UC-NRLF *B S3E BED 08- LIB R ARY * * UNIVERSITY OF CALIFORNIA. Received T^z^ > , 189/ . j| Accessions No. -4*2- f 33 Shelf No. OS* So Mo FHE3L3LEF A KEY TO THE TENTH EDITION OF BONNYCASTLE'S INTRODUCTION ALGEBRA; IN WHICH THE SOLUTIONS OF ALL THE QUESTIONS, THAT HAVE ONLY THE ANSWERS ANNEXED TO THEM, IN THAT WORK ARE HERE GIVEN AT LENGTH, IN A MANNER CONFORMABLE TO THE PRESENT STATE OF THE SCIENCE. By JOHN BONNYCASTLE, PROFESSOR OF MATHEMATICS IN THE ROYAL MILITARY ACADEMY, WOOLWICH. LONDON: PRINTED FOR J. NUNN ; LAW AND WHITTAKER; LONGNfAN AND CO.; CADELL AND DAVIES; JOHN RICHARDSON; BALDWIN, CRADOCK, AND JOY; SHERWOOD AND- CO. J AND JOHN ROBINSON. 1816. )8/| r %03$ T. B-ENdtEY and Son, Bolt Court , Fleet Street, Londo* CONTENTS. Page Practical Examples for computing the numeral Values of various Algebraic Expressions, or Com- binations of Letters I Addition , 2 Subtraction 3 Multiplication .- 4 Division 7 Algebraic Fractions 14 Involution . . , 26 Evolution 28 Irrational Quantities, or Surds 32 Arithmetical Proportion and Progression 4J Geometrical Proportion and Progression 48 Equations 49 Resolution of simple Equations ib Quadratic Equations 73 Questions producing Quadratic Equations 79 Of cubic Equations go Solution of cubic Equations 92 Of biquadratic Equations 96 IV CONTENTS. Page Resolution of Equations by Approximation. ...... 101 Of Approximation by Position ..-..,.. 104 Exponential Equations 7 108 Binomial Theorem Ill Indeterminate Analysis 120 Diophantine Analysis . 133 Summation and Interpolation of Infinite Series. . . . 156 Logarithms. 176 Multiplication by Logarithms ib' Division by Logarithms 177 Rule of Three by Logarithms ib. Involution by Logarithms, ......... 179 Evolution by Logarithms ....... 180 Miscellaneous Examples 181 Miscellaneous Questions , 183 Application of Algebra to Geometry 21? Page 29, Ex. 4, for 13# 2 read 13# 3 +,r 2 , and correct ac- cordingly. Page 47, Ex. 8, put = before the last expression for the answer. Page 59, Ex. lO, put = between y and the last expres- sion. Page 115, Ex. 11, for the numerator — 2.4.7^, in the line preceding the answer, read 2A.Jx*. 6 KEY TO BONNYCASTLE'3 ALGEBRA; Practical Examples for computing the numeral Values of various Algebraic Expressions, or Com* binations of Letters. JtvEauiRED the numeral values of the following quan- tities - } supposing a=(5, bzz5, c~4, d=l, and e=0. 1. 2a~ + 3bc -5d-72 + 60~5=127. 2. 5a z b — 10ab*-{-2e=:gOO~ 15004-0= -600. 3. 7d t + h-cxd+ezz252+5-4+Q~253. 4. 5\/ab-\-b*~- 2a&-e°— 5^/30 + 25 — GO-0 = #-7.0138/1. a , a— b 6 1 2 1 c d 4 14 2 6. 3 v / c+2v/2a+^— J=6-f2y / l6~l4. 7. as/a 2 -h £'.-|-3fov/ 3/ 3 * +3/ 3 Ex. 2. xP + x'y + xy'+y* x -y x 4 +x s y + x 2 y 2 + xy i — x*y — jt 2 2/ 2 — :n/ 3 — y* /r4 ■* -* * y 4 Ex.3. ^-f-ry + y* x A -\ r x 5 y+x' 1 y' 1 — ^ — - x*y* —xy 3 +xy+xy$+y* x 4 * +#y * -fy* Ex.4. 3**— 2^3/ + 5 3*-f-2#y — 3 3* 4 — 2:^4 5J 1 + 6dr 3 y-4.ry + 3 0.ry 2x 4 + 4x ? y-4xy+ lOn/ -4.:'-- IS ( 6 ) Ex. 5. 2a*-3tftf-f 4#* 5a?~<5ax~- 2x % — 1 2« 3 ar -f- 1 aaV— 24a# 3 10fl 4 — 27a 3 x+34&*x*—.l8az a --8* 4 Ex, 6. 5tf 3 -f 4ax* + Zcfx + a? 2x r -3ax + a* 10^ 5 + 8aa; 4 -f 6a*x 3 + 2a?z* — 1 Sox 4 — J 2aV— paV— 3a 4 # -f 5a** 3 + 4a 3 # 2 -|-3a 4 af-f-a s 10.r 5 -7a^ 4 - a¥~ 3aV+ a 6 Ex. ?. 3,r 3 + 2jry + 3^ 3 2ar 3 — 3ary+5?/ 3 6# 6 -f 4x & y 9 -f 6x 3 i/ 3 &**_ 5 j?y — tk 4 ?/ 4 -f- 2 1 #y -f ^t/ 5 + 153/ 6 Ex.8. jj 6 — ax 2 +bx--c x^—atf + bxP—cx* —dx*-\-adx 3 — dbx? + dcx _»_ e #3 _ aeari + i ex _ ec jc>-(a + d)x 4 +(b + ad+elx 3 ~(c + dl + ae)x* -]~(dc + be)x~-ec < t ) DIVISION. CASE t. When the divisor and dividend are loth swtpi? quantities. 16 i* . 12a*r a 3.r Ex. i. \6x 9 -~Qx, or ---- =c2,r: and — = t- n , —1 5 ay 9 , — IBa&y Ex.2. — \5ay z ^-Zay i ox -=r-52/;and 2 J J 3ay J -8cu t^ „ 2 f 4 | -2 5 5 Ex.3. — a--*-- «* = x- = — ^5 and 3 5 3 4 6' f 3 f * 5 f -1 = a,r -s a # = crxn 5 3 CASE II. When the divisor is a simple quantity, and the dividend a x compound one. 3 r 3 + 6*2 -f- 3 ax — i &v Ex. 1. Here — —.**+ 2x+a-5 „ J* 3a£c + \2abx — Qa*v Ex. 2. Here -^ ;=:c+4*~30. *» 3a& 40a 3 Z' 9 + 60a 8 #»— 17a£ Ex. 3. Here ZSL = _40a 2 £* — av ~60a£+l7. Ex. 4. Here : =3a£c-2?ca* 5a 5 +** Ex.5. Here &i ^—7=l4x\-3x~YI, or 7* + 1. ^ . ( 8 ) CASE III. When the divisor and dividend are hath compound quantities EXAMPLES FOR PRACTICE. Ex. 1. a— x)a q — 2az + a?(a— x a- — ar — ax-\-x' i r* ax + x* Ex. 2. x— a);r 3 -3ax» + 3a 7 x— * s (x'~2ar-f "* # 3 — ar 9 — 2a^ 2 + 3a 2 l r — 2az* + 2fl a tf a 2 ,r— a 3 dftr — a 3 Ex. 3. a + x)a? + 5a*x±5ax' i +x3(a x + 4ax.+ x* a 5 + a?x Aarx+Sax* Aatx -|- 4ax* atf + x 9 ax* + x 3 M •* ( 9 ) Ex. 4. y-8)^ 3 ^iey+2^-l7(2^-3y + 2 + -i r - 3y» + 2% 2J/-17 2y-lg Ex.5. *+l) 1 r»+i(^-a? s +«'-*+l —a 4 +1 •*+ 1 Again, *— l)z e -l (*»+**+ ^+^+^4. j or 6 — k r 5 *v -1 tft -*" &- - ! a»- -4f X~ 1 A'— 1 f ty/M/-^ ( 10 ) ~-4ax* + 6ox — 70a*x+105a s — 70a a #4- 1 05a? Ex 7. 2^+3^-l)^ 4 -9^ + ^- 3 ( 2/ - 3r+1 ' 4x* + 6x*-2x' i ' —6.x 3 — 7x*+6x — 6x>— 9x1— 3x 2x* + $x— 3 2tf a + 3*-I 2 remainder Ex. 8. ^-fl^ay-^+2fl 5 ^fl 4 (^ + (W - aa X+-OXS+ aV ax z -2a*x 2 + 2a i x ax 5 — aV+ a*x -cftf + aix-a* ~az* + a?x-a* Ex. O. 3#-0)6**~9<5 (2x* + 4x* + 8x+ 16 6^ 4 — 12JJT 3 ! 2^—95 i2* s — 24a? 24x a -96 24^—48^ 48*— 96' 48r— Q() ( 11 ) Again, a+jr^+ar 5 (a 4 — a*x+ 2a:4-3)32* 5 + 243 (16* 4 - 24.x 3 4- 3 6**-- 5 4* 4- 81 32.v 5 + 48X 4 — 48,r 4 4-243 — 4ar 4 ~72# 3 f24£ f 108x a 72a; 3 72**- -108**4-243 -]08a*~l62;r 162*4-243 1 62* +243 ( 12 ) Again, x - a)x 6 — a 5 (a: 5 -f ax 4 -f a\r 3 -{• a 8 * 8 + a A x + a* # 6 — ax 5 ax b — aV a 8 * 4 — a G a\r 4 — a** 8 a',?*— a a 4 * 8 — a 6 a 4 .z 8 — a 5 # a*x— a° Ex. II. b-if)h*~Zy* (is + l*y + ly*+y Z- 4 -% Vy-Zy* bhji-ly* — 2y 4 remainder. ( 13 ) Again. a + 2£)a 4 + 4a 2 £-h8£ 4 (a s -2a 2 £-f 4ab + 4ab*- a* + 2a 3 b [8£ 2 +S& 5 — 2aH + 4a?h -2a 3 b—4a?b* + 4a (2 b+4a*b* -J-4a 2 £-f8a£ 2 -f-4a 2 Z; 2 — 8a h + 4(2^+ 8ab 3 — 8ab*- — Sab*. - 8ab 5 -\6ab 3 + 8ab 3 + 8b* + 8ab 3 +8b* Ex. 12. x + a)x*+px+q(x+(p-a)-^+£?- +&c. x x 2 & + ax (p-a)x + q ' (p— a)x+pa— a 2 -pa+aP + q pa? -pa-- — x pa? x r * +pa? pa} x x 1 &e. &c. tiryim mF^a*+>I J,/?, /fa. ( 14 ) Again, :r— a)oP— px* + qx— r(* 2 -f (a -p)^-j-(a*— »ap-\-q) cF—ax* -f&c. (a— p)^ z + ^ (a— p)# 2 — (a 2 — «p)# (a 3 — ap-\-q)x. (a 2 — tfp -j- q)x—a{d 1 —ap + ^) + a 3 — tfp + aq &c. &c. ALGEBRAIC FRACTIONS. CASE I. To find the greatest common measure of the terms of a fraction. Ex. 4. Here **— a 3 )**— a 4 (v a? 4 — tf 3 # fl 3 ^— a 4 Divide by a 3 , then x — a) x 5 —a 3 (f^+ax+a 9 x 3 —ax* ax*— a 3 ax*—a?x cfx—a? a a x—a 3 Therefore x — a is the greatest common measure sought. -V ( 15 ) sCo -P^r^j Ex. 5. Here a 3 — (px—axt + ^tf—x* .... (a tf 4 — a 3 ^ — a*x* -f a at 3 a 3 # + a*x* — ax s •— J 4 Dividing the remainder by x a 5 +a\r—ax' 2 —x i ) a 3 — a 2 *— ar 2 -}-^! a 3 + a 2 ^— aA 2 — jr 3 —2^ + 2^ Divide by 2 x-, — a i + x*)a 3 + a z x—ax % — x s ( — a a 3 — a x* . - ci 2 x— X s Divide by X) a 2 — x*)-a*-fx q (— 1 — * 2 +ar 8 Therefore a 2 —* 4 is the greatest common measure sought. It frequently happens that the common measure of quantities of this kind is better found by resolving both numerator and denominator into their component fac- tors, in which it will be useful to remember that the dif- ference of any two even powers is divisible both by the difference and sum of their roots 5 and that the differ- ence of two odd powers is divisible by the difference of their roots, and their sum by the sum of their roots 5 thus, in the last example, a 4 — x* (a* + ^) (a 2 — x a ) a 3 — a?x— ax Li +x?~~ (ai—x^a— (a 2 — x*)x where it is obvious that cP—x 9 ' is a common measure of both terms ; and dividing by it, the fraction reduces to a* + x* - . > which is in its lowest terms; and consequently at—x* is the greatest common measure, as was found by the former rule. And the same method may be advantageously used in other examples of this kind. c2 ( 16 ) ' EXAMPLE 6. Here #*-f ay+a 4 )A' 4 +fli 3 -A- a\\ Divide " c^-a¥-^-2-fl 4 byajX 3 — or*— a 9 ,z— 2a 3 ) # 4 -f a V 2 + a 4 (x+a x 4—ax 3 —aW—2a 3 x ax 3 + 2ci 2 x' i + 2a 3 x + a* a* 3 — ay- -a 3 ,r--2a 4 3aV 2 -f 3a 3 x + 3tt 4 Divide by 3a q , and we have xt+ax-t-a^x 3 — ax*— a?x— 2a 3 (x—2a x 3 + ax* + a?x — 2ax*—2a' i x—2a 5 — 2ax i —2a' 2 x—2a 5 Therefore x 1 -f- ax -f a 2, is the greatest common mea- sure sought. Ex. y. Multiplying the denominator by 7, we have 7a*— 23ab + 6l>' i )35a 3 — \26a*b + 77ab*— 42b 3 *(5a-llb 35a 3 —115a 2 l> + 30ab z — lla z b + 47ab i —42b 3 Multiply by 7 — 77a z b+329ab*—2g4b 3 — 77a 2 b+253ab 2 - 66b 3 76ab*-+228P Dividing by 76b*, and we have a -3b)7a' i —23ab + 6b z (7a- 2b Ja <1 —2lab — 2ab + 6b* — 2ab + 6b z * Not b 2 , as it is printed in the work itself. tdj^U^ ^/uiAyisu^^^ ( 17 ) Therefore a — 3b is the greatest common measure sought. CASE II. To reduce fractions to their lowest terms. x*—a 4 (;r 2 -f a 2 )(x 2 -a 2 ) x* + a* Ex. 4. Here — = -— — = — — - x ° — aV x 3 (x 2 —a*) x z by dividing both terms by x a —a 2 . Ex. 5. 6a q + 7ax— 3# 2 )6a a + ]lor + 3* 2 (l 6a*+ 7ax—3x* Aax + Gx' 1 Divide by 2x : 2a + 3x)6a*+7ax— 3x*(3a—x da^ + Qax -2ax—3x 2 — 2ax—3x x ^ r ,6a 2 -f 7ax— 3x~ 3a— x : _ Therefore 2a + 3;r)- -■ -== ,thefrac- tion sought. 2^ 3 — 16jp~6 2(.i 3 -8ar — 3) 2 Ex. 6. Here — ; =— —„ -= - 3x 3 — 24* — CJ 3(* 3 — 8.T-3) 3 Ex. 7- There is an error in the printing of this example in the work itself -, it should have been Qx b + 2x 3 + 4x z -x-\-l ]5;r 4 — 2.z 3 -f ICto 2 — x+ 2 This being corrected -, multiply the numerator by 5, and we have nj^Ut^GJ-?^ pT x/ /L~m-sis*-*T~^Ts ' ^^y - ( 18 ) I5* 4 — 2* 3 +10a:*— # + 2)45.r 5 +iar 3 -(-20ar 2 — 5x + 5(3x 45x 5 —6x 4 + 30x 3 — 3x <2 + 6x 6x i —20x 3 + 23x 2 -llx-\-5 Mult, by 5 15x 4 -2x 3 +10x <2 -x + 2)3Qx A — 100x 3 + l\5x* — 55x + 25( + 2 30>r 4 - 4x 3 -\-2Qx* -2x 4* 4 — 9&r 3 +95^— 53#+2l Multiply the last divisor by 32, and we shall have — 96x 3 +95x*-53x + 2l)4SOx 4 — 04x 3 + 3202 2 — 32x + 64(—5x 480x i —4/5x 3 + 265x < >-\05x 411^+ 55x* + 73X + 64 Mult, by 32 -96,r 3 + 95a 2 — 53^ + 21)13152^+ 1760^ + 233(^ + 2048 (137 13152a: 3 — 13015x 2 + 7263^r -2877 14775^—4925^—4925, Dividing the latter by 4925, it becomes 3x— x— 1 ; which, by another opera t ion, exactly divides — 9&r 3 + 95a 2 — 53ar + 2l5 and therefore this is the common measure y 3ar 3 -+jr <2 +- 1 and the reduced fraction is 5x*-\-x +2* CASE III. To reduce a mixed quantity to an improper fraction.. 2x lxa—2x a—2x „ Ex.3. Here J = = Ans.. a a a Ex. 4. Here 5a- ( 19 ) 3x~ b 5a*—(3x—l>) 5* 2 —3x+$ a a a the fraction required. ax-\-x 2 2ax—(ax + x 2 ) ax—x* Ex. 5. Here x = - — — 2a 2a 2a 2x—7 l5x+2x — 7 \7x—7 Ex.6. Here 5+——= £ ~"^— '• 3o? 3a? 3# #— a— 1 a — (# — a — l) Ex.. 7. Here 1 = — * -= a a a — x-r-a-\-l 2a—x-\-\ ' = Ans. a a x—3 5x(\+2x)— (a?— 3) Ex. 8. Here l+2r- 5x 5QL 5x -f lQr 2 — x + 3 1 (to 2 -f- 4# + 3 fiqt 5 * Ans.. CASE IV. To reduce an improper fraction to a whole or mixed quantity. Ex- 2. Here (ax— x 2 )~xzza— x. Ans. Ex.3. Here (ah -2a*)+at>=z(h-~2a) + bzzl -~ Ex, 4. a — x)a^-\-x\a-\-x \a 2 — ax ax+ x 1 ' ax — x 2 - 2x* 2x* Therefore a-\-x-\ is the mixed number sought. ( 20 ) ivS _ new numerators. 6?X«= ah $ axc= ac common denominator. 2cx ah Hence and — are the fractions sought. ac ac Ex.3. Here aXc -=.ac -\ (a + b)lz=zab + l*f new numerators. b x czzbc common denominator^ tt ac j ab + b 2 . . Hence — and — - are the fractions sought. be be b Ex.4. Here 3xx3czz gcx \ 2bx2a~4ab > new numerators* dx2ax3czz6acJJ 1 X 3c x 2a=6flc common denominator. __ Qcx 4a b 6acd ' Hence- — ,- — - and — ; — are the fractions required; oac oac oac y Ex. 5. Here the three fractions, when reduced, are 3 2x , 5a + 4x -. — , and 4' 3 5 Therefore 3X3x5—45 -\ 2x x 4 x 5r=40r > new numerators. (5a + 4 ( r)x4x3=60*-f48,rJ 4x3X5=60 common denominator.. The fractions therefore are 45 40r ■ 60^ + 48^ ^tz> ~zzr an d st 60 60 60 fljiMt^tunx^ 2I > Ex.6. Here ax7x(a—x)=z 7a* — Jxal 3xX2x(a—x)=. 6xa— 6x 2 > numerators 2X7 X (a + ff) = 14a -f 14a: ) 2x7> < ( a ~" ,r ) ::::14a — 14a: com.denom* 7a 2 — 7xa 6xa—6x , and ■ ~~ 5 a — x a By the rule 3a: x (a—x) Xa—SaPx—Sax' 2 «% (a—x)X(a—x)5 =:5a?— 10ax + 5x 2 > rators - ( 22 ) Their sum rrlOa 2 — lOax + ScPx— 3a.r 2 + &r a , and 5 X (a— #) Xa=5a 2 — 5a.r common denominator. 10a 2 — l«+2«x+*n numerators . (a—x) x (a—x)=.d t —2ax-\-x°J Difference = Aax (a + x)x (a— 07) =a 2 —07 2 common denominator. 4&x ■**' Therefore — is the difference required. Ex. 8. This is the same, when properly arranged, as 2x+7 5x—6 4207+147 ax— x-\ j-« zz ax— x-\ - h 8 21 108 - 40o7— 48 8207 + 99 l68ax—l68x+82x + Q9 -=zax— x-\ - =— 168 1(58 108 l68ax-86x+Q9 . -*5 — the answer required. 168 i Ex Q. Here subtracting the first from the second, we llor—10 307-5 2O07— 10 307—5 have 07-1 s- r= 15 7 15 7 18207—70 45X—75 17807 + 5 105 105 105 the answer sought. f t-45«— rWl+SV^^ ( 24 ■ ) AaJt$£& Ex. 10. First to rind the difference of the fractions a— x . a + x and a(a + x) a{a—x) a(a—x)X(a—x)~a 5 —2a' 2 x + ax 3 1 a)a + x)xl« + 4=" 3 + 2a** + axS j numerators - Difference" = — 4a 2 x a{a+x) x a(a—x)=:a i — a~x 2 common denominator. Hence the second fraction subtracted from the first is — 4a 2 x — 4x a* — a % jfi m ~ a*—x z ' 4x and consequently a - — is the difference sought, CASE VIII. To multiply fractional quantities together. „ Is'x 5x —5x* . Ex. 4. , Here* — x T— "— r- Ans. 2 3b lb Ex. 5. Here — X-r— sc Ans. 5 2a 5a _ ^ XT 2# 4,r l a 8.r 3 a x . Ex.o. Here — x X = ! — Ans, 3 7 a + x 2la-\-2lx -^ t TT 2x 3ab 5ac Ex.7. Here — x ~ — x-r-r-^lo^a Ans. a c 2b Ex. 8. By reducing these to improper fractions they be- 2a 3 + bx 6a q x—b 1 2a*x + 6c Vb — 2a' 2 b — Q\v come x == — = * These points are placed here to denote such factors (a—yy=a b —5a 4 y+\0a 3 y 2 — \Oa*y 9 -h5ay*—y 5 INVOLUTION. RULE II. Ex. 3. Although the rule prescribes the rinding the coefficients separately, it must not be understood as ab- solutely necessary, being merely stated in those terms for the sake of perspicuity ; generally the whole opera- tion is performed in one line, thus. (a+x)*—a* + 4a 5 x-\ — a*x*+— ax 3 + — x* ■ 2 3 4 Or performing the divisions and multiplications (a + x)+=z a* + 4a*x + 6aW + 4ax 3 4- x* (y\t js^Mrl 1 > J l~» ( 28 ) And in th« same manner, we have (a—xyzna 5 — 5a 4 x+ lQa 3 x»AOa 2 x 3 + Sax*— x* Ex.4, Here(a + ar) 6 = a 6 + 6a b x+ 15a*x x) 3 + 3(a—bx) 2 c+3(a—bx)c 2 + c 3 Then involving the several powers of a—bx we have {a— bx) s = a 3 — 3a 2 l>x + 3ab 2 x 2 — Vx 3 3c{a—bx) 2 =z3a?c—6acZ>x+3cb 2 x 2 3c 2 (a— Ix) zz3c°a— 3cHx Whence by addition (a~bx+c) 3 =z A 3 + 3a°~c + 3c*a + c 3 —3a?bx--6acbx--3c' 2 bx + 3ab / 4a 1 Ex.6. Here^— -= 9r^ 3 V9V^\ / 'y i Srf (04/^tr^z^C^ ( 29 ) v + tj 5/ 8a3 V8V* 3 2a TT -32a 5 * 10 ^/-32Va 5 ^ 10 Ex/ 8. Here V =— zrr~ = 243 V243 — 2ax~ — 3 EVOLUTION. CASE II. To extract the square root of a compound quantity. Ex.2. a*-\ r 4a 3 x+6a <2 x*+4ax 3 + * 4 (a <2 +2ax+x* Ans. a* 2a* + 2ax) Aa 3 x + 6* 2 r 2 + 2aar 4a 3 ar+4tf 2 .* 2 2a 2 -f4a#+ l z 2 ) 2a 2 r 2 -+-4ar 3 -f-,r 4 2a 2 * 2 +4^ -J-* 4 Ex.3. ^ 4 -2^ + ^ 2 -|ar+JL(^ 2 -A--f-| Ans. a: 4 2a; 2 — x) — 2x 3 + lx p 44 + - 4a a 80 4 64* s P />* 8a 4 64a 3 * In this example the two last terms of the proposed quantity arc omitted for the convenience of setting up the work. ( 31 ) l 2+f) 1 | 1 + t 2+1-i) Til 2 +W) i-h EVOLUTION. CASE III. To find any root of a compound quantity, Ex.3. 4a 2 — 12a^ + 9^ 2 (2a 2 ~3x Ans, 4a 2 4a 2 )— 12aur 4a 2 — 12aar+9,r 2 Ex. 4. a 2 +2a£ + 2ac-f£ 2 +2fo + c 2 (a-f £-f- c 2a) 2a£ (a+l) 2 ~ a 2 +2a$ + #* 2a) 2ac (a + £ + c) 2 =a 2 +2a& + 2*c + £ 2 -f2£<:-|-c 2 Therefore a+b+c is the root sought, u ( 32 ) Ex.5, x 6 — 6x b + l5x* — 2Qx 3 + 15a 2 — 6x+l (x*—1x+l x 6 3x A ) —6x 5 (# 2 — 2#) 3 =# 6 — Gr 5 +I2# 4 - Sx 3 3x*) 3^ 4 — ]2^ (x*-2x+l) 3 =zx 6 —6x b +l5x*— 20# 3 +15£*-&r+l Ans. a 2 — 2,r+l Ex. 6. 16a*— 96a?x+2\6a , therefore ty— 27x* Ans. Ex. 5, Here (—2x) 4 =zl6x 4 , therefore £'l6£ 4 Ans. Ex. 6. Here (a 2 ) 6 =a ia , therefore #a 12 Ans. And (<•<*+ a/ £)*=«+£ + , . A/a-. a J 1 Again (—f=z —-= — /. v^a * JVbte /o Me above Case. Ex. l . Here 5/6= 3 )** Ans. Ex. 5. Here -, — reduces to the common denomi- n m n a tors and nm mn Therefore (<&*)»»» and (#")»» Ans. ( 35 ) CASE III. To reduce surds to their most simple forms. Ex.3. Here */ 1 25 v/( 81X3)= 81 V'l And 10/363 = 10\/(121X3)=110v f 3 Ans. igi/( 9X2)= 3V2 Difference 7 4/2 Ex. 2. Here ^320=V(64x5)=z4 1/5 And V 40= V( 8x5) =2 V5 Difference 2 \/5 Ex. 3. Here <• | = *4|= Ji/** 5 j id 5 And ^=^=1^15 4 Difference — -/1 5 * 45 V Ex.4. Here 2/1= 2 • | = Ex. 8. Mult fVl8 By 5 V20 Prod. =^ V36G:=f V(SX45)=5^45 Ex, 9. Mult. 2 \/3zz 2 %/27 By 13iV5=:13f^25 27 S/675 Answer. Ex. 10. Here 72|«*X 120fa*:=^X™x a T Xa ? 69649 X * ta ^1 ■— A = -2 — x ai 2Xa li = 8706--ai 2 Ans. 8 a Ex.11. Mult. 4+2^/2 By 2— a/2 8 + 4 \/2 -4^/2—4 8-— 4—4 the answer* Ex. 12. Mult. (tf+5)»=(a+^)'«» B7 (a-f-£)™ = (a+£) »»«. Product =(a+*>— CASE VII. 7o divide one surd quantity ly another. Ex. 5. Here ^^^2^/27 -2/(9x3) =6 a/3 Ans. ( 40 ) Ex.6. Here — ^ — -rr2V4, which will not reduce 2^18 lower. 3 2 23 3 60 Ex. 7. Here 5- -§--=?— x -=— 4 3 4 2 8 And •— -W^*/ l-- 4 /X=z- s /3 v 135 v 5 v 27 v 8l 9 V mI , 6q 1 60 . % Iherefore -^x-v / 3 = — ^3 is the quotient. »" 'l t.. 1 5 2 25 5 65 Ex. 8. First 3--~2-=. — x— =— ' . 7 5 7 12 42 v 3 V 4 V 3 3 y v 9 27 2 v. 1'3 ,. 65 2, 0'5 . Hence -x^S^^VS A*. r ■ T ■ 1 2 3 27 h^.Q. Here 4— r-2-:r-X-=— 2 3 2 8 10 i 4 « a 6 ' a* And v / ^-J-.ya^-^^a 3 ^ 3 = ~r = 'i = J? P (—)*; therefore ^g(£)* the answer. « 1rt tt 2 3 162 4 648 Ex. 10. Here32-~]3--: x — =— : 5 4 5 55 2/5 I l 3 S I And v'a-r- J5 V5-v /3 V*+ -N/3"" 2 =8 + 2*^5 Ans. 2STo^ to the above Case. ' --. 1 —2 * Ex.3. Here— -=a the answer.. a* — £ I Ex. 4. Here a *— — - the answer. Ex .5. Here — — =- — — — ^(a + a?)"" Ans. i a Ex.6. Here a(a 2 -:r 2 ) 3 :=-- — - Ans. CASE VIII. 7o involve or raise surd quantities to any pozver* Ex.3. Here (3 V3) 9 =(3 x3 T ) 2 =:9.3 T Ans. Ex.4. Here (17 V'2l)*=(l7x21 T )\=:;i7 2 x 21 =6069 Therefore 6069 is the answer sought 5 which, in* thls^ ease, is a whole number. e3 ( 42 ) Ex. 5. Here & VgfezQ x ^)^ x6^ Therefore — theAns, 3b Ex. 6. Multiply 3 + 2\/5 By ' 3+2 \/5 9 + V5 4-6v'54-4 s /25 Square =94-12^5 4-20=294-12 V^ Ex.7. Multiply s/x+3 y By V* + 3 */y ar4-3\/^y Square =x+6 \/xy + 9y Mult, by — 10^64-24 4th power =;49— 20 *=2-4\/12+24==2ff-. 8v^3, therefore a/(26— 8\/3) Ans. Ex.6. Here (4-V7) 2 = 16-8/7 + 7=23-8 s/7 Hence ^(23-8^7) Ans. Ex. 7. In examples involving cube root radicals it is useful to know the following form of the cube of a bi- nomial : viz. (a±b) 3 zza 3 ±b 3 -\-3ab(a±b) Hence (2 %/3 - 3 y 9) 3 =24 - 243 + 1 8 1/2J (2 jj/g -3*/9) = — 219 + 54(2 V3-3 3/9) Consequently y(-219 + 54(2 V3 - 3 #$>) is the general surd required. CASE XI. To extract the square root of a binomial surd. Ex. 3. Here si \a+\ v^^**=%/3+fv/36-20= V (3 + 2); and>>/|a -i^^^T/T^ V3— f \Z36~20 = V<(3-2) Hence a/(3 + 2) ± a/ (3 — 2) ~ V5 ± 1 Ans. Ex. 4. Here V £a +|^/^r3 =N /±l + 1 ^529^448== \/(~ + i> 2 2 2 Hence i /(^ + |± v / ( ^_2)-4± V7 Ans. * It may here be observed, that & denotes the quantity under the second radical after its coefficient has been introduced. Thus iii> the present example, bzzzo, because 2 V$—V'20 t ( 45 ) Ex! 5. Here ^_ N /(184-7) 3 and s/ia-^af^b- V 18~|Vl296~JKX)=rVa8-;) Hence */(l8 + 7)± V(13 - 7;=5± VH Ans. Ex. 6. Here = V(— +^)$andVfa-£v / ^--£ -s/ ^~ i V108Q S64=\/ (^-i^) Hence ^(y4-^±^~Y) = ^^±vi Aus, CASE XII. Tojind such a multiplier, or multipliers, as will make any binomial surd rational. Ex. 5. Given surd ^/5 — */x Multiplier v^-f- V* Product 5—x as required. Ex.6. Given surd */a-\- */t Multiplier ,/a— ^b Product a— h as required, Ex. /• Given surd a-\- ^b Multiplier a — */b Product a^ — b the answer. ( 46 ) Ex. 8. Given surd 1— Via Square of the terms =1 -f- 1/40? Product with sign changed = + l/2a Therefore 1+ ^2a + %f*a*2Z multiplier. Mult, by l-~\/2a 1+VU+ yAa* Product ss 1 — 3/8a 3 =l — 2a as required. Ex, 0, Given surd J/3 — $V2 Square of the terms l/g -f £ 1/4 Product with sign changed -f-£ V6 Whence 3/9+£V6+* V4 is the multiplier required. CASE XIII. To reduce a fraction whose denominator is either a simple or compound surd, to another that shall have a rational denominator. Ex.4. Here— ^.,-A-x^,: V74V3 V7-W3 V7-V3 V42- VI 8 Ans. 4 T O* Ex 5. Here 4 £ 3— \/X 3X—X^X 3+ ^/x 3+>/a? 'd — */x 9—x Ans, Ex.6. Here ■ . = 2- x ^ — t= a + ^fo a+ Vb ^fa—«/b a + b — 2^ab a — b Which is the answer sought. ( 4? ) Ex. /\ Here, by the preceding rule, the multiplier for the denominator is V4g+ 1/65 -f \/25. V7 - V V7 — V5 V49 + V35 + V25 _ the answer required. Ex. 8. Here V9+V10 V9+V10 V81- V 90 V»00 __ V3( V81 - V9O+VIOO) ysi — V90+ V100 ~~ 9+10 V243— V270 + V300 ; Ans. 4 4 V+-V5 Ex - 9, Here V4TW = vITvi * v^v5 _ 4(V4-V5) _ 4(V4-V5) 24- V5 _ — V-1-V5 ~" 2-V5 X 2 + V5 ~ 4(V4-V 5 )(2+<£) - 4(V 5-V4)(a +( /S) Ans. 4—5 ARITHMETICAL PROPORTION AND PROGRESSION. Ex. 3. Here the formula S=(a + /)x 9 becomes s=(l -f 1 000) X 500^500500. Ans. Ex.4. Here the formula s = {2a + (» — ))* $£ becomes $2*100x2* X-^ =202 X— -=]0l»=10201 2 JL ( *» ) Ex. 5. Here s=(a+/)^=(i +24) x 12:= 300 Ans. Ex. 6. Here the formula Z=a-f (n — l)c?, becomes /= 2+ (365 — l)2=2-f 728=730 Ans, YL 1 *? 1 Ex. 7. Here s=^2a— (* - 1)<* j-= 120-20 x -\ X — 2 3 2 20, 21 40 21 = (20— — )X— =— X— =140 Ans, 3 ' 2 3 2 Ex. 8. Here the 1st term is 1 -f 1 =2, and the last 100 + 100=200, the number of terms 100. Therefore (a-f /)- = (2 4 200)50=10100 yards, or 5 miles 1300 yards, the ans? GEOMETRICAL PROPORTION AND PRO- GRESSION. Ex. 3. .Here the 1st term T r Y)= 1 X fZTT"" 31 ^!"" 3 ^ 2 " 50048 08331 — — r=I , ■ 'Ans. 39380 I9093 Tbe answer in the Introduction was computed to 9 terms. Ex. 6\ Here a-=zl, r~2, 77=32. Whence ir*^l. 2 32 -i srr(a r.lX -=:2 S *— 1 = v r^ l 7 2—1 4294967295 farthings— 4473924/. 5*.3jc7. EQUATIONS. RESOLUTION OF SIMPLE EQUATIONS. EXAMPLES FOR PRACTICE. Ex. 1. Here'3# — 2 + 24z=31, by transposing Gives 3cr=31 +2 — 24=9 Whence or^| = 3 Ans. Ex.2. Here 4 —937=14 — 1 ly^ or l] I/— 9#= l 4— 4, or 2?/ = 10, Whence y= 1 ^ > =5 Ans. Ex.3. Here #+18=3,z — 5, or 3,r— a?:=Ll8 + 5, or 2#~23, whence ,r=li~. Ex.4. Here ar+|+|=ll Mult, by 6. ^+ar+^r==6(5 l or 1 lxsz<56, Whence orz^6 Ex. 5. Multiply the given equation by 2, and we have Ax — x-\-2~lOx — 4 3 whence 10jr+ l r—4o:=4-f 2, or 7^=6, wheoce#=~ 7 ( 50 ) Ex. 6. Mult. jA- 00 -- 30 -— £- by 60, gives 2t o 4 lO 30x-\-20x—15x=z42, or 35x=:42, 42 6 1 or #=: — -rr-rzzl- Ans. 35 5 5 Ex. 7. Mult. -2- +-=4 — , by 12, gives JL o 4 6#-f 18-f-4#=48— 3,r+15, or 45 6 13a"=45, whence or~ — zzz 3 — 13 13 Ex.8. Here 2-f- v /3^rrz ^(4 + 5^) being squared, Gives 4Y4\/3x + 3xz=.4-\-5x x Whence 4 ^3^=5^— -3x=2^ - Squaring, 48jr~4 i r 2 , or dividing by 4x, we have .r=:12. x* Ex. §. Here ar-f-a = , or a? 2 -f2aar-|-a 2 =:ar 2 Whence 2aar= — a 2 , or 2x~ — a, or #= 2 * * 2a Ex.10. Here v^+ VYa + #) = — r; — ; — r» rnult. by s/(a+x) \/(a + x) Gives A K /(ax-\-x' i )-\-a-\-x~2a } ox ^/(ax+x^—a—x, Hence by squaring ax-\-x*—a t —2ax-\-x <2 9 a Conseq. 3ax—a q , or 3x=a, or xzz- 3 _, ax— I a Ix hx—a Ex. 11. The equation — -— -f^n — — — 4 -J* o Mult, by 12 gives 3ax— $b + 4az=.6bx~ 4hx\Aa Whence 3ax— 2bx—3b, or ar(3a— 2£):=3Z; Consequently ar=— — ~? ( 51 ) Ex.12. Here VK+^^Vv^H^), by squaring Gives a 2 -}-^ =z \/(b* + x*), squaring again Gives a i + 2a q x' i + x*=:l>* + x* Whence 2a°-x 7, =:b*-- a 4 , or 2tf* Ex. 13. Here ^/(a+x)-^ \/(a—x)zz\/ax, by squaring Gives 2a + 2 \/(a?— x*)—ax, ax — 2a Whence \/ (a 2 +x 2 ) = — - — ; squaring again, m a 2 x a —4a <2 x+4a' 1 a 2 + x*= 4 Therefore 4x 2 =a i x' t —4a 2 'x, or dividing by x, 4x -zz-arx —4a 2 , or 4a q 4a 2 xzz , (and not , as given in the Introduction). a a Ex. 14. Here f- = &, becomes by reduction 1+a? 1— .r a — ax + a + ax la _. r»3E 6, Ol" 1— tf 2 ' 1— z* Whence 2azz.b — bx' 1 , ' 2 + x* b* Whence 4ax~b 2 ~- 4a 2 , or x=. a. 4a The answer required. ( 52 ) Ex. 10. Multiplying the given equation by 2, we have By squaring 2# 2 -f-2, v /(^ 4 -9a 4 )=4a l r 2 , or VX? 4 — 9a 4 )-2^ 2 — a; 2 ' Squaring again ar 4 — 9a 4 =4tf 2 j^ — 4a,r* + tf 4 , or (4ar— 4a ft )ar*=:9a 4 , or x=z^- 4-4a Ex. 17. Here v"(^+^) + VX a — x )=&> D y squaring, 2a + 2 A /(a 2 -* 2 ) = £ 2 , or A /(a 2 ~^)=f^ 2 ~a, • Whence a 2 — ar 2 =J^ 4 -.^ 2 a + a 2 And i^V^ : rM^ Ex.18. Given equation V(a + #)+ V (#—#)= #" By cubing both sides after the form of Ex. 7> Case x, Surds,, we have 2a + 3y(ffl*-s*)} V(a + *)+ V(*-*) |=^ But since V(a-f#) -f- V( fl — #) = #* t bis becomes 2a.+ 3 V(tf 2 — **) x £ = £ 3 , or 7,3 rt 3V(fl2-^) = -- jj or V(a 2 — ^) — t - ; whence id*,— .**= 33 (£ 3 — 2a) 3 .V/ftf-? 2«\ 3 2//> 3 V 3£ £3 2a Therefore ^=r /V /Ja 2 —( — - — ) 3 £ Ex. 19. Here ^4- tyx—A/ax, which divided by v^ Gives — — -f 1 = a/^ or = — 1 + va, or \/a — 3 , ( 53 ) Whence Jx{*Ja— \)~ ^a, or */x=z- Therefore x=j^~ Ex. 20. Here V~ + V^-77 = a > b Y squaring x— I #+1 x+1 x—1 „ # 2 — 1 Gives -— + + 2^- — - =" 2 #— 1 .r-t-1 a: 2 — 1 2o? 2 4-2 Or • -f2=a 2 -, or by multiplying «r 2 — 1 2x* -f 2 -f 2# 2 — 2 =.r 2 a 2 — a 2 , Whence 4.T 2 — <&&=■ — a 2 , or (a 2 — 4)# 2 =a a Therefore x= the answer. Ex. 21. This equation transposed, is */(a?-\-ax) -f V( a ~— ax)=aj squaring 2a 2 + 2V(a 4 — a 2 .r 2 )=:tt 2 5 whence ^(afi—cPx*) ±± — |a 8 • squaring agaia, a 4 — a% 2 =^a 4 , or a 2 ^ 2 =|a 4 , Whence #=* v^= IV 3 Ex. 22. By transposing the given equation, it becomes aV(l~^)~ VO* 2 — ^ 2 )=.r^/(^ 2 — 1) Multiplying by a 2 , and dividing by x, we have V(^ - * 4 )- V (~ - 1) - a/(* 2 -* 1) And again, by squaring both sides, <* 4 i a 2 /w« 4 a/* 2 T-a 4 + __i_2V (— -^)(-r-l)==a»-l .r* # 2 a? 2 • \r 2 < f3 ( 54 ) Rejecting the denominators 3* #* a 2 # 2 a,* v n ' fl 4 _ a 4jfl 4. ^2_ ^2^2^ 2 -y/(a 4 — fl 4 ^) (fl« — 3*) Dividing both sides by a 2 , o 2 — a?x*+l — x*z=2 v (I—^Xa 3 --* 2 ), or (1— ;r 2 )(a 2 - r J)=2^/(l— jr 2 )(a 2 — **) Squaring both sides, and dividing by (1— >r 2 )(a 2 + I) 2 . ... . 4 (a 2 — x°) we snail have « 2 -f j == — — 1 — x 1 Or (a 2 -f l) 2 (I~^ 2 )==4(a 2 - l r 2 ) Or (a a +l) 2 -(a 2 +i;^=4a 2 -4^ Whence *'£&=&& 4— (a^-fJ) 2 V I (« 2 +l) 2 -4 3 The answer in the Introduction is not correct. Ex, 23. Here ) = c 2 Whence A /(^H-«)(^+^)=|c 2 — fa— §Z> — x, Squaring again we have Where since x°~ and ,r 2 , (a-f-Z*)^ and — ( — #— £)# cancel each other, it follows that c 2 r=:i(c 2 — a— &) 2 — ab, or c*—a—l>. 9 ab c^-X-b a &+=.{ )*■—#, the answer. v 2c ' ( 55 ) Ex. 24. Here V + V = V a + x a — x as — x 9 - by squaring, _, Z- £ 4#c 4bc Gives — — J- + /-ft — ; b c Whence \- — — =0, or a + x a—x ab — bx4-ac + cx tf-x" 1 Therefore al — bx + ac + cx- 0, And hx — cx~ab + ac ab + ac b + c. Consequently #= =a(- ) b — c b —c The Resolution of Simple Equations, containing tzvo unknown Quantities, EXAMPLES FOR PRACTICE. Ex. L Here 4.r-f y=34, and 4y-hxzzl6. Multiply first equation by 4, and we have 4y -fl&r=136' Subt. 2d. Ay + ar== 1(5 ,, 120 Gives 15^—120, or Ess =8, 15 From 1st. ^=34— 4or~34-32:=:2. Whence 8 and 2 are the answer required. Ex. 2. Given 2x + 3y=zl6 ' 3x- ? + 3y=zl6\ r~2y~ll j (56 ) Mult. 1st. by 3 gives 6x + 93/ ~4Q 2d. by 2 gives 6x—4y—22 Difference 13y =26, or y~2 l6-3y 16-6 From the 1st. «r= — — - = =5. 2 2 ^. 2# 3y 9 - 3x 2y 6l Ex. 3. Given — +-£=—, and — + f£ = 5 T 4 20 4,5 120 1st. Equat. mult, by 20, 8x + 15y—Q 2d. Equat. mult. by 20, 15x + Sy~ — =10- Q \5y From the 1st of these xz=l - — — 8 From the 2d. x= ? " ? 15 9-15^ 101— 8y Hence JL-J^-A-JL, or 135— 225yzz8lf — 64y, or r6ly=53f,. Whence y=5a|-r-l6l=|, , Also * 9 ~ 15y = 2^ = £ 8 ~" 8 ,,2 Ex. - Given - + 73/^99, and § +7$s=5 D ♦ 7 7 Mult. 1st and 2d by 7, and we have x +49^=693 49^4- 3/=357 Mult, the latter by 49, and we have 240107 + 493/= 17493 Subt. #+-493/= 6Q3 Difference 2400^=16806, whence #=7, Also y=S57— 49#=357— 343=14. ( 57 ) Ex.5. Here - -12=^+8 2 4 •r-j-y x 2y—x Miilt. 1st by 4, and the 2d by 60 Then 2#-4S=?/ + 32 12o? + 12j/ + 2Or-480=30y- 15^-f- 1(520, Whence 2#— ^=80 1 47#-18y=2100J Mult. 1st by 18, gives 3&r— 1 8y =1440 Subt. it from 2d 4?x— 18t/=2J00 Gives 11^=660, or #=60 And 3/=r2#— 80=120— 80=40. Ex. 6. Given x4-y~ x. anrl *&. _ y a — L From the 1st x=s~y. 2d #=>v/(d+y 8 ) Whence s—yz= */(d+y*), or by squaring, **-2n/+3/ 4 =d-f2/ 2 , s 2 ~d Whence ~2sy — d—s*, or y= Also #=/ — y=j — =3 — J 2s 25 Otherwise, divide x^—y l2 =J By #+3/ =±#, And we have x—y=- t Whence, by adding and subtracting two latter, we obtain d t d 2x=s-\ — , and 1v = s , Whence x= and y— , as before. 2s u 2s ( 38 ) Ex. 7- I n tne fi fst of the two equations (x^y) \ a ;\ (x—y) : b, and x^—y^—c b(x+y)=a(x—y), or (b + a)y~(a—b)x, «-# , o (a -ft* Whence a= —x, and v == t t # f a + £ J (a + by\ Substitute this for y 2 in the 2d equation, and we hare x 2 (a + by—(a-b)*x 2 =c(a+by, or *«j(a+^)«— (a— ^) s j=c(a + *) 8 , or c(a + b)* xHAab) =c(a + l)*', whence a^g ^ , J , or v Aab 2 r *$ * Aab 4ab a — b , c 2 a£ Ex.8. Given ax-hby=cl dx+ey=f$ Malt . 1 st by J, dax + c% = Jc Mult. 2d by a, dax + aey—afy By subt. dby — aey=xzdc—qf f __ r . dc—af af—dc Whence v*= i? — - = — db —ae ac—db Mult. 1st by e, eax + eby^ec Mult 2d by b y bdx + eby — bf By subtraction eax— bdx=z ec— bf €C ~~~ VI u f C6 Whence x-=. - ~ — ea — bd bd — ea Ex. g. Given x+y==a, and x 2 —y 2 =b. Here, as in Ex. 6, divide x 2 —y*=b by x+y=za -, and we have x—y— - ( 59 ) fl a 2 +£ Whence by add. 2x=a-\ — , x = " a la And by subt. 2y—a , or y= 3 '■ a J 2a Ex. 10. ^ er £ ^ '<' xy~a\ f.\ xy = bS By add. a£+ 2ry-f y*=a + 6, or (=?<9 J Now, subtracting the latter from the preceding one, we have %=23. Also from the last y=29— z~2g— 23=0, And from the first x=53—y — z=53~- 29=24, That is #=24, y=6, and %=23. Ex.4. Given x+iy + }%=32* ix+iy+i ix+iV + i* \z=15 ^ |%=12 J ■ ,.s ( 60 ) Multiplying the first by 6, and the second, and third by 60, we have 6x+ 3y + 2%z=\Q2-\ 2(Xr+ \5y+12z—gOO > 15x+12y+Wz=720J Again multiply the first by 10, the second by 3, and the third by 4, give 60jc 4- 30y + 20*= 1 92a 6Qx + 45y + 36z=2700 60x + 48y + 40z=z2880 Subtracting now the first of these from the 2d, and the second from the 3d, we have - 15y + l6%=7801 3y + 4%=1S0J Mult, the latter by 5, 15y + 20z=g00 Subtract .... 15y + \6z=760 4x = 120, or x=30 180— 4% _ t 192 — 3?/ — 2% But ?/ = = 20, and,r=:~ -+ = 12 J 3 6 Therefore a=12, y=20, and s=30. Ex.5. Given 7x+5y + 2z— 7q 8x+7y + Qz=122 x + 4y-\-5z= 55 Multiply the first by 8*, the second by 7> and the third by 56, and we have 56x+ 402/ + 16%= 632^ 56x + 4Qy + 63z=z 854 - 56r + 224^ + 280* =.3080 J Subtract the first from the 2d, and the second from the 3d, and we obtain Qy + 47 *= 2221 ij5y+217% ==2236/ !} ( 61 ) Multiply the first of these by 175, and the'second by Q, and we have 1 57 5y + 8225%=38850 1575y+1953z = 20Q34 By subtraction 6272z=188l6, or k=3 : 222—472 : 55— 4y—5% But yx=z — - —9, and x~ 9 & 1 That is x=z4j y~9, and %=±3. Ex. (5* Given #+# x+y—a*] x-\-z~b > By addition 2x + 2y + 2z=a + b + c, or From which, subtracting each of the three given equations respectively, we have x= s^-fi^ — \c The values sought. SIMPLE EQUATIONS. Ex. 1. Let one of the parts ~x, Then the other will be = 1 5 — .r. And by the question 15—.*=^, or 60— 4x=3x, or 7x—6Q-, whence x— 6 f=z 8£ the one part 5 and 15 — x=6$ the other. Ex. 2. Let the value of the purse be x\ then -the money —7%i an d by the question 7^+^=20, or - 8.r=20, or #=*-£' 2s. 6d. the value of the purse, and consequently 17s. 6d. the money, contained in it. ( 63 ) £x. 3. Let the number of sheep — x ; then by the question, x + ,r + ftf+7£=500} or 2%v~z;49?2£j mult, by 2, 5cr=9§5 j Whence #= lii ±s lp7> tne number sought. Ex. 4. Let the length of the post =x; then by the question £# + |#+10=# 9 mult, by 12, 3x-\-4x+ 120=12#; Transposing 5.r= 120, or x=L^2 =24 feet, the answer required'. Ex. 5. Let the number of guineas =#; then .r— £.?— ^x=72, or mult, by 20, 20#— 5#-3*=1440$ whence 12^=1440, or a?= I4 - 4 ° = 120guineas. Ex. 6. Let b's share = x, then by the question, a's ghare —2x, c's share —3x, Consequently # + 2.r+3.r = 300, or 6jt = 300 5 whence «r= 2-~ = 50/. b's share, 2#— 100/. c's 6 share, and 3#=150Z. c's share. Ex. 7. Let the age of the wife at the time of the mar- riage —x, and that of the husband 3x\ Then after 15 years their ages will be x+ 15, and 3x-\- 15 5 and by the question 3#+ 15=2(r+ 15) ; or 3^+15=2^ + 30; Whence, by transposing, ,r=15, the age of the wife 5 and 3#=45 the age of the husband. Ex.8. Let the number sought =x-, then by the question 2(x 5\ x—5 will be the remainder, and — : ~ 40 ; 3 ' Whence 2x— 10=120, or 2,r=130, and conse- quently 111 =3= 65 the number sought. Ex. 9. Let the less number of voters =x, then th« greater number =#+ 120, and by the question #+,£+120=1296. ( 63 ) Whence 2x— 1 2g(3— 120~ 11765 conseq. X—58S for one candidate, and x4- 120=708 for the other Ex. 10. Let x represent the age of c, then 3x is the age of b, and 6x the age of a. Now by the question, cc+3x+6x — lO.r = 140 whence x=±±°- = J 4 c's age, 3^=42/ b's age,, and' &r=84 = a's age. Ex. 1 1 . Let the equal sum laid out by each be X' } then / leaves off with x+126, and b with x — 87 5 and by the question x+ 126=2(x-8/) -, or x+l r i6~2x—\74, Whence #=300/. the first stock ofeacb. Ex. 12. Let the price of the harness—* 3 then the price of the horse —2x, and the price of the chaise =6^j and by the question x+2x+6x=-60l. or yx=60; Whence x— 6 £—6t. 13s 4d. the value of the har- ness; 2#=13/. 6s. 8d. the horse 3 and 6x—40l. the chaise. Ex. 13. Let x denote the number of beggars; then by the question 3x— 8 was the number of pence he had about him; Which from the other part of the question may also be denoted by 2^ + 3 ; Whence 3:r— 8=2.r-f 3, or, by transposing, #= U, the number of beggars. Ex. 3 4. Let x denote the value of the livery; then $4-8 is the whole amount of his hire for the year,, or for 12 months. 7x4-56 Hence, as 12 : 7 :: .v-i-8 1 . the hire for 12 7 months ; but by the question the servant received 7x4-50 ;c-j-2j; whence =sx + 2^,or 7x4-56— 12x 1 2t + 32, ( 64 ) And by transposition 5x = 24, or x—' t £=z4±].= 4l. \6s. In the preceding examples only one unknown quan- tity has been employed, but it will be more convenient in many of the following questions to use two or more unknown letters according to the nature of the equation. Ex. 15. Here let x denote the son's share, and y the daughter's ; Then the value of their shares will obviously have to each other the same ratio as half a crown to a shilling \ tiiat is, as 5 to 2. Hence, then, we have x \ y \\ 5 \2\ And x -\-y = 560 i 5y From the first 2x=5y, or x—-^- bv ■ Whence from the second — -f-y— 560, or 5y + 2yzzzjy = 1120 ; where y = *J*2 = 1 60I, 5y the daughter's share, and x^— sx 400/. the son's share. 2 Ex. 16. Here it may be observed, that every number consisting of two digits is equal to 10 tjrnes the digit in the tens place plus that in the units. If therefore x be put for the former, and y for the lat- ter, the number itself will be denoted by IQr+yj and the number with the digits inverted by \Oy+x. tt u *• c I0>r + v= *x + 4v Hence by question, \ l0x + * + lQ=l( g + x Where the second equation gives gx-~ Qy=z\s, orx—^= — 2, or x=y — 2-, Which substituted in the 1st, gives 10(y — 2)+y=4 (y-a)+4y. And this, by multiplication and transposition, be- comes 1 Qy -\-y~4y — 4y =20 — 8, or 3y = 1 2, Whence y= l ^—4, and x—y — 2~2 ; Therefore the number sought is 24. ( 65 ) Ex. 1/. Let x represent the equal income of each >, Ax then by the questions a's yearly expenditure is — , and A.v * that of B — +50.; 5 j q x In four years, therefore, b spends \- 200 5 which exceeds his income in the same time (viz. Ax) by 100 ; hence we have the equation l6.v r-200=4r+100, r 5 1 6^+1 000 =20* + 500, Whence 4jr=500, or x= — = 125/. the income sought. Ex. 18. Let the number of persons in company be x, and the number of shillings each paid =y j then xy will the whole reckoning. , Now had there been three persons more in company, viz. (* + 3), each would have paid (y — 1) shillings 5 whence we have (x + 3)(y--\)=^xy; , And from tta other conditions of the question, (x-2)(y+])=xy. Whence, from actual multiplication these become xy + 3 y — x — 3 = xy xy ~-2y + x—2=xy And consequently by addition we have 2xy + y -5—2ry f Gr, cancelling the 2*3/ on both sides, y~ 5=0, 0Ty^5, the number of shillings each paid, And by subtracting the second equation from the first, 52/— 2s -1=0, 5y — 1 24 . Whence 2x=±5y — 1, or #= -~- — = — - ~ 12, the 2 2i number of persons in company. G 5 ( 66 ) Ex, lp. Let # denote the money he had about him ; then by borrowing x and spending one shilling he had left 2x— 1. Also, at the second tavern, after borrowing 2 t r— I ; he had 4x—2-, bat spending one shilling he had left 4*- 3. At the third tavern he borrowed 4# — 3, and then had Sx — 6 y and after spending one shilling, he had left At the fourth tavern j borrowing 8x— 7, he had l6x—\4; but after spending another shilling he had left l6x — 15 ; which by the question is equal to nothing. Whence \6x— 15=0, or #=f| =0j. l\\d. the money he had at first. Ex.20. Let x and y denote the two parts ; then by the question x-\- y—75, and 3x— 7y= 15 Mult, the first by 3, 3^ + 3^=22.5 Subtract the 2d, 3x — 7y= 15 And we have IO3/ — 210 itti 210 . , Whence y= =21 the least number, ? 10 ' And a'=75-~3/=54 the greatest number. Ex. 21. Let x= the whole quantity of the mixture ) then \x-\-25 was the quantity qf spirits, and \x— 5 the quantity of water. Which altogether made the whole x ; therefore by addition, \x+±x + 2Q=.x ; or by mult, by 6, 3x + 2x+120=6x; Whence, by transposing, j=120, andconseq. £#+25=85 gallons of spirits, -ir— 5=35 gallons of water. ( 6? ) Ex. 22. Let afc the number of guineas, and 3/= the number of moidores; Then 21#= the shillings paid in guineas, And 2Jy = the shillings paid in moidores. Now the whole number of pieces used being 100, and the number of shillings paid being 2400, we have X + y~ 1(;0 7 2\x + 27y — 2400 ) Multiply the first by 2/, and we shall have 27^+27t/=2700 The second 2lx + 2Jy = 2400 Hence by subtract. 62 = 300, or x=50, the number of guineas ; and consequently y= 100— #=50, the num- ber of moidores. Ex. 23. Let x be the number of days, then 14? miles will be travelled by one, and l6x miles by the other. Hence 30x—lQ7, or x= I ^J=6 d . I3 h -J. the time re- quired. Ex. 24. Let x to denote the weight of the body ; then \x-\-Q is the weight of the head, and by the question *=»|jf-f-9+9. or|jr=18 ; Whence x—3Q the weight of the body, ^r+9=27 the weight of the head, and 9 the weight of the tail j consequently 36 + 27-f9=rr72/Z^. the weight of the fish. Ex. 25. Let ^, ^, and |, represent the three parts re- quired, and the three latter conditions of the question will be answered ; For the first multiplied by 2, the second by 3, and the third by 4, will obviously be all equal to x, and there- fore equal to each other. Hence then it only remains to fulfil the equation xxx- -+-+-= 10, 2^3 4 Or multiplying by 12, to clear it of fractions, 6x + or by transposition 2y— a; =501 — y + 3.r=50J Mult, the latter by 2, and we shall have — 2y-\ &r=100$ to which adding 2y— x= 50, we have 5x=\50 f or,r=30/. the value of the first horse ; and from the second equation y=3x— 50= 40/. the value of the second. Ex. 28. Let x a's money, and y= b's. Then, when b gives a 5s7the latter will have x + 5, and the former y — 55 and if a gives b 5s. then a will have #— 5, and b y + 5. Now by the question {'^^{ylj). ( 69 ) Or multiplying and transposing 2x— y— 151 — x+3y=20J Multiply the latter by 2, and we shall have — -2x-\-6y—40, to which adding 2x— 3/= 15, the sum gives 5yz=55, ox.y = l\,fiz moneys 15 + V 15 + 11 „ , And 2*=15+y, or x= — - = — — = 13, bs money. 2 2 Ex. 29. Let x and y be the two numbers \ then by the question the difference is to the sum as 2 : 3, and the sum to the product as 3 : 5 - } that is x—if : x+y :: 2 : 3 jr-fy : xy :: 3 : 5 Which, by multiplying extremes and means, gives 3*~ 3y—2x + 2y'l 5x + 5y = 3xy J From the first z=5y, which substituted in the se- cond, gives 25y + 5y = 1 5y z , Hence, dividing by y, and transposing 15y=30-, or yzz2, one number, And x=5y=: 10 the other. Ex. 30. Let jt=3e the number of shillings he had at first} then by the question he lost Jay and therefore had 3x — left, to which he won 3s. 4 3 x 3x -\~ 12 After this he had h3= - — , of which last he 4 4 lost one-third, and had then two-thirds of it remaining, . 6ar + 24 ,.,,,. , . ,6^ + 48 viz. 'rr% to which adding 2s. won, he had « 12 ' 6 12 ( 70 ) Then losing one-seventh of this, he had -f-tbs of it, viz* 30r+288 , . L . t v . left •, which by the question was 1 2s. $ „ rt 3frr-}-288 Whence - =12, or 3&r+2S8= 1008 ; 84 And conseq. 3&r ~72b, or xz=20s. the money he had at the beginning. Ex.31. Let x be the number of leaps the grey- hound takes, and y the length of each; then by the question Ax 3 I AW x \ — , the number the hare takes, and 3 . 3 : 2 :: y : — , the length of each j But the hare being 50 of her leaps before the grey- Ax hound, he has to pass over -*-« +50 leaps of the hare. « o Ax Now each of these, viz. x and j-50, multiplied by their respective lengths, will obviously be equal - } viz. ^=( T+5 o)-| Or dividing each side by y, and reducing 8x 3xz=: - -f 100, or Qx— 8jt=300. 3 Whence ,r=300, the number of leaps the greyhound takes before he catches the hare. Ex. 32. Put iv, x, y, and z, for the four parts ; and let m be that quantity to which each part becomes equal after the operations expressed in the question are performed j Then we shall have w+x+y + xzzgo. ( ;i ) Also iu-L2~m* "ivzzm— 2 x — 2izm tf—m-f 2 2y —m 2 J Or - to *z~2m Whence adding together the four last equations w+x+y + %=4'm + ~ : ; or substituting 90 for zv+x +y + z from the first 4m±%=zQO, or 9771:1: ISO; whence mzz20; conseq. tvzz m~2zzl8'j x~m + 2—22', yzz%=lO; and z~2mzz40, the parts required. Ex. 33. Let the divisor rrar, and the dividend rry, then by the question 21*4-21:=: y \ #=42— yzr35 J Whencey~21 l r+ 21 ~/56, the number to be divided. The answer is wrong in the Introduction, being 1050. Ex. 34. Let x be the number of days the man would be in drinking it by himself; then J will be the quantity he drinks in a day, and T J S is the quantity the woman drinks in a day 5 Therefore the quantity both drink together is ~-\- J^ m But by the question, 12(J-h 1 I 5 ) = l, or multiplying by 30#, 3rJO-f-12x=30.r; whence, by transposing, 18#= 3dO, or ^?rz20, the number of days sought. Ex. 35. Let x be the number of men in the side of the less square, and x + I the number in the side of the greater; then «* will be the whole number of men in the former, and (ff+l)*=ff*-f2r+ 1 i* 1 the latter: Whence * 8 +384, and aS-2ff+l — 25 will each ex- press the whole number of men 5 from which we have ttois equation, or* + 2*— 34=^*4284, 01 2^=284+24=308 ( n ) Whence #=154; and consequently x 1 +284=24000, the whole number of men. Ex. 36. Let x, y, and %, be the number of days in which a, b, and c, respectively would finish the work ; then A will do - part of it in one day, b will do - part, and c - part. y * Then, by the question, we shall have 1 1 _ 1 x + y ~~8 x % 9 y $ * ~" io And consequently by addition £ 2 2__ 1 1 1 121 "x y "^"""i" g 10 ""300 Or by division, x\ \y %~~720' From this subtracting each of the three first equations, we have 1 31 720 7 - = - — , or %■=. = 23 — — days for c * 720 31 31 J 1 41 720 23 - ss -— -, or v= — 17 — = days for b y 720 * 41 / 41 J 1 49 720 t 34 - j - = , or x~ — - = 14 — =b days for a x 720 49 49 y ( 73 ) QUADRATIC EQUATIONS. EXAMPLES FOR PRACTICE. E*. 1. Given x* — 8x + 10~19; by transp. X*—8fc±i 9 Whence x =4 ±V(l6+9)=4±5, Therefore x =9, or — 1 Ex.2. Given x*—x — 4Qr~XJQ', bj transp. or 3 — ar=2L0 841 Whence #±=£± v^(*+210) ~£±v / — -> 4 1 2Q Therefore or == - ± — = 15, or — 14. Ex.3. Given 3s* + 2a?— ,9=76 ; I ■.;.-■■■ vasp. - 2 8.> 3:t a 4 ■ 2^=85, or a? 2 -f r #= — . 3 V 9 3 3 ™, « -1 ^ 16 1} Therefore #= ± — rr5, or *~ — 3 3 3 113 Ex. 4. Given - x 2 - • - ^ -f 7~ =8 j by transp, 2 d o -a* #=-, or x 9 #:£>— 2 3 8 3 8 Where *=i ± •(-+-) = ~± ^ 3 9 8 3 v afl ^, ^ 2 ^_7 1 5 fheretore ■ „r = - dt ~ s= 1*-, or — - 3 3 2 6 ( 74 ) Ex\ .5. Given -,r— - \/rc=22 - 2 1 Malt, by 2 : ,r — a/^= 44-; where f, 3 3 , ! _u ,, 1 , 133 x i , y/ 4oa ^=-±V(-+— ) = 5 ±v^(— ) Therefore V*=(* ± y) = 7, or ^, Conseq. *=7 2 =49, or (—)* = ^ Ex.~6. Given x+ tf(5x+10)=.8 By transposing \/(5x-\-10)=z8—x, By squaring 5x-\-10zz64—l6x + x' i , Where ^—21^= —54 j therefore x- — •+• /(— - U) - — ■+■ /— 2 ""^ 4 ° J ~~ 2"""^ 4 21 15 That is x— — ± — = 18, or 3, the answer. 2 2 Ex.7. Given (l0+x) z —(10 + x)*—2. Here, since the first index is double the second, the equation is a quadratic j therefore by the rule (10 f*J|£ ~ ±v'(* + 2) = 5±v/| =3, Whence (10 + ff) =2; ( 1 + x) *=:4> 10-f-o? = 16, and or— 6. Ex.8. Gi\ en 2t 4 — #H 9^=99 5 by transp. 1 3 2jr*— # 2 =3, or x 4 x 2 zz- 2 2 ( 75 ) Whence ,--±^(- j +-)=-±^, 6 ■ , 6 l „ Therefore a? 2 ^: -, and a?= \/ - = - yo v 4 4 2 Ex. 9. Given a; 6 + 20a? 3 — 10=59 ; by trans. a? 6 + 20a; 3 =69; whence a?~ — 10±v/(JOO-f-69) = -10±13=S or — 23 j and #= ##, or — ^ 23 In n Ex. 10. Given 3* —2a? +3 = 11* by transp. 2* a » 2 8 3# —2a? =8, or a: — ~4h*ss - 3 3 Whence ,»=I±V^ + f = i±4 5 J * • 6 x Therefore x =z -=2, and a?=s2 n 3 Ex. 11. Given 5 fy x— 3 \/a?=l 5, or 1 3 v'a?— 5%/x=z—l~', then I 5 i 4 By division ar 2 — -a? — 7 3 9 Where »*i J±V^S — )-S^*4 6 vv 3£) 9 ; 6 V 3G TOk 4 5,3 4 1 Whence a? 25 5 ± ~ ==■ — > or - , 6 6 3 3 r> / 4 N4 o 13 * Conseq.a?=(-)*=3£P ° r gT Ex.12. Given - aV(3 + 2a*) = - + ? x* 3 v y 2 3 ( ?6 ) 3 3 Mult, by - , and we have ay (3 + 2**) r: - -f x* Squaring, 3 ar 2 + 2^zr -4 -t- - a; 2 -f- ^ 3 Q Whence af 4 -r--af 2 = — ; therefore by the fule ■ —3 , , Q a 3 , 18 4 VV 10^J6 ; 4 V ld 3 3 1 .or a 2 r= ±-y% or #=-<•(— 3 + 3 v^2) 6 1 + ar 2 Ex. 13. Given x */( >#) = : mult, by Jx\ \r */x J v Xi/(6—x*) = 1 -f or 2 j or by squaring 6a 2 — ,r 4 :z:l+2a: 2 + a+ > Whence 2x+—4x z z=. — \, or a; 4 — 2a? 2 = — £ Therefore a? 2 = 1 ± y>( 1 ) — 1 ± W&j Consequently arss v*0 ~ ~ v^ 2 ). Ex. 14. Given - y'Cl— ^ 3 )=^5 multiplying by *r, we h a ve \/ ( 1 — a* 3 ) £ X s , or 1 — ^= a* 6 or ar^ + a^rrl 5 whence also we have, Ek. 15. Given ar^f — 1)= v^(ar 9 — £ s ). By squaring ax— x n ~=:x*— b 1 , or &z*— ax^sl*, ( n ) That .s a r-_^ = _,or a ;= i ±/( I g + -) Ex. 16. Given ^(1 + *— **)— 2(1+*-**)= - y Here (1-f.r— # 2 ) (l-f# — # 2 );rz , Therefore (1 +*-*«)*= i £ ^(-L 1 ±) ,■ 4~~ 16.18~*4 ■ 144"" 4 ~~ 12 "~* 3' ° r &' Consequently 1 +*-# 2 =(-)^= -, or (-) 2 = — o From the 1st. we have x 1 >-x-=. -, which gives 2"" VV 4^9 ; 2~6 V 35 From the 2d. we have £ 2 — tf^— , which gives 2 VV 4 + 36 ; 2 6 V Ex. 17. Given \/(# ) + 1/ (1 - -)=#, . Mult, by v'ftf )— i/i 1 )> and we have x # ar— 1=#V(* ) — ^/(l ) -, divide by x, x x 1 =^/(# )— /(l — )) and from the 1st, <3T X X S3 ( 78 ) xcz*/(x ) -f v"(l ) 5 therefore 1 +# s=2t/(r--), ©r which is the same k x (a>— _) — 2 t/{x->- — )-f- 3 =0^ or by extracting y^— -)— ImOj or or— - = 1, or #*— #= 1 , Consequent] j * = I±V(I-fi)r:~±^ ^5 Ex. 18. Gifen x — 2x +ff =6. By adding x to both sides, and transposing x , we have x — 2x -\-x rro? — x -f6, or (.v -at*) 3 — (# — o; n ) — 65 wherefore J 2 V V 4 * 2 V 4 That is a? -a =3, or x == » - ± \/(- +8), Or^ W = r ± -\/13 5 conseq.tf=;j/(~ ± y' — a/13) Which is the answer required- ( 79 ) QUESTIONS PRODUCING QUADRATIC EQUATIONS. QUESTIONS FOR PRACTICE. Ex. 1. Let .rand y represent the two parts; then we nave # + y = 4Q and x*+y q =8\8, From the 1st #=40— y, or # 2 =:l600-80y + y 2 By substit. 1600— 80y + 2y 2 ~818, or 2y 2 — 80y=— 782, ory 2 ^-40y=— 391 Whence y:r:20± V(400— 3<)])=:20±3=23, or if, But #=40— y = 17, or 23 3 Therefore 17 and 23 are the parts sought. Ex. 2. Let # represent the number sought ; Then by the question (10 — a?)#=21, or #* — 10#= — 21 Whence #=5 ±(25 — 2l)=5±2=7 or 3. Ex. 3. Let # and y be the two parts 3 then by the ques- tion we have x + y =2 241 Here #=24— y, which substituted in the 2d equat. Giv§s 35(24— 2y)=y(24— y), or 840— 70y=24y-y 2 , or y 2 — Q4y ~— 840, Whence y = 47±>v/(47 2 —840}=47 ±37z=10, Consequently #:=:24 — yzz24— 10=14. Ex. 4. Let one part =# 3 then the other will be 20— x, And by the question 20(20— #)~# 2 , or# 2 + 20#=40O Therefore #:=— 10± ^(100+400) = — 10-f 10^5, And the other gart =20— #;=30— 10^5. ( 80 ) Ek. 5. Let x and y represent the two parts, then #+^ = 60, and xy \ x°~+y*i \ 2 '. 5, or x + y=z60, and -^ t ^+y 2 mk Squaring the 1st ; x*-\-2xy+y*z=:3600 Subtracted* x*— 2^xy + y 2 =0 7200 We have 41^=3600, or xy=- =S00, From this we have xzs. — •: which substit. in the y 800 1st, gives f-y=00 J , or y*— 60y~ — 800, Whence y~30± */(900-800)=z30±10 = 40, or 20, Conseq. ^=60—40=20, or 60— 20— 40- In all quadratics of this "kind, in which x may be changed for y, and y for x, in the original equations, without altering their form, the two values of one of the quantities may be taken for the values of the two quantities sought. Ex. 6. Here, in order to avoid radicals, let us assume x z and y 2 for the two parts -, then by the question x 9 '-\-yZ=l46, and x—y=6. Which may now be solved the same as Ex. l. Ano- ther method is as follows : By squaring the second, ■ # 2 — 2xy + y*= 3d Subt. it from twice the iirst, 2# 2 -f 2y*^2Q2 And we have x*-\-2xy+ y*—25§ Hence by extracting x +y = l6 But x —y= 6 Whence by addition 2#=22, or or .1 = 18 And by subtraction 2y~ A, or y= 2 But the more direct solution is as follows : From the second equation x= — , which substituted y in the first, gives \-y = 20 ) oxy^— 20y— — 36 Whence t/=zlO± y'OOO— 36) = 10±8 — 18 or 2, Therefore 18 and 2 are the parts sought. Ex. 8. Let x and y be the two numbers, and conse- quently - and - their reciprocals. Then by the quest. 4 , 1 1 16 1 . x + y = - and - + - — — ; where the 2d becomes J 3 x y 5 ' l6xy , l6xy 4 20 5 a:4-w— 1 7 : whence - — -, or xy— — — — * 5 5 3 J 48 12 Consequently x ~— -; whi hich substituted in the first gives vty +y= 3* or5 + 12 y q = l6 y> or tf— : 6 v Whence 7=: - :£ a/ ( )—_ + -—-, or - * 3 Vl 9 12 ; 3~6 2' (T 1 5 Therefore - and - are the quantities sought. ( 82 ) Ex, g. Let x and y represent the two numbers, Then by the question #— y== 15, and -y-=y 3 , The second equat. gives xy—2y 3 , or x=2y% Whence by substitution in the first we have 1 15 2 V y — 15 j or 2/ 2 y— — > and hence 2d 2i 1 , ,, 1 15, 1 , 11 ^ = 7 ± V (-7H )=-±— —3, Consequently l r = 15+3/ = 15 + 3r=18. Here the two values of y are not those of x and y, because y is made to represent the less number, and can- not, therefore, be changed for x without altering the conditions of the question. Ex. 10* Let arand y be the two numbers y then by the question x—y=5 and x 5 — y 3 —\685, By the 1st x—5+y, or x 3 —\25 + 75y-\-l5y 2 +y 3 Consequently 125 -f 75y + 15y 2 +y 3 — y 3 — 1685,, That is, dividing by 15, y* + 5y = 104, — 5 25 — 5 Whence y= ± \/( f- 104), or y— h 21 2 4 2 — = 8; therefore x=5 + y— 5 + 8=13. 2t Consequently 8 and 13 are the numbers required. Ex. 11. Let x be the number of pieces, and y the shillings that each piece cost -, then by the question » xy = 675, and 4Sx=6y5+y. 675 From the 1st, y = -, whence by substitution we have 675 48*== 675+ — , or 48x*-675x=*675, ( 83 ) ' J. \. 225 225 Or x 2 ttoc— —r ; whence 16 lb 225 , .,225* 225 225 255 *= Jj. ± ^ ( 1F + 10 )= " 12 + "82 = 1 *' And 3/= = -Z-* =45, the shillings each cost, Ex. 12. Let or and y represent the two numbers; then by the question x(x-\-y) = 77, or -r 2 -f-#y=77> And y(x—yy=\2, or xy—y 2 =Y2. By subtraction we have jt 2 -j-?/ 2 ^ 65 5 whence y = \/(65—x' 2 ) ; which by substitution in the first gives ?*-\-x\/(65-x*)=77, or v*(65— s a )= / ~~' r . a 1 . • . 77 2 - 154;t 2 + :r 4 And by squaring, 05—x 2 = ■ , or 210 2^r 4 — 219^=— 5929, or x* -x 2 =—2Q64± 2i Whence x*= 0$ ± V(~- -2964§ ), or 219 23 242 196 , **= — ± T = t^*^ ? 4 \%Q Consequently x= «/60\, or */4Q=zJ ; and adopting the lattery = ^(65 — x*) = ^/ (65 — 49) -=4; that is, 7 and 4 are the two numbers required. Ex. 13. Let x represent the number of sheep, and y the shillings each cost, and consequently y-\- 2, what they sold for 5 then by the question we have And (*-15)(y + *)=1080 ] > the latter by multiplying, gives xy + 2x— + ab, JOv by reduction and transposition, 2as* + 2bs.zza 3 -ab, or i * ■ * « } 7 tf 2 2 Consequently jr= ± vC-~ h - « 2 A)v Where by substituting a~J5 and 1 = 85, we obtain 5=0: but r= zz = S 2s+a 27 Hence then x-\-y—6, and ^ — 8; from which are determined x=2, and 3/ =4 ; and therefore the numbers sought will be J, 2, 4, and 8. Ex. If). Let m+ft be one of the numbers, and tn—n the other 5 then we shall have («z-f?z) -f-(/»— 72) =11, or 2iK=U (w + tz)p -f- (m — nf =1783] =a. Now (^4-n) 5 = w 5 + 5m 4 ?7+ HfewW-f 10wV + 8«rt 4 -f 77% and (w — n) h ^=sm b — 57n*n-\- 10m 3 n 2 — 10?/2 2 /2 3 -}~57?m 4 ( 88 ) Consequently by addition we have 10////2 4 -j-20?rc 3 // a -f f ±nv>-=ia, or a— f lm b 72 4 -f-2mV 2 = 10/* Or substituting for m its value, found as above, in the second term, we have ; ■ 121 n a — 2m b n 4 -\ w 2 = 2 lOw — 121 , , T4641 a — 2m\ Whence n*= ± Ji — 1 ) 4 l 16 10m ' Where by substituting for a and m we derive 9 3 n-~ -, or«= -j consequently * m + »=5|+l§=7, and5f-l|=4, That is, the two numbers are 4 and 7. Ex 20. Let x — 3y, x—y, x + y, and x + 3y, be the four numbers j then by the question (#— 3y) (or—?/) (x + y) (x + 3y)~ 1 76935 =a, or (x 2 — 9y 9 -) (x 2 —y 12 ) =a } or .r 4 — 10# 2 j/ 2 -f9.y 4 =tf ; Or, since by the question (x + 3y) — (x + y)=2y =4 X or ?/=2 , this becomes ^—40^=176841 5 Conseq, * 2 :=20± ^(l /6841 +400)=441, Hence ,r ~ -v/44l=21, and #—33/ = 15, #— #==19. #-{-?/ = 23, and ,r-f-3z/— 27, the numbers sought. Ex. 21. Let x==: the number of hours march of the first detachment, and y the miles per hour ; then x-\- 1 will be the hours of the second, and y — j the miles per hour. Whence by the question we hay.e ( sg ) #y=39, and (x+\)(y — i)=39 3 or #?/ — $x + y — i~ 3p; Or since xy~ 3g, we have — 5^+y — £— O, or 4y — #= 1. 30 , 3Q Again, ars? — j whence 4?/ zsl, or 4j/ 2 — 39=y,~ 1 39 , * 1 , , 1 39 x 26 Or^-jJ/^-f 3 therefore y = - ± V (_ + -!!)=_ =34:. Consequently 3| and 3 miles per hour are their rates of marching. Ex- 22. Let x and y represent the two numbers 5 then by the question x q -\-xy=140Y y 2 —xy = 78 J By addition x 2 -f2/ 2 =218, or y= .^^IQ—x*) ^ , 1 r • 140 -J? 9 But by the first equation y= 5 140 #2 Whence = */(2\8—x*)', or by squaring and^ reducing 39600- 280^ + ^=218^— x\ or 2# 4 ~i98#= - 19(500, Whence, by dividing by 2, we have ^-249x 2 =-9800 5, That is * 2 = — ± V(^- -9800), or ife* ?^ ± 151 , viz. # 2 r=200, or 49) therefore by adopting the latter, '^, : 140—49 i > we shall have arz2j> and 3/ = • =^ 13 j which* are the numbers sought. i3> ( 90 ) Of CUBIC EQUATIONS. To exterminate the second term from a cubic Equation* Ex. 3. Given equation x s -6x^10, or x 5 — 6x q — 10=0 Here x~%-\- 2, r * S ==% 5 4-6V 2 4-12*4- 8 Therefore J -- 6x Q ^z — 6%*-24x— 24 £-10:= -10 Whence we shall have s. 3 — 12% — 26=0, or % s — \2zzz26 as required. Ex.4. Given equation y 5 — 1 5y* -h 8 ly— 243 ~0. Here y~ x + 5, r y 5 =zx 3 + \5%*+ 75X+V25 rvu c /~15w 2 = — 152 Q -1 50^—375- Therefore } +81 J = + 8 **+405 V— 243= -243 Whence we shall have x* + 6x— 88=0, or x s -\~6x=:S8 as required. 3 7 Q Ex. 5. Given eauation x s -f - x 2 + ~x ^ zzO, 1 4 8 16 Here xzzy , J 4 ( 91 > , , 3 . 3 < Therefore 7 ^8 3 . 3 3 8 y+ 6i 7 __7_ 8 * 32 -g 16 . n 3 Henee we have y 3 + — y - - = 0, or • 11 3 V + l6^~4 a8re ^ ulred - Ex.6. Given equations 2x 3 — 3.r 2 -f 4#-- 5z=0^ or 3 i 5 Here x=% + 2' r i X s zz. ■r+J* '+| 1 Therefore i < 3 n 2 3 -2* 3 "~ 2' 3 8 i + 2ff = + .2«+ 1 i — 5 _ -5 i ~2~~ "" 2 5 7 Whence we have % 3 H — * =0, or 4 4 z 5 -f - xz= - as required. In which equations the second term is wanting. { N ) SOLUTION OF CUBIC EQUATIONS. I Ex, 5. Given x 3 + 3x 2 — 6x=z8, to find x. Here xzziy — 1, s x 5 z=z y 3 — 3^f3y— 1 Therefore ^ 6r = 1 * JjJ+g Heduced equation ?y 3 — 93/— Consequently 3/nO 5 and dividing by y t we have- y*z=9, or^:=-J-3, or -—3; whence the three values of x are x=xy— 1 = 0^-1.= — 11 xzzy— 1= 3 — 1 — 2> as required. t ff=y-l = -3-U=-4) Ex. 6. Given # 3 + £*:=: 5 00, to find x. First #=% , 3' ;r 3 =% 3 — % a + -% 3 27 Therefore J 2 1 *.£ 9 -500= —500 Reduced equation % 3 — - % — 499|] =0, or; 3 1 a" » 3 --»=499lf Whence we have azz—- and £=499|f 5 conse- quently by our formula . ( 93 ) becomes Which last expression, being reduced, gives 4/^49-9629- v/(249'9^29 a - ~)\ or 81 * = V499'9257 + Voooi=7'950, Consequently x~%— T zz7 950- '333 zz7 6 \ 7 ans. Ex. 7. Given equation a? 3 — 4S.r~= — 200. First x=zz+l6, ^ = s 3 -f48x*+ 768%+ 4096 Therefore or x=z V(3 + 1)+ ^(3— 1), or x— #4 + 1/2 Ans. Ex. 9. The given equation a; 3 + 9^=6 is here al- ready in its reduced form ; whence we have immediately x=zV\3+ V(9+W)\+ 3 V\3-\/(9 + 27)l, or x~ 3/(3+ 6) + V(3— 6), or sri V9+ V-3, or ar= ?/9— 1/3 Ans. Ex. 10. Given x 3 -22ar=24, to find or. Here,, by the formula, we have 10648 N , 10648 x , ;y]i2-v(i44 — gL)}, Or by reducing and simplifying the expression- x=zV\!2+ rv /(144-27~ 10648)}, + o y\\2— ^(144-27-10648)1, or 3 x=z V[12+- -V(38S8- 1-0048)1+ 3 V{ 12— -a/,3888- 10648) |, or j?= V(i2+ ^V-6760) + V(12~- ~v/~676o> Which is also of the irreducible form, as in Example t ', and therefore cannot be resolved by the common for- mula. ( 95 ) 7 Ex.11. Given equation x 5 —17^ + 5 4j?= 3 50, t® find L r. This equation, by exterminating the second term, becomes 1 15 3 / 27 Whence, by the formula, we shall have Z =VS203^ + ,/[(203?±)«-(14±)»]$ Or by reducing and simplifying the expression, 1 *=;; V| 5502+ V '(5502*-127 3 )} + 1 r V{5502— v /(5502 2 -127 3 )}, or *~- V ^5502+^28223621^ + - Z/\5502- V28223621^ or ^=e r (V5502 + 5312'1776)+ - V(5502 — 3 3 5312-1776), or ^=-V'10814-1776 + -v / 189'9224z:9-28739U \y Consequently x=zz+ — = 14-954008 Ans. Note. Question 7 and 10 can only be resolved by a table of sines, or by infinite series $ for the method of doing which see my Treatise on Algebra, 2 vols. 8vo. 3613. ( 96 ) Ot BIQUADRATIC EQUATIONS Ex. 2. Here the given equation, viz. # 4 - 55X 2 -— 30*4-504=0, being of the proper form for solution, we have by the first rule, b^L—55, c= — 30, and d~3Q4. The cubic, or reduced equation, therefore, yiz. a 3 — (— b <2 + d)zz=z h 3 + - c 2 Id v 12 J 108 8 3 becomes ^-756 T L2=781H§| Where the root of this cubic being found =3 i'66666, or 31 -, our two quadratics are £tfegj.*$| + ^» a? =-(3l|— ") + ^J(31 j - 55 -g)'- 504 J --(^{3l| + ^|).— (8l|-f)-^(3lj -f ) 4 "504^ or i*^lQrp-aij and # 9 — 10*=± — 24. The first gives x-=z5±^/ (25 — 21), or *z=5, and f. And the second .rr:— 5± */(25— 24), or #=—5 and — 4$ That is, the four roots are 3, 7, —4, and ^ — ^. ( 97 ) Ex.3. Given equation x* + 2x 3 — 7$*— Bti&rr 12. First, in order to exterminate the second term, we have orzz%— |i whence *•=-=*♦— 3* 5 + | **— | z +Jg , 3 1 + 2^ = 42s 3 — 3s 2 + ~ as- T 2 4 -7^= -7~S+7z- 7 - -8* = — 82+ 4 4-12 = + 32 Reduced equation z 4 — 8- s*+ 14 — =0, or 2 16 I 280 225 17 . 8 Whence a -4 5 ±V( 1 |-- 1F )=-f± i = 4 4 ,25 5 — 5 Consequently »=rdfc\/-— — 4 — » or , and And therefore att»— f has the four following values, viz. __ 5 1 4 __ *"~" ~2~~ 2 Tr 2 ~T *$ 2 2 — 3 1 — 5 A ~ 2 "~ 2~" 2 ■-' : ~~ 2 2 ~"~ 2 ( 9® ) Ex. 4. Given x 4 - a* 3 + 14^ + 4^=8 to find x. First x— 2 + 2 Whence ,z 4 =* 4 + 82 s + 24% 2 + 32% + 16 - 8^ 3 = — 8% 3 - 4 8%' 3 — 96%— 64 -i- 14r 2 — + \4z 2 +56z + 56 + 4* = 4* + 8 — 8 == — 8 Reduced equation %*— 10%- — 4% + 8^l 0, From which we obtain the following cubic, 100 . -1000 16 80 J v 12 J 108 8 3 f -10-ur: 10 — . J 3 "* 27 — 4 The root of which cubic is y=. — 1*3333, or . Therefore the two quadratic equations are Or *?+£«=$, and* 2 — 2z=4 ; Whence the four roots, or values of 2, are *= — 1 + -s/3, and 2= — 1 — a/3 2 = 4-1+ v/* 5 * anc * *= + 1 — a/5 ; And since cc=z + 2, we have the four following va- lues of or, viz. x=3 + \/5, x=.3—\/5 x— 1 -f \/3, x= 1 — x/3, as required. r 99 ) The student will obs rve here, that as the quantifies-' under the two latter radicals, in the above quadratics, arise from the square of a negative quantity, their loots must be taken negative, and not affirmative. Ex. 3. Here the given equation, viz. a 4 — 17^—20^—6=0 is already in the proper form for solution, having 1-= — 17, CZZ-—20, and d= —6; whence our cubic is z 3 ~( o)z= -— -{ — a ', or v 12 I 108 8 3 1 53 x 3 -— 18 — s=-29 ; 12 ^108 The root of which cubic is s=2-3333, or srr 2f. Whence our quadratics are Or.^ 2 + 4or=— 2, and #*— 4x±?3, Therefore the four roots, or values of x> are x= — 2 -jV 2 > Xz=~2— a/2 #=4-2 -f */7, x— + 2— V7> as required. Ex.6. Given equation or 4 -2 7^-}- 162^4-356^-* 1200=0. 27 First x=z+ — 4 ^"; 2187 , 1Q683 531441 8 16 256 2187 5Q0 and d=— 3-, our cubic is * , x -1/28 144 % 3 — (12 — 3)z— — H 12, or V ' 108 8 % 5 — 9%— — 10 Where we have, from inspection, % — 2; whence the following quadratics, x*+ {**= — \s/Yl± ^(3-f- VS) X=z +1^121/(3—^/3) ( 101 ) Or the four values of x in numbers are - #=— 3-907378; x —-443277 x= 2858083 5 x = '6060i8 RESOLUTION OF EQUATIONS, BY APPROXIMATION. Ex.2. Given equation x q + 20xz= J 00. Here a few trials show the root to be nearly 4'lf let therefore #__ 4* 1 -+- % 5 Then x*~ 16*81 +8 2%-f x 2 ? lQQ 20,rr:82- -f20% J Therefore rejecting z*, we have 98 8fc-r-28-2x = 100, or 1-1Q ZSS =5-3= '042 5 282 And consequently a;=4i +%m4*l + 042z=:4 , 142 f Whence by repeating the operation, and assuming xzn 4*142 + %, foui other figures may be obtained which gives the value of #=4*1421356, as required. Ex. 3. Given x 3 + Qx*+ 4# — 80. By trials we find the root nearly zl2j assumingv therefore, xzz2 + z, we have .r 3 = 8+12» + 6k3+»3^ 9 l r 2 rz36+3.6x-f 9% 2 >== 80 4#= 8+ 4* J Whence, by rejecting the second and third powers of z, we have 28 52 + 52x=80, or %= — r-zL'5 52 And therefore #=2-5 = the root, nearly* k3 ( 102 ) Assume now xzz2'5+z, and we have ^=-I5-625 + 1875s + 7-5* b +« s ^ 9^=5fj-25 +45'z 4-9** >=8G 4x=z\0' -f A% J Whence .rejecting the higher powers of %, 81 '875 + 6775z- 80, or 1-875 0/-75 Where x=45 — ' 02? 6 —2' 4J 2 nearly; and if again we were to assume ,r=:2' , 472-fs, we should get four other decimals, which would give #-2*4721359. Remark. In the preceding examples we have involv- ed every power of x completely ; but it is obvious, that as the higher powers of z are always rejected, it is un- necessary to carry the expansions beyond the second terra, as shown in the following solution : Ex. 4. Given equation x* -3 Sx 3 -^ 210a : Q + 538a ^-f 289~0. Here, since the second term is equal to all the other terms of the equation, it is obvious that x is less than 38 j and by a few trials we find it to be nearly 30 j let there- fore xzz 30 + % ; then reserving only the first two terms of each expansion, we have x*= 810000+ 108000* + &c. — 38^= — 1026000— 102fj00x-&c. I -f2io# 2 = 189000+ 12600^-f &c. y =0 + 538*=: 16140 + 538.Z + &C. + 289 = 289 Whence - 10571 + 18538s=0, or *= — — =\5, ' 18538 And #— 305 nearly. Assume therefore x=30'5 + % ; and we shall have (rejecting the deci- mals as inconsiderable) ( 103 ) jr 4 r= 865305 + 1 13490* — 38^ 3 =-107S159~3078l6'x + 210ar 2 = 195352+ 12810s: + 538x =z 16409 -f 538* + 289 = 289 Whence — 744 f 1^02 2»-0 744 Or z = =-039; where a: =30*53 nearly : J9022 V 7' And by assuming again .r=30 53 + *, a still nearer ap- proximation may be obtained. But in equations of this kind, in which the coefficients are very large with re*- gard to the root, the approximations are always very slow, adding no more than one new figure at every ope- ration j whereas, when the coefficients are small with re- spect to the root, each operation doubles the number of figures last obtained. Ex.5. Given a^-f-6* 4 — 10a; 3 — 112,r 9 — 207# + *110 =0, to find x. Here x=4 nearly; assume, therefore, #=4 + *$ then x b = 1024+1280* + &c. + 6x A = 1536+1 536% + &c. - 10,ir 3 = — 640— 4S0*— &cJ — 112#°— — 1792— 89621- &c. J — 207x=— 828— 207*— &c.' + 110 =+ 110— * , 590 And —590+1233^=0, or ^=-4- ='47, 1233 Whence #=4-47 nearly. And by assuming ar=4'47+js> another approximation may be obtained, till at last we rind x=4'464 10l6l 5 but it would be useless to go through the entire operation in this place. This sign is misprinted in the Introduction. = ( 104 ) Of APPROXIMATION BY POSITION. EXAMPLES FOR PRACTICE. Ex. L Given ^4-10^ + 5^=2600. Here it is soon discovered that x is a little more than 11 j let us assume, therefore, x~l l'G and ,r=iri j Then by the rule 11-1 = x = 11-0 1367-631 = x 3 = 1331 1232-1 = + ]0s* = 1210 55'5 =+50?= 55 2655*231 Results • Therefore 25Q6 2655-231 1 1*1 260O 2596- 110 2596 Or 59231 : -l :: 4 : 00673 Whence ,r=iro-f- -00673 = 11-00673 Ans. Remark. In this example one of our suppositions approached so near the truth, that we have been enabled to obtain seven places of figures true, in a single opera- tion * which is a degree of approximation very seldom acquired with so little labour. Ex. 2. Given 2ar 4 — 1 6^ + 4 0^ 2 -30jr-f 1=0. Here x is nearly = lj assume, therefore, 2 = x * =z 1 ( 105 J Then, 32 == 2.r 4 = 2 -128 as —i6a 3 = — 16 4-160 = 4-40* 2 = -f40 — 60 = — 3(Xr = -f30 -f 1 +1 = + 1 4- 5 Results 4- 5 3 Therefore 2 1 — 3 O — 3 Or 8 : i :: 3 : -3 Whence #=13 nearly, assume, therefore, 1.3 = x = 1-2 Then 57122 = 2x* = + 4-1472 — 35-152 = — Idx 3 == -27*6480 4 67-6 = +40.1 2 = 4-57.6000 -39- = — 30# = —36*0000 + r = 4-1 =4- l'OOOO +-J602 Results — -g008 Therefore + • 1602 1-3 — '9008 1*2 —'9008 Or i-o6io : -i :: -9008 : -os5 Hence #=1*2 4- '085= T283 nearly $ and by repeat- ing the operation we obtain the still nearer approxima- tion 1-284724. Ex.3. Given x 5 + 2z 4 + 3z*-\-4x i + 5x= 54321. Here we find the value of x to be between 8 and 9 ? assume, therefore, ( 106 ) Then 32768 = x 5 = 59049 8192 55= 2.r 4 == 13 J 22 1536 = 3x 3 = 2187 256 = 4a? 2 = 324 40 = 5# == 45 42792 Results 74727 Therefore 74727 9 54321 42792 8 42792 Or 31935 : 1 :: 11529 i * Whence #=8*3 nearly. And by assuming a^S^ and 8*4, another approximation will be obtained; but the successi ve corrections are very small, and would occupy more room than can be devoted to a single example. Ex. 4. Given 1/(7 ** + ***) + y(20.z*— l(Xr)=28. Assuming here #==4 and x—5, we have 4 = X = 5 8 =3/( 7a 3 -Ma*)= 9-91596 1 6*7332= V(20a^— I0r)=21-21320 247332 Results 31-12916 Therefore 31-12916 5 28 2473332 4 24*7332 Or 6-39634 : 1 3-2668 : I Whence we shall have x—4'51 nearly. And bv repeating the operation #=4"51066j. ( 107 ) - Ex.5. Given i/l(lUx*--a? + 20)*} + )/\(196&—- a 2 +24) 2 }=114. Assuming x=7, and x=8, we have 7 = x = 8 47-906l= VK 144 ^~ * 2 + 20)^=46*4758 65-3834= v'K'S^ 2 — a?a + 24 ) 8 l == ^9- 2820 113*2895 Results 11 57578 Therefore 115.7578 8 114 113-2895 7 1132895 Or 2-4683 : 1 :: 7105 : -2 Whence a?==7'* nearly. Assuming now ,r=7 , 2, which is found too great 3 let us therefore take x=7'l and 7% and we have 7-1 = or = 7*2 65-9014 — / \(l96x*—x' i + 24)*i=66 4003 113-8805 Results 114*4002 1144002 7*2 114 113-8805 71 113-8805 Or -5197 : # i :: -1195 : -83 Whence x=7'l + '023—7*123 nearly. And if, for a new operation, there bef taken x=/. 123 and 7*124, we shall have x= 7.123883. And a further assumption of this kind will about double the number of decimals. ( 108 ) EXPONENTIAL EQUATIONS. Ex. 2. Given x* =2000, to find x. Here we soon find that x is nearly =5 j let us, there- fore, assume *=4*8, and #=4*9. Then the log. 2000=3 3010300; and Log. 4*8=06'812412 Log. 4-9=0-6901961 Mult, by 4.8 4.9 54499296 27249648 326995776 Results Therefore 3*38! 96 49 326995 4*8 62117649 27607844 3*38196089 3-30103 3-26995 Or -H201 : -l :: -03108 : -027 Whence #=4*8 + '027=4-827 nearly. And repeating the operation, by assuming x*= 4* 827 and #=4-828, we shall obtain x= 4 82782263. Ex.3, Here (6r)*=96; and our formula must there- fore be xx (log. 6 + log. o;)=log. 96. Where log. 96=1 98227 12, And x is obviously nearly =2 ; assume, therefore, 1=1-8; and #=1*9; then Log. 6 =0*77815 Log. 6 =077815 Log. 1-8=0-25527 Log. 1 •9=0-27875 1*03342 1-05690 1-8 1-9 826736 951210 103342 105690 1-800156 Results 2 0081 10 2 00811 1-86015 ( 109 ) Therefore 18 1*98227 L'86015 •14796 : -l :: -12212 : *os26, Whence x=V8 + *0826= 1*8826 nearly. Which rive figures may be doubled by another ope- ration, making #=1*8826432. Ex. 4. Given equation #*= 12.5456/89. Here, after a few trials, or from inspection in a table of powers, we find x is between 8 and 9, but nearer the latter than the former. Assume, therefore,'#-^8*6 and x—8'7. Then log. 123456789=80915149 Log. 8*6=0*9344935 log. 8*7— 0*9395193 • Mult, by 8.6 8.7 56069910 475988O 803668710 Results Therefore 8*l73tl 87 803668 8-6 65766351 75161544 8*17381791 8*09151 S'03668 •13713 : -l :: -05493 :*04 Whence x~ 8*64-040=8*640 nearly 3 And repeating the operation, by assuming x equal to 8*640 and 8641, x is found =8*6400268. JL Ex.5. Given x*— l=z(2x~x x ) x . This will be more convenient under the form (tf*—l)*=2z— x*. L ( no } Now, in order to find a first approximate assumption, it may be observed that 2x must be greater than x x , or 2 greater than x x — «. . The same will also be obvious if we put the equation (x x — %y under the form 4- x x — x z=^2. x Since then x is less than 2, but nearly equal to that number, let us assume #=1 '8 ; then ;r*=2'88G6, and (x x — x) x : L*. fa 0-6387 x Also x x ~ l = .1*6004 1st result 2-23pl And when x=l'7, then , (x x — x) x [647, and — — '— = x * Also x*— i = 1-4498 , (x* — x) x x*— 24647, and - — — = 0'3728 x 2d result 1-8226 Therefore by the rule, 2-2391 1-8 2 0000 1-8226 J 7 T8226 As -4i65 : i :: iff4 : -04; Whence #=17 + '04 = 174 nearly. And repeating the operation with the assumption «r=:174 and xzz\*75, we find #=1747933. Also, by another assumption, a still nearer approxi- mate value of x may be determined; and so on, to any degree of accuracy required. ( Hi ) BINOMIAL THEOREM. Ex. 5. Here the proposed binomial being ^(1 + 1)> or '■">■■'■'• ii ' . 1 w 1 (1 + 1) > we have />=!, 2= - = 1, — = - 1 7? 2 Whence 771 1111 tt 2 i 1 2i m—n 1—2 1 1 —1 — Ba= — X 2 X 7 = ^= C ' m,— In 1—4 —1 1 1.3 3n ~~ 2.4 1 2.4.0 m — 3n 1—0' 1.3 1 —1.3.5 4» 8 2.4 6 1 ~" 2.4.6-8 "" ' Where the law of continuation is sufficiently obvious ; and therefore we have v /(l + l)= [/ 2 1 1 1.3 1.3.5 1.3.5.7 2 2.4 2.4.0 2.4.0.8 2.4.0.8.10 i Ex. 0. Here we have to convert (8— l) 7 , or (2 s i — 1) T into a series. Make 2= a ; then it becomes (a 3 — :i * , i — 1 . m, 1 1) J : where p= a 3 , a= — -, and — = -, a 3 n 3 Whence ri • F» = (r)?a>(^) 3 = « = A, ( 112 ) m 1 <* — 1 —1 n 3 I a 3 3a s f»r*n __ ! — 3 — 1 — 1_— 1.2_ w — 2fz 1-6 —1.2 —1 —1.2.5 can x X — -- — '• = D, 3» 9 3.6V a 3 3.6.9a 8 Therefore y(a s -l)= 1 1.2 1.2.5 12. 5. S fl ~~37^~ 3.6a 5 "" 3.6.9^ ""3. 6.9. 12a""" € ' Or substituting 2 and its powers for a, we have V(8-1)=V7= 1 1 1.5 1.5.8 2 , . fee 3.2* 3.6.2 4 3.6 % 9.2 7 3.6.9. 12.a 10 Ex.7. Hereagain (243~3) T =(3 5 -3) T =z(a*-3) T , , —3 , m 1 by writing a~3 5 also p=a 5 , air -—-and — = -5 wteaiee a & n 5 *» la —3 —3 AQ= - X - X St =B, * 5 1a 5 5a 4 \ 7)*-n 1 — 5 —3 —3 — 4 3 2 « — B€tIZZ X X """■ ■ — C 2» ' 10 5** a 5 5.l0.a9 *"~ • *-2tt 1 — 10 — 4.3 2 —3 —4.0.3 s — ca X - Q x z= — ~d. 3» 15 5.10.a 9 a b 5.10.15a 14 Whence by writing 3 for a, and cancelling the like powers of 3, we have V(243 — 3)=z 3/240 — 3 _ J_ ___i i? 4.9-14 5.3 3 5J0.3 7 5.10.J53 11 5.10.15.20.3^ < m ) Ex.8. Here p=a, orz— , and — is -, a » 2 Therefore p 7I zz (a)« = a z zzA, r» 1 a\ ±x , # | — Ati= - x ~ X ■ = ± — a zzb, n 2 I a 2a ^__ w 3—2 ±_x J ±# — jt 2 E€i= — — - x -- — x = - : - a -=c, 'In 4 2a a 2.4a' . m—1n 1—4 — a* % ±x ±3x 5 | t ft= X a X — -fl =D, . 3/z 6 2.4a* a "" 2.4-6-a 3 W— 3JI J -6 ±3* 3 4 ±J7 -3.5^ 4 | 1 Whence (^±a) 2 = «* J 1 -4- -f — 4 + &c 1 I ^ 2a 2.4a 2 ~ 2 .4-6 a 3 2.4.6.8a ~ J :±& _ ?ra 1 Ex. Q. Here we have p=ce, 0= , and — zz - ^ a n 3 Whence p»r="(a)« = a 3 =a, * Aa - 3 X — « 3a bo — -— - X— - a 7 X — T7Tl a C > 2?z 3a a 3.6a 2 I_6 — 2# z ±6— ±2.5£ 3 I m—ln I— O —Lie nz^ = i:2.5 r_ _ ca= X ■ - , X — -a 3 =r> 3# 9 3.6a 2 a 3.0.(),a 3 l 3 ( H4 ) From which the law of continuation is obvious; and therefore we have (a±/>) T — « 3a 3.6a 2 ~ 3.6.(Ja 3 3.0.9.12a 4 ~ * Ex. 10. Here v= a> a= — , and — zz -, a n 4 Therefore - i (a)*=a =a, m 1 a - h —h$ — -A =-X— X' = -— a =B, 7f 4 1 a 4a w~w 1—4 — £ i —b —31 B a— — — X- — a x — a =c. ft An „ A O ,.2 °» 2?i 8 4a a 4.S.a tf!—.2n 1—8. — 3£* | — b -3.7P Cft=-7- X — — Q a * X — ■ = 3?z 12 4.8. a 2 ' a 4.8. J 2a* ~ Where again the law cf the series is discovered ; and we have (a — &)*= if * 3/; * 3./£ 3 3.;.1 1/; 4 - -v l 1_ ^4a ""4.8a 9 "4.8. 12a 3 " 4.8.12.10a 4 J In the three last examples we have repeated the frac- tional root of the first term in every line y but it is ob- vious, that this is not necessary, as we may leave it out of every term -, only remembering to introduce it at last as a general multiplier of the. series. Or we might have put onr examples under a different form, as in the following instance : ( 115 ) Ex. 1 1. Here the proposed quantity may be put un* — x % der the form a 3 x(l + -) ^j Rnd therefore, omitting a T for the present the multiplier a 3 , we have X 771 2 p = lj a= -, and — = -; whence a 7z 3 m m "%_ P^=1^Z=1 T =1=A, m 2 1 x 2x -Aft-- X- X -=T =B, n 3 1 a 3a ??? — 72 2 — 3 2# ^r__ — 2a^ __ __ B0 _ ___ x _ x __ ____ Cj m — 2n __2 — (5 —2^ £ _ 2.4a? 3 ~~3^~~ ° a "~ "IT X ~3J5a? X a ~^3~ia^ ""^ m,—3n 2— Q 2.4r 3 a? — 2.4.7.T 4 4» ' 12 3. 6.9a 3 a 3.0 9.12a 4 ' i 2, ,r — Whence (a + tf) 3 , ora 3 (l + - ) 3 = a 4 f 2# 2r* 2 4J 3 — 2.47JT 4 *i a **\ 1 + ^-JJtf + ^6J^ "3.6.9.1 2a 4 +&c ' /■ Which, by cancelling the multiplier 2, in the nume- rator and denominator, becomes f f 2a? .r 2 4# 3 4.7# 4 1 a X t l+ 3^~9^ ^W ~~ 9 2 .i2.15^ 4+&C '-) 1 w 2 Ex. 12. Here r^rl, g=— x and — n-, # 5 Whence we shall have m m 1 P^=:l^=t 5 =:l = A^ < u6 y. m -2 1 \-x — 2x n 5 1 1 ' 5 m—n 2 — 5 —2x — x ~2.3x t Ba== — — - X X = >~ c, 2?i 10 5 1 5.10 m — 2n 2—10 —3 — x -2.3 8r 3 caz: x - — — X = =.d, 3n 15 ' 5.10 1 5.10.15 Therefore (1— x) T =z 2x 2.3.r 2 _ 2 3.8# 3 2.3.8.13* 4 ~~~5 5.10 5.10.15 ••■ 5 10.15.20 ~~ 1 _I Ex. 13. Here , , ^ =(a±x) 2 : therefore (a±xy ' 7 ±x \ m —1 p=tf, a= , and — a= % a n 2 m m i Whence we have p n =a" —a 2 = a, And by omitting this factor in the subsequent ope- rations, m —1 ±x ~T.x — Aa= — - x — - = J — — b, n 2 a 2a m — n —1—2 ZLx +# +&** — bq= 7 — X ?— x - — = ■ , a =c, 2n 4 la a 2Aa l m —2n _ — 1— 4 -fr 2 ±3x _4l3.5# 3 __ _— ca = g— X — X — - 24l ^ -d,. TO — 3rc —1—6 +3.5X 3 ±x +3.5.7X* • 4w • 8 2.4. fra 3 a 24.6.8a" -I i Whence introducing the general factor a z ' 3 or — we hav« - — : — £ (a±aj ( t\1 ) ~o~~ T 1 1 + 27~~ CQ ~~ g ' *3J5a* x IT™ 3.6.9a 3 ***' ;//-3ff —1—9 qp47# 3 ±# -H 7-1 Or 4 ————— D ft — v^ n^ _____ 1 — ________________ — * ~ jg 4tt 12 3.6.9. a 3 a ""3.6.9. 1 2a 4 , Whence, introducing our two general factors ax -* zzaT, we have a(a±:x) 3 = .T( 1 _ F ± + i_t = p_i__t ,+__---- T& C.| (. ^3a 3.6a lj ^3.6.9.a 3.6.9. 12a 4 ^ ^ 1 — — Ex.15. Here \ r f ==(l+ar) ^consequently ,r m — 1 p=l, g= - and — = , I n 5 Whence ( 118 ) m -1 1 X — X — a a: n "~~1T x i x I = — =B ' m — n B* x 6x 9 6.113* 6.n \6x* 1 1 . *_ + _ Src, 5 5. 10 5. 10.15 5. lO. 15.20 Ex.10. Here ( ) — = -i X j (rt — x) z =JL±L i= =( fl + ar) -:a«-^) 2 . Whence, omitting, till the expansion is effected, the leading factor — #2 m [ a-\-x\ we have p = tf 9 , a~ — —■ and — = — -. a 1 n 1 And consequently i ^ m —„i _, i p» = (a 2 )« = ia?)— =a = - = a, a w —11 — .r 2 a? 2 «+*= 2 X « X -^- = ^- =B ' w — n —1-2 x 2 —x* 3x* na =z X-r-r- X 2« 4 2a? a 2 2.4^ ( H9 ) m — lti —1-4 3x* — x* 3.5.T* ca= X x es =_r», 3« 2.4^ a* 2.4.6V w — 3« -1-6 3.5a 6 — x* 3.5.7** 4* 8 2.4.6 ~ a* 2.4.6.8a* Or (a*-**) = 1 x 2 3x* 3.5x* a 7 2a 3 ^2.4a 5 X 2.4.6a? " IMult. by a + x x* 3x* 3.5.x* i H r H + z-r« 4- &c. 2a 2 2 -la 4 2*4.6a 6 x x 3 3x b 3.5x 7 xx* x* , 3.Z 4 So:* AnS - ^a^^i+^ + I^ + ^a-- 1 -^- Or, the expression in this example may be reduced to a more simple form by taking i__Y*=(iXt__LJ But in order to have the same answer as above, the terms of the result must be divided by a + x and its powers. Note. The coefficients of Example 14 and 15 are wrong in the answers in the Introduction, viz. the 3d, 4th, and 5th of the former, and the 4th and 5th of the latter. The unit is also omitted in the answer to the 16th. ( 120 ) INDETERMINATE ANALYSIS. PROBLEM I. EXAMPLES FOR PRACTICE. Ex. 1. Given equation 3x~8y — 16. 8y — it) 2y-l Here xzz. -£ ,. 2y— 5 4- — . 3 J 3 Let -2- — =:p ; then 2y- l=3/>, or 2y=3p + 1 3 3/>-f 1 />+ 1 Whence ?/ ~ -p-j 7 which latter will obviously be a whole number, provided p be taken any odd number. Assuming p=3, we have y=. -f- =.5 and a:.= -^— — =="8 , which are the least values soeght. 3 Ex. 2. Given equation 14r~5?/ + 7. Here it may be observed, that since 14jt and *J are both divisible bv 7, 52/ must be so likewise; consequent- ly y must be divisible by % Let therefore y = J /z ) nnd we have 5y=35z ; which substitute for5y 9 gives the equation 1^=35*4-7, or 2#=5s-f 1 j 5%+L z-fl Whence ar=- = 2z-\ —ivh. 2 2 z-f 1 Therefore — - — =p, or z^=2p + 1 ; whence #== 5 _I±2 i ^ =5/) + 3 j and y=:*=Up + 7 , Hence the general values of x and y are x=5p + 3, and yt=14/)-f7, where /> may be assumed =0, or any integer whatever -, the former value, viz. />=0, gives x=3, and 2/ =7 the least values. ( 121 ) Ex. 3. Given equation 27x=l600~ lGy. Here again it will be observed, that since both terms on the right-hand side are divisible by 1 6, the left-hand member, viz. 2Jx, must be so likewise ; which cannot be except x itself be divisible by 16 j make therefore x — l6z, and our equation becomes 27\6z~l60O-}6y, or 27z=z\00—y ; Whence ?/=100 — 2Jz ; and x~l(5%, where % may be assumed ut pleasure, providing 27 z is less ihun J 00. If z=J, then #=16, and y =73, z = 2, then #—32, and y—46, z—3, then #=48, and y = J 9, Which are the only four answers. Ex.4. Let 7# and 1 ly be the two parts required, then we have Jx+lly= 100. JOO— 117/ 24-3y Whence x — — = 14 - 2 y 4 -, 7 7 7 2 + 3?/ Make now £- = p 5 and we have 3y=7/> — 2, or 7/> — 2 />+* a . , P+l ?/ = — =2p — 1 -| -. Again make — — = 7, o o xi and we obtain p=z3q~—l. -ru f 7 ^~ 2 21?-/-2 * 1 Therefore 3/= _— = ^4^ = 7<1~3> A t 100— lly JOO— 77<7 + 33 And *= — ^ = -^ = 19-ny, Where it is obvious that q cannot be taken greater than 1; making therefore 2-18 + 36^ ^ Hence y= — — = - *- =9?- 4 > 2000— lo 7/ 2000—1177 4-52 and «•== 3 V '-12- =228—13? 9 9 Where ^ may be assumed at pleasure, providing only that 13^ be less than 228} it may therefore be any inte- ger from 1 to IJ; which latter therefore denotes the number of possible solutions. If £=1, then y=Qq-4=5, and #=228—13^=215 If q~1, then 3/ = 14., and #=202 If q—Z, then y = 23, and #=189 &c. &c. Hence the answer in the introduction is not correct. Ex.6. Given equation 11 x+5y =254. 254— 11# l-f# Here y— = 51— 2# ziivh,; 9 5 5 ' 1 4-# Whence making =/>, we have x~5p— 1, 5 254 — 11# 254 — 556+11 And y = » . - T =53-11/). ( 123 ) 53 Where/) must be assumed less than — ; that is e= any number from 1 to 4. " Hp=\, then x=5p— 1=4, and y— 53 — Up— 42 p=2, then #= 93 and y=3l p=3, then /r=l4; and #=20- p=4, then #= 19; and y== 9. Ex.7. Given equation l/^-f I9y + 21%=400. In questions of this kind, in which only the number of solutions is sought, the answer is more readily ob- tained from the following rule : Let ax+by=xc be any proposed indeterminate equa- tion, and find the value of p and q in the equation ap — bq ~\ ; then the number of possible solutions of the equa- tion ax-\-by=.c, is equal to the difference between the integral parts of the fractions — „ — . * In our proposed equation, by 'transposing 21x, we have 1 Jx -f 19^=400— 2\z ; and by giving to z the several values ], 2, 3, 4, &c. we obtain the following set of equa- tions ; p being =9, and q— 8 3 Equations. No. of Solutions. Q 37Q 8.37Q 17*-*- J 9^=358 3 19 17 9.358 8.358 19 17 ,~ , , ~ 9-337 8-337 17^4-19^=3375 * =1 17^4-19^=3165 17,+ 19^295 5 ^-?^£ = 1 19 17 9-316 8.316 """"nT 17 295 8.2§i 19 17" See, for other questions of this kind, my Treatise on Algebra, 1. and II. Vol. I. and II ( 124 ) 17 t r+lC)-r— 2/4 17x+igx—253 1/^+1^—232 17.t -\-10x— 211 \?x+ igx =zigo i7*-H 9^169 )tfx+ l|fe=448 17x+ 19^—127 17.r+igz=lQ6 17^r+lDr= 85 Ifx+IQ*** (34 17^+19^— 43 ijxfigzd 22 Total number of 9 274 8.274 19 9253 J 9 9.232 ~W ' 9.211 ~w~ 19 9169 19 9.148 19 9-127 19 9-iQS _ 19 9,85 17 8.253 ~~l7 8.232 = 1 =0 i7 8.211 8.190 " 17 8.169 17 8.148 T7" 8 127 8.106 = 1 = 1 .'9 0.64 17 8 8.5 >7 8.04 =0 — »9 9 43 9.22 17 8.43 8.22 ^0 W J7 tions — 10 ans Notf.. When any of the left-hand fractions are exactly equal to gjn integer, the quotient must be diminished by an unit. ( 125 ) Ex. 8. Given equation 5x-\-7y-\-\\%—224. Here % may have any value, from 1 to \g, which gives the following sets of equations $ also in the equation 5p — Jq=\ - 9 we have p—3 and q=2. Hence Equations. 5x + 7y=:2l3 5x + 7y = 202 5*-f 7^=191 5x+7y=180 5x+7y— 169 53 + 7y= 158 5x+7y-W 5x+ 7y=zl36 Sx + 7yi2l25 5x + 7y=zl\4 5x + 73/= 103 5x + 7yr=z 92 5x + 7y=z 81 No. of Solutions. 3.213 2.213 7 5 3.202 2.203 7 5 3.191 2.191 7 5 3. ISO 2.180 7 5 3.169 2.169 7 5 3.158 2.158 7 5 3.147 2.147 7 5 3.136 2.136 7 5 3.125 2.125 7 5 3.114 2.114 7 5 3.103 2.103 7 5 3.92 2.92 =6 =16 zr5 =5 =l5 —4 =:4 = 3 = 3 = 2 7 5 3.81 2.81. M 3 5tf+7#= 70} 5x+7y= 5Q; ( 120 ) 3.;o 2 70 7 5 3 59 2.59 — 1 7 5 3.4S 2,43 5x + 7y= 48 j — — =rl 7 5 3.37 237 5X^7^=37 5 — ~~ = 1 / 5 3 26 2.26 5x+7y= 265 — — =*i / 5 3.15 2J5 5z + 7y=z 15 5 < — . s 't 7 5 Total number of solutions rz 59 Ex. 9. Let # be the number of half guineas, and y the number of half crowns j then the number of six- pences in each of the former being 2 1 , and in each of the latter 5 ; also the whole number or sixpences being 800, we have the following equation : 2 1^ + 53/ = 800, Here the equation 21 p — 5^=1, gives p= I and qz:4. , 1-800 4.800 Whence by the rule — ~7. Ans. ? 5 21 Where, as in the preceding examples, the first quo- tient is diminished by unity being exactly integral. This rule, which is taken from Barlow's Theory of Numbers, is. much shorter than that which depends upon an actual determination of the several solutions; the latter is omitted here as presenting no difficulty to the student who has attended to the preceding solutions. Ex. 10. Let x represent the number of guineas I ' have to give, and y the number of louis-d'ors I am to re- ceive 3 then by the question, ( 127 ) Hence 17j/=2lJT— 1, or 3/=jrH — Z2it'& len 4; 4# i Make =j£> $ then 4.r= I7j&+I, or .r=4/>-f J Where, if we make- = q; then p=4q—l, p-\- 1 And consequently x=4p -f- =» 1//) — 4, Whence also we have y — x-\ =21p— 5 ; Where p may be taken any number at pleasure : if we take p— 1 -, then #=13, and y = l6, which are the least numbers ; viz. I must give 13 guineas, and receive 10 louis. Ex. 11. Let x, y, and z, be the numl^er of gallons of each sort respectively ; then by the question x+ y + s=100O 12x+15y+18z~17(x+y+z) By transposing the terms of the latter equation we have '— 5x—2y + z-. O But *4« y + 2= 1000 By subtract. 6x + 3y= 1000 Where it is obvious the answer cannot be obtained in integers, because the first side of the equation is divisi- ble by 3, and the other is not. We may therefore as- sume x or y at pleasure} taking for x the value given in the answer of the Introduction, viz. or= 111-, we have * ( 128 ) 1 1000— 6+HU 3331 1 #=111-: vm S £ = -1=111- 9 3 39 7 and «=5x + 2y=777- 5 11 7 That is 111- at 12s. : ill -at 15*. and 777- at 18*. 9 9 9 INDETERMINATE ANALYSIS. PKOBLEM II. Ex.3. Let. x= the number sought; then by the question .T— — 2 ^r—~3 ■ , and — — = whole numbers*. x — 2 Make— — =p ; then x=6p + 2. Substitute this value of a: in the second equation, and we have fy> + 2-3 6p-l 13 13 66—1 , 13y-f-l /7 + 1 Make-^--=?; then ]*== --*p-- = 2q+ IZ.-^. Let now ^— - =r$ then q=z6r— 1 3 and p=r — - — - — =13r— 2; and consequently x=6(l3r -~ 2) -f 2=78r— 10 3 where r may be taken at pleasure. If r=l, then #=68. E*. 4. Let a: be the number sought $ then by the oueation x — 5 x- 2 1 , and = whole numbers 7 9 2» ^ Make =*p, or «•= 7p + 5 . ( 129 ) Substitute this value for x in the second equation, and we have 7p + 5 — 2 7P + 3 , 9 9 TP+3 9q— 3 Ivlake — — = <7; and we have p— • 9 7 2? — 3 : =•2 + ~y— rrf * 27—3 7 r +3 Let now — =r, and we shall have 7= — 7 rfl = 3r + 1+ — — =wh. r.+ l Again, let there be taken = 5, or r~2s— 1. _ 7(21-1)+;* 9 (7j_2)-3 Then -7 = —7s— 2,and A= = 95—3 ; whence we have ^=7(9,9— -3^-f 5=:63s— 16; where s may he taken at pleasure. If xss I , then .r~47> and if j=2, xssl JO. &c. Ex. 5. Here by the question we have jff-rjifi' , a-— 27 and — = whole numbers, 39 5b r— 16 Make — =/>, or x—3gp + 16. This, substituted in the second equation, gives 3Qp+ 16-27 39p- 11 17/>+H * 56 50 r 56 •Make -^—^ ~q, or/) — L ~3?—l-f 56 z ' 17 * 5?4-6 5 ialr«=» — '7 17 ' a-7 + <5 , , , Bq + G lhen — - — ~wh. also make — - =r, then ( 130 ) 17r — 6 2r— 1 2 = ~3r— 1 H ~wA. 5 . 5 Make =5 ; then r= = 25+ — — =^A 5 / 2 2 Let therefore — = /, and we have 5=2/— 1 j Consequently r= -r-r- - =51—2, and '7(*-2)-g J 66(l7/-8)-U 5 17 =56f-27; and ar=39(56/—27) + l6— 2184/— 1037 j where / may be assumed at pleasure. When tzz\, then #=1147, tne least number agreeing with these conditions. Ex. 6. Here putting x for the number sought, we must have x — 5 x — 7 • x — 8 ' * x > and all whole numbers. 7 8 9 From the first, by putting zzp s we obtain x~7p -t-5', and this substituted in the other two, gives 7P-2 . 7P-3 . . . and -4—— s= whole numbers, 8 9 ™ * 1P—1 , 87 + 2 7 + 2 Make ££ =7 : thenar: L = an d z= — \ j Consequently the squares are 49, 25, and 1. Ex. 13. Let \x 2 — y, \x 2 , and \x 2 +y y be the three numbers in arithmetical progression ; Then we have to find x 2 } x z + y aad x 2 —y, rational squares 5 or # tt +yzz.m 2 i and x £ — 3/ = t* 2 . ( 13 a :=4lX4+l6=!S0; whence the ► three numbers will be 30j, 210 J, and 3<)0J. Aod if these numbers be multiplied by any square, number, the same conditions will obviously obtain ; Hence multiplying by 4, we shall have 120£, 840|, and >5'J0f, which are the numbers given jn the answer in the Intro- duction ; And if these last be again multiplied by 4, we shall ha*j 482, 3i>6 v 2, and 6242, which are all integral, and equally answer the conditions of the question. The question may le otherwise solved, as follows : Let x 3 y, and z, be the numbers, and assume oc+y — m* y + z~r* Then we shaU have z = f(—m 2 + n* + r*) Where, since the numbers are in arithmetical progres- sion, we have x + z~2y) or rfzzm' 1 — w a -f-f 3 ( 137 ) Therefore we have only to find That is, three square numbers in arithmetical progres- sion; Whence from Example 12, we shall have m 2 =(/> 2 — (f— 2pq)* r 2 = (/) 2 — f + 2pq¥ And consequently x=l(p*+f)*-4pq(p*-q*) z=zi(p+f)* + 4p q (p*-f) Where p and q may be taken at pleasure. Ex. 14. Let x 3 y 9 and %, be the numbers sought ; then by the question x^+y + z^in* y* + x + %~ q* % 2 -f x-\-y=. r 2 Assume x*-\-y + z = (x—n^-x 2 — 2nx + n 2 -, ■ Then we shall have x~- . 2n This value of x } substituted in our second and third equation s^ gives y + z — n q y+ 2— +*=? y-\-z — rc 2 * + — — +y= r * Hence, assume y% -f- ^ — — \-% — (y—pYz=zy f w = ],p=4, and xs±Ut> then #=1,3= — and am - j 3 9 which are the answers given in the Introduction, except that the first is omitted by the printer by mistake. Ex. 15. Assume 8#and ]5x for the two numbers \ then by the question (8^) 2 +(15^) 2 is to be a square - 7 But (8#) 2 +(l5tf) 2 =2Sg* 2 ; = :(l7;r) 2 ; therefore x may be any number at pleasure. If x— /2, we have 576 and 1080 for the numbers sought. Ex.l6. Let x and y be the numbers sought) then by the question we have to solve the equation x % + xy, or x(x+y)=m' 1 y a + xy, or y(x-{-y)~ » 2 Assume xz=p%, and y = q z , and these become p*(f + cf)-m* Which conditions will be fulfilled if weflnd /> a + ^ s = a square. Now we have already seen (Example 12) that for this purpose we have only to assume p=zr'—$* and q—2rs- ( 139 ) Therefore ,r=r^= (r 2 - s 9 ) 2 And y=^ 8 =4rV Where r and s may be assumed at pleasure. Ifr=2and j — 1, then #=9 and 3/ = 1 6. These how- ever are square numbers, which is not a necessary con- dition. Now it is obvious, that any multiple whatever of the same numbers will equally answer the purposes required in the question. We shall have, therefore, a more general solution by taking l r==/(r 2 — ^) 2 , and y = 4tr' i s 2 ' } where r, j, and t, may be assumed at pleasure. Ex. 17. Let x and y represent the two numbers 5 then by the question x 2 ~\- y 1 — 1 —m? a 2 -?/ 2 — 1-w 2 Assume x~y+ 1 ; then these two equations reduce to (y+i) Q +y*-i=iy*+2y=m* (;y + i) 2 -2 2 -i= 2y£=»* Where, putting the first under the form 2y(y-\- l)=w% and substituting 2t/ = ?i 2 , this becomes Consequently ^ + 1 must be a square 5 let, therefore y -f 1 =/ 2 3 then, since we have found 2y r= n Q , we have by subtraction y z=z w 2 — / 2 -f I And consequently x = w 2 — /> 2 -f-2 Where ?? and /» may be assumed at pleasure. If t?=4, and/>=3, then y=.8> and xzzQ, the numbers required. Ex. J 8. Without attending to the particular num- bers in the question, let us endeavour to resolve any given square number into two other square numbers. For this purpose, let a 2 represent the given square that is to be so resolved, and put x 2 and y* for the re- quired squares. ( 1^0 ) Then we have to satisfy the equation az-xt+y*, or Jn order to which, let us assume px qx ' p From which we have, by addition and subtraction, __px qx (p 2 +q 2 )x " q p " pq px qx ^ (p 2 —q*)v 2v=- q t n Whence, by multiplication and division, p*+q* Where the indeterminates p and q may be assumed at pleasure. But it is obvious that 2pq and p^+q*, as also p*—q* and p 2 + £ 2 , being incommensurate, the values of ar and y must be fractional, except a be divisible by.p z -\-q 2 . That is, unless # be divisible by the sum of two squares 5 and then the question will admit of as many integral answers as a has divisors of this form. Now by the question #=65 - } and 65 is divisible by 5=2*+ 1% by 13=3 2 -M 2 , and by 65=7* + 4% =^8 2 -H 2 . Hence we may assume /=2, and q=ly or ps=3 and £=2; or/> =7, and ?=4 $ or/=8, and 2=1$ ( m > Which give the four following solutions, 65 ? =3 6 a -f 63 5 =5() 2 -f33 € ^CO c 4-25 9 ==52 2 -f39 2 These being the only integral answers the question ad- mits of. Ex. ]Q. Let x*, y and *■—-, be the numbers required, and put the given number \Q—a 5 then by the question x't + a, y+a, and -^— -f a, are to be all squares, or Assume ^-|-fl=(^-i')^r 2 - 2rX + z z And t/-j-azrr 8 Then we have #=s: — - , And yzzr*>~n s Wherefore - =2r, and 4- + a = 4r* + a, vhieh y. *» must be a square. Let, therefore, 4/--f a-=:.(2r ^- $)*— An + Ars -\ $ % a s % In which case we shall have r= 4s And consequently y — ( )-— a Where s may be taken at pleasure, provided (— — )* .. be greater than a, ^ s Bat by the question a—lQj if, therefore, we assume s^\, we la ^ e 3/— C 4 ; — 9— ~> ~ — 2j — c2 =r 9 , and*=|-M) = ^ ( 142 ) Hence the three numbers are 5 25 *> 36' 1296 J 4 7 The question may be otherwise answered, thus: Let 4x*, and x i2 —a, be two of the numbers; then — — — will be tte third number 5 and by the question 4t*+o = m 2 4ar Here the second and third equations being squares, k only remains to find the first 4x* + a=?n' i . Assume 4x a +a=(2x+s)*=z4x' 1 + 4sx-{-s2 t a 5 s Whence x= — ~— 3 where s may be taken at plea- sure, provided only that S* be less than a. Ex. 20. Let x and y be the two numbers we have to find; then aP-\-y* + xy =m» Assume x* 4- xy + y*=z(x + r)*=ff a +2ra , -J-r* And we shall have #= r— - £ 2r-3/ Where r and ?/ may be taken at pleasure, provided y be greater than r, but less than 2r. If y = 3, and r—2, then .m 5 y~5, andm3, then xzzlG y=-7i andrzrfJ, thenar — 8 Ex. 21. If x, y, and %, are taken for the three num- bers, we have to find x 2 -f 23/ = □ y a -r-ff3=D . ( MS ) Or dividing by x*, we have in that case xx y* % X 1 X *>* y *L X 2 X Or if, in order to simplify the expression, we put - =3i m, and - ss», the above will become tfM+lrrrjnr 2 From the first jnn-zzr 2 — 1 j assume therefore fttrcr-f I 71 =r— 1 And there now remains to find (r+l) a +(r-l)=s» (r~l)M-(r+l)=f,or 7- 2 + 3r zzs 2 r 2 -r + 2=;* 2 Assume r 2 +3r==(r4-Tt>) 2 :z;r a -f-2m>-{-Tc* IV Whence r= : which substituted, gives «t' 4 W(3— 2w) 2(3 — 2™) 2 _ ^ • 7^ «_.X4 — * (3— 2™) 2 (3 — 2«u/) 2 (3-2w) 2 Or wH2m; s + 5w 2 - I2w+1&=Q Assume this =s (**>* + iv -f 2) 2 = w 4 -f 2^ -J- 5«it? li -f 4vo -f 4 Then — 12™ -{-18:=: 4™ +4, or — il _7 W "* 16 ~ 8" ( 144 ) And we shall have rzz • -— = — , 3 — 2w 80 Consequently m~ and nn , 4 J SO - 80 ' Or a™80, 3/1=129, and x — — 31, Which are three numbers answering the required conditions, and a similar process will give the three numbers, Q, 73, 328, all positive. Ex. 22. Find three squares,, a 3 , b 2 , c*j such, that their sum may be a square,, or ^-f^ + ^zri 1 That is, assume a* + b* -f c 2 — (c -f r) a 5 from which we shall obtain cz=. , 2r where a, b, and r, may be taken at pleasure, provided r 9 be less than a*-f- b*. This being done, let ax, bx, and ex, be the required squares 5 then cV 2 -f £V-f-c 2 ,r 8 =: 2 r 2 -f/>(2r-f m)— iv 2 Make r*-\-n(2r + m)=:(s— r)*— i 9 — 2$r-f r* __ T , s 2 — nm Then we have r — , 2(7* + *)' Which value of r, substituted in the latter equation, gives ( ~) +p\ -f-/» =™ 2 \2{n + sy yi n + s ' ' Which, by reduction, becomes {s x — nw)H4/^+«)(« + j) '' Or (s\— nmy-\- 4ps(s + m)(s + n) zzw 2 , Assume this ==}(<** — nm) — 2ps\ a ^z ( j«— « w)» - 4ps(s' 2 — nfn) -f 4/s 2 Then this, by reduction, gives (s + #/) (5 «f n) n 72/75— s 2 +ps J — im + n) And ar= Whence a^ain 5: f. ■ _ 2 r 2 (s 2 — 7Ztfz) 2 (7Z-f-j) 2r -f 77z 4(«s 2 •+ ^j) {n -f j)* (* 3 — nm)* 4s{s + m)(s + n) o ( 146 ) Now, if we take a=2, bz=6, and c=9, then d~ll, a 2 __^__6 j ___ c 9 p ""5 r== l2T' W ~~^" ~~72? an ^""^"" = 121' p—fai + w) 9-8 1 2209 :as= - = j-2~ , and #__ -~ , 2 12 1 242 " 0292O — 4418 whence ax 2x~ O2920 / « 13254 0Xz=:OXz=z ~ - 02920 19881 . , . , ex=ox= -—- the numbers required, J 62920 * Ex. 23. Let x, rx, and r*x, be the three numbers in geometrical progression ; then by the question x 4- rx— x ( I -f r) = i/ 1 r.r -f ^ e ^= Mr( 1 -j- r) = w 2 Dividing the second by the first, it is obvious that r must be a square 5 let, therefore, r~y*, and it only re- mains to find £{t+g*)z£ip* Which equation will obtain, if we make x=.nz^ and l+y*z=nuP Whence x=n(% 1 — id 1 ) -f y" + 1, Where all the indeterminates may be taken at plea- sure, if *=w, then in which y may be taken at pleasure. If t/=2, then x=5, and r=4} and the three numbers sought are 5, ^ and 80. ( 147 ) Ex. 24. Let x and y represent the two numbers ; then it is required to find x+l=m* y-fl = 72 2 x + y + lz=r Q Now here it is obvious that the three squares, r*, m 2 , and s 2 , are in arithmetical progression, their common difference being y. Let us, therefore, represent these squares as in Ex. 13 ; yiz. s*=(4pq — p*-2 q*)* Trf—Xp' 2 - -\-2q y + Qpf -f 1 =rc* Assume, therefore, in this case, 72=1 +4pq 3 And we shall have, by squaring and cancelling the like parts, 4fq — 1 2p*q*zz 1 6p V» Whence p=4q*+3q, ' in which expression q may be assumed at pleasure. Thus the general values of x and y will be determin- ed 5 viz. by first making p=z4q b + 3q, and then x=(p 2 +2q' i -2pq)' i -- 1 Where, by taking q~\, we have p=?7; whence ff= 1368 ( 148 ) and ^=: 840; which numbers answer the conditions 2 $ where #, b, and c, will be known numbers, and will possess the following pro- perties, viz. (c 2 +& 2 y±4alc{a+h), a complete sq. (c- + tf 9 )*±4al>c{a+b) 9 a complete sq. (e 2 + a+Z> 2 ) 2 ±4a£c()z*z. D a 2 (c 2 +a 2 ) 2 ±4abc{a+l)x*±n ^(c 2 + (a + b ) 2 £ ±4abc (a + fi#*** D ( m ) And it consequently only remains to find a(c*+b*) + x(c> + a*) + x{c* + x{c Ci +(a + #*)*]=== 4aZ>c(-— -— 4aoc(a+ b) Or putting c 9 + **=ra, ^-f-cfcrc, c 2 +(tf + £) 9 ~p, and 4tf£c(a-|-&)=^ w-f-w + p we shall have xz=l — . 771 . And 77ZX= — (zzz4-72-f p) nx = - (7w4-7Z+p) Which are the tforee numbers sought. If r=2, and 5=1 3 then a=3, b~5, and c=7. Also .m=z<*+lPzxJ4 i w= assuming other values for rand $> o3 ( 150 ) Ex. 2(3. Let x?, y*, and %*, be the required squares j thenwe have to find ,r 4 -f 3/ 4 -f % 4 = 7/? 2 Assume ar 4 + 2/ 4 + * 4 = H* 2 4-3/ 2 )--% 2 S 2 := Then 2* 2 ?/ 2 = 2s 2 * 2 -f 2z 2 # 2 jr 2 /y 2 And consequently s 2 af 2 +2/ 2 Therefore x*+y 2 must be a square; which it will be if we assume xzzfP—q*, and y=2pq ; for then x*+2/ 2 =.(/> 2 — ? s ) 2 + W^(/> 2 + 9«)* Hence the required squares will now be s 2 ^ 2 -? 2 ) 2 i/ 2 = 4p*f J (/-rY) 2 Where/* and =2, and <7=1, then 2/ 2 = 4/V =4 2 = 16 t__ (f—q*)*x4pY * t& 144 which numbers answer the required conditions; and ya* rious others may be found by giving different values to p and q. Ex. 27. Here, if x*, y*, and z 2 , are taken to repre- sent the three squares, we have to find **— 3/ 2 =D> or 7w 2 # 2 — % a =Q, or w* y 2 — z B = D, or •2 Hence, if there be assumed, as in Ex. 18, y= -~ — '-— .and x=— - — — * />* + ?* ***** ( 151 ) The two first equations will be resolved ; and there- fore it only remains to find tji %\ or -— — — =r 2 Or (p 2 -^) 2 (^+^) 2 -(p a + ^) 2 ( /9 -^) 2= =D ( pn* ^p^T^ttr 2 V >* t ( P^f-f 1 ? + ISP - ? V )*= D Assume p«=« 2 ^, and v a =^V, and this becomes Or dividing by 4= — '. and i>« — . ^ 4 5 Therefore the required squares are where x may be any number assumed at pleasure. If we take 07=153, we shall have (>97 3 , 185 a , and . J53», for the squares required. Ex.28. Without attending to the particular cube given in the question, let us endeavour to resolve gene- rally any cube v 3 into three other cubes $ viz, M us f \ such values for x, y, and z, that ar 3 -r3/ 3 -fz 3 :=: v 3 , or x 3 + y'zzv 5 — a 3 * ( 152 ) f For this* purpose assume x~u-\-w, and y — u — w and a=r- $ Then « 8 H-y s =2M(w 9 + 3«; 9 ) v 3 -Js , =2j(j» -j-3 r*) Whence 2m (ft* 4- 3w*) -2s (f+$#) Assume again u =.mt + 3np ivz=z nt — mp Then u(u*+3w*)=:(mt + 3np)(m' i + 3n a )(t*+3p' i ) Also assume s=at-\-3cp r—ct— ap Then ^(A 2 4-3?- 2 ) = (^ + 3cp)(^H-3c 2 )(/ 2 4-3/) Whence, and from the preceding equation, we have, after dividing by (f + 3p*), (mt + 3np){m' i + 3»*)==(a*-f3c/>)(tf* + 3c a ) 3c(a i + 3c i )-3n(m* + 3n*) Whence t= ~ ; » p Or taking the denominator —p, we shall have t=z3c(a* + 3c*)—3n(r7P + 3n*) p=m (m a -f-3rc 2 )— a(a 2 -f3 as re q ulred - tyi yS yjj If we assume c=l, m~l, ?:=0, and a=lj then t=z\2, and/>= — 3; which values give xnl5, y—§, z—12, and. v=18$ Whence l5 3 -J-g 3 -f 12 3 =18 3 , and 5 3 + 3 3 -f- 4 3 = 6 3 , or 5 3 3 3 4 3 + to =2 3 Conseq 1 25 aently , J , 64 and — =8. 27 Therefore 1, — , and 27 125 *are the cabes sought. 27 6 Ex.2y. Let x, y, and », be the roots of the required squares ; then we have to find x l+yl - z*=r* x r -j- z -.y 9 -** Where, by first assuming x=p 2 t «? y*/H«^ we shall have ** + z a ~y a =( p* — q % —7.pqY Where the two first conditions being fulfilled, it only remains to make a square of our third equation 5 which becomes, bv substituting for.r, y, and z, ( 154 In order to reduce this to a more convenient form, let p — (2 + m)q; then substituting this for p in the above equation, we have y* + 2?-x*=q*(ni A + W + 20/w 2 +l6m+l> where we have now to make the latter factor a square. For which purpose, assume its root =wz»4-tfwi+l 5 Then,, by squaring, we have m 4 + 2tfm 3 -f(a J + 2)^-f- 2am + l = Or, by making 2a= -f 16, there remains + 1 (5m 3 + 66m 9 = 8w 3 -f 20m a — 23 Whence 4(3m*= — 8m 3 5 wherefore m= , 15 But />=(*24-tft)y, or j»= q-, therefore we P 15 ^ have -=r : whence, if p = 15, q=— 4, we have q —4 x ' x=f + q* =241 3/=/+^-<=149 ^=^-^-^==269 Consequently 241 s , I4p 2 , and 269 s , are squares, hav- ing the required conditions; and others, might be found by giving different values to q and p. Ex. 30. Let (l+x) 3 9 (2-,r) 3 , and t/ 3 , be the re- quired cubes 5 and let a be any given number 5 then by the question (1-f.z) 3 — — a 3/ 3 — a= y 3 —a Whose sum 9— Qx+gx* 4- */ 3 — 3a is to be a square. Assume it :s=(3# — r) 2 ==9# 2 -— 6rr4-r», And we shall have 9 — 9*4 y s — 3tf= — 6nr + r* Whence oc— ( 155 ) r *_y*_9 + 3a 6r-9 Where y and r may be assumed at pleasure, provided 6r be greater than Q, and r* greater than y 3 . In our question a==f, and if we take r=4 and 3/=2, we shall have x= — ; 15' Whence (1+^)^(1 + ^)3^^- (2-x)3 = (2- -) 3 =(-) 9 ID 15 and y 3 — 2 3 = 8 Which are three cubes fulfilling the required conditions of the question. / Otherivtsc. Let x 3 , y 3 , b 3 , be the three required cubes, and a the given number; then it is obvious that the ques- tion only requires that x°-{-y z + b 3 — 3a=m' i . And here one of the cubes b 3 may be taken at plea- sure; also a being given, we may consider b 3 — 3a= + c } a given number; whence we have to rind x 3 + y 3 ±c=zm' i . Let x=zd-\-%, and #—/—», Then x 3 =d 3 + 3d 2 z + 3dz* + z 3 y 3 =f 3 -3f>% + 3fz*-z 3 By addition d 3 +/ 3 + 3 (d* -/*)* + 3 (d+f)z*±c=-m % 3(^+/K + 3(^ 2 -/ a )% + ^ 3 +/ 3 ±c=w a Which latter expression may always be made a square, provided 3(d+/)=a square. Assume d~2, and/=l, then 3(d+/)=9, and the above becomes 9^ + 9x + 9±c=w ( 1^6 ) Let this =s (3 x +*■)*= 9**+ frz-f r \ Then 9s-f-9±c=6r;s-f r% and 9±c— f* or-g which is similar to the preceding expression. SUMMATION AND INTERPOLATION INFINITE SERIES. PROBLEM. Any series being given, to find the several orders of differences. Ex. 3. Here 1, 3, 6, 10, 15, 21, given series. 2, 3, 4, 5, 6, first diff. 1, 1, 1, 1, second diff. Ex. 4. Here 1, 6, 20, 50, 10.5, 196, &c. giv. series. 5, 14, 30,55, 91, 1st dirt. - % 9, lft, 25, 36, 2d diff. 7, 9, 11, 3d diff. 2, 2, &c. 4th diff. 1111 1 „ Ex. 5. Here -, -, -, — , — , &c. the given series. 2 4 8 16 32 ° •— , — , —7r f -, &c. 1st diff. 4 8 16 32 -, — , — , &c. 2d diff. 8 10* 32 — *sS , &c. 3d diff. 16 32 -^, &c. 4th diff. &c. &c. ( W ) PROBLEM II. Any series, a, b, c, d, e, &c. being given, to find the first term of tjie nth order of differences. Ex. 3. Here a, b, c, d, e, Sec. are respectively 1, 3, 9, 27, 81, &c. also w=8. Whence #— nb+ jJ -c - v U+kc. V2 1.2.3 ~~? zzl-8b + 28c—56d + 70e—56f+28g—8h + i = 1 — 24 + 252- 1512 + 5670— 13608 + 20412— I74p6 + 656lz:32896— 32640^256 Ex. 4. Here a=z 1, b= ~, c = -, d— -, Sec. 2 4 8 Also »rz5 j whence W(«— 1) 72(72— l)(tf — 2) e — tf + Wfc 9, 11, 1st cliff. 2, 2, 2, 2, 2ddiff. 0, 0, 0, 3d diff. ( 158 ) Therefore d'zzZ, d"=2, d"'—0, and n=I5j Whence n-l («-l)(«-2) . (n-l)(n-2)(n-3)- 1 J .2 T 1.2.3 #"+£c. = 1 + 14d' + 9lGT= 1+42 + 182 = 225 Answer. Ex.4. Here 1, 8, 27, 64, 125, 216, &c. 7> 19, 37, 61, 91, &C. 1st diff. 12, 18, 24, 30, &c. 2d diff. 6, 6, 6, &c. 3d diff. O, O, &c. 4th diff. Whence, since d'=7, d"=12, d'"z=6, and 72=20, we have t n-l „ (*-l)(n-2) 7// (w-l)( w-- 2)(w--3 ) T 1 T 1.2 1.2.3 d"'+&c. =« 1 + 19^'+ i7i^' , +9^9^ w — = 1 + 133 + 2052 + 5814=8000 Ans. Ex.5. It will here be sufficient if we take merely the denominators of the proposed fractions, viz. 1, 3, 6, 10, 15, fee. 2, 3, 4, 5, &c. 1st diff. 1, 1, 1, &c. 2d diff. 0, 6, &c. 3d diff. Whence, d'zzl, and(» +^(«+a)»fa> jL 3 3 Ex.5. Here 1, 3, 6, 10, 15., &c. giv. series.^ . 2, 3, 4, 5, &c. lstdiff. , I, \ } >j &c. 2d diff. 0, 0, &c. 3d diff. Consequently a- 1, d'zz2 > cT=:3, Whence we have 1.2 T 1.2.3 »(»-!> i »(n-l)(,-2) ' 1.2 T 1.2.3 «'i fc ft*' = Ans. 6 1.2.3 P 2 ( I*) ) Ex.6. Here- J, 4, 10, 20, 35, kc. 3, 6, 10, 15, &c. istdiff. m 3, 4, 5, &c. 2d difT. 1, 1, &c. 3d difT. Whence a=l, (1=3, gT=3, tf'szl'i and therefore »(»-0 m ri(n-\)(n-2) 1/2 1.2.3 . n(n -l)(? i-2)(7i-3) £ 1.2.3.4 « 2 1.2 3 n(n~i)(n —2)(n— 3) * *' 1.2.3.4~~ ' 3 ?i 2 ~ 7Z 7?(?2 2 — 3 W + 2) 72(?2 3 — 6?l* -f- 11/2— 6) - 2 •+■ - + - 2.3.4 ... 7i(» 3 + 6/2 2 -f llW+6)_J?2 w-fl it + 2 fl-f 3 7=1 ~ 2X4 ~~T X ~ X T~ x ~i~~ Ans. as required. Ex. 7. Instead of w terms of the proposed series, let us find the sum of n+ 1 terms of the series 0, 1, 16, 81, &c. which will reduce the magnitude of the successive differences. Then we shall have 0, 1, 16, 81, 256, &c. 1, 15, 65, 175, &c. 1st difT. 14, 50, 110, &C. 2d difT. 36, 60, &c. 3d difT. 24, * &c. 4th difT. Whence o=0, jikl, 31, 211,/81, 2101, &:c. 1st d iff. 30, 180, 570, 1320, &c. 2d difF. 150, 390, 750, &c. 3d cliff.- 240, 360, &c. 4th difF. 120 &c-. 5th dirT. Whence <*=0, 'Aekl) d"=30, d'"=150, d /v =z240, and PROBLEM V. The series a, I, c, d, e, &c. le'ing given, whose terms are an unit's distance from each other, to find any interme- diate term by interpolation, Ex. 2. In order to save the trouble of reduction, take the logarithms of the numbers \ in which case we shaJl have Terms. Logarithms, d' ', d" , d" f > d w a, — = -1-69S9700 86002 h ^-= -17075702 I 1671 84331 66 c, — = -17160033 52 1605 82726 $8 4, i=s -17242759 53 81 179 1547 e, — = - 17323938. 54 Here the 1st terms of the differences are d' = 86002, d"=;l67l, d'"—6Q, d iv =8 3 ,and the distance of y, the: term to be interpolated being 2|, we have ,, oc(x-> 1) ... x(x—\)(x—2) v „ n = a +-d' +™d"+—d?"~ - & ^28 15 32 == — 1.6989700—215005 + 3133 — 21 — 1 m - |.72015 9 4=log, of ^- iyjjL Ans. ♦ ( 163 ) Ex. 3. Given the natural tangents of 88 s 54', 88° 55', 88° 56', 88° 5f, 88° 58', 88° 5$ and Sg% to find that of 8 1*58' 18". Tangents, d' d' f d" d» d* d vi a = 520806/3 801436 Z> = 52'882109 25042 826478 1IQ3 c=537085S7 26235 76 852/13 1296 7 rf=54-56l30O 27504 83 I 880217 1352 8 e=55'44!5l7 28856 91 909073 1443 / = 56 350590 30299 939372 £ = 57-289962 Again (88° 58' 18")— (88° 54') = 4' 18"=4' 3, Whence ar=4 ,, 3 5 and consequently^ 4 ., , 43X3'3 ,„ , 4'3 X 3-3X2*3 _„, y,=a + 4'3d + d H JF + 4*3 X33x 2-3X1-3 r 4 3X3'3X&c '3X-'7d' -d iv + 120 720 And consequently, by collecting the terms 52-080673 3-446175 17767* 6489 134 5571 1 144 = tangent 88° 58' IS'' the answer required. ( 164 .) PROBLEM VI. Having given a series of equidistant terms, a, b, c, d', e, &c. whose first differences are small, to find any inter* mediate term by interpolation. Ex.2. Given the cube roots of 45, 46, 47, 48 and 4Q, to find the cube root of 50. Here the number of terms are 5 ; and against 5 in the tablet we have a— 5b + 10c—10d + 5d— /=0, or /=a— 5b+ 10c— 10^+50. Num. Roots. V45 = 3*556893 =a 3/46 = 3*584040 = b V47 = 3-608826 = c V48 = 3--634241 = J V49 = 3-659306 =e Hence a= 3556893 5Z>= 17915240? K)c=36-08S260 10^=36-342410 5^=18-296530 Sum +57941683 54-257650 Sum —5-1-257650 3-684033 = the cube root of 50. Ex. 3. Here, following the same process as before, we have d=log. 50= 1 -6989700 &=log. 51 = 17075702 c=log. 52=1-7160033 d=log. 53 required. e=log. 54=1-7323938 /=log. 55=17403627 £=log. 56=17481880* ( *« ) Also the number of terms being 6, we have from the tablet a — 6b+\5c — 20d+\5e— 6/+£=0, or , _ a— 6b + 15c+.15 +/)=20 6875974 Difference =34*4855 1 7 1 Therefore 34'4855171-i-20=l72427585 = the log. of 53, as required. EXAMPLES FOR PRACTICE. Ex. 1. Here, by the differential method, 2, 5, 8, 11, &c. given series 3, 3, 3, &c. 1st diff. Now rc=100, a=2, and gT= 3 j therefore n(n—\) L 100x99 j=na + ^ V =2X100+ — x3, or 1.2 j 2 s=200+ 14850= J5050, the sum required. Ex. 2. Here by the differential formula, 1, A, 9, 16, &c, 3, 5, 7, &c. 1st diff. 2, 2, 2d diif. Wlience #=1, rf =a, oT=2, and 7z=50^ ( 166 ) Therefore j=7zcH — -J + — — -d 3 01 12 1-2-3 ' = 50+ 1225d' + 1 (jGOOrf", or 5=50 + 3675 + 39200=42925. Ex.3. Let, — ^— =5=l + 3^-f 6.r 2 +10^+ &c. (1— %y Then *=(i- ar) 3 x (1 + 3x+6x x + 10o; 3 4- &c.) Which, by actual multiplication, is == 1 ; therefore »=sl, and $= rp as required. Ex.4. Let - — ^—=5=1+ 4x + 6x*+20x 5 + 35a? 4 - (I-*) 4 . + &c; Then*=(l — ar) 4 x(l+4# + 6ar 9 +20;r 3 4-35^+&c)==I as appears by the actual operation ; Whence x=l, and xas —> the sum required, ' . (1-ar) 4 ^ Ex. 5. Let »= - + - + - + - + - + &c. Then *-ls= « + 1 >A ^£4 •£ 4 &c. And by subtraction, / 2 2 2 2,2. 13 ^ 3.5 T 57 7*9 911 or =2( h + • H H &e.) v l-3 T 3-5 X.57 79 9'H Consequently by division, 1 1 1 I „ 1 1-3 7, 3-5 5-7 7*9 a which is the sum required. {! ( 167 ) Ex. 6. Let the given, or proposed series l'2 + 3*4+5-6 + 7*8 + &c. be put under the form (l 2 +l) + (3 2 + 3) + (5* + 5)+&c. or -l 2 + 3 9 + 5 9 + 7 2 + & c . 1 +3 +5 +7+ &c . Then, by differencing in the first, we shall have oT=8, d"=8, a=], andn=40j _ TT1 n(n—l) ., n(n--l)(7i — 2) .„ Whence na+ — -d '+ -± - -&'zz 2 2-3 And consequently 4+780d' + 9880d r '= 4 + 6240 +79040 =85284 Also 1 +3 + 5 + 7+ &c. = 1600 Therefore, by addition, the sum required =86884. The answer given in the Introduction agrees with the series r2 + 2'3+3'4 + 4'5 + &c. Ex. 7. Here the given, or proposed series 2x—\ 2x— 3 2x— 5 2x—7 . H V + + &c. 2x 2x 1x 1x mav be separated into the two 2x 2x 2x 2x h — -\ H + &c. 2x 2x 2 2x -"(1 + 3+5 + 7 + 9 + &c.) The sum of w terms of the former of whkh = /?, and n q the same number of the latter = — -, 2x ( 168 ) ft 2 2nx—n* ,2x-n x . Whence n — — = . -^m ( ), the sum re« 2x 2x v 2x ' quired. Ex. 8. Let 3= 1 J 1 ^ + &c 1-2.3 T 2-3*4 3 -45 4*5 '6 Then *~ ^ = ill + 3^F5 + l-fe ^^H^Wfej 13 3 3 Whence - = H 1 — - + &c. b 1-2-3-4 2-3'4'5 T 3'4'5'G T by subtraction 3 • .1 1 11 Consequently -f H 5- -f ■ — f ' l'2-3'4 <% 23-4-5 34*5 6 ^ 45 6* -f &c. = - -j-3= — , the sum required. 6 18 4 Ex.9. Here the denominators of the fractions are figurate numbers of the third order, and if we divide the proposed series by 6, it will become 1 1 1 1 1'2<3 2*3-4 345 4*5.6 1 1 I . 4'5 Assume %= -j 1 1 + &c. 3-2 23 34 ' 45 Then % _I = .J- + J-- + JL + _L +&c. 1 23 34 T 4-5 56 And by subtraction, I- __ ?_ ?LI 2 _ 2 , & 2~ ri-k * 2*3*4 ;*.a;4 # 5\*'.^-" _ /r-1)* " r n V— 1 * (r-l)* ; (r*-l)* r- -1 the sum of n tefcis as required; where p= — . ,™ 11111,1 1 And therefore by subtraction, 3 4_ 4; 4 4 4 ~ 2-6 + 7? + Flo + Tia + c * Whence £| | = ig + ^ + ^ + ^^. The infinite sum required. Again, let »=I + i + I + I + -L + ...i- c m > «£ 3 1,111 1 Thenz--™-f 17+ ^ + 7^+"-7T 4 b 8 10 12 2# 3 1 1 113 1 And * - i + ^T^ + ^+5 = ■ S + 5 + To f 12 T 14^ 2?z + 4' Whence by subtraction, 3 _ _1_ _JL_ _ 4^ J, _4_ 4 4*" 2/2-J-2 2/2 + 4 ~~*2(j 4-8 6'10 S'J2. -f &c. to » terms 5 Consequently by division, 111 1 1 l _^ j. (-&c. to n terms. 2 6 T 4'8 T 6 10 8 12 3 1 1 572-f3;z* ~" HJ &<»-fl) 8(»-f-2) " 10?. e -f48?2i32 the sum of 72 terms, as required. Ex. .3. Let^iH-I + i + ^+Scc. And by subtraction, 13 3,3 3 ' . 3 36 + 6-9 9*12 I2<15 + ' 1 l * l J.) Therefore -X 7 = — = ~-f -g— 4- —-7; -f 3 4 12 3*8 012 9'i6 r— -f &C. 12-20 ( W ) A • 1 111 1 1 . Again, let x = - 4- - 4 [- -—^Scc. . . — Then x- - = - 4- - 4- j \~ kc. . . — 3 6 Q T 12 T 15 3rc . , 1 1 1111 1 And *-- + _ — = +-+—+-- + 3 3»-f3 Op 12 15 3« + 3 "Whence by subtraction, IS 1 1 1 1 :; "~ m l = 1- 7 h 1 h&c. Pterins 3 3(w-f-l) 3'2^ (5*3 9'4 12*5 And dividing by 4, * 11 1 ,rr: + ^ 1 ? H h & c< to « terms. 3-8 6'12 piO 12-20 *•->.;! 1 « "~ 12 ^ 12(»+1) ~" 12(« + 1) ~ 5 ' the sum of w terms as required. Ex. 14, Let *= - 4- - -f 1 h&JC 2 7 r J2 17 Then % = - 4- .— 4 |~ — 4-&c. 2 7 l'i ^ 17 2^ r ! And by subtraction, 15 5 5 5 , c - as 1 j j, ^ &c . 2 27 7*12 12-17 1/-22 Whence, multiplying by ~, we have o 6 6,6 6 o -. . . 3 +. 4 1 4- &c ad inf. = ~ 2-7 7'i2 J^ 12J7 17*22 5 Now the general term of the series is 6___ (5« -3) (57/4-2) ' (173 ) And therefore, to find the sum of all the terms beyond this, we need only assume s= 1 1 h &c. 5w + 2 5^4-7 5/2 + 12 rr.. 1 1 1 * * Then x — as + \- -f &c, 5»+.2 5w + 7 5n+12 5»+17 And by subtraction 15 5 5» + 2 ~ (5n-f-2)(5»-f7) (5« + 7)(5»+12) Or multiplying by -, we have tne sum of » terms, 5 5(5n + 2) 5w + 2 Ex. 15. This series may be put under the form 1 5 g 13 6> 1.2.3.4 + 3 A. 5. 6 + 5.6.7.8 +8C ° ') ' Let us therefore assume 1 1 , 1' 1 ■' « 1<2 34 5'6 x 7:8 m , 1 1 I 1 1 Then a = ■ -\ — — -f ■ ^ h&c. - 2 34 T 5'6 : _ 7'8. 9'10 „ J;? . 10 ■ 18 26 11 By subtraction j- +■ — = &c.= - J 1.2.3.4 3.4.5.0 5.0.7.8 2 Or by dividing by 2, 5 g 13 1 ; .14.3.4 x 3.4 5.6 ?-, 5.6.7.9 T 4 a3 1 5 2.3 A 1 ~&8 and 9 ( ,174 ) consequently ! )3 i 3.4,5. 1 910 6 ' 5.6.7.8 1 rft 3.6 12.12 -f &c.)=— , or ' 24 4- &"c.= — inf. sum. 24 Again, if the above assumed series be carried beyond the nth term, it becomes 1 1 (27! + 2)(2*-j-l) ' (2w + 4)(2» + 3)" Whence as above < = inf. sum 12(2w-r-2)(272-M) after n terms -, consequently 11 n n 24 1 2(2n + 2) (2* +1) " 2(3 + 6?*) 4(6 -f6») = sum of v terms. Ex. 16. Assume 1 1 1 1,1 3 5^7 9 11 ^ l —l l 1 1 l e Then x __ = ,__ + ___ + ___ +&c . 18 12 16 20 24 Subtract** - =_ - — *T£-"£1J + 77^ + &c. Dividing by 4, 2 3 I 5 _ 1 . - &c.= — inf. sum. 3.5 57 79 9.H 12 Again, if the above assumed series be continued be- yond the nth term, it will be * =S — 3 + 2» + 6 + 2n ~7 + 2n + 9 + 2« + ( 175 ) And therefore, as before. ± will be the ' 4(3+2/2) infinite sum of the terms after the nth 5 Consequently — dt — = sum of n terms H \ 12 4(3 + 2n) required, the ambiguous sign being + when n is odd and — when n is even. Ex. 17. Assume 3 4-5 6 *= + > + + -*- + &c. 1.2 2.3 3.4 4.5 3 4 5 6 7 o Then * m H h — - -f ~~- + &c. 2 2.3 3.4 4.5 5.0 T 3 5 6 7 8 Subtracting, - = _+_ + _+_ -f &c. inf. sum. ; Again, if the above assumed series be carried beyond the wth term, it will be n + 3 w + 4 n + 5 t *; — rrr; — rrt ~r (»+l)(» + 2) (n + 2)(w + 3) (« + 3)(w-M) Whence, as before, the first term 72 + 3 (w +"!)(» + 2) will be the infinite sum of the terms past the *th. Therefore 3 w + 3 2"" (n+l)(n + 2)' or 3 2 1 + 2 n+l . w + 2 sum of rc terms. ( w > LOGARITHMS. MULTIPLICATION BY LOGARITHMS, Ex. 7. Log. of 23 14 = T3643634 log. of 5 062 ~ 07043221 Prod. 11 7' V34J 2 0686855 Ex. 8. Log. of 40763 = 6102661 . log. of 9B432 = 0-0931363 Prod. 40.12383 1-6034024 Ex. 9. Log. of 498*256 = 2.6974525 log. of 4T2467 == 1*6153892 Prod. 20551-41 43128417 Ex. 10. Log. of 4«26747 =: 0604954 1: log. of -012345 =—20914911 Prod. '04971016 -2-6964452 Ex. 11. Log. of 3i 2567 = 0-4949431 log. of -02868 rr — 2*4575791 log. of M2379=— l-0920856v Prod. -01 109705 —2.0452078 Ex. 12. Leg. of 2876-9 = 3*4589248 -1067 4 = -1 0283272. .09876 2= —2*9945899 •003159 8 =z —3'4Q965g6 Prod. '0958299 — 2-98 15015 ( 177 ) DIVISION BY LOGARITHMS. Ex.?. Log. of 125= 209091OO log. of 3 72S zzz 3-2375437 Quot. '0723379 —2-8593663 Ex. 8. Log. of 1728-95 = 32377825 log. of 1106/8 s& 00440613 Quot. 1562-144 3-1937212 Ex. 9. Log. of 10 23674 as 1-0101617 log. of 4.96523 = 06959303 Quot. 2-061685 0-3 1422-24 Ex. 10. Log. of I99567 =43000888 log. of -018235 =^26833623 Quot. 413719 66167265 Ex, 11. Log. of -067S59- —2'b3 16075 log. of 1234'5y= 30915228 Quot. '0000549648 -5*7400847 RULE OF THREE BY LOGARITHMS. Ey.,5. Comp. log. of 12-678 =8 8969493 log. of 14-0(J5--=ri4S1397 lo£. of 100 079 = 2 0042311 Arts. 112 0263 2-0493201 ( m ) £x.6. Comp. log. of 1-9864= 9. 7019333 Jog. of -4678= — 1'6700602 log. of 50'4567 = 1-7029274 Ans. 1188262 1-0749209 Ex. 7. Comp. log. of -09658 = 1 J 0151 127 log. of -24958=— 1-3972098 log. of -008967 =z~ 3 9526472 Ans. -02317234 — 2 364g6Q7 Ex.8. The first part of this example is misprinted, it should have been a third proportional, not a mean proportional, Comp, log. of '498621 = — 103022294 log. of 29587 = 0-471100.9 log. of 2.9087 =x 04711009 3d prop. 17*55623 1*2444312 Whence 17-55623, the answer required. Comp. log. of 12 796 =. 8*892925$ log. of 324718 = 05115063 log. of 3'24718 *= 05115063 3d prop. -8240212 -1-Q159384 ( *79 ) INVOLUTION BT LOGARITHM*. Ex. 5. Log. of 6.05987 == 07824533 2 Ans. 3672203 1 -5649266 Ex. 6. Log. of 176546 as — 1*2468579 3 Ans. -005502674 —3740573? Ex.7. Log. -076543 = -2-8839055 4 Ans. '00003432594 -5-5356220 Ex.8. Log. of 2*97643 =2 OA?36g57 5 Ans. 233-6031 2*3684785 Ex.9. Log, of 21-0576 == T3234089 6 Ans. 871&7340 7'9404534 Ex. 10. Log. of 1-09684 = 00401432 7 Ans. r909864 02810024 ( 180 ) EVOLUTION BY LOGARITHMS. Ex.7. Log. of 365 5674 - 2)2-5C)2(}6*;4 Ans. 19*11981 1-2814837 Ex.8. Log. of 2*987635 - 3) 0-4753270 Ans. 1-440265 0*1584425 Ex. 9. Log. of '967845 4) — 1 '9858059 Ans. -9918624 - I'9964515 Ex.10, Log. of -098674 7) -2-9942027 Ans. 71831-46 —1-S5031 47 Ex. U. Log. of 21 as 1-3222193 log. Of 373 ss 2-5717088 -2*7505105 Multiplying by 2 Dividing by 3)— 3*5010210 Ans. '146895 =3 -~l-l670070 Ex.12. Log. of 112 = 2*0492180 log. of 1727 = 3 2372923 — 2-8119257 Which being multiplied by 3, and then 'divided by 5, gives log. —1-2871554, Ans. -1937ns ( 181 ) MISCELLANEOUS EXAMPLES. Ex.1. Log. of 2 = 3010300 log. of 123 bb 2-0809051 2) —2 2111249 Ans. -1275153 — 1*1055624 Ex.2. 0-log. 3*14159 3)— 1*5028505 •6S27842 — 1*8342835 Ex. 3. Log. '00563 = —3*7505084 •07=t2<5> theref. 7 100) -16-2535588 Ans. '6958321 — 1*8425355 The student will observe, that 84 is borrowed in this example to make the l6up to 100 according to the rule. 174 ~ { V X id' * 3 ~ V * 52 Hence 0— log. 6 6) — 1-2218487 fOx -1-8703081 Log. of 7 V —2*7611180 Ans. '04279825 -2 6314261 ( 182 ) Ex. 5. Log. of \ . flog, of | log. of -0J |log. T 7 T Ans. -00116^ 2 = >713 om btract L58636 - 7247622 1875096 -1-1549020 - 1 -8079400 -20791 812 -1-9345684 -3-0665916 Ex. 6. Log. of i I log. of U log. -03 1 log. 15j -1*C457574 -1-8595867 -2-4771213 039J9479 -37764133 log. of y- i log. 12J. log. -\g * * log. J7{- 0-8653014 0-3621199 -1-2787536 0*3084076 Fr Su 08145825 -37764133 08145S25 Ans. -0009: Ex. 7. Log. of | i log. 19 !Nat. numb. ' Log.* . ilog.^ -4-96 18308 -1-2208187 0-6393768 — 1-8601955 -I756962O 0-516G615 Nat. numb. —0*2730235 ( 1S3 > Log. of 2'5ggS5S2 *fc 0'4\4g4Q6 log. of ~ = 15017437 19166933 In the same manner we find H- OH— iVV28|) z=lO'2230796 Ans. 49 38712 r693ul37 MISCELLANEOUS QUESTIONS. Ex. 1. Let x be the number of minuies after 8, or the number the minute hand is before it overtakes the hour hand, after the former is at 12, and the latter at 8. x Then — will be the number of minutes that the 12 hour hand has advanced in the same time. And by the question 1 II 40 A x=x, or — ar=40, or 12 12 40X12 480 7 x= ss =43 — ; 11 11 11 viz. the time was 8h. 43m. 38 T \ sec. Ex. 2. Let x represent the digit in the place of 10\s, and y that in the units ; then will \0x -f#— the number itself, and \Qy-\-x the number formed by the inverted digits. Hence, by the question, - x*~y*=i\Ox + y 1 10* +y + 36 =10y-f x$ y From the latter we have 9.r— 9?/=— 36, or #~-#=-*4 ( 1S4 ) or y=x + 4 5 whence it appears that y is greater than x, and our first equation becomes Or, by substituting the above value of y, we have (.r + 4) 2 — ,r 2 =10.r + ,r + 4 That is 8^+16=10^+^ + 4, or 3^=12, or #=4, and i/=,r + 4=S ; Consequently 10x+y= 10X4 + 8=48, the number sought. Ex. 3. Let x and y represent the two numbers j then by the question *-y \ x+y '• 2 : 3 \ or 1 x+y t ^ : : 3 : 5 j J 3.r— 3y = 2x + 2yl 5x-\-5y=z3xy J From the former of these x~5y, which substituted fjr x in the latter, gives 25 y + 5y -1 5y*, or 30^ = 1 5y* Whence dividing by I5y we have y = 2, and conse* quently t rrr53/=10. Ex. 4. Let # be the number of games won, and y the number lost 5 then by the question #+ y~20l 9 2x — 3y — 5) Multiply the first by 3, and we have 3x+3y~6Q Add the latter 2x— 3y= 5 and we have 5x =65, or or ~ 13, games won And consequently 20— #=20— 13=7> games lost. ( 185 ) Ex. 5. Let x be the number of yards in the four sides ; ' then 3x is the number of feet ; and therefore by the question 3#— 150 is the number of palisades, As is also #-f- 70; and consequently 3^—150 — ^-1-70, or 2^=220, or 07=110; Therefore 3x— 150=.r-f 70= 180, the number sought Ex. 6. Let x be the number of hours in which b will fill it; then - will be the quantity thrown in by Bin an hour, and — is the quantity a throws in. Also since the two together will fill it in 12 hours, — will be the quantity the two throw in in an hour. 11 1 Whence - -| __ — , or a? 20" 12 60 3x _ 5x 60x 60x ~ 6Qx' x—3v =60 ; therefore 2x~ 60, or #==30. Ans. Ex. 7. This is not properly an algebraical question ; but the best method of solution is as follows: Dividing each effect by the time it is produced in, We have -, - and -, for the momentary effect of each * f g 1 r a * C t agent, and therefore — f- - -f- -, the momentary ef- * J g 7 feet of the three agents; consequently d-r-{--\- — f- _) e f g will be the time in which all three will produce the ef- fect d, K3 ( 186 ) Ex- 8. Let x be the required number ; then by the question (# + 3) : 0+19) •• (*+19) ' (x + 5l) Or, since by the nature of geometrical progression the product of the means and extremes are equal to each other, we have C* + 3)C* + 5) = (ff+l9)Vor a? a + 54ff+153=ff a + 38,r + 3(>l,or 54^—38^-208, or l6xz=20S, or oc=:\3 Ans. Ex. 9. Let r be the ratio, then since 3 is the first term, 3 \ 3r : ; 3r 2 : 24 are the first four proportionals, and 3r and 3r 2 the two means sought. Hence, since the product of the extremes and means are equal, we have 9^=72, orr 3 r=8, or r=2; and therefore 6 and 12 are the means required. In the same way, in the second part, we have 3 : 3r t: 3r 2 I 3r* :: 3r 4 : 96, Whence also we have 3rX3r 4 =3 X96, or 9r 5 =288, or r 5 — 32, or r=2, Therefore 6, 12, 27, and 48, are the four mean pro- portionals. Ex. 10. Let #, rx 9 r^x, r s x, r*x, r b x, be the six pro- portionals. Then by the question x + rx + r < *x + r 5 x + r 4 x + r b x=3\5 7 and x +r b xzzi65 ) By subtraction rx+ r^x + rSx + r+x — 150 Now, by the rules for geometrical progression, our equation may be written r 6 —l and the latter x—315 *) r-1 t r 4 -l ( rxz=l50J ( 187 ) By dividing the former of these by the latter r 6 ^l __ 315r __ 2lr r 4 — 1 r7 ;. 150 ~~ TcT Reducing the latter, by dividing both terms by r 2 — 1, we have r 4 -f-r 2 +l 21 r — 13 , or ' r 2 +i 10 21r And by adding r 2 to both sides ' r 4 + 2r2+]= _2L( r « +1 ) +r i >or (r 2 +l) 2 --^(r 2 +l) = r 2 Hence by completing the square ~ 21r I x 441r 2 n 441r 2 (Y 2 +l) 2 fr 2 +l) + = r 2 + V y 10 v ' 400 400 A v>'. ; a 21r, . N ; 4llr 2 841r 2 v -r t 10 v y ^ 400 400 21 2Qr And by extracting the root (r*+ 1) r= , or 25 r 2 r=-l 10 Or r= — ±i/(- 1) = — ± — =2 3 20 v v 400 ; 20 20 But from our first reduced equation we have _315(r— 1) __ 315 : Therefore 5, )0, 20, 40, 80, and 100, are the pro- portionals sought. ( 188 . ) iG IW Ex. 1 1. Let x and y be the two numbers; then by the question x + y~ 1 1 -4- - = * 2/ From the 2d. x-\-y=bxy=za> or ,ry= j Squaring the 1st, «®-f 2o?y +y 2 jra a 4* Subtract . 4^= — We have *»— 2sy +y a =a«— y a 2 Z,-4a Or *-y=S — j Repeating again x+y~—a, the 1st question, ,, T , . , i. • M cfih—4a We* have by addition 2:r=ra-fv — j , or 1 1 a J *=2* + 2 ^^" 4 > And by subtraction 2y=a — \/(- — j #) or y=i a -|vf<«*-4) Ex. 12. Let # be the number of men employed at first •, then x-\-l6 will be the number in the second in- stance; and since the time in performing the work is re- ciprocally as the number of men employed, we have x : id :: x+\6 : 24 256 Whence 24x~\6x+25d, or x=z =32, the num- ber of men at first, and 32+ 16=48, the number during the second part of the time. Hence 32x24x l£=ll52$.\ . ., A An€ ( 189 ) Ex. 13. Let x and y be the two n a rubers y then by the question, we have x*+y~ 62 1 From the first y=62 — x*, and consequently s/ ft -62~— 124^ + tf 4 This substituted in the second gives 62 e — l24^+x 4 + l r=l/6 or, .r 4 — 1 24**= J 76— 3844 - or or, x*-124x~ + x=. — 3663 Whence x will be found =7> an d consequently z/ = 62 - ^r a = 13 3 but the question cannot, I believe, be in any manner reduced to a quadratic form, at least while it is considered under the general form xz+y — a, and y*-\-x~b. Ex. 14. Let x denote the number of feet in the cir- cumference of the less wheel, and y the circumference of the greater wheel. Then will be the number of revolutions of the x less wheel, and the number of revolutions of the V greater. 360 360 ) * y ~ Whence by the question "< 360 300 From the first, 360(y—x)z=6xy From the second, 30O(y — ar)=4(ff + 3)(y + 3) The latter of which by multiplication gives 360y—360x—4xy + 12i + I2y + 06 or 348y — 372x=4xy + 36 ( m ) And the former, by division, 60xy — 60a£=z xy Mult, this by 4, 240j/ — 2AQxiz Axy Subtract 348# — &?2xzz 4xy+-36 And we have --1082/-f 132#= — 3(5 Whence '£j&£^&& 132 11 Substitute this value of x in the equation 60y — GOxtssxfi \ And we have . 540y~180 93/ 2 — 3y 6f%— 54(ty-f iQOsrJty*— 3y, or a?/ 2 — 123^ = 180, or 41 1^ y~2Q 5 whence and consequently ^"""" == — ■ = 12 j that is, * the circumference of the greater wheel zzl5, and the cir- cumference of the less =12. Ex. 15. Here by the question x + y=al mx -f ny = & J Multiply the first by m, and we have wx-\- my —ma Subtract, mx-\- ny=z: b 1 .. ma — b Aud we have {m—n)y—7?ia—b i oxyzz m __ n ( 101 ) Again multiply first by n, and we obtain nx + nyzz.na Subtract mx-\-nyzz b And we have x(n — m)-=.na — b, or x(m—n)z=.l—7ia Whence or= . m — n Ex. 16. Here the several portions of wine drawn off were, 10 gals. remains 74 10.74 . ■ . 10.74 74* remains 74 == 84 84 84 1074 2 . 74 2 ]0'7*J 2 74 3 W t • remains -, zz — - ^^4 9 84 84 2 84 2 10.74 3 . 74 3 10.74 2 74 4 remains — r ; sc — - 84 3 84 2 84 3 84 3 744 Therefore — - =r 50*569 gallons remaining. Ex. 17« Let x represent the number of persons, and y the number of pounds each received, then xy is the whole sum divided. Now by the question (x-3)x(y + 150)—xyl (x + 6)x(y—l20)=zxy ) or xy — 3y~\- \50x-450-xy 1 xy + 6y — \M)x-720=xyl' ° V 150x—3y—4501 — 120ar + r3?/=720j Multiply the first equation by 2, and we have 3GQx~6yz=zg0O Add the 2d 120* -f 6^=720 Whence 180^=1620, or x=Q, the number of persons 5 300*- Q00 And consequently 3/= -r-^ — - = 300, the sum (1 9 2 ) «ach received j and 9x300=2700/. the whole sum di- vided. Ex. 18. Since the reduced value of the i6 pieces is 8/. 8s. and the part taken from them is \6x2s. 6d.=.2t. the original value was 10/. 8s. or 208s. j consequently 208 —-=13.?. the original value of each. Ex. 19. Let x be the number of ounces of tin, and y the number of ounces of copper. Then by the question, x + 3/zzlOOl 4fcr+5|y=505J Multiply the second equation by 4, and the first by 17. Then 17 . o? + ^9=32=^ J Assume xy=p ; then a^-f ?/ 3 = a 3 -Sap * 3 +y»=(a 3 -3ap) 3 -3/> 3 (a s — 3ap), Or a°— 9a 8 7p + 27a 5 /) 2 — 27a 3 /} 3 -3a 3 p 3 -f-9a//z=^ 5 s ( 19* ) Or in numbers, by transposing, &c. 40 80 3 3 Which equation is resolvable into the factors (^_ 4 p + 4)(jB»-— p+—)=0 Whence by the solution of these two quadratics we have 14 a/136 . 14 \/136 /)=2, p=2, p= — + — ^-, and £= ~ — - But #-h y=2, and xyzzp ; Whence x—yzzidb V (4— 4/>)=2 y^l — p) Or a?=zl±v/(l— p,) andj/=l+ V(l—^)5 And substituting here the above values of j>, we have the following solutions, viz. ar=l-f V—l, and#=l — \/— 1 r — 11 \/136, o^l-f-V^— 3~i*and -11 ^136 2/^i-v/J— — ^ — jr -U \Z136, * l+\/l-^-+— r-l* and o o ,—11 ^130, The two latter of which are the only real answers ; the others being imaginary. — Note. The divisor 3 is omitted in the answer in the Introduction. Ex. 23. Let x and yx represent the two numbers 5 then by the question yx*= t/V— x l , and y*x* + x*=y 3 x 3 - x 3 ( m ) Whence, dividing both equations by x" } they reduce to y=y q -i From the first of which we have y a -y= l > or y =\ + W5 And from the second 2/ 3 ~l 1 + V5 2 V Consequently yx=l J5 X (i + f v r 5) = J(5 + *f$) ; That is, |v/5 and i(5+ aA5) are the numbers sought, Ex. 24. Let or and y be the two numbers 5 then The geometrical mean = \/xy The arithmetical mean -=.\x-\-\y 2xy The harmonical mean = — ,— -. *+# Therefore by the question \x+\y-*/xy~\3 2iy From the first of which equations we have x+y~26 + 2i/xy And from the 2d, x 4-?/ zz - J */xy—\2 2xv Consequently 26 + 2 */xy=z — - ; W h ence 26 \/xy + 2xy — 3 1 2 — 24 v^ — 2*y Or 2 V^/ —312, or \fxy zzl56, and xy =24336 ( 196 Substituting the value of */xy in the first equation, and repeating our last, we have x + y — 3 r 3$ xy =24336 And by squaring the first of these x C2 + 2xz/-f 2/ 2 =ll4244 - Axy = 97344 B y subt rac t . x- — 2xy + y' 2 = 1 6QOO By extraction x—y = 130 Also # + 7/ = 338 3 Whence by addition 2x=46S, or #=234 And by subtraction 2^=208, or #=104. Ex. 25. Here *fy +^ r = $^ x°y* + y 6 x*=7) By squaring the first equation, we have #y-f 2x 4 y i -f y 6 x 2 =^Q Subt. the 2d, x*y* + y*a* = 7 2xy t ==:2, or ;n/ = I. Hence, dividing the first by xy, and the second by x Q y <2 , we have x~ + y q =3 x A -\-y^^y From double the last 2x A -f 2y 4 = 14 Subt the square of the first x 4 + 2x*y*+y*=Q And we shall have x* — 2x Q y 2 + y 4 ~5 Or by extracting x~— y~zz\/5 And by repeating x"-\- y' 2 — 3 Therefore .by addition and subtraction y : __3 2 -r 1 2 A/5 l f 3 2 — 1 2 \/5 ( 197 ) j us are = 23) 2 =J67> =385) Whence, extracting these by the rule for binomial surds, we have #=£(1+V5), and 3/=§(l — v^). Ex. 26. Here the equations are x + y -\-z = xy + xz + yzz xy% to find x, y, and z. From what has been said in the Introduction, relative to the doctrine of equations, it is obvious that these num- bers are the coefficients of a cubic equation which has its three roots equal to the several values of x, y 3 and z 5 whence w r e have at once x 3 — 23 JT 2 -f 167^—385=0. The three roots of which equation, by the rules for cubics, are found to be 5, 7, and J 1 , the numbers sought. Ex. 27. Here the given equations are xyz = 231=a xyw~ 420= £ xz%v~ 660=: J 2/%w=1540=c Multiplying these into each other, we have x 3 y 3 w 3 z 3 — ahcd or x y w z = \/abcd Whence, dividing this last equation by each of the given equations, we have %/abcd a %/alcd 2/ abed d l/abcd c . s 3 =20 = 11 = 7 = 3 ( 199 ) Ex. 28. Here the equations are x-\-yz—384:Zza y + xz^237~b %-\-xy=z\Q2= c From the first x~a—yz; which substituted in the se- cond and third, gives y + az—y z^zzb z + ay—y q z =c Also from the first of these two equations b — a% az — b Whence, substituting this value of y in the latter, we aH—ab xfaz—b)* ***** ^^^r^T^riy^ or z(z*-~l)* + a(az--b)~-z{az-l>) Ci ==c(z q --l)' i Which, by multiplying and involving the several fac- tors, becomes z s — cz i—2z s +(2c + ab)z ci —(h* + a (m' 2 + m) a(» + » 2 ) =c ( m* -f- nm) From the first of these we shall have 71 = — ' a Which value of n, substituted in the second equa- tion, gives * Wnfi+m)-a\xilm*+V + a)m-a} =c{m ^ m) a And by reduction, b*m i +(ab + 2b*)m i + (l>*--ab--ae)Tri i -- a(a + 2b + c)m+ fl 2 =0 5 Or in numbers, 82 i 65331 8Q208 11664 nrA m 3 tt— m 2 -— m =* * -. 23 4761 ' 4761 4761 From which is obtained m=3 \ b(m* + m)-a 20 a . Consequently n= = — , y= */( ) =3, x=my=g, and z=ny =20, as required. Ex.30. Given **4=Vy +3/ 2 = 5 -> ar 4 -r-^V+3/ 4 = 11 J Here, dividing the latter by the former, we have x*-zy+y*= - x^-\-xy+y-z= 5 By addition x 2 -\- y 2 = — By subtraction xy= — ( 200 ) Also, by adding and subtracting double the latter from the former, we have 36 28 64 x*+2xy + y' i = 1 da — _ j _ 36 __ 28 __ _8_ - — 2.T2/4-2/ — 2 ~ — — — 10 64 Whence tf-j-i/ = a/ — Q And a?— y = V — r - 10 1 64 1 8 Consequently *=-*/- + -,/--, a ^ l > 64 * • 8 And v= ~ v a/ — ; y 2 V 10 2 V 10* Or by reduction 2 , 1 2 1 y = - a/10 — - a/5 Which are the values of x and y, as required. Ex. 31. Here the given equation x^ n —2x 3n + x n =a May be put under the form (x zn -x n )*— (#**— x n ) =a, Which being now a quadratic, we find x* n -x n =i± a/(| + «) And this is again a quadratic, from which we derive Therefore, by extraction, ( 201 ) Ex. 32. This question is misprinted in the Introduc- tion ; it should have been as follows : It is required to find by what part a people must in- crease annually, so that they may be doubled at the end of every century. Let a represent the number of people, and _ the first x year's increase ; then at the end of each following year the numbers will be a 1st year a+- ' x a a 2d a + 2- + — x x 2 3d a -f3--f- 3~ -f -^- X X 2 X 5 . . a „ a a a .1th a + 4- + 6— +4 — + — X X* X 3 X* a a a a a 5th a + 5- + 10— + 1.0 — +5 — + — XX 2 X 3 X* x b 100th a + J »- +B _ +/ ,_+&c.»_ + ^ Where m, ?i, p, &c. are the coefficients of the binomial U + l) xo °. Which expression may therefore be written «(1+ -) , X which by the question is to be equal to 2a ; whence (1 + V°°=2, or ±="^2-1* x x That is- = -00695, or x~ * — -r— s 144 nearly ( 202 ) The annual increase must therefore be i+^th part of the population. Ex 33. It is obvious that the least number of weights that can be used to weigh 3/6. is two, viz. \lb. and 3lb. and if to these we add a gib. we shall be able to weigh all the weights, Q±l, g±2, g±3, 9 + 4, viz. asfaras 13/$. * Increasing again our weights by 3x9lb.= 2Jlb* we shall be able to weigh 27±1, 27±2, 27+3, &c. 27 ± 13, that is, to 40lb. ; and in the same manner by the addition of three times the last weight, viz. 81, we can weigh 81±1, 81 ±2, 81±3, 81±4, &c. 81±40. Therefore 1, 3, 9, 27, 81, &c. are the weights re* quired.* Ex. 34. Let w, x, y, and z, be the four numbers -, of which let w be the greatest ; then we have to find zv^=x' 2 -\-y 2 -hz% V 2 Assume y*—2x%, or x= —> „ T 2% Mi 4z* Where y and z may be taken at pleasure. If y=z6, and z~2-, then x~g, and w=ll, which are the- least whole numbers. Ex. 35. Let x be the number sought j then by the x—5 x~4 x — 3 x—2 , #— 1 question -g~, -y-, -j-, -j- , and — ^-, are to be all whole numbers. % 5 Make — — =/>; thenar=6/? + 5 5 substitute this in the 2d, and we have * The most general method of solving questions of this kind is by means of the ternary scale of notation.— See Barlow's Theory of Numbers, chap, 10. Then t^=-^— + i/ 3 + » 2 ( 203 ) 6/> + l p + 1 _£ —p-f. £ =zo, ovp-=5q— 1, and conse- 5 5 quently #=30^— 1. Substitute this in the 3d, and we have . 30?— 4 , a . — ± — — =zwh. or 7<7 — 14-— =wfl. 4 /Y 2 Whence ^=2r; consequently o;=60r — 1. This, substituted in the 4th and 5th equations, gives whole numbers ; therefore the general value of x=60r — 1, and if r=l ; whence x=60, the least value sought. Ex.36. Let x be the year required; then if #+ g, x + 1, and x + 3, be divided respectively by 28, 19, and 15, the remainders will be the cycles of the sun, the golden number, and the Roman induction. Hence #+■9—18 #+1—8 ,#^3 — 10 and 28 19 15 0r£ _9 x _ =1 andf _7 28 19 15 must be all whole numbers. Let X -^- =/), then x~2Qp + 9; 2.0 and substituting this value in the second equation, it be- 28*> + 2 . . 28p + 2 9P + 2 comes — - zzwh* : whence — - — p+ — 19 !9 19 = a whole number. T 9P + 2 , 19<7— 2 — 2 Let- =j theno;=-i^=l7p+-^ 2p— I . 5?-|-l = it>/$ j assume — - — zzq 3 then />:=: — ~wh. Or - —whzzr; whence q~2r— \, Where r may be taken at pleasure 5 if r=l J , then qszl, and p=:3 5 whence #zz — = =52, and yzz 153, the least numbers that answer the conditions of the equa- tion. ( 205 ) Ex. 33. Let x be one of the equal sides, y the base, and % the perpendicular of the first triangle ; and x' s y\ and %, the corresponding lines in the second triangle. Then 2x+y is the perimeter of the first, and 2x +y' the perimeter of the second. Also v — the area of the first, 2 and - — the area of the second : 2 ' Whence, by the question, we must have 2x+y—2x'+y f and #*= j^V Also jf*= ^ — h ^*, and x 2 — ^- +z* 4 4 JL 2 Then *=(**- ^) *=2rw 2 -f w// + w e ) 2 4- (HH- 2wtz)* y'=zr li — p*=2tvi L2 i-7nn-\- -/z*) a - 2^^2a*a) f Where 7/?- and n may be takeii at pleasure. If tw = 2 and w=l, then or- 58, ^mSO, #'=74 and v'=48 5 or since all these numbers arc divisible by 2., we have V=37, j/'=24l squired. Ex. 39. Let or, 3/ and % represent the base, perpendi-. cular and hypothenuse, of the first triangle. x', y, and %', those of the second, and x'' y y ', and z , those of the third 5 then we have to find x 2 -f y ~—z 2 } x 2 +y 2 — *' 2 k all rational squares* x'^ + y'" 1 -**' A\ so xy *2r xy ' zzx"y" And in order to fulfil the three first condition?, Let x =m 2 — -n °~, and yzz2m?i 1 #' =7n' 2 — ri % , and y'—2mn' x'—iri'i—ri"*, and y // =:2m // n // Then it remains to find (m' x -n z )X2m n zz{m' *—?i' *) X2m' n' (?;z 2 — 72 /2 ) x 2m'n / -zz (m" 2 -- ra" 2 ) x 2m"n" Which equations may be resolved into the factors {?» -f« )(w — n ) mn ~(m' + ri)(m—n)mn / (ra" -f n'){m' — n") m'n" zz {jn -f ri) (m —ri) mn' Where it is only necessary so to equate the iac tors of ( 207 ) each of these equations, that the reduction of them may- give the same ratio to two of the quantities. For which purpose, let m + nzz2?i and m — in' \ then m — n~2(m-<-ny and n = 2ri — in! But since the product of the preceding factors are equal, we must have m' + n' In —m ss , or 7 ' n=5m Again, in the second equation we may assume m 1 -\-ri'~on' n'-=^ m' and 2{m"— ri')^ m—rf . oil — ni Then m 2 Where again, because the product of the factors are equal, wc must have 3n' — in! m'-\-n' — 2 — ■* ~~3~~ ' ° r 5m 7 " ' the same ratio as before. Assuming therefore, ra'— 7 ar *d w'=5, we have wz=7 and tz=3; w"~8 and »''*=? i Whence * =^w 2 — ?z 2 ^40; y ~2m n = 42 or' = m' 2 — ^ /2 =245 z/' =2raV = 70 x"=m"*—n"*= 15 j \j"—2m'n"— 1 12 Ex. 40. Given 3;* = 1-2655 to find jr. Here a few trials show that x is between J 3 and 1*4. „ n .- t lo£. 1*3 „ / Where, if t=T3, then - ■ ° f - = 'OS764Q locr. 1-2655 == '102262 Error —'014613 ( 20S ) V4 log. 1-2055 = -102262 i'.J-j, , los;. 1*4 And if .r—i-4^ then — = '104377 1*4 Error + -002115 Hence -016728 : -1 :: '014613 : -0876 Therefore 13 + 0876= 1 '3876 Ans. Ex, 41. This equation is better put under the form £ - -6 a? Or log. yXx— log. crX3/=rlog. *(5= — 1-77815. Which by a few trials shows us that x is between 4 and. 5, and y between 5 and 6 ; let us, therefore, take one of those quantities y as constant, and correct the other #. Here then, assuming y=.5 % 5, let us take x=4*6 and 47j then First, if #=4*6 ; log. 5 5 X 4 6= 3-405^65 log. 4*6X5' 5^3-6451 58 — 1760507 log. -6 =--1778151 Error —017644 Secondly lfx—4 7, then log. 55x4 7=3:479704 log. 47x5-5=3.696538 — 17831 fit) log. -6 —1 778151 Error -f 005015 Whence -022659 : 1 :: 017644 : 09 Therefore r=4*69 nearly. ( 20p ) By a similar process we may find y~5'5l nearly ; and repeating the operations again on x and y, we have and q=5, then m=30aad 2*==1 680 x3 ( 210 ) the numbers sought ; and various others might be found by giving different values to p and q. Ex. 44. Here x and y being taken to denote the numbers sought, we have to find x* + y*=U Assume x 2 -f 3/ 2 i=(r3/— x) 2 ; then we have x* J ty 2 —r ll y*—2ryx+x*, or y =zr~y —2rx f or 2r Whence x 3 + y 3 =z- /-£- + w 3 ■•'* 8r 3 "* which is to be a square, or ( ^-iy + Srs ; 2r * 2ri 2 And consequently ?/ = -— — — — -, Where r and s may be taken at pleasure. 4s* 3s 2 If r=2, then v= — and 0?= — ; so that taking szrQI, we have y~364 and #—273, the numbers required. *Ex. 45. Let x 2 and i/ 2 be the numbers sought ; then the two latter conditions will be fulfilled, and it will only remain to tin d x*-\-y*zz a cube. For which purpose let x—rz, and y=s% j then rV+/s 2 - a cube. trrrr % * Assume rV- -f s*z % z=. — ; v J Then *« v s (r* + A-«) v r=?t; 3 (r a -f j-) ( 211 ) Where r, s, and*/, may be assumed at pleasure. If y=2, jziI, and i»=l, then #=10 and y-—5; conse- quently ^ 2 =100, and y°-z=z25 ) are the numbers sought; And/ by giving different values to r, s and % an in- definite number of other answers may be found. Ex. 46. This is the same question, except a little va- riation in the enunciation, as Ex. 2f3 of theDiophantine Pioblems, where we found generally ( />2 r/ 2V x=z(p cl -f) > fiJ/. zzz%pq±l LL Kr * ; ' . v (p*+f) Hence, if each of these be multiplied by (p 2 +? 2 ), p i —q 4 9 2pq( k p 9 +^) and 2pq(p 2 —q*) will be the inte- gral roots in the present question ; p and q being taken = any unequal numbers whatever. If r-2, and s=zl 3 then 15 4 , 12 4 , and 20 4 , are the biquadrates required. Ex.47. Let ax 2 , ay* and -^-, represent the three numbers in geometrical progression] then by the ques- tion (y 2 — x*)a= Q Or y*—x' l —am' i and y^—x^—an* Whence v a -f a 2 = — - szw'* Assume, therefore, y=p 2 — ^ e , and x~2pq 5 where/? and 9 may be taken at pleasure. Jf p=2, and gasl* then 3/=:3, and x=4; and the numbers are 10a, Qa and — — , or 250a, 144a and 8.1 a* ( 212 ) y* — X* a being 3C : — - — ; where m 2 may be any square factor whatever of y^—x*. In the present instance let m~\, then «=7, and the required numbers are 1792, 1008 and 56?. Ex. 43. Let x, y, and z, denote the three numbers, and assume x+y~ a 2 , x + zzz.b*, #-{-%=c*$ then by subtraction all squares. y- We have, therefore, only to find such values of a, b and c, as will satisfy the latter conditions ; for in that case the former must have place. But such values of «, b and c t have been found in Ex. 27 Diophantine Problems, viz. 697 2 =485809, 185 2 =34225, and 153^=23409. Hence, considering these quantities as known, we have by the common rules x— %=a 2 — c 2 y x-yzzlP—d*' V a 2 + ^ #— =2483121- 2 2 a * + c *_£3 y= 2 — =2374961 %= T =2140871; Or multiplying each by 4, in order to avoid fractions, we have #=993250, y=g4ggs6, and % = 85&350. Which numbers answer the conditions of the ques- tion ; and various others may be had by rinding diffe- rent values for &, b, and c. Ex. 49. Let x and y be the numbers sought j then we have to find x +y —a square, x*+yz-=. a biquadrate. ( 213 ) First, in order to make x (1 + y 2 = a square, assume a?=p* — £*> and ?/=2/>^ Then shall x* + y iz =^( /> 2 -f ? 2 ) 2 But when L r 2 4-2/"±= a biquadrate, p*-irjp must be a square ; assume therefore, again, pssr* — s^i and q=2rs, Then we shall have # 4 -f 3/ 2 ==(p*-f-<7 2 ) 2 =(r Q -f *) 4 , a biquadrate as required. And it now only remains to find x + y=r* + 4r 5 s— 6r% s — 4rj 3 -{-5 4 = Q. Hence, in order to redace this to a more convenient Form 3s Yor solution, substitute r= K; and the above for- mula, after multiplying by 16, reduces to s*— 2g6sH-\-408sH*+ 1 605p 3 -f 1 6/ 4 = D Again, assume the formula = (**+ 148 f/>— 4/)*= * 4 — 296^* + 21 8965V— 1I84^ 3 4- 10V 4 , Then, by cancelling the like terms in both, this reduces to 2189&— 11S4*=408$4- I6O/5 Whence j 1344 _ 84 7~ 21488 -" 1343 Assume, therefore, £=84, and /=J343, and we shall 3s have r= — 4-/= 1469, and consequently x—r*— 6r*s*f-te 45654860277(51 ^=4r^ — 4r5 8 ^: 1061652293520 for the numbers sought C 214 )■ Ex.50. The solution of this question is intimately connected with that of Ex. 48 j for if we here call w, x y y and % % the four numbers, and at the same time make wjz.x-^y + z, we shall have to find W — X-=zy + %= □ w—y=;x-]-z=z □ ^— z—x+y~\2 Also .r— ■?/= JJ, #-— s = □> ^nd ?/ — * — Q • It is, therefore, only necessary that „r, y> and *, may be Such numbers that the sum and difference of every two of them may be a square, which are the conditions of Ex, 48, where we found the three numbers to be x=. 993250 a= 856350 And consequently ws=27Qg586 Which numbers answer the conditions of the question. And if, in our 48th Example, we had taken a=2l65, b~2067 and c=2040, we should have found x =2399057 y =2288168 * =1873432 and iv=z6560657 which are the numbers given in the answer in the Intro- duction. Ex. 51. Here the proposed series may be put under the form 111 I - -f 1 f- — , 3 9 27 ' 81 1 1 1 + 9 + 57 + £T' 1 . 1 &c. 3 1 &c. 3 1 = 2*9 &c. 3 1 ~~2 27 &c. 3 1 ~2 X 8i < %n ) Whence, by the rules and formulas for geometrical pro- gression, the whole sum is 3 11 11 x 3 3 1 3 = -(--1 1 H Y &:c.) =- x -X-=- 2 l 3 T 9 T 2; T 81 ' 2 23 4 Ex. 52. This may be separated into the two series 3 2/ 243 a 16 3 4+6i + T^ +&c - =: y x 4 o 81 /2Q x 16 9 12 Q 3 Whence the whole series as-- — - = -. Answer. 7 7 7 Ex. 53. This question does not properly belong to any of the cases of series that have been treated of in the Introduction ; the reader is therefore referred for the mode of solution to Stirling's Mcthodus Differentialis, or to Euler's Analysis lnfinilorum^ Part I. Ex. 54. Here, by the rules for arithmetical progres- sion, the 72th term =5 + (n—> 1) Theivf,:? j iO-f {n— \)\- — 7 (n + Q) is the sum re- quired. 2 2 Ex. 55. The 25th term of the progression 1, 2, 4, 6, 16, &c. =. 2 24 =:l67772l6 5 therefore the 25th term of the pro- posed series, is 216777216; that is, the 16777216th power of 2. Now the log. of 2zz0 3010300 Mult, by 16777216 Gives 5050445.3324800 for the logarithm of the 25th term 5 consequently the in- ( 216 ) dex being 5050415, the number of integers will be 5050446. Ex. 56. This series is the same as (2*-2) + (4*-4)-f (6 2 -6) -h&c. viz. r4(i 4 -h2* + 3 2 + 4* + &c. 10 o 4 )- 12(1 4-2 +3 +4 -f &c. 100) But the sum of the former (n+ 1)tz(2«+ 1) 4x 101 X100X201 :4X 6 = 1353400, And the sum of the latter z=2x . = 10100. 2 Whence 1353400—10100=1343400, the sum required. Ex. 57. Here the general form of the series being (w+l)(w)(2n+l) q and n being =50, we shall have 51x50X101 , . , =42925, the sum requirea. Ex. 58. By the differential formula we have 35, 72, 111, 152, &c 37, 39, 41, &rc. 1st diff. 2, 2, &c. 2d dirt. Whence a=35, i\ 7 , d"—2, n—25i And consequently 1*2 1-2-3 35x25-1-12x25x374-25x8X23 = 875+1110(5 + 4600=165/5, the sum sought. ( 217 ) APPLICATION OF ALGEBRA TO GEOMETRY. MISCELLANEOUS PROBLEMS. PROBLEM I. Let A3CD be the given semicircle ; ab, its diameter j g, its centre; and cdfe, the required square. and it a e Then, since df=*3e, we have fg:=ge. Let therefore ab=<£, or cg=| Whence x=< i /%d*=:ld\/3, the side of the inscribed triangle. Again, produce en to. p, and ae both ways to g and n; and draw si, ik, perpendicular to ef and eg. ( 219 ) Then It is obvious from the proposition above referred to, and Euc. (iv. 3) that kf=£he, and hf=^hi. Let now hi (the side of the circumscribed triangle) ~y, and we, shall have H£ fi =EF a -f hf s j and since EF=fc?, H2— d f . and the above becomes d*3=$d*+iy*, or3/ 2 =3a 2 , or y — d\/3', the side of the circumscribing triangle. Ex. 4. It appears, from Euc. iv. lO, that the side of an equilateral and equiangular decagon inscribed in a circle, is found by dividing the radius of the circle into extreme and mean ratio, the greater part of which is the side of the decagon E Hence calling the radius oB=rr, and the greater part 00=*, we must have r(r—x)~x*, or £ 2 -f rxz=r* j Whence *=: — |r+|V5r a , or xzz\r( - 1 + */5) That is, bc or ab in the above figure = £r(— -1-f- \/5) Produce bo to e, and join ec > then by Euc. i. 4?, Ec 2 =EB 8 -BC 2 =r 4 (|-f |-v/5), or i£=ri/(4+|V5) Again, as eb ! ec :: ec t KD=r(£-f.iv'5)j and E3 ! bc !I bsg I BD=r(J— ^a/5), Whence dc — ^Xdb) =V ^(1 + ^5) (%L-%l/g) =^^(10—2^5), or AC=£ck/j 10—2^. the side of the pentagon required*. ( 220 ) Ex. 5. Let *— the length of the rectangle, and y= the breadth; then 2x + 2y= peremiter, andx^= area. Now since the side of the square =a, its perimeter =4a, and its area =a*, we have xy = {a*\ 2x + 2y=4a,S Whence aP+2xy+y*=:4a* Subtract Axy = 2a fi We have x 2 — lxy + y*—2a\ or x—y = u s /2 Rut x+g=2a Whence by addition and subtraction x^a + \a*S2=.a(\ + \ s /2) the length and breadth as required. Ex. 6, Let abc be the given equilateral triangle, and bisect tbe two sides ab, Ae, by the two perpendiculars DO, EO; C Then shall the point o be the centre of the inscrib- ed circle; and od its radius. Also, if ao, ob be joined, they will bisect the angles A and b, and o w ill therefore be the centre of the cir- cumscribed circle, and oa its radius, Euc. (iv. 4) and (iv.5.) ( m ) Aga'iB, if od be produced, it follows from propositions, that it will pass through c, and bisect angle acb: Therefore the triangles acd and aod are similar 5 and since ad^Jac, therefore do=|ao j Let now ab— s the given side, or ad=§s; also do^zx, and consequently ao=z2jtj then ao 9 ~ad 2 -|-do^ or 4 l z ,2 =Js 5 + x 2 ; whence 3x 2 ^is^ or ff==$i Vy=-fA/3~ 2'8868 and 2x=zs\/}zz j -f \/3zz5'77 r dC> the two radii required. Ex.7. Let abcd be the rhombus, and ac, db> its two diagonals, intersecting each other in e. D_ _C Also let the perimeter -~4p, that is each side of the rhombus ~/>, and the sum^of the two diagonals zzs. Then since the diagonals of parallelograms bisect each - other, CE-r-EB=fs. And because the three sides of the triangle dec, and ceb, are respectively equal to each other, the angles at e are each equal to a right angle. Therefore calling ACira? and r>B=y, or cb~ \x t and eb~^5 we have x+yzzs 7 From 8 times the latter 2.v*-r-2y 2 =8/>* * u 3 ( 222 ) Subtract (*-hy)*= x*+2xy + y*—s\ And we have x*—2xy -f y q = 8/> 2 - s*, or x-y ez 3 w 2 ) Or since $=8, and p=3 ; these become *=4+ 1^8=4+^2 2/=4~|v/8=4~/2 Which are the two diagonals required. Ex. 8. Here the three sides of the right angled trian- gk are x 9 *, x* x , x x ; and since the square of the longest side is equal to the sum of the squares of the other two (Euc. i. 47), we have a^=^* + x ix ; or x 4x — x 2 *~l Whence x**=l + \ V5 = l 618034 And ^=v/r6l8034==l-272020; x* x X«*"* Therefore = J 029085 the area. Ex. 9. It is a well known geometrical theorem, that the diagonals of a parallelogram bisect each other 5 and that the sum of their squares is equal to the sum of the squares of the four sides of the parallelogram. 3^ -.C It therefore we represent the given parallelogram by ( 223 ) the figure abcd, and make its side Bc = a, oc=^ DB=rf, and ac=.v, we have from the above theorem ar a +flF=2a«-f 2#*, or j*=2a*-f-2fl*-d*, or a: =rv / (2fl* + 2i«— rf 2 ) Which is the diagonal required. Ex. 10. Let abc be the proposed triangle, en its per- pendicular, and ad, db the segments, of which the dif- ference is given. Let that difference =xrf, the sum of the sides =$, and perpendicular cd=^. C Also put AD=i/-f-|^ and DB=y— { 9 =2?/?j then Squaring again both sides, and actually performing the multiplication of the firs* two factors, we have ( 224 ) And by involving and collecting the terms (2m + 2f-^)y^rn z -[pd?-- T \d*—f, wi .m^-klM*— r " d*-p\ Whence y = */( — - — £-) In which formula, substituting the values of j>, d, and m 9 we have y~472£; Whence y + ±d— 720= ad y — | f /— 225 = de f AC= v / (ad*+cd»)=780' Consequently < ^=y(i>B*+CD*)=37^ C ABr=AD-fDB = 945 Which are the thtee sides required. Ex. 11. Let abc be the required triangle, and ap, ee, and cd, the three given lines bisecting the three sides cb, ac, and ab. Make af=g, be=^ cd=c, also cb=jt, AC^y, and ab=x. Now it is a well known property of triangles, that " double the square of a line drawn from any angle of a triangle to the opposite side, together with double the square of half that side, is equal to the sum of the squares of the other two sides j" that is 2« 2 + ^ 2 =3/ 3 +%* 2P+ %y*=zx 9 + z* ( 225 ) Or By addition l|3/ fi -f.ih 8 +li^=2(^H^4-t 3 ) or i,H* a +* s — $(a s -M s -f c*)i Whence, subtracting each of the above equations frosa this last, we have 4y«=4(a»+ £S-f-c°~)-2Z> 8 or *-*/!( -f **+** + <*) Where, by substituting the given values of a } b, and c, viz. «zzl8, b—24, c=30, we have #=34-176, y=28 844, and s=20, which are the sides required. £x. 12. Let abc be the proposed triangle, of which the base ab is given =50*=2&. C E 1) Then, since the area is also given =790, the perpen- dicular = 790 25 zzp is also known. Make now aeiz half base =£, and ed=^ then ( 223 ) ad= Zr-fir, and db=&— x ; also Euc. (i. 47) Whence, calling the given difference —10=d, we have VI (£ + *)*+/*} -V{(£— x)*+t*\=d Squaring both sides, and transposing -f2/> l -d 2 ; Or, in order to simplify the operation, writing 2l>*+ 2p*— &=z2m, we have s/\(b+x)*+p*\x\/\(l>-x)*+p*\=z*+m. Again, squaring both sides, and performing the mul- tiplication on the two first factors, we have (b*—x*)* + 2p\b*+x*)=x* + 2x' i M+m*. Squaring and collecting the terras £ 4 — 2b*x> + 2p*b* f 2ffkfh 2x i m + /w 2 , or (2/>» - 2b* - 2m)x*~ m*— &*— 2p*b* Whence x=\/{-^- ■ — - — ) Where, by substituting any numeral values of p, b, and w, the answers for the sides will be found. But in the present question they do not agree with those given in the Introduction, a press error having been committed in printing the question. Ex. 13. Let abc be the proposed triangle, ab its base z=zlQ4=zb; ik the diameter of the circumscribing circle =200=d, drawn parallel to ab -, and dc the Dis- secting line =£fj6-— a. Then we shall have Ht = GK.~£ (ik— ab)=3, And consequently AH=GB= \^(lO X OK) = S/197 X 3 = V^l z=C. Let now cd be produced to meet the circle in e j Then, because cd bisect the angle acb, it will bisect the arc aeb, and therefore the perpendicular elm will pass through the centre l ; Consequently EM=100-f-v'519=e is also known, as is also mn=100- ^519~ f. Now let dezzx; then, since the two triangles enc and mub are similar, we have me : de :: ce : ne, or * . e : x :: a+x : d- } Whence x*+ ax—de, orar= |-V ( \-de), which thus becomes known ; and consequently the rectasgle ldXde, or ax {-7 r* V(— 4- and consequently v== — =/( , the sides required. n n The numerical values of which may be found by sub- stituting those- of a, b and d in the original equa- tions. Ex. 14. Let abc be the proposed triangle, AE=a, and DC=&, the two given lines. Also let x and y represent the sine and cosine of the a/)g!e bac respectively -, then by trigonometry we have I JL x zees bae, and V ==cw bcd. Also X \ b \\ */-~r- I - V—7T~ =AC, J iZ 1 M ( 229 l+y ) and y \ a\\ s/ h l+x : 2 l + x Whence V \+y Again by trigonometry, sina a 1+y y v 2 as - a/ or by' liana 1 + ta« c a , and 1 — to 2 a l+ta7i*a' Putting therefore tan bac=*, and substituting 2t - l — /* #= -, and y= :> 1+* 2 1-H* We have = -— , -v/2 b(\— t* or 0(1 + OU — * 9 ) = 2a*v% or ,s + ^ + ( i±W2 )<=li Which is a cubic equation, whence the value of/ may be determined ; viz. the tangent of the angle bacj and Hence also the angles bdc and be a become known, and consequently the sides ab=35'80737, kc=47 , 40728 and ac=59 , 41143, as required. Ex. 15. Let abc be the proposed triangle. c , Then the base ab=8, or ae=~ base zr4=Z', cd=4, =/>, and ac+uc = 12=s$ and make alsoED=# # x ( 230 ) Then AD—b + x, and db=£— x; and consequently 4/H* + *)* + P*}=ac V\(b-xf+t>*\=&C; Whence by the question Squaring both sides, and transposing -2p* h Or, in order to simplify, writing s*— 2b*— 2/> 2 =2ra Squaring again, and collecting the terms (b*-x <2 )*+2p*(bz + x*) +p*=m*~2mx* + x* Or fr - 2b 2 x* + 2p~b*+ 2p 2 x z ~m- - 2mx* 5 Therefore xH2p"—2b" + 2m)=m^ 2f> z b°~—b*—p* Or o:= A /(- m*—2p 2 b' 2 —l 4 p* ) Whence op, and consequently ad and db become known; and lience also ac= V(AD 2 -f cd 2 ), and cb=V(db 2 + cd*), are determined. Jn the present case, p—4, b=4, m=40 f whence fv/5j therefore ac =6 -f- 1 / 5, and bc=(5 — |a/5. r — 1 Ex. 1(3. Let abc be the proposed triangle. c Then ABnl5=2/;, or ae=£ base=7^ = £, and 45 H bb 6 =/>, the perpendicular; also the ratio Ac ! cb ;: 3 I 2, or m : n, and edzzot. ( 231 ) Then, as in the preceding example, V\(b— jr) 2 -fp 2 S=CB And by the question ; V\(b+x)*+p*\ : fy(s-*xy+p\ : m : n, or {t*+^*4-|f} : Uh-x)*+?l :: ?* 2 : »*, Whence 71W + 2n*l-x-\-n*x* + tfp' 2 zzm*b 9 — 2m*bx + m*** (to — t* 2 )* 2 — 2^(w 2 -|- « 2 ).r= (?2 2 — m Q )P+ (?j 2 — m 2 )/* 1 Or **-2*( 1— )*=_£«_.<>* Where, by substituting the proper values of I, p, m and rc, the numeral value of x may be determined, and hence those of ac and bc. Ex. 17. Let abc be the proposed triangle, and make the perpendicular cd = 24=/>, ce the line bisecting the angle ACBzr'i^ — /', and cf, the line bisecting the base, =40=r. C F ED Then (Euc. 1.47) ed= > v /(ce 2 -cd 2 ) = 7=?« > Also FD= ^/(FC 2 — CD 1 )=32=W| ( 232 ) And in order to simplify, let EF-zzq. Also, let half the base af=fe= 1 z , j then AE = Jr-f<7, EB=zX—q ; AD~X-\-72, DB = X—7l} Hence ac= v'{(ff+») 2 47> a $ BC= a/\{x— ?2)Hp 2 ! And from (Euc. vi. 3), we have ac ; bc :: ae : eb, or Whence \(x + ny+p*\x(n- q y=\(x-ny\+p*x(x+qf Which, by multiplying, cancelling, &c. becomes 72^(tf 2 -f f) =zqx{a*+ rf -f/> 2 ) Where *= £g|g= gg» or the base of the triangle $ which, by substituting tbepro- 250 per numeral values of y, 72 and p, gives —~\/\4 5 from which and the given lines the other two sides are readily obtained. Ex. 18. Let abg be the proposed right angled trian- gle, and o the centre of its inscribed circle 3 and let co — Ao=2=d, and ac=10=A. C ( 233 ) Produce co to d, and let fall upon it the perpendicular ad ; which put =x. Then, since co and ao, bisect the two angles c and a, and these two angles together are equal to a right angle, it follows that the two angles oac and oca = half a right angle. But the outward Z. of any triangle, being equal to the two inward opposite Z.s, Z.aob=Z-Oac + Z-Oca. Whence also AOD=half a right angle; and since d is a right angle, dao is also = half a right angle. Therefore DO = AD=.r, and ao3=v/2# 2 =#a/2; and consequently co=x^/2-\-d, and cd=# + x ^2 + */= (l+ K /2)x + d. Now AD a +DC*=AC 2 ; Or ff a +$0+ -N/2)^r + c/p=A 2 , or }l + (l + or (A 2 + 72 2 — w 2 ) 2 og= */\rf— ■ 4# Whence ag, gc and og, being known, we have also ab=ag-t-og=5*87447, and bc = gc+ og = 8'08004. Ex. 19. Let abc be the required circle, AB=a and dc=& the two given lines, kl the required diameter, and ef=c, the given distance. Also draw the two perpendiculars eh, eg ; and call fg=#, and hf or eg=^ 5 then a?+y*=zc* (Euc. 1. 47,) &+y)&-y) = (it>+x)&-x). (Euc. in. 35.) The latter of which gives ar—y^^z^b 2 — %ar And, as before, ^ a +i/ 2 =c 2 , ( 235 ) Then by addition x?3xt^b*±ria*~\- lc*, and by subtraction ?/ 2 = la 2 — l^-j-fc 2 . Now join ed ) .then .ed 2 =eg 2 |gd q , or ED= v /(fa 2 + j£ 2 + {c 2 )= radius, or KL= v /|a 1 ~f-|^ 2 + 2c 2 )= diameter. Ex. 20. This question, of which the figure is as fol- lows, does not require the assistance of algebra. 15 For bd = V(hb 2 +hd 2 )= \/(400-r- 14400)= ^14800 Whence ed and be are each zz\/14S00j Consequently ec= \/(ed 2 — dc°-) = V{ 14800— 6400) = ^8400=20/21 j and ae= a/(be2— ab 2 )zz a/( 14800— 10000)= ^4800=40^/3, as required. Ex. 21. Let acbd be the given trapezium, Where ad=6, db=4, cb=5, and ca=3. Then draw the diagonal ab, and Jet fall upon it the ( 236 ) two perpendiculars ce and df, and make CE=p and df=//; also put the required diameter =#. Then by (Euc. vi. c) />#=5X3:=:15 p'.* =4x6=24, or x(p+p') = 39 In the same manner calling q and q' two perpendicu- lars falling on the diagonal cd, we have ^^=3x6rrl8 q'x=4 x5 = 20, or x(q + q) =38 Whence (p+p') : (? + »/ 2. J 39 Y 38 Now the three sides of each of the triangles abc, and adb being known, the perpendiculars ce and df are readily determined. Let, therefore, these be denoted as above, by p and />', and we shall have from our first 30 equation, a?= - „ the diameter sought. P+P Ex. 22. This question, of which the figure is as fol- lows, is nothing more than having the three sides of a triangle given to find the radius of the circumscribing circle. Let therefore abc be the given triangle, of which ac=25=#, ab=30=£, cb=20=c, and o the re- quired centre. From which the perpendicular cd is readily found. For, by putting ad=o?, and DB=y, we have x 1 — y*z=a 2 -c* x-\-y = b -y~— z— Whence x — vzz By addition # ^ a + a 2 -c 2 CD= ^{a' 2b And hence (£ 2 -+**-c 2 ) 2 Whence Diam. = 4*' Again (Euclid vi, c.) Diam. Xcd=;acxcb, 25X20 __ 80 25 ■r^ 3023711^ and consequently oa, ob, or oc= 151 18558 the distance sought. Ex. 23. Let abc be the proposed semicircle, ecf the given equilateral triangle, whose area zrlOOzza, and whose perpendicular cd, which is the radius of the cir- cle > is required. A. E D f B Let ce, the side of the triangle sxj then Ki>=s|i. and cd = V(.r 2 - £**) zr \x *J3 . But DCXEDr=| l rx|^\/3~100, the area, Whence fa?V3 = 100, or *£= •— = 20 Vj a/3 Consequently \& ^3 = 103/3 = radius, and therefore 20 V3 the diameter. Ex. 24. Here the rectangle of the perpendicular and diameter of the circumscribing circle of any triangle, is -equal to the rectangle of the two sides from which the perpendicular is drawn. (Euc. vi. c.) GD Whence the present question reduces to this, i. e. given the perpendicular —/>, the radius of the inscribed circle og, oe or opzzr, and the product of the two sides = k/>, to find the sides. Let, therefore, the segment at>zzz-\-x, andDB=x — x. Then (Euc. i. 47.) ac=^/{(s + ^)H/>^ AB = 2% Also, because ab xcd=(ab + bc+ ac) Xog, we have v / S(^ + ^) 2 -r-/> a }X v /{(%-^) ,2 +/}=2R/ > ,and ( 239 ) Whence, squaring the latter equation, and substituting for the double rectangle, we have 4 (ft — rV% a 2x 2 + 2r 2 -f-2p 2 + 4Rp= vr ' . r Or a?= -~ — L#i_tfi— y— 2r/> 2(/?-r> , „ or by putting — — lzzm, and/> 2 -f 2r/)=?z Also, squaring the first equation, and multiplying, In which, substituting for a? the value found above, we have (* 9 — m**;f »)» + 2^(* a + m% , -ra)=4R-^ 9 — p 4 j That is, 3i 4 (l-w) a + 2«(l ~7?7)* 2 -{-2/> 2 (l + ™)s a +?r— 2p*/i =4r*/?' 2 ~/> 4 3 which reduces to 2??(1 — /tt)-f2/(l-f w) , __4r 2 /— p 4 — n* + 2pn * ("l— vif f (1 -m)> Where, by re-establishing the value of ?7, the second side of the equation becomes =0, and there remains ,_ 2k(pi-1)-2/(m+1 ) (772-1)* Or, by substituting for ?/* and n their respective va- lues, we have r\/(2R/> — 4r/— r*) />— 2r wu „ _27V(2R/-4Rr-r 3 ) Whence 2x= — p— 2r the base; from which the other sides may be determined. Ex. 25. Let abc be the proposed triangle, in which \B=2a, and cn=a. ( 240 ) Then assuming AD=a + #, and DB=a~ ( r, or —(x-~a) we have AC= \/(2a* + 2ax + x'>) BC=V(2a 2 -2^+^) And by the question *" I i (2

) 3 3 And this, by extracting the roots, gives AC=a(2-r-| \/6) Bc=a(2— 4V6) Which are the two sides required. T. Bensley and Son, -*£2222 Bolt Couxt, Fleet Street, London, >^