S.'-M i f/ / - ' V OF THE UNIVERSITY EDUCATION LIBHj, K. ^ ^v ^ h Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/algebratreatiseOOgreerich £^Sf- >^M^ PEACTICAL TREATISE ON ALaEBRA, DESIGNED FOR THE USE OF STUDENTS IN HIGH SCHOOLS AND ACADEMIES. BY BENJAMIN GREENLEAF, A.M., Arxnou OF toe " national arithmetic," ktc. Si'irt^ Kmprobeti Stereotype 23oitiou. BOSTON: PUBLISHED BY ROBERT S. DAVIS k CO, NEW YORK : Geo. F. Cooledge & Brother, and D. Bcrgess & Co. PHILADELPHIA: Lippincott, Grambo & Co. ST. LOUIS : John Hals all. 1854. Entered according to Act of Congress, in the year 1852, by BENJAMIN GREENLEAF, In the Clerk's OfEce of the District Court cf the District of Massachusetts. Entered according to Act of Congress, in the year 1853, by BENJAMIN GREENLEAE, In the Clerk's Office of the District Court of the District of Massachusetts. GREENLEAF'S SERIES OF MATHEMATICS. 1. MENTAL ARITHMETIC, upon the Inductive Plan; designed for Primary and Intermediate Schools. Revised and enlarged edition. 144 pp. 2. INTRODUCTION TO THE NATIONAL ARITHMETIC ; Ok, COMMON SCHOOL ARITHMETIC. Improved stereotype edition. 324 pp. 3. THE NATIONAL ARITHMETIC, designed for advanced scholars in Common Schools and Academies. Improved stereotype edition. 360 pp. COMPLETE KEYS TO THE INTRODUCTION AND NATIONAL ARITHMETIC, containing Solutions and Explanations, for Teachers only. 2 vols. 4. PRACTICAL TREATISE ON ALGEBRA, for High Schools and Academies, and for advanced Students in Common Schools. New edition, revised and stereotyped. KEY TO THE PRACTICAL ALGEBRA, containmg the Answers, and full Solutions and Explanations, for Teachers only. Stereotype edition. 278 pp. EDUCATION LIBBf stereotyped by nOBART & IlOBBIJSrS, nrrw England type and stereotype fodndert, R O ST o N. PREFACE. The following Treatise is designed to present a system of theoretical and practical Algebra. It is intended to be both elementary and comprehensive, and adapted to the wants of beginners, as well as those who are advanced in the study. In the course of his labors the author has consulted the most approved European treatises on the subject, and availed himself of whatever he thought might add to the interest and usefulness of his work. It has been the aim of the author, throughout his investiga- tions, to give to it a practical character, that those who study it may know how to apply their knowledge to useful purposes. The demonstrations connected with the several Roots, will greatly aid those who wish for a complete and thorough knowl- edge of Evolution in Arithmetic. The method of solving Cubic Equations by completing the square, the author believes, will be very useful. This method will not apply to all problems ; but, wherever it will apply, it not only very much abridges the labor, but the result is perfect accuracy, which is not always the fact by the common method of approximation. The Table of Logarithms at the end of the work, will be often found convenient and useful. The examples, of which a large number have been placed under each Kule, are intended to be neither too numerous nor too difficult; and all who may use the work, either by themselves or in connection with a class, are advised to solve all the prob- lems, in the order in which they are given. No labor on the part of the pupil will be productive of more intellectual and practical benefit. The answers to several questions have been designedly omitted, that the pupil may try his skill as upon an original problem. IV PREFACE. One who has a thorough knowledge of Arithmetic, will find the study of Algebra a most pleasing, and, generally, not a difficult task. As a mental exercise, it is admirable for its efiect upon the logical powers of the mind, assisting one to think and reason closely and conclusively. As Mr. Locke has remarked, in his Essay on the Human Understanding, "Nothing teaches a man to reason so well as Mathematics, which should be taught to all those who have time and opportunity, not so much to make them mathematicians, as to make them reasonable crea- tures " BENJAMIN GREENLEAP. Bradford, January 23, 1852. ADVERTISEMENT TO THE STEREOTYPE EDITION. In revising this work for a second edition, the author has made such changes and additions as he believed would better adapt it to its purpose. Every part of it has been carefully and critically examined, and many portions have been entirely re-written. In a few cases, where improve- ment in that respect seemed desirable, the arrangement of articles has been somewhat altered. The new articles which have been inserted, will, it is hoped, add ma- terially to the interest, as well as to the value, of the treatise. The theory of Equations has been more fully developed, and illustrated by a variety of carefully prepared examples. A brief space has been given to Indeter- minate Analysis, a subject which, though usually omitted in elementary works on Algebra, the author believes to be one of no small practical importance. It gives the student the command of a class of problems which cannot possibly be solved by the rules of Arithmetic, nor by the more familiar principles of Algebra. In the revision of the work, the author has availed himself of the suggestions of several teachers who have used it as a text-book since its first publication ; and he would take this opportunity to express his gratitude for their kindness. April 26, 1853. CONTENTS. SECTION I. PAGE Definitions and Notations 9 Explanation of the Signs -i*;- . 10 — 15 Practical Examples 15 SECTION II. Addition 16 SECTION III. Subtraction 21 Simple Quantities 22 Compound Quantities 28 SECTION IV. Multiplication 28 Multiplication by detached Coefficients 35 SECTION V. Division 37 Division by detached Coefficients , 43 Questions to exercise the foregoing Rules 45 SECTION VI. Fractions 46 Definitions of different Fractions 46, 47 To find the greatest Common Measure of the terms of a Fraction . 48 To reduce Fractions to their lowest terms 50 To reduce a Mixed Quantity to the form of a Fraction 52 To represent a Fraction in the form of a Whole or Mixed Quantity 53 To reduce a Complex Fraction to a Simple one 54 To reduce Fractions to a common Denominator 56 Addition of Fractions 59 To add Fractional Quantities 59 Subtraction of Fractions 61 1# VI CONTENTS. PAGB To subtract one Fraction from another 61 Multiplication of Fractions 63 To multiply Fractions together 63 Division of Fractions 65 To divide one Fraction by another 65 Negative Exponents 66 To free Fractions from Negative Exponents 67 SECTION vn. Equations 68 SECTION VIII. Problems 78 SECTION IX. Equations or the Fikst Degree, containing two Unknown Quantities 89 Elimination by Addition and Subtraction 89 Elimination by Comparison 91 Elimination by Substitution 92 SECTION X. ELI>nNATION WHERE THERE ARE THREE OR MORE UNKNOWN TeRMS INVOLVED IN AN EQUAL NUMBER OF EQUATIONS 95 Equations or the First Degree, containing several Unknown Quantities 98 SECTION XI. Negative Quantities 108 Problem of " The Couriers " • . 106 Theory of Indetermination Ill SECTION XII. Theorems in Negative Quantities . . . *. 113 SECTION XIII. Involution 121 SECTION XIV. Evolution, or the Extraction of Koots 125 Evolution of Polynomials 126 Evolution by Detached Coefficients 130 Extraction of the Square Root of Numbers 131 Cube PvOOt 132 Cube Root by Detached Coefficients 134 CONTENTS. VII SECTION XV. PAGE Surds, or Kadical Quantities 138 To EXTRACT THE SqUARE RoOT OF A BiNOMIAL SuRD 160 SECTION XVI. IlMAGINARY QUANTITIES 162 SECTION XVII. Quadratic Equations, or Equations op the Second Degree . . 164 The Theory of the Lights and Attraction 168 Application of the above Principles 169 Affected Quadratic Equations 172 Problems in Quadratic Equations 186 Examples of one or more Unknown Terms in Quadratic Equations 192 SECTION XVIII. Cubic and Higher Equations 198 SECTION XIX. Ratios 202 Comparison of Ratios 203 Compound Ratios 204 Proportion 205 Approximation of Ratios 207 Problems for Proportion 216 SECTION XX. Arithjietical Progression 218 Table of twenty different Cases in Arithmetical Progression . . . 231 SECTION XXI. Geometrical Progression, or Progression by Quotient . . . 282 Table of twenty different Cases in Geometrical Progression . . . 240 SECTION XXII. Harmonical Progression 241 SECTION xxni. Infinite Series 242 SECTION XXIV. SiBiPLE Interest 244 SECTION XXV. Discount at SniPLE Interest 248 VIII CONTENTS. SECTION XXVI. PAGE Partnership, or Company Business 250 Partnership on Time, or Double Fellowship . 253 SECTION xxvn. Indeterminate Analysis 257 SECTION XXVIII. Variations, Permutations, and Combinations 266 SECTION XXIX. Logarithms 270 Multiplication of Logarithms 281 Division by Logarithms 282 Involution by Logarithms 286 Evolution by Logarithms 286 SECTION XXX. Compound Interest . • 289 Discount and Present Value at Compound Interest 294 SECTION XXXI. Deposits 295 SECTION XXXII. Exponential or Transcendental Equations 299 SECTION xxxin. Annuities 803 SECTION XXXIV. Involution op Binomials 807 SECTION XXXV. Binomial Theorem 311 Indeterminate Coefficients 814 SECTION XXXVI. Summation and Interpolation op Series 817 SECTION XXXVII. Cubic Equations, containing only the Third and Second Powers 824 Cubic Equations, containing only the Third and First Powers 329 Problems in Cubic Equations, &c 332 Miscellaneous Questions 334 Table containing the Logarithms of Numbers from 1 to 10,000 345 ALGEBRA SECTION I. DEFINITIONS AND NOTATIONS. Article 1 • Algebra is the art of computing by symbols. 2. In Arithmetic we represent quantities and perform cal- culations by figures, each of which has a known and definite value. 3. In Algebra we employ the letters of the alphabet, and other characters, the value of which is either known or un- known, according to the conditions of the problems. 4. Those quantities whose values are given are called known quantities ; and those whose values are not given are uTiknown quantities. 5. The symbols used to denote known quantities are, gener- ally, the first letters of the alphabet in the small or Italic character, as «, 3, c, d^ &c. ; and those used to denote unknown quantities, the last letters, as w, x^ y, z. 6. In addition to the above, which are the more common symbols, capital letters may be used, as A^ B, C, D, &c., or letters of the Greek alphabet, as «, ^, y, d, s, 'c, &c. In ex- tensive operations, the use of these, or some other suitable characters, is sometimes very convenient. 7. Sometimes quantities are expressed in Algebra, as in Arithmetic, by figures instead of letters. 8. When a quantity is doubled or trebled, or multiplied any 10 ALGEBRA. number of times, the number of multiplications is usually ex- pressed by a numerical figure or figures. Thus, let a denote a certain quantity, and la will denote twice the same quantity, 3a three times the same quantity, &c. 9i The figures or letters prefixed to any symbol, and denoting the number of times the quantity represented by the symbol is taken, are called the coefficieiit. Thus, in 4a, lb, and 4:ax, the coefficients are 4, 7, and 4(2. 10, A quantity which has no figure prefixed to it is considered as having a unit for its coefficient. Thus, a is the same as \a. 1 1 , Quantities represented by the same symbol or letters, and of the same power, are called like quantities ; and those repre- sented by difierent symbols or letters, or by the same letter of different powers, unlike quantities. Thus, Za, 4a, and ba, are like quantities ; and Sa, 43, and 5c, unlike quantities. In like manner, Sab, Aab, and Qab, are like quantities ; and dab, 4ac, bdc, and Qmn, are unlike quantities. 12, Besides the symbols and figures used to denote quantity, there are certain signs, which are used to express the difi"erent relations between quantities, and the operations to which these quantities are subjected. These signs are the same as are often employed in Arithmetic, but, in Algebra, they are indis- pensable. lej, The sign = is that of equality, and denotes that the two quantities between which it is placed are equal to each other. Thus, a=2b signifies that a is equal to 23. 14t The sign -}- is called plus, and signifies addition. Thus, a-{-b signifies that a is to be added to b. 15. The sign — is called minus, and signifies subtraction. Thus a — b signifies that b is to be subtracted from a. 16. Sometimes both the signs -|- and — occur before the same quantity, as a:^x, in which case they signify that the quantity may be either added or subtracted, or that it is doubt- ful which operation is to be performed. 17» The sign X signifies multiplication. Thus, «X^ denotes DEFINITIONS AND NOTATIONS. 11 that a is to be multiplied by h ; and ay^by^cy^d, that the quan- tities a, b, c, and d, are to be multiplied together. This sign is read into. Thus, ay^b is to be read, a into b. Sometimes a single point is substituted for X* Thus a.b signifies that a is multiplied by b. 18, The sign -r- signifies division. Thus, a-r-b signifies that a is to be divided by b. 19, Division is also represented by placing the divisor under the dividend, in the form of a fraction. Thus, - signifies that a is to be divided by b ; and -, that a — b is to be divided by a-\-b a^b. 20, The sign ]>, standing between two quantities, denotes that the one before it is greater than the one after it. Thus, a'^b signifies that the quantity a is greater than the quantity b. 21, On the other hand, the sign <[ denotes that the quantity before it is less than the one after it. Thus, bc U_40 [ 18— 10 _40 I4_ • 4^»-10^"^ V5a-12(^/ ~'10"^aJ30-12~~I0"^aJ9~ 4|. 11. 2a'+3^c-5=127. 16 ALGEBRA. 12. ba''b—10ab'-\-27e=z—Q00. 13. 7a''-{-{b-c)X{d—e)=2bS, ^ . alP' , a — h , ,-„ „ 14. — Xd ; — \-27a^e=^Q^. c a 15. S^c-\-2a^{2a-^b-d)=M. 16. a/s/{a'-\-e)-\-Sbc{a'—b'')=mQ. 17. Sa^b-j-W {c''—A/2ac-{-c')—oe=D42. SECTION II. ADDITION. Art. 5^» Addition in Algebra is the connecting together of several quantities by their appropriate signs. 57. The operation consists in collecting into one term all the like quantities, and so arranging the several terms, thus obtained, as by signs to indicate the proper sum of all the quantities, both like and unlike. 58. Addition in Algebra embraces three cases. I. When the quantities are alike, and their signs alike also ; as, a, Sa ; or, — b, — 43. II. When the quantities are alike, and their signs unlike ; as, 33, -53. III. When the quantities are unlike, some having like and others unlike signs ; as, Sa, 43, — ix. Case I. 59. When the quantities are alike, and their signs alike. Rule. Add together the coefficients belonging to the like quantities, and place their sum before the common letter ADDITION. 17 or letters^ with the common sign prefixed ; and the result will be the sum required. Thus, let it be required to add together 3a3, Idby 8^3, the operation will be as follows : — Zab 'Jab Sab ISab. The reason of this rule is obvious ; for, since ah, whatever be its value, must represent the same quantity in every instance, it is evident that 3 times, 7 times, and 8 times the same quan- tity, will make 18 times the same. In like manner, let it be required to add together — 7b, — 53, and — 63. -73 -53 —63 —183. EXAMPLES. (1) (2) (3) (4) (5) 3a 7A — Sax 4:xy' Za- If 4a bk —4:ax xf 4a- Sif Qa Sh — ax Zxf Qa— y" a Sh — 2ax Ixf a — 6?/^ 5a h 24A - —lax 2xf Vlxf 5a— 2y^ 19a -VJax 19a-142/2 (6) (7) (8) (9) (10) 7x 14a3c ^y — \mn bh-\- x 4^ lla3c y — 3??m h-\-2x 11a; babe y — mn 2h-i-4.x 9x 4a3c oy — Yhnn h-i- X X abc 4y — mn Ih-^-Qx 18 ALGEBllA. 11. Add 7a, 11<2, <2, 4a, Qa, and 3a together. Ans. 32a. 12. Add 4Ji, Qk, k, k, llh, and 7k together. Ans. dOk. 13. Add together {Sa'—b), (7a'— 43), and (a'—b). A?is. lla^— 63. 14. \Yhat is the sum of 3/v/a-, 4/\/a-, a/o^, 7a/ a*^, and 2a/ a'? ^^w. 17/v/«'- 15. Add together 3/s/ci-{-b, Qa/o^S, hj a-\-b^ and 12/\/a+3. ^?Z5. '22/s/a-{-b. Case II. 60. When the quantities are alike, and have unlike signs. Rule. Add all the affirmative coefficients into one sum, and those that are negative into another ; then subtract the less of these results from the greater, and prefix the sign of the greater to the difference, annexing the common letter or letters. Required the sum of -j-7a:r, — Aax, — 3aa;, -\-Ylax, — ax, and -\-ax. 7ax — 4:ax — Sax 17 ax — ax ax VJax. We find the sum of the plus quantities to be 25a:r, and the sum of the negative quantities — ^ax ; and the difference be- tween those coefficients is 17, which we prefix to ax; thus, 17a:;::. The reason of this Rule is obvious, when we consider that two equal quantities, the one with a positive and the other with a negative sign, exactly cancel each other, so that their sum is nothing. Of course, then, when two like quantities of opposite signs are not equal, the difference between them must be the proper sum, which will be positive or negative according to the affirmative or negative character of the larger quantity. ADDITION. 1 EXAMPLES, (1) (2) (3) (4) (5) 7a — G»^ Aax ISn 7 a— mp^ — Sa 771 — Sax n a-{- Gmp'^ a -11m ax — 20?z — 11a — S/Tzp^ — 5a bm —7 ax 6?i 8a+ll777/ 11a — m ax 8 71 — da — 7777j9'^ a 20m 12ax — 71 18a — lb77ip' 12a S?7l Sax 771 14a — 977ip' (6) (7) (8) (9) (10) 6y 477171 —Sxy - Apn Sxy — 77iy - 72/ 7n7l 7xy pn 3z7/-j- ?7iy 4:y S77271 —4:xy p7i — 11a; 7/ — 18777^^ -112/ 1877271 — ^y -llp7i — 4iXy-\- 977vp 9y 777171 9xy 7pn — 8Z7/ — 3777^^ — ^y Sm7i -Sxy p7l 12a:7/-}-12777> 19 11. Add 4+a^a;, Q—a^x, 3+6a':i', 15— 5a^:7;, 3+a^:c, and e+Ta'^a: together. Ans. 37+9a'a:. 12. Add 14aa;— 6?/, 7ax-]-y, bax — 7y, 9ax — lly, and Sax-\Sy together. A7is. ASax — 20y. 13. Add 3a— 43+6c, 7a+113— 3c, 8a4-^— 7c, and a— 113 +15c together. Atis. 19a— 33-j-llc. 14. Add 16^;'^— 5?/— 16, Sx^+^y^-b, x^'-^Sf-ST, x'—y^ +7, Qx''-{-77/—ll, and 2x''—Sy'—21 together. Atis. 29a:--f5?/^— 83. 15. Add 5a— 3, 33-|-8c, 4a— be, ba—bb—c, 7a— Qc, and lla+43— 7c together. Ans. 32a4-3— 16c. Case HI. 61. "When the quantities are unlike, some having like and others unlike signs. It is evident that unlike quantities cannot be united into one ; or otherwise added than by means of their signs. J 20 ALGEBEA. Thus, for example, if a be supposed to represent 20, and h 12, then the sum of a and h can be neither twice 20 nor twice 12, but it must be 20+12=32, that is, a-^h. 62i Hence, to add unlike quantities, we have the following KuLE. Collect all the like quantities together^ as in Case II., and write down those that are unlike, one after another, with their proper signs. 63i When several quantities are to be added together, it is immaterial in what order they are written. Thus, a-{-b — c, a — c-\-b, — c-\-a-\-b, are equivalent expres- sions. (1) ^ ^JlXAMPLES. (2) Zax 7 a-\-7m A^mn -L 6 a — 7x -W Axy — bm Ixy nn — Qy'^-{-7xy. ^x +82:?/ 2>ax-\-A7s lU-\-12xy+2m—Ax, (3) (4) / (5) 9^V 14:ax — 2x'^ 4fl^2:— 130+3a:^ —"Ix'^y ^ax"^-}- Sxy 5^2 +3aa:+9ar2 Saxy Sy^ — Aax 7xy—Ax\^^' -4.xf dx'' +26 ^x—AO —^x" 6. Add together a-\-x and y — c. Ans. a — c-\-x-\-y. 7. Add 3a+3— 10, c—d—a, _4c+2£z— 35— 7, and Ax'-^-b —18771 together. An^. 4^^—23— 12— 3c— ^+4:i;'^—18??i. 8. Add 7a— by^, S/s/x-\-2a, by^ — a/x, and — 9a-{-7/s/x to- gether. Ans. 14:a/x. 9. Add 4m7i-\-Sab — 4c, Sx — 4:ab-\-2mn, and o?n' — 4p to- gether. Ans. 6 mn — ab — 4c+3a:+3m'^ — 4:p. 10. Find the sum of Sa^-\-2ab-\-4:b\ ba''—Sab-\-b'', —a''-\-bab —b-, lSa^—20ab—ldP, and Ua''—Sab-{-20b-. Ans. 39a''—2'iab+bb\ a -^ / SUBTRACTION. 21 11. Find the sum of \.x^—^c^—hax^-\-^d^x, ^d^-\-%x^-\-^^cix^ ^1a^x-Vlx^-\-V^a£--\^d}x, \Zax'—'nd}x-\-\U\ %d}x—1^a' -\-Vlx\ and ^\(i~x—1x^—Z\ax~—lx\ Ans.—7x^—a\ 64. Coefficients, whether figures or letters, that are common to several terms, may be connected with them by a paren- thesis. (12) (13) - Add amx-{-2dy hy-\-4cmx ^cx — ^dy my-\- Mx Zax-\-by 'hiy — ^mx {am-\-'lc-{-^a)x-\-{b—d)y. {Ji-\-'m-\-^n)y-\-{M—bm)x. 14. Add 4:ax — b7ny, odx-\-77iy, and 7mx-\-4?ny, together. A71S. {4ia-{-dd-\-77n)x-{-{7n — 7n)y. 15. Add ohz — t)x, A.mz-\-7ix, and bai — ^px, together. Atis. {^h-\-^m-\-ha)z-\-{7i—b—4:p)x. to -/5- - r SECTION III. a o ~ ^ ' ^ * SUBTRACTION. Art. 65 Subtraction is the taking of one quantity from another, or the method of finding the difierence between any two quantities or sets of quantities of the same kind. 66. If it be required to subtract 10 — 7 from 12, we might first subtract 7 from 10=3, and take the 3 from 12=9 ; or we might take the 10 from 12, and the remainder 2 must neces- sarily be increased by 7 to produce the correct result. If from a we wish to subtract c — d, we first subtract c, and it gives a — c. This quantity, since we have taken d too much from a, is too small by d. Therefore d must be added, thus, a — c-}-d. 22 ALGEBRA. 67. If a simple quantity is to be taken from another simple quantity, it is only necessary to write them one after the other ; thus, if 8 is to be taken from 15, it may be expressed thus, 15_8=7. If it were required to subtract h from c, it should be written thus, a — h ; but if we were to subtract a — h from cA^d^ it is evident that if only a were to be taken, it would be written thus, c-\-d — a. But this evidently gives a result too small ; for a was to be lessened by h before the subtraction. Therefore, as the remainder is too small by 5, we must add l to the remainder, which will give c-f-^ — a-^l ; for it makes no difference in the result whether the minuend be increased or the subtrahend lessened. Subtract 7—4 from 13. Taking 7 from 13 leaves 6 ; but 6 is too small, for the 7 should have been lessened by 4, and we must either subtract the 4 from the 7 before the operation, or add it to the remainder; and, if added to the remainder, the expression will be thus, 6-}-4=10. 68. We therefore see the propriety of the following Rule. Change the signs of all the quantities to be subtractedj and proceed as in Addition. SIMPLE QUANTITIES. (1) (2) (3) (4) (5) (6) From +7a -16^ +17^ —2% -j-153 —6c Take +2a — bx +8^ -18^ + ^^ — c 4-5a —11a; + 9d —llg + 85 —5c The above questions are performed as in Arithmetic, the minuend being the larger number, and having the same sign as the subtrahend. (7) (8) (9) (10) (11) (12) (13) From — Sa -\- 7x +18?/ —35 — 7c +8(7 —(Jh Take —15a +14a; -\-20y —7b —15c +11(7 —Sh + 7a — 7x — 2y +45 + 8c - 3(7 -\-2h J OrJ i SUBTRACTION. * 23 In these examples the minuend is taken from the subtrahend, and all the signs in the subtrahend are changed. (14) (15) (16) (17) (18) (19) (20) From +27« — 63 —7c +% ~r'^'^^^ — bx — 7y Take —13a +185 — c —9g -Ibk +17a; —Iby +40a —243 —Qc 17g +2Qh —22^ + 87/ In these examples we change, mentally, all the signs in the subtrahend, and then proceed as in addition. These questions may all be proved, as in Arithmetic, by adding the remainder to the subtrahend. COMPOUND QUANTITIES. 69. The same rule must be observed in subtracting compound quantities as in simple quantities ; that is, all the signs of the quantities to be subtracted must be changed, the signs + to — , and the signs — to -}- ; we then proceed as in addition. (1) OPERATION. From ab-{- cd — ax — 7 ab-\- cd — ax — 7 Take 4:ab—Scd-{-4:ax—lb — 4:ab-\-ocd—4:ax-\-lb —Sab-{-4:cd—bax-\- 8 70. It is a better way for the pupil to conceive the signs in the subtrahend changed, but to let them remain without alter- ation, otherwise it might be difficult to correct errors that might arise in the operation. (2) (3) From 7x+5?/— 3a— 6A 1abc—llx-\-by— 48 Take x—ly-]-ba-\-llh llabc-\- 3a;-}-7?/-j-100 Qx^l2y—^a—VJk _4a3c— 142-— 2?/— 148 (4) (5) From 14^—42+ 9?/ -{-x ^x—babc—Qh—b\ Take —U~lz-\-41y—Vlx 19a;— 7a3c— 8A-|- 1 17y^_j_32_32?/-j-18z _10:r-j-2a3c+2/^— 52 24 • ALGEBRA. (6) (7) From "dxy—a^ — SA^— y' 7x''—a'b'-{- ly'-^-W Take —xij—a' ^-7/^'--10^/^ x'-^a?b'—\ly'— h' (8) (9) From b^ax^~-2>f -\-la^—l Sor^-f y'-\- V7A+ 5 Take 3V«^'+ y" — 5dz^+7 4:cL3y5_ ^7^_ 6 2V«^'— 4r+12aL8 4a;^+4y^+2V 7^+11 10. From 3«— 55+6/^— ^ take a-]-b—ld. Am. 2a—Qb-\-m-]-M. 11. From Zlx'—Zy''-^ab take IT^'^+S?/^— 4a3-]-7. J[?z5. 14jc^— 82/^4-5a^— 7. 12. From 5/+14i— 9^^ take _3/+73— 15^. Am. 8/-|-73-|-6^. 13. From lla—lb^c take ^+7^— 3c+ll. ^?z^. 10a— 143+4c— 11. 14. From 77i^-\-3n^ take — 47?^- — 67^^-["71''^• ^715. 5^2+97^3— 71.r. 15. From 31a— 15a;— 7 take 2a—2bx^if. Am. 29a-\-l0x—f—7. 16. From abc^ — xy^ take — Qabc^-^Sxy^ — 7k. Am. labc^—^xf-\-7h. 17. From llcA'— 5 take 5c/i'— 5+47^:. Ans. 6cA- — 47a;. 18. From m7i?-\-j£t take — lm7i?-\-4:^x—y''\ Am. 8m?i'+Z;?— 48a;+2/\ 19. From 47a^/z— 37+962^^ take 7abh. Am. 40a3A— 374-96?/". 20. Take 7^;^+ ^771 from 8a;y+17. Ans. xry^ — hm-\-\l. 21. Take 5^^— 3c+597?i from 1131 Am. 6^2.^ 3c— 59771. 22. Take 6a— 33— 5c from 6a+33— 5c+l. Am. 63+1. SUBTRACTION. 25 23. Take 4:lx''-{-7f+abc from m\ Ans. 7)-^ — 41a;' — l^f — ahc. 24. Take x^ from —Vlx^-\-\^y—a^h. Am. --\W^\^y—a-\-h. 25. Take a—h from a-^h. Ans. -f 1h. 26. From %xz take a:z— 7/i— 57?^^-f-7. Ans. 8:1:2+7^4- 5??z3— 7. 27. From Il/i7?i-|-8?z2 take x^—f. Ans. Wlim-\-^7i} — :>?-\-if' 28. From a-\-h take a — 3, and gj — 3, and — a4-^- Ans. +23. 29. From gj — h — c take — a-\-h-\-c and fz — l-^-c. Ans. a — h — 3c. 71. When similar quantities that are to be subtracted have literal coefficients, the operation may be performed by placing the coefficients with their proper signs within a parenthesis, and then subjoining the common quantity ; thus. From ay—h From ax^-\-gy'^ Take dy — c Take bx^ — hy"' {a-d)y+c-h {a-b)x'-}-{g+h)f. 72, If a set of quantities enclosed in a parenthesis is com- bined with others by means of the sign + , the parenthesis can have no effect upon the result, and may, therefore, be retained or not, at pleasure. Thus, a-{-{b-\-c) is evidently equivalent to a-\-b-\-c ; for it can make no difference whether b and c be first added together, and their sum then be added to a, or the sum of the three quan- tities, a, b, c, be taken at once. Again, x — y-\-{b — z) will amount to the same thing as x — y -\-b — z; for it is immaterial whether b — z he added to x — y at once, or b be added to it first, and from the result z be subtracted. The subtraction of a polynomial may be indicated without performing the operation, by inclosing the quantity to be sub- tracted in a parenthesis, and prefixing the sign — . 26 ALGEBRA. If we wisli to subtract 7a—^x-\-Qy from 11a — 2x-{-Si/, it may be indicated thus (ll; "^ ^ ^^ -\-14:axy — Sy'^~\-1277iy 2Sa^xy—lQay'^-\-14:axy—'8y'^-\-24:amy-{-12my. (4) (5) (6) 2x''-{-y Sa-\-A7n Sa—2b x~-\-y 2a — 27n 2a — bb 2x'-]-7?y 6«2+8a7?i 6^2—4^3 \.2x'y-{-y'' _6a?7i— 8?;i^ —15^3+1031 2x'-\-^x'y-\-7f. 6a^-{-2am-Sm\ Qa''—ldab-\-10b\ 34 ALGEBRA. (7) 20"— Za b—4 b'' —12a'b-\-lba%''-\-24:d'b'-^ab' —lQa'b"-\-2^a%^-\-Z2ab'—U^ U'—22a'b—^la?h'-\-'i'^(rb'-{-2Qab'—U\ 85. When positive and negative terms balance each other in the product, they should be cancelled. (8) (9) a— X 1 -{-X a^-{-a^x-{-ax'^ l—x-]-x'^—x^ —a^x—ax^—x^ -\-x—x^-\-x^—x'^ a? —x^. 1 —x\ 86, The continued product of factors is often expressed in one line. 10. (1+2-) {l+x') (l—x-{-x''—x'')=l—x\ 11. {a-\-2x) {a—3x) {a-\-4:x)=a'-\-Sa'x—10ax''—24:x\ 12. Required the continued product of oa—x, 2a-^4x, and 4:a-2x. Am. 24:a'-\-2Sa:x—^Qax''+^xK 13. Multiply 2>x^—2xy—y' by 2x—^. Ans. 6z^ — lQx'^y-\-'Qxif-\-^y^. 14. Multiply x^-\-2x-\-l by x'—2x-\-^. Ans. 2:^-1-4.^+3. 15. Multiply a-\-b — c by a — b-\-c. Ans. a'—b''-\-2bc—c\ 16. Multiply ^a—2b by —2a-\-^b. Ans. — 6a*^+16a3— 851 \ 17. Multiply ba'-'^ab-\-W by 6^-53. A71S. 30a^—43a'5+39«^.=— 20^*1 18. Multiply a'-\-ab-\-b'' by a-b. Ans. a?—b\ MULTIPLICATION. 35 19. Multiply o^—x" by o>—x\ Ans. a^—2a'x'-\-x\ 20. Multiply 'Ix^'—Sxy-^-Q by 3x^+3a:?/— 5. Ans. 6x'—dx'y-\-Sx-—9xy-i'^^xy—^0. ■ 21. Multiply ba'—4:ax-\-Sx'' by 2a'—Sax—4:X-. Ans. 10a'-2Sa'x-2orx''+7ax'—12x\ ^ 22. Multiply 2a'—Sax^4:x" by ba'—Qax—2x\ Am. 10a'-27a'x-\-S4:a'x'—l^ax'—Sx\ 23. Multiply a'-Sa'+Sa-l by a'^_2«-|-l. ^"■"" ^7W. a'-ba'+lOa'-lOa'+ba-l. 24. Multiply «'"—«'' by 2a—a'\ Am. 2a"^^—2a"+'—a'"+''-\-a^\ 25. Multiply a^—a^x-\-a^x^—ax^~^x^ by o;-}-^- MULTIPLICATION BY DETACHED COEFFICIENTS. 8T. The coefficients of the polynomials should be arranged according to the successive powers of the letters, increasing or decreasing by a common diflFerence ; and, when this common difference is wanting, its place should be supplied by zero. The following examples will illustrate the above : 1. Multiply a'-{-2a+l by a''—2a-{-l. 1+2+1 1-2+1 1+2+1 _2-4-2 +1+2+1 1+0-2+0+1. In adding the coefficients of the partial products, we perceive that the second and fourth places are a zero : but the letters must be written with their powers regularly ascending from left to right ; and, where zero is the coefficient, the value of the quantity is nothing. Thus, a'-i-0a^—2ar~{-0a-{-l=a'—2a'-{-l, because zero is the coefficient of the second and fourth terras. 36 ALGEBRA, 2. Multiply x'—x" by x^-\-x. 1+0-1 1+0+1 1+0-1 _|_l+0-l 1+0+0+0-1. With the letters and their powers added, it will be The second, third, and fourth terms are of no value. 8. Multiply 3^3- 4a3H6^'by 1d?—U\ 3+0-4+6 2+0-4 6+0— 8 + 12 -12- 0+16-24 6+0-20+12+16-24. We now annex the letters with their proper powers, decreas- ing by a constant common difference, thus : 6a^— 20«^^^+12a^3^+16a3^— 243'. 4. Multiply 2a?-^ab''-\-bb^ by 2a^-U\ 2+0— 3+ 5 2+0- 5 4+0— 6+10 _10_ + 15-25 44_0— 16+10+15— 25. Affixing the letters with their powers, we have, ^a^-\-^a'b—l^a'b''-\-10a'b^-{-lhab'—'lbb'= 5. Multiply 5a'— 3a^+^ by 2a'-{-a\ Ans. 10fi^9+5a'— 6a'— a'+a\ 6. Multiply 3.t3-2.7:-2 by 2:^-3. Ans. 3:r'-lla:^— 22'-+6a:+6. DIVISION. 37 7. Multiply ?/^-f-y— 3 by if—y. Arts. y^J^y^-^f-o/J^^. 8. Multiply x''-{-x''-{-x^-\-z'-\-x-\-l by x—l. Ans. x^ — 1. 9. Multiply a"^— 2a34-452 by d}-^1db-\-W, Ans. a'^^a%''-\r\U\ 10. Multiply 3a'+3a'3+3^23'^+3a^^+33^ by 7a— 7^. A.m. 21a'— 2W. 11. Multiply x^-}-x-y-\-xif-^7^ by x — y. Ans. x^ — y^. SECTION V. DIVISION. Art. 88: Division is the converse of Multiplication, and is performed like that of numbers. Its object is to find how many times one quantity is contained in another; or to find what quantity, multiplied by a given quantity, will produce another given quantity. The product of like signs, as in the rule of Multiplication, produces +, and unlike signs — . Case I. 89t When the divisor and dividend are both simple quan- tities. If dhc be divided by a, the quotient will be he ; because a multiplied by he will produce ahc. If ^ahc be divided by 2a^ the quotient will be 2hc ; because 2a multiplied by 2hc will produce 4(z3c. If 9^:r be divided by 3a;, the quotient is 33 ; for 33 multiplied by %x is ^hx. From the above illustration we derive the following Rule. Write the divid.end over the divisor, in the manner of 4 38 ALGEBRA. a fraction^ aiid reduce it to its simplest form by cancelling the letters and figures tJiat are common to all the terms. Or, divide the coefficient of the dividend hy the coefficient of the divisor, and cancel the letters common to the divisor and dividend. EXAMPLES. 1. Divide ^ab by 2a. ^~=Sb ; or, Qab-^2a=Sb. za 2. Divide 12abcxy by 4:bx. .^ — — — -=Sacy ; or, 'i.2abcxy-r-4:bx=3acy. 3. Divide m?iop by op. Arts. mn. 4. Divide 7(z5??z by a7?i. Ans. lb. 5. Divide 14:xyz by 7x. An^. 2yz. 6. Divide lOabcd by ^bcd. 7. Divide 9mnx by ox. 8. Divide 17 ab by ab. 9. Divide Adqrst by 7qt. 10. Divide 20hm7io by 4:?io. 90. Powers and roots of the same quantity are divided by subtracting tbe index of the divisor from that of the dividend. Thus, if we wish to divide a^ by a^, we subtract the index 3 from the index 5, and set the remainder 2 over the a; thus, a^. This process is evident from the fact that a^=aaaaa, and a^ =aaa, and aaaaa divided by aaa gives 0^=0^. 11. Divide 4:a'b* by 2ab\ -^=2a'^^%- or, AaW-i-2ab''=2a'b\ Aa¥ 12. Divide 7a^ by (r. Am. 7a\ 13. Divide Ga'b'cd by Sab. Ans. 2a%cd. 14. Divide Try by r^. An^. 7r'^p^. 15. Divide 60/?/' by 30/. Ans. 2pY. 16. Divide \2axY by 4^2:1 Ans. 3?/. 17. Divide 96r\vfV by 48.a^c — a-\-Qah. 4. Divide 1()a'bc-\-lbabd'—l^a^be by bab. Ans. 4a-c-{-3d^—2ae. 5. Divide 15x^+30^^2/^ ^J ^^' -^^• 6. Divide lax'^yz^ — 14iXyz-\-21xy'^ by Ixy. Ans. 7. Divide p^mq-\-p^m — p'^mc by p^. Ans. 8. Divide Atxz—St'^z+z'^ by z. Ans. 9. Divide 12a-''~Sa'b-\-lQa'x-10a-^y by 2a\ Atis. Ga-^ — 4:b-\-Sax—ba~*y. Case III. 92. When the divisor and dividend are both compound quan- tities. 40 ALGEBRA. Rule. Write down the quantities in the same manner as in the division of numbers in Arithmetic, arranging the terms of each quantity so that the highest powers of one of the letters may ftand before the next loicer. Divide the first term of the dividend by the first term of the divisor, and set the result in the quotient, ivith its proper sign. Multiply the whole divisor by the terrfythus found ; and, having subtracted the result from the dividend, bring down as manij terms to the remainder as are requisite for the next operation, which perfm'm as before ; and so proceed., as in Arithmetic, till the work is finished. 1. Divide a^-^'lah^W by a^b. 2 I o I r 7.2/^+^ divisor. a^-\-1ab-\-b\ — pj- \a-\-b quotient. c^-\- ab ab^b"" ab-\-W 2. Divide c^-\-^a}x-{-hax^-\-x? by a-{-x. a-\-x a^4-ba^x4-bax'^-\-x^l —r \a' '-{-4:ax-{-x\ a^-\-a^x ^a^x-\-bax^ Aa^x-^-^ax^ ax^-{-x^ ax'^-\-x^. 3. Divide a'-\-^a'b''-\-lW by a'—2ab-^^b\ a'-2a%-]-4.a~P 2a^b-\-lW ^a^b—^a^b^-^'^ab^ 4:a%''—Ub^-\-l^b' DIVISION. 41 It may be verified that a^-^2ab-{-4:b'^ is the true quotient, by multiplying it by the divisor. It should also be observed, that in every stage of the proceeding, the terms involving the highest powers of a have been placed first on the left. 4. Divide 4x'—9a'x'-{-Qa'x-a' by 2x''—^ax-{-a\ Ax'—Qax^-\-2a''x'' Qax^ — lla^x^'-^-Qa^x Qax^ — da^x~ -\- Zcc'x 9Si If the divisor be not exactly contained in the dividend, the quantity that remains after the division is finished must be placed over the divisor at the right of the quotient, in the form of a fraction. 5. Divide c^ — x^ by a-\-x. a-{-x '-x'f- ^a^—ax-\-x'^- a^-{-a^x 2x a-\-x —d^x- — drx- -x' -ax^ ax^—x^ ax'^-\-x^ -2x\ Oil The operation of division may be considered as ter- minated when the highest power of the letter, in the first or leading term of the remainder, is less than the first term of the divisor. The division of quantities may also be sometimes carried on ad infinitum, like a decimal fraction ; in which case a few of the leading terms of the quotient will, generally, be sufficient to 4# 42 ALGEBRA. indicate the rest, without its being necessary to continue the operation. 6. Divide a by a-\'X. ( X x' x' a a" a a~\-x — X x" — X a a x^ a' x^ x' a?' ~a?' 7. Divide a by a — x. 8. Let c^ — 2ax-{-x^ be divided by a — x. Ayis. a — x. 9. Divide a?—2,d'h^Zab''—b^ by a-b. Ans. a^—lab^y. 10. Divide 'ia^-^a-b-^alJ'-^-W by "la-h. Am. ^a^—Zh^. 11. Divide Zb^-\-ZaW—^a%—^a^ by a^b. Ans. —^a^J^W. 12. Let 2aV— Sdiz-f 2 be divided by "lax—l. Ans. ax — 2. 13. Divide Ila'-'IW by la-lb. Ans. Za'-\-Za^b-\-Za%^-\-Ub^+U\ 14. Divide x'-y'-\-2y''z'—z' by x'-\'y''—z\ Ans. x"^ — y'^-^-^^- DIVISION. 43 15. Divide l-\-a by 1—a. Am. l-|-2a4-2a2+2a^-f 2a^+, &c. 16. Divide Sx''—lby''-\-2dyz—2xy—Sxz—Qz''hj2x—Sy+z, Ans. 4:X-\-by — Q>z. 17. Divide 6a;^— 96 by 3a:— 6. A7U. 2z'+4z'+8^-fl6. 18. Divide a^J^a%''-{-a'b'-\-a'b'+b^ by a'^a^b+a?b''-\-ab^ +b\ Ans. a'—a%^a'b''—ab^-\-b\ DIVISION BY DETACHED COEFFICIENTS. 95 1 As the pupil has seen in Art. 87 that the operation of many questions in Multiplication is facilitated by using de- tached coefficients, he will readily perceive that the same prin- ciple will apply to Division. The terms of the divisor and dividend are to be arranged according to the power of the letters, and zero must be inserted in the terms that are wanting. The first literal term of the quotient is obtained by dividing the first letter of the dividend by the first letter of the divisor ; and the letters belonging to the other terms are written in the same order, as they are found in the divisor and dividend. EXAMPLES. 1. Divide a3+3a'3+3a32+^»3 by a^b. / l+l coefficients of the divisor. \l-|-2-f-l coefficients of the quotient. 1+1 2+3 2+2 1+1 1+1. a^-7-f^=(2^, first literal term of the quotient. The others will therefore be a3+3^ and these terms annexed to the coefficients will be a-+2«5+3l 44 ALGEBRA. 2. Divide x^ — if by x~ — y^. 1+0-1 l+0+0+0-l(-^ 1+0-1 1+0-1 1+0-1. x''-^x^=.x^^ first literal part. The other regular parts are xy-\-y^. Having prefixed the coefficients, it will be x'^-\-Oxy +2/^; but, as the coefficient of the second term is zero, the term has no value. The correct answer will therefore be x^-\-y'^. 3. Divide 3a;^— 48 by 3a:— 6. 3+0+0+0-48(j^- 3-6 6 6-12 12 12-24 24-48 24-48. x'^-^x=.x'^, first literal part. The succeeding terms will, there- fore, be x^-{-x-\-x^. Hence the true quotient will be, x^-\-2x^-\- 4a;+8. 4. Divide \—a? by l-\-a. Ans. 1 — a-{-(r — a?-\-a'^ — a^-\-a^ — (^. 5. Divide 3?/^+3a:?/^ — ^x^y—^x^ by x-\-y. A7is. oy'^ — 4:X~. 6. Divide a'-^a'b-Sa'b'+l^aP-Sb' by a"-i-2ab-2b\ Am. a'^—5ab-\-4:b\ 7. Divide m^—b77i'n-^107n^n~—10m'^n^-{-5?n?i^—n^ by tti^ — 2m7i-{-7i-. Atis. m^ — 2,7ri?n-\-omnr — n^. 8. Divide a''H-"_|_a"+i^'«-i__a"'->Z,''+i_^"»+" by ^"-^+3"^^ Ans. a"+i— 3"+^ MISCELLANEOUS QUESTIONS. 45 QUESTIONS TO EXERCISE THE FOREGOING RULES. 1. What is the sura of the following quantities : 12a-|-5c-j- 17^-1-133, Sa-\-12b-^lbd-^8c, llc + lba-l-2Sb-^10d, and U-f- 3«-|-20Z»+18c ? A7i^. 38a+683+42c+46rf. 2. Add together 5(Z-|-35— 4c, 2a—bb-^(5c-}-2d, a— 4b— 2c -]-3e, and 7«+43— 3c— 6e. A?is. 15a— 2^— 3c-[-2^— 3c. 8. Find the sum of 3a2+2ai+4^-, bar—^ab-\-Qb\ —4a:-{-bab —b\ lSa:'—20ab—l%\ Ua'—2ab-\-20b\ and —^Qa:'-\-24:ab— 10b'. Ans. 4. Kequired the sum of 5a^5— ITa^^c- 15^V-[-5, —4:ab-{- 8a;'bc—10bY—4:, —^a'b—Sa'bc-\-20b'c'—S, and 2arb-\-12a'bc-\- 53V4-2. ^7^5. 5. Add the following quantities: a-^b-\-c-{-d, a-\-b-\-c — 2c?, a-\-b—2c-{-d, a—ob-{-c-{-d, —a-\-b-\-c-{-d, and a— b— 2c— 2d. Ans. 4:a. 6. Multiply x'-\-2x-{-l by x'—2x-\-S. Ans. x'-]-4:X^S. 7. Multiply l—x-\-x''—x^ by 1+a:. A7is. l—x\ 8. Multiply l_2a;+3a;^— 4^'3+5:c^-6:i'^+72;^-82-^ by 1-f 2x-\-x''. Ans. l—9x^—Sx'. 9. What is the continued product of (i-{-b, a — b, a'^-\-ab-{-b'^, and a^—ab+P ? Ans. a'—b\ 10. Multiply x'-Jroax"+^a'x-i-a^ by x'—dax''+8arx—a\ Ans. x'—dd'x'-\-Sa'x''—a\ 11. Multiply a^'-'-^b'"-' by «"+'— 3"+^ J[?Z5. a'"+"-\-a''+^b"'-^—a'^-^b"+'^—b'"-^". 12. Divide rc^ — a^ by a; — a. Atis. x'^-\-ax^-\-d^x^-\-a?x-\-a^. 13. Divide a:* — ^x^ — ^xy — if' by x^-\-'^x-\-y. Alls, rr^ — ox — y. 14. Divide x^—4:X^-\-Qx'^—4x-\-'l by a;^— 2:r-{-l, and x^— 2a^x^-\-lQia^x — Iba^ by x^-\-2ax — 3^^^, and find the difference of their quotients. Ans. 2x — 2ax — l-}-5a^ 15. Divide a;^— 16aV4-64a^ by x—2a. Ans. 2;^+2az*+4aV— Sa^:^-— IGa^a;— 32al 46 ALGEBRA. SECTION VI. FRACTIONS. Aet. 96. Algebraic Fractions are similar to vulgar frac- tions in Arithmetic ; they express a part, or parts, of a quantity or a unit. D7o They consist of two parts, the numerator and denom- inator, the former being written above the line, and the latter beloiv it ; and these, when taken together, are the ter^ns of the fraction. 98 1 The denominator shows into how many parts the quan- tity or unit is divided ; and the numerator, how many of these parts are represented by the fraction. 99. A proper fraction is one whose numerator is less than its denominator ; as, a—h 7 a-\-d' """^ 8* 100. An hnpi'oper fraction is one whose numerator is equal to or greater than its denominator ; as, a b-\-c 7 -, or — - , or K. a — c 6 101. A mixed quantity is a whole number or quantity, with a fraction annexed, with the sign either plus or minus ; as, a m , a m ^„ ~-\-y, or X, or y-\ — , or x , or 7f. b ' ^ n -^ ' c n ^ 102. A compound fraction is a fraction of a fraction; as, - of- of- ; or, I off of -^V 103. A complex fraction is a fraction having a fraction in its numerator or denominator, or in both : as. FRACTIONS. 3 4 a a 4 7 b c — d 14X' 11' -, or c 7^ 12 ^ p 47 104. The value of a fraction depends on the ratio which the numerator bears to the denominator. 105. The value of a fraction is not changed by multiplying or dividing both numerator and denominator by the same quantity. 106. The greatest common measure of two or more quantities is the largest quantity that will divide all of them without a remainder. 107. The least common multiple of two or more quantities is the least quantity that can be divided by them all without a remainder. 108. A fraction is in its lowest terms when no quantity, ex- cepting a unit, will divide both of its terms. 109. Quantities are said to be prim£. to one another when their greatest common measure is a unit. 110. Frime factors 0^ quantities are those factors which can be divided by no quantity but themselves or a unit ; thus, the prime factors of 35 are 7 and 5. 111. A composite quantity is that produced by multiplying two or more quantities together. 112. A fraction is, in value, equal to the number of times the numerator contains the denominator. 113. A fraction is increased in value either by multiplying its numerator or dividing its denominator. 114. A fraction is diminished in value either by dividing its numerator or multiplying its denominator. 48 ALGEBRA. Case I. 115. To find the greatest common measure or divisor of the terms of a fraction. Rule. Arrange the two quantities according to the order of their powers, and divide that which is of the highest dimensions by the other, having first cancelled any factor that may be con- tained in all the terms of the divisor, without being common to thnse of the dividend. Divide this divisor by the remainder, simplified as before, and so on for each successive remainder, and its preceding divisor, till nothing remains ; and the last divisor will be the greatest common measure or divisor required. If any of the divisors, in the course of the operation, becoTne negative, they m.ay have their signs changed, or be taken affirm- atively, without altering the truth of the result ; and, if the first term of a divisor should not be exactly contained in the first term of the dividend, the several terms of the latter may be midtiplied by any number or quantity that loill render the division com- plete. ^ EXAMPLES. CX-\-X^ 1. Find the greatest common measure or divisor of -5 — ; — ^. ° ac-\-a^x cx-\-x^)a^c-{-(i^x c-\-x) a^c-{-a^x{cr ac-\-a~x. As X is found in both terms of the divisor, we divide those terms by x before the operation. The greatest common measure of both terms we perceive is c-\-x ; that is, it will divide them both without a remainder. Thus, c-{-x) .^ ^ z=- ac-j-ax a 2. Required the greatest factor of FRACTIONS. 49 x'yilx'^lrx —2bx'—2b''x) x-\-b)x^-]-2bx-\-b\x-\-b. x'^-^- bx bx-\-h' bx-i-b'' We cancel 2bx in both terms of the second divisor, as it is common to both. As x-\-b is the last divisor, it is the greatest factor or com- mon measure of the quantities proposed. 3. Required the greatest common divisor of Sa^ — 2a — 1, and 4a3_2a-— 3«-}-l. 3a2_2a— l)4a'— 2fl'^— 3a+l(4a o O 12a'-Sar—4:a 2a'—ba-{-^y6a~—2a-l 2 Qa"— 4a— 2(3 6a2_15a4-9 lla-11 a-l)2d'—5a-{-^2a- 2a'-2a -3 — 3a4-3 — 3a+3. As 11 is common to both terms of the third divisor, it is cancelled; therefore a—1 is the greatest common factor of both quantities. 4. "What is the greatest common divisor of x^—a^, and x"^ — a~ ? ,' _ o: A?is. X — a. 5 50 ^ — Y^i ALGEBRA. 5. What is tlie greatest common factor o^ x^ — 1, and ax-\-a? Ans. x-\-\. 6. Required the greatest common factor of?/"' — a;^ and 'if — ifx — yx^-\-x^. Ans. if — x\ 7. Required the greatest common measure of a^ — crx-{-ax'^ —x^^ and a^ — x^. Ans. d^ — (j^x-\-a'Ji? — x^. 8. Required the greatest common factor of a'' — x^^ and (jC'-\- a?x^. ,/ ^^ ' ' Arts. d^-{-x^. Case II. 116. To reduce fractions to their lowest terms. Rule. Divide the terms of the fraction by the prime factors common to both. Or, divide both terms of the fraction by their greatest commoii divisor. IIT. That fractions after reduction have the same value as before, is evident from the fact that their numerators retain the same ratio to their denominators; for equi-multiples and sub-multiples of any two numbers have the same ratio to each other as the numbers themselves. Letters or numbers common to all the quantities in each term of the fraction may be cancelled. EXAMPLES. ^ ^ , ^abc . . 1. Reduce tttt-. to its lowest terms. ba^bd 4:abc 2abX^c 2c — -2— z= . Ans. Qd^bd 2abX.Sad oad' In this operation we find 2ab to be the largest factor in both 2c terms ; it, therefore, may be cancelled, and the answer is ^—z. 2. Reduce —, — — to its lowest terms. admny abxv bx . ^=r - — . Alls. admny dmn FRACTIONS. 51 In this question we find a and y common to both terms ; and, hx they beina; cancelled, the result is -; . dmm 3. Reduce ^— to its lowest terms. Ans. — . mTwp-qx px 4. Eeduce , .,,., to its lowest terms. Ans. ■=-. 1 2(2771 4^3 5. Reduce r^r- — to its lowest terms. A71S. -^r^. l^bcm DOC 6. Reduce ^^ „ „ to its lowest terms. Atis. -— . oQd'x^ 9ax _, _,' QaTi' . . .an 7. Reduce —-7-- to its lowest terms. A71S. -prr- obo7i bo „ ^ ^ bQbTTi^x^- . - , . 14:mx^ 8. Reduce -77^ — 5- to its lowest terms. Atis. —-7- — . ibb7nxr 11) ^ _, , V^ah-cde^ . . a b 9. Reduce _,. .,. , - to its lowest terms. Ans. Iba^cde 4aV 10. Reduce „ , ^■, — — -, to its lowest terms. X -\-YJ)X-\-b- In performing this question, we first find the greatest common measure of the two terms of the fraction, which is x-\-b ; wo then divide both terms by it. Thus, , A x^ — Wx x~ — hx . ^ J x'-^2bx~\-b^ x-\-b -.-. -r. 1 Qa^-{-bax — Qx^ . , ■ 11. Reduce ^. , ' ^ — ^, to its lowest terms. ba^-\-loax-{-ox' Sa,—2x ^a-\-'2x' 12. Reduce — — —, to its lowest terms. . 1 a — x^ Ans x^ if 13. It is required to reduce —. — ^ to its lowest terms. x^ — y^ Ans. d'-\-x^' x^-\-xhf-\-y'^ x'-\-y\ 52 ALGEBRA. Case III. 118i To reduce a mixed quantity to the form of a fraction. Rule. Multiply the integral part by the denominator of the fractional part ; to this product annex the numerator of the fraction, prefixing to it the sign of the fraction ; under the whole write the denominatoi' of the fraction. EXAMPLES. 1. Reduce 71- to a fractional form. -^ =-^. Ans. ^ 5 5 2. Reduce a-\ — to the form of a fraction. e ayCe-\-b ae-4-b ' = — —. Ans. e e 3. Change a-{ to a fraction. ayCm4-a~ — b' am-\-a- — b^ 7n rti 7Yi-\—n 4. Chano;e a ^^— to the form of a fraction. e ayce — m-\-n ae — m — n -— ^ — = . Ans. 7 5. Reduce x to the form of a fraction. m x'yCra — a — b mx — a-\-b . = . Ans. 171 m b^—cd 6. Chancre a-\ to the form of a fraction. n an-\-}p- — cd Ans. ■ n A.7i^ I 5fl^ 7. Reduce Ix ^ — to the form of a fraction. o . 56z — A.11^ — 5a Ans. :x . FRACTIONS. 53 8. Keduce 15a -. to tlie form of a fraction. 4:771 Am. ; ■ — . 4:771 'J Q ']Jl 9. Reduce 7a — b -. to the form of a fraction. 471 2^an — 4b7i — *Je-{-7ri Am. 7 ■ — . 4n 10. Change llm — 4.7i-\-^ -— ^ to the form of a fraction. ° 6m — lii^ . 33??z- — 2277171^ — 12vin-\-'^7i^-\-a^ 3m — 27i'^ 11. Reduce ^x^-^-by^ — ,, to the form of a fraction. JiX — ^y lQx^^l()xy''—24xY—^^y'—d^—a' ^^- 2x-lf • Case IV. 119» To represent a fraction in the form of a whole or mixed quantity. Rule. Divide the numerator by the deTwmmator for the inte- gral part, and write the remainder, if any, over the denomiTiator for the fractional part ; annex this to the integral part, and it will represent the quantity required. EXAMPLES. 27 1. Change -^ to a mixed quantity. 8 27 :27-i-8=3#. Ans. 2. Chano;e -— to a whole number. o 21 j^=88^11=8. Am. 3. Change " - to a mixed quantity. ax-]~a' ■.ax-\-d^ -r-x=a-\ — . A7is. X ' x 5# 64 ALGEBRA, 4. Change — - — to a mixed quantity. ab — a^ ~- ,. ^ c^ . — - — ■=.ao — a-^o^=a — -. Ans. b 5. Change — to its equivalent mixed quantity. Ans. (f- — axA^x^ a-\-x 6. Chancre — ; — to a whole number. Atis. x~ — xii+il^, ^ 1? 7. Change ~ to a whole number. Ans. x''~~\-xy-\-y^. X y 9 Q^X' X 8. Find a mixed quantity equivalent to . Ans. x^ . a Case V. 120# To reduce a complex fraction to a simple one. Rule. If the numerator or denominator, or both, be ivhole oi mixed quantities, reduce them to improper fractions. Then multiply the denominator of the lower fraction into the numerator of the upper for a neio numerator and the denominator of the upper fraction into the numerator of the lower for a new denominator ; or, invert the denominator of the complex fraction vjhen reduced, and place it in a line with the numerator ; then tnultiply the two numerators together for a new numerator , and the two denominators together for a new denomirmtor . All fractions in this proposition must be reduced to this form, a 3 c 4 -, or -, before they can be solved by the above rule. Now, b 5 every fraction denotes a division of the numerator by the denominator, and its value is equal to the quotient obtained by such a division. Hence, by the nature of division, we have, FRACTIONS. 55 c a b ah d c d cd' Bj the preceding rules we are enabled to show all the vari- ations that can possibly happen in preparing fractions, and also the method of reducing them to their lowest terms. EXAMPLES. 7 7 1. Reduce f to a simple fraction. ^='^X.i=i^- -A.ns. 7-1 2. Keduce ^ to a simple fraction. 71 29 *¥ ~_£^ 8J- -1-7. Li. liT 59v 2 — 58 29 i^T^o ^1 17 4'AtT — WQ — in"* -^^t^' 7 3. Reduce - to a simple fraction. 7 ^ 1 1 1/^1 1 '"-'- Ans. 8 4. Reduce -^ to a simple fraction. 3. 3. 5 5 '¥ "4 a g3 _2X 5^2 7 lTf5 ?f5' -^'t^* 5. Reduce — -- to a simple fraction. b b a \ a m-\-n m-\-n b m-\-n bm-\-bn 6. Reduce to a simple fraction. y a " ^^ =%xA- = ^. Am. a xy-{-a 1 xy'\-a xy-\-a ^ y y 56 ALGEBRA. 7. Keduce to a simple fraction. m X n , h ac-\-b c c ac-4-b n acnA-bn . = — !— X = ! — . A71S. m nx — m c nx — m cnx — cm ^ n n 8. Reduce ;^ to a simple fraction. Atis. \^. X 9. Heduce — ^r- to a smiple traction. Ans. „, , . . , , % ^ 634-4?/ n ^"~3 %m—n 10. Reduce to a simple fraction. Aiis. — ^ — 2/— ^'+2 11. Reduce — — to a simple fraction. „ o i ^ Case VI. 121. To reduce fractions to a common denominator. Rule. Multiply each numerator into all the denomiruitors except its oivn for a new numerator ^ and all the denominators together for a common denominator. Or, find the least common multiple of all the denominators, and it will be the denominator required. Divide the common multiple by each of the denominators, and multiply the quotients by the respective numerators of the fractions, and their products will be the numerators required. IIRST METHOD. 1. Reduce yV? i? and |, to a common denominator. 5X 8x4=160, numcHkr for tV=^|£. 7X12X4=336, numerator for I =ff|-. 3X12X8=288, numerator for f =§||. 12x 8x4=384, common denominator. FRACTIONS. 57 Equimultiples of the terms of a fraction express the same value as the fraction itself. The terms of -^^ are each multi- plied by 8 and 4. Hence 4^f £^ has the same value as ^^. The same may be observed of |- and f . 2. Reduce -, -, and — , to a common denominator. a n «X^X^=^<^^i=numerator of -= — . h bdn cX^X^=^c?^== numerator of -=--7-. a bdn ^X^X^=^^w=uumerator of-=^-^- 71 bdn i'Kdy^n=hdn-=.QommoB. denominator. SECOND METHOD. 3. Keduce |, y^^-j ^^^ h to a common denominator. 4 )8, 12, 4 2, 3, 1; 4x2x3=24, common denominator. 24 12 4 3x7=21, numerator for |- =f^. 2x5=10, numerator for •r2-=^2¥* 6x1= 6, numerator for ^ =2^4. 4. Reduce -7-, — , and 7—, to a common denominator. 4x X hx X)4:X, x^, '^x 4 )4, X, 8 1, X, 2; a: X^X'^X 2= 8:?;^ common denominator. Sx"" lax Ax x' Sx 2xX,(i=2.ax, numerator to 7-=^ ,. 4x Sx^ 8 X^=83, numerator to —=;::-—. x^ Sx^ \f1 1/7 '7* xyJoa=.^ax^ numerator to „ — -— r-. 5. Reduce |-, ^^^-j ^"^^ h to a common denominator. Ans A6 21 J.8 68 ALGEBRA, 6. KeducGy^y, -j^g-, ^, and 7, to a common denominator. Ans. 7. Heduce f of 7|- and -^j- of 5 to a common denominator. Ans. 8. Reduce f of -^j of 17 and J- of 19 to a common denom- inator. Ans. 9. Reduce - and -^r of ;|- to a common denominator. 5 "^ ^ 7 1 /I710 8 6 13 240_ yi/Z.6. -rs-T-^n-j TITBIT' mmon denominato 33^:^ — ocx^ 4:bmy — 4icmy axy 10, Reduce — , — , and •; , to a common denominator. y X b — c Ans. 11. Reduce — , --, and , to a common denominator X X — 2 y Ans. bxy — cxy ' bxy — cxy ' bxy — cxy d , to a common denominator. y axy — 2<2?/ bxy doi? — Zo? — IdxAr^'^ x^y — ''Ixy x^y — '^xy x\j — ^Ixy 12. Reduce , -, and ^,, , to a common denominak)r X y — 2 18 18g?/+18%— 36^—363 h\x—\Ux x''y—7xy—2x''-\-14: c ISxy — SQx ^ISxy — 36:c' ISxy — 36^; Ans. 13. Reduce - — ^, -, -, and ^, to a common denominator. b — 6 b X X — ^abx^ — 20(z3z abx^ — Zay? — habx-\-\hax Ans. • b''x'—3bx^—bb~x-{-15>bx 3 V— Sbx'—bb^x+lbbx b''dx—Udx—Wd-\-\bbd ab''x—b^x—^abx-\-Wx . Wx'—Ux'—bb'x-^lbbx b'x'—dbx'—bb''x-]-lbbx 1 -, to a common denominator. 2/— 3 x^y — 3^;^ xy"^ — "^xy ay — oa ax xy — Zx ' xy — 3a; ' xy — 3a;' xy — 3a;* 14. Reduce x, y, -, and -^ to a common denominator. X y—3 Atis. 15. Reduce a, b, c, d, and -, to a common denominator. o ab b" be bd a '^'' Tl'l' T' ~b' FRACTIONS. 59 X y 3?/ 16. Chano^e - and -:p to a common denominator. 771 1 2 m—n . 6a; . Atis, -k — and Smy 'S?ny X 5J- 17. Cliano;e — and — to a common denominator. 7^ X "&■■ 2x~ , 80 Ans. — r- and —;r-. lOZ iDX Case VII. ADDITION OF FRACTIONS. 122. To add fractional quantities. KuLE. Reduce the fractioris to a common denominator^ and write the sum of the numerators over the common denominator. EXAMPLES. 1. Add 1^, y^j, and J-^, together. [lere 7x12x16=1344 ^ ^X 8X16= 640 > the new numerators. llX 8x12=1056) 3040 :1|^. Ans. And 8x12x16 = 1536, the common denominator. 2. What is the sum of -, -, and - ? b d f Here aXdXf=cidf \ cX^X/=c^/" / *"G new numerators. eX'^Xd=ehd ) And by^dycf^=bdf, the common denominator. ™ „ adf . cbf , ebd adf-\-cbf-\-ebd 60 ALGEBRA. ... . . . Sx' , , ,2ax 3. Add tlie following quantities, a 7- and b -\ . c Sx"" ab—Zx^ , , lax hc-\-'lax b b c c ah — Zx^yc^c^iahc — ocx~ ) ,0 . ^ , ( numerators. bc-Ylaxy^b^b'c^-^abx ) by^c =:bc^ common denominator. abc—2>cx' b\-[-2abx abc—Zcx'-\-Wc^1abx , _ , be be be 2ahx — Zcx^ . ^ . Ans. be 4. Add together g^, ^, and ^, 84gV+8Q^7?z?z"4-105a^^e ^^' U^adn^ 5. What is the sum of ^, -jV, and |-? Aois. Iff. 6. What is the sum of f, y\, and f ? Ans. 2^2_7^. 7. What is the sum off, f, |, and -J^l Ans. 2^|. 8. What is the sum of 8f , 3f , and 7| ? Ans. 20i. 9. What is the sum of | of 7^, and -/_ of 13 ? Ans. 13/^. 10. What is the sum of f of 1, and J- of f ? Ans. -fi. 11. What is the sum of | and -|- ? Ans. f|. 12. What is the sum of f of ^ and f of ^ ? Ans. if §9. .« -r^. i , ^3:r , 2:c . ^exA-^ax 13. Find the sum of -j— and ^r— . ^?w. 4fi^ 3e * ' Viae XXX ^ 47a: 60" .. T.. -. 1 ^4a , ft-3 . 23^^-21 15. Find the sum 01 - and — ; — . A7is. 14. Find the sum of k, -7, -. Ans. ^^. o 4 7 4 • 28 lo. Find the sum of 47?z, — r: — , and 3 9^ + 24772 + 871 + 1 Ans. o FRACTIONS. 61 17. What IS the sum oi — ^ — , — - — , and — ? o A 139a-8 Am. — ^TT — . 3 3 ^^ 18. Add — -— and tocrether. a^ a^h a-h ^^ ^« a a — b a-\-b — -, — — , and a—o c-\-d c — a 2a'c-\-ac'—ad'-2ahc-{-2abd—2b'd 19. Add ■ — -, — ^— „ and — -^^ together. ac-—ad^-\-bd^—bc^ o/^ All 3^- c , 4ac— 2^c+9a5^— 93^ 20. Add to -. A71S. -jr^-, Q ,3 , QZ4 • a—b 2a— b babe — \)abc-\-6bx ~2~ ""3" Case VIII. SUBTRACTION OF FRACTIONS. 128. To subtract one fraction from another. Rule. Reduce the fractions to a commoii denominator, sub- tract the numerator of tlie subtrahend from the numerator of the minuend^ and write the diffei-ence over the common denomi- nator. EXAMPLES. 1. From I take -/y. =77) 4X 9=36 ) ^' And 9x11 = 99, the common denominator. Here 7x11=77 . 'le new numerators. Whence U—U=U' ^^^' a 1 c 2. From - take -. b d Here ay^d=-ad . C the new numerators. Xd=^ad ) Xb=:bc ) And bXd^bd, the common denominator. .^_ ad be ad — be . Whence r-.— 73= — 73 — . Arcs, bd bd hd 62 ALGEBRA. 3. From | take j\. Ans. |f . 4. From f take |. Am. f. 5. From 7f take 4|-. A?is. S^. 6. From Q^ take | of 5. Ans. 2§^. 7. From 8| take f of 17^. ^?w. 1}. 8. From f of ll^-^ take -iJ- of 3|. A?is. If Jg-. 9. From | of 13f take /^ of 7^. Ans. 9^\. 10. From f of 7 take ^ of 17^. Ans. 2^^. 11. From ^ take i. ^7^.. ^^2_7_. 12. Take — ;- from -A.. ^tw. ^ fi^-j-l fi — 1* ' «^ — 1* 13. From — - take — -. - Ans. — — . 5 7 35 ^, ^ 3a— 23 , 2a— U 14. irom — rr — take 3c 5^. l^ab—Qac—10b''+1 2bc Ibbc. • 15. Required the difference of -^ and — . A?is. -— -. 16. Subtract ^ from ^±^. ^72^. ^^^ a;-j-2/ "" '2; — ?/' ' a;"^ — y^' -ihT c( 1 ^ — ^ n o fl^+3 , ^ 2a 17. Dubtract a — irom da-] r— . ^7W. la-\---. ad a 18. Subtract z — from Ix ^ — . Ans. (Sx-{-a-\-j;. Z a b 19. From ^ ' ' take ^^ — -^. Am. 4. g;-!-^ « — ^ "^ . ~X . hd'-\-2^ah-k'U^ 20. From take -. Am. —^ i — . a — h a-\-h ba'^—bb^ FRACTIONS. 63 Case IX. MULTIPLICATION OF FRACTIONS. 121 1 To multiply fractions together. Rule. Multiply the numerators together for a new numer- ator^ and the denominators for a new denomi7iator. Whe?i the numerator of one of the fractions and the denom- inator of the other can he divided by some quantity which is common to each of them^ the quotients may he used instead of the fractions themselves. Also, ivhen a fraction is to he multiplied hy an integer, it is the same whether the numerator is multiplied hy it or the denom- inator is divided hy it. If an Integer is to he multiplied hy a fraction, or a fraction hy an integer i the integer may he considered as having unity for its denominator. A mixed quantity should he reduced to an improper fraction. Powers or roots of the same quantity are multiplied together hy adding their indices. EXAMPLES. 1. Multiply f by I. iXl=U- Ans. ^ ^^ ^ . . a . m a m am . 2. Multiply - by — . _X-=7— • -^^s. ^ '' h n h n hn ^ ,r 1 . 1 cibc , mh ahc mh ah . 3. Multiply — by—,. — X7-i=— ,- Ans. mn hcd mn oca tia Note. — The Z>, c and m, are cancelled in both factors. ^ ,^ , . , 3a , ^ 3a 2m ^am . 4. Multiply — - by zm. =^X-^i-=;;T-- ^1^^^- ^ ^ Ihc -^ 7hc 1 The 5. Multiply a+— by -. hy a^-j-hy a^-\-hy m c^m-\-hmy . a-\ — ^= * X — -^ • -ii-ns. a a a n an 64 ALGEBRA. 6. Multiply ^^^ by ^-^^. ^ ^ x+y ^ x+y m-\-7i 77i-\-n m'^-\-2mn-\-Tt^ 7. Multiply ^: by 4^. Ans. ^. mm innd mttd 8. Multiply ^ by g^. . ^,«. ^^. y. Multiply —- by ^rj-r., Ans. ^^^ ,„ , . - — by — -o. Ans. -j-, hy ay^ ahy^ 11. Multiply 5^ by g^,. An.. 1. When the multiplier and denominator of the fraction are the same quantity, they cancel each other. 12. Multiply -^ by "Imn. 13. Multiply ^ by lla3. -I A T./r 1 . 1 4(2CC? , 14. Multiply — — by xy. xy 15. Multiply —^ by 17 ab. 16. Multiply ^^ab by --;^-. 17. Multiply -^^ by — . irni^ mri^ 18. Multiply';^" by il^. ^ ^ ihy ^ m-rt 19. Multiply- by—. Am. —r^. ^ '' b -^ h bh Ans, , Sab. Ans. Qmn. Ans. 4acd. Am. 2)hm. Am. X. 5 ab^ mhi^ A?is. ^. Am. ^m+n FRACTIONS. 65 Case X. DIVISION OF FRACTIONS. 125. To divide one fraction by another. Rule. Multiply the denominator of the divisor by the nu?ner- ator of the dividend for the numerator, and the numerator of the divisor by the deiwminator of the dividend for the denominator. Or, invert the divisor, and proceed as in multiplication. Or, divide the numerators by each other, and the denominators by each other, when this can be done without a remainder. Mixed quantities should be changed to improper fractions. EXAMPLES. - -r, . . , « , 3a Y 71 ^. Alls. 7^~'^. To free fractions from negative exponents. Rule. TraTisfer the letters which Jmve Tiegative exponents in the numerator to the denominator^ and those vjhich have Tiegative exponents in the denominator to the numerator, and then change the sign of the exponent. Note. This rule implies the multiplying of all the terms of the numer- ator and denominator by the same quantity. Therefore, by Art. 121, the value of the fraction is the same. EXAMPLES. 1. Free the fraction _^ from negative exponents. A d~e Ans. -^73. 1-3 /v,-2 m n — 7) 2. Free the fraction — _^ from negative exponents. X "Ij z Ans. my^z XTV'jf 68 ALGEBRA. 3. Free the fraction — ^ _^ from negative exponents. Tit iL C Am. ^—f- — . 4. Free the fraction _.^ _^^ _^ from negative exponents. Ans. 93'' ^ 1 9^4 ^ O I 5. Free the fraction — ^- from negative exponents. Ans. 6. Free the fraction ,1,3 _^,_^ 3 from negative exponents. We'd' Am. — 2^-. SECTION VII. EQUATIONS. Art. 126. The doctrine of equations is that branch of Algebra which treats of the method of determining the values of unknown quantities by means of their relations to others that are known. This is effected by making certain algebraic expressions equal to each other ; which formula, in that case, is called an equation. 127. The terms of an equation are the quantities of which it is composed ; and the parts that stand on each side of the sign = are called the two members, or sides, of the equation. Thus, if x=LaA^h^ the terms are x^ «, and h ; and the mean- ing of the expression is, that some quantity .t, standing on the left side of the equation, is equal to the sum of the quantities a and 3, on the right side. EQUATIONS. 69 128. A simple equation is one which contains only the first power of the unknown quantity ; as, X rc-|-X-\-o Qox-\-Wl Transposing, ^ . ,, =6.r-j-12— 6a;— 7=5. bx-{-6 ' Multiplying by 6a:-|-3, 63z4-117=30a;-f-15. Transposing, 63a;— 30a;=15— 117. Keducing, 33a;=— 102. D ividing, x=. — oj\. _. ^. 2ar4-8^ 13a;-2 x 7x x+lQ ^ _ . 30. Given —^-^,^^-^-=-—^ to find x. Multiplying all the terms by 36, it being the least common multiple of 9, 3, 12, and 36, we have Sx^M- :,^ i 12a;=21a;-a;-16. J. IX — O-j -r. , . rn 468^;— 72 Keducing terms, oU=— — — -. 17a; — o2 Multiplying by 17.2;— 32, 850.r-1600=468a;— 72. Reducing terms, 382a:=:1528. Dividing by 382, a:=4. Ans. EXAMPLES. 1. Given 5a;-}-22— 2a;=31 to find x. Ans. a;=3. .2. Given 4— 19a;=14— 21a; to find x, Ans. a:=5. 3. Given 24a;— 12=240— 12a; to find x. Ans. a==7. 4. Givenl5a;-|-7a;— 10=12a;-}-90tofinda;. Ans. a:=10. 5. Given 7a;-l-2a;=12a;— 36 to find x. Ans. a;=12. EQUATIONS. 77 6. Given 12a:— 3a;— 2:?;=63 to find x. Am. a;=9. 7. Given a:-|-r+^=87 to find x. Am. a:=60. ' 4 ' 5 108. X X 8. Given a:— -r+13=;^-["40 to find x. Am. x= 9. Given ^-{-^=:^-|-22 to find x. A7is. a:=120. 10. Given a;--+20=^4-^+26 to find x. Am. x=b6. 7 2 ' 4 ' 11. Given 3a:+^4-15=|4-41 to find x. Aiis. a;=8. 4 2 12. Given a; ^^^-=8 to find x. Am. a;=28. b 3a:— 11 5a:— 5 97— 7a: 13. Given 21-j — — = — j — to find x. JLO o Ji Alls, a: =9. 3^ 5 2a: 4 14. Given x-\ — =12 — ^^^—- — to find x. Atis. a:=5. 2 o 15. Given n:c-^-^-^-^=.20x-^-o to find x. tJ 2 J.72^. a::^2. 16. Given 9a: ^ — I ^ — =12a: ; 13 to find x. 2*3 4 Am. xz=7. 17. Given a:-f ^+^-t-^+^=2a:4-17 to find x. jU O 4: Alls, a: =60. 18. Given -=ih-{-c to find x. Arts, x X ' ' b-\-c' 19. Given 8a:— 40=0 to find x. A?is. x=b. 20. Given a-\ — =b-\-c-^- to find x. Aiis. x=. X X a — h — c „^ „. 3a:-3 , . 20-a: 6.a:-8 4a:-4 ^ „ . . 21. Given x }-4= — ^ = 1 ■= — to find the value of a:. Arts. 2'=: 6. 7# 78 ALGEBRA. 22. Given ax^-\-hx=7nx^-\-nx to find x. Ans. x=: a — m 23. Given aa;+7?z=3a:-4-% to find a:. Atis. x= . a — b ox X hciTTi c) 24. Given =m — c to find x. Ans. x=—^^ . b c oc — b 25. Given =i\hx-\-n to find x. Ans. x=z- ^-- — . a m Im — Ibam 26. Given — ; =a — b to find x. b c . 4ab — ac-l-abc — Pc Ans. xz= — b—c CiX d 27. Given \-de=Sx to find x. b—c e Atis. x= cde^ — bde^ — bd-{-cd A71S. x= — - — —T^ — ^7 . oce-\-ae — obe ' — to find a c 2bcm-\-2bc?i — 4ab — 2ac — abc 4x—a 2x-{-2a 2x-\-2a zo. Given ox ; i z=?n-\-n ' — to find x. 3 ' 4 ' c 113c— 8c+43 6a:+18 ,^ 11 — 3:r ^ ,^ 13—:?; 21— 2a; 29. Given -^-4-1 ^^=5.-48-^^ ^^ to find tlie value of :r. Ans. 2:=10. ^^ ^. 4x4-3 7a;— 29 8a;+19 , , , oU. Given — f-- =-?;= — tt^ — to find the value of x. 9 Dx—12 18 Atis. x=Q. SECTION VIII. PROBLEMS. 1. A gentleman stated that his age was twice that of his old- est son, and that the sum of their ages was 72 years. Kequired the age of each. PROBLEMS. 79 Let X z= the age of the son. Then 2x = the age of the gentleman. Therefore, a;-|-2:c=72, the age of both. Or, 8a;=72. Dividing, x=24, the age of the son. 2:c=48, the age of the gentleman. Proof, • 24+48=72. 2. What number is that, to which if ^ of it be added, the sum will be 99 ? Let X = the number. 4:X Then, — -|-2-=99. Clearing effractions, 4a;-|-7:r=693. Collecting the terms, lla:=693. Dividing, x= 63, the number. 3. A's and B's estates are valued at $3240, but B's is only | the value of A's. What is the property of each ? Let X = A's estate. 7x Then, -— = B's estate. o Therefore, a:+5=3240. o Clearing effractions, 8a;+7:r=25920. Or, 15:^=25920. Dividing, a:=:1728, A.'s estate. 7X172 8 =1512, B.'s estate. o 4. If § of a certain number be added to ^ of it, the sum will be 98. Required the number. Let X = the number. Then, T+i=^^- Clearing of fractions, 4x-{-Sx=bSS. Or, 7:i;=588. Dividing, x= 84. A?is. 80 ALGEBRA. 5. A certain gentleman divided his property, consisting of $5300, among his four sons. A, B, C, and D. He gave $350 more to B than A ; he gave C $400 more than B ; but he gave D twice as much as he gave A and B. How much did each son receive ? Let X = A's share. Then rc+350 = B's share. And a;+ 350+400 = C's share. And 2(2z+350)=4:r+700= D's share. Therefore, a:+a;+350-|- a: -{-850 + 400+4a;-|- 700=5300 Reducing, 7a:+1800=5300. Transposing, 7a;=5300— 1800. Reducing, 72:= 35 00. Dividing, x== 500, A's share. 500+350= 850, B's share. 850+400=1250, C's share. 2(500+850)=2700, D's share. Verification, 500+850+1250+2700=5300. 6. Divide $70 among James, John, and Charles; give John twice as much as James, and give Charles twice as much as John. Let X = the sum given to James. Then 2x = the sum given to John. And 4a; = the sum given to Charles. Then, by the conditions of the question, x-{-2xi-4:X=70. Or, 7:r=70. Dividing, x=10, the sum given James. 2:?;=20, the sum given John. 42'=40, the sum given Charles. Verification, 10 + 20+40=70. 7. Two men found a purse containing $144, and it was agreed that B should have $30 more than A. How many dollars did each receive ? PROBLEMS. 81 Let X = the sum A received. Then x-^dO = the sum B received. Therefore, x-{-x-{~^0=lU. Or, 2a:+80=144. Transposing, 2:c=144— 30. Or, 2x=lU. Dividing, x= 57, the sum A received. a:+30= 87, the sum B received. Verification, 57+87=144. 8. My horse and chaise are worth $336, but the chaise is worth five times as much as the horse. What is the value of each? Let X = the value of the horse. Then 6x = the value of the chaise. And, a:-f5a:=336. Or, 6z=336. Dividing, x= 56 = value of the horse. 5:r=280 = value of the chaise. Proof, 56-1-280=336. 9. What number is that whose third part exceeds its fifth by 12? Let X = the number required. X Then its third part will be -. o X And its fifth part, — . Therefore, ^_f=12. Multiplying all the terms by 15, we have, 5z— 32-=180. Or, 22:=180. Dividing, x=z 90, the number required. 10. John Smith's oldest daughter is 15 years old, and his youngest daughter is 11 ; he has $1728, which he wishes to give them. How shall he divide this sum, that each may de- posit her share in a bank which pays 6 per cent, simple interest. 82 A L a E B K A . SO that both shall have an equal sum when thej are 21 years old? Let X = the sum the youngest receives. And, 1728 — x = the sum the oldest receives. Then, a:+a;X.06XlO=1728— a:+(1728-a:X.06x6). Or, a:+.63:=1728— a:-f622.08-.36:i:. Transposing, 2.96:r=2350.08. Dividing, a:=$793f^, the youngest receives. $1728— $793ff =$934^-, the oldest receives. Let the pupil prove this question. 11. A man being asked the value of his horse and saddle, replied that his horse was worth $114 more than his saddle, and that § the value of the horse was 7 times the value of his saddle. What was the value of each ? Let X = the value of the saddle. And :?:-}- 114 = the value of the horse. Then, f (:c+114)=7a:. Or, 2:c+228=21a; Transposing, 19z=228. Dividing, a:=$12, value of the saddle. $12-{-$114=:$126, value of the horse. 12. A can reap a field in 7 days, B can reap it in 5 days. In what time can they both reap it together ? Let X = the days they would reap it together. A would reap j- of it in a day, and B would reap -i of it in a 1 1 12 day ; therefore in one day both together would reap s:+^=Qr of it. But, by the conditions, the field was to be reaped in x days. Therefore, — : 1 : : 1 day : x days. oO 12a: Multiplying extremes, -^ = 1. Multiplying by 35, 12:c=35. Dividing, x= 2j-^ days. Ans. 13. I have two carriages ; the value of one is five times that of the other, and the value of my horse is equal to both of my PROBLEMS. 83 carriages. The worth of them all is $300. What is the value of each ? Ans. First carriage $25, second carriage $125, horse $150. 14. A gentleman being asked his age, replied that his was twice that of his wife, and that his wife was three times as old as his daughter, and that the sum of their ages was 120 years. Required the age of each. { Gentleman's age, 72 years. Ans. } His wife's age, 36 years. ( His daughter's age, 12 years. 15. A man met 4 beggars, to whom he gave 77 cents. To the first he gave twice as many as to the second ; to the third, as many as he gave to the first and second ; and to the fourth, as many as he gave to the first and third. What sum did he give each ? Am. First 14 cents, second 7, third 21, fourth 35. 16. A drover has a lot of oxen and cows, for which he gave $1428. For the oxen he gave $55 each, and for the cows $32 each, and he had twice as many cows as oxen. Required the number of each. A7is. 12 oxen, 24 cows. 17. A gentleman, at his decease, left an estate of $1872 for his wife, three sons, and two daughters. His wife was to receive three times as much as either of her daughters, and his sons to receive each one half as much as one of the daughters. Re- quired the sum each received. Atis. Wife $864, daughters $288 each, sons $144 each. 18. A boy bought apples, oranges, and pears ; he gave two cents a-piece for the apples, three cents for the oranges, and four cents for the pears. He had twice as many oranges as apples, and three times as many pears as oranges. The sum he expended was $2.24. How many did he buy of each kind ? Ans. 7 apples, 14 oranges, 42 pears. 19 Let 85 be divided into two such parts that one of them shall be four times as large as the other. A71S. 17 and 68. 84 ALGEBRA. 20. Divide $100 among A, B, and C, so that A may have $20 more than B, and B $10 more than C. Am. A $50, B $30, and C $20. 21. A prize of $1000 is to be divided between A and B, so that their shares may be in the proportion of 7 to 8 ; required the share of each. Ans. A's share $4G6|, and B's $533^. 22. What number is that whose 3d part exceeds its 5th part by 6f ? A71S. 48. 23. A laborer agreed to serve for 36 days on these condi- tions; that for every day he worked he was to receive $1.25, but for every day he was absent he was to forfeit $0.50. At the end of the time he received $17. It is required to find how many days he labored, and how many days he was absent. A71S. He labored 20 days, and was absent 16 days. 24. Out of a cask of wine, which had leaked away ^, 13 gallons were drawn, and then being gauged it was found to be half full. How many gallons did the cask contain ? Ans. 78 gallons. 25. Divide 30 into two such parts that f of the one shall exceed f of the other by 6f. Atis. 18 and 12. 26. Wliat two numbers are those whose difference is 3, and the difference of whose squares is 51 ? Atis. 10 and 7. 27. Three men, A, B, and C, trade in company ; A put in a certain sum, B put in twice as much as A, and C put in three times as much as both, and they gain $864. What is each man's share of the gain ? Ans. A's $72, B's $144, C's $648. 28. James and William have between them 44 apples, and James says to William, if you will give me 12 of your apples, your number will then be only f of mine. William replied, if you will give me 12 of yours, your number will then be only § of mine. Required the number of each. Ans. James had 24 apples, and William 20. —/^ PROBLEMS. 85 29. Let 112 be divided into two such numbers that the gi'eater shall be to the less as 9 to 7. Ans. G3 and 49. 30. Let 19 be divided into two such parts that three times the greater shall be equal to four times the less. Required those numbers. Ans. lOf and 8|. 31. There are two numbers whose sum is 24, and if 7 be added to the larger, and 4 to the less, their ratio will be as 4 to 3. Required those numbers. Ans. 13 and 11. 32. The difference of two numbers is 4, and 7 times the larger number is equal to 11 times the less. Required those numbers. Ans. 11 and 7. 33. A merchant has two kinds of grain, one at $2.50 per bushel, and the other at S2 per bushel. He wishes to make a mixture of 80 bushels, that shall be worth $2.10 per bushel. How mjanj bushels of each sort must he use ? '"' "'/^ ^" ^ ' Ans. 16 bushels at $2.50, and 64 at $2. 34. A man having lost ^ of his money, found he had $96 left. Required the sum he had at first. An^. $128. 35. J. Jones found a certain sum of money, which was equal to ^ of what he possessed ; but having spent $40, the remainder was f of the sum he found. Required the sum he at first possessed. -f^-d^-Oz. %2 Ans. $36f_. 36. In my school | of my pupils study grammar, § of the remainder read, 10 spell, and the remainder, which is ^ of the number that read, study navigation. Required the number of pupils in the school. A')is. 70 pupils. 37. A gentleman lent a certain sum of money for 3 years at' 5 per cent, compound interest ; that is, at the end of each year he added ^'^ to the sum due. At the close of the third year he lost $15.25, but then there remained due to him $2300. Re- quired the sum lent. A^is. 82000. 38. A spendthrift spent -i- of the fortune left him by his father, and he then earned $124. Soon after he lost in specu- lation § of his property, after which he gained $274. His 8 86 ^ ALGEBEA. property was now valued at ^, wanting $86, of his original estate. What was the sum left him by his father ? Am. $1720. 39. A asked B how much money he had. He replied, if I had 5 times the sum I now possess I could lend you $60, and then -1- of the remainder would be equal to ^ the dollars I now have. Required the sum which B had. "■" ^ ^. A^is. $24. 40. A gentleman left an estate of $1862 for his three sons. He gave his youngest $133 less than his second son, and to his oldest son he gave as much as to the other two. How much did each receive ? Ans. Youngest son $399, second $532, oldest $931. 41. A, B and C, found a purse of money, and it was mutu- ally agreed that A should receive $15 less than one-half, and that B should have $13 more than one quarter, and that C should have the remainder, which was $27. How many dollars did the purse contain ? . Ans. $100. 42. Lent my good friend S. Jenkins a certain sum of money, at 6 per cent., which he kept until the interest was ^ of the principal. The sum then due was $500. Required the sum lent. A71S. $350. 43. A certain man added to his estate \ its value, and then lost $760. But he afterwards gained $600. His property then amounted to $2000. What was the value of his estate at first ? Ans. $1728. 44. James said to John, I have 40 shillings more than you. Yes, replied the other, and ^ of yours is equal to \ of mine. Required the number of shillings that each had. Arts. James 72 shillings, and John 32. 45. A merchant bought a number of barrels of flour, and having sold half the number and 4 barrels more to A, and |- of the remainder wanting 4 barrels to B, he had 20 barrels re- maining. Required the number the merchant bought. Ans. 1 36 barrels. //> PROBLEMS. 87 46. What number is that from which, if 7 be subtracted, ^ of the remainder will be 5 ? Ans. 37. 47. It is required to divide 44 into two such numbers that | of one of them shall be 6 more than f of the other. .^ - /: -n ^ / k I' -. % Ans. 24 and 20. 48. It is required to divide the number 43 into two such parts that one of them shall be 3 times as much above 20 as the other wants of 17. Required the numbers. ' /' Ans. 29 and 14. 49. John Jones can reap a certain field in 10 days, but, with the help of his oldest son, he can do it in 8 days. How long would it require his son to perform the labor himself ? Ans. 40 days. 50. A engaged to reap a field for 90 shillings, and he could perform the labor in 9 days ; but he took in B as a partner, and they supposed it would require 5 days for both to perform the labor, but they finished it in 4 days. How much, in justice, must A pay to B ? Ans. 50 shillings. 'tJl. I have two horses, and a saddle worth $30. Now, the saddle and first horse are worth f the second horse, but the saddle and second horse are worth three times the first horse. Required the value of each. Ans. First horse $60, second horse $150. 52. A gentleman let f of his money at 5 per cent., and the fv, remainder at 6 per cent., and his interest amounted to $180. What were the sums lent ? Am. $1200 at 5 per cent., $2000 at 6 per cent. 53. A can do a piece of work in 12 days, B can do the same work in 10 days, and C can perform it in 8 days. How long would it require A and B to do it ; how long A and C ; how long B and C ; and how long A, B and C, to perform the labor ? An^. A and B b^^ days, A and C 4| days, B and C 4|- days, A, B and C, 3/^ days. OO ALGEBRA. 54. Lent $TSO, at 6 p^r cent , for 5 rears. Wlmt principal will aniomit to the sum in 4 years, at 10 per cent ? Ans, $724,284. 55. Lent mj neighbor Jenkins 8270 for 4 Tears, at 6 per cent, ; some time afterwards. I borrowed of him $500, at S per cent. How long shall I keep it, to balance the faTor ? -z:^^ K S-x -^ ^^ 7«? X ^K ' Ans. 1|^ years, b'o. A fox is pnrsned by a greyhoimd, and is 60 of her own leaps before him. The fox makes 9 leaps while the greyhoiind makes bnt 6 : but the latter in 3 leaps goes as far as the former in 7. How many leaps does the greyhound make before he catches the fox ? Ans. The greyhonnd makes 72 leaps, and the fox lOS. 57. A gentleman gave in charity $-r6 : a part thereof in eqnal portions to five poor men, and the rest in equal portions to 7 poor women, Xow. a man and a woman had between them $8. What was given to the men. and what to the women ? Ans. The men receired $25, and the women $2L 5S. A man has two farms, and his stock is worth $18o. !Now, the stock and his first farm is worth once and two-sevenths the value of the second farm, but the stock and the second farm is worth once and five-eiirhths the value of the first faim. "What is the value of each &rm ? Atis. First farm, $384; second farm, $441. 59. A certain clock has an hour hand, a minute hand, and a second hand, all turning on the same centre. At 12 o'clock all the hands are together, and point at 12, How long will it be before the second hand will be between the other two hands, and at equal distances from each ? Also, before the minute hand will be equally distant between the other two hands ? Also, before the hour hand will be equally distant between the other two hands ? Jjis, 60^^^j seconds, i31f|4 seconds, 594-f seconds. GO. TThat ntimber is that, the treble of which, increased by 12, t'^hnll as much exceed 54 as that treble is less than 144 ? Ans. 31. EQUATIONS 89 SECTION IX. EQUATIONS OF THE FIRST DEGREE, CONTAINING TWO UNKNOWN QUANTITIES. Art. 149. When the problem contains two unknown quantities, there must be two independent equations involving them ; and from them an equation may be deduced, which shall contain only one of the unknown quantities. The process by which one of the unknown quantities is thus removed is called elimination; and this may be performed in three ways. First, by Addition and Subtraction. Second, by Comparison. Third, by Substitution. 150. Elimination by addition and subtraction. EXAMPLES. 1. Given j p , ;: PY ( to find the value of a; and y. 1. By first condition, 2. By second " 3. Multiplying 1st by 2, 4. Multiplying 2d by 1, 5. Subtracting 3d from 4th, 6. Dividing 5th by 9, 7. Multiplying 1st by 5, 8. Multiplying 2d by 2, 9. Adding 7th and 8th, 10. Dividing 9th by 27, VEKIFICATION. Sx—2y= 11. Qx-\-^y= 67. 6a;— 47/= 22. Qx-{-5y= 67. 9y=z 45. y= 5- lDz—10y= 55. 12a:+10z/=134. 272:=189. x= 7. 3x7—2x5=21-10=11. 6x7+5x5=42-1-25=67. 8^ 90 ALGEBRA. ( 5:r-[-42/=23 ) 2. Given | Qx—Sv=12 ) *^ ^^^ *^^ value ofx and ?/. 1. By the first condition, bx-\-4:y= 23. 2. By the second, Qx—By= 12. 3. Multiplying 1st by 6, 302;+24?/=:138. 4. Multiplying 2d by 5, 30^—152/= 60. 5. Subtracting 4th from 3d, 39?/= 78. 6. Dividing 5th by 39, y= 2. 7. Multiplying 1st by 3, 15x-\-12y= 69. 8. Multiplying 2d by 4, 24:x—12y= 48. 9. Adding 7th and 8th, 39a:=117. 10. Dividing 9th by 39, VERIFICATION. x= 3. 5x3+4x2=15+8=23. 6x3—3x2=18-6=12. 3. A says to B, if \ of my age were added to | of yours, the sum would be 19^ years. But, says B, if f of mine were sub- tracted from I of yours, the remainder would be 18^ years. Required the sum of their ages. X 2?y 1. By first condition, -+ -^= 194-. 5 o 2. By the second, -^ /= 18^. 7x_2y 3. Clearing the 1st of fractions, Sx-{-10y= 290. 4. Clearing the 2d, 35a:— 162/= 730. 5. Multiplying 3d by 35, 105a;+350?/=10150. 6. Multiplying 4th by 3, 105a;— 48?/= 2190. 7. Subtracting 6th from 5th, 398?/= 7960. 8. Dividing 7th by 398, y= 20. 9. Substituting 20 for y in the 3d, 3a;+200= 290. 10. Transposing and uniting, Sx== 90. 11. Dividing 10th by 3, x=z 30. VERIFICATION. 30 . 2X20 5^ 3 =6+13^=19^ TT* I><20^2Gi-8=18i. EQUATIONS. 91 From the operation of the preceding examples, we deduce the following Rule. Multiply or divide the given equations by such num- bers or quantities as will make ike term that contains one of the unkrwwn quantities the same in each of them; then add or subtract the tivo equations thus obtained^ and there will arise a new equation with only one unknown quantity in it, which may be resolved by Art. 147. 151 1 Elimination by comparison. 4. Given 5 ^ """^^ - . [ to find the values of a: and y. ( bz — z?/=14 ) 1. By the first condition, 1x-\-^y=Vl. 2. By the second, 5a;— 2?/=14. 3. Transposition of the 1st, 2x=.VJ—^y. 4. Dividing the 3d by 2, x= — ^^. 5. Transposition of the 2d, bx=:l^-\-2y. 14+2?/ 6. Dividing the 5th by 5, x= — i— ^. As things which are equal to the same are equal to each other, we therefore infer that — ^— ^, in the 4th, is equal to — ^^—^ in the 6th ; because they are both equal to x. 7. Therefore, 17 -2>y 14+22/ 2 5 • 8. Clearing of fractions, 85- -152/=28+42/. 9. 10. Transposing 8th, Dividing 9th by 19, 192/=57. 2/=3. 11. Substituting 3 for the value ofy in the first equation, we have, 22:=17-9. 12. By reduction, 2a;=8. 13. Dividing 12th by 2, a;=4. 92 ALGEBRA VERIFICATION. 2x4+3x3= 8-f9=17. 5x4-2x3=20-6=14. Hence the following EuLE. Observe which of the unknown quantities is least m- volved^ and find its value in each of the equations, as i?L Art. 148. Let the two values thus found be made equal to each other, and there will arise a new equation, with only one unkrwwn quantity in it, whose value may be found as in Art. 147. 152i Elimination by substitution. 5. Two boys playing marbles, the older said to the younger, if you had three times as many marbles as you now possess, the sum of yours and mine would be 19. But the younger replied, if twice the number of mine were subtracted from four times as many as you have, the number would be 20. Required the number of marbles that each possessed. Let X represent the marbles of the elder ; And y the number of the younger. 1. Then, by the condition of the question, x-\-Zy=\^. 2. And 4a:— 2y=20. 3. Transposing the 1st, :r=19 — 'dy 4. Putting the 3d into the 2d, 4(19— 3?/)— 22/=20. 5. Then, 76— 12?/— 22/=20. 6. Transposing and reducing, 2/=4. 7. Putting the value ofy into the 1st, :c-|-12=19. 8. Transposing and reducing, a:=19 — 12=7. Alls. The elder had 7 marbles, and the younger 4. VERIFICATION. 7+3x4= 7+12=19. 4X7-2X4=28- 8=20. By the above method of operation, we deduce the following BuLE. Find the value of either of the unknown quantities iii that equation in lohich it is least involved ; then substitute this value in the place of its equal in the other equation, and there EQUATIONS. 93 will arise a new equation, with only one unknown quantity in it ; the value ofivhich may he found by Art. 147. EXAMPLES. Ans. 2::=4; y=^. 7. Given < ^ ^ J Required x and y. Ans. a:=5 ; y-=-1. 8. Given \ } Required a; and y. Ans. x=.l ; y=^^. 9. Given < o _or \ I^equired x and ?/. Atis. x=S : ?/=2. _„ „. ( 12x-}-Sy=llQ) T, . , 10. Given jo o \ Required z and y. Ans. x=6 ; y=7. ^^ ^. { llx-\-3y= 124) 11. Given i ^ n rr>{ to find X and ?/. ( 2x—Qy=z—oQ ) ^ Ans. a:=8; y=Vl. -lo n- (9a:+42/=58) ^ , 12. Given s « ' ^^ f to find x and ?/. ( 32;-|-2?/=26 ) ^ ^?w. a;=2; ?/=10. 13. Given ] „ ^ o^^ c to find the value of a; and y. ( ^x—1y= 80 ) -^ Ans. a:=12; y=&. ( lx—2y= 6 ) 14. Given < ^ , _ r. ^ c to find the value of x and y. ( 2x-{-2y=i 24 ) ^ Ans. x=2; 2/=10. 15. Given \ ^ „^ o„ [ to find the value ofx and y. ( b^;— 222/= — 30 ) ^ ^?z^. a:=10; ?/=5. ( 2x4- 3?/= 47 ) 16. Given \ ' -^ ^.^ C to fiii V to find the value of x and y. [ x-{-12y=UQ\ Ans. x=2; y=12. 22. Given 23. Given r^_7^_ 7 10"~ X ;+32/=134 4 ' ^ i ► to find the value of x and y. Ans. a;=56; y=4:0. !, , f to find the value of x and ?/. mx-\-ny=d ) . 3^ — 7ZC «^ — 77ZC Ans. x=- ; y= 1— . om — an an — om 24c. Given a h :in \-Y-=n \c a ■ to find X and y. . ahcm-\-acdn bcdn — abdm Ans. x= :rT-7-- — ; y= ad-\-bc ad-\-bc EQUATIONS 95 25. Given [|-12=|+8 5 ^3 4 ^ to find the value of X and ?/. Am. a:=60; 2/=40. SECTION X. ELIMINATION WHERE THERE ARE THREE OR MORE UN- KNOWN QUANTITIES INVOLVED IN AN EQUAL NUMBER OF EQUATIONS. Rule. Find the values of one of the unhioivn quantities in each of the three given equations, as if all the others ivere knmjvn ; then put the first of these values equal to the second, and either the first or second equal to the third, and there ivill arise two neio equations with only two unhioivn quantities in them, the values of which may he found as in Art. 147, and thence the value of the third. Or, the unknown quantities may be obtained by multiplying each of the three equations by such quantities as ivill make one of their terms the same iji all of them ; then, having sub- tracted any two of these resulting equations from the third, or added them together, as the case may require, there will remain only two equations, which may be resolved by the former rules. Or, toe may find the value of one of the unknown quantities in that equation in which it is least involved, and then substitute this value for that unknown quantity in all the other equxitions, and, proceeding in the same way with these equations, we obtain the other unknown quantities. 1. Griven EXAMPLES. a:-{-37/-[- z=47 -+ ^4- -=10 2^ 3^ 4 > to find the value of x, y and z. 96 ALGEBRA. 4. From the 1st equation, x=41 — y — 2z. 5. From the 2d, x=4:7—Sy— z. 6. From the 3d, ^,-=20—?^- -. 3 2 7. Equal values of x in 4th and 5th, 41— 2/— 2z=47— 37/— z. 8. Value ofy in 7th, 2/=^i^- 9. Equal values of x in 4th and 6th, 41-2/-2z=20-?^-^. ^ 3 2 10. Value of 2/ in 9th, 2/=63-^. 11. Equal values of 2/ in 8th and 10th, ^±5=63——. 12. Keducing, z=12. 13. Substituting for z its value in 8th, ?/= — - — =9. 14. Substituting for y and z their values in 4th, 2:=41— 9— 24=8. 1 / 5a:+4?/-2z=28 \ ^ ^ , on- o)in o \ A on (to nnd the value of x, y, 2. Given 2 < 10a: — 6?/+4z=30 > , -^ Q / o . a \ and 2r. 3 \ Zx-\- y — z= 9 ; Subtracting the 2d from twice the 1st, we have, 4. 142/— 8z=26. Subtracting the 2d from 5 times the 3d, 5. Il2/-9z=15. Subtracting 14 times the 5th from 11 times the 4th, 6. 382=76. 7. 2=2. Substituting for z its value in the 5th, 8. Il2/-18=15. 9. ?/=3. EQUATIONS. 97 Substituting for y and z their values in the 3d, 10. 2.T-f 3— 2=9. 11. a:=4. { Zx— y-2z= ^ 3. Given < 6z-}-2?/-|-3z=45 > to find x, y, and z. I 4:X-\-Sy— ^=-31 ) A71S. 2:=4; y=Q; z=o. / 8a;-%-7z=-36 ^ 4. Given < 12:c — ?/— 3z= 36 > to find x, y, and z. ( Qx—2y— z= 10 ) A71S. x=4:; y=G', z=2. r 7x-\-4:y— z= 7S \ 5. Given < 4x — by—Szz= — 21 > to find x, y, and z. ( a:— 3z/-42=— 37 ) Am. x=S; y=:7 ; z=Q. / x^y=SO \ 6. Given < x-\-z=2b > to find x, y, and z. ( y-i-z=W ) Am. a:=20; y=\^\ zz=b. ( Sx—4:y=:24:— z \ 7. Given < 6a: -j- y= 2:-{-84 > to find x, y, and z. ( a;-f80=32/+4z ) ^7w. a:=12; 2/=20; z=8. f ,^_^=23" 2^3 4 8. Given -j -—=^-4 — =12 \-io find x, y, and 2r. 3 4^ 2 ' U^2 3 Am. a;=36; 2/=24; z=12. C3w-f x-{-2y— z=22^ J 4^;— ?/+3z=35 9. Given i . , ^ / .^ ^ 4tu-\-2»x—2y =19 Uw _}-42/-f-2z=46 to find z^, a:, ?/, and z. yl?;„«. 7/=4; 2;=:5 ; ?/=:6 ; z=7. 98 ALGEBRA. EQUATIONS OF THE FIRST DEGREE, CONTAINING SEYERAL UN- KNOWN QUANTITIES. EXAJSIPLES. 1. A says to B and C, give me half of your money, and I shall have $55. B replies, if you two will give me one third of yours, I shall have $50. But C says to A and B, if I had one fifth of your money, I should have $50. Required the sum that each possessed. Am. A=$20, B=$30, C=$40. 2. A merchant has three kinds of sugar. He can sell 3 lbs. of the first quality, 4 lbs. of the second quality, and 2 lbs. of the third quality, for 60 cents ; or, he can sell 4 lbs. of the first quality, 1 lb. of the second quality, and 5 lbs. of the third quality, for 59 cents ; or, he can sell 1 lb. of the first quality, 10 lbs. of the second quality, and 3 lbs. of the third quality, for 90 cents. Required the price of each quality. A71S. First quality, 8 cents per lb. ; second, 7 cents ; third, 4 cents. 3. A gentleman's two horses, with their harness, cost him $120. The value of the worst horse, with the harness, was double that of the best horse ; and the value of the best horse, with the harness, was triple that of the worst horse. What was the value of each ? Atis. Harness, $50 ; best horse, $40 ; worst, $30. 4. Find three numbers, so that the first with half the other two, the second with one third of the other two, and the third with one fourth of the other two, shall each be equal to 34. A71S, 10, 22, and 26. 5. Find a number of three places, of which the digits have equal difi"erences in their order ; and, if the number be divided by half the sum of the digits, the quotient will be 41 ; and, if 396 be added to the number, the digits will be inverted. A?is. 246. EQUATIONS. 99 6. A farmer has a large box, filled with wheat and rye ; seven times the bushels of wheat is equal to four times the bushels of rye, wanting 3 bushels ; and the quantity of wheat is to the quantity of rye as 3 to 5. Required the bushels of wheat and the bushels of rye. Ans. Wheat 9 bushels, rye 15 bushels. 7. A says to B, if 7 times my property were added to -f of yours, the sum would be $990. B replied, if 7 times my prop- erty were added to ^ of yours, the sum would be $510. Re- quired the property of each. Atis. A's, $140 ; B's, $70. 8. If j- of A's age were subtracted from B's age, and 5 years added to the remainder, the sum would be 6 years ; and if four years were added to -i of B's age, it would be equal to -j^L of A's age. Required their ages. Atis. A's, 98 years ; B's, 15 years. 9. "What fraction is that, if 1 be added to its numerator, its value is ^ ; or, if 1 be added to its denominator, its value is |- ? Am. /^. 10. A says to B, if ^ the difference of our ages were sub- tracted from my age, the remainder would be 25 years. B replies, if ^ of the sum of our ages were taken from mine, the remainder would be -^ of yours. Required their ages. Ans. A's, 30 years; B's, 20 years. 11. There are two numbers, and if ^ of their difference were taken from 4 times their sum, the remainder would be 62 ; but the difference of their sum and difference is equal to f of the larger number. Required the numbers. Atis. 12 and 4. 12. Three men reckoning their money, says the first, if $100 were added to my money, it would be as much as you both possess. Says the second, if $100 were added to my money, I should have twice as much as you two have. Says the third man, if $100 were added to mine, I should have three times as much as you both have. How much money had each man ? Am. First, $dj\, second, $45y\, third, $63^^^. 13. A, B and C, speaking of their ages, A said that the sum 100 ALGEBRA. of their ages was 90. B replied, that if his age were taken from the sum of the other two, the remainder would be 30. G said, if his age were taken from the other two, the remainder would be ^ his age. Required their ages. Am. A's, 20; B's, 30; C's, 40. 14. There are 4 men, A, B, C and D, the value of whose estate is $14,000 ; twice A's, three times B's, half of C's, and one fifth of D's, is $16,000 ; A's, twice B's, twice C's, and two fifths of D's, is $18,000 ; and half of A's, with one third of B's, one fourth of C's, and one fifth of D's, $4000. Re- quired the property of each. Ans. A's, $2000; B's, $3000; C's, $4000; D's, $5000. 15. Find four numbers, such that the first, together with half the second, may be 357 ; the second, with 4- of the third, equal to 476 ; the third, with ^ of the fourth, equal to 595 ; and the fourth, with -i- of the first, equal to 714. Ans. First number, 190; second, 334; third, 426; fourth, 676. 16. If I were to enlarge my field by making it 5 rods longer and 4 rods wider, it would contain 240 square rods more than it now does ; but, if I were to make its length 4 rods less, and its breadth 5 rods less, its contents would be 210 square rods less than its present surface. What are its present length, breadth, and contents ? A71S. Length, 30 rods ; breadth, 20 rods ; contents, 60-0 square rods. 17. A person exchanged 12 bushels of wheat for 8 bushels of barley, and £2 16^. ; ofiering, at the same time, to sell a C3rtain quantity of wheat for an equal quantity of barley, and £3 155. in cash, or for £10 in cash. Required the prices of the wheat and barley per bushel. Ans. Wheat at 8 shillings, barley at 5 shillings, per bushel. 18. A farmer, having 89 oxen and cows, found, after he had sold 4 oxen and 20 cows, he had 7 more oxen than cows. What number had he of each at first ? Ans. 40 oxen and 49 cows. 19. A and B driving their turkeys to market, A sa^s to B, EQUATIONS. 101 give me 5 of your turkeys, and I shall have as many as you. B replies, but give me 15 of yours, and then yours will be f of mine. What number of turkeys had each ? Ans. A 45 and B 55 turkeys. / 20. It is required to find two such numbers, that if i of the first be added to -^ of the second, the sum shall be 25 ; but, if ^ of the second be taken from |- of the first, the remainder will be 6. Ans. 48 and 36. 21. What fraction is that, if 5 be added to its numerator, its value is 2, but if 2 be added to its denominator, its value is ^ ? A71S. |. 22. B says to C, if 3 years were taken from your age and added to mine, I should be twice as old as you. C replies, if 3 years were taken from your age and added to mine, our ages would be the same. Required their ages. A?is. B's age 21, C's age 15 years. 23. It is required to find two numbers, so that § of the first ' added to f of the second shall be 15|, and if | of the second be subtracted from |- of the first, the remainder shall be 5-f |-. Am. 10 and 12. 24. It is required to divide 50 into two such parts that f of the larger shall be equal to f of the smaller. Ans. 32 and 18. 25. A gentleman, at the time of his marriage, found that his wife's age was to his as 3 to 4 ; but, after they had been married 12 years, her age was to his as 5 to 6. Bequired their ages at the time of their marriage. A71S. The man's age 24, his wife's 18 years. 26. A farmer hired a laborer for ten days, and he agreed to pay him 12 shillings for every day he labored, and he was to forfeit 8 shillings for every day he was absent, and he received at the end of his time 40 shillings. How many days did he labor, and how many days was he absent ? jhis. He labored 6 days, and was absent 4. 0# 102 ALGEBRA. 27. A gentleman bought a horse and chaise for $208, and | of the cost of the chaise was equal to f the price of the horse. What was the price of each ? Ans. Chaise, $112; horse, $96. 28. A and B engaged in trade, A with $240, and B with $96. A lost twice as much as B ; and, upon settling their accounts, it appeared that A had three times as much remaining as B. How much did each lose ? ^-9/. -x Ans. A lost $96, and B lost $48. 29. Two men, A and B, agree to dig a well in 10 days, but, having labored together 4 days, B agreed to finish the job, which he did in 16 days. How long would it have required A to complete the labor ? - - Ans. 9§ days. 30. A merchant has two kinds of grain, one at 60 cents per bushel, and the other at 90 cents per bushel, of which he wishes to make a mixture of 40 bushels that may be worth 80 cents per bushel. How many bushels of each must he use ? Am. 13^ bushels of 60 cents, 26| of 90 cents. 31. A farmer has 30 bushels of oats, at 30 cents per bushel, and which he would mix with corn at 70 cents per bushel, and barley at 90 cents per bushel, so that the whole mixture may consist of 200 bushels, at 80 cents per bushel. How many bushels of corn, and how many of barley, must he mix with the oats ? A71S. 10 bushels of corn, and 160 of barley. 32. A drover sold 6 of his oxen and 8 of his cows, and he then found he had twice as many oxen as cows. But after he had sold 10 more of his oxen, he found he had 2 more oxen than cows. How many had he of each at first ? Ans. 30 oxen and 20 cows. y 33. Four times the larger of two numbers is equal to six times the less, and their sum is 15. Required the numbers. A71S. 9 and 6. 34. A and B can perform a piece of work in 6 days, A and C in 8 days, and B and C in 12 days. In what time would each NEGATIVE QUANTITIES. 103 of them perform the work alone, and how long would it take them to perform the work together ? A71S. A would do the work in 9| days, B in 16 days, C in 48 days, A, B and C together, in 5^ days. 35. A gentleman left a sum of money to be divided among his four sons, so that the share of the oldest was ^ of the shares of the other three, the share of the second 4- of the sum of the other three, and the share of the third \ of the sum of the other three ; and it was found that the share of the oldest exceeded that of the youngest by 814. What was the whole sum, and what was the share of each person ? A?is. "\Miole sum, S1'20 ; oldest son's share, SIO ; second son's, S30 ; third son's, S24; youngest son's, S26. SECTION XI. NEGATIVE QrAXTITIES. Akt. 1.53. The student will sometimes find that, on account of his misconception of the question, he has added a quantity which should have been subtracted, or that he has subti-acted a quantity which should have been added. This may be illustrated by the following EXAMPLES. 1. The length of a ceiiain field is a, and its breadth is b; how much must be added to its breadth that its contents may be m ? Let X = the quantity to be added to its breadth. Then b-\-x=ih.e breadth. And a{b-\-x)=m, the contents, ab-\-ax=m. m b-\-x=-. a m X=z 0. a 104 A L G t: B 1{ A . 2. Let the length of the field be 10 rods, and its breadth 6 rods ; how many rods must be added to its breadth, that the contents of the field may be 80 square rods ? Let X = the quantity to be added to the breadth. Then, by the above formula, x= b=~—Q==2 rods, the quantity to be added. VERIFICATION. 10x6+2=80 square rods. 3. Let the length of the field be 10 rods, the breadth 8 rods; it is required to find what quantity must be added to the breadth that the contents may be 60 square rods. By the formula, 77Z 60 x= Z»=:t7:— 8= — 2 rods. a 10 We perceive by the above that it is — 2 rods which are to be added, and not -\-2 rods; but we add quantities together in Algebra by simply writing them one after the other, with their respective signs, so that — 2 added to -|-8 becomes 8 — 2=6, the answer, which is the same as subtracting -\-2 from -|-8. And, in general, adding a minus quantity brings the same result as subtracting a plus quantity of equal value, and vice versa. VERIFICATION. 10 X^ — 2=60 square rods. Ans. 4. Suppose the field to be 10 rods long and 8 rods wide, it is required to ascertain how much must be subtracted from its width that its contents may be 60 square rods. To subtract a minus quantity is the same as to add a plus quantity. If, therefore, we change the sign of x in the formula first obtained, x will then express how much is to be subtracted. Thus, — x= 3, a or, x=0 =:b — :r^=2 rods. a 10 NEGATIVE QUANTITIES. 105 VERIFICATION. 10x^—2=60 square rods. 6. If the field were 10 rods long and 8 rods wide, how many rods must be taken from its width that its contents may be 100 square rods ? By the formula, x=^b = b — TTr= — ^ rods. a 10 That is, — 2 is to be subtracted from -{-8 ; or, as we perform subtraction in Algebra by changing the sign of the subtrahend, and thus annexing it to the minuend, we have 8— (— 2)=8+2=10; so that, in general, subtracting a minus quantity is the same as adding a plus quantity of equal value. 6. John Smith, at the time of his marriage, was 50 years old, and his wife was 40. When will his age be twice that of his wife ? Let a:=the time. Then, 50-}-a;=2x40+^. 50+:r=80+2^. And, a:==50— 80=— 30 years. As the answer is — 30 years, it is evident that he is not myw twice as old as his wife, but 30 years ago his age was twice hers. VERIFICATION. 50-30=40—30x2. 20=20. 7. J. Jones is 40 years old, his wife 30. AVhen will they both be of the same age ? Let a:=the time. Then, 40+ x=.Z^-\-x. 40—30= x—x. And, 10=0. The answer being zero, it is certain they never will be of the same age, but that one will always be 10 years older than the other. 106 ALGEBRA. 8. What fraction is such, that if 2 be added to its numerator its value is -I, or if 2 be added to its denominator its value is ^ ? A71S. -12' 9. What fraction is such, that if 7 be added to the numerator its value is nothing, but if 2 be added to its denominator its value is infinite ? . — 7 Atis. — ^. 10. What fraction is such, that, if 4 be added to its numer- ator its value is nothing, but if 10 be subtracted from its denominator its value is 1 ? — i^. THE COURIERS. P 1. Two couriers set out at the same time from A and C, and travel towards each other until thej meet. The distance from A to G is m miles. The first courier travels a miles per hour, and the second b miles per hour. How far from A and C will they meet ? A B C D Let us suppose them to meet at B. And let x = the distance A B. And y = the distance B C. Then x+y = A G = m. As the first travels x miles at the rate of a miles per hour, to find the time he will travel this distance, we say, X As a miles : x miles : : 1 hour : - = the time the first cou- a rier will travel the distance A B. y And, as b miles : y miles : : 1 hour : ^ hours = the time ^ b the second courier will travel the distance B C. As both couriers set out at the same time, and arrive at the same time at C, Therefore -=7-. a b And x—^, NEGATIVE QUANTITIES. 107 If we substitute this value of x in the j&rst equation, we have ay And 'iy-\-by=bm. bm Hence y = a-\-b Substituting this value of y in the equation x=-^, we have a bm abm am b a-\-b ah-\-h' a+b The values of x and y in the above equation are both posi- tive. Therefore, whatever value we may assign to «, b and m, it will answer the conditions of the question. This may be illustrated by the following question : 2. Two men, A and B, set out from two places, distant from each other 144 miles, and travel towards each other. A goes 12 miles an hour, and B four miles an hour. How far must each travel before they meet ? By the above formulae, x= — r-T= -..->,. = 108 miles, the distance A travels. a-\-b 12-f4 bm 4x144 And 7/ = — --==-————== 36 miles, the distance B travels. -^ a-\-b 12-1-4 VERIFICATION. 108 + 36=144 miles. 3. If the couriers were to set out at the same time from A and B, and travel towards C, both going the same direction, the first going a miles per hour, and the second b miles per hour, and the distance A B being m, how far would each travel before they meet, suppose at a point C ? F A B C D Let X = the distance A C. And y = the distance B C. Then x-y=AC—BC=AB=z7n. 108 ALGEBRA. By performing the same operation as in the first question, "w^e find a 3' and x=-f-, o Therefore ~- — 7iz=m. b -^ And ay — ly=.hin. bm Whence y-. a — b Substitute this last value of y in the former equation, and we have ay a bin dbm am b b a — b ab — l? a — b Here it is evident that the values of x and y will not be positive, unless a be greater than b ; or, in other words, unless the courier which sets out from A travels faster than the one that sets out from B, he will never overtake him. 4. Suppose the first courier to travel 9 miles per hour, and the second 6 miles per hour, and the distance A B to be 18 miles, and it was required to find how far each would travel before the one overtook the other. Then «=9, ^=:6, and 7?z=18. And, by the first formula; am 9x18 p ^ -i i t , r. • ii -=o4 miles, the distance the first courier woukl a—b 9—6 travel. " And, by the second formula, bm 6x18 „^ ., , ,. , , . 2/= ;=-?T — 77-=oo miles, the distance the second courier a — b 9 — o would travel. We perceive, by the above operation, that the point C, where NEGATIVE QUANTITIES. 109 the couriers meet, is 54 — 36=18 miles further from A than B is, which is equal to the distance ?//. 5. Again, let a=Q, ^=9, and 7?z=18 ; or, suppose the first courier sets out from A and travels 6 miles an hour, and the second sets out at the same time from B and travels in the same direction towards C at the rate of 9 miles per hour. What distance will each travel before thej meet ? Bj the first formula, am 6x18 . ,, ^ ,, \ = — rfo miles, the first travels. d—b 6—9 By the second formula, bm 9x18 C-. M 1 x= ,=-7^ — 7^= — &4 miles, the second travels. a — b — 9 Here the values of x and y are both negative. Now, how shall we interpret this result ? AYhat is the meaning of the negative sign, in this case ? To understand this, we must observe that we began by sup- posing the parties to be travelling towards C, and any motion in this direction would have been indicated in this example, as it has been in the preceding examples, by the sign -f-- But, when the sign -\- is taken to indicate motion in one direction, the opposite sign — must indicate motion in the opposite direction. Hence the minus sign, resulting as above, indicates that the parties, in order to meet, must travel, not towards C, as we at first supposed, but in the opposite direction, towards F, a point 36 miles from A, and 54 miles from B, where they will meet. 6. Again, let fi^=6, b=Q, and ??z=18; or, we will suppose the couriers both to start at the same time from A and B, and both to travel in the same direction towards C, and both trav- elling at the same rate of 6 miles per hour, the distance A B being 18 miles. What distance will each travel before they meet ? 10 > 110 ALGEBRA. By the first formula, x=. r, or =-7r» •^ G^ — o a — a U _6X18_108 °^ ^— "0=^" "0"* By tne second lormula, ?/= -, or =-7r> *' a — h a — a 6x18 108 As both couriers are travelling in the same direction, and at the same rate, it is certain they will never meet, but the dis- tance between them will continue the same. ■«v. mi ^ 1 • '3;?7^ 108 154. Thereiore, the expression —t~ or — tt-j or any quantity with zero for a denominator, is the symbol for infinity ; for it is well known that the value of a fraction depends on the number of times the numerator contains the denominator, or the number of times the denominator may be taken from the numerator, until nothing shall remain. It is certain that, if a be greater than 3, however small the difference, the couriers will eventually meet ; but, if the differ- ence between a and h be less than any assignable quantity, then X and y may be considered infinite. Again, let a=.h^ and 7?2=0. am «X0 Then x=i And y a—b 0' Im ^XO a—b ~0' From the above we infer that x and y are equal, and that each is equal to the other. Thus, x=.x. This is an identical equation, and the values of the unknown quantities cannot be known by it. And, as 7?z=0, it is evident, that as both couriers start from the same point, and travel at the same rate, and in the same NEGATIVE QUANTITIES. 11.1 direction, they will always be together, and therefore cannot meet. We say, therefore, that the ^, in this case, is an expression of an Indeterminate Quantity, because that x and y may be any quantities whatever. But it is not true that the expression % is always the sign of an indeterminate quantity. 155. In fractions, when the numerator and denominator have a common factor, and which in some cases becomes zero, and makes the fraction assume the form of §, but which, without that factor, has a definite value, the expression is not inde- terminate. The following fractions are examples of this kind : n(m — 71) Now, if 77«=?7, the value of the quantity is §. But, on examination, we perceive that both the numerator and denominator have the common factor m — n. Therefore, by dividing both terms of the expression by m — n^ lTl(.lTi-\~7l\ it becomes — ^^ , which, if 77z=7z, is equal to 2w. n ^ x—\ The value of the expression -, if we divide both terms by x — 1, is 1 ; but, if a;=l, the value is a Again, let a:= m — n Then, if ??z=7z, the value of a:=^. But, if we divide both terms by the common factor m — ?i, its value is ')r^-\-mn-\-r^^ and then, on the supposition that m=ni its value will be 3w^ INDETERMINATION. 156. In investigating the theory of indetermination, we find many curious results and apparent absurdities. This will appear evident by investigating the following prob- lems. 112 ALGEBRA. 1. If it be admitted that a=l and x=l, it may be shown that 1 is 2 and 2 is nothing, or any assignable quantity. Let a=x. Multiplying both terms of the equation by x, we have ax=x\ Subtracting a^ from both members, Q o O ax — a=x' — a. Resolving both terms into factors, a[x — a) = {x — a)[x-\-a). Dividing by x — a, a=x-\-a. Substituting a for its value x, a=a-\-a. Dividing both terms by c, 1=1-1-1=2. Again, we have found above that x"^ — a^=:ax — d^. Dividing both terms by the common factor x—a, we have ax — a^ x-j-a= . X — a Now, as X and a by the supposition are each equal to 1, we see that ivi— r 14-1= ^^ ^, ^ 1—1 And 2=g. Thus it appears that we have dearhj proved that 1 is 2, and 2 any assignable quantity. Q. E. D. The fallacy is this, that if nothing be divided by nothing the quotient is any assignable quantity. This principle may be further illustrated by considering the following identical equation. Let 16=16. Resolving into terms, 12-|-4=12-}-4. Transposing, 4—4=12—12. Resolving second term into factors, 4 — 4=3(4 — 4). Dividing by 4 — 4, 1=3. NEGATIVE QUANTITIES. 113 Thus it appears that 1 is 3 ; and, in the same manner, a unit may be proved to be any definite number. From various articles in the foregoing section, we infer the following : 1. If zero be multiplied by zero, or any assignable quantity, the product will be zero. 2. If zero be divided by zero, the quotient may be zero, or any assignable quantity. 3. If zero be divided by any quantity, the quotient will be zero. 4. If any quantity be divided by zero, the quotient will be infinity. 5. If any quantity be added to or taken from infinity, the result will be infinity. 6. If zero be multiplied by infinity, the product may be any quantity. 7. K infinity be divided by infinity, the quotient may be any assignable quantity. 8. One infinity may be infinitely larger than another. SECTION XII. Theorem I. AuT. 157. If a binomial be multiplied into itself, the pro- duct will be equal to the sum of the squares of both terms, plus twice the product of the terms. XoTE. — The theorems in the following section may be illustrated h^ diagrams, and it would be well for the pupils to draw them. When a number or quantity is multiplied into itgelf, the product is a square. EXAMPLES. 1. Multiply a-}-b into itself. ALGEBRA. VERIFICATION. 8+4=12 12 8+4=12 12 64+32 144 32+16 114 a-\-b a^-\-ab ab^b'' d'J^^ab^b^ 64+64+16=144. We perceive, by the above operation, that the square of any Hnomial may be readily obtained. 2. Multiply Za-\-'lb into itself. 3aX3a+2x3flX2^+23x23= 9a^+12a3+43l 3. Multiply x-\-2y into itself. Ans. x'+ixy-^-^y'. 4. Multiply dab-\-?n into itself. Atis. 5. Multiply by-\-4:X into itself. Aiis. 6. Multiply 2m-\-o?i into itself. A7is. 7. Multiply 7fZ+2e into itself. A?is. 8. Multiply 2?^+3^^? into itself. Arts. 9. Multiply 5«-+23 into itself. Ajis. 10. Multiply 1+^ into itself. A?is. 11. Multiply 3+^ into itself. A?is. 12. Multiply 2+J- into itself. A?is. Theorem II. 158. If the sum of two numbers or quantities be multiplied by their difference, the product will be equal to the difference of their squares. Multiply a-{-b into a — b. a+b a—b 8—4= 4 4 VERIFICATION. a-\-b . 8+4=12 12 ci'-^ab 64+32 48 -ab—b'' —32-16 cr -b"" 64 -16=48. NEGATIVE QUANTITIES. 115 Theorem III. 159. If the difference of two numbers or quantities be mul- tiplied into itself, the product will be equal to the sum of their squares, minus twice their product. EXAMPLES. . Multiply a- a — b a — b -b into a- -b. VERIFICATION. 12-8= 9 12-3= 9 9 9 a^ — ab -ab-\-b'' 144-36 -36-f9 81 a^—2ab-\-b\ 144—72+9=81. 2. Multiply 2>a—2b into 3a— 23. Ans. ^d'—12ab-\-^b\ 3. Multiply bm — ?i into bm — ?^. 4. Multiply Aab—x into ^ab—x. 5. Multiply ^a'—b^ into ^d'—b^ 6. Multiply a;* — y^ into x'^ — if. Note. — If the square of the difference of two numbers be subtracted from the square of their sum, the remainder will be equal to four times their product. Thus, {a-\-hf-{a-bf={dJ^2ab^b'') — [d—2ab-\-b'') =4.ab. Theorem IV. 160. If twice the product of two quantities be subtracted from the sum of their squares, the remainder will be equal to the square of their difference. (^a^^P)—2ab=ar—2ab-\-b\ But this expression, by Problem 3d, is the square of their difference. VERIFICATION. Let 9 and 3 be the two numbers. Then (92+32)-(2x9X3)=(9— 3)^ (81+9)-(54)=36. 90-54=36. 36=36. 116 ALGEBRA. Theorem Y. 161 1 If there be two quantities, one of which is divided into any number of parts, the product of the two quantities will be equal to the product of the undivided number into the several parts of the divided number. Let the two quantities be a and 3, and let h be divided into three parts, c, c?, and e. Then 3=c+^+e. And ab^=^ac-\-ad-\-ae, VERIFICATION. Let the two numbers be 12 and 10, and let 10 be divided into the parts 5, 3, and 2. Then 10=5+3+2. And 12X10=12X5+12><3+12X2: 120=60+36+24. 120=120. Theorem VI. 162. If any quantity be divided into two parts, the square of this quantity will be equal to the sum of the products of this quantity into its two parts. Let a represent the quantity, and h and c the parts into which it is divided. Then «=:3+c. And aX«^=«^(^+c). VERIFICATION. Let 12 be divided into two parts, 9 and 3. Then 12=9+3. 12x12=12(9+3). 144=108+36=144. Theorem VII. 163. If any quantity or number be divided into two parts, the product of the whole and one of the parts will be equal to THEOREMS. 117 the product of the two parts, plus the square of the aforesaid part. Let a represent the whole quantity, and h and c the parts. Then a^=^h-\-c. Multiplying both sides of the equation by 3, we have YEEIFICATION. Let 12 represent the number, and 9 and 3 the parts into which it is divided. Then 12=9+3. Multiplying both parts of the equation by 9, we have 9x12=9(9+3). 108=81+27=108. Theorem A^III. 164i If any quantity be divided into two parts, the square of the whole quantity will be equal to the squares of the two parts, plus twice their product. Let a represent any quantity, and h and c the parts into which it is divided. Then a=.h-\-c. By squaring both sides of the equation, we have VERIFICATION. Let 9 be divided into two parts, 6 and 3. Then 9=6+3. By squaring both parts of the equation, we have 9-=(6+3)l 81=36+3G+9=8L Theorem IX. 165. If any number or quantity be divided into two equal parts, and into two unequal parts, the square of one of the equal parts will be equal to the product of the two unequal 118 ALGEBRA. parts, plus tlie square of half the difference of the two unequal parts. Let a represent one of the equal parts, and h and c the two unequal parts. Then CL=-^' And 2a=3-]-c. We now add — 43c to both sides of the equation. And 4«2_43c=— 43c4-3'+2ic+cl d^— bc=. 4 52_23c+c^ 4 VERIFICATION. Let 12 be divided into two equal parts, 6 and 6 ; and into two unequal parts, 9 and 3. And 36=27+?14i±^. ' 4 36=27+9=36. Theorem X. 166. If any quantity, 2a, be divided into two equal parts, and if any quantity, b, be added to 2<2, the product of 2a-{-b into />, plus the square of a, will be equal to the square of a-\-b. Then, by the proposition, 2a-\-b will be the whole quantity. Multiplying by b, we have b{2a-\-b)=2ab+b\ By adding a^ to each member of the equation, we have b{2a+b)+a'=a'+2ab-{-b\ Therefore b{2a+b)-\-d'={a+b)\ THEOREMS. 119 VERIFICATION. Let a=10, and b=2. Then 2(2xl0 + 2)+10'^=(10+2f ; 144 = 144 Theorem XI. 167. If any quantity be divided into two parts, the square of this quantity, and one of the parts, will be equal to twice the product of the whole quantity and that part, plus the square of the other part. Let the whole quantity be denoted by a, and the parts by b and c. Then a=b-\-c. And a — c=b. d'-\-c'=2ac-\-b\ VERIFICATION. Let 12=3+9. Then 12'^+9^=2xl2x9 + 3^ And 144+81=216+9. 225=225. Theorem XII. 168. If any quantity be divided into any two parts, four times the product of the whole quantity into one of the parts, plus the square of the other part, will be equal to the square of the quantity which consists of the whole and the first-mentioned part. Let a represent the quantity, and b and c the two parts into which it is divided. Then a=i+c. Multiplying both members of the equation by Ab, we shall have 43xa=43X(^+c). 4a5=4i^+45c. 120 ALGEBRA. We now add & to both members. Or ^ab-^e={c^nf. VERIFICATION. Let a=12, and b=9, and c=3. Then 12=9+3. And 4x12x9+3^= (3+2x9)'^. / ? _ ^ t - '^ ^j 441 = 441 ^^ Theorem XIII. 169. If any quantity be divided into two equal parts, and also into two unequal parts, the sum of the squares of the two unequal parts will be double the square of half the quantity, plus twice the square of the quantity which consists of the diiference oetween half the quantity and the larger of the unequal parts of the quantity. Let 2a represent the quantity, and a = one of the equal parts, and b = half the difference between the equal and un- equal parts. Then a-\-b =: the larger part. And a — b = the less part. And [a-\-by'-{-{a — by = the sum of their squares. But (d'-{-2ab-\'b'')-i-[ar—2ab+b')=2a'-\-2b\ And 2a'^-\-2b^ = twice the square of half the quantity, plus twice the square of half the difference between the equal and unequal parts ; that is, the difference between half the quantity and the larger of the unequal parts. VERinCATION. Let 10=7-1-3; 10-^2=5; 7—5=2. Then (5-}-2)^+(5-2)'=2x5'+2x2^ And 49 + 9=50+8. 58=58. INVOLUTION. 121 Theorem XIV. 170. If any quantity, 2a, be divided into two equal quantities, a and a ; and, if any quantity, 3, be added to 2«, the square of 2a+^, plus the square of ^, will be equal to twice the square of (2, plus twice the square of «+^- Now (2a+3)2+3-=4a^+4a^+3^+^'=4a-+4a3+2i\ But 4c-+4a^4-2^»'=2a-+2(fl+i)-l Therefore (2a+3)^4-3==2a^+2(a+3)l VERIFICATION. Let G=10; and 3=4. Then (2x10+4)^+4^=2(10'-) -1-2(10+4)1 And 576+16=200+392. Therefore 592=592. SECTION XIII. INVOLUTION. AuT. 171. Involution is the raising of powers from any pro- posed root ; or, the method of finding the square, cube, biquad- rate, &c., of any given quantity. 172. A. pmver is the product of any quantity multiplied into itself a certain number of times, and the degree of the power is denoted by an exponent written over the root. Thus a^ is the third power of a, and a is the root. 173. The exponent^ or index., shows how many times the root has been used as a factor. Thus, G^X^X«X«=^S and a:X^=^^- 174. When a quantity is written without any index, its index is uniformly considered a unit. Thus, a=a^, and x=x^. There- 11 122 ALGEBRA. fore, to raise any quantity to any required power, the pupil will see the propriety of the following Rule. Multiply the index of the quardity by the index of the power to which it is to be raised ^ and the result will be the paiver required. Or, multiply the quantity into itself as many times, less one. as is denoted by the index of the power, and the last product will be the answer. 175. When the sign of any simple quantity is -|-, all the powers of it will be + ; and when the sign is — , all the even powers will be -}-, and the odd powers — , as is evident from multiplication. EXAMPLES. 1. What is the fifth power of a.? Ans. a^. 2. What is the third power of ax ? Ans. a^x^. 3. Required the square of a^x. Atis. c^t^. 4. Required the cube of — 3a^ Aiu. — 27fl^^ 5. Required the fourth power of —aV^&. Ans. a%^c^^. 2ax^ 4a~x^ 6. Required the square of — ^7—. Ans. . 7. Required the fifth power of 2ab''x\ A7is. S2a'b'°x''. 8. Required the sixth power of f a^z^. Ans. y^/^a'lr'^. 9. Required the third power of 2a~^. Ans. S^r^. 10. Required the fourth power of — 3m~^. Ans. 81m~-^ 11. Required the ?7ith power of a". A7is. a'"". 12. Required the fourth power of 2a;"'. Ans. I62:'''". , Sa'b^ ^ 27a'b' 13. Required the third power of -j — 5-. Ans» „ 3 ,^ . 176. Polynomials are involved by multiplying the quantity by itself as many times, ivanting one, as there are units in the exponent of the power. INVOLUTION. 123 14. Let a-}-b be raised to the fifth power. a-\-b a^-\-ab -{-ab-^-b"- ^aJ^lYz=za^J^1ab^V' a-\-b a?J^2a%-\-ab'' J^a^b^2ab''-^P {a+bf=za?-]-Zd'b-\-Zab''+b^ a-\-b a'^^a?b-]-^a%''-\-ab^ \.a^b-]-^d'b''-\-^aW-\-b^ (^a-\-by=a'-\-4:a?b-\-U''b''-\-^ab^-\-b' a-{-b a'j^4.a'b^QaW-\-4:a%^-^ab' j^a'b-\-^a%^-\-Qa%^^^ab'+b' {a-\-bY=a'-\-^a'b^l^a?b''-\-10a%^-\-bab'+b^ Kequired the third power of a — b. (^a—bf=a—b a — b d^ — ab^ —ab+b'' {a^bf=a^—2ab+b^ a — b a^—2a'b-{-ab'' —a'b+2ab''—b^ {a—bY=a?—U%+Zah^-b^ 1st power. 2d power. 3d power. 4th power. 5th power. 1st power. 2d power. 8d power. 15. Required the fifth power of x — 2y. A71S. x'—10x'y-^^0xy—S0xy-\-S0xy'S2y\ 124 ALGEBRA. 16. Required the third power of a — b-\-l. 17. Required the second power of 2x'^ — Sx-{-4. 18. Required the sixth power ofx — 2. Ans. x'—12x'+QQx'—lQ0x''-{-240x''—ld2x+G4:, 2x-7j-^ 19. Required the second power of . Ans. ^ W—2\bd^\U' 20. Required the fourth power of «'"—«". Ans. tt^"'—4a^'"+"-f6«""'-^'^"—4a"'+''" +«■'". 21. What is the second power of 2x^—ox-\-^ ? Ans. 4x^—\2o?-^\\x'—Zx-\-\. 22. What is the third power of a-|-23— c ? Ans. a^-\-^a^b—^a\-^\2ay'—\2abc^Za(?^W—\2b''c^Uc^ -c\ 23. What is the fourth power of a-\-b-\-c-\-dl Ans. a'-\-4a%^^a%''-^4ab^-^b'-\-4a\-\A2a%c-\-\2ab\-\-4b^c ■\-\aH^Vla%d-Y\-2abH-^Wd^U^c~^\2ab& + We -f \2a^cd ■\-24abcd-\-\2b\d 4- 6aV _|_ i2a^^2 _|_ 532^2 _^ 4^^3 _^ i2ac'^ -f- \2acd^-\-4ad'^-\-4b&-\-\2bc'd^\2bcd^-\-4bd^^c'-^4&d -f 6c*'d^^ + 4c^^+^^ 24. What is the second power of x^-\-27?-\-x-\-2 ? Ans. a;«+4a;^+6:?;^+8:c^+9:c2_}_4^_|_4^ a b a^ b'^ 25. What is the second power of- — -? Ans. —^ — 24--. u a a 26. What is the third power oix} — x — 1 ? Ans. x^—Zx^-\-hx?—Zx—\, 27. What is the third power of a — b — 2c' — d^ ? Ans. a^— 3a'A+3a32— ^3— 6aV+12aJc^— 6^V— 3a'^3+6aW' -Z})'d^-\-Vlac'-\-\2aed^^Zad'—Vlbo' — Ubc'd'- Ud' — 8c' — l2c'd^—Qed'—d\ EVOLUTION. 125 SECTION XIV. EVOLUTION, OR THE EXTRACTION OF ROOTS. Art. 177 1 Evolution is the reverse of involution, being the method of finding the roots of any given quantity. It will, therefore, be necessary to trace back the steps of the operation in involution. Hence, to find any root of a monomial, we adopt the fol- lowing Rule. Extract the required root of the coefficient for the coefficient of the answer^ and the root of the quantity subjoined for the literal part of the answer. 178i If the quantity proposed be a fraction, its root will be found by taking the root both of its numerator and denom- inator. 179. The square root, the fourth root, or any other even root of an affirmative quantity, may be either plus or minus. Thus, A/a^=+a or — a ; and /s/b^=-\-b, or — b. But the cube root, or any other odd root of a quantity, will have the same sign as the quantity itself. Thus /^a^=a ; /^ — a^=—a^ and /s/ — a^-= — a. The reason why -\-a and — a are each the square root of a*^, is obvious; since, by the rule of multiplication, (+a)X(+«) and ( — a)X( — ^) are each equal to cr. 180. In the case of the cube root, fifth root, &c., of a nega- tive quantity, the rule is equally plain ; since, by multiplying, we have (•— o)X(— a)X( — a) = —a^. It may also be stated here that any even root of a negative quantity is unassignable ; or, as it is usually called, imaginary. Thus, a/ — (^ cannot be determined, as there is no quantity, either positive or negative, that, when multiplied by itself, will produce — d^. 11^ 126 ALGEBRA. EXAMPLES. 1. Find the square root of 9a^ Here A/Va^=A/^X/s/a^=^Xci=Ba. Am, 2. "What is the cube root of 8a;^? Here /^'^=/^X/^x^=^Xx=2x. Ans. 3. It is required to find the square root of -^-. Here a^b'^ fj c^lp- ah — — -L — Au'i ^a'b' 4. What is the cube root of ^^„ 3 3 Here — Sa'b' /^X^ 64a« 8a^ ' 16a« 64a« ■^a 4a^"^8^ 128^7 64^«"^"64^ bx^ bx'' 64^« 128^ , &c. 6. What is the square root oi x'—2x^^Zx'—2x-\-l ? Ans. a? — x-\-\, 7. What is the square root Q'ix^—1x^^x''-{-l7?—1x^^\ ? Alts, y? — o?-\-\. 8. What is the square root of a'-^Aa%-^\^a%''^\1db^^W ? Ans. d'^2ab-\-m. 9. Extract the square root of a* — 2a?-\-2d' — ci-\-^. Ans. a' — a+2"- 10. What is the square root of 4aV — \1(]^x^-\-\^o}x'—^(V'x -fa"? Ans. 2ax''—Sa''x+a\ 11. What IS the square root 01 -r- — ■qTT'Q'a • a 2b 130 ALGEBRA. EVOLUTION BY DETACHED COEFFICIENTS. 1. What is the square root of 4:X^—4:X^J^lSx'^—6x-{-91 4—4+13—6+9(2—1-1-3= 4 2x''—x-{-d. 4_1)_-4_|_13 -4+ 1 4-2+3)12—6+9 12-6+9. 2. What is the square root of 9a;'— 242:*+12a;'+16;r^— 16a: +4? 9+0-24+12+16—16+4(3+0-4+2= 9 3a;3+0a;2— 4a;+2= 6+0— 4)+0— 24+12+16 Sx^—4:X-}-2. _24— 0+16 6+0—8+2)12+ 0—16+4 12+ 0—16+4. 3. What is the square root of 4a;«— 4a;5+12a;'+a;'— 6a;+9 ? 4_|_0+0— 4+12+0+1— 6+9(2+0+0-1+3= 4 2a:^+0a;3+0a;2— a;+3= 4+0+0-l)+0 + 0-4+12+0+l 22;^— a;+3. _|-0+0— 4— 0—0+1 4+0+0-2+3)12+0+0-6+9 12+0+0-6+9 The pupil will perceive that the 5th power of x in the second question, and the 3d, 6th and 7th power of x in the third question, are wanting; therefore their place in the operation must be supplied by zero. 4. What is the square root of 4a^-16fi^'+24a'^— 16a+4? Ans. 2^2— 4a+2. EVOLUTION. 131 5. What is the square root of 4x'''—12x^—12x^-\-9x''-]-lSz _|_9? Ans. 2a;^— 3x— 3. 6. "What is the square root of 16x'-\-2Ax'-\-^9x''-{-Q0x-]-100 ? Am. 4x'^-j-3a:4-10. 7. What is the square root of 9a;'— 12a;^+10:c^— 28:c^-f 17x^ — 8:c-|-16? Ans. ox^—2x^-{-x—4. 2 1 8. What is the square root of 771^^+2771— 1 1 — rj Ans. m-\-l . m EXTRACTION OF THE SQUARE ROOT OF NUMBERS. 184. As numbers are not expressed in the same manner as algebraic quantities, it is evident that the same rule for ex- tracting the square root of algebraic quantities will not apply to extracting the roots of numbers without additional con- siderations. But, if the foregoing rule be assisted by the " Method of Pointing," it will enable us to extract the square root of numbers. 185. Since the square root of 1 is 1 ; the square root of 100 is 10 ; the square root of 10000 is 100 ; the square root of 1000000 is 1000, &c., it is evident that the square root of a number of figures less than three must consist of only one figure ; that of a number more than two figures and less than five, of two figures ; that of a number more than four figures and less than seven, of three figures, and so on. Whence it follows, that, if a dot be placed over every alternate figure, beginning at the unit's place, the number of such points will be the same as the number of figures in the root. The same rule may be extended to decimals, by first making the number of decimal places even, and then commencing at the unit's place and pointing towards the right hand over every alternate figure, as before ; and the number of such points will be the same as the number of decimal places in the root. 132 ALGEBRA. EXAMPLES. 1. Extract the square root of 273529. ARIl'HMETICAL FORM. SYMBOLICAL FORM. 273529(523 273529 (500-|-20+3 25 500^=250000 102)235 2X500+20=1020)23529 204 20400 1043)3129 2X(5004-20)+3=1043)3129 3129. 3129 The pupil will perceive that both these operations are per- formed by Art. 182. 2. Extract the square root of 45796. Ans. 214. 3. Extract the square root of 106929. Ans. 327. 4. Extract the square root of 36372961. Am. 6031. 5. Extract the square root of 22071204. Am. 4698. 6. Extract the square root of 33.1776. Ans. 5.76. 7. Extract the square root of .9409. An^. .97. 8. Extract the square root of .0029997529. Am. .05477. 9. Extract the square root of .001234. Am. .035128+. 10. Extract the square root of 32176552.863844. Am. 5672.438. CUBE ROOT. ISO, Investigation of a rule for extracting the Cube Root of a compound algebraical quantity. Since {a-\-bf=a?-\-Za'b-\-''6ab'^-^h^, we must have the cube root of the latter quantity = o,-\-b ; and our object is to deter- mine how it may be deduced from it. Now, the first term a of the root is the cube root of a?, and the first term of the proposed quantity ; hence, taking away a^ we have 2>a^b -\-Zab'^ -\-h^ left to enable us to find b; but Za'b-^'6ah'-\-b^=^{^a'-^oab-]-y)b. It is, therefore, manifest that h wiU be obtained by dividing the first term of the re- mainder by three times the square of a; and, to complete the EVOLUTION. 133 divisor, we must add to Sa^ three times the product of the two terms, or Sab, and also the square of the last, b"^. Thus, the second term being found, the repetition of a similar process will evidently lead to the root, whatever number of terms the ex- pression may contain. Hence the following Rule. Arrange the terms according to the powers of some letter, and extract the root of the first term, which must be a cube, or some power of a cube ; place this root in the quotient, subtract its cube from the first term, and there will be no re- mainder. Bring down the three next terms for a dividend, and put three times the square of the root just found in the divisor^s place, and see how often this is contained in the first term of the dividend, and the quotient is the next term of the root. Add three times the product of the two terins of the root, plus the square of the last term, to the term already in the divisar^s place, and the divisor will be completed. Multiply the complete divisor by the last term of the root ; subtract the product from the dividend, and to the remainder connect the three next terms, and proceed as before. EXAMPLES. 1. Find the cube root o? a^-]-Za%^Sab''-{-b\ a'-\-U%-{-Zab^-\-b\a-^b. U''+Ub-\-b^)-\-Sa'b-\-Sab''-\-b 2. Extract the cube root o^ x^—'Sx^-{-bx^—Sx—l. x^—Sx^J^bx^—Sx—l{x^—x—l. —82-^+82:^— x^ 82:^_62:^4-32:-f-l)-82:^+62-3— 82:-! — 32:^-1-62:5-32:— 1. 12 134 ALGEBRA. The first divisor is found thus : And the second thus : S[x'—xY-\-3{x'—x) {-l)-^{—lY=Sx'—Qx^-\-Sx-{-l. 3. Extract the cube root of x^—Qx^-\-lbx'—20x^-\-lbx^— 62'4-l. x^—Qz'^ldx'—20x'-\-lbx''-Qx-{-l{x''—2x-\-l, x' Sx'—Qx^+4:x'')—Qx'-\-lbx'—20x -Qx'-^12x'— 8x Sx'—12x^-\-lbx''—Qx+l)Sx'—12z'-\-'^bx''—QxJrl Bx'-123^+lbx''—6x-{-l. 4. Extract the cube root of x^-\-dx'^-\-27x-{-27, 5. Extract the cube root of 1—6 2/+ 12?/^— 8?/^ 6. Extract the cube root of «^— 6a^+40a^— 96a— 64. Ans. a^—2a—4:. 7. Extract the cube root of a^-{-Sa'b-{-Bab''-{-b^-^Sa'c-\-Qabc -\-Sb''c-{-Bac'^-i-Sbc^-{-c\ Am. a-f^+c. BY DETACHED COEFFICIENTS. 1. What is the cube root of x'-\-Qx'—^^x^-\-^^x—U ? l_|_6+0-404-0+96-64(l+2-4 ■ rx3 = 3) 6 (1+2)3= i_|_6_[-l2+ 8 1^x3 = a)_i2_48 1+6-1- 0-40+0+96—64, Hence, 1+2— 4=.t'^+22:— 4. Ans. EVOLUTION. 135 2. What is the cube root o^ Sx^—BQx' -{-^4x^—27x^1 8-]-0-364-0+54-|-0-27(2-[-0-3. 2-X3 = 12)4-0—36 84-0-36+0-1-54 -f-0— 27. Hence, 24-0— 3=2z^4-0ar^— 32:=2:c3— 3z.' 3. What is the fourth root of a;*4-8a;^-f-24a;24-32:r4-16 ? Am. x-\-2. 4. What is the cube root of x^—2>x^y-\-'^x^'f — ?/^? Ans. x^ — y. 187. Reasoning analogous to that employed in Art. 185 will show, that, if a point be placed over every third figure, begin- ning at the unifs place, the number of points thus placed will be the number of digits in the cube root ; and attention to Art. 186 will furnish the following operation : 1. Extract the cube root of 1860867. . a -\- b -\-c 1860867(1004-204-3=123. a? = 1000000 = first subtrahend. So" = 30000)860867 = first remainder. da'b = 600000 ^ab'' = 120000 b' = 8000 728000 = second subtrahend. 3(^4-3)2 = 43200)132867 = second remainder. S{a+bYc = 129600 3(^4- 3)c2 = 3240 c^= 27 132867 = third subtrahend. This process is the origin of the Rule given on page 248 of the Author's National Arithmetic, to which the pupil is re- ferred. 136 ALGEBRA SYMBOLICAL FORM. 1860867(100+204-3 (100)^= 1000000 [=123. 3(100)2+3(100)2+(20)^=36400)860867 728000 3(100+20)2+3(100+20)3 + 3^=44289)132867 132867. 2. What is the cube root of 31255875 ? A7is. 315. 3. What is the cube root of 37259704 ? Ans. 334. 4. What is the cube root of 116930169 ? A7is. 489. 5. What is the cube root of 508.169592 ? Am. 7.98. 6. What is the cube root of .724150792 ? Am. .898. 188, To extract any root of a compound algebraical quantity. Since {a-}-x)'"=a"'-]-7na'^~^x-\- &c., it is obvious, that when the quantities are properly arranged, and the first term of the root is found, the second term of the ?nih. root will be obtained by dividing the second term of the proposed quantity by ?na""~\ or by m times the first terra, raised to the (vi — l)th power. And, if the root thus found be raised to the Tnih. power, and the result be subtracted from the quantity proposed, and the process be repeated when necessary, any root of a compound quantity may be determined. The similarity of the processes employed in this and the pre- ceding articles will be immediately noticed, it being observed in the former, the complete powers of a monomial, binomial, tri- nomial, &c., are subtracted from the proposed quantity by 07ie, two^ three^ &c., operations; whereas, in the latter ^ the subtrac- tion of the same quantities is efi'ected at 07ice. Hence the following General Rule. 1. Arrange the terrm so that the highest power shall stand in the first term,, and let the 7iext higher occupy the second place. 2. Firid the root of the first ter7n, and place it in the quotient ; EVOLUTION. 137 and, having raised this root to the required power, subtract it from the first term, and then bring down the second tern for a dividend. 3. Involve the root last found to the next inferior power, and multiply it by the index of the given power for a divisor. 4. Divide the dividend by the divisor, and the quotient will be the next term of the root. 5. Involve the whole root thus found to the required power, which subtract from the given quantity, and divide the first term of the remainder by the same divisor as before. 6. Proceed in this manner for the next term of the root, and so proceed until the work is finished. See page 255 of the Author's National Arithmetic. EXAMPLES. . 1. Required the square root of a*— 2a^a;-f3aV — ^a^-^-x"^, a'—2a'x-{-Sa''x^—2ax^-\-x\a''—ax+x\ 2a')—2a'x a'—^a^x^a^x"- 2a2)2aV a'—la^x-^Za^x'—^ax^-^-x'. 2. Required the cube root of a;^-j-6:c^— 40:c^+96a:— 64. ajS-j-e^:^— 40:?;^+962:— 64(:c^+2:c— 4. x^ %x')M' a;6_|.6:c^_[_12:^44_8^3 3:c^)— 12.^^ a:6-|_6a;^— 40^:34-962:— 64. 3. Required the fourth root of \^x''—^^xSj^2VoxY—'21^^'if ^^\y\ 12# 138 ALGEBRA. Wx'—mx'y+21Qxy—21Qxf+Sly\2x—dy. 16x' S2x')—mx'y lQx'-9Qx'y-lr21Qxy—21Qxf+Sh/. 4. Required the cube root of w^ — Q?n^-{-4cOm^ — 96m — 64. Atis. iri} — 27a — 4. 5. Required the fifth root of 322;'— 8 0^:^+8 0:^:^—4 0x^+1 0:e — 1. Ans. 2x — 1. SECTION XV. SURDS, OR RADICAL QUANTITIES. Art. 189t Surds, or radical quantities, are roots whose values cannot be exactly obtained, being usually expressed by means of the radical sign, or fractional indices ; in which latter case the numerator shows the power to which the quantity is to be raised, and the denominator its root. I 2 Thus, a/S^ or 3^, denotes the square root of 3. s/d\ or a^, m is the cube root of the square of a; and c", or s/oJ^t is the wth root of the ?7zth power of a. 190. The quantity /y/^, or a/3", is an irrational quantity or surd, because no number, either whole or fractional, can be found, which, when multiplied by itself, will produce either 2 or 3 ; but their proximate values may be found, to any degree of exactness, by the common rule for extracting the square root. Problem I. 191. To reduce a rational quantity to the form of a surd, or radical quantity. RADICAL QUANTITIES 139 KuLE. Raise the quantity to a power corresponding to the index of the surd to which it is to be reduced^ and over this neio quantity place the radical sig7i, or proper index, and it ivill be the form required. EXAMPLES. 1. Let 5 be reduced to the form of a square root. Here 5x5=5^=25; whence V^- -^^^« 2. Reduce 2x'^ to the form of the cube root. Here {2xY=Sx' ; whence ,^^/^, or {Sx'fy or S^x'^. Ans. 3. Let — 2x be reduced to the form of the cube root. Here {—2xY= — Sx^; therefore 4^— 8^1 Ans. 4. Let Sa^ be reduced to the form of the square root. x^ 5. Let — be reduced to the form of the cube root. Am. 1^. 6. Reduce x^ to the form of the fifth root. An^. x- x—y x" 7. Let be reduced to the form of the fourth root. Am. (-^\^. \{x-yf) 8. Let (x — y"') be reduced to the form of the square root. Alls. ({x—y'^f\^. 192. If a rational quantity be joined to a surd, it may be reduced to the form of a surd by raising the rational part to the required power, and multiplying it by the surd. 9. Let 5 a/ 7 be reduced to a simple radical form. 5^^=V5x5X^^=A/^X^/7=>^/^75■. Ans. 10. Let ^Asfa be reduced to a simple radical form. 3Va=V3x3xV«=/>/^ ^ns. 140 ALGEBRA. 11. Let 3/^3 be reduced to a simple radical form. 3A^=/^3x3x3X/v^=/V^XAf3=/s^/^. Aiis. 12. Let ^/s/a be reduced to a simple radical form. A?is. a/J. 13. Let 4/^yF be reduced to a simple radical form. 4 Atis. 16' 14. Let Z/ym be reduced to a simple radical form. Ans. /^x^—x''—x-\-i' — i— , or ( -y. Ans, X—l \X—1J 16. Let — U— 2 be reduced to a simple radical form. A... Jt' Problem II. 193. To reduce quantities of difierent indices to others that shall have a given index. Rule. Divide the indices of the quantities given by the index under which the quantities are to be reduced, and the quotients will be the new indices for those quantities. Then, over the quantities with their new indices place the given index, and they will be the equivalent quantities required. EXAMPLES. 1. Reduce 4^ and 8"^ to other quantities of the same value, each having the common index ^. RADICAL QUANTITIES. 141 Here J-i-i=^Xf=l=3» t^® fii'st index. And- ^-j-^=^Xf=3=^' ^^® second index. Whence (4^)^=4^ ; and (8^)^=8^ Atis, 194. The truth of this rule will be evident; for if 4 be raised to the 3d power, and the 6th root extracted, that root will be equal to the square root of 4. Thus, 4x4x4=64; ..^/154=2 ; /v/4=2. And, if 8 be raised to the 2d power, and the 6th root extracted, the result will be equal to the cube root of 8. Thus, 8x8=64; /y64=2; ^8=2. 2. Reduce 3^ and 5^ to the common index ^. Ans. a^2^a79; ^=,^/1M. 3. Reduce a^ and /T6. 2. Reduce 4^ and 6*" to equal quantities, that shall have the same index. 1 and ^ = 3-^2- and ^. Therefore 4:^==:4^^={A')^'^={2b6)'^'^ or V2F6. Arts. And 6^=6^^=(6^)i^=(216)'^'2- ^r jy^TO: Ans. 2 _1 3. Reduce 2^ and 3- to equal quantities having a common iJidex. jins. /^OB'and /^yST. -i -1 4. Reduce a^ and b^ to equal quantities having a common Mex. ^^. /y^and/^/^ 5. Reduce x'" and ?/" to quantities having a common index. Ans. '""V^ and "^"A/y^. Problem IV. 196. To reduce surds to their most simple form. Rule. Resolve the given quantity into two factors, ojie of which shall be the greatest corresponding power contained in it, and set the root of this power before the remaining factoi', with the proper radical sign between them. Note. — When the given surd contains no factor which is an exact power, it is already in its most simple form. Thus a/ 15 cannot be re- duced lower, because neither of the factors 5 or 3 is a square. examples. 1. Let a/IB be reduced to its most simple form. RADICAL QUANTITIES. 143 We divide 48 into two factors, 16 and 3, 16 being the great- est power of the required root. We therefore extract the square root of 16, and write its root, 4, before the other factor, having the sign prefixed to the surd. Thus V48=a/T6X3==4V3. Ans. 2. Let /v^TOB be reduced to its most simple form. In this question we find the factors of 108 to be 27 and 4, 27 being the largest possible factor of which the cube root could be extracted. The operation, therefore, is Thus a7108=4/27X4=3/^: A?is. 3. Let a/To be reduced to its most simple form. Am. 5v^- 4. Let /v^SO be reduced to its most simple form. A71S. 2^. 5. Reduce /s/TTc^ to its simplest form. Here V 27aV= V 9^^^* X ^clx= //9a^ X /s/^ax= ^ax^A/^ax. 6. Reduce /^fh^^cc'or} to its simplest form. Ans. ^ax/^2a'^x. Problem V. 197. When any number or quantity is prefixed to the surd, that quantity must be multiplied by the root of the factor, as in Art. 196, and the product must then be joined to the other part, as before. EXAMPLES. 1. Let 2/\A32 be reduced to its most simple form. Here 2^32=2^/16x2=2 x4/v/2=8/v/2: A7is. In performing this question we first find the factors of 32, which are 16 and 2. We then extract the square root of 16, and multiply its root, 4, by the number prefixed to the surd, and find the product to be 8, to which we subjoin the surd 2. 19S. This and all similar questions might have been per- formed by squaring the number prefixed to the surd, and then 144 ALGEBRA. * multiplying this number by the surd. Let this product be divided into two factors, as before, and the square of the former prefixed to the latter will give the answer. Thus, 2/v/32=V2X^X32=V128=a/134x2=:8a/2. Ans, 2. Let b^^/^4: be reduced to its most simple form. Here 5^24=5A/8x3=5x2/^^=10A^'3: Or 5a^24=a^/^X5x5><24=/>^3000=^1000x3=10.^/^ 3. Reduce 2/^/W to simple terms. Ans. 4/v^ Problem VI. 199. A fractional surd may be reduced to a more convenient form by multiplying both the numerator and denominator by such a number or quantity as will make the denominator a com- plete power of the kind required, and then proceeding as be- fore. [Art. 198.] EXAMPLES. 1. Let a/^ be reduced to its most simple form. 2. Let /^/~% be reduced to its most simple form. 3. Let /\/y be reduced to its most simple form. Ans. |a/T4. 4. Let /^/J be reduced to its most simple form. Ans. i/V^. 5. Let />/J be reduced to its most simple form. Ans. |/v^. EXAMPLES TO EXERCISE THE FOREGOING RULES. 1 . What is the most simple form of a/1% ? An^. b/s/W. 2. What is the most simple form of V 80aV ? Ans \axsPSx. RADICAL QUANTITIES. 145 3. What is the most simple form of /^/T^^aWc^l Atis. 2>ab\/ lac^. 4. What is the most simple form of 7/\ASD ? Ans. 28^/57 5. What is the most simple form of |A/f ? Ans. -^^a/^. 6. What is the most simple form of xxa/^ ? Ans. -^-j/^/T. 7. Let /\/96a-r^ be reduced to its most simple form. Ans. 4iax/s/Wx, 8. Let ^/v^^SB^F+Glp be reduced to its most simple form. Ans. ^/^(lx'-\-^f). Problem VII. 200 1 To add surd quantities together. I. When the radicals are similar, annex the radical part to the sum of the coefficients. EXAMPLES. 1. Add 7/v/2 to 5 V2: Ans. 12^/57 2. Add b/^~ah to Z/s/~ab. Ans. S/s/ab. 3. Add a/s/xy to hfy/xy. Ans. {a-\-b)/s/lcy. 4. Add "J^d^—y to y/s/d'—y. Ans. (7+?/)/y/a^ZI^. II. When the radical parts are dissimilar, make them similar by Art. 197, and proceed as above. But, if the surd part cannot be made the same in all the quantities, they can only be added by the signs + and — . 5. Add a/T5 and a/Z2 together. First VT8=V 9x2=3/v^. And /v/^=,v^l6x~2=4v^. Then 3V^+4V2=7>\/2: Ans. 6. Required the sum of />^375 and /v^I92. First a^^^7B=a^125x3=5a^. And /^/Tm=^ 64x3=4^/3: Then 54/T-|-4.^y3=9.^. Ayis. 13 146 ALGEBRA. 7. Required the sum of a/27 and /\/4H. ^^- 7V^. 8. Required tlie sum of />/S(J and .a^T^. Am. 11 V^- 9. Find the sum of V^I^ and VioF. A7is. 15a/3T 10. It is required to find the sum of /^/40 and y^l35. il;i5. 5/v^. 11. Find the sum of 4^/^ and 54/128. ^?i5. 324/2. 12. Find the sum of ^/fand ^/S- ^^- S^ 13. Required the sum of %s/ d^h and 5a/16^. Problem VIII. 201 1 To find the difiierence of surd quantities. Rule. When the radicals are, or have been made, similar, aiinex the common radical part to the difference of the rational parts. But, if the quantities have no commo7i surd, they can be sub- tracted only by changing the sign of the subtrahend. examples. 1. From VMTtake ^./W. First //320=a/(I4x^=8V5T And ^/~8D=A/I(Jx5=4^/5: Then 8V^-4^5'=4a/5: Ans. 2. Find the difference between /v/T28" and a/^. First A7I2B=4/(5ix2=44/2. And 4/~54=4/27x2=34/5: Then 44/^-34/2=a/2: Ans. 3. Required the difference between 2.a/^ and /\/TK. Am. 7a/2T 4. What is the difference between 24^213 and 3a/3^ ? ^7W. 24/5. RADICAL QUANTITIES. 147 5. llequired the difference of VTo and a/?S^ A7is. a/3] 6. Required the difference of /v/256 and /^82. Am. 2a/4: 7. Required the difference of a^ and a/^ . Am. ^/s/~^. 8. Required the difference of /^^ and a^/^^ Ans. -j2^/^/T5. 9. Find the difference of f A^o^and f ,y~ab. Problem X. 205 • To divide one surd quantity by another. E,uLE. When the surds are of the same ki?id, JiTid the quotient of the ratioTial parts^ and the quotients of the surds, and tfie two joined together^ with the common radical sign between them, will give the whole quotient required. But, if the surds are of different kinds, they must be reduced to a common index, and be divided as above. The quotients of different powers or roots of the same quantity are found by subtracting their indices. EXAMPLES. 1. Divide Q/s/m by 3V^ Here?^^=2^A2=2^/4X^=(2x2)A/B=4V3: A71S. Sa/S 2. Divide ^^/IM by 2a/^. Here?^^^=4^Arg=4V^X2=(4x3)V2=12V^. Ans. 2a/1 3. Divide 8^"512 by 4^. Here ?^^^=2/>^/^5^=2/y64x4=8A^ Ans, 4a/2 13# 150 ALGEBRA. 4. Divide 12 times the cube root of 280 by 3 times the cube root of 5. Here ^^"^^^^ =4:,yM=4:y^/^yO = SaTT. Am. 3/^/5 ^ 5. Divide 6a/51 by 3a/2. Am. 6V^. 6. Divide 4;.^/T2 by 2^0^: Ans. 2,^. 7. Divide 4/,/Mhj 2^/W. Ans. 2VW: 8. Divide 64/100 by 3,^/5'. Ans. 2a^'^. 9. Divide /v/20+VT2 by a/5+V^ ^^5. 2. 10. Divide 32fAA'by IBfA^ Ans. ^^[-Y, 206, Since the division of surds is performed by subtract- ing their indices, it is evident that the denominator of any fraction may be taken into the numerator, or the numeratoi into the denominator, by changing the sign of its index. EXAMPLES. 1. Let - be expressed by a negative index. a 1 2. Let — be expressed by a negative index. 3. Let — be expressed by a negative index. 4. Let — be expressed by a negative index. ^2— 1 — ^ • RADICAL QUANTITIES. 151 5. Let a ^ he expressed by a positive index. _ 1 a~^ 1 6. Let be expressed by a negative index. Ans. {a-{-x)~2. 7. Let a{a^ — x^)~'^ be expressed by a positive index. 1 Ans. — ^ -i. 8. What is the value of — ? or or- Whence it follows that a" is a symbol equivalent to unity ; consequently 1 may always be substituted for it. This, how- ever, has been demonstrated in a previous article. Problem XI. 207 • To involve or raise surd quantities to any power. h Let a° represent a surd quantity; then, by Art. 204, its square will be Therefore, to involve a surd to any required power, we adopt the following KuLE. When the surd is a simple quantity, multiply its index hy 'ifor the square, 3 for the cube, SfC, and it will give the power of the surd part, which, being annexed to the proper power of the rational parts, ivill give the whole power required. If the surd be a compound quantity, multiply it by itself the requisite number of times. EXAMPLES. 1. What is the square of 3a^ ? ^a^^^=U^=^^a\ Ans, 152 ALGEBRA. 2. What is the cube of f a/3 ? Here (|V3)'=/y//57=^VV(9X3)=|^/^. Ans. 3. Required the square of 3/^3. Ans. 9/^/Vi. 4. Required the cube of 17/v^^ Aiis. 103173^21. 5. What is the fourth power of ^/v/B"? Am. -jJ^. 6. Required the cube of y\/F' Am. Sa^/W. 7. Required the third power of ^V^ Ans. ^/s/^. 8. Required the fourth power of ^/«/2T A7is. ^. 1 "* 9. What is the mth. power of a" ? Am. a". 10. Required the square of 2-{-a/W. Am. 7+4/v/^ r I. ^ 11. What is the -th power of a'' ? Am. a?\ Problem XII. 208. To find the roots of surd quantities. Rule. When the surd is a simple quantity, multiply its index by ^ for the square root, hy ^for the cube root, c^c, and it will give the root for the surd part, which being annexed to the root of the rational part, will give the whole root required. The truth of this rule may be illustrated by the following EXAMPLES. 1. What is the cube root of the square root of 64 ? The square root of 64= VM=6#=8. And the cube root of 8=a/B=8^=:2. Ans. 209. The same result would have been obtained if we had multiplied the index (^) of the given quantity by the index of the required root (^), the product of which is 4X-^=i 5 and if we had considered this (-^) the index of the root to be extracted of the given quantity 64, the operation would have been thus : /y^=2. Ans., as before. RADICAL QUANTITIES. 153 2. Required the cube root of the square root of a. ^2 /\ ^=a^. Atis, 3. Required the fourth root of /v/BT 3^^^=3^. Am, 4. What is the square root of O/vAo"? Here (9^3)2=9^X3*'^^=9^X3^=34/^ 5. What is the square root of 10^ ? 10^=1000 ; VT000=10V1U. Ans. 6. What is the cube root of ||^/\/^? Atis. ^/^/~a. 7. What is the square root of ^f a^ ? Atis. %c^/sj~a. Probleji XIII. 210. To find factors that shall cause any surds to become rational. I. When the surd is a monomial, multiply it by the same quantity, with an index such as when added to the index of the given quantity will make it a unit. The quantity A/a~or a- is made rational by multiplying it by \J a or a-. Thus, f/a^s/a^ or arY^a^z=a. -1 2. And it will be rational if a^ be multiplied by a^, thus, i 2 4 \ Also, if a^ be multiplied by a^ it will be rational; thus, k 1 EXAMPLES. 2 1 1. What factor will make x'^ rational ? Ans. a;"^. 2 5 2. What factor will make y^ rational ? ^7i5. y^ . 3. What factor will cause ar^ to become rational ? 154 ALGEBRA. II. When the surd is a binomial or residual quantity, and both the terms are even roots, to find a factor that will make the quantity rational. In Art. 158 we have shown that the product of the sum and difference of any two quantities is equal to the difference of their squares ; therefore, when one or both of the terms are even roots, we multiply the given binomial or residual by the same quantity, with the sign of one of its terms changed. Note. — It is sometimes necessary to repeat the operation. EXAMPLES. 1. To find a multiplier or factor that shall make 4-{-/vA5 rational. Griven surd, 4-{-/,/5 Multiplier, 4— />/5 16-f-4^/^ Product, 16 — 5=11 rational quantity. 2. Find a factor that shall make /s/a-\-/sfh rational. ^ — b a — b rational quantity. 3. What factor will make l-|-/vA5~i*^tional ? 1-V^ l+VB — V o — 6 — 8= — 2 rational quantity. R A U I C A L Q U A N T I TI K S . 155 4. What factor will make a/o — a/1 rational ? /V^-HVI 5— a/o -f-V^-1 5 — 1=4 rational quantity. 5. Find multipliers that shall make A/b-\-/^/^ rational. />/B+a/3 a/o-^"B a/5 -a/B 5-Vlo +A/I5-3 5 — 3=2 rational quantity. 6. What multiplier will make /\/o — />/rFrational ? a/o— a/^ a/O+a/S 5 — a/o^ + a/ox — a; 5 — X rational quantity. III. A trinomial surd may be rendered rational by changing the sign of one of its terms for the multiplier. EXAMPLES. 1. To find multipliers that shall make /\/7+a/B— a/^ rational. 156 ALGEBRA. 7+V21-A/14 +/vA21+3-a/(3 8+2a/2I -8+2a/21 -64-16a/21 +16^/2I+84 84 — 64=20 rational quantity. 2. Find a factor that will make a/^— a/1— V^ rational. a/^-a/1-VB 8~a/B-a/21 +//24-/s/a-3 4-2V3 4+2V^ 16-8a/3 +8/v^— 12 16—12=4 rational quantity. QUESTIONS FOR EXERCISE. 1. Find a multiplier tliat shall make a/^ — V^ rational. Atis. V5+/v/^. 2. Find a multiplier that shall make a/7+a/(j rational. Ans. /v/7— V^- 3. Find a multiplier that shall make .^/ 10— a/2 rational. A77S. a/TU+a/2'. RADICAL QUANTITIES. 157 4. Find multipliers that shall make /s/a-\-/s/h-\-A/c rational. Am. s/a—h/h—f>/^, and {a—h—c-^-^sTbc). 5. Find multipliers that shall make />y^—/v/lL rational. Am. (/4^+^1)(a/^+/\A). Problem XIV. Art. 211. To reduce a fraction, whose denominator is a surd, to another that shall have a rational denominator, without changing its value. Rule 1. When the proposed fraction is a simple one, multiply each of its terms by the denominator. 2. If it he a compound surd^ find such a multiplier hy the last Art. as will make the denominator rational, then multiply both the numerator and denominator by it. EXAMPLES. 1. Reduce to a fraction whose denominator shall be />/a b ^,/\/~a J/s/fl . rational. X— — = -• Am. fs/a /sfa ^ 2 Reduce to a fraction whose denominator shall be rational. -^X— ~=^^^- ^^^s- ^a f^ « 2 3. Reduce the fraction — to another whose denominator Vo shall be rational. 2 2 _/v/5 2/v/5 . = — ^X — r=— ^. Am. /\/5 a/5 V^5 ^ 3 4. Reduce to a fraction whose denominator shall /s/^-a/2 be rational. ^^3 3 VH-a/^ 3a/F+3^/2 Here = — —z X — ^ — sj^sj2 j^-fsf2 a/o+/V^ ^-^ 3VF+3V2-^VF-fV2-^^^^^ ^^. 3 1 14 158 ALGEBRA. 5. Extract the square root of |. Here F^^^^^^^^^^_^^^, ^,,. 6. Reduce r 7= to a fraction whose denominator shall be rational. Here V^ ^ V^ 3+V^ 3V2+2_3V2+2_ 3-^ 3-V^ 3+V^ 9-2 7 7. Reduce — — to a fraction that shall have a rational denominator. ^^w. ^ . 8. Reduce 7 7^ to an equivalent fraction having a ra- 3 — a/1 tional denominator. Ans 2 * 5 9. Reduce the fraction to an equivalent fraction a/5-V^ having a rational denominator. Ans. ^ . 10. Reduce the fraction to an equivalent fraction having a rational denominator. Ans. = . 1 11. Reduce ^ . — — to a fraction that shall have a ra- V^+a/7 . , , . A a/^— a/7 tional denominator. JiJis. — x- — • # RADICAL QUANTITIES. 159 3 12. Reduce the fraction to an equivalent fraction that shall have a rational denominator. Am. J5 =V^+V ^• o Problem XV. 212* To change a binomial, or residual surd, into a general surd. lluLE. Involve the given hiTWinial, or residual, to a power corresponding with that denoted hy the surd; then write the radical sig7i of the same root over it. EXAMPLES. 1. It is required to reduce 24-/>/S to a general surd. Here, (2-{-VF)^=4+4V"5+B=74-4V^. Therefore, 2+VE=V(7+4V3). 2. Reduce /v^+a/^ to a general surd. Here, (V2+V3f=2+2v^4-3=5 + 2VI>. Therefore, a/^+V3=a/(o-|-2V^- 3. Reduce ^/2-\-/^/~i to a general surd. Here, (/^+..yi)^=6+64/^-j-6^. Therefore, a/2+a^=aJ^+a^+>^). 4. Let 3 — a/5 be reduced to a general surd. Ans. //(U-GVo). 5. Let /v/2'+ 2/\/^ be changed to a general surd. Ans. /s/|2(3+SV^3). 6. It is required to change 4 — vT to a general surd. Ans. //(23— 8VT). 7. Let 1 /sf^—o^s/^ be changed to a general surd. Ajis. /^^/(786-13234^-f567^¥). 160 ALGEBRA. Problem XVI. TO EXTRACT THE SQUARE ROOT OF A BINOMIAL SURD. 213. A binomial surd is one in which one of the terms, at least, is irrational ; as a-\-\/F, or fJ~a-\- k/T. To extract the square root of (z-fV^s we put /s/ {a-\- j\fh)=^Tn-\-n. And fs/ {a—^/V)^=m—n. By squaring both of these equations. We have a-\-fs/T=irr^-\-%nn-\-7^. And a — /s/b=7)r — 2mn-\-7i?. By addition, la =2^^ -^2n^. And a=z7rt^-{-7i^. Multiplying the two first equations together, We have //(«+VT)X V(fi^— //^) = (^+^)X(w— ^). And //(^^ — h)=m^ — tz^ Having both the sum and difference of m} and n^, we obtain, by addition and subtraction, the following equations : >^-=°+^f-^-l, and „.^'^-Vf-^). Therefore, ^=v(2±^f^"^), Consequently, V{a+V*)=V(^i^^— ) + / a-V(a'- ^)\ And V(a-V^)=v(^-±^^-^')-V(^-=^-=^). It is certain that both a and /^{a^—l) must be rational, in order that the expressions within the parentheses may bo RADICAL QUANTITIES. 161 rational, in which case each of the above values will be either two surds, or a rational and a surd. The above formula) will apply to any particular values for a and h ; observing that if h be negative, the signs of h in the formula) must be changed. EXAMPLES. 1. What is the square root of 11-|-a/T2 ? Here, a=ll, and i=72. Therefore, And v(°-^^--l))=v( "-^f '-^^' )=V2. Therefore, V(ll4-/v/7^)=34-V^ 2. What is the square root of 10— a/M? Let a=10, and 3=96. Then v("-±^?^)=v(i^^°^^)=^ And v( -^-^) )==v( ^"-^f-^^' )=2. Therefore, ^(\S)-s/M)=s/^—% 3. ^Vhat is the square root of 6-f-/y/2D? Ans. l-|-/\/5. 4. What is the square root of 6-f-2V~5? Ans. s/h-\-\. 5. What is the square root of 12-|-2^"3o ? Ans. V^-fVi'- 6. Required, the square root of 36±10^Tr7~ Ans. SrhVlT. 7. What is the square root of 7— 2VT0 ? Ans. ^o— /s/2. 8. What is the square root of 1+4^"^^? Ans. 2-1-^/^=:^, or 2-,v^:=^. 14^^ 162 ALGEBRA. SECTION XVI. IMAGINARY QUANTITIES. Art. 214. As every algebraical symbol hitherto considered, whether it be affected with the sign -}- or — , when raised to an even power gives a positive result, it follows that no even root of a negative quantity can be either positive or negative. The even roots of negative quantities having, therefore, no sym- bolical representation in accordance with the views of Algebra, so far as we have yet considered it, can only be indicated or expressed by means of the radical sign, or corresponding fractional inder. Hence arises a new sjjecies of symbolical expressions, called Imaginary or Impossible Quantities. Thus the square root of — a^ is neither -\-a nor — a, but is written V —d\ and is equivalent to />/<^^X( — l)=/>/^V — 1 =dz«V'— Ij which is said to be impossible, or imaginary, in consequence of involving the symbol a/ — 1. By Art. 78 we learn that the product of real quantities, that have like signs, is always plus ; and, if the signs are unlike, the product is minus. We, therefore, infer, that the product of two imaginary quantities, that have the same sign, is equal to minus the square root of their product, considering them as real quantities. Hence, (+a/— «)(+/%/— ^)= — sj cL--=i—a. (—a/ — a)(— V — a) = — /\/(2-=— a. (— /\/— a)( — js/ — h) = —/s/ab. 215. If the two imaginary quantities have different signs, then, it is evident, their product will be equal to plus the square root of their product, considering them as real. Thus, (-|-V^Z^)(— V^=^) = -|-V^. IMAGINARY QUANTITIES. 163 EXAMPLES. 1. Multiply 4^^=B by IsT^. 2. Multiply ^-^rsT—^ by 3— V— o- 3— //"=F 12+3V^^ 8. Multiply 3V^^ by 7a/^^. ^^- — 21v^. 4. Multiply — 7V^=¥by -3.^^=^. ^tw. -21^/T2: ' 5. Multiply 44-a/^=^ by V^^- ^^^5- 4V'=^-/\^T5'. 216. If one imaginary be divided by another, having the same signs, the quotient is equal to plus the square root. But, if the imaginaries have different signs, it is evident that their quotient will be equal to minus the square root of their quotient. EXAMPLES. 6. Divide 6V^=S by 2v'=i: Am. 3Vf . 7. Divide Isf^-^ by — 5/v^^^. Ans. — |/>/o. 1 8. Divide —/sf—l by — 7v^^^. Ans. +^^7tt- 9. Divide -\-,sJ —a by -\-s/ —h. Ans. ■\-/sJj' 1) a 10. Divide —s/ —a by —s/—h. Ans. -\-\/-j. h 11. Divide 44- V"^^ by 2— a/^=^. Ans. 1+//^"^- 12. Divide 1-f V^l by 1— V^=^I. Ans. sf^^. 13. Divide 2V^^ by — 3^^=^- ^^^- — f V|^ 164 ALGEBRA. SECTION XVII. QUADKATIC EQUATIONS, OR EQUATIONS OF THE SECOND DEGREE. Art. 217. A quadratic equation is one in whicli the un- known quantity rises to the second power. Quadratics are of two kinds: those which contain only the square of the unknown quantity are called pure quadratics, and those which contain both the first and second powers of the unknown quantity are called affected quadratic equations. The following are examples of pure quadratics : EXAMPLES. 1. Given 4:x'—7=29 to find x. Conditions, 42;^— 7=29. Transposing, 4a;2=29+7=36. Dividing, x'^=9. Extracting square root, x=-\-^. 2. Given ax'^-\-b=c to find x. Conditions, ax'^-{-b=c. Transposing, ax^=c — b. Dividing, x^=zc — b. Extracting square root, a;=-4- a \c—b 4 a Hence, to find the value of the unknown term, we have the following Rule. Transpose and reduce the equation, so that the un- known quantity may be positive, and the first member of the equation. Divide both members of the equation by the coefficieiit of the unknown quantity ; then extract the square root of both members. QUADRATIC EQUATIONS. 165 3. Given 5a:'+5=3a;-+55 to find x. Conditions, 5:^;^-f-5=3a;^-{-55. Transposing, 5z- — 3a;'^=55— 5. Reducing, 22;-= 50. Dividing, x'^=2b. Extracting square root, x — 1 5. 4. Given 2a;2+ 8=3.^2—28 to find x. Conditions, 3.i"— 28=2a;2+8. Transposing, Sz^— 2a;'=28 + 8. Reducing, x^=oQ. Extracting square root, a: = -l-6. 5. Given 7a;2— 5=32:^+11 to find x. Am. a;=±2. 6. Given 4x'-\-lb=7x''—4:17 to find x. Ans. x=±:\2. hx~ 7. Given 32-^+7=^+35 to find x. Ans. 2:= ±4. 8. Given ax^-\-n^m — c to find x. Ans. a;= + 9. Given x- — ab=.d to find x. Ans. x=-h\/d-\-ab. 10. A lady bought a silk dress for £8 15^., and tlie number of sbillings she paid per yard was, to the number of yards, as 4 to 7. How many yards did she purchase for her dress, and what was the price per yard ? Let X = the number of shillings paid per yard. Ix Then -J- = the number of yards. 7^2 And the price of the whole, ~-r-= 175 shillings. Clearing of fractions, 7a;-=700. Dividing, a;'^=100. Extracting the square root, x=.\^s., price per yd. Ix Therefore, —=17^ yards. A7is. 166 ALGEBRA. 11. I have 10 acres of land. If it were a square field, what would be the length of one of its sides ? Ans. 40 rods. 12. A and B lay out money on speculation ; the amount of A's stock and gain is $27, and he gains as much per cent, on his stock as B lays out. B's gain is $32 ; and it appears that A gains twice as much per cent, as B. Required the capital of each. Am. A's capital, $15; B's, $80. 13. There are two square fields, the larger of which contains 25,600 square rods more than the other, and the ratio of their sides is as 5 to 3. Required the contents of each. Ans. Contents of the larger, 40,000 square rods. Contents of the smaller, 14,400 square rods. 14. I have three square house-lots, of equal size ; if I were to add 193 square rods to their contents, they would be equivalent to a square lot whose sides would measure each 25 rods. Re- quired the length of each of the sides of my three house-lots. Ans. 12 rods each. 15. A farmer has a square field, and the number of rods round it is -^h the number of square rods of its contents. Re- quired the number of acres in the field. A7is. 10 acres. 16. John Smith has a field, which is a right-angled parallel- ogram ; its sides are in the ratio of 4 to 3 ; a diagonal, passing from one corner to its opposite, is 100 rods. Required the contents of the field. Ans. 30 acres. 17. Two workmen, A and B, engage to work for a certain number of days, at different rates. At the end of the time, A, who had been absent 4 days, received 75 shillings ; but B, who had been absent 7 days, received only 48 shillings. Now, if B had been absent only 4 days, and A 7 days, they would have received exactly alike. How many days were they engaged for, how many did each work, and what had each per day ? Atis. They were engaged to work 19 days. A worked 15, and B 12 days ; A received 5 shillings, and B 4 shillings per day. QUADRATIC EQUATIONS. 1(37 18. Two numbers are to each other as 4 to 5, and the sum of their cubes is 1512. What arc those numbers ? Ans. 8 and 10. 19. A bushel measure contains 2150f cubic inches, and I wish to make a box that shall contain 50 bushels. Its lengt] is to be to its breadth as 3 to 1, and its height f its breadth. What are its dimensions ? A71S. Length 108.84-f , breadth 36.28+, and height 27.21 + inches. 20. What must be the dimensions of a cubical box that shall contain 100 bushels ? A?is. Height, length, and breadth, 59.9+ inches. 21. Two numbers are to each other as 3 to 7, and the differ- ence of their cubes is 2528. What are those numbers ? Atis. 6 and 14. 22. Bought a house-lot for S5184. Its length is to its breadth as 3 to 1. I gave as many dollars per square rod as the lot is rods in breadth. What were the dimensions of the lot ? yins. 36 rods long, 12 rods wide. Problems. 23. Let 771 be divided into two parts, whose squares shall be to each other as 7i to p. Let X = the greater. And m — x = the less. Then x"^ : (wi — xY : : n : p. Multiplying extremes, px^=.7i{7n — xY. Evolution, x,s/p=dz/^yn{j}i—x). Reducing, xs/p^=m\fn—x,,/n. Transposing, X\/~p-\-Xs/Ti^mspri. -p.. .,. ms/n Dividmg, 2-= the greater. \fpArs/n Subtracting, m = the less. sfp-^sj^i \fp-\-fs/n 168 ALGEBRA. If we take the minus sign, we have x/>/p=. — s/n{m — X). Multiplying, x^=—m/s/n-\-x/s/~n. Transposing, Xfy/p—x/s/n= — m^7i Changing signs, X/^i—x/sf^^^m,s/n. 771 //m-=fsJT{d — x). Reducing, x/^ m=^ds/~e^xs/~e. Transposing, x/s/mA^xs/ e=.d\/~e. JDividmg, x-=.- Substituting the value of d^ e and ?/^, we have 240,000 V BO 2146624.8 V80+V1 8.94427+1 tance from the earth. 15 =215865.4 miles, = the dis- 170 ALGEBRA. 240,000— 215865.4=24134.6 miles, = the distance from the moon. / If we take the negative sign, we shall find the point beyond the moon where the attraction of the two bodies will be equal. Taking the minus sign, xs/m= — /s/~e{d — x). Reducing, x\/m==i — dfsfe-\-x/s/~e. Transposing, xjsfe — x/s/Tn=.d\/~e~. Uiviamg, x-=. — . Substituting the values of d, e and m, we have 2146624 8 x= .^-^ ^—=270,210 miles from the earth's centre, and, a/oO — V 1 therefore, 270,210—240,000=30,210 miles beyond the moon. 26. Required the distance from the earth, in a direction tow- ards the sun, where a body would remain at rest, the distance of the earth being 95,000,000 miles from the sun, and the quantity of matter in the sun being 333,928 times greater than that of the earth. Let S represent the quantity of matter in the sun, E the quantity of matter in the earth, and D the distance between the earth and sun, and x the required distance from the sun. Then, substituting these letters for those in question 23, we have the following formula : d/\/J~ 95,000,000V3M;928 ^^^^33^^33^. V ^33,928+ vr 95,000,000—94,835,885=164,115 miles. Am. 27. The diameter of Venus is 7700 miles, its distance from the sun is 68,000,000; the diameter of the earth is 7912 miles, and its distance from the sun, as stated above, is 95,000,000 miles. How much greater, therefore, is the intensity of light at QUADRATIC EQUATIONS. 171 Venus than at the earth, and what is the comparative quantity that each receives from the sun ? A71S. The intensity of light at Venus is 1.95-|- tunes greater than at the earth. Venus receives from the sun 1.84-f- times more light than the earth. 28. Mercury is 37,000,000 miles from the sun. How much greater, therefore, is the intensity of light and heat at Mercury than at the earth ? Ans. ^-f-^^^ times. 29. Jupiter is 490,000,000 miles from the sun, and its diam- eter is 89,000 miles. Saturn is 900,000,000 miles from the sun, and its diameter is 79,000 miles. How much more light, therefore, do we receive from Jupiter than from Saturn, when they are in opposition to the sun ? Let a = the distance of Jupiter from the sun. b = the diameter of Jupiter, c = his distance from the earth. d = the distance of Saturn from the sun. e = the diameter of Saturn. h = his distance from the earth. The distance of these planets from the earth is obtained by subtracting the earth's distance from the sun from their dis- tance from the sun. The surface of Jupiter is to the surface of Saturn as the squares of their diameters ; and as the quantity of light which a planet receives from the sun is as the square of its diameter di- rectly, and inversely as the squares of its distance from the sun, Therefore, if b'^ = the surface of Jupiter, and e^ = the surface of Saturn, and a and d their respective distances from the sun, then the intensity of light at Saturn will be to the intensity of light at Jupiter as -,, is to — . And as the light which each of these planets gives to the earth is in intensity inversely as the squares of their distances from the earth, therefore, if— = the quantity of light at Saturn, and — = d" a^ 172 ALGEBRA. n the quantity of light at Jupiter, then -r—^ == the quantity of light which Saturn gives to the earth, and -y^ = the quantity which Jupiter gives. Therefore, to find how much more light we receive from Jupiter than from Saturn, we use the following proportion : Therefore, x= ., „ „ . If we substitute for these letters their numerical values, we shall have 900^X805^X89^ X=i. 79'^X490'X395' 810000X648025X7921 =17.7+. Am. 6241X240100X156025 That is, we receive more than seventeen times as much light from Jupiter as we do from Saturn. In the above operation, we have cancelled the ciphers in the distances and diameters of the planets, AFFECTED QUADRATIC EQUATIONS. 219. An afiected quadratic equation is one containing the first power of the unknown quantity in one term, and the square of that quantity in another term. Every equation of this kind, having any real or positive root, will fall, when properly reduced, under one of the four following forms : 2. x^—ax^z b, where 2=-}-^riiV ( jH"^ )• 3. x'^-{-ax=z—b, where 2-= — -±V(t — ^)« , d , f CL' \ 4. x-—ax=—h, where x=-{---±:/\/ \t~^)' QUADRATIC EQUATIONS. 173 220t No exact root can be taken of a binomial ; but, if the first term of a binomial be a square of the unknown quantity, and the second term the quantity itself, with 1, or any other quantity, for its coefficient, the square of half the coefficient of the second term, added to the binomial, will make the whole quantity an exact square. This may be illustrated by the fol- lowing examples. Let x^-\-4:X be the binomial, then 2 is half the coefficient of the second term, and its square is 2x2=4. This we add to the binomial, and the result is :c--f 4a;-[-4, and this quantity is an exact square, and its root, by Art. 183, is x~\-2. If the binomial be x^-\-ax, and we add to it the square of half the coefficient of x, — , the sum will be x^-\-ax-}~j-, the exact root of which is x-{--. Again, if the binomial be x^ — Zahx, we have only to add the Q„2I2 square of half the coefficient of a;, which is — — , to the bino- Q 2/2 mial, and the sum will be an exact square, x^ — Zahx-\ . For ( x-—oabx-\ — j- y =x — . 221. If, therefore, there be any binomial whose first term is an even power of the unknown quantity, and the second term half that power, and we add the square of half the coefficient of the second term to the binomial, the result will be an exact square. 222. To solve an affiicted quadratic equation, we adopt the followino' O Rule, Bring all the unknown terms to one side of the equa- tion, and the known terms to the other, observing so to arrange them that the term which contains the square of the unknown quantity shall be positive, and stand first in the equation, and the term luhich contains the first power of the unknown quantity the second term of the equation. 15=^ 174 ALGEBRA. Divide each side of the equation by the coefficient of the un- known square. Add the square of half the coefficient of the second term to each side of the equation^ and the unknown side imll he a complete square. Extract the square root of each side of the equation, and from the result the value of the unknown quantity may he obtained. Given x^-\-^x=^^i to find the values of a:. Here, by the question, 7? -\-%x-=.%\. Completing the squares, 2;^-}-8:r-|-16=84-f-16=100. Extracting the square root, a:-f-4=10. Whence, rr=10— 4. And, x=.\i. Ans. In solving this question, we first add the square of half of 8, that is, 16, to both sides of the equation ; we then extract the square root of x^-j-^^+l^' ^^<^ ^^^ ^^ result to be a:+4, and the square root of 100=10. Therefore, a;-j-4=10, that is, a:=10— 4=6. Ans. 223, It may also be demonstrated, by the following diagram, that if the square of half the coefficient of the second term be added to the first member of an equation, it will be a complete square. Let x represent one side of the square ABCD ; then x^ will represent this square. To this square we must add 82:, and this quantity must be ap- plied equally to the two sides AB and BC, or the figure would not be a square. Therefore 4x, which is half of 82:, will be applied to either side, d If this quantity, 4a:, be divided by x, the quotient, 4, will repre- sent either of the distances EA or BGr. Having added the two equal parallelograms EABF and BGHC to the square ABCD, we find our figure needs the small square FBGL to complete the square. The contents of this must be equal to the product of QUADRATIC EQUATIONS. 175 FB and BG, that is, 4 multiplied by 4, or the square of 4 = 16 ; but 4 is half the coefficient of the second term. We add this quantity to x^-^Sx, and the sum is ar^-{-8:c-j-16, and its square root is x-\-4:, by Art. 182. 224. A quadratic may be solved by the following Rule. Having transposed the unknown terms to one side of the equation, and the known to the other, multiply each side by 4 times the coefficient of the square of the unknown quantity. Add the square of the coefficient of the first power of the un- known quantity to both sides of the equation, and the unknown side will then be a complete square. Extract the root of both members, and the value of the un- known quantity is obtained as before. EXAMPLES. 1. Given Sa^'^+^a:— 7=88 to find the values of a:. Conditions, ^x^-\-^x—l^'^'^. Transposing, 3a:2+4a:=88+7=95. Multiplying by 4 times 3, 36f2+48a:=1140. Completing the square, 36a;24-48a;+16=1140+16=1156. Evolving, 6:?:-l-4=±34. Transposing, 62'=±34— 4=30, or— 38. Dividing, x=b, or —6^. 2. Given 2a;'^— 10a;-f 7=— 5 to find the values of a:. Conditions, 2^^— 10a;4-7=— 5. Transposing, 2x^ — 10a;=— 5 — 7 =—12. Multiplying by 4 times 2, 16a;-— 80:c=— 96. Completing the square, 16a;^-80a:+ 100=— 96+100=4. Evolving, 4a;— 10=±2. Transposing, 4a;=±2+10=12, or 8. Dividing, a;=3, or 2. 3. Given 3a:--f-5a;— 8 = 34 to find the values of a;. Arts. xz=zo, or — 4|. 176 ALGEBRA. 4. Given z'^-{-Qx-{-4:=22 — x to find the values of a:. Atis. x=2y or — 9. 5. Given Sx'^ — 72:4-6 = 171 to find the values of x. Atis. x=5, or — ^^-. Yj^x 350 6. Given j-lOz — 20=175 to find the values of x. X Ans. x=7, or — 5. 7. Given :c~ — 6a; -{-12=4 to find the values of x. Atis. x=4c, or 2. 8. Given Sx'^-i-S2x=2Q0 to find the values of x. Conditions, Sx''-{-S2x=SQ0. Dividing, a;'^-}~4:c=45. Completing the square, a;--j-4:i;-|-4=45-{-4=49. Evolving, 2:-}-2=±7. Transposing, x=zt7 — 2=5, or — 9. 9. Given a;"— 8a;-{-50=:98 to find the values ofx. Conditions, 2:''— 8.^+50=98. Transposing, a;'— 82:=98— 50=48. Completing the square, x"^ — 82:-}-16=48-|-16=64. Evolving, 2;— 4 =±8. Transposing, a:=±8-f-4=12, or —4. 10. Given x'-\-ax=zb to find the values of a;. Conditions, x'^-\-ax=b. Completing the square, x'^-}-ax-\—T-=b-{'-r-. a Evolving, ^-{-■^=± Transposing, a:=i: ('4)- 11. Given ox"^ — 3a;-j-6=5^ to find the values of a:. Conditions, 3a;'— 3a:-)-6=5^. QUADRATIC EQUATIONS. 177 Transposing, ^z^ — 3a:=5^— 6= — f. Reducing, x^ — x= — f. Completing the square, x^—x-\-l= — |-|-i=-}-^Jg-. Evolving, x—^=dzi' Transposing, 2-=±^-|-2-=§' o^' h 2 12. Given | — ^+20^=42f to find the values of x. Conditions, — f_|_20i=42f. Transposing, ^-|=42| -20^=22^. Clearing of fractions, x^ — ^=44^. 2x1 1 400 Completing the square, a:- — -}--==44^-J-^=-Tr— . 1 20 Evolving, x—-=^—=±^. o o Transposing, a;=zt:6|4-^='<''5 oi' — 6|. 13. Given c2:'^-|-Z'J!;=c to find the value of x. Conditions, ax^-\-bx=^c. Dividing, x'^A =-. ^ , . , ^ bx b'^ c b'- Completing the square, x'^-\ [--— ^= — b-r-^- €L "iCti Qj ^Cb \/c P\ b Evolving and transposmg, a:=± ( — f-j^ J — — 14. Given ax" — bx-{-c=d to find the values of x. Conditions, ax'^—bx-{-c=d. Transposing, ax'^ — bx=d — c, -p.. .,. 2 ^^ ^— ^ Dividing, x' = . za 178 ALG1]BRA Completing the square, a;- [--— = \"a~%' Ci 'iCL CI ~rfl/ Evolvmg, ' '-2-a=^4 \~^'^^r Transposing, ^=2^=^ J l^T +47^' Z> 1 f Reducing, ^=^-±9- [4a(c?— c)+5']. 225. If the equation contains two powers of the unknown quantity, and the exponent of the one is double that of the other, it may be resolved like a quadratic. Thus, 15. Given x'^-\-4:X^=117 to find the values of .t. Conditions, x^-\-Ax'^z=117. Completing the square, a;^-}-42;'-}-4=117+4=121. Evolving, a:'4-2=±ll. Transposing, 2;'^= ±11— 2=9, or —13. Evolving, x=o, or a/ — 13. 16. Griven x^ — 6a;^=16 to find the values of 2-. Conditions, a;''— 6a;^=16. Completing the square, x^—Qx^-\-d=lQ-\-9=i2b. Evolving, x'' — 3=zb5. Transposing, 0:^=^54-3=8, or —2. Evolving, x=2, or /^—2. 17. Given ^=22^ to find the values of ^. J. Conditions, ^ rr =22^. 2xi Clearing of fractions, x ^ =44^. QUADRATIC EQUATIONS. 179 2x^ 1 1 400 Completing the square, x — f--=44^-f--^— -. Evolving, 4 1 20 Transposing, J. 20 1 21 ^ 19 3 Involving, a:=49, or +^-. 18. Given 3a;^"- -2:i'"=25 to find the value of a:. Conditions, 3:f2"— 2a;"=25. Dividing, , 2x'' 25 "- 3 - 8 • 2a;" 1 25 1 76 Completing the square, x"" ^4--=— -f-=— . Evolving, 2;._-=__==___. 1 2VT9 i+2vnr9 iransposing, 5;"=-+ — - — == — —- . liVOlvmg, 2'^ I — ! — J " . 19. Given /i/4a:-|-10=12 to find the value ofx. Conditions, /s/ 4iX -{-1^=11. Squaring both sides of the equation, 42:-|-16=144. Transposing, 4a;=144— 16=128. Dividing, a;=:32. 20. Given /y2a:4-3-f-4=7 to find the value of a:. Conditions, A^2a;4-3+4=7. Transposing, /i^2z-{-3=7 — 4=3. Involving both sides, 22:-f-3=27. 180 ALGEBRA. Transposing, 2xz=27 —S=24t. Dividing, a:=12. 21. Given V V2-\-x=2-{-^/^to find the value of x. Conditions, a/ 12-\-x=2-\-a/x'. Squaring both sides, 12-}-^=4-|-4/\/^4-a:. Transposing, &e., 8=:4V"£ Dividing, 2=a/x'. Involving, 4=a:. 22. Given V^H" 40=10— V^ to find the value of a:. Conditions, a/ x -{-4:0=10— a/x. Squaring both sides, re 4-40=100— 20/%/^ 4-a;. Transposing and reducing, ' 20/y^"i=60. Dividing, , a/'^= ^• Involving, xz= 9. 23. Given /^x — a=/s/x — ^/s/a to find the value of x. Conditions, a^ x—a=i/\/ x—^a/ a. Involving, x—a=x—A/ax-{-j. Transposing, Vtzz=flj-j"T=x* Involving, ax=— Tr- io Dividing by a, ^~T6 * .8 24. Given Sx^ ^=_592 to find the values of x. 8 Conditions, So:^— -^ = -592. 8 Changing the signs, &c., — 3a:^=592. 8 62;^ 1184 Multiplying by f , x^ =—^^, QUADRATIC EQUATIONS. 181 ^ , . ,, 8 Gx^ 9 1184 . 9 5929 ComiDietmg the square, X'' —-{-—= Extracting the root, x^ — -=-1 — --. 5 ' 25 5 ' 25"" 25 * 4 3 77 5==^"5- 4 77 3 74 Transposing, x'-^:=:±:-^-\-p=lQ, or — —. / 74\3 Evolving, x=S, or f — ~ J^. 21 25. Given V^^'+l+^//bx^\^=A/5x-\-2 to find the value of x. Conditions, A/bx + 10=A^ bx-{-2. Squaring both sides, bx-{-10=bx-{-4:Ay bx-\-4:. Transposing, &c., 6=4/\/5^- Dividing, S=2a/5x. Involving, 9=20a;. Dividing, &c., a;=/^. 29. Given ^^/^±3^^^/^_^ to find the value of a:. />/^-}-3 A/x^-\-^b s/lc-^la sfx-^^a (Jonditions, — -— = — -=. * fs/x-^-h /sj x-\-'^h Multiplying both sides of the equation by s/lc-\-h and sfx-\-Zh^ we have Reducing, &c., (2a— 2^»)XA/V=2a^. ah Dividing, /s/x= a — b' / ab \- Involving, x=z\——^j. 1 1 fl I 4 y ). Given — I — = -+ \-^r^A — - to find the value of a:. ^ ,. . Ill, Conditions, — I — = ^ — 4- x^ a Sa^ \ 2 2 I 4 12 11 14 9 Squaring both sides, ~A 1 — :,=— + -— t,H — r- ■ ^ x^ ^ ax a' «- ' \a-x^ x^ QUADllATIC EQUATIONS. Transposing, &c., Multiplying by a:, Squaring both sides, L+i +1=1+1 x^ ^ ax (V- d^ X' Reducing, &c., 4__8 ax X?' Dividing, &c., 1_2 a X Transposing, &c., x=.2a. 183 31. Griven x=^ a^-\-X/^b^-{-x^—a to find the value of a:. Ans. x= — -. . 4a 32. Given — -z — =^!^ to find the value of a-. a/x X Ans. x= 1—a 83. Given a;2+12a:— 16=92 to find the values of a:. A71S. x=Q, or — 18. 34. Given a:^— 3a;=10 to find the values of 2;. Am. x=b, or — 2. 35. Given a;^— a;4-3=45 to find the values of a:. Ans. x=7, or —6. 36. Given bx'^-}-x=4: to find the values of x. Ans. x=-=; or — 1. 37. Given 2x'^ — a;=21 to find the values of a:. Ans. xJ-, or -3. 38. Given 5a;-+6a;— 3=60 to find the values of a:. 21 Ans. a:=o, or — ::-. 39. Given (a:— 12)(a;-|-2)=0 to find the values of a:. Ans. a:=12, or —2. 184 ALGEBRA. 40. Given S:^^— 14a;+15=0 to find the values of x. Ans. x=Z, or 1|. 41. Given ax^ — bx:=zc to find the values of a;. ^,„. ,^^±V(f+M. 42. Given 4:c-— 6:?;=108 to find the values of x. Ans. x= 6, or — 4^-. 14— a; 43. Given Ax — r-=14 to find the values of x. x-\-l 7 Ans. x=4:, or — -. 4 10 14 2x 22 44. Given — =~ to find the values of x. X X- 9 o 21 A71S. 2-=d, or ~. 45. Given a;-|-^5:r-fl0=8 to find the values of x. Ans. a;=18, or 3. 46. Given x -|-V10^-+6=9 to find the values of:?:. Ans. 2-=25, or 3. 47. Given 3:i;2-f 22'— 9=76 to find the value ofx. Ans. .^=5, or — 5§. 48. Given x'—l^x=—2b to find the value of a:. Ans. x=zb. 49. Given 32:'— a:— 140 = to find the value of a;. Alls. x=7, or -2jf . 7a: 50. Given bx^~\——=:7x^ — 51 to find the value of 2:. Ans. 2:= 6, or — 54-. due of 2'. Ans. 2'=: 4, or ^. 52. Given ^+202:=32:2— 80 to find the value of a:. An^s. 2:=:10, or — 2f . 53. If 2;'^-f-82:=65, what are the two values of a: 1 Ans. 2:=5, or — 13. 42: — 4 51. Given 22;^ ^-— — =72; to find the value of 2*. QUADRATIC EQUATIONS. 186 54. If 6a;2— 2;=92, what are the two values of a; ? 23 Ans. x=4:, or — ^. 55. If dx^-\-4:X=M0, what are the two values of a; ? A?is. a:=10, or — 11^. 56. If a;2— 10a;=— 21, what are the two values o^x? A71S. x=7, or 3. 57. If bx^—-=7S, what are the two values ofx? Ans. x=4:, or — S^^j- 58. If lla;2— 100a;=— 201, what are the two values ofx? Ans. a:=3, or Qj\. 59. If 32;2— 17a;=2a;2+84, what are the two values oix? A71S. a;=21, or —4. 60. Given a:+16— 7VFR^=10— 4.V^^+TB to find the values of x. Ans. 2*=9, or —12. 61. Given 9a;-fVT6p+3BP=15x^— 4 to find the values 4 1 of a:. Ans. x=-, or —^. o o 124- 8a;^ 62. Given x= — ^^^- to find the values of x. -^ X — 5 Ans. a; =9, or 4. 63. Given /'a:2--'y' + ('a2-3Y=^ to find the value of a:. Arts. a;=dz'2 -. — • 2 64. Given a:— 1=2H to find the values of a:. x^ Ans. a; =4, or 1. 65. Given /s/l?^^=x—h to find the values of x. b Ha?-b' Ans. x=- ^2^^~12^ A/4a:+2 4— V^ 66. Given -7- — ;=i= p— to find the values of x. 4-|-Va; sf^ Ans.. ^,,or'^. 16* {See Key, p. 119.) 186 ALGEBKA. 67. Given a/^ — 1/s/x—x=.^js/ x to find the values of a:. Ans. a;=4, or 1. 68. Given /s/^-\-/\/x^=lQ/s/x to find the values of a:. Ans. a;=2, or — 3. X sJ X 69. Given -=22|-j — — to find the values of x. .o S61 Atis. a;=4y, or — — . 70. Given = ?\t=0 to find the values of a:. Am. a:=49, or 25. 71. Given x^-{-x'^=7t)Q to find the values of a:. Ans. a:=243, or —28^. lies of a:. Ans. a:=4, or /^4d 15 73. Given V5-|-^~I"V^= / - to fi^^^i t^^ value of a;. -4?w. a:=4. 74. Given /s/ x-{-V2-{-/s/ x-\-12=Q to find the values of x. A71S. 4, or 69. n 75. Given a;" — 2ax^=b to find the values of x. Am. x=: (a± V^"+^)''- 8 4. 5a^^ 76. Given 3a;^ ^=—592 to find the values of a:. 2_ 72. Given x^ — a;^=56 to find the values of a:. Ans. x=S, or ( — p- J . Problems. 1. A merchant bought a number of pieces of two kinds of silk, for £92 3^. There were as many j^ieces bought of each kind, and as man}'' shillings paid per yard for them, as a piece of that kind contained yards. Now, two pieces, one of each kind, together measured 19 yards. How many yards were there in each ? QUADRATIC EQUATIONS. 187 Let X = the number of yards in one piece ; it -will also equal the number of pieces, and also the number of shillings per yard ; and 19— a; = the number of yards in the other piece. Therefore, ar^-|~(l^ — xf=i the value of both kinds. And a;3+(19— a:)''=1843. Or 57a;-— 1083a;4-6859=1843. By transposition, 57a;-— 1083a:=— 5016. Or a;2— 19z=-88. ^ 1 . , o -in , 361 361 ^^ 9 Completing the square, a;-— ly^c-j — j- =—7 oo=t- 19 3 Evolution, X — -^=--±i-^. 3,19 ,, „ a:=±-+— =11 or 8. 19_2;=:8 or 11. Both values answer the conditions of the question ; therefore there were 11 yards in one, and 8 in the other. 2. The plate of a looking-glass is 18 inches by 12, and is to be framed with a frame all parts of which are of equal width, and whose area is to be equal to that of the glass. Bequired the width of the frame. Ans. 3 inches. 3. A grazier bought as many sheep as cost him £60, out of which he reserved 15, and sold the remainder for £54, gaining two shillings a head on them. How many sheep did he buy, and what was the price of each ? Ans. 75 sheep, at 16 shillings each. 4. A merchant sold a quantity of flour for 839, and gained as much per cent, as the flour cost him. AYhat was the price of the flour? Aiis. 830. 5. There are two numbers, whose difi"erence is 9, and whose sum multiplied by the greater is 266. What are those num- bers ? Ans. 14 and 5. 6. A and B gained, by trade $18; A's money was in the 188 ALGEBRA. firm 12 months, and he received, for his principal and gain, $26. B's money, which was $30, was in the firm 16 months. What money did A put into the firm ? Ans. $20. 7. A merchant bought a quantity of flour for $72, and he found that if he had bought 6 barrels more for the same money, he would have paid $1 less for each barrel. How many barrels did he buy, and what was the price of each ? Alls. He bought 18 barrels, at $4 per barrel. 8. A square court-yard has a gravel-walk around it. The side of the court wants 2 yards of being 6 times the breadth of the gravel-walk, and the number of square yards in the walk exceeds the number of yards in the perimeter of the court by 164 yards. Kequired the area of the court. Atis. 256 square yards. 9. Given —— — rQ=^ to find the values of a:. Ans. x=2, or ^. 10. Given x^ — 2s^-^x=lS2 to find the values of x. Ans. x= ^ . 11. Given 9x-\-/s/l^x'^-{-'Si5x^=l^x^ — 4 to find the values of a:. Ans. x=^, or — i. 12. It is required to find two numbers, the first of which may be to the second as the second is to 16, and the sum of the squares of the numbers may be equal to 225. A?is. 9 and 12 QUADRATICS WITH TWO OR MORE UNKNOWN TERMS. 1. Given x-{-y==10 \ ^ , , A J T /? ( to find the values of x and y. And xy=lb ) ^ (1.) First equation, a:-|-2/=10' (2.) Second equation, xy==lQ. (3.) Squaring the 1st, x'^-\-2xy-}-2f=100. (4.) Multiplying (2) by 4, 4xy =64. QUADRATIC EQUATIONS. 189 (5.) Subtracting 4th from 3d, x^-2xy-^f=Z^. (6.) Evolving 5th, a:— 2/=±6. (7.) The 1st, x^y=\^. (8.) Adding Gth and 7th, 2a;=16, or 4. (9.) Subtracting Gth from 7th, 22/=4, or 16. (10.) Dividing the 8th by 2, a:=8, or 2. (11.) Dividing the 9th by 2, 7/=2, or 8. Hence, a;=8 or 2, and 2/=2 or 8. This method may be adopted whenever the sum and product of two unknown quantities are given. 2. Given x — 2/=3 And xy 2/=3 \ =10 ) to find the values of x and y. (1- (2. (3. (4. (5. (6. a- (8. (9. (10. (11. First condition. Second condition. Squaring 1st, Multiplying 2d by 4, Adding 3d and 4th, Evolving the 5 th, The 1st, Adding 6th and 7th, Dividing 8th by 2, Subtracting 7th from Gth, Dividing 10th by 2, X — y=^. xy=\S). a;- — ^xy-{-y^-=.%. 42:?/=40. a;2^22'^+?/-=49. x-y=?.. 2:r=10, or -4. a;=5, or —2. 22/=4, or —10. 2/=2, or —5. Hence, a:=5 or — 2, and y=2 or — 5. We may proceed in the same manner whenever the difi"erence and product of two unknown quantities are given. 3. Given a; +?/ = 20 ) And or-X- - '^OS ) *° ^^^ ^^ values of a; and y. (1.) First equation, (2.) Second equation, (3.) 2d multiplied by 2, .T+7/=20. :r+?/-=208. 2a;24-22r'=416. 190 ALGEBRA. (4.) Square of the 1st, (5.) Subtracting 4th from 3d, (6.) Evolving 5th, (7.) First equation, (8.) Sum of 6th and 7th, (9.) Half of the 8th, (10.) Subtracting 6th from 7th, (11.) Half of 10th, a;2-j-2:i:?/4-7/2=400. x" — 2a:?/-j-?f=16. ^~2/=±4. a;4-2/=20. 2a:=24, or 16. a:=12, or 8. 2y=16, or 24. yz= 8, or 12. Hence, 4. Given x — ?/ = x=\% or 8; 2/=8, or 12. And x^-\-y 3 ) 2=117 ) to find the values of x and y. (1- (2- (3. (4. (5. (6. (7. (8. (9. (10. (11. First equation, Second equation, The 2d multiplied by 2, Square of the 1st, x—y=.Z, 2a;2+22/'=234. Subtracting 4th from 3d, x^-^lxy^y^'^ll^. Evolving the 5th, The 1st, Sum of the 6th and 7th, Dividing 8th by 2, Subtracting 7th from 6th, Dividing 10th by 2, X — 2/ = 3. 2:r=18, or —12. a:=9, or — 6. 2?/=12, or —18. 2/=6, or —9. Hence, a;=9, or —6 ; 2/=6, or —9. 5. Given ^^^1^=10 ) . , ^, , . , 2 no \ to nnd the values or x and y. And X- — y (1.) First equation, (2.) Second equation, (3.) Square of the 1st, (4.) Sum of 2d and 3d, (5.) Half the 4th, (6.) Square root of 5th, V^:q:^=l0. a:2-2/2=28. a;2+2/2=100. 2a;2=128. a:-=64. x=%. QUADRATIC EQUATIONS. 191 (7.) Subtract 2d from 3d, 2y'-=72. (8.) Half the 7th, 7f=^(). (9.) Square root of 8th, y=^' Hence, x=S, and y=Q. And r'+2/'^=35 ) \J1. >Aj n n ( to find the values of x and v. And X- — y'=.\2 ) ^ Ans. x=4:; y=2. 8. Given x -\-y = 6 ) . , ., . o r^r, c to nnd the values ot x and v. And x--{-y-=2Q ) ^ Ans. x=b, and y=l. 9. Given x'^-i-f-=74: K . , . , . , . , ^ c to nnd the values oi x and ?/. And x—y = 2 ) ^ Atis. x=i7, and 2/= 5. A 1 , -, ^ ( to find the values of x and y. And .r -f?/ = 17 ) ^ ^7W. 2:=10, and ?/=7. 192 ALGEBRA. 11. Given a:-— ?/-=85 K ^ i ^ i r. ^ . , . -, «. ( to nnd the values of x and y. And x + y=17 ) ^ ^715. :i*=ll, and y=Q. ]^2. Given 2^ ?/ =2 ) ' . T o thereiore, — is increased by the b{b-^n) b{b-]-n) ' b ^ addition of n to both terms. COMPOUND RATIOS. 241 . Ratios are compounded by multiplying their antecedents together to form a new antecedent, and their consequents to form a new consequent. The resulting ratio is called the su7n of the compounding ratios. RATIOS. 205 Thus, the ratio of « : ^ is compounded with the ratio of c : d by multiplying the antecedents a and c together for a new ante- cedent, and the consequents b and d together for a new conse- quent, and the resulting ratio ac : bd is the sum of the com- pounding ratios a : b and c : d. If the ratios 4 : 7, 6 : 11, and 7 : 9 are compounded to- gether, the resulting ratio is 4x^X7 : 7X11X9, or 168 : G93, which, reduced to its lowest terms by dividing both terms by 21, becomes the ratio 8 : 33. 242» When any ratio, a : b, is compounded with itself twice, thrice, or any number of times, denoted by n, then the resulting ratios are a^ : b^, a" : W^ d^ : Z**, &c., and are called the dupli- cate, triplicate, quadruplicate, &c., ratios of the primitive. As the indices or exponents, 2, 3, and ?z, express the number of times the ratio of a : ^ is compounded of itself, they are called the measures of these ratios. Since the index may be any quantity, either integral or frac- tional, let the fraction be -, -, -, — , &c. ; then, 2 d 4 m The ratio of a- it cc ^T 1 " " am b~ is the square root of the ratio of a : ^. i^ is the cube root of " " o1a:b. 1 b^ is the fourth root of " " of a : b. b~i is the mih. root of " " of a : b. 243. The ratios of a^ : b^, a^ : b^, a^ : b'^, &c., are also called the subduplicate, subtriplicate, subquadruplicate, &c., ratios of a to b. TKOPORTION. 214. Proportion consists in the equality of ratios. a c Thus, if the ratio of o^ : ^ is equal to that of c : ^, or j=-j^ then a, b, c, d, are said to be proportional. The numbers 3, 12, . . o o 1 T 4 1 4, 16, are proportionals, for --=zj, and — =-. IS 206 ALGEBRA. This equality of ratios is expressed by writing the four quan- tities thus, or ac=b'^. Conversely. If the product of any two quantities is equal to 208 ALGEBRA. the square of a third, the third is a mean proportional between the other two. 7 Let acz=.lr\ and, dividing both by ^c, we obtain -=-, or h c a : b : : b : c. Prop. III. Of four proportionals, any three being given, the fourth may be found. Let a : b : '. c '. d ; then ad=bc. T-T i'C , ad ad . be Hence, a=— -; b= — ; c=— -; d=—. a c b a Hence, of three proportionals, any two being given, the third may be found; for ad=.b'^^ therefore b=/sfad^ «=^-j and d a Proi*. IV. Quantities which have the same ratio to the same quantity are equal to one another, and conversely. a c Let a : b : : c : b, then t=t l ^i^d, multiplying each by by we obtain a=c. Conversely. Quantities which are equal to one another have the same ratio to the same quantity. Let a=:c, and let 3 be a third quantity; then, dividing both by b, we obtain ■5-=T' therefore a : b : : c : b. Prop. N . Ratios that are equal to the same ratio are equal to one another. Let a '. b : : e : f, and c : d '. : e : f; then, also, a : b : : c '. d. (L S C S CL C Since -=7;, and -,=-7, it is evident -7=-^j ^"^^ therefore a : b of a J b d : : c '. d. Or let 2 : 4 : : 8 : 16, and 3 : 6 : : 8 : 16. Then 2 : 4 : : 3 : 6 ; for -,~ and ?=^. 4 2 2 Prop. VI. Iffour quantities are proportionals, they will also RATIOS. 209 be proportionals when taken inversely ; that is, the second will have the same ratio to the first that the fourth has to the third. Let a '. b '. : c : dy then b : a : : d : c. Since by Prop. I. bc=ad, And, dividing by ac, we obtain -=-, a c Hence, b : a : : d : c. Prop. VII. If four quantities are proportionals, they will also be proportionals when taken alternately ; that is, the first wiU have the same ratio to the third that the second has to the fourth. Let a : b : : c : d ; then, also, a : c : : b : d ; As j=-ji if we multiply each quantity by -, we obtain ah cb , . , - . . c 3 , _ , , 7— =—7-; which, reduced, is -=-, therefore a : c : : b : a. be cd c a This may be illustrated by numbers ; thus, Let 2 : 4 : : 3 : 6, then 2 : 3 : : 4 : 6 ; 2 3. . . .4 As 2;=pj if we multiply each side of the equation by •^, the 2434 8 12 24 result will be ^X^=gXg; j^— 18' s'^G' ^^^^^^'^^^ ^-^ : : 4 : 6. Prop. VIII. If four quantities are proportionals, they will also be proportionals when taken jointly ; that is, the sum of the first and second ^ill have the same ratio to the second that the sum of the third and fourth has to the fourth. Let a '. b : '. c : d, then a-\-b : b : : c-\-d : d. d C CL C Since -7= -j, we add 1 to each quantity, and obtain -4-1=- d b a , -I (Z-\-b c-{-d ,0 , , ,77 4-1, or — - — = — -— , thereiore a +c» : b : : c-f-d : d. b d This, also, may be made evident by taking numbers ; thus, Let 2 : 4 : : 3 : 6, then 2+4 : 4 : : 3+6 : 6. 18# 210 ALGEBRA. 2 3 Since ■7=tt> we add 1 to each number, and obtain 4 D 2 , , 3 , _ 2+4 3+6 -+1=^+1, or -^=^-. Therefore, 2+4 : 4 : : 3+6 : 6. Prop. IX. If four quantities are proportionals, they will also be proportionals by separation ; that is, the difference between the first and second will have the same ratio to the second that the difference between the third and fourth has to the fourth. Let a : b : : c : d, then a — b : b : : c — d : d. a c Since -7=3, we will subtract 1 from each quantity, and we o a , , . a - c - a — b c — d obtam - — 1=-^ — 1, or — ; — : — 7—. b d b d Therefore, a — b : b : : c — d : d. This demonstration may be illustrated by numbers ; thus, Let 4 : 2 : : 6 : 3, then 4—2 : 2 : : 6—3 : 3. 4 6 Since 75=0? we subtract 1 from each term, and we have 4 -,_6 4-2_6-3 2 3 '°^ 2 "~~3~* Therefore, 4—2 : 2 : : 6—3 : 3. Prop. X. If four quantities are proportionals, they will also be proportionals by conversion ; that is, the first term will have the same ratio to the sum or difference of the first and second, that the third has to the sum or difference of the third and fourth. Let a : b : '. c : d ; then a : a-±Jj : c : : c+ic?. Since -=-, and, by Prop. VIII. and IX., — = — = — r—, invert these b d b d fractions, and we have r= -', and, by multiplying the a-^^o c -\~ ct one by -, and. the other by its equal -, we obtain --Xt= b a a-hb b d c a c ,, „ . , , ,X jj or r-= 3, thereiore a : :?=T» T== DEMONSTKATION. I aXeXi cXgXl h d' f Ilk m' bXfXk dXhX^ Wlience —-=z-^—, therefore, aei : hfk : cgl : : dhm. hjk dhm q. e. d. ILLUSTRATION BY NUMBERS. 6x10x14. Prop. XIY. If there are any number of quantities more than two, and as many others, which, taken two and two in order, are proportionals, then, by equality, are the extreme terms in the former series proportional to the extreme terms in the latter. Let «, ^, c, d^ be any number of quantities, and let e, /, g^ A, be as many others. Let 2 : 3 : : 4 : 6 4 : 5 : : 8 : 10 6 : 7 : : 12 : 14 Then 2X4X6 : 3X5X7 :: 4X8X12 Whence 48 105 : : 384 : 840. Let a : b '. : e : f, b : c : : f : g, c : d : : g : h, Then, also, a : d h. CI- a e b f Since T=T> -=-5 ^nd , b f c g d DEMONSTRATION. C g h , we obtain, by multiplying the 1 o • 1 '^^c efg a e alternate tractions tos-ether, —-=-—-, or -^=t > thereiore, a : a ° bed jgh d k . e : h. Let 2 3 4 ILLUSTRATION BY NUMBERS. 3 4 12 6) 8 ( Then 2 : 12 : : 4 : 24. 24 By multiplying the alternate fractions, we have 2X3X4 : 3X4X12 : : 4x6x8 : 6x8x24. Whence, 24 : 144 : : 192 : 1152, or 2 : 12 : : 4 : 24. RATIOS. 21o Prop. XV. If there are any number of quantities more than two, and as many others, which, taken two and two, in cross order, are proportionals, then inversely, by equality, are the extreme terms in the first set proportional to the extreme terms in the second. Let a, h, c, d, be any number of terms, and e, jT, g, h, as many others, and Let a : h : : g '. h \ b : c : : f : g > Then, also, a : d : : e : k. c : d : : e : f ) DEMONSTRATION. Since -=^, -=-, and -=-%, by multiplyino; the alternate b k c g d f "^ r J o fractions together, we obtain ahc gfe a e bed hgf d It Therefore, a : d : : e : h. ILLUSTRATION BY NUMBERS. Let 2 o 3 : 4 4 : 3 Then, 2 : 3 : : 8 : 12. 2X3X4 : 3X4X3 : : 8x6x8 : 12x8x6. Whence, 24 : 36 : : 384 : 576. ^j dividing the first two terms by 12, and the last two by 48, we obtain 2 : 3 : : 8 : 12. Prop. XYI. When four quantities are proportionals, if the first and second are multiplied or divided by the same quantity, and also the third and fourth by the same quantity, the resulting quantities will be proportionals. Let a '. b '. \ c '. d ; then, also, ma : mb : : nc : iid. DEMONSTRATION. CL C Since T=y "^^ multiply both terms of the first by m, and both terms of the last by 7i, and we obtain — -= — . ; mo 7id 214 ALGEBRA. Therefore, ma : mh : : nc : 7id, where m and n may be any quantities, either integral or frac- tional. ILLUSTBATION BY NUMBERS. Let 2 : 4 : : 3 : 6. Now, if we multiply the first two num- bers by 7, and the last two numbers by 9, their products will be proportionals. Thus, 2X7 : 4X7 : : 3x9 : 6x9=14 : 28 : : 27 : 54; and if any other numbers were taken instead of 7 and 9, the products would be proportionals. Prop. XVII. When four quantities are proportionals, if the first and third are multiplied or divided by the same quantity, and also the second and fourth by the same quantity, the re- sulting quantities will be proportionals. Let a : b : : c : d, then, also, ma :71b:: mc : nd. DEMONSTBATION. a c 'tn Since -7=-j-, multiply both these quantities by — , and we obtain — y-= — r, therefore, ma : Tib : : mc : 7id, where m and no nd n may be any quantities, either integral or fractional. ILLUSTRATION BY NUMBERS. Let 12 : 4 : : 18 : 6, and we will multiply the first and third by 2, and the second and fourth terms by 4. "^ Thus, 12X2 : 4x4 : : 18x2 : 6x4=24 : 16 : : 36 : 24. It is evident these terms are proportionals ; 24 36 12 12 ^'' 16=24' '' -S=-S- And if we divide the first and third terms by 3, and the second and fourth terms by 2, their quotients wiU be proportionals. Thus, 124-3 : 4-^2 : : 18-^3 : 6-7-2. Or 4 : 2 : : 6 : 3. RATIOS. . 215 4 6 Whence, o"^^* If any other numbers be taken for multipljing or dividing, the result will be the same. Prop. XVIII. If four quantities are proportionals, the like powers or roots of these quantities are also proportionals. heta:b::c:d; then, also, a"' : h"" : : e" : d'". (Z c Since -=3, raise each of these fractions to the power ex o d t) = ( l ) ' "^^^ 1^'^1~^' therefore, c"' : IT- : : c'" : d"^, where m may be any quantity, either integral or fractional. ILLUSTRATION. Let 2 : 3 : : 4 : 6, then 2^ : 3'^ : : 4^ : 61 If we raise each of these terms to the third power, the result will be 2X2X2=8 : 3x3x3=27 : : 4x4x4=64 : 6x6x6=216. That 8, 27, 64, and 216, are proportionals, is evident from 8 64 the fact that ^-=7tt:^, and, beino; reduced to their lowest terms, 27 210 ° 27~27' Prop. XIX. Of any number of quantities in continued pro- portion, the first has to the third the duplicate ratio, to the fourth the triplicate ratio, to the fifth the quadruplicate ratio, &c., of that which it has to the second, or of that which the second has to the third, &c. Let a '. b : : h : c : : c : d : : d : e i : e : f : : &c. &c. Then a : c : \ a' : h'\ or in the duplicate ratio oi a '. h. a : d : : a' : h"\ or in the triplicate ratio of a : h. a : e : : a^ : Ij\ or in the quadruplicate ratio of a : b. DEMOXSTRATION. 1st. a : b : : b : c, OX, by Prop. XVIII., a^ : h- : : h' : c" ; but, by Prop. II., h'z=zac, therefore, a^ : P : : ac : c-. 216 ALGEBRA. ov a^ : P : : a : c, hence a : c : : a^ : b^ ; also, a^ : ac : : b^ : c^ ; therefore, a : c : : b^ : c^. 2d. a : c : : a^ : b-; but c : d : : a : b ; therefore, a : d : : a' : b^ : : P : c^ : : c' : (f. 3d. But d '. e '. \ (T \ W \ therefore, a : e : : ft^ : ^-^ : : 34 . ^4 . . ^4 . ^4 . . ^4 . g4_ The above may be easily illustrated by numbers. PROBLEMS FOR PROPORTION. 1. Divide 50 into two such parts that the greater, increased by 3, shall be to the less, diminished by 3, as 3 to 2. Let X = the greater number, and 50 — x the less. Then a:+3 : 50— a;— 3 : : 3 : 2. Multiplying extremes, 2x-\-Q=zlbQ — 3z — 9. Transposing, 5:r=135. Dividing, a;=27, the greater. And 50-27=23, the less. 2. What number is that to which if 3, 8, 12, and 20, be severally added, their sums shall be j)roportional ? Let X = the number. Then, a;+3 : x-j-8 : : x-{-12 : 2:+20. Multiplying extremes, a;2+23a:+60=2-+20:e+96. Transposing, 23z— 20^'=96— 60. Dividing, 2:=12. A7is. VERIFICATION. 12+3 : 12+8 : : 12 + 12 : 12+20=15 : 20 : : 24 : 32. 3. If Mars, when in opposition to the sun, is 49,000,000 miles from the earth, and the quantity of matter in the earth is 11 times greater than that in Mars, at what distance from the earth, in a direction towards Mars, will a body remain at rest ? See Art 218. Let X = the distance from the earth. Then 49,000,000— a; = the distance from Mars. And let ^=49,000,000. Then, a;2 : {a—xY : : 1 : 11. RATIOS. 21' Multiplying extremes, llx'=a^ — 2ax-\-2^. Transposing, 10x^-{-2ax=a'^. 2 Reducing, x--{-~z=—. Completing the squares, x-+-^^+—=-+—=-^-^. a Evolving, x-\-—=—Ayna\ a .-^-^ a Transposing, &c. a:=— a/11 — — . And, by supplying the value of a, we have 10^ ('ll(49,000,000)A-iM^M5?=ii,351,430miles. Ans. 4. There are two numbers which are to each other as 5 to 3 ; and, if 4 be added to the greater and 8 to the less, they will then be to each other as 6 to 5. What are the numbers ? Ans. 20 and 12. 5. Divide the number GO into two such parts that their pro- duct shall be to the difference of their squares as 2 to 3. A71S. 40 and 20. 6. I have two square house-lots, which, together, contain 208 square rods; and the area of the greater is to the area of the less as 9 to 4. IIow many more square rods are there in the greater than in the less ? Ans. 80 square rods. 7. The product of two numbers is 12, and the difference of their cubes is to the cube of their difference as 13 to 4. \Vhat are the numbers ? A7is. 2 and 6. 8. Divide the number 100 into two such parts that G times their product shall be to the sum of their squares as 24 to 17. What are those parts ? Ans. 80 and 20. 9. There are two numbers, whose pAduct is 35, and the dif- ference of their squares is to the square of their difference as G to 1. What are the numbers ? Aiis. 7 and 5. 19 218 ALGEBRA. 10. There are two numbers in the triplicate ratio of 4 to 1. whose mean proportional is 32. What are the numbers ? Ans. 25 G and 4. 11. Divide 20 into two such numbers, that the quotient ot the greater divided by the less shall be to the quotient of the less divided by the greater as 9 to 4. What are those numbers ? Am. 12 and 8. 12. Divide 26 into three such parts, that the first shall have the same ratio to the second that the second has to the third, and that the first term shall be ^ the third term. A.ns. 2, 6, and 18. SECTION XX. ARITHMETICAL PROGRESSIOIf. Art. 251. An Arithmetical Progression is a series of num- bers or quantities, increasing or decreasing by a constant difierence. It is sometimes called Fj'ogression by Difference. 252t The constant difi'erence is called the Common Difference^ or ratio of the progression. Ratio here used is an Arithmetical rate. Thus, let there be the two following series. (1) (2) (3) (4) (5) (G) (7) (8) First series, 1, 4, 7, 10, 13, 16, 19, 22=92. Second series, 30, 26, 22, 18, 14, 10, G, 2=128. 253. The numbers which form the series are called the terni^ of the progression. 254. The first is ci^led an ascending series of progression, where the first term is 1, the common diiference 3, the number of terms 8, the last term 22, nnd the snui of tlie series 92. ARITHMETICAL PROGRESSION. 219 255. The second is called a descending 5eWes of progression, where the first term is 30, the common difi'erence — 4, the number of terms 8, the last term 2, and the sum of the series 128. 256. The first and last terms of the progression are called extremes^ and the other terms are the means. 257. The number of common difi"erences in any number of terms is one less than the number of terms. Hence, if there be 8 terms, the number of common differences will be 7, and the sum of the differences will be equal to the difference of the extremes. We therefore infer, that if the difference of the extremes be added to the first term, the sum will be the last term ; also, if the difference of the extremes be taken from the last term, the remainder will be the first term. 258. Also, if the sum of the common differences be divided by the number of common differences, the quotient will be the common difference. To illustrate this, we will examine the following series : (1) (2) (3) (4) (5) (6) (7) 2, 5, 8, 11, 14, 17, 20. Here the first term is 2, the last term 20, the number of terms 7, and the common difference 3. Now, if we had only the first term, number of terms, and common difference, to find the last term, we should have only to add the difference of the extremes to the first term. The common difference is 3 ; and, as there are 7 terms, the number of common differences is 6. The difference of the extremes will, therefore, be 6X^=18, and the last term will be 2+18=20. Hence, having the first term, common difference, and number of terms given, to find the last term, we have the following Rule. Multiply the number of terms, less one, by the common difference, and to the product add the first term. 220 ALGEBRA. Again, if we invert the terms, we have (1) (2) (3) (4) (5) (6) (7) 20, 17, 14, 11, 8, 5, 2. Here we have 20 for the first term, — 3 for the common dif- ference, and 7 for the number of terms, to find the last term. 6X— 3=— 18 ; 20—18=2 the last term. The pupil will perceive that 18 is a negative term; and to add a negative term to a positive is to write their difference. Again, we have given the extremes 2 and 20, and number of terms 7, to find the common difference. Here the number of common differences is 6 ; for we have before shown that the number of common differences is always one less than the number of terms; therefore, 18-7-6=3, the common difference. 259» The principles of an arithmetical progression may be well illustrated by literal terms. Let a be the first term of an ascending series, and d the common difference ; then the second term will be a-\-d, and the the third term a-\-2d, and the series will be (1) (2) (3) (4) (5) (6) a, a-{-d, a-\-2d, a-\-od, a-{-4:d, a-\-bd. If it be required to form a descending series, when the first term is a and the common difference — d, it will be thus : (1) (2) (3) (4) (5) (6) a, a — d, a — Id, a — 3^, a — 4c?, a — bd. 260t It is evident that the last term in both series is equal to the first term with the common difference repeated as many times, wanting one, as there are terms in the series. Hence, if n represent the number of terms, the following will be the formula to find X, the last term. h=^a-\-{n — V)d. EXAMPLES. 1. If the first term be 7, the common difference 4, and the number of terms 20, required the last term. i=a+(7i-]y=7+(20-l)4=83. Am. ARITHMETICAL PROGRESSION. 221 2. If the first term is 3, the common difi"erence 5, required the 50 th term. jL=a+(w-l)6^=34-(50— 1)5=248. Ans. 3. If the first term is 90, the common difi"erence —7, re- quired the 10th term. X=a+(7i-l)(-^)=90+(10-l)(-7)=27. Ans. 4. If the first term is f , the common dijQFerence 1^, what is the 20th term ? i=fl+(7z-l)^=f+(20-l)l^=26TV. Ans. 5. If the first term is 18, the common difi"erence —4, what is the 10th term? X=a+(?i-l)(— ^)=18-f(10— 1)(-4)=-18. Ans. 261 . The formula for obtaining the first term, a, is obtained from the former by transposition. Thus, if X=a+(?z — l)d, then, by transposition, a=L — (n — l)d. 6. If the last term is 25, the number of terms 6, and the common difierence 2, required the first term. fl=i-(7i—l)^=25-(G— 1)2=15. Alls. 7. If the last term is 50, the common difierence 6, the number of terms 10, required the first term. a=L—{7i—'[)d=^0—(iO-l)Q=-4. Ans. 8. If the last term is 27, the common diff'crence 2^, number of terms 10, required the first term. a=L— {7i-l)d=27 -{10— 1)2^=U. A?is. 262. The formula for obtaining the common difi"ercnce, d, is obtained from the first by transposition and division. Thus, L=a^{?i—l)d. Then, by transposition, L — a={n — \)d. And by division, —=d, •^ 71 — 1 L — a, Changing terms, d= r- . 222 ALGEBRA. 9. If the extremes are 6 and 30, and the number of terms 13, what is the common difference ? , L — a 30 — 6 ^ . n — 1 Id — 1 10. If the extremes are ^ and 15j, and the number of terms 11, what is the common difference ? , L—a 15|— f ^ d= -=-^--i=li Am. n — 1 11 — 1 263. The formula for obtaining the number of terms may be obtained from the first formula. Thus, L=a-Y{n—V)d. By transposition, L — a={n — V)d. By division, — ; — =?2 — 1. -^ d By transposition, — - — \-\=.7i, T Changing terms, n= — j-1. 11. If the extremes m'e 3 and 39, and the common difference 2, what is the number of terms ? L-a . . 39—3 n=—, [-1= — \ f-l=19- ^-^is. rl V. 12. If the first term is 5, the last term 89, the common dif- ference 7, required the number of terms. L — a , ^ 89 — 5 ^ TO . n=— — ^^="7 [-1=13. Alls. Having, therefore, any three of the four terms given, the other may be found, as we have demonstrated above, by the following yORMULiE. (1.) To find the last term. L=a-j-{n — l)d. ARITH.AIET1CAL PROGRESSION. 223 (2.) To find the first term. a=L — [n — l)d. (3.) To find the common difference. J L — a d= =. n — 1 (4.) To find the number of terms. L — a ■1. d When the series are descending, the unknown difi"erence is a minus quantity in the 1st and 2d formula) ; thus, — d. 13. A man travelled 10 days; the first day he went 8 miles, the second day 13 miles, and thus increased his distance each day 5 miles. How far did he travel the last day ? Ans. 53 miles. 14. John Smith's family expenses for the first year were $500 ; but, after he had been married 12 years, he found his last year's expenses to have been $1325. By how much did he increase his expenses yearly ? A?is. $75. 15. A man set out from Boston to travel into the country ; the first day he travelled 12 miles, the second day 9 miles, the third day 6 miles, and thus continued to travel each day 3 miles less than the preceding. How far did he go the tenth day ? Ans. — 15 miles. 261. To find the sum of the series. ARITHMETICAL SERIES. (1) (2) (3) (4) (5) (G) Let 2, 5, 8, 11, 14, 17, be the series. And 17, 14, 11, 8, 5, 2, same series inverted. 19, 19, 19, 19, 19, 19, sum of both series. LITERAL SERIES. (1) (2) (3) (1) (5) (G) Let a, a-{-d, a-{-2d, a-\-od, a-{-4:d, a -{-5c? bo a series. And a-\-^d, a-{-4d, a-{-Sd, a-\-2d, a-{-d, a same series inverted. 2a-\-bd, 2a-[-bd, 2a-\-bd, 2a-[-bd, 2a-\-bd, 2a-f 5^, sum of both series. 224 ALGEBRA. We perceive, from the above aritlimetical and literal series, that the sum of the extremes is equal to the sum of any two of the means equally distant from each extreme ; and that, by adding the two series in their present arrangement, we have the same number for the same successive terms ; also, that the sum of both series is twice the sum of either series. Therefore, if 19, the sum of the extremes in the arithmetical series, be multiplied by 0, the number of terms, the product will be the sum of both series. Thus, 19x^=114, sum of both series. Therefore, 114-f-2=57 will be the sum of either series. Again, 2a-]-^d is the sum of the extremes in the literal series; and, if this sum be multiplied by 6, the number of terms, the product will be the sum of both series. Thus, {2a-{-M)Q=12a-{-S0d, sum of both series. And (12a+30*i) -^2=Qa-{-l^d, the sum of either series. Therefore, in all cases, we find that the simi of the series is equal to the sum of the extremes multiplied by half the number of terms ; or, the number of terms multiplied by half the sum of the extremes. If, therefore, the sum of any series be denoted by S, the first term by a, the last term by L, and the number of terms by w, the following will be the formula for obtaining its value : Therefore, if the extremes and the number of terms are given to find the sum of the series, we adopt the following EuLE. Multiply /lalf the sum of the extremes hij the number of terms. The two following formulae, or equations, contain five quan- tities : a, the first term of a progression ; i, the last term ; d, the common difierence ; n^ the number of terms ; and »S, the sum of the series. If any three of these be given, the other two may be ob- tained. (1.) L=a+{n-\)d. (2.) S=(^^^». S^f^^' AKITHMKTICAL PKOGEESSION. 225 265 1 The pupil will find that twenty different cases may arise which may be solved by difierent combinations of the above equations. To find n in the last equation. 2 By multiplication, ^Sz={^L-\-a)n. By division, -^^-- — =7^. L-\-a 2S Therefore, 7i-. 'L-{-a If, therefore, the extremes and the sum of the scries are given to find the number of terms, we divide twice the sum of the series by the sum of the extremes. 16. Let the extremes be 3 and 39, and the sum of the series 399, to find the number of terms. 2S 2x399 ^- . L-\-a o\}-\-6 266. To find the last term, L, from the second equation. By multiplication, 2S={L-\-a)n, By division, — =L-{-a. 2S By transposition, a=L. n 2S By transposition of terms, L=- a. Therefore, having the first term, number of terms, and sum of the series, given to find the last term, we divide twice the sum of the series by the number of terms, and subtract the first term from the quotient. 267 » To find the first term, a, from the second equation. 226 ALGEBRA. Multiplying, 2S={Li-a)n. 2S Dividing, — =L4-a. n Transposing, L=a. Chandnsf terms, ' <2= L. ° n Therefore, having the last term, number of terms, and sum of the series, given to find the first term, we divide twice the sum of the series by the number of terms, and subtract the last term from the quotient. 17. Let the last term be 39, number of terms 19, and the sum of the series 399, to find the first term. 2S 2X399 OQ o 4 a= X(= — o9=D. Atis. n 19 268 1 To find the common diflerence, d^ from the 1st and 2d equation. We find the value of X, in the first equation, to be h=.(i-\-{n — \)d. Substituting this value of L for iS in the 2d equation, and then transposing, we have ^ 2S—2an o= . ?l(7l — 1) 18. If the first term is 5, the number of terms 15, and the sum of the series 285, what is the common difference ? Ans. 2. 19. If the first term is 3, the number of terms 19, and the sum of the series 399, what is the common difference ? Alls. 2. 20. If the first term is 7, the number of terms 8, and the sum of the series 100, what is the common difference ? A?is. li. Problems. 1, The first term is 5, the common difference 3. What is the 7th term ? A?is. 23. ARITHMETICAL PROGRESSION. 227 2. The first term is 3, the common difference 4^. "What is the 5th term ? Ans. 20^. 3. The first term is 18, the common difference ^. What is the 7th term ? Am. 19J . 4. The first term is 7, the common difference 2^, and the number of terms 5. Required the last term. Ans. 17. 5. The first term is ■'^, the common difference |. What is the 10th term? Am. 7^^-. 6. The first term is 0, the common difference Ij-. What is the 20th term? Am. 28^. 7. The first term is 10, the common difference — 2. What is the 4th term ? Am. 4. 8. The first term is — 8, the common difference — 3. What is the 10th term ? Am. —35. 9. The first term of a descending series is 85, common dif- ference 7. Required the 10th term. Ans. 22. 10. The first term is 3i, the common difference 2|-. What is the 5th term, and the sum of the series ? Atis. 124-, and 39^. 11. The first term in a descending series is 2^, the common difference is ^. What is the 10th term, and the sum of the series? An^. {, and 13|. 12. The first term is a, the common difference is d. AVhat is the Tzth term ? A7is. a-\-d{?i — 1). 13. What is the sum of the odd numbers from 1 to 100 ? Am. 2500. 14. If the first term is 4^-, the common difference 3^, and number of terms 8, what is the sum of the series ? Aiis. 134. 15. If the first term is 7, the common difference — 4, and the number of terms 6, what is the sum of the series ? A?is. —18. 16. If the first term is 5, the last term 19, and the number of terms' 6, what are the other terms of the progression? Am. 7|, lOf, 13|, IQi. 228 ALGEBRA . 17. If the extremes are —9 and 18, and the number of terms 5, what are the other terms of the progression ? Atls 2i 4i llJ- 18. If the last term of an ascending series is 20, the com- mon difference 5, and the number of terms 8, what is the sum of the series ? Ans. 20. 19. There is a number consisting of three digits in arith- metical progression, whose sum is 12; and, if 896 be added to the number, the digits will be inverted. What is the number < Am. 246. 20. There is a certain island 50 miles in circumference. Two men, A and B, set out to travel round it. A goes 10 miles each day. B goes 2 miles the first day, 5 miles the second day, and 8 miles the third day, travelling each day 3 miles further than the day preceding. How far will A and B be apart the 8th day ? Ans. 30 miles. 21. John Smith and John Jones set out from Boston for the city of Washington, the distance being 440 miles. Smith started 5 days before Jones, and travels 15 miles per day. Jones travels 25 miles the first day, 23 miles the second day, and 21 miles the third day, travelling each day 2 miles less than the preceding. How far apart will Smith be from Jones at the end of the 20th day, and how far will each be from Washington ? Ans. 135 miles apart. Smith 140 miles from Washington. Jones 275 miles from Washington. 22. If the first term is ^, the common difference — -^-, and the number of terms 20, what are the last term and the sum of the series? . ( Last term, — 2f. ( Sum of the series, — 21§. 23. If one extreme is ■^, the common difference — yL, and the sum of the series — 1^, what is the number of terms ? Ans. 12. 24. If the first term is -/g^, last term 2J-, and the sum of the series 37, what is the number of terms ? Ans. 24. ARITHMETICAL PROGRESSION. 229 25. If the first term is 3, the last term 17, and the number of terms 29, what arc the terms of the series ? Am'. 3, 3^, 4, U, 5, 5i, &c. 26. The sum of the series is 16^, the number of terms 10, and the common difference -], to find the first term. Ans. 4. 27. The first term of an arithmetical series is — 5, the com- mon difierence 1^; what is the 9th term ? Ans. 7. 28. What are the three means between — 1 and 15 ? A71S. 3, 7, and 11. 29. The first term is 1^, number of terms 10, and the sum of the series 6|. What is the common difference ? A?is. — ^. 30. There are three numbers in arithmetical progression (vhose sum is 10, and the product of the second and third is 33^. What are those numbers ? Atis. — 3^, 3^, and 10. 31. The number of terms of an arithmetical progression is equal to J- the common difTerence, the last term is equal to 4 times the first, and the sum of the series is equal to f the square of the first term. What are the scries, and the sum of the series ? , The series, 20, 32, 44, 56, 68, 80. Atis. ■ i Sum of the series, 300. 32. There arc four numbers in arithmetical progression whose sum is 28, and the sum of whose squares is 216. What are those numbers ? Ans. 4, 6, 8, and 10. 33. Find three numbers in arithmetical progression whose sum is 9, and the sum of whose cubes is 99. Ans. 2, 3, and 4. 34. What are those four numbers in arithmetical progression the sum of the sc[uares of whose first two terms is 34, and the sum of the squares of the last two is 130 ? An£. 3, 5, 7, and 9. 35. A certain number consists of three digits, which are in arithmetical progression ; and, if the number be divided by the sum of its digits, the quotient will be 274, but, if 396 be added 20 230 ALGEBRA. to the number, the digits will be inverted. Required the num- ber. Ans. 579. 36. What are those four numbers in arithmetical progression the sum of the squares of whose extremes is 90, and the sum of the squares of the means is 74 ? Ans. 3, 5, 7, and 9. 37. What are those four numbers in arithmetical progression whose sum is 14, and whose continued product is 120 ? Atis. 2, 3, 4, and 5. 38. There are four numbers in arithmetical progression, the product of whose extremes is 112, and that of the means 120. What are the numbers ? Ans. 8, 10, 12, and 14. 39. A and B, 165 miles from each other, set out with a design to meet. A travels one mile the first day, two the second, three the third, and so on. B travels 20 miles the first day, 18 the second, 16 the third, and so on. How soon will they meet ? Atis. 10 days, or 33 days. 40. There are four numbers in arithmetical progression, whose continued product is 1680, and common difference is 4. Re- quired the numbers. Ans. 14, 10, 6, 2. 41. Five persons undertake to reap a field of 87 acres. The five terms of an arithmetical progression, whose sum is 20, will express the times in which they can severally reap an acre, and they all together can finish the job in 60 days. In how many days can each, separately, reap an acre ? A71S. 2, 3, 4, 5, 6 days. 42. A gentleman set out from Boston for New York. He travelled 25 miles the first day, 20 miles the second day, each day travelling 5 miles less than the preceding. How far was he from Boston at the end of the eleventh day ? A?is. 43. Suppose a number of stones were laid a rod distant from each other for twenty miles, and the first stone a rod from a basket. What length of ground will that man travel over, who gathers them up singly, returning with them, one by one, to the basket? Atis. 128,060 miles, 2 rods. ARITHMETICAL PROGRESSION. 231 There are twenty diflferent eases in Arithmetical Progression, all of which are exhibited in the followinc^ Table. No. Given. a, d, n a, d, S a, 71, S d, 71, S a, d, n a, d, I a, I, n d, n, I Requir'd. Formulae. I =a-\-{n — \)d. I I =^-JLd±^'ldS+{a-^df. n n 2 S=^?^[2a+(7^— 1)^]. S=- 1d ' S=ln{^1l—{ii—\)d). 9 I a, ?z, I 10 12 18 14 15 IG 17 18 19 20 a, n, o n ! «, z, s n, I, S I — a n — 1' d: d = 2S-2an n{7i — 1) ' 2S—l-a 27il-2S d= n[n — 1)* d, 71, I d, 71, S d, I, S 71, I, S a, d, I a, d, S a, I, S d, I, S a=l—{ii—\)d. S [ii—\)d 71 a=.id±^ {l-]-U)-—2dS. 2S a= L 71 71 I— a 7l = ± V ('2a—dY-ir^dS—2a-\-d 2d 2S 71= 2lJ^d±/s/ {2l-j-d)'—MS 2d ■ 232 ALGEBRA. SECTION XXI. GEOMETRICAL PROGRESSION, OR PROGRESSION BY QUOTIENT. Art. 269t When there are three or more numbers, such that the same quotient is obtained by dividing the second by the first, and the third by the second, and the fourth by the third, &c. ; or, such that they increase or decrease by a constant multiplier, they are said to be in Geometrical Progression, and are called a Geometrical Series. Thus, (1) (2) (3) (4) (5) (6) (1.) 2, 6, 18, 54, 162, 486 = 728, sum of the series. (2.) 486, 162, 54, 18, 6, 2 = 728, sum of the series. The first is called an ascending series, and the second a de scending series. In the first the quotient or multiplier is 3, and it is called the ratio. In the second the ratio is ^. The first and last terms of a series are called the ex- tremes, and the others are the means. 271. It will readily be perceived, in either of the above series that the product of the extremes is equal to the product of any two of the means equally distant from the extremes. Thus, 2x486=6x162=18x54=972. 272a If there are only three terms, the product of the ex- tremes is equal to the square of the second term. 273. It is evident, by examining either the above series, that any term may be obtained by multiplying the first term by the ratio as many times, wanting one, as there are terms required. If, therefore, the 1st term is 2, and the ratio 3, and we wish to obtain the 6th term, we have only to multiply the 1st term, 2, by the ratio 3, five times. Thus, 2X3X3X3X3X3=486, the 6th term. GEOMETRICAL PROGRESSION. 233 The above may be generalized in the following manner : Let a = first term of a series. L = the last term. r = the ratio. n = the number of terms. S = the sum of the series. (1) (2) (3) (4) (5) (G) Then a, ar, ar^, ar', ar^, ar^, &c., may represent any geometrical series ; and, if r, the ratio, is considered as more than a unit, the series is ascendiiig ; but, if r is less than a unit, the series is descending. The exponent of r in the second term is 1, in the third term 2, in the fourth term 3, in the fifth term 4, and so on ; there- fore, the exponent of r in the last term will always be one less than the number of terms. The exponent of the nih. term in the above series would therefore be Gir""^ 274. If, therefore, in any series the number of* terms be denoted by ti, and the last term by L, the following will be the formula for finc^ng the last term : (1.) L=ar'-\ And L=r"~^j when the first term is a unit. In the above equation we have four quantities, a, L, r, and ?i ; <^nd, if any three of them be given, the others may be obtained ds follows : To find a, the first term, we divide both terms of the above equation by 7-"~^, and transpose the terms ; and we have (2.) «=^- To obtain r, the ratio, we divide the terms of the 1st equa- tion by a, extract the (7i— l)th root, and transpose the terms ; and we have To find 71, we shall show when we come to treat of exponential quantities. 20^ 234 ALGEBRA. EXAMPLES. 1. If the first term is 7, the ratio 3, and the number of terms 5, required the last term. i=«r"-i=7(3)4=:567. Am. 2. If the first term is 1, the ratio 5, and the number of terms 5, what is the last term ? i:=r"-^=54=625. A71S. 3. If the last term is 405, the ratio 3, and the number of terms 5, what is the first term ? L 405 , , 4. If the last term is 8, ratio 5, and the number of terms 4, what is the first term ? Z 8 8 ^ ^.n-l 54-1 125 5. If the first term is 5, the last term 1215, and the number of terms 6, what is the ratio ? /L\-L- /1215\J_ ^, i ^ , r={ - l"-i= — ^— V'-i =243^=3. A71S. i!)^-iT} 27 6. If the first term is ^, the last term — — , and the number of o2U terms 4, what is the ratio ? r= 7. If the first term is y^g, the last term 64, and the number of terms 6, required the ratio. Ans. 4. 8. If the last term is 135, the number of terms 4, the ratio 3, what is the first term ? A71S. 5. 275. To find any number of geometrical means between any two given numbers. n-l IJ^ In the 3d formula, we found r= — . GEOMETRICAL PllOGRESSION. 235 If we let m represent the number of means, then ?n-\-2=i7i , for the number of terms is always two more than the number of means. Therefore, ( — \'-J= (^ - \ "'+i. Consequently, ?•= f — j"'+^ 276. Having, therefore, the extremes given to find any num- ber of means, we divide the greater extreme or number by the less extreme, and extract that root of the quotient denoted by the number of means plus 1. This root is the ratio ; and having the ratio, the means are readily obtained. EXAMPLES. 9. Find two geometrical means between 6 and 162. lG2-r-6=27 : /^27=3, the ratio ; 6x3=18, the first mean ; 18x3^54, the second mean. 10. What is the geometrical mean between 18 and 882 ? 882-^18=49 : V49=7, the ratio; 18x7=126, the geo- metrical mean. 11. Required the five geometrical means between 1 and 64. Am. 2, 4, 8, 16, 32. 12. A has a piece of land, which is 18 rods wide, and 288 rods long. Required the side of a square piece that shall con- tain an equal number of square rods. Ans. 72 rods. 277. To find the sum of all the terms of a geometrical series. Let the following be the series : (1.) 2, 6, 18, 54, 162. By examining this series, we find the first term 2, the ratio 3, and the last term 162. If we multiply each term in the series by the ratio 3, we obtain (2.) 6, 18, 54, 162, 486. It is evident that the simi of this last series is three times the 236 ALGEBRA. former ; therefore the difference between them will be equal to twice the sum of the first series. Thus, From 6, 18, 54, 162, 486, second series. Take 2, 6, 18, 54, 162, first series. — 2 486=484, difference of the series. From the above operation, it appears that 484 is twice the sum of the first series ; and, therefore, 484-7-2=242 is the sum required. Bj examining the process, we perceive that 242 is obtained by multiplying the last term of the first series, 162, by the ratio 3, and subtracting from the product the first term 2, and dividing the remainder, 484, by 2 a number which is one less than the ratio. Hence the propriety of the following KuLE. Multiply the last term by the ratio, find the difference between this product and the first term, divide this remainder by the difference between the ratio and unity, and we have the sum of the series. 278. We may generalize the above, as follows : Let a represent the first term of a geometrical series, r the ratio, L the last term, n the number of terms, and S the sum of the series. Then (1.) S=a-\-ar-\-ar^-\-ar^-\-ar^-\-ar^. We next multiply each term of the above equation by r, and we have (2.) Sr=ar-\-ar'^-{-ar^-Yar'^-{-ar^-{-ar^, By subtracting the first equation from the second, we have Sr — S=ar^ — a. Dividing by r — 1, we have the formula for finding the sum of the series ^ ar^—a af'—a (?'"— 1) iS= =-, or ^r-, or a =— . r — 1 r — 1 r — 1 If the ratio is less than a unit, we transpose the terms, thus : ^ a—ar"^ a—ar"" (1— r") 6=^j , or — , = a— . 1 — r 1 — ?• 1 — r GEOMETRICAL PROGRESSION. 237 270i The index of the ratio is always equal to the number of terms. By the above formulas, we have a method for finding the sum of the series without the last term, which may be expressed by the following Rule. Raise the ratio to a power wTiose exponent is equal to the number of terms ; inultiply this power by the first term, find the difference between this product and the first terra, and divide this remainder by the difference between the ratio and unity. If we substitute the value of L as found in Art. 274, we shall have ^ Lr — a o= — . r — 1 A rule for this formula would be the same as in Art. 278. 13. If the first term is 7, the ratio 3, and the number of terms 5, what is the sum of the series ? SJ'-^JJ^^^UI. Ans. r — 1 o — 1 14. If the first term is 9, the ratio f , and the number of terms 4, what is the sum of the series ? „ a—ar'' 9— (9x(^)^ . , 15. If the first term is 144, the ratio 1.06, and the number of terms 4, what is the sum of the series ? Ans. 629.945. 16. If the first term is 9, the ratio |-, the number of terms 6, what is the sum of the series ? Am. llJ-§f^. 17. If the first term is a, the ratio r, and the number of terms n, required the sum of the series. . ar^—a «(?•"— 1) r — 1 r — 1 18. If the first term is 1, the ratio 2, and the number of terms 7, what is the sum of the series ? Am. 127. 238 ALGEBKA. 19. If the first term is 5, the ratio 10, and the number of terms 7, what is the sum of the series ? Aiis. 5555555. 20. If the first term is 4, the ratio ^, and the number of terms 5, what is the sum of the series ? Arts. 5f^. 21. If the first term is 5, the ratio -i, and the number of terras 5, what is the sum of the series ? Ans. Gy^^. 22. A gentleman agreed with another to board him for 9 days ; he was to pay 3 cents for the first day's board, 9 cents for the second day, 27 cents for the third day, and so on, in this ratio. What was the amount of the bill for the gentleman's board? Ans. $295.23. To find L, r, and a, from the following equation. r — 1 MultijDlying by r — 1, Sr — S=L7' — a. Resolving into factors, S{r — l)=Lr — a. Transposition, Lr=S{r—l)-\-a. Division, L= . ?• To find r from the above equation. r — I Multiplying by r — 1, Sj' — S=Lr — a. Transposing, Sr — Lr=S — a. Dividing by S — L, ^= c — f To find a from the above equation. r — 1 * Multiplying by r — 1, Sr — S=Lr — a. Transposing, a=Lr — (r — 1)S. 23. If the first term is 3, the ratio 2, and the sum of the series 93, what is the last term ? A7is. 48. a EOM ETHICAL PROGRESSION. 239 24. Insert three geometrical means between l and 128. Am. 2, 8, 32. 25. If tlie first term is 2, the last term 4374, and the number of terms 8, what is the ratio ? Ans. 3. 26. If the ratio is 2, the number of terms G, and the great- est term 128, what is the least term ? Ans. 4. 27. If the first term is 3^, the ratio f , the number of terms 8, what is the last term, and what is the sum of the series ? Atis. Last term -fW^^' ^"^ *^® ^^^ ^^ series ^^^ij-^- 28. If the first term is 1, the last term 64, and the number of terms 7, what are the ratio, and the sum of the series ? A71S. Ratio, 2 ; the sum of the series, 127. 29. If the last term is 64, the number of terms 7, and the sum of the series 127, what are the ratio, and the first term ? Ans. Ratio, 2; the first term, 1. 30. If the first term is 2, the ratio 4, and the number of terms 12, what are the last term, and the sum of the series ? Ans. Last term, 8388608; sum of the series, 11184810. 31. The product of three terms in geometrical progression i.s 64, and the sum of their cubes is 584. What are those num- bers ? Ans. 2, 4, 8. 32. There are four numbers in geometrical progression, the second of which is less than the fourth by 24, and the sum of the extremes is to the sum of the means as 7 to 3. Required the numbers. A?is. 1, 3, 9, 27. 33. It is required to find four numbers in geometrical pro- gression, such that the difierence of the two means shall be 14, and the difference of the extremes 49. A?is. 7, 14, 28, and 56. The following are the two fundamental equations from wliich the twenty difi"erent cases are exhibited, — r — 1 and which are found in the followino; 240 No 13 14 ALGEBRA. TABLE. Giv( 1 a, r, n 2 a, r, S 3 a^ n^ S 4 7', 71, S 5 a, r, 71 G a, ?*, I 7 a, 71, I 8 7\ 71, I 9 a, 71, I 10 a, 71, S 11 a, I, S 12 71, I, S r, 71, I r, 71, S 15 r, I, S 16 71, I, S 17 a, 7', I 18 a, r, S 19 a, I, S 20 r, I, S Requir'd. Formulae. 7' l{S-l)"-^=a{S—aT-\ r"— 1 S= ar — a S= S= S= Ir — a r-1 ^" ^."— 1 o;?*" — 7'S=a — S. S — a a=- (r-l)S a=- -— . r"— 1 a=lr — (r — 1)S. a{S-a)"-'=l{S-l)"-\ loo;./ — loo;. a , ^ 71=^^, \-l. log. r 71 = log. [a-\-{r—l)S]—\og. a log. r loo;. ^ — \ocr.a 71= log.{S-a)-\og.{S-l) +1- 71 log.Z— log. pr— (r— 1)S] log. ?• HARMONICAL PROGRESSION. 241 The last four cases in the preceding table can be performed only by the aid of logarithms, as they belong to exponential or transcendental equations. They will, therefore, receive atten- tion in their proper place. SECTION XXII. HARMONICAL PROGRESSION. Art. 280i Three numbers are said to be in harmonical pro- gression when the first is to the third as the difference between the first and second is to the difference between the second and third. Thus the numbers 3, 4, 6, are in harmonical proportion. For 3:6:: 4—3 : 6-4. Or fl, b, c, are in harmonical proportion when a : c : : h — a : c — h. Thus, if the length of three strings of a musical instrument be as the numbers 3, 4, G, they will sound an octave 3 to 6, a fifth 2 to 3, and a fourth 3 to 4. 281. Four numbers arc in harmonical proportion when the first is to the fourth as the difference between the first and second is to the difference between the third and fourth. Thus the numbers 5, 6, 8, 10, arc in harmonic proportion. For 5 : 10 : : 6-5 : 10-8. Strings of such lengths will sound an octave 5 to 10, a sixth greater 6 to 10, a third greater 8 to 10, a third less 5 to 8, and a fourth 6 to 8. 282. Any number of quantities, a, Z*, c, rf, e, &c., are in har- monical progression if « : c : : a — h : h — c; b : d : : h — c : c — d ; c : e : : c — d : d — e, &c. 21 242 ALGEBRA. 283. The reciprocal quantities in harmonical progression are in arithmetical progression. Thus, if a, b, c, d, e, &c., are in harmonical progression, 1 1 1 1 1 p .„ , . -, -, -, -, -, &c., will be m arithmetical progression. a b c a e ^ ° SECTION XXIII. INFINITE SERIES. Art. 284. An infinite decreasing geometrical series is one whose ratio is less than unity, and the number of whose terms is infinite. To find the sum of an infinite series decreasing in geometrical progression. We have already found. Art. 277, that the sum of a descend- ing series in geometrical progression may be ascertained by the following formula. _, a — «?•" ^ a ar"" o=— , or o=: 1 — r 1 — r 1 — r 285. Now, if ?'" be a fraction less than a unit, it is evident that the greater the number ?z, the smaller will be the quantity r". If, therefore, a great number of terms of a descending series be taken, the quantity r" will be very small ; and, if we suppose n greater than any assignable number, then the quan- tity, or its value, may be considered as nothing = 0. Hence the latter part of the formula, — , should be 1 — r omitted, and it will stand a Thus, S=-. \-r The rule, therefore, for finding the sum of the series, is as follows : Rule. Divide the first term by the dijfcrence between unity arid tJie ratio. INFINITE SERIES. EXA3IPLES. 243 1. AYhat is the sum of the infinite series, 1, ^, ^, ^V' ¥i' &c.? 1 1 =14. Atis. 1 1 2 ^ 2. What is the sum of 1, ^, ^i, |, &c., to infinity ? J.715. 2. 3. What is the sum of the series, 8, |, ^\, yf y, &e., carried to infinity ? Aiis. 10. 4. Find the value off, 4, ^, yV' <^^'» *^ infinity. J.7Z5. 1*. 5. Find the value of 4, 1, ^, y^g, &c., to infinity. A?is. 5i. 6. AYhat is the exact sum of 1, yV, i^^' <^c-» ^^ infinity ? ^?w. 1^. 7. Find the exact value of the circulating decimal .444, &c., to infinity. .444, &c.=^+y^^+y^VTT. tlie ratio being J^- jt _* ICF TO" _^ VIO :tO * Atj^ 1 1 9 [See National Akithmetic, page 128.] 8. What common fraction will exactly express the value of the repeating decimal .454545, &c. ? .454545=^^^+y^VW+inJxr*^TjTr> t^e ratio being y^. 4 5 4 5 Ta^ l^*^ 45 VlOO 4500 5 /I77C •^ 10 loTJ 9. What common fraction is the exact value of the decimal .571428? A?2s. f 10. What common fraction is the exact value of .857142 ? Atis. f. 11. AVhat is the exact value of .53 ? •5o=y^ and TTTTr~riooo I iooott? &c. 3 1- TD^y TO'TJ ;? v/ 1 30 1 . 5 I l 8 A-ntf TU TTT 12. What is the value of .138 ? A?is. -^^. 13. Find the ratio of an infinite series whose first term is 8, and the sum of the series 10. Atis. 4-. 244 ALGEBRA. 14. Find the ratio of an infinite series whose first term is |, and whose sum is IJ-. Ans. J-. 15. Find the first term of an infinite progression of which the ratio is \, and the sum 10. Ans. 8. SECTION XXIV. SIMPLE INTEREST. Art. 286. Interest is the compensation which the borrower makes to the lender for the use of a certain sum of money for a given time. Pi'incipal is the sum lent. Rate per cent, is the sum agreed on for the loan of $1, or $100, for one year. Amownt is the sum of the interest and principal. Legal interest is the rate per cent, established by law. Let p = principal. r = rate per cent., written in hundredths. t = time in years. a = amount. i or a — p = interest for the given time. Hence, if r be the interest of one dollar for one year, it is evident that the interest of ^ dollars will be p times r=pr. And ifpr be the interest of p dollars for one year, it is cer- tain that for t years it will be t times as much, = ^^ifr, and that p~\-ptr will be the amount, and i or a — p will be the interest. 287. Hence, having the principal, rate per cent., and time given, to find the interest and amount, we have the following formulae : Formula for the interest, i=ptr. Formula for the amount, a=p-\-ptr. SIMPLE INTEREST. 245 From the preceding formulae we have, for finding the interest and amount, the following KuLE. Multiply the 'princiiml by the rate per cent., considered as a decimal, and this product by the time in years, and the result is the interest. If there are months and days, let the months be considered as fractions of a year, and the days as fractions of a month. By adding the interest to the principal, zve have the amount. [See National Akithjietic, page 164.] EXAMPLES. 1. What is the interest of $740 for 4 years, at 6 per cent. ? i=;??r=740x.06x4==$177.60. Ans. 2. What is the interest of $380 for 10 years, at 5 per cent. ? Alls. $190. 8. What is the interest of $890.75 for 3 years, 6 months, at 8 percent.? ^725. $249.41. 4. What is the interest of $17.18 for 5 years, 2 months, 10 days, at 4i per cent. ? Ans. $4.02. 5. What is the amount of $144 for 3 years, at 8 per cent. ? a=;7+;?r^=144+(144x.08x3)=$178.56. Ans. 6. What is the amount of $800 for 6 years, 1 month, 12 days, at 6 per cent. ? Ans. $1093.60. 7. What is the amount of $670.18 for 3 years, 7 months, 20 days, at 9 per cent. ? Ans. $889.66. 288. Having the amount, time, and rate per cent, given, to find the principal. By transposing, &c., the last equation, we have a From which we have the following KuLE. Mzdtiply the time by the rate per cent., and add 1 to the product ; ivith this sum divide the amount, and the quotient is the principal. 21^' 246 ALGEBRA. 8. Keceived $472 for a certain sum that had been on interest, at 6 per cent., for 3 years. What was the sum lent ? a 472 ^Ar^r^ A ^ l+tr 1+(3X.06) 9. What principal will amount to $570 in 10 years, at 5 per cent.? Am. $380. 10. What principal will amount to $1140.16 in 3 years, 6 months, at 8 per cent. ? Ans. $890.75. 11. Lent a certain sum for 5 years, 2 months, 10 days, at 4^ per cent., and received interest and principal $21.20 ; what was the sum lent? tI?!^. $17.18. 12. My friend borrowed of me a certain sum, which he kept 3 years, and for which I charged him 8 per cent., and received interest and principal $178.56. What was the sum I lent him ? Am. $144. 13. Keceived as interest and principal $889.66 from a friend to whom I had loaned a certain sum for 3 years, 7 months, and 20 days, at 9 per cent. What was the consideration of his note? Am. $670.18. 289. Having the amount, principal, and rate per cent, given, to find the time. By transposing and reducing the last equation, we have the following formula for finding the time, t. rp rp From the above formula we have the following Rule. Divide the interest hy the jrroduct of the principal multiplied by the rate per ce7it., and the quotient is the time. [See National Arithmetic, page 181.] 14. How long will it require $300 to amount to $372, at 6 per cent. ? T . . «-^ 372-300 ^ Let t= =-7T7: — Frr77:=4 years. Ans. rp .06X300 -^ 15. In what time will $380 amount to $570, at 5 per cent. ? Ans. 10 vears. SIMPLE INTEREST. 247 16. Lent, at 8 per cent., $890.75, for which I received $1140.16 ; for how long time was the money lent ? A?is. 3 years, 6 months. 17. For $17.18, which was loaned at 4J- per cent., there was received $21.20. For how long time had it been lent ? Atis. 5 years, 2 months, 10 days. 18. The interest and principal, on a certain sum, at 9 per cent., are $889.66; and the interest is $670.18 less than the amount. How long was the money at interest ? A71S. 3 years, 7 months, 20 days. 19. A has B's note, dated January 1, 1851, for $320, at 9 per cent. When will the note amount to $353.60 ? A?is. March 1, 1852. 290« Having the principal, interest and time given, to find the rate per cent. By transposing the last formula, we obtain the following for finding r, the rate per cent. Thus, a — p i " or pt jjt The pupil will perceive that the amount is known when the interest and principal are given. What is the rate per cent, for $300, that it shall amount to $372 in 4 years? a-p 372-300 „. ^ r= =-— — — --= .06, or o per cent. pt 300x4 ^ Hence we deduce the following Rule. Divide the interest by the product of the principal inulliplied hy the time, and the quotient is the rate per cent. 20. If $380 amount to $570 in ten years, what is the rate per cent. ? Aiis, 5 per cent. 21. Lent $890.75, for 3 years, 6 months, and received for the amount $1140.16. 'WTiat was the rate per cent. ? Ans. 8 per cent. 22. If $17.18 amount to $21.20 in 5 years, 2 months, and 10 days, what is the rate per cent. ? Ans. 4:^ per cent 248 ALGEBRA. 23. If the interest of $670.18 for 3 years, 7 montlis, and 20 days, be $219.48, what is the rate per cent. ? Ans. 9 per cent. 24. John Smith, Jr., gave me his note, dated January 1, 1848, for $144 : but he having been unfortunate in business, I agreed, May 7, 1851, to give him up his note for S153.64.8. What per cent, did I receive ? Ans. 2 per cent. 25. My tailor informs me that my " freedom suit " will re- quire 7^ square yards of cloth ; but the cloth I am about to purchase will shrink 5 per cent, in width, and 4 per cent, in length, and the cloth is 60 inches wide. How many yards must I purchase ? Ans. 4 yards, 38{| inches. SECTION XXV. . DISCOUNT AT SIMPLE INTEREST. Art. 291 . Discount is an allowance for the payment of any sum of money before it becomes due, and is the difference between that sum and its present worth. The present worth of any sum due some time hence is such a sum as, if put at interest, would, in the time for which the dis- count is to be made, amount to the sum then due. To find the worth of any sum due at any time hence : Let S = the sum due. p = the present worth. t = the time in years. r = the rate per cent, considered as so many hun- dredths. We have before shown, in Art. 287, that a=p-^ptr. DISCOUNT AT SIMPLE INTEREST. 249 We now substitute S for a, and consider p to represent the present worth ; and, by transposing the equation, find S from which we deduce the following Rule. Multiply the time ly the rate per cent., add 1 to the product, and divide the sum on vjhich the discount is to he taken by this su.m, and the quotient is the present vjorth. If the present worth is taken from the sum due, the re- mainder is the discount. [See National Arithmetic, page 187.] 1. What is the present value of $500, due 4 years hence, at 6 per cent. ? ;?=,-^-=--4^^-=$403.22+. Ans. ^ l-\-tr 1 + (4X.0G) ^ By transposing the quantities in the above formula, we may obtain the values of 5, t, and r. 2. What is the present worth of $372, due 4 years hence, at 6 per cent. ? Ans. $300. 3. What is the present worth of 8133.20, due 20 months hence, at 8|- per cent. ? An^. $117.09, 4. What is the discount on $21.20, due 5 years, 2 months, 10 days hence, at 4i per cent. ? Ans. $4.02. 5. A has B's note, dated January 1, 1851, for $353.60, to be paid March 1, 1852, without interest. What was the value of this note at the time it was given, if 9 per cent, discount is allowed? Ans. $320. 6. Which is worth the most, A's note for $144, due 10 years hence, at 6 per cent., or B's note for $176.40, due 8 years hence, at 12 per cent. ? Ans. 7. A legacy of $1725 is due one year hence. What is its present value, at 15 per cent. ? Am. $1500. 8. James Brown has S. Smith's note for S162, payable 6 250 ALGEBRA. months hence ; but Brown, being obliged to raise money, sold the note for $150. What per cent, did he allow ? Ans. 16 per cent. 9. Bought a farm for $590, for which I was to pay in a cer- tain time, without interest ; but, by making prompt payment, I was allowed a discount of 6 per cent, for the whole time, and paid only $500. How long was the time allowed for payment ? A71S. 3 years. 10. Bought a horse for $200, and gave my note, payable in 60 days. What ready money, at 15 per cent., will discharge the debt? Ans. $195.12+. 11. What is the present worth of $1827, due 100 years hence, at 6 per cent. ? Ans. $261. SECTION XXVI. PARTNERSHIP, OR COMPANY BUSINESS. Art. 292i Partnership is the association of two or more persons in business, with an agreement to share the profits and losses in proportion to the amount of the capital stock con- tributed by each. EXAMPLES. 1. Three men. A, B and C, enter into partnership for two years, with a capital of $1600. A puts into the firm $300, B $500, and C $800. They gain $320. What is each man's share of the gain ? Let a: = A's gain. Then, as each man's share of the gain will be in proportion to his stock. And -— - = B's gain. 8a; — = C's gain. PARTNERSHIP, OR COMPANY BUSINESS. 251 And x+^-i-^ = S320. o o Sx+bx-\-Sx = 960. 16a; = 960. X = 60 = A's gain. ::^ = 100 = B's gain. o ^ = 160 =r C's gain. o VERIFICATION. 60-f-100-|-160=$320. Or, let m, n, and 'p represent A, B, and C's stock, and a the sum gained. Also, let :6- = A's gain. Then, it is evident that each man must receive according to his capital. That is, as A's stock is to his gain, so will B's stock be to his gain, &c. nx Therefore, m : x : '. n : — = B s. m px ^ And m : X : : p : — = C s. m Then, x-] \-^ = a. m m And 7nx-\-nx-\-px^=am. ^, , am 320x300 „,.„ ,, . Therefore, a:=— —=—-—4^--— -=$00, A's gam. m-\-7i-{-p oUO-f-oOO-j-oOO Then, by the principle above stated, a7?i an 320x500 ^, «i • • • 77 • = =8100, B s gam. '''•ra-\-n-\-p' m+n-\-p 300+500+800 ^ ' ^ And, am ap 320x800 „^ ^, m-\-7i-{-p m-\-n-\-p oUU+oUU+oUU 252 ALGEBRA. •VERIFiaA.TION. am an ap " [m~{-n-\--p)a =a=$320. m-\-n-{-p 7n-\-n-\-p m-\-7i-\-p m-\-n-\-p Therefore, to find the gain or loss on any man's stock, we deduce from the above formulae the following Rule. Multiply the whole gain by each man^s stock, and divide the product by the whole stock. 293. Having each man's gain, and the amount of stock given, to find each man's share in the stock. . 2. A, B, and C, while in trade, gained as follows. A gained $50, B $70, and $90. The amount of their stock in trade was $4200. What was the amount of each man's stock ? It is evident that each man's stock was in proportion to his sain. Let a;=A's stock. Ix Then ~=B's stock. 5 9a; And --=C's stock. 5 Therefore, :.+-^4-:^=4200. 5:^_|_7x+92-=21000. 21:r=21000. a:=1000. A's stock. 5=1400. B's stock. 5 ^=1800. C's stock. 5 4200. Proof. If we change the symbols of the first question, putting m, 7^, and p, for the gain of each man respectively, and a for the stock, wc obtain the followinor formuhx) for tindinji; • the amount of each man's stock : PARTNERSHIP, OR COMPANY BUSINESS. 253 ma 50x4200 m-\-n-{-p 50+70-1-90 na 70x4200 m-\-n-]-p 50-j-70-f90 pa 90x4200 = $1000. A's stock. $1400. B's stock. = $1800. C's stock. m-^n+p 504-70-1-90 Hence, for finding each man's stock, we have the following Rule. Midtiply the whole stock hy each man^s gain, and divide the product hy the lahole gain. 3. Two men, M and N, engaged in trade. M put in $500, and N $750. They gained $120. What is each man's gain ? Ans. M gained $48, N gained $72. 4. Q and X hired a field for $120, which they used for a pasture. Q put in 11 cows, and X 15 cows. What sura should each man pay ? Ans. Q pays $50.76if , X pays $69.23^5. 5. A and B purchased a factory for $17,000. A paid $10,000, and B the remainder. They gained $1500. What sum should each receive ? Ans. A $882-iV, B $0171^. 6. A, B, and C engaged in trade, with a capital of $6000. They gained $240. A's share of the gain was $100, B's $80, and C's $60. What part of the stock did each own ? Ans. A $2500, B $2000, and C $1500. 7. A, B, and C hire a pasture for the season for $100. A put in 5 horses, B 7 oxen, and C 9 cows. Two horses eat as much as 3 oxen, and 4 oxen eat as much as 5 cows. What part of the expense must each pay ? Ans. A pays $34.56^Y_^ B pays $32.25i^^, and C pays $33.17f if . 8. Three men. A, B, and C, agreed to reap a field that was 40 rods square for $32. A reaped a part that was 25 rods square, B reaped 400 square rods, and C the remainder. What sum did each receive? Ans. A $12.50, B $8, C $11.50. PARTNERSHIP ON TIME, OR DOUBLE FELLOWSHIP. 9. A, B, and C engaged in trade. A put in $2000 for 4 months, B put in $3000 for 8 months, and C put in $4000 for 22 254 ALGEBRA. 12 months. They gained $780. What is each man's share of the gain ? Let w, n, p, represent each man's stock, a the whole gain, and t, f, t'\ the time each man's stock was in trade. It is evident that each man's stock gains not only in proportion to its sum, but also in proportion to the time it is in trade. For $2000 will gain four times as much in four months as it would in one month, and $2000 for four months is the same as $8000 for one month. We must, therefore, multiply each man's stock by the time it was in trade. It is therefore evident, that as A's gain is to B's gain, as A's stock multiplied by his time is to B's stock multiplied by his time, &c. Let a:, ?/, z = A, B, C's gain respectively. X : y : : mt : nt', 7it'x y= — — = B s gam. Then Multiplying extremes, &c., And Multiplying extremes, &c., 2r= mt z : ; pt"x And Multiplying by mt Therefore, mta ntx X- lilt pf'x mt : pt". = C's gain. •a. mt mt mtx-\-nfx -\-pf'xz=mta. mt-{-7it'+pt' But 2000X4X780 ^ 2000x4+3000x8+4000x12 * A'^gaia y- nt'x mt ' . _ , ... nt' mta nt'a And by substitution, 2/=-,X,;;^^q:^,q:^= ^^^_^^^,,_^^,= ggggX8x780 2000X4+3000X8+4000X12 ^ plf'x And And by substitution, mt PAllTNERSHIP ON TIME. 255 pf mta pf'a mt mt-\-nt' -\-pt" mt+nt'-{-pt" 4000X12 X780 __ . 2000X4>3000X«+4000X"12" ^ O s gam. The above equations, by dividing the numerators each into two factors, may be expressed by the following proportions : 7nt-\-nf -\-pf' : mt : : a : x. mt-{-nt'-{-pt" : nf : : a : y. mt-\-nt'-\-pt" : jyt" : '. a : z. Hence the following arithmetical E-ULE. Multiply each marCs stock by the time it was continuea in trade, aiid then say, As the sum of all the products is to each man's product, so is the whole gain or loss to each man's gain or loss. [See National Arithmetic, Sec. LVI.] 10. A commenced business January 1, 1850, with a capital of 83000. May 1, 1850, he took B into partnership, with a capital of $4000. January 1, 1851, they had gained $340. What was each man's share of the gain ? Ans. A's gain $180, B's gain $160. 11. A, B, and C traded in company. A put in $300 for 10 months, B put in $400 for 8 months, and C put in $600 for 2 months. They gained $120. What is the gain of each ? Ans. A's gain $48.64^1, B's $51.89/^, C's $19.45f f. 12. Three men, A, B, and C, hire a pasture in common, for which they are to pay $76.80. A put in 24 oxen for 12 weeks. B put in 25 oxen for 12 weeks, and C put in 30 oxen for 6 weeks. What sum ought each to pay ? Am. A $28.80, B $30, C $18. 13. John Jones hired a house for one year for $500, with the privilege of admitting two more families if he pleased, with the understanding that all the occupants should have equal priv- ileges in the house. At the end of three months he took in John Smith, and at the end of 9 months Richard Roe. What share of the rent should each pay? ^7?.^. Jones $291f , Smith $166|, Roe $41f . 256 ALGEBRA. 14. Two men, A and B, hired a coach in Boston to go to Worcester, the distance being 42 miles, for $20, with the privi lege of taking in two persons more. Having rode 80 miles, they take in C ; and on their return from Worcester, when within 20 miles of Boston, they take in D. What ought each man to pay for his accommodation in the coach ? Am. A $7.46^1^, B $7.46^f j, C $3.88f||, D $1.19^l2_ 15. A and B engage in trade. A puts in a dollars for b months, B puts in c dollars for d months, and they gain e dol- lars. What share of the gain shall each receive ? A71S. —r-, — 7 As gain. ——■ — - B s gain. ab-\-cd ^ ab^cd ^ 16. A, B, and C engage in trade, with a capital of $1911. A's money was in the firm 3 months, B's 5 months, and C's 7 months. They gained $117, which was so divided as that the ^ of A's gain was equal to -^ of B's and ^ of C's gain. What was each man's stock and gain ? ( A's stock $693^2_VV' ^'s S623|ffi|, and C's $594/^3^^. \ A's gain $26, B's gain $39, and C's gain $52. 17. If 12 oxen eat 3^ acres of grass in 4 weeks, and 21 oxen eat 10 acres in 9 weeks, how many acres would 36 oxen eat in 18 weeks, the grass to be growing uniformly ? Atis. 24 acres. 18. Three men engage in partnership, for 20 months; A, at first, put into the firm $4000, and at the end of 4 months he put in $500 more ; but, at the end of 16 months, he took out $1000. B, at first, put in $3000, but at the end of 10 months he took out $1500, and at the end of 14 months he put in $3000. C, at first, put in $2000, and at the end of 6 months he put in $2000 more, and at the end of 14 months he put in $2000 more ; but, at the end of 16, he took out $1500. They had gained, by trade, $4420. What is each man's share of the gain ? A71S. A's gain, $1680 ; B's gain, $1260 ; C's gain, $1480. • INDETERMINATE ANALYSIS. 257 SECTION XXVII. INDETERMINATE ANALYSIS. Art. 294, In the common rules of Algebra, such questions are usually proposed as require some certain or definite answer ; in which case, it is necessary that there should be as many inde- pendent equations, expressing their conditions, as there are unknown quantities to be determined ; otherwise the problem would not be limited. But, in other branches of the science, questions frequently arise that involve a greater number of unknown quantities than there are equations to express them ; in which instance, they are called indetermiTiate^ or unlimited problems, being such as commonly admit of an indefinite number of solutions ; although, when the question is proposed in integers, and the answers are required only in whole positive numbers, they are in some cases confined within certain limits, and in others the problem may become impossible. Note. — The rule of Alligation belongs to Indeterminate Analysis. See the Author's National Akitumetic, page 275. EXAMPLES. 1. Let 5a;+3?/=49. It is required to solve the equation, and find all the integral and positive values of x and y which are possible. (1.) By transposition, 3?/=49 — bx. (2.) Dividing as far as possible, 2/=lo — X- a By changing the fraction, for the sake of convenience, to a positive quantity, (3) 2^=16-20;+^. Since we consider only the integral values of y, the fraction must be a whole number. 22=^ 258 ALGEBRA. Let ?z=tliat number. Then 7i=^±l. o x=Sn — 1. Substituting this value ofx in (3), We have, y=lQ—2{Sn—l)+n. Or, 2^=18 — K>n. We have now the values of x and y in the terms of n, which must be whole numbers. By trying various values for n, we shall find all the possible values of x and y. Let 72=1, and x= 2, and y=13. 71=2, " x= 5, " y=S. w=3, " x= 8, " 7j=d. n—4:, " a:=ll, " y=—2. This last value of y, being negative, is not allowed by the conditions of the question. The equation, therefore, admits of only three sets of answers. 2. How can $100 be paid with 100 pieces, using eagles, dollars, and " nine-pences," each of the latter equalling one- eighth of a dollar ? Let a:=eagles, 2/= dollars, z=nine-pences. (1) Then, 2+?/+z=100. (2) And 10.i'+7/+|=100. (3) Multiplying (2) by 8, SQx+Sy-{-z^SQO. (4) Subtracting (1) from (3), ' 19x-\-7y=7Q0. (5) Transposing, 77/=700— 7U2\ (6) Dividing, 2/=100— 12:c-f-^. (7) Let 7i=~. 79n INDETERMINATE ANALYSIS. 259 (8) 7n=bx. In (9) ^=T- (10) By substitution, 7/=100- Let n = b, it being the smallest number that will give an integral value to x ; and we find x=7, and y=21, and z=^72. Again, let 7i = 10, the next smallest number that will make X a positive whole number, and we find a:=14, and y a negative quantity ; and so with every value of 7i that can be assumed, except 5. The question, then, admits of but one answer ; that is, 7 eagles, 21 dollars, and 72 nine-pences. The answer might have been obtained by eliminating y, instead of z. (1) Thus, x+y-\-z=100. (2) And 102--|-^+|=100. (3) Subtracting (1) from (2), 9x- -T=«- (4) Multiplying, 7z=72x. (5) Dividing, z=10x-^^. (6) Let • 2x G) Multiplying, 77i=2x. (8) Dividing, 7n x=—. Let ?z=2, it being the least number that will make x a wholo number, and Xz=.7, and z=72, and 7/=21. If we suppose ?z=4, it being the next larger number that will make x an entire number, then a:=14, and £:=144, which is impossible, by the conditions of the question. It is, therefore, certain that no numbers but 7, 21 and 72, are correct. 260 ALGEBRA. 3. Let x-{- y-\- z= 41 ) to find all the integral and pos- And24:r-}-197/-j-102=741 ) itive values of a:, y, and z. (1) Conditions, x-\-y-\-z=4:l. (2) And 24^-\-197j-\-10z=74:l. (^ (3) Transposing (1), z=41—x—y. y (4) Transposing &c. (2), z= -. (5) Values of (3) and (4), 4:l—x—y=^-^^—^:^^. (6) Multiplying, 410— 10:c— 10?/=741— 24a:— 1%. (7) Reducing, %+14a;=331. (8) Transposing and dividing, y= =SQ—x-{-— — —- Changing the signs in the last term, so as to make (9) X positive, y=^Q—x — . (10) Let St-^=^- (11) Multiplying, bx—7=9n. (12) Dividing, a:=^^i^. o (13) Substituting this for the value of x in the equation (9), we have 9^ 17 y=oQ^ n. (14) Multiplying, 5y=1^0—9n—7—^n. (15) Reducing, by=17^—14:n. Let n= 2, then 2/=29, and x=: 5, and z= 7. n= 7, " ?/=15, " xz=14, " z=12. n=12, " 7/= 1, " x=2S, " 2:=17. Another solution of the above question : (1) Let x-]-y-{-z= 41. (2) And 242;+19?/-|-10^=741. INDETERMINATE ANALYSIS. 261 (3) Eliminating the 2-'s, we have 14z=243— 5y. (4) Dividing, 1" , ^-% (5) Let 5—5?/ (6) Multiplying, 1472=5—5?/. 0) Transposing, 5?/=5— 1471. (8) Dividing, . , 147Z (9) Reducing, &c., 7/=l-372+|. "We might use the first value of ij ; but, to do what it is con- venient to do in some cases, let us introduce a second auxiliary quantity, to represent the fraction in the 2d value of y. (10) Let m=^. (11) Multiplying, 5?n=n. (12) By substitution, y=^ — 147?2. (13) And ,=17+!zi£=I^. (14) Therefore, z=ll-]-bm. The value of y requires that m should be zero, or negative. Let us first suppose the value of 7)i to be 0. Then, 7/=l— 14(0)=1. And z=17— 5(0)=17. And 2=41-1-17=23. Let — 1 be taken for the value of ???. And ?/= l-(— 14)= 1+14=15. r=17+5(-l)=17— 5=12. .^=:41_12-15=14. Again, let — 2 be taken for the value of m. And 2/=l-14(-2)=29. 262 A I. a E 15 R A . And ;^=:17_|_5(_2)=7. a:=41— 29— 7=5. Again, let — 3 be taken for the value ofm. Then, 2/=l-14(-3)=43.\ This value of y is more than the united values of x, y, and z by the conditions of the question. The three values of m (0, — 1, — 2), then, are the only ones •which will give integral and positive values for all the quantities. The reason for using the second quantity [m) was to avoid fractions in the values of y and z. Three or four successive auxiliary quantities may be used advantageously in some cases. 295. To find two square numbers whose sum shall be a square. 4. Let x^-\-y''=z\ Then x-=z" — y^=[z-\-y){z — ?/). Multiplying both sides by m, we have mx^=m{z-{'y){z — y). Assuming, mx=z-{-y, and x=7n{z — y), We have, z-\-y=m-{z—y). Therefore, [m^-\-'V)y={m^ — Vjz=.{mr — l){7nx—y) = * (tr- — V)mx — {jrr — \)y. Therefore, 27n~y= {m- — 1 ) mx. And 2my x= „ ^,. To obtain whole numbers without fractions, let y=:m^ — 1; then we have x=i27n, and z=7?i'^-\-l. That is, the general forins of the three numbers will be a;=2??z, y=:^in^ — 1, and 2:= If ??z=l, we have x-=. 2, ?/= 0, and z=2. 772=2, " x= 4, y= 3, " 2:=5. m=3, " x= 6, y= 8, " c=10. ?7Z=4, " xz=^ 8, 2/=. 15, " z—Vl. 771=5, " a;=10, 7/=24, " 2=26. The pupil will perceive that the values of x and y may INDETERMINATE ANALYSIS. 263 represent the base and perpendicular of a right-angled triangle, and z the hypothenuse. 296 • To find two numbers the sum of whose squares is given. Bj substitution, we have 6. Find the values of x and y which will satisfy the equation In these equations, any number may be assigned for the value of 77Z. If ?7Z=1, we have a;=10, and 2/=0. m=2, (( x= 8, " 2/=6. 7?Z=3, (( x= 6, " ?/=8. 7?Z=4, (i 80 150 a 32 126 , m=S, " :r=:— , " 2/=— , &c. 297. To find two square numbers v/hose difi"erence shall be a square number. 6. Let X' — y-=:z"; therefore [x-\-y)m[x — y):=mz'; whence, assuming x-\-y=mz, and m{x — y)=z, we have x-\-y=m-[x — 2/), and {7rr-{-l)y=(m^—l)x. Therefore, x=l —, — - hj, and if yz=m-—l, then will x=^m^ -\-\, and z=2m. If ??i=l, we have x=. 2, y= 0, and 2r=2. m=2, " x=: 5, y= 3, " z=4. w = 3, « :f=10, 7/= 8, " r=6. ??z=4. " x = VJ, y=\b, " r=8, &c. "We might assume a fractional value for m. z 264 ALGEBRA. 298. If the diflference of the two squares be given, we have the following formula for ascertaining their value ; m{x-\-y)=m-z, and m[x — y)=z. Whence, 2?nx={77r-\-l)z, x={ \z, and y= ( — — J 7. What values of x and y will satisfy the equation x'^—if =242? He. .= (^)h ana ,= (^).4, where the values ofm may be assumed at pleasure. lfm=l, we have x=24:, and y= 0. m=2, " a;=30, " ?/=18. m=S, " :r=40, " ?/=32. 7?2=4, " 2:=51, " 2/=45, &c. 8. The difference between the squares of the ages of two persons at one period was 45, and at another it was 159. Re- quired the age of each. Ans. At the first period their ages were 9 and 6, and at the second 28 and 25. Or, at the first period they were 23 and 22, and at the second 80 and 79. EXAMPLES. 1. How many pounds of sugar, at 11 cents per lb., shall be mixed with another kind, at 5 cents per lb., that the mixture shall be worth $2.54 ? Ans. 19 lbs. with 9 lbs. ; 14 lbs. with 20 lbs.; 9 lbs. with 31 lbs. ; and 4 lbs. with 42 lbs. 2. A person divides 65 shillings among 15 persons, men, women, and children. The share of a man is 7 shillings, that of a woman 3 shillings, and that of a child 2 shillings. How many persons were there of each class ? Atis. 6 men, 5 women, and 4 children. 3. A gentleman has two farms, valued at $2000. The best is worth $21 per acre, and the other $17 per acre. How many acres are there in each farm ? INDETERMINATE ANALYSIS. 265 Ans. The first may contain 92, 75, 58, 41, 24, or 7 acres; and the second may contain 4, 25, 46, 67, 88, or 109 acres. 4. I purchase wheat at 17 shillings and barley at 11 shillings a bushel, and expend in all £27 2^. How man}^ bushels of each do I purchase ? Atis. 6 of wheat and 40 of barley ; or 17 of wheat and 23 of barley ; or 28 of wheat and 6 of barley. 5. It is required to divide 100 into two such parts that one of them may be divisible by 7, and the other by 11. Atis. The only parts are 56 and 44. 6. In how many ways can a debt of $25 be paid with $2 and $3 bills ? Ans. Four ways. 7. I wish to mix corn at 70 cents per bushel with wheat at $1.90 per bushel. How many bushels of each must be taken to amount to $9.20? Ans. 5 bushels of corn, and 3 of wheat. 8. It is required to find the least whole number which, being divided by 17, shall leave a remainder of 7, and, when divided by 26, shall leave a remainder of 13. Ans. 143. 9. A person wishes to purchase 20 animals for £20 ; sheep at 31 shillings, pigs at 11 shillings, and rabbits at 1 shilling each. In how many ways can he do it ? Ans. He can buy 12 sheep, 2 pigs, 6 rabbits; or 11 sheep, 5 pigs, 4 rabbits ; or 10 sheep, 8 pigs, 2 rabbits. Note. — The question will admit of only these three answers. 10. It is required to find two numbers, one of which being multiplied by 7, and the other by 13, the sum of the products shall be equal to 71. Note. — This question does not admit of an answer in whole numbers. No value can be given to the auxiliary unknown quantity (/i)j which will render x and y both integral and positive. 11. It is required to find two numbers the sum of whose squares shall be 1225. Ans. The only positive and integral numbers are 21 and 28. 12. The difierence of the squares of two numbers is 1521 ; what are the numbers? Ans. 52 and 65, or &c. &c. 266 ALGEBRA. SECTION XXVIII. VARIATIONS, PERMUTATIONS, AND COMBINATIONS. ■ . ^ ' '' Art. 299. The different arrangements that can be made of any number of quantities, taking a certain number at a time, are called Variations. Thus, if a, b, c, be taken two together, the variations will be ab, ba, ac, ca, be, cb. And if a, b, c, d, be taken three together, their variations will be 24. Thus, abc abd acb acd adb adc bac bad bca bed bda bdc cab cad cba cbd cda cdb dab dac dba dbc dca deb. If all the quantities are taken together, their variations are called Permutations. Thus the permutations of <2, Z*, c, are abc, acb, bac, bca, cab, cba. The permutations of 1, 2, 3, are 123, 132, 213, 231, 312, 321. The different collections that can be made of a number of things, taking a certain number of things together without re- garding their order, are called Combinations. Thus the com- binations of a, b, c, taken two together, are ab, ac, be. Each combination will supply as many corresponding varia- tions as the number of things it contains admits of permutations. VARIATIONS. Let V = the number of variations required. n = number of different things. 7' = number of things taken. The following, therefore, will be the formula for obtaining the number of variations of 7i thino-s, taken r to<:!fether. VARIATIONS. 267 The number of variations of n things, taken r together, is n{n—\){7i—1) [7z— (r— 1)]. Let a, h, c, d, &c., be the n things; then the number of variations which can be made, taking them singly, is n. Let 71 — 1 of these things, namely, b, c, d, &c., be taken singly ; then the number of their variations is n — 1 ; and, if a be placed before each, we shall have 7i—l variations of ?i things, taken two together, in which a stands first. Similarly, we shall have n — 1 such variations in which b stands first, and simi- larly for all the n things; hence there will be, on the whole, {n — 1) variations of n things, taken two together. Again, taking n — 1 of these things, namely, b, c, d, &c., their variations, taken two together, will be n{n — l){n — 2) ; and pro- ceeding as before, there will be, in the whole, {n—l){?i — 2) variations of n things, taken three together. Similarly, their variations, taken four together, will be n{n — 1) (ji — 2)(?z— 3). Hence, if Fj, Ts* ^3> &c-» K? denote the varia- tions of 7i things, taken 1, 2, 3, &c., r, together, we have Fi=w, Vo=n{n-l), V^=n{n—l){7i-2), &c. V,=?i{7i-l){7i-2){7i—B) [7i-{r—l)]. From the above we infer that the permutations [jj) of n things are their variations taken all together ; therefore, by writing 7i for r, we shall have p=:n{n—l){n—2) {7i—{n—2) ){n—{7i—l) ) = n{7i—l){7i-2) 2.1=1.2.3 71. 1. How many changes can be rung with 7 bells out of 10 ? V=7i{n—l){7i—2) (7^— (r— 1)). As there are 10 bells, 71=10 ; and as they are taken 7 at a time, r=7, and r— 1^6; therefore, 7i — (r — 1)=10 — 6=4. Hence F-=10.9.8.7.6.5.4=604800 changes. A7is. 2. How many words can be made with 4 letters out of 5 ? Atis. 120. 3. How often can 4 boys change their places in a class of 8 so as not to preserve the same order ? Atis. 1680. 268 ALGEBRA. PERMUTATIONS. 300* When a and h are different, their permutations are ab^ ha ; but, when (2=^, they become aa. Let a recur "p times ; ^, q times ; c, r times ; and P be the number of permutations required. Then, if all the a's be changed into different letters, they will form 1. 2. 3 j9, permutations ; and, out of each of the P permutations, we should form 1. 2. 3 permutations. In like manner, if all the Z''s were changed to different letters, they would form 1. 2. 3 q permutations; and, therefore, there would be P. (1. 2. 3 jp. 1. 2. 3 q) permutations. Now, when all the quantities have become different, the number of permutations is 1. 2. 3. 4 n. by Art. 299. Therefore, P. (1. 2. 3 j9. 1. 2. 3 q. 1. 2. 3 r. &c.) =1. 2. 3 n. 1. 2. 3. . . ?^ Whence, P= 1. 2. 3. . . p. 1. 2. 3. . . q, 1. 2. 3 r, &c.* 4. In how many ways may the word enunciation be written ? In this word there are 11 letters, of which 3 are tz's and 2 are i's ; therefore, ?z=ll, _p=3, q=l. „ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11 Hence, P= ^ ^ o ^^ ^ = 3326400 ways. Atis. 5. In how many ways may the word algebra be written ? Ans. 2520. 6. How many different numbers can be made with the follow- ing figures, 1225555 ? Aiu. 105. 7. How many variations may be made of the letters in the word zapliTiathpaaneah 1 Ans. 454053600. COMBINATIONS. 301 . The different collections that can be made of a number of things, taking a certain number together, without regarding their order, are called their Combbiations. COMBINATIONS. 269 Thus, the combinations of «, h, c, taken two together, are ah, ac, he. Each combination will supply as many corresponding varia- tions as the number of things it contains admits of permu- tations. Each combination of r things supplies 1. 2. 3 r varia- tions of r things ; hence, if C, be the number of combinations of 71 things, taking r together, the following will be the formula. C, (1. 2. 3. . . r)=V,=n{n-l)[n—2) {n-{r-\) ). iherefore, t^=z— ^ ,^ ^ ^ • 1. 2. o. . . . r 8. Into how many diflferent triangles may a decagon be divided, by drawing lines from the angular points ? Note. — The number of triangles 'will be equal to the number of lines that can be drawn by connecting 7 at a time of the 10 angles, with each angle ; taken 7 together, ^ n{n—l){n-2) {n—{r—\)) 10. 9. 8. 7. 6. 5. 4 1. 2. 3 r ~ 1. 2. 3. 4. 5. 6. 7 = 120. Ans. 9. How many different combinations can be made with 5 letters out of 8 ? Ans. 56. 10. From a company of 12 persons, it is proposed to ascertain how many parties, of ten each, can be selected, and no two parties to be composed of the same individuals. How many parties can be selected ? Ans. 66. 11. A company of soldiers consists of 40 men, and 6 of them are selected every night to mount guard ; on how many nights can a different guard of 6 sentinels be made ? Aiis. 3838380. 12. How many different numbers can be made out of one unit, two 2's, three 3's, and four 4's, supposing all the figures to be in every number ? Ans. 12600. 13. What is the total number of combinations of 16 things, taken 1, 2, 3, &c., at a time ? Aiis. 65535. ^ \ 0, 1, 2, 3, 4, 5, 6, indices, or logarithms. 16, 32, 64, geometrical progression. 0- \l J 270 ALGEBRA. SECTION XXIX. /^^ ^- LOGAEITHMS.^^ „><. ^--^ . <-cy>^ Art. 302. Logarithms are a series 6f numbers in arith- metical progression, answering to another series of numbers in geometrical progression. ( 0, 1, 2, 3, '(1248 ■c^^ejLf ' ' ' ' .. (0, 1, 2, 3, 4, 5, 6, indices, or logarithms. ( 1, 3, 9, 27, 81, 243, 729, geometrical progression. 0, 1, 2, 3, 4, 5, indices, or log. 10, 100, 1000, 10000, 100000, geomet. prog. From the above, it is evident that the same indices may serve equally for any geometrical series ; and, consequently, there may be an endless variety of systenis of logarithms to the same com- mon numbers, by only changing the second term, 2, 3, or 10, &c., of the geometrical series of whole numbers ; and, by interpolation, the whole system of numbers may be made to enter the geomet- rical series, and receive their proportional logarithms, whether integers or decimals. It is also aj)parent, from the nature of these series, that, if any two indices be added together, their sum will be the index of * The invention of Logarithms is due to Lord Napier, Baron of Mer- chiston, in Scotland, and is properly considered as one of the most useful inventions of modern times. A table of these numbers was first publislied by the inventor at Edinburgh, in the year 1614, in a treatise entitled Canon Mirificum Logarithmorum , ^Yhich was eagerly read by all the learned throughout Europe. Mr. Henry Briggs, then professor of geom- etry at Gresham College, soon after the discovery went to visit the noble inventor ; after which, they jointly undertook the arduous task of com- puting new tables on this subject, and reducing them to a more convenient form than that which was at first thought of. But, Lord Napier dying soon after, the whole burden fell upon Mr. Briggs ; who, with prodigious labor and great skill, made an entire canon, according to the new form, for all numbers, from 1 to 20000, and from 90000 to 101000, to 11 places of decimals, and published it in London, in the year 1G24. LOGARITHMS. 271 that number which is equal to the product of the two terms in the geometrical progression to which those indices belong. Thus the indices 2 and 3, being taken together, make 5 ; and the numbers 4 and 8, or the terms corresponding to those indices, being multiplied together, make 82, which is the number answer- ing to the index 5. In like manner, if any one index be subtracted from another, the difference will be the index of that number, which is equal to< the quotient of the two terms to which those indices belong. Thus the index G, minus the index 4, is 2 ; and the terms correspond- ing to those indices are 64 and 16, whose quotient is 4, which is the number answering to the index 2. For the same reason, if the logarithm of any number be mul- tiplied by the index of its power, the product will be equal to the logarithm of that power. Thus, the index or logarithm of 4, in the above series, is 2 ; and, if this number be multiplied by 3, the product will be 6, which is the logarithm of 64, or the third power of 4. And, if the logarithm of any number be divided by the index of its root, the quotient will be equal to the logarithm of that root. Thus, the index or logarithm of 64 is 6; and, if this number be divided by 2, the quotient will be 3, which is the log- arithm of 8, or the square root of 64. The logarithms most convenient for practice are such as are adapted to a geometrical series increasing in a ten-fold ratio, as in the last of the above forms ; and are those which are to be found, at present, in most of the common tables on this subject. The distinguishing mark of this system of logarithms is, that the index or logarithm of 10 is 1 ; that of 100, 2 ; that of 1000, 3, &c. In decimals, the logarithm of .1 is — 1, and that of .01 is — 2, that of .001 is — 3, and so on. The logarithm of 1 in every system being 0, it follows that the logarithm of any number between 1 and 10 must be and some fractional parts, and that of a number between 10 and 100 will be 1 and some fractional part, and so on for any other number whatever. And, since the integral part of a logarithm, usually called the Index or Charac- 272 ALGEBRA. teristic, is always thus readily found, it is commonly omitted in the tables ; being left to be supplied by the operator himself, as occasion requires. 303. Another definition of Logarithms is, that the logarithm is the index of that power of some other number whiuh is equal to the given number. So, if there be iY=r", then n is the loga- rithm of N ; where n may be either positive or negative, or nothing, and the root, r, any number whatever, according to the difi"erent systems of logarithms. When 7z is = 0, then iV is = 1, whatever the value of r is, which shows that the logarithm of 1 is always in every system of logarithms. When n = 1, then N = r ; so that the radix, r, is always that number whose logarithm is 1, in every sys- tem. When the radix r = 2.718281828459, &c., the indices n are the hyperbolic, or Napier's logarithm of numbers, N ; so that ?i is always the hyperbolic logarithm of the number JV, or (2.718281828459)". 804. When the radix ?• = 10, then the index n becomes the common or Briggs' logarithm of the number N ; so that the common logarithm of any number 10" or iV is n, the index of that power of 10 which is equal to the said number. Thus, 100, being the second power of 10, will have 2 for its logarithm ; and 1000, being the third power of 10, will have 3 for its logarithm. Hence, also, if 50 = iqi-gosot^ ^^^^-^ jg 1.69897 the common logarithm of 50. That is, 10 has been raised to the 169897th power, and the lOOOOOd root has been extracted, which is fomid to be 50, nearly. And, in general, the following decuple series of terms, namely, 10\ 10^ 10-1, 10-2, jo-3, 10-4, 10, 1, .1, .01, .001, .0001, 1, 0, -1, -2, -3, -4, for their logarithms, respectively. And from this scale of num- bers and logarithms the same properties easily follow, as above mentioned. 305. To compute the Logarithm to any of the Natural Num bers, 1, 2, 3, 4, 5, &c., we have the following lOS 10^ 10^ or 10000, 1000, 100, have 4, 3, 2, LOGARITHMS. 273 Rule. Take the geometrical series, 1, 10, 100, 1000, 10000, 1728x144x6 Log. 1728 = 3.237544 Log. 144 = 2.158362 Log. 6 = 0.778151 Log. 36=1.556303 Arith. Com. =—2.443697 Log. 18=1.255273 " = -2.744727 Log. 12=1.079181 " = -2.920819 Am. 192=2.283300 11. "What is the value of a; in the following equation? 48x.75x72x.0625 X-. I .027X120 Log. 48 = 1.681241 Log. .75 =—1.875061 Log. 72 = 1.857332 Log. .0625 =—2.795880 Log. .027=— 2.431364 Arith. Com. = 1.568636 Log. 120= 2.079181 " " =-3.920819 Ans. 50=1.698969 12. What is the value of x in the following equation ? 654X320X-3691 a:=- Am. 2192.28. 87 X 9 X. 045 13. What is the value of :c in the following equation ? .69x7.5x32.71X.003 x=^- Ans. .000813. 87X8908X.0008 14. Multiply three hundred twenty-seven ten-thousandths by three hundred twenty-seven thousand. A7is. 10692.9. 15. What is the product of one thousand and twenty-five, multiplied by three hundred twenty-seven ten-thousandths ? Ans. 33.5175. 16. Multiply .0716 by 1.326. A7is. .0949416. 17. Multiply .0009 by .009. Aiis. .0000081. 286 ALGEBRA. INVOLUTION BY LOGARITHMS. Rule. Take out the logarithm of the given number from the table. Multiply the logarithm thus found by the index of the poiver proposed. Find the number answering to the product, and it will be the power required. Note. — In multiplying a logarithm "with a negative index by an affirm- ative number, the product will be negative ; but that which is to be carried from the decimal part of the logarithm will be affirmative : and, therefore, their difference will be the index of the product, and is always to be made of the same kind with the greater. EXAMPLES. 1. What is the square of 2.579 ? Logarithm of 2.579 = 0.411451 o Ans. 6.651 = 2. What is the third power of 32.16 ? Logarithm of 32.16 = Ans. 33261.9 = 3. Required the fourth power of .09163. Logarithm of .09163 = Ans. .000070494 = Here 4 times the negative index being - the difference — 5 is the index of the power. 0.822902 1.507316 3 4.521948 -2.962038 4 -5.848152 — 8, and 3 to carry, 1 EVOLUTION BY LOGARITHMS. Rule. Take the logarithm of the given number out of the table ; divide the logarithm thus found by the index of the root ; then the number answering to the quotient will be the root. When the index of the logarithm to be divided is negative, and does not exactly contain the divisor without some remainder, increase the index by such a number as will make it exactly divisible by the index, carr tying the units borrovjed, as so many EVOLUTION BY LOGARITHMS. 287 terhs, to the left-hand place of tJie decimal, and then divide as in ivhole numbers. EXAMPLES. 1. What is the square root of 365 ? Logarithm of 365 = Ans, 19.10409 = 2. What is the third root of 12340 ? Logarithm of 12340 = Ans. 23.108 = 8. What is the seventh root of 6 ? Logarithm of 6 = Ans. 1.2917 = 4. Find the tenth root of 9. Logarithm of 9 = Ans. 1.245 = 5. Find the square root of .083. Logarithm of .083 = Ans. .28809 = 6. Find the cube root of .00059. Logarithm of .00059 = Ans. .083872 = 2.562293(2 1.281146i. 4.091315(3 1.363771f. 0.778151(7 0.111164f 0.954243(10 0.095424^^^. -2.919078(2 -1.459539. -4.770852(3 -2.923617. Here the divisor 3, not being exactly contained in —4, it is auo-mented by 2, to make up 6, in which the divisor is con- tained just 2 times; then the 2 thus borrowed, being carried to the decimal figure 7, makes 27 ; which, being divided by 3, gives 9, &c. 7. What is the value of x in the following equation ? -( 27x38xl5.61\^ .36X1^37" 288 ALGEBRA. Log. 27 =1.431364 Log. 38 = 1.579784 Log. 15.61 =1.193403 Log. .36=- -1.556303 Arith. Com. =0.443697 Log. 1.37= 0.136721 " =—1.863279 4.511527 3 13.534581(4 Ans. 2419.05=3.383645 8. Find the value of x in the following equation. 37 /14.21x.00208\^ . -.qo.qq "=223 \ im ) • ^-^••1^2438. 9. What is the value of x in the following equation ? 7 /144\^ /703\^ 10. Find the value of x in the following equation. 345 /872x.0065\5 "=4i7- {Am<}]^\\ ■<■«■ 128.!!. '-f COMPOUND INTEREST. 289 SECTION XXX. COMPOUND INTEREST. Art. 312. Compound Interest is interest charged not only on the principal, but also on the interest of preceding years. Let p = principal. r = rate per cent., considered as a decimal, or hundredths. t = time in years. A = amount. Then 1 + r will represent the amount of $1, or 1£, for one year. And^ (\--\-r) will be the amount of any principal (7;) for 1 year. The amount for two years will be p (1-fr) . (l-j-7-) = p[\-\-rY', the amount for 3 years will be p(\.-\-rY . (1+7") =p[\-\-rf\ for 4 years it will be^(l+r)^ . {l-{-r)=pil-\-rY. Hence, for any number of years, it will be p[l-\-r)"', or pil-^ry. Putting A for amount, we have the following formula fo7 ascertaining the amount of any principal at any rate per cent, for any definite time, at compound interest. A=p{l+rY. This equation contains four quantities, A, p, r, and t ; any three of which being given, the other may be obtained., ' -^^V Thus, we have the following , .u-") FOEMUL^. (1.) A=pil+rY. ^3 , .=(£)4_i. log. (1+r) From the first formula, the pupil will perceive the following 25 200 A L G E B K A . Rule maj be deduced for finding the amount of any sum at com- pound interest. lluLE. Add 1 to the ratio, then raise this sum to a pmver lohose ex'ponent is equal to the time, vmltijjly this pmver hy the jn'irucvpal, and. the product is the amount. By logarithms the operation is much facilitated, especially when the time is of much length. EXAMPLES. 1. What is the amount of $78.39 for 8 years, at 6 per cent, compound interest ? OPERATION BY THE FIRST FORMULA. ^==jt7(l-l-r)'=78.39(l+.06)l Log. (l-j-r)=1.06 = 0.025306 Multiply by if =8, 8 (l_f-r)'=(1.06f == 0.202448 Log. |;=78.39 = 1.894261 74=$124.94. A?is. = 2.096709 2. What is the amount of $144 for 6 years, 9 months, at compound interest, at 5 per cent. ? Log. (l+r)=1.05 = 0.021189 Multiply by t, 6 (l+,-)'=(1.05)' = 0.127134 Loo-. ?,=:144 = 2.158362 Log. of am.ount for 6 years = 2.285496 Log. (1.0375) = 0.015988 .4=8200.21. Ans. = 2.301484 We have just found the logarithm of the amount for 6 years, and to this we have added the logarithm of 1.0375, it being the amount of $1 for 9 months, at 5 per cent. COMPOUND INTEREST. ""J 3. What is tbe amount of $500 for 9 years, at G per ct (j per annum, the interest to be paid semi-annually ? As the time, ?, is to be calculated in half-years, and as r is considered the interest of $1 for one year, therefore 2t will T represent the time, and - the interest of $1 for half a year. •J The formula will therefore be A=p(\ V7,)^"=500(l+.0a)i^ Log. n 4-!^^ = 1.03 = 0.012837 Multiply by 18 half-years, 18 Log. ("l+^'V"' = 0.231066 Log. ^=500 = 2.698970 ^=$851.21. Ans. = 2.930036 4. What principal, at compound interest, will amount to $4000 in 10 years, at 6 per cent. ? This question must be performed by the second formula. A 4000 P Log. 1 06=0.025306 (1+r)' {IMf"' 10 0.253060 Arith. Com. = —1.746940 Loc.^ yl=4000 = 3.602060 ^=$2233.57. Ans. = 3.349000 5. At what rate per cent, must $2233.57 be, at compound interest, to amount to $4000 in 10 years ? This question should be performed by the third formula. \p/ V'2-233.5Ty 2PA / ALGEBllA. m /ooo ^2233.57 / . = 3.602060 = 3.349000 / 0.253060(10 Jog. (l+r)=1.06 ^ - 1 = 0.025306 .06, that is, 6 per cent. A'tis. 6. In what time will $2233.57, at compound interest, at 6 per cent., amount to $4000 ? This question is solved by the fourth formula. T (^\ T (^^\ _^' \p) _ ^' V2233.57y _ Log. 4000-log. 2233.5 7 ^"Log. (l+r)~ Log. (1+.06) ~" Log. (1+.06) ' Log. ^=4000 = 3.602060 Log. j9=2233.57 = 3.349000 0.253060 Log. (l+r)=1.06 0.025306 ^, ^ 253060 ,^^ Thereiore ^==————==10 years. Ans. The value of this fraction can be ascertained by logarithms. Thus, Log. 253060 = 5.403223 Log. 25306 = 4.403223 1.000000 t=. 10 years, as before. 7. What will $16 amount to in 30 years, at 5 per cent, com- pound interest ? A71S. $69.15. 8. "What will $2000, at compound interest, amount to in 11 years, at 8 per cent. ? Ans. $4663.31. 9. What will $27.18 amount to in 8 years, 3 months, at 4 per cent, compound interest ? Ans. $37.56. COMPOUND I N T E 11 E S T . 293 10. What is the compound interest of $1728 for 8 years, months, at 6 per cent, per annum, the interest to be paid every 8 months? Ans. $1138.74. 11. What is the amount of $18.29 for 8 years, 8 months, 12 days, at 4 per cent. ? Atis. $25.73. 12. What sum, at compound interest, will amount to $800 in 7 years, at 5 per cent, compound interest ? A7is. $568.54. 13. What sum will amount to $500 in 9 years, at 6 per cent. per annum, the interest to be paid every 3 months ? Am. $292.54.5. 14. At what rate per cent, will $800, at compound interest, amount to $1609.76 in 12 years ? A7is. 6 per cent. 15. In how many years will $3726 amount to $5007.43, at 3 per cent, compound interest ? A?is. 10 years. - 16. How many years will it require for any sum to double itself, at 6 per cent, compound interest ? Let 2p= the amount. Then, 2p=p{l-{-ry. Arid 2= (l+r)'. Loo-. 2 ■i. o Log. (l-j-r)* Log. 2 Lon Or/ m—o7i where P is the first term of a binomial, Q the second divided by the first, — the index of the power or root, and A, B, C, &c., the terms immediately preceding those in which they are first found, including their signs -j- or — . 332. This theorem may be applied to any particular case, b}" substituting the numbers or letters in the given example for P Q, m, and 7i, in either the above forrauh-c, and then finding the result according to the rule. When the index of the binomial is a whole number, the series will terminate, as observed under the article Involution ; but when it is a negative or fractional number, as in the following examples, the series will proceed on ad mjinitian, and will become more convergent the less the second term of a binomial is with resnect to the first. >C I - ^=x. • -~->^ - ■ J- YX '' 312 ^ ALGEBRA. EXAMPLES. . iJ^. 1. It is required to convert {ar-\-xY into an infinite series. X 071 Let P=aP, Q=—, —=h-> 0^ 7^i=l and n^2. a- 71 m TO Then P^=(a2)~=(a2)2-^^^^^ 71 '^ 1 a- 2 "^ which, it is obvious, that the signs are alternately -|- and — , and the exponents, both in the numer- ator and denominator, increase continually by 1, that of x in the numerator being always 1 less than that of b in the de- nominator. 2 2. Expand — — r into a series. a'-{-2ax — x- , 2x 5a;2 12:r3 , „ Am. 1 \ — ^4-, &c. a a- a^ 3. Expand /y/ {or — x-) into a series. 316 ALGEBRA. , ^ , 1+2:?; . 4. Jiixpand - — ' into a series. 1 — X — X- Ans. l4-32:+4:?;2+72;^+lla;4-|-18^-^+, &c. This is a recurring series, in wtiicli each, of the coefficients, after the second, is the sum of the two preceding ones. 5. Expand /s/ (1 — a) into a series. ' 2~'2:4~2A6~2X6:8"'2.4.6.8.10~' ' 1—x 6. Expand rr— into a series. 1 — za: — 6x" Ans. l^x-\-bx''+12,x^-\-41x^-\-\'l\a^-\-mbx\ &c. 7. What is the expansion of (a—b)^ ? ^/ 5 3^2 3.7^^ 3.7.11^-^ \ ^7Z5. a 1^1 4^-4:8^2-4x12^3-4^^12.160'^"' ^''- ; 8. It is required to expand [a-\-x)~^. 1 2x , 32;2 4x^ Ans. -5 r-^ — 2 r+, &c. 9. It is required to expand ^^j^^' icx.-hl^ 1 62: , 24x^^ mx^ „ Ans. -o— — ri -r T-, &c. 9 ' 10. It is required to find the expansion of {c^xf 2 4x , Qz^ Sx^ , , ^^^- 72—73+-^—^+' &c. 11. It is required to find the expansion of — rTTTTs' , 1/' 6/^ , 24^*2 8O33 , ^ Ans. - 1 -j ^ — H, &c. 12. What is the value of r in a series ? 1 3; 3x^ 3.52;3 3.5.7a;* • "^ 2^.3+2:4^5 2X6^^"^2.4.6.8^«"~' ^'^' SUMMATION OF SERIES. 817 SECTION XXXVI. SUMMATION AND INTERPOLATION OF SERIES. Art. S34. The Summation of Series is the method of finding a terminated expression equal to tl\e whole series. Interpolation is the method of finding any term of an infinite series, without producing all the rest. DIFFERENTIAL METHOD. 335. The Difi"erential Method consists in finding, from the suc- cessive diflferenccs of the terms of a series, any intermediate term, or the sum of the whole series. Problem I. 336. To find the several orders of differences. Let a-{-b-\-c-\-d-]-e-{-, &c., be any series; subtract each term from the one following it, and the diff"erences — ci-{-b, — b-{-c, — c-\-d, — d-\-e, &c., will form a new series, called the Jirst order of differences. Again, subtract each term of this new series from the one that follows it, and the differences a — 2Z»-|-c, b — 2c-\-d, c — 2d-\-e, o;c., will form another series, called the second order of differences. Proceed in like manner for the third, fourth, fifth, &c., order of differences, until they at last become equal to 0, or are carried as far as is required. 337. When the several terms of the series continually in- crease, the differences will all be positive ; but, when they decrease, the differences will -be alternately negative and posi- tive. 1. Required the several order of differences of the series 1, 6, 20, 50, 105, 196, &c. 1, 6, 20, 50, 105, 196, &c., the given series. 5, 14, 30, 55, 91, &c., 1st differences. 9, 16, 25, 36, &c., 2d 7, 9, 11, &c., 3d 2, 2, etc., 4th 0, &c., 5th 27^ 318 ALGEBKA. 2. Required the several order of dififerences of the series of P, 2\ 32, 42, 52, &c. 1, 4, 9, 16, 25, &c., the given series. 3, 5, 7, 9, &c., 1st differences. 2, 2, 2, &c., 2d 0, 0, &c., 3d 3. Required the several order of differences of the series of cubes, l^ 2\ 3^ 4^ 5^ Ans. 4. Find the order of differences in the series ^, ^, -^, j'^, ^^2-, &c. ^?w. Problem II. 338. To find the first term of any order of differences. Let d', d", d"\ d"", &c., represent the first terms of the 1st, 2d, 3d, 4th, &c., order of differences ; then d'z=z — a-\-b, d"=a —2h-\-c, d"'=-a-\-Sb—Sc+d, d""=a—^b^Qc—U+e, &c.; from which it is obvious that the coefficients of the several terms of any order of differences are respectively the same as those of the terms of an expanded binomial, and are obtained in the same manner ; for the terms that are subtracted are actually added, but with contrary signs. Hence we infer that d", or the first difference of the nth. order of differences, is -\-a^nb-{-7i . — ^ 71 — 1 n — 2 - - . 1 . 1 o 11 c^=-n. — ^— . — p~"^±» ^^'j *o n-{-l terms ; m which formula the upper signs must be taken when n is an even number, and the under when n is an odd number. 5. Required the first of the fifth order of differences of the series 6, 9, 17, 35, 63, 99, 148, &c. Let a, h, c, d, e,/, &c. = 6, 9, 17, 35, 63, 99, 148, &c., and 71=5. Then n{n — l) n{n—l){7i—'l) n{7i—l)[n—'l){n—Z) -a + nb ^ c-\ — d ~— e ^ ^(,,_l)(,,_2)(,^_3)(7^-4) ^•^.^•'^•^^ 2.3.4.5 •" ' 2 ' 2.3 SUMMATION, ETC., OF SERIES. 319 494— 491=+3. Ans. 6. Required the first of the sixth order of differences of the series 3, 6, 11, 17, 24, 36, 50, 72, &c. Ans. —14. Problem III. 339. To find the wth term of the series a, b, c, d, e,f, &c. As we have found in the last problem that d'= — a-\-b, there- fore bz=a-\-d\ and, in the same manner, wc find c=a-{-2d'-\-d", d=a-\-U'+M'-{-d"', e=a^W-{-U"-{-W"-^d"'\ &c. ; whence the nth. term is n—1 n—1 n—2 ?^— 1 7i—2 n—S =ai~^d'Jr-^- . -^d -f--^ . -^ . -^-d" +, &c. 7. Required the 7th term of the series 3, 5, 8, 12, 17, &c. 3, 5, 8, 12, 17, &c., the given series. 2, 3, 4, 5, 1st difference. 1, 1, 1, 2d difference. 0, 0, 3d difference. Here d'=2, d"=\, d"'=0, and n=7. Therefore a-{-'^d'-\-'^ . !^^"=3+^ . 2-f I^.llll?. 1=3+12+15=30= the 7th term. 8. Required the 9th term of the series 1, 5, 15, 35, 70, &c. A71S. 495. 9. Required the 10th term of the series 1, 3, 6, 10, 15, 21, &c. Ans. 55. Problem IV. 310. To find the sum of 7i terms of the series a, b, c, d, e, &c. If we add the values of a, b, c, &c., as found in the last problem, we obtain 2a-\-d'=a-\-b, 3a+3c^'+(^":=a+3+c, 4a+ 320 ALGEBllA. Qd'-j-4d"+d'"=a-j-b-}-c-\-d, &c. "^ATierefore it is evident that the sum of n terms must be na+n.^-d-^n.^-.-^d +n.-^.-^.-^d +, &c. 341. When the differences become at last = 0, any term, or the sum of any numbers, can be accurately found ; but, when the differences do not vanish, the formulae in this and the preceding problem give only an approximation, which will come nearer the truth as the differences diminish. 10. Required the sum of 8 terms of the series 2, 5, 10, 17, &c. Here n=S, a=2, d'=o, d"=2, and d"'==S). Hence, na-]-7i . -^j-d'-^n . —-— . ^d"=^ . 2-|-8 . ^ . 3+ 8. 1. 1. 2=16+84+112=212= the sum of 8 terms. 11. Required the sum of 100 terms of the series 1, 2, 3, 4, 5, &c. Here 1, 2, 8, 4, 5, 6, &c., given series. 1, 1, 1, 1, 1, &c., 1st difference. 0, 0, 0, 0, &c., 2d difference. Here ?2=100, a=l, and d=X. n—\, .^. . ..^ /lOO— 1' iia +,i.!i_^=100-| 100.(^1^^) 1=5050. Am. 12. Required the sum of 12 terms of the series, 1, 4, 10, 20, 85. Ans. 1365. 13. Required the sum of n terms of the series 1-, 2-, 3-, 4-, h\ 6-, 7-, &c. Here 1, 4, 0. 16, 25, 36, 49, &c., given series. 3, 5, 7, 9, 11, 13, &c., 1st difference. 2, 2, 2, 2, 2, &c., 2d difference. 0, 0, 0, 0, &c., 3d difference. SUMMATION OV SERIES. 321 Leta=l, rf'=3, and^"=2. Then na-^--^-^ d'-\ ^ — a = ^ . 14. Required the sum of 7i times of the series 1\ 2\ 3^ 4^ 5^ 6-^ &c. ; 1, 8, 27, 64, 125, 216, &c. Here 1, 8, 27, 64, 125, 216, &c., given series. 7, 19, 37, 61, 91, &c., 1st difference. 12, 18, 24, 30, &c., 2d difiference. 6, 6, 6, &c., 3d difiference. 0, 0, &c., 4th difiference. Let a=l, d'=l, d"=l'2, d'"=G. Then ^. . ^(^^-1)^. , n{n-l){?i-2) ^^,, , ni7i-l){?i-2){n-S) ^ ^^, "^ 2 "^ 2 : 3~ "^ 2.3.4 7n{n—l) 12?i{7i—l){7i—2) Q7i(n—l){7i—2){n—3) =n-f- -; 1 ^ 5 1 n — o 4 =71 -j ^ \-2n^ — b?i- 4-471 -f 2 ' ' ' 4 471 Un^-'Un Sti"^— 24?i-+16?^ yz^— Gti^+Utz-— 67t _ T"' 4 * 4 ' 4 "~ 7i^+2;z"-f?z- 7i-(7i-fl)2 . , — ■ — = — ^^ — ■ == sum 01 n terms, as required. 4 4 ., ^ 15. What is the number of cannon-shot in a square pile, the bottom row consisting of 25 shot "^ ? - Ans. 5525. 16. I have 10 square house-lots, whose sides measure 5, 6, 7, 8, 9, &c., rods, respectively. What is their value, at 25 cents per square foot ? A?is. $67,041,561. * Shots and shells are generally piled in three diflferent forms, called triangular, square, or oblong piles, according as their base is either a triangle, a square, or a rectangle. A square pile is formed by the continual laj'ing of square, horizontal courses of shot, one aboye another, in such a manner as that the sides of their courses decrease by unity from the bottom to the top row, which ends also in one shot. / y 1 € '/ ■\ ALGEBRA . 822 ' '' 17. There are 5 cubical blocks of marble, whose sides meas- ure, respectively, 2, 3, 4, 5, and 6 feet? What is their value at $2.75 per cubic foot ? Am. $1210. 18. What is the number of shot in a square pyramidical pile, whose side at the base contains 100 shot ? Ans. 338350. 19. What is the sum of 20 terms of the series 1^, 2'^ 3", 4^ 5^ 6^ &c. ? Am. 44100. 20. What is the sum of 20 terms of the series l^ 2*, 3^ 4^ h\ OS &c. ? Am. 722666. Problem Y. 312, To find a fraction that will express the value of a geometrical series to infinity. In Art. 284 we find that the sum of an infinite series is obtained by the following formula : c ^ 1 — r and, by this formula, we may find the sum of algebraic series. EXAMPLES. 1. What is the sum of the series \-\-a-\-a-'-\-d'-\-a''^ &c., carried to infinity ? , 1 ^ Am \—a By the above formula, the first term of the series will be the numerator of the fraction, and the denominator is obtained by subtracting the second term from the first. 2. What fraction will express the exact value of the series 14-5+25 + 125, &c., to infinity ? 1 Alls. 1-5* 3. What fraction will express the infinite series 1 — a-\-(r — a'' -\-a^—a', &c. ? ,1 l+<2 A -itn n • . ll l)h Irh 4. What fraction will express the series — ! — zr-\ — ^+> & to ^ a a^ a < Ji infinity? :; Ans. -. /h ./ --, . •< — - 11. <(L^.^ SUMMATION OF S E R I F, ;> T 323 . 1111 5. What is the sum of the series — j — --j — ^-j — -, + , etc., to X X" x" x^ infinity ? Atis x—\ 6. What fraction will express the series l + 24-4-[-5^+16, &c., to infinity ? Ans. — — . 1 x'^ x:^ x!' 7. What fraction is equal to the series ^-| — ; =+, &c., a a' a' a' . , , Gj to infinity ? Ans ar-\-x'' 8. What fraction wiU express the value of l-j-l + l+l? &c., -I to infinity ? Ans. z. — r-. O o X" X 9. Express by a fraction the value of the series x4-'—-\- — -f--:,4-) &c., to infinity. Ans. — '-■-. a' a — X X X" X," 10. What is the value of the series 1 \-—,——-\-^ &c., a a~ a' to infinity ? Ans. . '' a~{-x 111 11. Required the sum of the series t^4"4To4"^7t4"' <^^-' i.A l.o 0.4 continued to infinit3^ Ans. 1. This question may be performed by separating the factors of the denominators so as to form two series, and then subtractinrr the less from the greater, as follows : Let l-|-2~j~3~{~4' ^^' = '^^ greater series. And 2~f"3~l~4» ^^- = ^^® -^^^^ series. Then 1 = the sum of the series. XoTK. — Another method may be found in the Key. 12. Required the sum of the series =— r-f— r-^4--^^+T-^4-j ^ l.i~2.o' 3.0^4.7 ' &c., to infinity. Ans. -^. 324 ALGEBRA. SECTION XXXVII. CUBIC EQUATIONS, CONTAINING ONLY THE THIRD AND SECOND POWERS. Art. 343» Any numerical equation, containing only the third and second powers of the unknown quantity, and having one rational root, may be reduced by rendering both of its members perfect squares, and extracting the square root of both sides : completing the operation by former rules. The only difficulty lies in multiplying the equation by such a number that, after adding to each side the fourth power of the unknown quan- tity, and the second power with a coefficient easily determined, both sides will be perfect squares. This multiplier must be ascertained by trial ; for, though a general formula might be given for obtaining it, yet it would be so complicated as to be of no practical use. It may be either an integer or a fraction, and is positive or negative according to the sign of the known quantity. Though there always is such a multiplier whenever the un- known quantity has one rational value, yet, when the numbers are very large, or the equation is very complicated, it may not be readily found, and the process of trial may become too tedious to be of service. Whenever the equation does not con- tain too large numbers, the pupil will find little difficulty, if he thoroughly understands the following Rule. Divide both sides of the equation by the coefficient of the unknown cube, if it have any expressed. Place the third power of the unknown quantity on one side of the equation, and the second power, with the known quantity, on the other. Mul- tiply both sides by the number nearest to unity which will 7)iake the known quantity a positive square; or, which is the same thing, separate the known number into two factors, one of which shall be the greatest square contained in it. and multiply both sides by the other factor. Multiply the last equation by 4 ; add the fourth power of the CUBIC EQUATIONS. 325 unknown quantity^ and the second power, with a coefficient equal to the square of half the coefficient of the third power, to each side ; and extract the square root of both sides, if possible. By taking like signs of the two members of the equation in evolving, we shall obtain one root ; and, by taking unlike signs, the other two may be found by quadratic equations. But, if that member of the equation which contains the knoion quantity is not a perfect squure, substitute 1, 9, 16, f , ^, \, ^, or some other square number, in the place of 4, a7id proceed as above, till, by trial, a number is found which will accomplish the object. Note. — 1. The sum of the three values of the unknown quantity should always be equal to the coefficient of the second power in the original equation, after dividing by the coefficient of the cube, and placing it on the same side with the known quantity, opposite the positive cube ; hence, if two values were known, the other might easily be found. 2. When one of the values is known, the others might be found by the usual method ; bringing all the terms of the original equation to the same side, and dividing by the difference between the unknown quantity and its known value, reducing, by quadratic equations, the equation thus pro- duced. But the three values are here given directly, by using the different signs in evolving, thus rendering the solution shorter, and more satisfac- tory. It is evident that, in extracting the square root of an equation, both sides may be considered positive, or both negative, or either one positive and the other negative. Thus the square root of the equation 4a2_8a6+462=c2-f-2cc?-f(^^ is -f(2a— 26 )=-[-( c-{-^), or —{2a— lb) =—{c-\-d), or -i-(2a— 26)=— (c+c^), or _(2a— 26)=-f (c-f-^i). But, if both sides take like signs, the result will be the same, whether they are both positive or both negative, as the signs of both sides of an equation may always be changed; while, if they take unlike signs, a different equa- tion will be produced, it making no difference which side is positive. Hence, there are but two results that can be obtained, and we have pre- ferred to express them, in the following examples, by the same method as in quadratic equations ; prefixing the sign i; to the right-hand member of the equation produced by evolution. 3. By observing whether the root of the known quantity is greater or less than half the coefficient of the second power on the same side, if we also notice the sign, we may usually know whether the multiplier we have used is too small or too large. When there are two rational values of the unknown quantity, of course the third will be rational, and there will be three different multipliers, which will answer our purpose, thus giving three different solutions for the same example. 28 826 ALGEBRA. EXAMPLES. 1. What are the values of x in the equation 7? — a:^=4 ? Here the multiplier, which would make the known quantity a perfect square, is unity ; therefore we transpose, and multiply by 4, 42;3=42;2-j-16. /4\2 Adding x'^ and ( - J a:^ to each side, a;^-f"4^+4a;^=^*"l~^^^"t"l^' Evolving, a;2+2a^=:±(a:^+4.) Taking the positive sign and cancelling, 22:= 4. Dividing, a: =2. Taking the negative sign, 7?-\-^x:=. — :ii? — 4. Transposing and dividing, a:^-{-a;= — 2. By quadratics, xz=. . Hence a:=2, or . Ans. The sum of their values, 2 -J j — . — , is 1. 2. What are the values of a; in the equation 4a;^-|-10a:^=9 ? Conditions, 4a:^-j-10a;^=9. 5 9 Dividing by 4 and transposing, a:^= — h^^+t* 9 - being a square, multiply by 4, 4a:^= — 10a;^-{-9. Adding x" and i-\ x\ x'^^A7?-\-^x''=x''—Qxr-\-^, Evolving, a;2+2a:=±(a:2— 3). Taking the positive sign and cancelling, 22:= — 3. 3 Dividing, x:^ — -. Taking the negative sign, x'^-\-2x=sz—x'^-\-Z. CUBIC EQUATIONS. 327 3 Transposing and dividing, x--\-x=-, A Bj quadratics, xz=. w^- A Hence 2;=—-, or . Am. The sum of these roots is . 2 3. Given 32;^— 2:^2^931 to find the values of a:. (1.) Conditions, 3r''— 2:^2=931. 2^2 . 931 3 "^ 3 * (2.) Dividing and transposing, 0? /Q X rru . . 931 . 49 (d.) ihe greatest square m -5- is — - ; o 1 ^, „ 931 49 19 therefore ___=__X— . Multiplying (2) by -^, Multiplying by 4, 3' 3 ~~ 3 ' 9 • 76z'^_ 152a;2 70756 ' 3 ""~3 ' 9~* Adding a;'' and ^--j x% 2;^+ _^-{— ^—=2:^4— ^4-_^_. 17 1 • , , 38a: / , 266\ Evolving, x''-\-—=:±(x' + —). Taking the positive sign and cancelling, 38:c=266. Dividing, x=7. m T .u r • o , 38^ , 266 Taking the negative sign, ci^-\ — —z=—x- ^. 3 3 Transposing and dividing, ar-| — j— = — -^. o o ■R ;i .• — 19±V — 1^35 J3y quadratics, x= . 6 XQ.^,^ 1*^35 2 Hence x=7, or =^ — . The sum of these is tt. o 3 328 ALGEBRA. 4. Given a;^=12a;^— 81 to find the values of ^. Conditions, a;^=12:r2_8i. By multiplying both sides by —1, — 81 becomes a positive square, — a;^= — 122;^-|-^1« We find that neither 4, 9, 16, nor 25, will answer our purpose, and we multiply by 36, — 36r^=— 432a;24-2916. Adding a;4 and (--\ x\ z''—m3?-\-Z2^^=zx^—\^^x'-{-2^1Q. Evolving, a;2_i8^__j_(2;2_54). Taking the positive sign and cancelling, — 18^:= — 54. Changing signs and dividing, a: =3. Taking the negative sign, x^ — 18a;= — x^-\-b^. Transposing and dividing, a? — 92;=27. Completing the square, x^^9x-]—^z=27-\—j-=—^. Evolvmg, ^""2^~~2~* Transposing, .=^^. 9rh3A/2l Hence a;=3, or . The sum of these is 12. 5. Given x'-^-x^^z — 4 to find the value of :r. Ans. —2, or — . • 6. Given 7x^=x'^-\-^Q to find the values of x. Ans. x=Q, or 3, or — 2. 7. Given x^—4:X^= — 9 to find the values of x. Ans. x=S, or ^ . 2 8. Given 2:c^=99— 5:^2 to find the values of a:. Ans. x=6, or . 4 9. Given 4:c"+102;-=125 to find the values of.?:. A?is. x=2i, or i:!±|v^. CUBIC EQUATIONS. 329 10. Given a:^=8a;^-}-363 to find the values of x. Ans. a:=ll, or . 11. Given 372;-=7a;^-|-144 to find the values of a:. Ans. x=4:, or 3, or — If. CUBIC EQUATIONS CONTAINING ONLY THE THIRD AND FIRST POWERS. Art. 344, Any numerical equation containing only the third and first powers of the unknown quantity, and having one rational root, may be reduced by multiplying both sides of the equation by the unknown quantity, and adding the second power to each side, with such a coefiicient as, after adding a number readily determined, will make them perfect squares. The only difficulty lies in finding this coefficient, which must be ascer- tained by trial ; though, by adopting the following rule, it can readily be found, unless the equation is so complicated, or the numbers so large, as to render the operation tedious. In this and also the preceding case, the rule might perhaps be so framed as to obtain the roots without reducing the coefficient of the cube to unity, the two methods bearing somewhat the same relation to each other as the two in quadratic equations. But we have preferred to use fractions occasionally, rather than render the rule more complicated. Rule. Divide both sides of the equation by the coefficient of the unknown cube^ if there be any expressed. Place the two powers of the loihnmvn quantity on one side, and the hncnvn quantity on the other, and multiply both sides by the unknown quantitij ivith such a sign as shall render the fourth power positive. Separate the coefficient of the first power of the unknown quantity in the equation, thus produced, into two factors, and add the second power, with a coefficient equal to tJie square of one of these factors, usually the smaller, to each side. If it make the * Until the factors are found, it is sometimes better to give the known quantity and the first power a common denominator, even though the former might be reduced to a whole number. 28=^ 330 ALGEBRA. coefficient of the square, on the same side as the fourth power, equal to the other factor, add the square of half this coefficient to each side, and extract the square root of hoth members, completing the operation by former rules. But, if the above coefficient be not equal to the other factor, separate the same number into two other factors, or perhaps ex- change the same, and proceed in the same way till the right ones are found. « Note 1. — The sum of the three values of the unknown quantity should always be 0, as there is no second power in the original equation ; hence, if two are known, the third will be equal to their sum, with the sign changed ; and there must always be one positive and one negative value, the other being sometimes positive and sometimes negative. 2. We obtain the three values by the same method as in the preceding case, prefixing the sign -j- to the right-hand member of the equation in evolving. Taking the positive sign, we obtain either one or two values, and the negative sign gives the remaining values or value. When one of the values is known, the others might also be found by bringing all the terms of the original equation to the same side, and dividing by the difference between the unknown quantity and its known value. 3. When two of the values are rational, the third will of course be rational ; and there may be three different methods of separating into factors, each of which will answer the purpose, thus giving three different solutions of the same equation. EXAMPLES. 1. Griven o;^ — %x=l to find tlie values of a;. Conditions, o? — 3a;=2. Multiplying by x, x'^ — Sx"=2x. Separating the coefficient of 2:?; into factors, 2x1' Adding (iyx"^ to each side, ' x^—2x^=x'^-\-2x. Add (1)2, x'—2x''-{-l=x''-\-2x-^l Evolving, x'^—l=±:{x+l). Taking the positive sign, and transposing, x^ — :r =2. By quadratics, x=2, or — 1. A?is. The sum of these is 1 ; hence the other value is — 1, and the equation has two equal roots, — 1, and — 1. CUBIC EQUATIONS. 331 2. Given l()x=x^-}-S to find the values of x. Conditions, 10x=3:^-\-d. Transposing, — ci^-{-10x=B. Multiplying by — x, x* — 10x^=—Sx. Separating into factors, 3=3 Xl* Adding (3)'^;^ to each side, x^ — x'=dx'^ — dx. Since coef. of x^ = the other factor, add {±y% x'—x^-i-i=9x''—dx+l. Evolving, x^-—}=±{dx—^). Taking the positive sign, and cancelling, x-=Sx. Dividing, x=S. Taking the negative sign, x^ — ^= — 3z-j-J. Transposing, x'^-\-dx=l. By quadratics, .==^^ . Hence, a;=3, or . The sum of these is 0. 3. Given 4r'-|"^^=l^^ *o find the values of x. Conditions, 4^:3+3^=182. Dividing by coefficient of 7?^ ^^~1""T^^~T~* Multiplying by x, a;*-[-_=__. . 182. ^ ^ 182 14 ^„ Separating —^ into factors, ~r^^~T^^^' Adding {-t-)^" to each side, a;^-f-132;^=:— j — j — -— . Since coefficient of x^ = the other factor, add [-^^ , ^•^+132;-+(^-j =_+_+^_ ) i3y Evolving, ar+— =±(^-2-H-j. 7.r Taking the positive sign and cancelling, 2;-=-^. 832 ALGEBRA. Dividing, ^=o- Taking the negative sign, a;^-|— — -= — — •. Transposing, rc2_|_ __ — \'^ 7x 49 49 159 Completing the square, a;2_|_ i ___][3_|_ _. — ^ , . ,7 V"^=T5^ Evolving, ^+|=zfc ^ . ^ . * _-7±aA15^ Transposing, :c=: r . The sum of these values is 0. 4. Given x^—7x=:Q to find the values of a:. Atis, x=S, or — 1, or — 2. 5. Given :r^=37a:-|-84 to find the values of x. Ans. x=7, or — 3, or — 4. 6. Given 2x^-\-7x=z4:74: to find the values of x. Ans. x=Q, or ^ . 7. Given 9a;3=169:i:+280 to find the values of a:. 7 8 Ans. xv=o, or — ^r, or — -. o o 8. Given x^ — 3:r=322 to find the values of a:. Atis. x=z7, or . Problems. 1. There is a cubical block of marble ; and if 50 be added to the number of square feet in half its surface, it will be equal to the number of cubic feet in its contents. What are the solid contents of the block ? Let X = the side of the cube. Then, x^ = the contents of the block. And x^ = the superficial contents of one side of the block. CUBIC EQUATIONS. 333 Then, Zx^ = the superficial contents of one-half the surface of the block. Therefore, :r^=3a;2-|-50. Multiplying both sides by 8, 82r^=24:c2_j_4oo. Adding x"^ and square of 4:r, x'^-{- ^x^-{-lQx^—x^-\-A.^x^-\-A.^^. Evolving, a;2-f4;^:=2;2^-20. Cancelling, .4x=20. Dividing, a:=5. Therefore the contents, a;^=125 cubic. 2. A gentleman having asked a lady her age, she replied, that if 29 times the square of her age were subtracted from twice its cube, the remainder would be 225. What was the lady's age ? Let a:= lady's age. Then, 2r'— 29a;2=225. Transposing, 23?=2^x'^J^ 225. Adding :c'* and the square of ^, x'^-\-2x^-{-x^z=x'^-\-^{)x^-\-21b. Evolving, x^-\-x = a;^ -f 1 5 . Cancelling, x=lb years. 3. A boy, being asked what he gave for his books, replied, that if 51 times the square of the number of dollars he gave* for them were subtracted from 6 times the cube of the number, the remainder would be 900. What was the price of the books ? A71S. $10. 4. A man, being asked how many dollars he had in his pockets, replied, that if three times the cube of the number he had in his pockets were added to five times the square of the number which he had, he should have 272. Kequired the number he had in his pockets. Ans. $4. 5. A boat has been sailing two hours, with a light breeze, against a strong current ; nineteen times the number of miles it has sailed is equal to the cube of that distance, added to thirty miles. How far has it sailed ? Ans. It has gained either 3 miles or 2 miles, or it has lost 5 miles. 334 ALGEBRA MISCELLANEOUS QUESTIONS. 1. Multiply 7a/^T3V^— V^"' by Oa//. Ans. 2. Multiply a^-i-P by a-^—b-\ Ans. ar'^W—cC-h-''. 3. Multiply ar-\-}f by «-2«+^-". 4. Multiply tsfax by — a/«^. Ans. — ax. 5. Divide — a by — 3a. J.7i5. \. 6. Divide a"""* by oJ", Am. a-"". 7. Divide a^-\-z^ by a+^' ^^^' ^"^ — a^x-}-a^x^ — az^-\-x*. 8. Multiply y'^-j-x'" by 2/—^. -4?Z5. 2/"'+^+a;'"2/— ^2/'"— ^'""^^. _ -^ . . , 1 wfl . wV , - , wV:c"~^ 9. Multiply -^—^-^^^^ by :r''+7ia2:"-^4—^— . Ans. l+-r^. 10. Divide 1 — x^ by 1—x. Ans. lJ^x-^x^+x^-\-x^+a^-\-x^-\-x^, 11. Multiply 3/^G>=a^ by 4VF=^. ^7z^. 12^'(^— 3a:c«— 2aV+3aV+6aV+a^a:^— a^a;2— 6aV — 3aV+2a^2:+3a'':r2— «^) „ _. 41-35Z 7-2:^2 l+3a: 2:^-2^ , . , ,, 12. Given — -7- — — =-_i_ __ ^ to find the 105 14(0;— 1) 21 6 value of a:. Ans. a:=:4. 13. Given a/x-\-9=1-{-/s/x to find the value of a:. Atis. z=16. 14. Given .^ — — r^=l — = — itt^ — r-r^^ to find the value 1— 2z 7— 2a: 7— 16z+4a;- of a:. Ans. 2;=— |^. 15. Given (A/^+28)(A/FF^)=(V^+3t3)(A/^+¥) to find the value of x. Atis. a;=4. y^\ ~ ^ ~ '}l. ^ ' MISCELLANEOUS QUESTIONS. 335 16. Qiven {x — l)/s/''lx—x^=-^ to find x. Ans. x= 7-^—. 17. Given x—2/s/ x+2=\-\-,^ x^—6x-{-2 to find x. Ans. 2:=9±4VT, or ^ — . 18. Given aJ a-\-x=l^ x^ — bax-{-b^ to find the value of a;. Ans. x=—- — . 7a 19. Given P=a^-\-bx to find the value of x. Ans. x= — - — . 20. Given ^{x—a)—l{2x—U)=l^a-\-llb to find the value of a:. Am. a;=25a-{-243. 21. Given i^+-^,+^?^±l?=5£f+^ to find the value of a:. . ab Am. X a-\-b 22. Given (^+3:F+(^-^)' ^^^ ^^ ^^^^ ^j^^ ^^1^^ ^^^^ (<2-|-^) — (a— a:)2" 2a3^ Ans. x=- — -. 1+3 23. Given ^ ^^ ^ _|_ ^^ -^ L ^^^t to find the value of a:. x'^ x^ Am. 2;=4(a — 1). 24. Given ^-+3^— (-— 3V=c- to find the value of 2:. Ans. x=- 25. A gentleman travelled 252 miles. The first day he rode 4 miles, the last 128, and each day's journey was double the preceding one. How many days was he performing the journey ? Atis. 6 days. 26. A gentleman dying left his sons an estate of $13,187.50. He bequeathed to his youngest son $1000, to the oldest $5062.50, 836 A L G E B 11 A . and ordered that each son's portion should exceed the next younger by the ratio of 1^. How many sons had he ? A71S. 5 sons. 27. The first term in a geometrical progression is 3, the last term ^, and the sum of the series 4f . What is the number of terms ? Ans. 4. 28. The first term in a geometrical series is ^, the ratio 7, and the last term 3361f . What is the number of terms ? Ans. 6. 29. What are the three arithmetical means between -^ and ^ ? 30. Required the sum of 200 terms of the series 1, 3, 5, 7, 9, &c. Am. 40,000. 31. The first term of an arithmetical series is — 7, the tenth term is 12. What is the sum of the series ? Atis. 25. 32. If a man travel 20 miles the first day, and 15 miles the second, and so continue to travel 5 miles less each day, how far will he have advanced on his journey the 8th day ? Ans. 20 miles. 33. The first term of an arithmetical series is 5, the number of terms 20 ; what must the coromon difi'erence be, that the sum of the series shall be 123^- ? Ans. -^Yu- 34. If a man travel 20 miles the first day, 19 the second day, I82V tbe third day, and so on in a geometrical progression, in how many days will he have travelled 400 miles ? Ans. ^. 35. A merchant, having mixed a certain number of gallons of wine and water, found that if he had mixed 6 gallons more of each, there would have been 7 gallons of wine to every 6 gallons of water ; but, if he had mixed 6 gallons less of each, there would have been 6 gallons of wine to every 5 gallons of water. How much of each did he mix ? Ans. 78 gallons of wine with 66 of water. 36. A person bought 2 cubical stacks of hay for £41 ; each of them cost as many shillings per solid yard as there were linear yards in a side of the other, and the greater occupied MISCELLANPOUS QUESTIONS. 837 9 square yards of ground more than the less. What was the price of each ? Atis. £25 and £16. 37. A certain man owes $1000. What sum. shall he pay daily, so as to cancel the debt, principal and interest, at the end of the year, reckoning simple interest at 6 per cent. ? Am. $2.81974. 38. A and B travelled on the same road, and at the same rate, from Portland to Boston. When A was at 50 miles' dis- tance from Boston he overtook a drove of geese, which were pro- ceeding at the rate of 3 miles in 2 hours ; and, two hours after- wards, met a stage-wagon, which was moving at the rate of 9 miles in 4 hours. B overtook the same drove of geese when he was 45 miles distance from Boston, and met the stage-wagon exactly 40 minutes before he arrived within 31 miles of Boston Where was B when A arrived at Boston ? A?ts. 25 miles from Boston. 39. A gentleman has two sons, John and Nathan. John is 10 years old, and Nathan is 15. He wishes to divide $1000 between his sons, in such a manner that each, by depositing his share in a savings' bank which pays 5 per cent, compound in- terest, shall have the same amount in the bank when he is 21 years old. What smn shall each deposit ? Am. John, $439.30 ; Nathan, $560.70. 40. My garden is 100 feet square, and I wish to raise its surface 2 feet with the soil taken from a ditch with which I intend to surround it. This ditch is to be 5 feet deep, and out- side the garden ; what should be its width ? Ans. d.l-\- feet. 41. A engaged to reap a field for $10, which he would do in 10 days ; but after he had labored 2 days he engaged B, by whose aid he supposed he could finish the field in 3 days. But, B proving to be a very inefficient workman, A was obliged to hire C the last two days, who proved to be a superior laborer ; the field was completed in 5 days. Now, if he had not hired C, and A and B had completed the work themselves, B would have received $1.08J-| in addition to his services for his 3 days' 29 338 4 ALGEBIIA. labor. How long would it have required B and C, each, to reap the field ? Alls. B could have reaped it in 11^ days, C in SJ-f days. 42. A man travelled 105 miles, and then found that if he had not travelled so fast by 2 miles an hour, he would have been 6 hours longer in performing the same journey. How many miles did he go per hour ? Atis. 7 miles. 43. The difierence between the hypothenuse and base of a right-angled triangle is 6 feet, and the difference between the hypothenuse and perpendicular is 3 feet. What are the sides of the triangle ? Ans. 15, 12, and 9 feet. 44. In a parcel which contains 24 coins of silver and copper, each silver coin is worth as many pence as there are copper coins; and each copper coin is worth as many pence as there are silver coins, and the whole is worth 18 shillings. - How many are there of each ? Ans. 6 of one, and 18 of the other. 4^ The income of a certain estate is to be sold for a term of 7 years. A offers to pay $300 down^ and $300 at the end of each year; B ofi"ers $800 doiDn, and $250 at the end of each year;_C offers $1300 doion^ and $200 at the end of each year; D will pay $2500 " cash down.'''' Which has made the best offer, if interest is to be reckoned at 6 per cent, compound interest ? / Value of A's offer, $1974.71.4. )b'^ Ans. \ B's offer, $2195.59.5 ; C's offer, $2416.47.6. 's offer, $2500. Hence D's offer is the best. 46. A gentleman being asked the age of his two sons, replied, that if the sum of their ages were multiplied by the age of the elder, the product would be 144 ; but if the difference of their ages were multiplied by that of the younger, the product would be 14. What was the age of each ? Ans. 9 and 7. 47. The sum of two numbers is 20, and the sum of their cubes is 2060. What are the numbers ? Ans. 9 and 11. 48. If the product of two numbers be added to the square of the larger, the sum will be 112 ; but, if the square of the less MISCELLANEOUS QUESTIONS. 339 be taken from their product, the remainder will be 12, Re- quired the numbers. Ans. 8 and 6. 49. AVhat number is that which, being added to twice its square root, equals 24 ? Ans. 16. 50. If a man owe $2000, what sum shall he pay daily, so as to cancel the debt, principal and interest, at the end of the year, reckoning the interest at 6 per cent, ? Ans. $5.0394. 51. I have 84i- square feet of plank, that is 3 inches thick. T How large a cubical box can be made from it ? Alts. Each side measures 48 inches. 52. From 62f|- feet of plank, that is 2h inches thick, I wish to make a box whose length shall be four times its width, and whose height and width shall be equal. What are its dimen- ' sions^ Ans. Length 8 feet, width and height 2 feet. 5&. There was a cask containing 20 gallons of wine ; a cer- tain quantity of this was drawn off into another cask of equal size, and this last filled with water, and afterwards the first cask was filled with the mixture. It now appears that, if 6f gallons of the mixture be drawn off from the first into the second cask, there will be equal quantities of wine in each. What was the quantity of wine drawn off at first? Ans. 10 gallons.. 54. Aftor A had travelled for 2f hours, at the rate of 4 miles an hour, B set out to overtake him ; and, in order thereto, went four m\[Q§ and a half the first hour, four and three-quarters the second, five the third, and so on, gaining a quarter of a mile every hour. In how many hours would he overtake A ? Alls. 8 hours. 55. The sum of the first and second of four numbers in geo- metrical progression is 15, and the sum of the third and fourth is 60. Required the numbers. Ans. 5, 10, 20, 40. 56. The sum of the squares of the extremes of four numbers in arithmetical progression is 200, and the sum of the squares of the means is 136. What are the numbers ? ^^: Am. 14, 10, 6,2. ■^^SSO ALGEBRA." ^/''^'^^ 57. A tailor bought a piece of cloth for £147, from which he cut off 12 yards for his own use ; he sold the remainder for £120 5^., gaining 5 shillings per yard. How many yards were there, and what did it cost him per yard ? A71S. 49 yards, at £3 per yard. 58. In a mixture of rye and wheat, the difference between the quantities of each is to the quantity of wheat as 100 is to the number of bushels of rye, and the same difference is to the quantity of rye as 4 to the number of bushels of wheat. How many bushels are there of each ? A71S. 25 bushels of rye, and 5 of wheat. 59. It is required to find two numbers, such that the product of the greater into the square root of the less shall be equal to 48, and the product of the less into the square root of the greater may be 36. A71S. 16 and 9. 60. If the difference of two numbers be multiplied by the greater, and the product divided by the less, the result will be 48; but, if the difference be multiplied by the less, and the product divided by the greater, the result will be 3. What are the numbers ? Atis. 16 and 4. 61. Find two numbers, such that the square of the greater, multiplied by the less, shall be equal to 448; and the square of the less, multiplied by the greater, shall be 392. Atis. 8 and 7. 62. If two numbers be each multiplied by 27, the first pro- duct is a square, and the second the square root of that square; ^^' but, if each be multiplied by 3, the first product is a cube, and the second the cube root of that cube. What are the numbers ? ji 7 Atis. 243 and 3. '^ 63. A farmer has two cubical stacks of hay ; the side of one is 3 yards longer than the side of the other, and the difference of their contents is 117 solid yards. Kequired the side of each. Atis. 5 and 2 yards. 64. A gentleman started from Boston for New York ; he trav- elled 20 miles the first day, 18 miles the second day, and 16 MISCELLANEOUS QUESTIONS. 341 miles the third day, so continuing to travel two miles less each day than the former. How far was the gentleman from Boston at the end of the twentieth day ? A7is. 65. A certain farm is a parallelogram, and a diagonal line from one corner to the opposite is 60 rods, and the longer side is to the shorter as 4 to 3. Required the contents of the farm. Ans. 10 Acres, 3 Hoods, 8 Poles. 66.' A gentleman asking a lady her age, she replied. If you add the square root of it to half of it, and subtract 12, there will remain nothing. Required her age. A^is. 16. 67. What number is that to which if 1, 7, and 19 be sever- ally added, the first sum shall have the same ratio to the second that the second has to the third ? Atis. 5. 68. The sum of two numbers is 12, and they have the same ratio to each other that their difierence has to 40. AVhat are the numbers ? Ans. 2 and 10. 69. There are two numbers whose product is 54, and the greater is to the less as their sum is to 10. What are those numbers ? A7is. 9 and 6. «^ 70. Divide 20 into two such parts that the square of the greater shall be to the square of the less as 9 to 4. What are those parts ? A?is. 12 and 8. " 71. Let 24 be divided into two such parts that the quotient of the greater divided by the less shall be to the quotient of the less divided by the greater as 9 to 1. A?is. 18 and 6. 72. Divide 14 into two such parts that their squares shall be to each other as 9 to 16. Am. 6 and 8. 6 73. Divide 12 into two such parts that the sum of their squares shall be to the difference of their squares as 5 to 3. Ans. 4 and 8. 74. There are two numbers, whose product is 12, and the sum of whose cubes is to the cube of their sum as 91 to 343. What are the numbers ? Ayis. 3 and 4. 29^ 342 ALGEBRA. 75. The product of two numbers is 120 ; and, if the greater be increased by 8 and the less by 5, the product of the two num- bers will be 300. What are the numbers ? Atis. 10 and 12. - 76. A, B, and C, make a joint stock ; A puts in $60 less than B, and $68 more than C, and the sum of the shares of A and B is to the sum of the shares of B and C as 5 to 4. What did each put in ? \ ; -«j -^ -; , A71S. A put in $140, B $200, and C $72. 77. A and B engage in speculation, with different sums ; A gains $150, B loses $50. Now A's stock is to B's as 3 to 2 ; but, had A lost $50, and B gained $100,^then A's stock would have been to B's as 5 to 9. What was the stock of each ? Ans. A's, $300 ; B's, $350. 78. Find two numbers in the ratio of 5 to 7, to which two other numbers, in the ratio of 3 to 5, being respectively added, the sums shall be in the ratio of 9 to 13, and the difference of the sums shall be 16. ( First two numbers, 30 and 42. ( Last two numbers, 6 and 10. 79. A merchant mixes wheat, which costs 10 shillings per bushel, with barley, which costs him 4 shillings per bushel, in such proportion as to gain 43|- per cent., by selling the mixture at 11 shillings per bushel. Required the proportion. Ans. He must mix 14 bushels of wheat with 9 of barley. 80. A and B can dig a cellar in a days, A and C can do the labor in b days, and B and C can do the same in c days. In what time would each perform the labor, and how long would it require A, B, and C, to complete the work ? . . . 2abc ^ _ . 2abc , „ . A?is. A in — — days, B in , . , days, C in ac-^bc — ab ab-\-bc — ac days, and A, B, C, in — ~ days. ab-\-ac—bc *^ ' ' ' ' ^^_j_^c4-^c 81. A and B made a joint stock of $833, which, after a suc- cessful speculation, produced a clear gain of $153. Of this B had $45 more than A. What did each person contribute to the ,., stock? ^7w. B $539, and A $294. MISCELLANEOUS QUESTIONS. 343 82. A gentleman having asked a ladj her age, she modestly replied, that if she were four years younger, and he were four years older, his age would be twice that of hers ; but, if she were four years older, and he were four years younger, their ages would be the same. What was the age of each ? Ans. Gentleman's age, 28 years ; lady's age, 20 years. ALGEBRA APPLIED TO GEOMETRY. 83. Suppose a tree, 48 feet in height, to stand on a hori- zontal plane. At what height from the ground must it be cut off, so that the top of it may fall on a point 24 feet from the bottom of the tree, the end, where it was cut off, resting on the stump ? Ans. 18 feet. 84. A certain man, owning a farm lying in a circle, gave it in his will to his wife, four sons, and four daughters, as follows : to his sons he gave four circles, as large as could be drawn within the circumference of the farm ; to his daughters he gave the four spaces lying between the son's circles and the circum- ference of the farm, and to his wife he gave the part remaining in the centre, which contained just one acre. How much did the whole farm contain, how much did each son have, and how much did each daughter have ? t The farm contained 21 Acres, 1 Kood, 12 Poles. Ans. } Each son had 3 Acres, 2 Roods, 25^- Poles. ( Each daughter had 1 Acre, 1 Rood, 274- Poles. 85. A gentleman has a garden in the form of an equilateral triangle, the sides whereof are each 100 feet. At each corner of the garden stands a tower; the height of the first tower is 40 feet, that of the second 45 feet, and that of the third is 55 feet. At what distance from the bottom of each of these towers must a ladder be placed, that it may just reach the top of each tower ; and what must be the length of the ladder, the ground of the garden being horizontal ? Ans. From the foot of the ladder to the base of the first tower, 63.273+ feet; second tower, 59.820-f- feet; third tower, 50.779+ feet; length of the ladder, 74.856+ feet. 344 ALGEBRA. 86. If c be tlie hjpotlienuse of a rigbt-angled triangle, b the base, and a the perpendicular, it is required to find the segments made by a perpendicular drawn from the right angle to the hypothenuse. . b'^-\-c' — a? ^ a?-\-c^ — b'^ 'Ic ^ 2c * 87. From a point within an equilateral triangle, there are drawn three perpendiculars to the several sides ; the length of the first is 20 feet, the second 30 feet, and the third 36 feet. Kequired the length of the sides of the triangle. Ans. 49.652-f feet. 88. A sphere of gold, whose diameter is one inch, weighs 10 ounces, and each ounce is valued at $16. What is the value of 5 spheres of gold, whose several diameters are 1, 2, 3, 4, and 5 inches? Am. $3600. 89. There is a loaf of bread, which is half a sphere, whose diameter measures 12 inches. How thick must the crust be baked, that the remainder shall be half the contents of the loaf? Ans. .8038-1- inch. 90. There are two towers of unequal heights, situated on a plane, near each other. A line extending from the base of the less to the top of the larger is 100 feet ; and a line from the base of the larger to the top of the less is 80.27-{- feet; a per- pendicular let fall from the point where the lines cross each other, to the surface of the plane, is 32 feet. Required the height of the towers, and their distance from each other. ( Height of the larger tower, 80 feet. Ans. } Height of the less, 53^ feet. ( Distance between the towers, 60 feet. 91. There is a conical glass, 6 inches deep ; the diameter at the top is 5 inches, and it is \ full of water. If a ball 4 inches in diameter be put into this glass, how much of its axis will be immersed in the water ? Ans. .546 inch. 92. How many balls 1 inch in diameter can be put into a cubical box whose sides measure each one foot in the dear ? Ans. 2151 balls. TABLE, CONTAINING THE LOGARITHMS OF NUMBERS FROM 1 TO 10,000. Numbers from 1 to 100 and their Logarithms, with their Indices. No. Log. No. Log. No. Log. No. Log. No. Log. 1 0.000000 21 1.322219 41 1.612784 61 1.785330 81 1.908485 2 0.301030 22 1.342423 42 1.623249 62 1.792392 82 1.913814 3 0.477121 23 1.361728 43 1.633468 63 1.799341 83 1.919078 4 0.G020G0 24 1.380211 44 1.643453 64 1.806180 84 1.924279 5 0.698970 25 1.397940 45 1.653213 65 1.812913 85 1.929419 6 0.778151 2G 1.414973 46 1.662758 66 1S19544 86 1.934498 7 0.845098 27 1.431364 47 1.672098 67 1.82G075 87 1.939519 8 0.903090 28 1.447158 48 1.681241 68 1.832509 88 1.944483 9 0.954243 29 1.462398 49 1.690196 69 1.838849 89 1.949390 10 1.000000 30 1.477121 50 1.698970 70 1.845098 90 1.954243 11 1.041393 31 1.491362 51 1.707570 71 1.851258 91 1.959041 12 1.079181 32 1.505150 52 1.716003 72 1.857332 92 1.9G3788 13 1.113943 33 1.518514 53 1.72427G 73 1.863323 93 1.9G8483 14 1.14G128 34 1.531479 54 1.732394 74 1.869232 94 1.973128 15 1.17G091 35 1.544068 55 1.7403G3 75 1.875061 95 1.977724 IG 1.204120 36 1.556303 56 1.748188 76 1.880814 96 1.982271 17 1.230449 37 1.568202 i)i 1.755875 77 1.886491 97 1.986772 18 1.255273 38 1.579784 58 1.763428 78 1.892095 98 1.99122C 19 1.278754 39 1.591065 59 1.770852 79 1.897627 99 1.995G35 20 1.301030 40 1.602060 00 1.778151 80 1.903090 100 2.000000 Note. — In the following part of the Table the Indices are omitted, as they can be very easily supplied by the directions given in Section xxxs., p. 270, on Logarithms. 346 LOGARITHMS N. 1 1 2 3 4. 5 1 6 7 8 9 |D.| 100 000000 000434 0008G8 001301 001734 0021G6 002598 003029 003461 003891 432 1 4321 4751 5181 5609 6038 6466 6894 7321 7748 8174 428 2 8G00 9026 9451 9876 010300 010724 011147 011570 011993 012415 424 3 012837 013259 013680 014100 4521 4940 5360 5779 6197 6616 420 4 7033 7451 7868 8284 8700 9116 9532 9947 020361 020775 416 5 021189 021G03 022016 022428 022841 023252 023664 024075 4486 4896 412 i ^ 5306 5715 G125 G533 6942 7350 7757 8164 8571 8978 408 7 9384 9789 030195 030600 031004 031408 031812 032210 032619 033021 404 8 033424 033826 4227 4628 5029 5430 5830 6230 6629 7028 400 i 9 742G 7825 8223 8620 9017 9414 9811 040207 040602 040998 397 |110 041393 041787 042182 042576 042969 043362 043755 044148 044540 044932 393 1 5323 5714 6105 6495 6885 7275 7664 8053 8442 8830 390 2 9218 9606 9993 050380 050766 051153 051538 051924 052309 052694 386 3 053078 0534G3 053846 4230 4613 4996 5378 5760 6142 6524 383 4 G905 7286 7666 8046 8426 8805 9185 9563 9942 060320 379 5 060698 0C1075 061452 061829 062206 062582 062958 063333 063709 4083 376 6 4458 4832 5206 5580 5953 6326 6699 7071 7443 7815 373 7 8186 8557 8928 9298 9668 070038 070407 070776 071145071514 370 8 071882 072250 072617 072985 073352 3718 4085 4451 4816 5182 366 9 5547 5912 627G 6640 7004 7368 7731 8094 8457 8819 363 120 079181 079543 079904 080266 080626 080987 081347 081707 082067 082426 360 1 082785 083144 083503 3861 4219 4576 4934 5291 5647 G004 357 2 6360 6716 7071 7426 7781 8136 8490 8845 9198 9552 355 3 9905 090258 090611 090963 091315 091667 092018 092370 092721 093071 352 ! 4 093422 3772 4122 4471 4820 5169 5518 5866 6215 6562 349 1 5 6910 7257 7604 7951 8298 8644 8990 9335 9681 100026 346 6 100371 100715 101059 101403 101747 102091 102434 102777 103119 3462 343 i "^ 3804 4146 4487 4828 5169 5510 5851 6191 6531 6871 341 8 7210 7549 7888 8227 8565 8903 9241 9579 9916 110253 338 9 110590 110926 111263 111599 111934 112270 112605 112940 113275 3609 335 130 113943 114277 114611 114944 115278 115611 115943 116276 116608 116940333 1 7271 7603 7934 8265 8595 8926 9256 9586 9915 120245 330 2 120574 120903 121231 121560 121888 122216 122544 122871 123198 3525 328 3 3852 4178 4504 4830 5156 5481 5806 6131 6456 6781 325 4 7105 7429 7753 8076 8399 8722 9045 9368 9690 130012 323 5 130334 130655 130977 131298 131619 131939 132260 132580 132900 3219 321 G 3539 3858 4177 4496 4814 5133 5451 5769 G086 0403 318 7 6721 7037 7354 7671 7987 8303 8618 8934 9249 9564 316 8 9879 140194 140508 140822 141136 141450 141763 142076 142389 142702 314 9 143015 3327 3639 3951 4263 4574 4885 5196 5507 5818 311 '140 146128 146438 146748 147058 147367 14767G 147985 148294 148603 148911 309 i 1 9219 9527 9835 150142 150449 150756 151063 151370 151676 151982 307 i 2 152288 152594 152900 3205 3510 3815 4120 4424 4728 5032 305 ! 3 5336 5640 5943 • 6246 6549 6852 7154 7457 7759 8061 303 1 ^ 8362 8664 8965 9266 9567 9868 160168 1604G9 160769 161068 301 5 161368 161G67 161967 162266 162564 162863 3161 3460 3758 4055 299 1 ^ 4353 4650 4947 5244 5541 5838 G134 6430 6726 7022 297 j 7 7317 7613 7908 8203 8497 8792 9086 9380 9674 9968 295 1 8 170262 170555 170848 171141 171434 171726 172019 172311 172603 172895 293 1 '^ 3186 3478 3769 4060 4351 4641 4932 5222 5512 5802 291 289 1150 176091 176381 176670 176959 177248 17753G 177825 178113 178401 178689 ! 1 8977 9264 9552 9839 18012G 180413 180699 180986 181272 181558 287 1 2 181844 182129 182415 182700 2985 3270 3555 3839 4123 4407 285 3 4G91 4975 5259 5542 5825 6108 6391 6674 6956 7239 283 4 7521 7803 8084 8366 8647 8928 9209 9490 9771 190051 281 5 190332 190612 190892 191171 191451 191730 192010 192289 192567 2846 279 G 3125 3403 3G81 3959 4237 4514 4792 5069 534G 5623 278 7 5900 6176 G453 6729 7005 7281 7556 7832 8107 8382 27G 8 8657 8932 9206 9481 9755 200029 200303 200577 200850 201124 274 9 201397 201670 201943 202216 202488 2761 3033 3305 3577 3848 272 [n. 1 1 1 2 1 3 4 II 5 1 G 7 8 1 9 1 D. OF NU3IBERS. 847 N.| 1 1 1 2 1 3 4 II 5 1 6 1 7 1 8 1 9 iL>.| lliU 204120 204391 204GG3 204934 205204 205475 20574G 20G01G 20G2bG 206556 271 1 G82G 7096 7365 7634 7904 8173 8441 8710: 8979 9247 269 2 9515 9783 210051 210319 210580 210853 211121 211388 211654 211921 267 3 212188 212454 2720 2986 3252 3518 3783 4049 4314 4579 266 4 4844 5109 5373 5638 5902 6166 6430 6694 6957 7221 264 5 7484 7747 8010 8273 8530 8798 9060 9323 9585 9846 2G2 C 220108 220370 220631 220892 221153 221414 221675 221936 222196 222456 2G] 7 2716 2976 3236 3496 3755 4015 4274 4533 4792 5051 259 8 5309 5568 5826 6084 6342 6600 6858 7115 7372 7630 258 9 7887 8144 8400 8657 8913 • 9170 9426 9682 9938 230193|25G 170 230449 230704 230960i231215i231470! 231724 231979 232234 232488 232742 255 1 2996 3250 3504 3757 4011 4264 4517 4770 5023 5270 25:; 2 5528 5781 6033 6285 6537 6789 7041 7292 7544 7795 252 o o 8046 8297 8548 8799 9049 9299 9550 9800 240050 240300 25U 4 240549 240799 241048 241297 241546 241795 242044 242293 2541 2790/249 5 3038 3286 3534 3782 4030 4277 4525 4772 5019 5266 248 6 5513 5759 0006 6252 6499 6745 6991 7237 7482 7728 24G 7 7973 8219 8464 8709 8954 9198 9443 9687 9932 250176 245 8 250420 250664 250908 251151 251395 251638 251881 252125 252368 26101243 9 2853 3096 3338 3580 3822 40G4 4306 4548 4790 503l!242 180 255273 ^5514 2557551255996 256237 256477 256718 256958 257198 257439 241 1 7G79 7918 8158 8398 8637 i 8877 9116 9355 9594 9833 239 2 260071 260310 260548 260787 261025 261263 261501 261739 261976 262214 238 3 2451 2688 2925 3162 3399 3636 3873 4109 4346 4582237 4 4818 5054 5290 5525 57611 5996 6232 6467 6702 69371235 5 7172 7406 7641 7875 8110 8344 8578 8812 9046 9279 234 6 9513 9746 9980 270213 270446 270679 270912 271144 271377 271G09I233 7 271842 272074 272306 2538 2770 3001 3233 3464 3696 3927 232 8 4158 4389 4620 4850 5081 5311 5542 5772 6002 C232 230 9 190 6462 278754 6692 6921 7151 7380, 7609 7838 8067 8296 85251229 2789821279211 279439 279667 279895 280123 280351 280578 28080622S 1 281033 2812611281488 281715 281942 282169 2396 2622 2849 3075227 2 3301 3527 3753 3979 4205 4431 4656 4882 5107 533222G 3 5557 5782 6007 6232 6456 6681 6905 7130 7354 7578 225 4 7802 8026 8249 8473 8696 8920 9143 9366 9589 9812 223 5 290035 290257 290480 290702 290925 291147 291369 291591 291813 292034222 G 2256 2478 2699 2920 3141 3363 3584 3804 4025 4246 221 7 4466 4G87 4907 5127 5347 5567 5787 6007 6226 6446220 8 6665 G8S4 7104 7323 7542 7761 7979 8198 8416 8635 219 9 8853 9071 9289 9507 9725 9943 300161 300378 3005951300813 218j 200 301030 301247 3014G4 301681 301898 302114 302331 302547i302764i302980 2171 1 319C 3412 3628 3844 4059 4275 4491 4706 4921 5136 21g| 2 5351 55G6 5781 5996 > 6211 6425 CG39 6854 7068 7282 215; 3 7496 7710 7924 8137 8351 8564 8778 8991 9204 9417 213 4 9630 9843 31005G 310268 310481 310693 310906 311118 311330:311542:212; 5 311754 3119GG 2177 2389 2600 2812 3023 3234 34451 36562111 G 3867 407S 4289 4499 4710 4920 5130 5340 555l| 57G0|210| 7 5970 6180 6390 6599 6809 7018 7227 7436 764G 7854209' 8 8063 8272 8481 8689 8898 1 9106 9314 9522 9730 9933:208! 9 ;320146 320354 320562 320769 320977,321184 321391J321598 321805 322012 207] 210 322219 322426 322633 322839 323046 323252 323458 323665 323871 324077 2001 1 4282 4488 4694 4899 5105 5310 5516 5721 5926 6131 205 2 6336 G541 6745 6950 7155 7359 7563 7767 7972 8176 204 3 8380 B583 8787 8991 9194 9398 9601 9805i330008|330211 203 4|330414 330G17 330819 331022 331225 3314271331630 331832 2034 2236 202 5 2438 2640 2842 3044 3246 3447 3649 3850 4051 4253 202 C 4454 4655 4S5G 5057 5257 5458 5658 5859 6059 6260 201 ■3 6460 6660 6860 70P0 7260 7459 7659 7858 8058 8257 200 S 8456 865C 8855 9054 9253 9451 9650' 9S49;340047j34024G 199 9'340444'3i0f)42 '34081l'341039 '341237, ']41435;341G32 341830 1 2028 1 « 1 2225 198! N.l .. 1 I ! 2 1 3 1 4 II 5 1 G 1 7 1 9 J^ 348 LOGARITHMS N.| 1 2 1 3 1 4 II 5 1 ^ 1 ^ 8 1 y ii>.| 220 342423 342G20 342817 343014 343212 343409 34360G 343802 343999 344196 197 1 4392 4589 4785 4981 5178 5374 5570 5766 5962 6157 196 2 6353 6549 6744 6939 7135 7330 7525 7720 7915 8110 195 3 8305 8500 8694 8889 9083 9278 9472 9666 9860 350054 194 4 350248 350442 35063G 350829 351023 351216 351410 351603 351796 1989 193 5 2183 2375 2568 2761 2954 3147 3339 3532 3724 3916 193 6 4108 4301 4493 4685 4876 5068 ^260 5452 5643 5834 192 7 G02G 6217 6408 6599 6790 6981 7172 7363 7554 7744 191 8 7935 8125 8316 8506 8696 8886 9076 9266 9456 9646 190 9 9835 360025 360215'360404 360593 360783 360972 361161 361350 361539 189 230 361728 361917 362105 362294 362482 362671 362859 363048 363236 363424 188 1 3G12 3800 3988 4176 4363 4551 4739 4926 5113 5301 188 2 5488 5675 5862 6049 6236 6423 6610 6796 6983 7169 187 3 7356 7542 7729 7915 8101 8287 8473 8659 8845 9030 186 4 9216 9401 9587 9772 9958 370143 370328 370513 370698 370883 185 5 3710G8 371253 371437 371622 371806 1991 2175 2360 2544 2728 184 6 2912 3096 3280 3464 3647 3831 4015 4198 4382 4565 184 7 4748 4932 5115 5298 5481 5664 5846 6029 6212 6394 183 8 6577 6759 6942 7124 7306 7488 7670 7852 8034 8216 182 9 8398 8580 8761 8943 9124 9306 9487 9668 9849 380030 181 240 380211 380392 380573 380754 380934 381115 381296 381476 3816*6 381837 181 1 2017 2197 2377 2557 2737 2917 3097 3277 3456 3636 180 2 3815 3995 4174 4353 4533 4712 4891 5070 5249 5428 179 3 5600 5785 5964 6142 6321 6499 6677 6856 7034 7212 178 4 7390 7568 7746 7923 8101 8279 8456 8634 8811 8989 178 5 91GC 9343 9520 9698 9875 390051 390228 390405 390582 390759 177 G 390935 391112 391288 391464 391641 1817 1993 2169 2345 2521 176 7 2097 2873 3048 3224 3400 3575 3751 3926 4101 4277 176 8 4452 4627 4802 4977 5152 5326 5501 5676 5850 6025 175 9 6199 6374 6548 6722 6896 7071 7245 7419 7592 7766 174 250 397940 398114 398287 3984611398634 398808 398981 399154 399328 399501 173 1 9674 9847 400020 400192 400365 400538 400711 400883 401056 401228 173 2 401401 401573 1745 1917 2089 2261 2433 2605 2777 2949 172 3 3121 3292 3464 3635 3807 3978 4149 4320 4492 4663 171 4 4834 5005 5176 5346 5517 5688 5858 0029 6199 6370 171 5 6540 6710 6881 7051 72L1 7391 7561 7731 7901 8070 170 G 8240 8410 8579 8749 8918 9087 9257 9426 9595 9764 169 7 9933 410102 410271 410440 410609 410777 410946 411114 411283 411451 169 ■ 8 411620 1788 1956 2124 2293 2461 2629 2796 2964 3132 168 9 3300 3467 3035 3803 3970 4137 4305 4472 4639 4806 167 260 414973 415140 415307 415474 415641 415808 415974 416141 416308 416474 167 1 6641 6807 6973 7139 7306 7472 7638 7804 7970 8135 166 2 8301 8467 8633 8798 8964 9129 9295 94G0 9025 9791 165 3 9956 420121 420286 420451 420616 420781 420945 421110 421275 421439 165 4 421604 1768 1933 2097 2261 2426 2590 2754 2918 3082 164 ■ 5 3246 3410 3574 3737 3901 4065 4228 4392 4555 4718 164 G 4882 5045 5208 5371 5534 5697 5860 6023 6186 6349 163 7 6511 6674 6836 6999 7161 7324 7486 7648 7811 7973 162 8 8135 8297 8459 8621 8783 8944 9106 9268 9429 9591 162 9 9752 9914 430075 430236 430398 430559 430720 430881 431042 431203 161 270 431364 431525 431685 431846 432007 432167 432328 432488 432649 432809 161 1 2969 3130 3290 3450 3610 3770 3930 4090 4249 4409 160 2 4569 4729 4888 5048 5207 5367 5526 6685 5844 6004 159 3 6163 6322 0481 6640 6799 6957 7116 7275 7433 7592 159 4 7751 7909 8067 8226 8384 8542 8701 8859 9017 9175 158 5 9333 9491 9648 9806 9964 440122 440279 440437 440594 440752 158 6 440909 4410(;6 441224 441381 441538 1695 1852 2009 2166 2323 157 7 2480 2637 2793 2950 310G 3263 3419 3576 3732 3889 157 8 4045 4201 4357 4513 4669 4825 4981 5137 5293 5449 156 9 5604 5760 5915 6071 6226 6382 6537 6692 6848 7003 155J N 1 1 2 1 3 4 II .-> 1 G 7 « 1 !) 1 1).| OF NUMBERS. 349 N.I 1 1 2 3 1 4 II 5 1 ♦; 7 8 1 'J |JJ.| 280 447158 447313 447468 447623;44777» 447933 448088 448242 448397 448552 15j 1 870G 8861 9015 9170 9324 9478 9633 9787 9941 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OF NUMBERS. 351 N. 1 1 2 3 1 4 !| .5 1 6 7 1 « 1 9 1 1). 40U 602060 602109 U02277 602386 6024941 602003 602711 602bl9 602'J2b, 603036,108 1 3144 3253 3361 3469 3577 3686 3794 3902 4010 4118 108 2 4226 4334 4442 4550 4658 4766 4874 4982 5089 6197 108 3 5305 5413 5521 5628 5736 5844 5951 6059 6166 6274 108 4 6381 6489 6596 6704 6811 6919 7026 7133 7241 7348 107 6 7455 7562 7669 7777 7884 7991 8098 8205 8312 8419 107 6 8526 8633 8740 8847 8954 9061 9167 9274 9381 9488 107 7 9594 9701 9808 9914 610021 610128 610234 010341 610447 610554 107 8 610660 610767 610873 610979 1086 1192 1298 1405 1511 1617 106 y 1723 1829 1936 2042 2148i 2254 2360 2406 2572 2678 106 410 Gl'2784 612890 612996 613102 613207 613313 613419 613525 613630 613736 100 1 3842 3947 4053 4159 4264 4370 4475 4581 4686 4792 106 2 4897 5003 5108 5213 5319 5424 5529 5634 5740 5845 105 3 5950 6055 6160 6265 6370 6476 6581 6686 6790 6895 105 4 7000 7105 7210 7315 7420 7525 7629 7734 7839 7943 105 5 8048 8153 8257 8362 8466 8571 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30^ ^54 LOGARITHMS N.| ■ 1 1 2 3 1 4 1 6 1 6 1 7 1 8 1 y 1 D. 580 763428 763503 763578 763653 763727 763802 763877|763952| 764027 764101 75 1 4176 4251 4326 4400 4475 4550 4624 4699 4774 4848 75 2 4923 4998 5072 5147 5221 5296 5370 5445 5520 5594 75 3 5669 5743 5818 5892 5966 6041 6115 6190 6264 6338 74 4 6413 6487 6562 6636 6710 6785 6859 6933 7007 7082 74 5 7156 7230 7304 7379 7453 7527 7601 7675 7749 7823 74 6 7898 7972 8040 8120 8194 8268 8342 8416 8490 8564 74 8638 8712 8786 8860 8934 9008 9082 9156 9230 9303 74 , 8 9377 9451 9525 9599 9673 9746 9820 9894 9968 770042 74 9 770115 770189 770263 770336 770410 770484 770557 770631 770705 0778 74 59U 770852 770926 770999 771073 771146 771220 771293 771367 771440 771514 74 1 1587 1661 1734 1808 1881 1955 2028 2102 2175 2248 73 2 2322 2395 2468 2542 2615 2688 2762 2835 2908 2981 73 3 3055 3128 3201 3274 3348 3421 3494 3567 3640 3713 73 4 3786 3860 3933 4006 4079 4152 4225 4298 4371 4444 73 5 4517 4590 4663 4736 4809 4882 4955 5028 5100 5173 73 6 5240 5319 5392 5465 5538 5610 5683 5756 5829 5902 73 7 5974 6047 6120 6193 6265 6338 6411 6483 6556 6629 73 8 6701 6774 6846 6919 6992 7064 7137 7209 7282 7354 73 9 7427 7499 7572 7644 7717 7789 7862 7934 8006 8079 72 GOO 778151 778224 778296 778368 778441 778513 778585 778658 778730 778802 72 1 8874 8947 9019 9091 9163 9236 9308 9380 9452 9524 72 2 9596 9669 9741 9813 9885 9957 780029 780101 780173 780245 72 3 780317 780389 780461 780533 780605 780677 0749 0821 0893 0965 72 4 1037 1109 1181 1253 1324 1396 1468 1540 1612 1684 72 5 1755 1827 1899 1971 2042 2114 2186 2258 2329 2401 72 6 2473 2544 2616 2688 2759 2831 2902 2974 3046 3117 72 7 3189 3260 3332 3403 3475 3546 3618 3689 3761 3832 71 8 3904 3975 4046 4118 4189 4261 4332 4403 4475 4546 71 9 4617 4689 4760 4831 4902 4974 5045 5116 5187 5259 71 610 785330 785401 785472 785543 785615 785686 785757 785828 785899|785970 71 1 6041 6112 0183 6254 6325 6396 6407 6538 6609 6680 71 2 6751 6822 6893 6964 7035 7106 7177 7248 7319 7390 71 3 7460 7531 7602 7673 7744 7815 7885 7956 8027 8098 71 4 8168 8239 8310 8381 8451 8522 8593 8603 8734 8804 71 6 8875 8946 9016 9087 9157 9228 9299 9369 9440 9510 71 6 9581 9651 9722 9792 9803 9933 790004 790074 790144 790215 70 7 790285 790356 790426 790496 790567 790637 0707 0778 0848 0918 70 8 0988 1059 1129 1199 1269 1340 1410 1480 1550 1620 70 9 1691 1761 1831 1901 1971 2041 2111 2181 2252 2322 70 620 792392 792462 792532 792602 792672 792742 792812 792882 792952 793022 70 1 3092 3162 3231 3301 3371 3441 3511 3581 3651 3721 70 2 3790 3860 3930 4000 4070 4139 4209 4279 4349 4418 70 3 4488 4558 4627 4697 4767 4836 4906 4976 5045 5115 70 4 5185 5254 5324 5393 5463 5532 5602 5672 5741 5811 70 5 5880 5949 •6019 6088 6158 6227 6297 6366 6436 6505 69 6 6574 6644 6713 6782 6852 6921 6990 7060 7129 7198 69 7 7268 7337 7406 7475 7545 7614 7683 7752 7821 7890 69 8 7960 8029 8098 8167 8236 8305 8374 8443 8513 8582 69 9 8651 8720 8789 8858 8927 8996 9065 9134 9203 9272 69 630 799341 799409 799478 799547 799616 799685 799754 799823 799892 799961 69 1 800029 800098 800167 800236 809305 800373 800442 800511 800580 800648 69 2 0717 0786 0854 0923 0992 1061 1129 1198 1266 1335 69 3 1404 1472 1541 1609 1678 1747 1815 1884 1952 2021 69 4 2089 2158 2226 2295 2363 2432 2500 2568 2637 2705 68 5 2774 2842 2910 2979 3047 3116 3184 3252 3321 3389 68 6 3457 3525 3594 3662 3730 3798 3867 3935 4003 4071 68 7 4139 4208 4276 4344 4412 4480 4548 4616 4685 4753 68 8 4821 4889 4957 5025 5093 5161 5229 5297 5365 5433 68 9 5501 5569 5637 5705 5773 5841 5908 5976 6044 6112 68 N. 1 2 3 4 5 6 7 8 y D. OF NUMBERS. 355 N. 1 2 1 3 1 4 1 5 6 1 7 1 8 1 9 ID.I 640 80uiao 806248 806316 806384 806451; 806519 806587 806655 806723 806790! 68 ( 1 0858 6926 6994 7061 7129 7197 7264 7332 7400 7467| 68 2 7535 7603 7670 7738 7806 7873 7941 8008 8076 8143 68 3 8211 8279 8346 8414 8481 8549 8616 8684 8751 8818 67 4 8886 8953 9021 9088 9156 9223 9290 9358 9425 9492 67 5 95 GO 9627 9694 9762 9829 9896 9964 810031 810098 810165 67 810233 810300 810367 810434 810501 810569 810636 0703 0770 0837 67 7 0904 0971 1039 1106 1173 1240 1307 1374 1441 1508 67 8 1575 1642 1709 1776 1843 1910 1977 2044 2111 2178 67 9 2245 2312 2379 2445 2512 2579 2646 2713 2780 2847 67 G50 812913 812980 813047 813114 813181, 813247 813314 813381 813448 813514 67 1 3581 3648 3714 3781 3848; 3914 3981 4048 4114 4181 67 2 4248 4314 4381 4447 4514 4581 4647 4714 4780 4847 67 3 4913 4980 5046 5113 5179 5246 5312 6378 5445 5511 66 4 6578 5644 5711 5777 5843 6910 5976 6042 6109 6175 66 5 6241 6308 6374 6440 6506 6573 6639 6705 6771 6838 66 6 6904 6970 7036 7102 7169 7235 7301 7367 7433 7499 66 7 75G5 7631 7698 7764 7830 7896 7962 8028 8094 '8160 66 8 8226 8292 8358 8424 8490 8556 8622 8688 8754 8820 66 9 8885 8951 9017 9083 9149 9215 9281 9346 9412 9478 66 660 819544 819610 819676 819741 819807 819873 819939 820004 820070,820136 60 1 820201 820267 820333 820399 820464 820530 820595 0661 0727 0792 66 2 0858 0924 0989 1055 1120 1186 1251 1317 1382 1448 66 3 1514 1579 1645 1710 1775 1841 1906 1972 2037 2103 65 4 2168 2233 2299 2364 2430 2495 2560 2626 2691 2756 65 5 2822 2887 2952 3018 3083 3148 3213 3279 3344 3409 65 G 3474 3539 3605 3670 3735 3800 3865 3930 3996 4061 65 7 4126 4191 4256 4321 4386 4451 4516 4581 4646 4711 65 8 4776 4841 4906 4971 6036 5101 6166 5231 5296 6361 65 9 542G 5491 5556 5621 6686 6751 6815 6880 6945 6010 65 670 82G075 826140 826204 826269 826334 826399 826464 826528 826593;826658 65 1 6723 6787 6852 6917 6981 7046 7111 7175 7240 7305 65 2 7369 7434 7499 7563 7628 7692 7757 7821 7886 7951 65 3 8015 8080 8144 8209 8273 8338 8402 8467 8531 8595 64 4 8660 8724 8789 8853 8918 8982 9046 9111 9175 9239 64 5 9304 9368 9432 9497 9561 9625 9690 9754 9818 9882 64 6 -9947 830011 830075 830139 830204 830268 830332|830396 830460 830525 64 7 830589 0C53 0717 0781 0845 0909 0973 1037 1102 1166 64 8 1230 1294 1358 1422 1486 1550 1614 1678 1742 1806 64 9 1870 1934 1998 2062 2126 2189 2253 2317 2381 2445 64 680 832509 832573 832637 832700 8327641 8328281832892 832956 833020 833083 64 1 3147 3211 3275 3338 3402 3466 3530 3593 3657 3721 64 2 3784 3848 3912 3975 4039 4103 4166 4230 4294 4357 64 3 4421 4484 4548 4611 4675 4739 4802 4866 4929 4993 64 4 5056 6120 5183 5247 5310 5373 6437 6500 6564 5627 63 5 5691 5754 5817 6881 5944 6007 6071 6134 6197 6261 63 6 6324 6387 6451 6514 6577 6641 6704 0767 6830 6894 63 7 6957 7020 7083 7146 7210 7273 7336 7399 7462 7525 63 8 7588 7652 7715 7778 7841 7904 7967 8030 8093 8156 63 9 8219 8282 8345 8408 8471 8534 8597 8060 8723 8786 63 690|838849|838912 S3S975 839038 839101 839164 839227 839289 839352'839415: 63 I 1 9478 9541 9604 9667 9729 9792 9855 9918 99S1 840043 63 1 2 840106 840169 840232 840294 840357 840420 840482 840545 840608 0671 6-3 3 0733 0796 0859 0921 0984 1046 1109 1172 1234 1297 63 4 1359 1422 1485 1547 1610 1672 1735 1797 1860 1922 63 5 1985 2047 2110 2172 2235 2297 2360 2422 2484 2547 62 6 2609 2672 2734 2796 2859 2921 2983 3046 3108 3170 62 7 3233 3295 3357 3420 3482' 3544 3606 3669 3731 3793 62 8 3855 3918 3980 4042 4104 4166 4229 4291 4353 4415 62 9 4477 4539 4601 4664 4726 4788 4850 4912 4974 5036 62 N. 1 1 2 3 1 4 1 5 6 1 7 1 S 9 1 D. 356 LOGARITHMS N. 1 2 3 1 4 1 5 6 7 8 1 9 D. 700 845098 845160 845222 845284 845346 845408 845470 845532 845594 84565b| 62 1 5718 5780 5842 5904 5966 6028 6090 6151 6213 6275 62 2 6337 6399 6461 6523 6585 6646 6708 6770 6832 6894 62 3 6955 7017 7079 7141 7202 7264 7326 7388 7449 7511 62 4 7573 7634 7696 7758 7819 7881 7943 8004 8066 8128 62 5 8189 8251 8312 8374 8435 8497 8559 8620 8682 8743 62 6 8805 8866 8928 8989 9051 9112 9174 9235 9297 9358 61 7 9419 9481 9542 9604 9665 9726 9788 9849 9911 9972 61 8 850033 850095 850156 850217 850279 850340 850401 850462 850524 850585 61 9 0646 0707 0769 0830 0891 0952 1014 1075 1136 1197 61 710 851258 851320 8513811 851442 851503 851564 851625 851686 851747 851809 61 1 1870 1931 1992 2053 2114 2175 2236 2297 2358 2419 61 2 2480 2541 2602 2663 2724 2785 2846 2907 2968 3029 61 3 3090 3150 3211 3272 3333 3394 3455 3516 3577 3037 61 4 3698 3759 3820 3881 3941 4002 4063 4124 4185 4245 61 5 4306 4367 4428 4488 4549 4610 4670 4731 4792 4852 61 6 4913 4974 5034 5095 5156 5216 5277 5337 5398 5459 61 7 5519 5580 5640 5701 5761 5822 5882 6943 6003 6064 61 8 6124 6185 6245 6306 6366 6427 6487 6548 6608 6668 60 9 6729 6789 6850 6910 6970 7031 7091 7152 7212 7272 60 720 857332 857393 857453 857513 857574 857634 857694 857755 857815 857875 60 1 7935 7995 8056 8116 8176 8236 8297 8357 8417 8477 60 2 8537 8597 8657 8718 8778 8838 8898 8958 9018 9078 60 3 9138 9198 9258 9318 9379 9439 9499 9559 9619 9679 60 4 9739 9799 9859 9918 9978 860038 860098 860158 860218 860278 60 6 860338 860398 860458 860518 860578 0637 0697 0757 0817 0877 60 G 0937 0996 1056 1110 1176 1236 1295 1355 1415 1475 60 7 1534 1594 1654 1714 1773 1833 1893 1952 2012 2072 60 8 2131 2191 2251 2310 2370 2430 2489 2549 2608 2668 60 9 2728 2787 2847 2906 2966 3025 3085 3144 3204 3263 60 730 863323 863382 863442 863501 863561 863620 863680 863739 863799 863858 59 1 3917 3977 4036 4096 4155 4214 4274 4333 4392 4452 69 2 4511 4570 4630 4689 4748 4808 4867 4926 4985 6045 59 3 5104 5163 5222 5282 5341 5400 6459 5519 6578 5637 69 4 5696 5755 5814 5874 5933 5992 6051 6110 6169 6228 69 5 6287 6346 6405 6465 6524 6583 6642 6701 6760 6819 69 6 6878 6937 6996 7055 7114 7173 7232 7291 7350 7409 69 7 7467 7526 7585 7644 7703 7762 7821 7880 7939 7998 69 8 8056 8115 8174 8233 8292 8350 8409 8468 8527 8586 69 9 740 8644 8703 8762 8821 8879 8938 8997 9056 9114 9173 59 869232 869290 869349 869408 869466 869525 869584 869642 869701 869760 59 1 9818 9877 9935 9994 870053 870111 870170 870228 870287 870345 69 2 870404 870462 870521 870579 0638 0696 0755 0813 0872 0930 58 0989 1047 1106 1164 1223 1281 1339 1398 1456 1515 68 4 1573 1631 1690 1748 1806 1865 1923 1981 2040 2098 68 5 2156 2215 2273 2331 2389 2448 2506 2564 2622 2681 68 6 2739 2797 2855 2913 2972 3030 3088 3146 3204 3262 58 7 3321 3379 3437 3495 3553 3611 3669 3727 3785 3844 58 8 3902 3960 4018 4076 4134 4192 4250 4308 4366 4424 58 9 4482 4540 4598 4656 4714 4772 4830 4888 4945 6003 58 750 875061 875119 875177 875235 875293 875351 875409 875466 875524 8755821 68 ( 1 5640 5698 5756 5813 5871 5929 5987 6045 6102 6160 68 2 6218 6276 6333 6391 6449 6507 6564 6622 6680 6737 68 3 6795 6853 6910 6968 7026 7083 7141 7199 7256 7314 58 4 7371 7429 7487 7544 7602 7659 7717 7774 7832 7889 58 5 7947 8004 8062 8119 8177 8234 8292 8349 8407 84C4 57 6 8522 8579 8637 8694 8752 8809 8866 8924 8981 9033 67 7 9096 9153 9211 9268 9325 9383 9440 9497 9555 9612 67 8 9669 972t 9784 9841 9898 9956 880013 880070 880127 880185 57 9 880242 880299|880350 880413 880471 880528 0585 0642 0099 0756 67 N. 1 1 2 3 4 5 6 [ 7 8 9 D.| OF NUMBERS. 357' N. 1 2 1 3 ^ 1 5 1 6 1 7 1 8 9 D. 760 880814 880871 880928 880985 881042, 881099 88115G 881213,881271 881328 57 1 1385 1442 1499 155G 1613 1670 1727 1784 1841 1898 57 2 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 57 3 2525 2581 2638 2695 2752 2809 2866 2923 2980 3037 57 4 3093 3150 3207 3264 3321 3377 3434 3491 3548 3605 57 5 36G1 3718 3775 3832 3888 3945 4002 4059 4115 4172 67 G 4229 4285 4342 4399 4455 4512 4569 4625 4682 4739 57 7 4795 4852 4909 4965 5022 6078 5135 5192 5248 5305 57 8 53 Gl 6418 5474 5531 5587 5644 5700 6757 5813 5870 57 9 592G 5983 6039 6096 6152 6209 6265 6321 6378 6434 66 770 88G491 886547 886604 886660 886716 886773 886829 886885 886942 886998 56 1 7054 7111 7107 7223 7280 7336 7392 7449 7505 7561 56 2 7C17 7674 7730 77B6 7842 7898 7955 8011 8067 8123 56 3 8179 8236 8292 8348 8404 8460 8516 8573 8629 8685 56 4 8741 8797 8853 8909 8965 9021 9077 9134 9190 9246 56 5 9302 9358 9414 9470 9526 9582 9638 9694 9750 9800 56 6 98G2 9918 9974 890030 890086 890141 890197 890253 890309 8903 G5 56 7 890421 890477 890533 0589 0G45 0700 0756 0812 08 G8 0924 56 8 0980 1035 1091 1147 1203 1259 1314 1370 1426 1482 56 9 1537 1593 1649 1705 1760 1816 1872 1928 1983 2039 56 780 892095 892150 892206 892262 892317 892373 892429 892484]892540 892595 1 56 j 1 2651 2707 2762 2818 2873 2929 2985 3040 3096 3151 56 2 3207 3262 3318 3373 3429 3484 3540 3595 3651 3706 56 3 37G2 3817 3873 3928 3984 4039 4094 4150 4205 4261 55 4 4316 4371 4427 4482 4538 4593 4648 4704 4759 4814 55 5 4870 4925 4980 5036 6091 5146 6201 6257 6312 5367 55 6 5423 6478 5533 5588 6644 6699 5754 6809 6864 5920 55 7 5975 6030 6085 6140 6195 6251 6300 6361 6416 6471 55 8 6526 6581 6636 6692 6747 6802 6857 6912 6967 7022 55 9 7077 7132 7187 7242 7297 7352 7407 7462 7517 7572 55 790 897627 897682 897737 897792 897847 897902 897959 898012 898067 898122 55 1 8176 8231 8286 8341 8396 8451 8506 8561 8615 8070 55 2 8725 8780 8835 8890 8944 8999 9054 9109 9164 9218 55 3 9273 9328 9383 9437 9492 9547 9602 9656 9711 9766 55 4 9821 9875 9930 9985 900039 900094 900149 900203 900258 900312 55 5 900367 900422 900476 900531 0586 0640 0695 0749 0804 0859 55 6 0913 0968 1022 1077 1131 1186 1240 1295 1349 1404 55 7 1458 1513 1567 1622 1676 1731 1785 1840 1894 1948 54 8 2003 2057 2112 2166 2221 2275 2329 2384 2438 2492 54 9 2547 2001 2655 2710 2764 2818 2873 2927 2981 3036 54 800 903090 903144 903199 903253 903307 903361 903416 903470 903524 903578 54 1 3633 3687 3741 3795 3849 3904 3958 4012 4066 4120 54 2 4174 4229 4283 4337 4391 4445 4499 4553 4607 4661 54 3 4716 4770 4824 4878 4932 4986 6040 6094 5148 5202 54 4 5256 5310 5364 6418 6472 6526 5580 6634 5688 5742 54 5 6796 5850 6904 6958 0012 6066 6119 6173 6227 6281 54 6 6335 0389 6443 6497 6551 6604 6658 6712 6766 6820 54 7 6874 6927 6981 7035 7089 7143 7196 7250 7304 7358 54 8 7411 7465 7519 7573 7626 7680 7734 7787 7841 7895 54 9 7949 8002 8056 8110 8163 8217 8270 8324 8378 8431 54 810,908485 908539 908592 90864G 908G99 908753 908807 9088G0 908914 9089G7 54 1 9021 9074 9128 9181 9235 9289 9342 9396 9449 9503 54 2 9556 9610 9663 9716 9770 9823 9877 9930 9984910037 53 31910091 910144 910197 910251 910304 910358 910411 910464 910518! 0571 53 4 0624 0G78 0731 0784 0838 0891 0944 0998 1051 1104 53 6 1158 1211 1264 1317 1371 1424 1477 1530 1584 1637 53 G 1690 1743 1797 1850 1903 1956 2009 2063 2116 2169 53 7 2222 2275 2328 2381 2435 2488 2541 2594 2647 2700 53 8 2753 2806 2859 2913 2966 3019 3072 3125 3178 3231 53 9 3284 3337 3390 3443 3496 3549 3002 3655 3708 3761 53 N. 1 1 2 1 3 1 4 1 5 6 1 7 1 8 1 '9 1 D. 358 LOGARITHMS N.l 1 1 1 2 3 4 1 5 1 6 7 8 1 9 |D.| 820 913814 913867 913920 913973 914026 914079 914132 914184 914237 914290 53 1 4343 4396 4449 4502 4555 4608 4660 4713 4766 4819 53 2 4872 4925 4977 5030 6083 5136 6189 5241 5294 6347 53 3 5400 5453 5505 6558 5611 6664 5716 6769 6822 6875 53 4 6927 5980 6033 6085 6138 6191 6243 6296 6349 6401 53 5 6454 6507 6559 6612 6664 6717 6770 6822 6875 6927 53 6 6980 7033 7085 7138 7190 7243 7295 7348 7400 7453 53 7 7506 7558 7611 7663 7716 7768 7820 7873 7925 7978 52 8 8030 8083 8135 8188 8240 8293 8345 8397 8450 8502 62 9 8555 8607 8659 8712 8764 8816 8869 8921 8973 9026 52 830 919078 919130 919183 919235 919287 919340 919392 919444 919496 919549 52 1 9G01 9653 9706 9758 9810 9862 9914 9967 920019 920071 52 2 920123 920176 920228 920280 920332 920384 920436 920489 0541 0593 52 3 0G45 0697 0749 0801 0853 0906 0958 1010 1062 1114 52 4 1166 1218 1270 1322 1374 1426 1478 1530 1582 1634 52 5 1686 1738 1790 1842 1894 1946 1998 2050 2102 2154 52 G 2206 2258 2310 2362 2414 2466 2518 2570 2622 2674 52 7 2725 2777 2829 2881 2933 2985 3037 3089 3140 3192 52 8 3244 3296 3348 3399 3451 3503 3555 3607 3658 3710 62 9 37G2 3814 3865 3917 3969 4021 4072 4124 4176 4228 52 840 924279 924331 924383 924434 924486 924538 924589 924641 924693 924744 52 1 4796 4848 4899 4951 5003 6054 6106 5157 5209 5261 62 2 5312 5364 5415 5467 5518 6570 6621 5673 5725 6776 62 3 5828 5879 6931 5982 6034 6085 6137 6188 6240 6291 51 4 6342 6394 6445 6497 6548 6600 6651 6702 6754 6805 51 5 6857 6908 6959 7011 7062 7114 7165 7216 7268 7319 51 6 7370 7422 7473 7524 7576 7627 7678 7730 7781 7832 61 7 7883 7935 7986 8037 8088 8140 8191 8242 8293 8345 51 8 8396 8447 8498 8549 8601 8662 8703 8754 8805 8857 51 9 8908 8959 9010 9061 9112 9163 9215 9266 9317 9368i5l| 850 929419 929470 929521 929572 929623 929674 929725 929776 929827 929879 51 1 9930 9981 930032 930083 930134 930185 930236 930287 930338 930389 51 2 930440 930491 0542 0592 0643 0694 0745 0796 0847 0898 51 3 0949 1000 1051 1102 1153 1204 1254 1305 1356 1407 51 4 1458 1509 1560 1610 1661 1712 1763 1814 1865 1915 51 5 1966 2017 2068 2118 2169 2220 2271 2322 2372 2423 51 6 2474 2524 2575 2626 2677 2727 2778 2829 2879 2930 51 7 2981 3031 3082 3133 3183 3234 3285 3335 3386 3437 51 8 3487 3538 3589 3639 3690 3740 3791 3841 3892 3943 51 9 3993 4044 4094 4145 4195 4246 4296 4347 4397 4448 51 860 934498 934549 934599 934650 934700 934751 934801 934852 934902 934953 60 1 5003 5054 6104 6154 6205 5255 6306 6356 6406 5457 50 2 5507 5558 5608 5658 5709 6759 5809 6860 6910 6960 60 3 6011 G061 6111 6162 6212 6262 6313 6363 6413 6463 50 4 6514 G564 G614 6665 6715 6765 6816 6865 6916 6966 50 5 7016 7066 7117 7167 7217 7267 7317 7367 7418 7468 50 G 7518 7568 7618 7668 7718 7769 7819 7869 7919 7969 50 7 8019 8069 8119 8169 8219 8269 8320 8370 8420 8470 60 8 8520 8570 8620 8670 8720 8770 8820 8870 8920 8970 60 9 9020 9070 9120 9170 9220 9270 9320 9369 9419 9469 50 870 939519 969569 939619 939669 939719 939769 939819 939869 939918 939968 60 1 940018 940068 940118 940168 940218 940267 940317 940367 940417 940467 50 2 0516 056G 0616 0666 0716 0765 0816 0865 0915 0964 50 3 1014 1064 1114 1163 1213 1263 1313 1362 1412 1462 50 4 1511 1561 1611 1G60 1710 1760 1809 1859 1909 1958 50 5 2008 2058 2107 2157 2207 2256 2306 2355 2405 2455 50 6 2504 2554 2603 2653 2702 2752 2801 2851 2901 2950 50 7 3000 3049 3099 3148 3198 3247 3297 3346 3396 3446 49 8 3445 3544 3593 3643 3692 3742 3791 3841 3890 3939 49 9 3989 4038 4088 4137 4186 423G 4285 4335 4384 4433 49 N. 1 1 2 3 4 1 6 6 7 8 9 D. OF NUMBERS. 359 N.| i 2 3 4 1 .-, .; 1 7 1 b 9 ,I».| am 944483 944532 944581 944631 9446«0 944729 944779 J44b2>5 9448771 ^44927 49 1 4976 6025 5074 5124 5173 5222 5272 5321 5370 5419 49 2 5469 5518 5567 5616 5665 5715 5764 5813 5862 5912 49 3 5961 6010 6059 6108 6157 6207 6250 6305 6354 6403 49 4 6452 6501 6551 6600 6649 6698 6747 6796 6845 6894 49 5 6943 0992 7041 7090 7140 7189 7238 7287 7336 7385 49 6 7434 7483 7532 7581 7630 7679 7728 7777 7826 7875 49 7 7924 7973 8022 8070 8119 8168 8217 8266 8315 8364 49 8 8413 8462 8511 8560 8609 8657 8706 8755 8804 8853 49 9 8902 8951 8999 9048 9097 9146 9195 9244 9292 9341 49 l9 890 949390 949439 949488 949536 949585 949634 949683 949731 9497801949829 1 9878 9926 9975 950024 950073, 950121 950170 950219 950267 950310 49 2 950365 950414 950462 0511 0560 0608 0657 0706 0754 0803 49 3 0851 0900 0949 0997 1046 1095 1143 1192 1240 1289 49 4 1338 1386 1435 1483 1532 1580 1629 1677 1726 1775 49 5 1823 1872 1920 1969 2017 2066 2114 2163 2211 2260 48 6 2308 2356 2405 2453 2502 2550 2599 2647 2696 2744 48 7 2792 2841 2889 2938 2986 3034 3083 3131 3180 3228 48 8 3276 3325 3373 3421 3470 3518 3566 3615 3663 3711 48 9 3760 3808 3856 3905 3953 4001 4049 4098 4146 4194 48 900 954243 954291 954339 954387 954435 954484 954532 9545801954628 954677 48 1 4725 4773 4821 4869 4918 4966 5014 5062 5110 5158 48 2 5207 5255 6303 5351 5399 5447 5495 5543 5592 5640 48 3 5688 5736 6784 5832 6880 6928 6976 6024 6072 6120 48 4 6168 6216 6265 6313 6361 6409 6457 6505 6553 6601 48 6 6649 6697 6745 6793 6840 6888 6936 6984 7032 7080 48 6 7128 7176 7224 7272 7320 7368 7416 7464 7512 7559 48 7 7607 7655 7703 7751 7799 7847 7894 7942 7990 8038 48 8 8086 8134 8181 8229 8277 8325 8373 8421 8468 8510 48 9 8564 8612 8659 8707 8755 8803 8850 8898 8946 8994 48 910 959041 959089 959137 959185 959232 959280 959328 959375 959423 959471 48 1 9518 9566 9614 9661 9709 9757 9804 9852 9900 9947 48 2 9995 960042 960090 960138 960185 960233 960281 960328 960376 960423 48 3 960471 0518 0566 0613 0661 0709 0756 0804 0851 0899 48 4 0946 0994 1041 1089 1136 1184 1231 1279 1326 1374 48 5 1421 1469 1516 1563 1611 1658 1706 1753 1801 1848 47 6 1895 1943 1990 2038 2085 2132 2180 2227 2275 2322 47 7 2369 2417 2464 2511 2559 2606 2653 2701 2748 2795 47 8 2843 2890 2937 2985 3032 3079 3126 3174 3221 3268 47 d' 3316 3363 3410 3457 3504 3552 3599 3646 3693 3741 47 9201963788 963835 963882 963929 963977 964024 964071 964118 964165 964212 47 1 4260 4307 4354 4401 444S 4495 4542 4590 4637 4684 47 2 4731 4778 4825 4872 4919 4966 5013 5061 5108 5155 47 3 5202 5249 5296 5343 5390 5437 5484 6531 5578 5625 47 4 5672 5719 5766 5813 6860 5907 5954 6001 6048 6095 47 5 0142 6189 6236 6283 6329 6376 6423 6470 6517 6564 47 6 6011 6658 6705 6752 6799 6845 6892 6939 6986 7033 47 7 7080 7127 7173 7220 7267 7314 7361 7408 7454 7501 47 8 7548 7595 7642 7688 7735 7782 7829 7875 7922 7969 47 9 8016 8062 8109 8156 8203 8249 8296 8343 8390 8436 47 930i968483 968530 968576 968623 968670 968716 968763 968810 968856i968903 47 1 8950 8996 9043 9090 9136 9183 9229 9276 9323 9369 47 2 9416 9463 9509 9556 9602 9649 -9695 9742 9789 9835 47 3 9882 9928 9975 970021 970068 970114 970161 970207 970254 970300 47 4 970347 970393 970440 0486 0533 0579 0626 0672 0719 0765 46 5 0812 0858 0904 0951 0997 1044 1090 1137 1183 1229 46 C 1276 1322 1369 1415 1461 1508; 1554 1601 1647 1693 46 7 1740 1786 1832 1879 1925 1971: 2018 2064 2110 2157 46 8 2203 2249 2295 2342 2388 2434 2481 2527 2573 2619 46 9 2666 2712 2758 2804 2851 2897 2943 2989 3035 30S2| 46 N.| 1 1 2 1 3 1 4 II 5 6 1 7 1 8 9 1 D. 860 LOGARITHMS OF NUMBERS. N.| ' 1 1 2 1 3 1 4 1 5 6 7 8 9 "HoTI diOdTdVlS 973174 97322U 973266 J73313 973359 973405 973451 973497 973543 46 1 3590 3636 3682 3728 3774 3820 3866 3913 3959 4005 46 2 4051 4097 4143 4189 4235 4281 4327 4374 4420 4466 46 3 4512 4558 4604 4650 4696 4742 4788 4834 4880 4926 46 4 4972 5018 5064 5110 5156 5202 5248 5294 5340 5386 46 5 5432 5478 5524 5570 5616 5662 5707 5753 5799 5845 46 6 5891 5937 5983 6029 6075 6121 6167 6212 6258 6304 46 7 6350 6396 6442 6488 6533 6579 6625 6671 6717 6763 46 8 6808 6854 6900 6946 6992 7037 7083 7129 7175 7220 46 9 7266 7312 7358 7403 7449 7495 7541 7586 7632 7678 46 950 977724 977769 977815 977861 977906 977952 977998 978043 978089 978135 46 1 8181 8226 8272 8317 8363 8409 8454 8500 8546 8591 46 2 8637 8683 8728 8774 8819 8865 8911 8950 9002 9047 46 3 9093 9138 9184 9230 9275 9-321 9366 9412 9457 9503 46 4 9548 9594 9639 9685 9730 9776 9821 9867 9912 9958 46 5 980003 980049 980094 980140 980185 980231 980276 980322 980367 980412 45 G 0458 0503 0549 0594 0640 0685 0730 0776 0821 0867 45 7 0912 0957 1003 1048 1093 1139 1184 1229 1275 132f> 45 8 1366 1411 1456 1501 1547 1592 1637 1683 1728 1773 45 9 1819 1864 1909 1954 2000 2045 2090 2135 2181 2226 45 960 982271 982316 982362 982407 982452 982497 982543 982588|982633 982678 45 1 2723 2769 2814 2859 2904 2949 2994 3040 3085 3130 45 2 3175 3220 3265 3310 3356 3401 3446 3491 3536 3581 45 3 3626 3671 3716 3762 3807 3852 3897 3942 3987 4032 45 4 4077 4122 4167 4212 4257 4302 4347 4392 4437 4482 45 5 4527 4572 4617 4662 4707 4752 4797 4842 ,4887 4932 45 6 4977 5022 5067 5112 5157 5202 5247 5292 5337 5382 45 7 5426 5471 5516 5561 5606 5651 5696 5741 5786 5830 45 8 5875 6920 5965 6010 6055 6100 6144 6189 6234 6279 45 1 ^ 6324 6369 6413 6458 6503 65481 6593 6637 6682 6727 45 970 986772 9868171986861 986906 986951 986996 987040 987085 987130 987175 45 1 7219 7264 7309 7353 7398 7443 7488 7532 7577 7622 45 2 7666 7711 7756 7800 7845 7890 7934 7979 8024 8068 45 3 8113 8157 8202 8247 8291 8336 8381 8425 8470 8514 45 4 8559 8604 8648 8093 8737 8782 8826 8871 8916 8960 45 5 9005 9049 9094 9138 9183 9227 9272 9316 9361 9405 45 6 9450 9494 9539 9583 9628 9672 9717 9761 9806 9850 44 7 9895 9939 9983 990028 990072 990117 990161 990206 990250 990294 44 8 990339 990383 990428 0472 0516 0561 0605 0650 0694 0738 44 9 0783 0827| 0871 0916 0960 1004 1049 1093 1137 1182 44 980 1 2 991226 991270,391315 991359 991403 991448 991492 991536 991580 991625 44 3 1669 1713 1758 1802 1846 1890 1935 1979 2023 2007 44] 2111 2156 2200 2244 ' 2288 2333 2377 2421 2465 2509 44 3 2554 2598 2642 2686 2730 2774 2819 2863 2907 2951 44 4 2995 3039 3083 3127 3172 3216 3260 3304 3348 3392 44 5 3436 3480 3524 3568 3613 3657 3701 3745 3789 3833 44 6 3877 3921 3965 4009 4053 4097 4141 4185 4229 4273 44 7 4317 4361 4405 4449 4493 4537 4581 4625 4669 4713 44 8 4757 4801 4845 4889 4933 4977 5021 5065 5108 5152 44 9 5196 5240 5284 5328 5372 5416 5460 5504 5547 5591 44 990 995635 995679 995723 995767 995811 995854 995898 995942 995986 996030 44 1 6074 6117 6161 6205 6249 6293 6337 6380 6424 6468 44 2 6512 6555 6599 6643 6687 6731 6774 6818 6862 6906 44 3 C949 6993 7037 7080 7124 7168 7212 7255 7299 7343 44 4 7386 7430 7474 7517 7561 7605 7648 7692 7736 7779 44 6 7823 7867 7910 7954 7998 8041 8085 8129 8172 8216 44 6 8259 8303 8347 8390 8434 8477 8521 8564 8608 8652 44 7 8695 8739 8782 8826 8869 8913 8956 9000 9043 9087 44 8 9131 9174 9218 9261 9305 9348 9392 9435 9479 9522 44 9 9565 9C09i 9652 9696 9739 9783 9826 9870 9913 9957 43 N. „^> LJ_L.2-. 3 1 4 1 ^ 6 7 ^ 8 1 9 ID.j 7 -\ K-^U Oc 2- - / 1 1 ai ^ ^^-"z ^7 "!)■ \.j/ V^ / V— / I H5 77022 1853 Educ • Lib.