-(,Sx-y'=2ox'' a cube. Let 25a7''=aV, then x=:.\aK Let a=5, then x=5, and the numbers are 100 and 75. 6. To find three numbers in arithmetical progression, such that the sum of every two of them may be a square. Let x—y, X, and K-f j represent the numbers, then 2a;, 2x— !/, ind 2x-\-y must be squares. Assume 2x=r^-)-s', and y=2,rs, then the second and third will be squares, and it only remains to make r''-\-s'^ a square, which will be accomplished by making r=7n^—n^, and s=2mn. This gives 2x={m^+n'y, and the three numbiirs are l(m^-\-n-y —imT.(m^—n^), iim'+n^y, and i(m2+7i')=-|-4'""(™'— »')• If m=9 and n=l, the numbers are 482, 3362, and 6242. 28 so APPENDIX. This question may also be readily solved by assuming the three numbers 2x' — y, 2a;^, and ^x''-\-y, and then putting y=4x — 1 7. To divide universally any given whole number, N, into as many different square numbers as it contains units. Let ax — 1, hx — 1, ex — 1, &o., continued to N terms, represent a series of roots, the sum of whose squares is to be N. Let each of these be squared separately, and put the sum of all the coeffi- cients of x', that is, a^+^^+2^-]-, &c., =m, and those of x, that is, twice the sura of a, h, c, &c., =n, we shall then have mx'' — nx -1-N=N; from which x= — To apply this to a particular ques- m tion let it be required to divide the number 4 into 4 square numbers. Let a, h, c, and d=2, 3, 4, and 5, then m=a'~\-h'-\-c'-\-d- =4-f9+16+2r3=54; n=2a-\-2b+'2c+2d=i+Q+8+l0=28 .-. ^=38=1 4 and ax—1, bx—l,cx—l, and dx—l=-}^, -IJ, 3|, and |2, and the numbers are (oV)^ (2l)^ (M)'. and (|?)^ or 1 2Jd 84 J „-,j 1 84_9 VHV Tin' 72a> ''"" 7J9 ■ 8. Find three cube numbers, whose sum shall be a cube. Put x^, y^, and z' for the three cubes, and let their sum .-. 2/'=3x22+3x2;2. Put x:=pz, then y^=2p^z^-\-'ipz^=:z^(Sp^-\Sp). It is now required to make 3p^+3? ^ cube, which it is when p=l ; consequently if we make 2^8. p::=l, and y=6 ; .-. the three cubes are 1^ 6', nnd 8', whose sum is 9'. By making t= any multiple of 8, we may obtain as many integral solutions aa we please. The learner should observe that it is generally important to assume the numbers so as to satisfy as many of the conditions as possible. 0. Find two numbers, such that their sum and difference shall be squares A/ts. v^-^1, and 2v, or and 4, &.c. 10. Find two numbers, such that if each be added to the square of the other, the sum shall be a square. Aiis. ' \ and 8«+l , whicli are found by assuming 4x and x — 1 for the numbers D+1 DIOPHANTINE ANALYSIS. 33] 11. Fiiiil two numbers, such that the difference of Uieh- cuhca may be a square number. Ans. "i and ^"'T"'" 12. Find a number, such that the sum of its square and cube may be a square. Ans. v'^—l, or 3, 8, &c. 13. Find two numbers, such that if to each of tliem, and to their sum and diflerence 1 be added, each of the four sums may be a square. Ans. 168, and 120. Let x^-\-ix and x'' — 2a; represent the numbers. 14. Render 2x2 — 2 a square. Ans. ,r='" "^ ' »j2 — 2' First assume x=y-\-l. 15. Find two numbers, such that the difference of their squares may be a cube, and the difference of tlieir cubes a square. Ans. lOi)*, and 6i>* Assume tlie numbers equal to x^-\-2a^, and a;' — 2a^. 16. Find a number, such that if 1 be added to its double and triple, each of the results may be a square. Let 2x--\-'2x represent the number, then x= — J^ _ v'^ — 6' Ans. 40, 3960, &c. 17. Required three numbers, such that the sum of all three, and the sum of every two of them may be a square number. Assume 4.t;, x^ — 4a;, and 2a:-|-l for the three numbers. Ans. -3(v=-l), a's(«^-l)=-§(«=-l), and J(«=+2). 18. Find two numbers, such that their difference may be equal lo the difference of their squares, and that the sum of their , ■ 2i>— 2 , v''—2v Equares may be a square. Ans. , and •u' — 2 i;2 — 2 ■ 19. Find two square numbers, whose sum shall be equal to tlieii product. Ans. ^J^jtll', and (^+1)'^ q^ 35, and f 5 &c. iv' — 1)^ 41)2 " ii" 20. To divide a number vi^hich is the sum of three square num- bers in arithmetical progression, into three other squares which shall also be in arithmetical progression. A-ns. The numbers will be found by dividing one-third cf the given number into two squares, a- and i^, by Example 1, and taking (a — by, a''-\-V, and {a-\-hy a.s the required numbers. iiU2 APPENDIX. 21. To find three whole nuiiibers in arithiiielical progression who.se common difference shall be a cube, the sum of any two diniiniohed by the third a square, and the sum of the root.s of the required squares a square. Ans. 26980713761144832, 51885988002201600, 76791262243258368. PROPERTIES OF NUMP, ERS. Art. 35. Definitions. Even numbers are those which are divisible by 2. Odd numbers are those which, when divided by 2, leave a remainder 1. An even number is represented by the formula 2n, and an odd number by the formula 2n-{-l. A prime number'has no divisor except itself and unity. Numbers are prime to each other when they have no common divisor, except unity. M Art. 36. If — be a fraction, and if d be the greatest com- mon divisor of M and N, so that M.=ad, and N=M, then = ^; and ^ is the fraction _ in its lowest terms, and a is N /' i N prime to b. Tliere can be no other fraction " when p is prime to q, 1 which shall be equal to . ^ ; for if so, M and N would have two N greatest common measures, which is absurd. Art. 37. I. Every nwnher N may he expressed by ike formula N=}ra-|-J-. For if the number N be divided by q, and if n he the quotient ind r the renuiinder, then by the principles of division, N =qn-\-r ; q is called the mndiilu!:, and by giving different values to q, differ- ent forms of numbers may be obtained. It is evident that r can not e.\ceed q — 1. t R O P H R T I E S OF NUMBERS. S.'?ri Art. 38. II. Every number i> of one of the forms, tin, O" 3n±l. 0-jniparing this with the general formula N=qn-\-r, we have q=S, r=0, 1, or 2. .-. N=3m, or 3n+l, or 371+2; but 3n+2=3ra+3— l=3(re+l)— l=3n— 1. .•. every number is one of the forms 2n, or 3n±l. Art. 39. III. Every square number is of OTie of the forms An, or 47i-l-l. Every number is either 2n, or 2ra-j-l. If N=27i ; N2=4re'=4n', if n''=?i'. If N='ira+1; N==4K=+4™+l=47i(n+l)+l=4re'4-l, if /t(re+l)=ra'. Hence, N^ is of the form 4?!, or 'H-|-l ; that is, every_ square number is either divisible by 4, or, when divided by 4, leaves anity for its remainder. Art. 40. IV. The difference between the squares of any two odd lumbers is divisible by 8. Let M and N be the numbers, M being ]>N. Also, let M=2m+1, and N=2ra-| I, .-. M'— N2=4(m2— re=)+4(m— ra) =4(m+ra)(m — ra)-|-4(m — re)=4(m — n)(m-{-n-{-l'), '; .-. m=3, n=2, and r=l; .-. number of divisors =4x3x2=24. Akt. 5S. XV. To find, a number N Qial shall have a given num- ber of divisors. Let d represent the given number of divisors, and resolve it into factors, as d=lyiuy.v. Take m=t — 1, n^u — 1, r=v — 1 &c., and let a, b, c be any prime numbers whatever, then N=a'" . b" . cT, &LQ,., as is evident from the preceding proposition Ex. Find a number that shall have thirty d'visors. First, 30=2x3x5; .-. m=2— 1=1, 7i=3— 2=1 r=5 — 1=4; •. N=:a6V is the requireu number If a=2, J=3,c=5; then 2x32x5''=11250. If a=.5,6=3,c=2; then 5x3 = X2''=720. If a=5,i=2,c=3; then 5x2 = x3-'=1620. If a=3,i=5, c=2; then 3x5 = X2^=1200. Each of these numbers has thirty divisors, and in the same manner various other numbers might be found having the same property, by giving iz, i, and c other prime values. PROPERTIES OF NUMBERS. 337 Art. 52. XVI. To find the sum of the divisors of N=a"' . b" . c^ Since every divisor of N is contained in the product, (Art. 50), vl+a+o', &c., +a")(l-|-i-)-J2, &c., +i")(l+c+c', &c., +c"), and since by the rule for summing a geometrical series, (Alg. Art. 297), 1+a+a^ +a'»=?_l_^, a — 1 IJfh + h^ -ft"=!i_ i, &C. . . the sum must be ( ^- ) I H^ ) | ? — '^^ ] . Ex. 1. Find the sum of all the divisors of 360. 360=23x32x5; therefore, Ex. 2. Find the sum of the divisors of 28, the number itself being excluded. Here 28=2x2x7=22x1'; .'. the sum of the divisors is ihe number 28 itself, the required sum is 56—28=28. A perfect number is one which is equal to the sum of all its divisors (not including itself). Thus 28, which is equal to l-f-2-j-4+7-f 14, the sum of its divisors, is a perfect number. Otlier perfect numbers are 6, 496, and 8128; there are onlv eight perfect numbers known. EXERCISES. 1. Prove that n^ divided by 4 can not leave 2 for a remainder, n being any whole number. 2. Prove that no number can be a square which has any f na of the numbers 2, 3, 7, 8 for its last digit. 3 Prove the following properties of a square number . 29 338 APPENDIX. (1). A square number can not terminate with an odd numbef of cyphprs. (2). II a square number terminates with 5, it must termina e wiUi 25. (3). No ''quare number can terminate with two figures the same, exceot they be two ciphers, or two 4's. 3. If each of the quantities a, b, n, be a whole number, show that jSfl-f- n — l)i|- is always a whole number. 4. Show thai, x^ — 5x^-^ix is divisible oy 120, when x is any positive wbole number. Suggestion, x^ — !ix^-\-4x=x^(x^ — 4) — x(x^ — 1). 5. Prove that if any square number be divided by 12, the remainder is a square number, that is, that it is 1, 4, or 9. 6. Find the number of divisors of 1000. Ajis. 16. 7. Find the number of divisors of 2160, and also their sum. Ans. 40, and 7440. 8. Prove that the product of two difTcrcnt prime numbers can not be a square. SCALES OF NOTATION. Art. 53. To explain the different systems of notation. Def. Notation is the method of representing numbers by symbols; and it comprises different scales dependent upon the number of the symbols or figures employed. In the common system of notation, each figure of any number increases in value in a tenfold ratio in proceeding from right lo left. Thus .5432 is equal to 5000+400+30+2 =5x1000+4x100+3x10+2 =5xl0'+4xl0=+3xl0'-i-2 The figures 5,4,3,2 are called di^/te, and the number 10, eccording to whose powers they proceed, is called the radix of the Bcalc. It is purely conventional that 10 should be the radix; tho choice of it has probably arisen from the circumstance of our SCALES OF NOTATION. 339 Qa\ing' ten fingers on the two hands. There may bo any number of different scales, euch of which has its .own radix. When the radix is 2, the scale is called Binary; when 3, Ternary; when 4, Quaternary; when .'3, Quinary; when 0, Senary; when 7, Septenary ; when 8, Oclary ; when 9, IVonary ; wlien 10, Denary ; when 11, Umlenary ; when .2, Duodenary ; and so on. If 5432 represents a number in the Senary system, whose scale is 6, it may be represented thus, 5x6'+4x6=+3x6+2; or, inverting the order, 2+3x6+4x62+.5x6^ And generally, if the digits of a number be a„, a,, mj, uj, &c., reckoning from riffhl to left, and the radix be r, the number will be represented by Or, if there be n digits, by reversing the order of the terms, the number will be expressed by a„_,)-"-'4-a„_s,?-"-2+o„_gr'"-3+. . . . -]-a,r+a„. In any scale of Notation every digit is necessarily less than r, and the number of the digits, including 0, is equal to r. Also, in any number, the liighest power of r is less by 1 than the numha' of digits. Cor. Hence the digits including the cipher, in the Binary scale are 1, 0. Ternary " " 1,2,0. Quaternary " 1,2,3,0. Quinary " 1,2,3,4,0. And so on. In the Duodenary scale it will bo necessary to add two charac- ters to represent ten and eleven ; we, therefore, for ten put I for eWen e. i .-. Duodenary digits are 1,2, 3, 4, 5, 6, 7, 8, 9, t, e, 0. Akt. 54. To express a given number in any proposed scale Let N be the number, and r the radix of the scale Then if a„, «.,, a^, &c., be the unknown digits 'N=a„-\ra,r+a2r^-\-a^r^-\-, &c. If N be divided by r, the remainder is a„ , if the quotient be divided by r, the rem. is a, ; if this quotient be divided by r, the rem. is a^, and so on, till the last quotient is 0. The last remaindei will evidentlv be the figure in the highest place. 5 329 5 65, 4; 5 13, 5 2,3 340 APPENDIX. TJierefore, all the digits a„, a,, a^, a^, <^e., are the successive remainders obtained by dividing the given number N, and the succes- sive quotients, by r tlie radix of the proposed scale. Ex. Transform 329 in the common scale, into the quinary scale whose radix is 5. 329 1" remainder a^^i, 2'' " o,=0, 3" « 0^=3 1)7T 4" " 03=2- .•. the number required is 2304. To verify this result we must have 4+0x5+3x52+2x5'=329, which is found to i*. correct. By the same method a number may be transformed from any given scale to any other of which the radix is given. But in per- forming the division, it must be recollected that the radix is not 10, but some other number. In general, it is best to change the number to the denary scale, and then from that to the proposed Bcale. Ex. Transform 32.56 from a scale whose radix is 7 to a scale whose radix is 12. Observing that the digits in 3256 increase from right to left in a sevenfold ratio, the division by 12 is performed thus, =4, =1. 12 3256 12 166, 4; .-. 1'' remainder 0,,= 12 11,1 2'' remainder a,= 0, 8 .-. S"* remainder a^ = .■. the required number is 314. Or thus, 3256 in the septenary scale is 0+5x'7 +2x7-^+3 X'7'=1168. 12)1168 12) 97, 4=a„, 12) 8. l=a,. Ans. 814 0, 8=0,. AsT. 55. In any scalp of Xotalion whose radix is r, a number N when (lit'idirl hy r — \,leavis the same remainder as the sum of its digits leaves when divided In/ r — 1. SCALES OF NOTATIOS. 341 For, let N=a-f ir+cr2-(-(/H+, &,c., =Kr— l)+c(r2— l)+ — Recurriny — Reversion of --■'••• •♦*---- 22t — 225 Uontiuued Fractions- .... ..».«.'., ,._... 229 Logarithms -■• 232 Exponential Equations • - . . - 237 Interest and Annuities 242 General Theory of Equations 2J/ Transformation of Equations .••" -i Equal Roots 23 Sturm's Theorem * 257 Kesolufcion of Numerical Equations — Rational Roots * 261 Horner's Method of Approximation 273 Additional Examples in Higher Equations - 283 Approximation by Double Position ^.., 280 Newton's Method of Approximation 292 Cardan's Solution of Cubic Equations - ^.. 293 Reciprocal Equations — 295. Binomial Equations • -«• 298 APTENDIX. Indeterminate Analysis • SOO Diophantino Analysis 317 Properties of Numbers 332 ^JcaJes of Notation •-• -..► 338 Algobraio Paradox* -.-- ,....., 34j, ECLECTIC EDUCATIONAL SERIES. RAY'S MATHEMATICS. 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