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Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/completealgebrafOOolnerich OLNEY'S NEW SERIES THE COMPLETE ALGEBRA. FOR HIGH SCHOOLS, PREPARATORY SCHOOLS, AND ACADEMIES. BT EDVV^ARD OLNEY, Professor of Mathematics in the University of Michigan, NEW EDITIOIT NEW YORK SHELDON & COMPANY, 8 MURRAY STREET. OLNEY'S NEW SERIES EMBRACES THE FOLLOWING BOOKS. FIRST LESSONS IN ARITHMETIC, | „ „ , „ . PRACTICAL ARITHMETIC. [ Tvv,.-Book Series. This Series has more examples, and at less price, than any ever published. SCIENCE OF ARITHMETIC. (A strictly High School Text-Book^ Send for full Circular of Olney's Arithmetics. FIRST PRINCIPLES OF ALGEBRA. COMPLETE ALGEBRA. (Newly electrotyped in large type.) NEW ELEMENTARY GEOMETRY. Copyright, 1870, 1875, 1878, 1881, by Sheldon <&* Co, Smith & McDougal, Ei.ectrotypbbs, 82 BeekmuD iSt., N. V. ^^^ ^i^^^l^ 'i'^ J ^l^ t ^i^ This book is a new edition of the author's Complete Algebra. The changes made in the original work are not sufficient to preclude the use of the old edition in classes with the new. The matter of the new edition runs article for article, and, with very inconsiderable differences, page for page, with the old. The changes made consist in the breaking up of the longer definitions and rules into short paragraphs, for the convenience of teacher and pupil in the class-room, the simplification of a few of the more abstruse demonstrations, the omission of quite a large number of ** suggestions " to teacher and to pupil, and the substitution of simpler problems for a few that teachers generally have thought too difficult. These changes, together with the larger and more open page, and the more beautiful typog- raphy, it is thought will render the book still more accept- able to teachers than it has hitherto been. In effecting this revision the author has had the aid of several accomplished teachers who have been familiar with the practical workings of the book in the class-room. Of these. Prof. C. H. Churchill, of Oberlin College, has read the entire book and offered suggestions freely. With refer- ence to the changes made in the problems, the author has 543474 IV PBEFACE. been guided by the opinion of a large number of practical teachers who have used the book for years. The sanie thoroughness in the discussion of principles, comprehensiveness, philosophic accuracy, and clearness of statement, and the same careful adaptation to training the pupil to think clearly and express his thoughts with pro- priety, characterize this edition as won for the first its great popularity. This book is, therefore, not a mere child's book, but is designed for pupils who have the knowledge of Arithmetic usually considered requisite for a commence- ment of this study. For less mature pupils the author has prepared another volume. This treatise is designed to lay the foundation of a good mathematical education. Algebra develops the mathemat- ical language, and is the great mathematical instrument. The Literal Arithmetic (see pages 1-210) is the basis of all mathematical training. Without a good measure of ability to handle the various forms of literal expressions — the more complex as well as the more simple — it is impossible to become a mathematician. Hence an unusual amount of care is bestowed upon this part of the work. A mere elementary knowledge of the simpler forms of numerical equations, is indeed easily obtained, but it is of very little worth. A thorough mastery of the elements of Algebra as presented in this volume is one of the most valuable acquisitions which can be secured in the schools, whether we consider its value as a mental discipline, or as a founda- tion for more advanced work. The attempt to train the pupil to methods of reasoning, rather than in mere methods of operating, has given char- PREFACE. T acter to the presentation of every topic. Propositions are clearly stated at the outset, and demonstrations are given in form, and with the rigor of a geometrical argument. That there is some defect in our methods of instruction, in this regard, must be painfully evident to every one who has been called to examine large numbers of our youth in this study. The author has examined for admission to college, from 25 to 150 different students from all parts of our coun- try, each year, for the last 27 years, and he has almost invariably found little or no knowledge of the processes as arguments, even when a good degree of skill in the use of the processes had been attained. Perhaps a majority of those examined could multiply the square root of 2 by the cube root of 3, but scarce one in 50 could develop the pro- cess in a logical form, or, in most cases, give any rational account of it. Now, it need not be said that, in a course of education, this is a fundamental defect; it is failure just where success is vital. The processes of a mathematical science are of comparatively little worth to a great majority of those who study them ; the development of the reasoning powers to which such studies are addressed, is of the high- est importance to all. By teachers who cannot appreciate these truths, this book will very probably be misunderstood ; but to such as do feel the force of them, the author appeals with the fullest confidence, not indeed that his book will meet the exigency, but that it will be welcomed as an effort in the right direction, and as a help in remedying this radical defect. The Introductory portion is found by many an adequate discipline for the younger and more immature pupils before VI PREFACE. entering upon the treatise proper, while the Appendix makes the volume sufficiently comprehensive for many of our colleges. Grateful for the favor with which the original work, as well as all he has written, has been received ; the author has been stimulated to perfect, as far as possible, the different members of the Series, with the hope of rendering them still more useful. EDWARD OLNEY. University op Michigan, Ann Arbor, July, 1881. AN INTRODUCTION TO ALGEBRA .... 1-35 INTRODUCTION TO THIS BOOK 1-7 Section l A BRIEF SURVEY OF THE OBJECTS OF PURE MATHEMATICS AND OF THE SEVERAL BRANCHES 1-4 Section n. LOGICO-MATHEMATICAL TERMS 5-7 *(0> PART I. LITERAL ARITHMETIC. Chap tei^ I. FUNDAMENTAL RULES. Section i. NOTATION 8-27 Section ii. ADDITION 38-40 vitt CONTENTS. ECTION in. SUBTRACTION 40-49 Section iv. MULTIPLICATION 50-62 SECTfON V. DIVISION 63-76 FACTORING. Section i. FUNDAMENTAL PROPOSITIONS 77-88 Section n. GREATEST OR HIGHEST COMMON DIVISOR . 88-98 Section in. LOWEST OR LEAST COMMON MULTIPLE . 98-101 Chaptei| iil FRACTIONS. Section i. DEFINITIONS AND FUNDAMENTAL PRINCI- PLES 102-106 CONTENTS. IX ECTJON II. REDUCTIONS 107-117 Section in. ADDITION 117-121 Section iv. SUBTRACTION 121-124 Section v. MULTIPLICATION 124-129 Section vi. DIVISION 13C-127 POWERS AND ROOTS. Section i. INVOLUTION 138-153 Section ii. EVOLUTION 154-177 Section hi. CALCULUS OF RADICALS 178-192 Section iv. COMBINATIONS OF RADICALS 192-210 CONTENTS PART II. ALGEBRA SIMPLE EQUATIONS. Section i. EQUATIONS WITH ONE UNKNOWN QUANTITY 211-252 Section n. SIMPLE, SIMULTANEOUS, INDEPENDENT EQUA- . TIONS WITH TWO UNKNOWN QUANTITIES 253-271 ECTfON in. SIMPLE, SIMULTANEOUS. INDEPENDENT EQUA- TIONS WITH MORE THAN TWO UNKNOWN QUANTITIES 272-280 ChAPT1I| II. RATIO, PROPORTION, AND PROGRESSION. Section i. RATIO . . 281-285 Section h. PROPORTION 285-293 Section m. PROGRESSIONS 294-313 CONTENTS. XI BUSINESS RULES [OF ARITHMETIC] ECTION I. PERCENTAGE 314-31G Section n. SIMPLE INTEREST AND COMMON DISCOUNT 317-326 ECTION III. PARTNERSHIP 327-330 ECTION IV. ALLIGATION 330-334 <^HAP TEI\ lY. QUADRATIC EQUATIONS. ECTION I. PURE QUADRATICS 335-344 Section ii. AFFECTED QUADRATICS 344-356 Section hi. EQUATIONS OF OTHER DEGREES WHICH MAY BE SOLVED AS QUADRATICS .... 357-366 xii CONTENTS. Section iv. SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE BETWEEN TWO UNKNOWN QUAN- TITIES 367-380 €haptei| ¥. LOGARITHMS 381-390 M/>tfj«, with a f/ootl knowledf/e of Arithmetic, will be able to commence at page 1 of the body of the work. 2 INTRODUCTIOl^. i/ irt t^gebreo we often use letters to represent, or stand for, numbers. The following exercises will show how: 6. Two times a certain niimber -{-'^ ivaiQ^ ihQ same nii7n- ber 4- 4 times the same number, are how many ; i. e., how many times that number ? 7. Let a stand for a certain number. How many are 4 times a-fS times « + 2 times a ? Ans., 9 times a. 8. If we let 771 represent the number of needles in 1 paper of needles, how many needles are there in 5 papers -j- 3 papers + 1 paper, or how many times m are 5m's, 3m's, and Im? If m represents 25, how many are ^m ? 2, Thus we may use any letter to represent any number, ojily so that it always means the same number in the same exercise. 3, When a letter is used to represent a number, the figure which tells how many times the jvumber repre- sented by the letter is taken, is just written before the letter, the woi^d ''times " being left out. Illustration. Ba means 3 times a, 45 means 4 times 5, 7m means 7 times m, 105a; means 105 times the number represented by a*, what- ever that number may be. 4, A Coefficient is a number placed before a. letter to tell how vvany times the letter is taken. A coefficient is therefore a factor. Thus in 5«, 5 and a are two factors of the number represented by 5a. The letters are called Literal factors. The figures are called the Numerical factors. Thus in 5a, 5 is the numerical factor. Letters may be used as coefficients as well as figures. If no figure stand before a letter, the letter is taken once, or its coefficient is said to be 1. Thus, m means one time w, and its coefficient is 1. 9. Mention the numerical coefficients and the literal fac- tors in the following expressions : HOW LETTERS REPRESENT NUMBERS. 3 (1.) 6a-\-5b + 3fu — 4fi. (3.) a—b—VZc. (2.) 3« — 15m + 22m— 4^ + ^— ^. (4.) d4-2c— m-|-4w— «. Queries. — Of what is 12 the coefl5cient in 12aj? Of what is x the literal factor? Of what is 13 the numerical factor? 10. What is the sum of 10a-{-ba-[-7a-\-2a^* If a stand for 13 what is the sum ? A?is., 10a -\- 6a -\- 7 a -[-2a = 24« = 24x13 = 312. Query. — If the a in 10a meant one number, in the 5a another number, the a's in la and 2a still other nuinlK-rs, could you answer this exercise in the sanu: way? You could not answer it at all. The a must mean the same number all the time, in the same example. 11. 3a + 2a4-5rt + 8rt, are how many times «? Arts., 18^. How much is this if a is 6 ? Ans*, 108. How much if a is 11? Ans.,1^%. 12. Eleven times 8, minus 5 times 8, are how many- times 8? 13. 11a— 5a are how many times a? Eleven times any number, minus 5 times the same number, are how many times that number ? 14. \2x—lx = how many times .r? How much is this if X represents 3? II x represents 2\? Ans. to the last, 11. 15. 6b -\- 4b -{■ 10b — 12b = how many times b ? How much is 6b-\-4.b + 10b—12b, if b = 3f ? 16. How much is 3?/i + 8m— 47W + 67?i— 5?w— 2m? How much is this if /w = -f ? Solution. Sm + Sfn are 11m, 11m— 4m are 7m, 7m4-6m are 1dm, 13w_57w are 8;w, Sm—2m are 6m. Hence the answer is 6m. If 7« = I, 6m = 6 X I = 4. ♦ The pupil is presumed to be acqnaioted with the use of the signs. They are explained in the body of the full treat i(>c', dimply as a part of the science. 4 INTRODUCTION. 5, In such examples as the last you can add together all the quantities with the + sign into one sum, making in this case 17w, and all those with the — sign into another, making in this case 11m, and then subtract. Thus, 17m —11m is 6m, the same result as before. This is, generally, the better way to solve such examples. 17. How much is 10a:— 4a;— 2:r-f 3a;— 8a;-|- 11a;? How much is it if a; is 3 ? 18. How much is 10a:— ISa* ? Suggestions. — Of course we cannot take 15a; out of 10a:. But we can take lOaj of the 15a; from the first lOu, and there will then remain 5x of the 15a;, which cannot be taken out of the 10a;. We indicate this by writing it —5x. This means that oa* was to be subtracted, but that we had nothing to take it from. 19. How much is 3a; — 6x — 2a:? Ans., — 4a:. Query. — What does this answer mean? 20. How much is 12a-{-3a—5a—20a? How much is this if a = ^. Query. — What does the answer mean? 21. How much is 2 times 3 times a certain number, as 5 ? A?is., 6 times the number. 22. How much is 5 times 7m ? that is, 5 times 7 times a number which we will represent by m ? Ans., 35 times m, or 35m. 23. How much is 6 times Sa? 7 times 3a ? 10 times lb ? 9 times 8y ? 9 times 8 times a number are how many times that number ? 24. How much is 3 times 7m, and 4 times 8m, minus 2 times 22m, if m represents 6 ? Ans., 9m, or 54. HOW LETTEE8 REPRESEin: NUMBERS. 5 25. What is 10a divided by 2 ; that is, what is J of 10a ? 272; divided by 9 is how much ? Ans. to the last, 3x, 26. How many times a number is 10 times that number, divided by 2 ; that is, i of 10 times a number ? 27. How much is J of 48a: ? 25a; divided by 5 ? ^ of 11a;? 11a; divided by 11 ? 7x divided by 7 ? 28. Divide 10a; by 5, then add 3a;, then multiply by 2, then subtract 4a;, then divide by 3 ? What is the result ? . M€^M>M COMBINATIONS OF LETTERS. 6*. When tiuo lettei's representing numbers are wriir- ten side by side, their product is indicated. Thus, ab means the product of the two numbers repre- sented by a and b, or a x b, Sabc means 3 times the product of the numbers represented by «, b, and c, ov Sxaxbxc. [Note. Instead of saying, as above, the number represented by a, we usually simply say " the number a," or, " «," without using the word number at all. Thus we say 3 times the product of a, 6, and c] 7, If we want to represent the product of a number represented by a letter, as a, by itself a certain numJber of tijnes, instead of uTiting aa, or aaa, etc., as we might, we write €^, a^, etc. Thus ¥ means the same as bbbb. d^ is read "a square ; '' a\ " a cube ;" ¥, " b fourth power ;" a^, " x fifth power ;" etc. [Do not read &* "h fourth,''^ but "& fourth power." You will hereafter learn that 6'" is *'5 fourth," and is quite a different thing from b*.] • IinDKODUCTIOK. The little figure placed at the right and a little ahove the letter is one form of what is called an Exponent. Thus in m"^, 2 is the exponent of m, and m" means m times m. In 6aj', 3 is the exponent of x^ and 6x^ means Qxxxxxx. [An exponent does not always indicate a product. It does this only when it is a whole number. Thus a~^ does not mean aaa ; and in a% the expcment f has quite a different meaning from the 3, or —3, above. These other meanings of exponents will be ex- plained later in the work.] Ex. 1. How much is 4a^h, ii a — 2, and b = 5? How much is Sa^ly^x, if a = 3, b = 2, x = S? How much is a^by, it a = 2, b = 1, tj = S? 2. How much is aW + 2ay — by, if a = 4, b = 3, y = 2 ? How much is 3a'^by^—2ay'^-\-5b, the letters having the same values as before. How much is 5by—2ai^-{-4:a^y^ —2a? 3. How many times ab^ is 4:ab^ -\- 2ab^ — 3ab^ ? How many times a^y is 10a^y-\-4:a^y—6a^y—a^y ? 4. How many times amy is 4 times 3amy ? How much* is 6 times 2amy? 4 times 'ila^b^c^? 10 times 5. How many times ax is -J of 20ax? | of S5ax? W2ax divided by 3 ? How much is J of 72a^a^? 125^5^2 divided by 25 ? 18ny^ divided by 9 ? S. We have learned in arithmetic that representing numbers by the figures 1, 2^ 3, 4, etc., is called Arabic or Decimal Notation. In like manner, representing numbers by the small letters of the alphabet, as a, b, c, d, x, y, etc., is called Literal Notation. * This means the same as the preceding question. COMBINATIONS OF NUMBERS. 7 The pupil will see that this Literal Notation is altogether a diflferent thing from the Roman Notation, in which the seven capital letters, I, V, X, L, C, D, M, are used. Because the Literal Notfition is so much used in Algebra, it is often called the Algebraic Notation. But this notation is just as much used in some other branches of Mathematics, as in Algebra. f>, A Term is a group of letters combined together by multiplication, or division, or both. In the expression 5a'a; — 3a + r- , oa'j;, 3a, — , and -^, are terms. A Monomial is an expression consisting of but one term. A Binomial is an expression consisting of two terms connected by the sign +, or — . A Trinomial lias three terms. A Polynomial is an expression with more than one term. 6. Point out the monomials, binomials, trinomials, and polynomials in the following : 2ax—^b^, 6xy—0cd-\-a—2yy dahnHy, (?—d}, a + m, a + b + c—d, 226a^t^c^d^, abcd,a^b, ab, c-x^y + ax, x^+y'^, \W + ^xy, Q?-y\ Sahnxy,a^-5x+d, 10, Similar Terms are terms which have the same literal factors affected with the same ei^onents. 7. Point out the similar terms among the following: 3ax% 2ax, -5^2^, ax, lUx\ 16cy\ -12(^y% Sa^x^, -5cy% Qcy^ lOax^, c^y^ Sa^x, dax^, 61a^x. 11. Positive Terms are terms which have the sign -|-. Negative Terms are terms which have the sign — . When no sign is expressed, 4- is understood. nrrRODucTioiir. HOV\/ NUMBERS ARE ADDED IN THE LITERAL NOTATION. 12. Rule. — I. Write the expressions so that similar terms shall stand in the same column. II. Comhine the teriyis in each column and tvrite the result underneath with its own sign. The polynomial thus found is the sum. Ex. 1. Add 5ax—2cy, 3ax-{-^cy, cy—2ax, —^ax—dcy, —ax + 6cy, and 2ax-{-2cy. Operation. — Having written the numbers so 5ax — 2cy that similar terms fall in the same column, we 3^^ 1 ^^y may begin to add with any column we choose. 9^1. * Adding the right-hand column we find it makes ' ^ + 7cy, and write this sum underneath the column ^^^ '^^^ added. In like manner the other column makes — ax -{- 6cy Sax (or +dax), which, as it stands first, we write 2ax ~\- 2cy without any sign, as 4- will then be understood. o ^ , w~ The sum is Sax + ley. (3.) 10am — Sdy^ + 2«% — 6am + 4:dy^ — lOa^x ^a?n — 8dy^ — 2d^x lam — 13dy^ + Ga^x — 9am -f 2dy^ — ^urcr am + 18dy^ — a^x ^am — IWx bed + 2a + bxy ♦ W^exx no sign is expressed before a term, + is understood. (2.) bed — 2a + ^xy 2cd + 3a — J>xy 8a — 2xy Qcd + Uxy — 3a — 7xy + 11a 4- xy Acd — 15a StJBTRACTlON IN THE LITERAL NOTATIOK. 9 4. Add 5x—'6a-\-b-^7 and —4:a—dx-\-2b—d. Sum, 2x—7a-^3b—2. 5. Add2a + 3Z>— 4c— 9 and5a— 3^>H-2c— 10. 6. Add Sa-^2b—5, a + bb—c, and 6fl— 2c + 3. 7. Add 6a;?/— 12a^, —4:X^ + Sxy, Lx^—'Zxy, and — 3:^?/ -f-4a;l aS'^^w, ^xy—^y^. 8. Add 3«2 + 4^c— e2-|-l0, _5a3_|-6*c + 2e2— 15, and _4a2_9^c_10^>2_^21. 9. Add 5a;^«/— 3ca;, ^7?y, Ibex, Idcy—ab,^ llab—12cx —^x^y, and — 10«J— 13c^. 10. Add 'lax-^-^cy — bax—cy-\-3cy—ax-\-llax—by-\-3c. 11. Add —3coi^-\-bay, ax—Sy, and b—dx^ ECTION IVo HOW NUMBERS ARE SUBTRACTED IN THE LITERAL NOTATION. IS, Rule. — Change the signs of the terms in the subtrahend, or conceive them to he changed, and add the result to the minuend. Ex. 1. From bby—^a^-\-Zofi subtract 2by-\-Za^+3^. Operation. bby — W-\-30l^ Minuend. — 2by — Zcfi — X^ Subtrahend with signs changed. 17 fv 2 I o^ i "^^ Remainder pought, which is the snm of the mintw oOy — vCl^ -f- /C.'^ \ end and the subtrahend with its signs changed. 2. From 3ax + bl/^y^—2m^ take Sax—bY—^f^^- Bern., 6^/+wi«. 10 INTBODUCnON. 3. From 10a—'db-j-2c—a^ subtract J--5c + a^. Rem., 10a-4:b-\'7c'-2a^. 4. From 12xy—3c^ + ab take Qxy-\-c'^—2ab. 6. From bc^y—3ab take mx—2c^y. Rem., Hc^y—Sab—mx, 6. From x^— 11 xyz-i- 3a take —6xyz-\-'7—2a—bxyz. Rem., x^-\-5a—7. 14, When there is a term in the subtrahend which has no similar term in the minuend, we see that this term appears in the remainder with its sign changed. 7. From 6a(^—2by^ + 4:X—3cy take —2aci-\-3b^y-dx —3cy-k-m. Rem., Sa(P—2by^—3b^y-\-7x—m. 8. From Sm^x^—3a—4:b take a^—b^. Re7n., 8m8ir3— 3«— 4Z>— a2-i-52. 9. From a^-\-2ab + b^ take a^—2ab-\'W. 10. From aH ^ take a^—1^. 11. From 3cm — y take 2b— 3c, 12. From x^^2xy^l take ^y. HOW NUMBERS ARE MULTIPLIED IN THE LITERAL NOTATION. 15. To multiply two Monomials together. Rule. — /. Multiply the numerical coefficients. To this pj^oduct affix the literal factors, affecting each letter iiith an exponent equal to the sum of the exponents of that letter in both terms. MULTIPLICATION IN THE LITERAL NOTATilON. 11 //. If the signs of the terms are alike, the sign of the product is + ; if unlike, — . Ex. 1. Multiply 6aa^ by da^x\ Operation. — The product of the numerical coefficients 8 and 5 is 15. To thi' we affix the letters a and x; and 5^,1 as a has an exponent 1* in the multiplicand, and 2 in the ^n2r< multiplier, we give it an exponent 3 in the product, and X an exponent 5 for a like reason. The product is there- ^^^ -^ fore 15 aV. 2. Multiply lOmn^ by S?nhi^. dxy by ix^if. Hex by Sex. 2a by a^. 3. Multiply — 5fl« by 6b. Product, —30a^. 4. Multiply — 3«2a; by —2ah?y. Product, (ja^x^. 5. Find the products of the following : —la^x by 2axi/; lOc^nix^ by —dm^x; 9a by 4^» ; —am by —xy; Sc'^d^ by —ab; —5xy by —x^y^; dax^hy —5a^; —Hxy by —10a^y\ 5abx by —ca^; ax by cy. 16, To multiply two factors together when one op both are Polynomials. Rule. — Multiply all the terms of the TnultipUcand hy each term of the multiplier, and add the pro- ducts. Ex. 1. Multiply 2a^x—Uy-\-^m by ^a^m. Operation. — It is immaterial 2a^x — ^by-\-^m where we write the multiplier, 2a^l^m but we may as well write it as in ~ ~ ^ „_„ " arithmetic. So also it makes no ia^V^'nx-Qa^lhny + ^a?l>'w? difference whether we begin at the « When no exponent is expressed, 1 is always understood. 1^ * tKTRODUCTlOK. right hand or the left to multiply. It is cu8to7uary to arrange the letters in a term alphabetically ; thus we write ^a^'V-mx instead of ^"^a^xm^ or any such foriii. There would be really no difference in the value of the terms, however, in whatever order the letters came. 2. Multiply 6ax—5a^x-i-Saa^ by 2a^x^; 2my—3cy^ by jm'c^; 4:ab—3cd-{-x hy ay ; a-\-b~c hj x\ lOa^m^x- \-Smx^ by 6«y, 3. Multiply Sa^x^—2ay^ + 6xy by a^—2xy. Operation. Sa'^x' — 2ay^ 4- 5xy a''—2xy Product by a\ 3a V — 2aY + 5a'«y Product by —2xy, — Qa^x^y + ^ax y* — lOx^y^ Sum of partial prod's., 3a V — 2aY + ^a'xy—Qa^x"y-\-4:axy* — lQx'^y^ There being no similar terms in these partial products, we can add them only by connecting them with their proper signs. 4. Multiply x^—2xy-{-y^ by 2x—3y, Operation. x^—2xy+y^ 2x—3y Product by 2a;, 2x^—4:x'^y + 2xy^ Product by —dy, — 3a-V + 6ry"— 3y° Entire product, 2x^^7x'y+Sxy^—Sy^ 5. Multiply a-{-n hy a—n. Prod., a^—n\ 6. Multiply a'^+a^ + a^ + a-r-l by a^—l. 7. Multiply a^—2ah + J^ by a^-\-2al)-\-l^. 8. Multiply m-\-7i by m-\-n. 9. Multiply m—n by m—n. 10. Multiply 4:X^—9y^ by x-{-2y. DIVISION IN THE LITERAL NOTATION. 13 11. Multiply together x—3, x—b, a:— 4, and x—1. Prod., a:4_i9a,^+i3ia45_389a;^42o. 12. Multiply x^-\-y^-\-z^—xy—xz—yz by x-\-y-^z. Prod., 7fi-\-^-\-z^—3xtjz. E€TION ¥L HOW NUMBERS ARE DIVIDED IN THE LITERAL NOTATION. 17. To divide Monomials when dividend and divisor consist of the same letter affected with exponents. Rule. — Subtract the exponent of the divisor from that of the dividend. The common letter with this difference as an exponent is the quotient. If the signs of the divisor and dividend are alike, the sign of the quotient is + / if unlike, —. Ex. 1. Divide «» by a\ QiwL, aK The student will observe that the product of the divisor and quotient must always equal the dividend. In this case, «' x a^=a\ 2. Divide —x'^ by x^. Quot., —a^. 3. Divide — m^ by — m^. Quof., m. 4. Divide J« by —b*. Quot., —P. 5. Give the quotients in the following cases : y'^-^y* ; —x^-i-x^y cfi-, — flS; — c^h — c^. 18. To divide one Monomial by another when there is no letter in the divisor which is not in the dividend, and no exponent in the divisor is greater than the exponeni of the same letter in the dividend. 14: INTEODUOTION. Rule. — Divide the numerical coefficients, and to the quotient annejc the literal factors, affecting each with an exponent equal to its exponent in the dividend minus that in the divisor, and sup- pressing all factors whose exponents are 0. If bhe dividend and divisor have liTce signs, the quo- tient will have the sign + ; if unlihe, — . Ex. 1. Divide Ibl^x^y by ^hxy. Quot., Wx, 2. Divide 21a^m^ny^ by —'la^mny. Quot.^ —Sm^y. 3. Divide — 105ay by —21aK Quot., 5ayK 4. Divide —ISmn'^x by Cjmnx. Quot., — 3w. Give the quotients in the following : 5. 12aY -^ SayK 6. — 64ay _i_ iQay, 7. Slx^y^ -^ - 2Wy\ 8. — ZbaWT? -. UV^x^. 9. 24my -, 8my. 19. General rule for dividing one Monomial by another. Rule. — Write the divisor under the dividend in tlie form of a common fraction, and then reduce this fraction to its lowest terms hy cancelling all factors common to both numerator and denom- inator. The sign of the quotient is determined as in the last case. 10. Divide l^a^a? by 4«W Operation. 18aV-7-4<*V = -t— ,-v • Now in 18 and 4 there is 4a^a:^ a, common factor 2 which can be cancelled ; the a^ in the numerator DIVISION IN THE LITERAL NOTATION. 15 will cancel two factors of a from a* in the denominator, and leave a factor a in the denominator, and in like manner a?' in the denom- inator cancels x^ from the x^ in the numerator, leaving x therein. Hence, 18aV _ 9ar ' 4rtV ~ 2a' 11. Divide 12my by Idmhj. Operation. 5m ' 12. Divide ^Sa^3?f by —Z^a^u^y'^. ^'""•' -2' 13. Divide ^^()}^a^y by — 40Z^a;*. Quo., l 14. Divide Qax by 4J)y. 5-| a^t^ + a^ by fl2— aJ + Jl 7. Divide lOac^-'dd^-i-^a^ + ^i^ + Sab + Sbc hj 2b+a + 'Sc. 8. Divide x^—y^ by ic— y. Operation. a5*y-y* xY-y' «y' a;y*— y* 9. Divide x'^—y'^ hj x-\-y. 10. Divide ^a^—b^ by 2rt— J. 11. Divide 20ax^-{-4:a^-^x^—2ba^x^ by 2«3-f-2a;8— 5fl^. 12. Show that (a^-b^)^(a^^ab + i^) = a—b. E€T50M YII. A LITTLE ABOUT FACTORING. 22. The Factors of a number are those numbers which multiplied together produce it. Thus 3 and 4 are the factors of 12, because 3x4 = 12. 18 INTRODUCTION. So a-\-b and a — b are the factors of a^—H^, because (a+d)x{a—b) = a^—l\ Try it, and see. Ex. 1. What is the product of a and l ? What are the factors of ab ? 2. What is the product of 3, x, and y ? What are the factors of ^txy ? 3. What does a^ mean? What are the factors of «8V Whatof «3^2? OihaW^ Suggestion. — Such an expression as 5a^Z>'^ may be resolved in a great variety of ways : thus 5, a, «, a^ b, and h are its factors ; also, 5, a^, and 5^ ; also, 5, a'\ a, and 6^ ; also, 5a, a'', & and 5, etc. 4. What is the product of a and x-\-y ? What are the factors of ax-\-ay? 6. What is the product of Sa and a—b? What are the factors of Sa^—Sab? Of 2a—2ab ? THE SQUARE OP THE SUM OF TWO NUMBERS. Ex. 1. What is tjie product ot a-\-b and a-\-b? What are the factors of a^ + 2ab + ^ ? 2. What is the product of x-\-y and x-\-y'^ What are the factors of x^ + 2xy + y"^ ? 3. What is the product of 1+a; and l-\-x't What are the factors of 1 + 2a; + ic^ ? 23» We see from the last examples that the square of the sum of two numbers equals the square of one of them, -\- twice the product of the two, -\- the square of the second. Thus {a-\-b)x {(t-\-b) is the square of the sum of the two numbers a and b, and is equal to a^-\-2ab + lP. This principle, and those in 24 and 25, are of great im- portance in factoring. A LITTLE ABOUT FACTORING. 19 THE SQUARE OF THE DIFFERENCE OF TWO NUMBERS. Ex. 1. What is the product of x—y and x—y ? What are the factors of x^ —"Ixy -\; fi 2. What is the product of m^n and m—n'i What are the factors of w^—'lmn-^-rfi ? 3. What is the product of \—x and \—x't What are the factors of l-'2x-\-a^? 4. What is the product of 2—x and 2—x? What are the factors of 4— 4a; + a;2 ? 24. From these examples, we see that the square of the differetice of tiuo iiumhers is equal to the square of otie of them, — tivice the product of the two, + the square of the other. Thus {x—y) x {x—y) =x-—'lxy-\-y^. THE PRODUCT OF THE SUM AND DIFFERENCE OF TWO NUMBERS. Ex. 1. What is the product of x-^y and x—y't What are the factors of x^—y^'t 2. What is the product of «4-J and a—h'- What are the factors of a^— ^ ? 3. What is the product of l + a; and \—x'> What are the factors of \—x^'i 4. What is the product of %-\-x and %—x'i What are the factors of 4— rc2V 2*^, We see, from these examples, that the product of thc: sum and difference of two numbers is equal to the difference of their squares. Thus {x + y)x {x—y) =:.7?—y\ Ex. 1. What are the factors of 2«— 25? 2. What are the factors of ^a^—^a^x ? 3. What are the factors of c^+^crf+f?^? ) INTRODUCTION. 4. What are the factors of o^ — 'lam-{-tr? ? 5. What are the factors of c^—c^ ? 6. What are the factors of a;^— 2a; + l ? 7. What are the factors of 9— ic^? 8. What are the factors of a2^2a+l ? MCTION VIIL HOW OPERATIONS IN FRACTIONS ARE PERFORMED IN THE LITERAL NOTATION. 2(y. For the various operations in fractions in the literal notation^ the ordinary rules of arithmetic for the corresponding cases apply, only, that the fundamental operations of addition, subtraction, multiplication, and dinsion are performed by the preceding rules. TO REDUCE FRACTIONS TO THEIR LOWEST TERMS. ^7. What is the rule for this operation in arith- metie ? Ex. 1. Reduce —^ — ^—^ to its lowest terms. Result, ^ 105W 2. Reduce ^ ^ • „ to its lowest terms. 3. Reduce -tt^^- ^— „„ to its lowest terms. Suggestion. — Divide numerator and denominator by 4 t to equivalent fractions having a common denominator. Also, — ^ — , , ^' x + y' x—y' 3 Results of the last, ^^^33^ , ^^^ZT^f ' ^^^ZT^f TO ADD FRACTIONS. 31. Repeat the rules of arithmetic for this purpose. 31a; Ex. 1. Add %, %, and f- Sum, 2. Add |; and ^^. Sum,'-^^ 3. Add ^^ and ^^ Sum, ^^• 4. Add ^-±\, and ^J. Sum, ^Jt^- 5. Add \-^^, and i±^. 5«.m, ^^. FRACTIONS IN THE LITERAL NOTATION. 25 TO SUBTRACT FRACTIONS. 32, What is the rule given in arithmetic for this purpose. Ex. 1. From — take -r* Rem., ^7^- rt TT_ « X 1 * ^ 2« — 35 2. From - take ^' 3. From x—\ 3 take x + 2 5 4. 5. From From 1 x—y 1-x^ take ta,ke 1 x-\-y 1+x^ 6 Qfim 2a:-ll 15 Rem. 2y Rfim -^a? l-fa:2 ^ l—x^ -""''' i_a4 MULTIPLICATION OP FRACTIONS. 33, How is a fraction viultiplied by a whof-e number ? Ex. 1. Multiply ll by Sal Prod., ^'. 2. MuUiply 3^^ by a + J. Pro^., gl|. 3. Multiply \^^ by l-o:. Pro^., 1^. 4. Multiply g^ by 3a. Prorf., ?|. 5. Multiply 3x by y- P?W., -r-* 26 INTRODUCTION. 6. Multiply -y— ^f-7T9 by a + ^. Prod,, ^ — 7. Multiply 106^2 by ^. Prod., ?5^. 8. Multiply I by 6 ; by 8 ; by 10. 9. Multiply ^ by 3 ; ^^ by 5. 34, Oive the rule for multiplying one fraction by another. Ex. 1. Multiply I by A. Prod., g. 2. Multiply J by -• Pro(7., t^- !j If 3. Multiply — — by --^— • Pro^Z., 5a ^ 2— a ' 10a— 5a2 4. Multiply — -!^- by -— • Prori., ^ 5 7 7 5. Multiply -j^ by — j- /'«rf.. -^-. DIVISION OP FKACTIONS. 33, How is a fraction divided by a whole number f Ex. 1. Divide g by 3x«. Qmt, ~ 2. Divide ^ by 2a2j. Quot., ,^- HOW PROBLEMS ARE SOLVED IN ALGEBRA. 27 3. Divide by a—o. Quot., 36, How is any quantity divided by a fraction ? Ans. By midtiplying it by the divisor inverted. Ex. 1. Divide 6 by % Quof., -• 2. Divide ^ by -• Quot., — • 3. Divide— by— ^. 0^^o^., -y-^- • 4. Divide i by Quot., ax-\-ay—x — y cm + dm — en — dn 5. Divide f-^- by -^. Q^lot., ^. 6. Divide — ^ by ^,-^. 0^.o^ -3^. ION IX. HOW PROBLEMS ARE SOLVED IN ALGEBRA. Ex. 1. John is 3 times as old as James, and the sum of their ages is 32 ; how old is each ? 28 IKTRODUCTTON". Solution. — To solve this by Aritlimetic, we reason thus: Since John's age is 3 times James's, both their ages together make 4 times James's age, and this is 32 years. Now 4 times James's age = 32 years. Hence, James's age is i of 32, or 8 years ; and John's age being 3 times James's, is 3 x 8, or 24 years. To solve it by Algebra, we proceed as follows : Let x represent James's age ; then, since John is 3 times as old, dx will represent his; the sum of their ages will be Bx + x. Now since the sum of their ages is 32, 3.i' + x=S2; or, 4a; = 32. If 4a; = 32, a; = ^ of 32, or x = S. Hence, James's age is 8, and John's being 3a;, is 3 x 8, or 24. S7. The expression 3ic+a; = 32 is what is called an Equation ; and it is by the use of equations that we solve problems in Algebra. 2. A merchant said that he had 72 yards of a certain kind of cloth, in three rolls. In the first roll, there were a certain mimber of yards ; in the second, 3 times as many as in the first ; and in the third, twice as many as in the first. How many yards were there in the first roll ? Suggestions. The equation is a? + 3a; + 2a; = 72 Now, X + 3a; + 2a; is 6a;, hence 6a; = 72 And if 6a; = 72, a-, or la;, is | of 72, a- = 12 Queries. — What does the a; stand for? Answer. The number of yards in the first roll. In this problem which is the most, a;4- 3a; + 2a;, or 72 ? To start with, do you know how much x is ? Then is it a Jcnoicn., or an unknown quantity, at the outset ? 38. The answer to a problem is represented in the beginning of the solution by one of the latter letters of the alphabet, usually x, if there is need of but one letter, and is called th(^ Unknown Quantity. 3. A boy on being asked how old he. was, replied, "If you add to my age 3 times my age, and 5 times my age. HOW PROBLEMS ARE SOLVED IN ALGEBRA. 29 and subtract twice my age, the result will be 49 years. How old was he ? Suggestions. Tlie equation is x + ^x-\-5x—2x = 49 Hence, since x + Sx-\-5x—2x is 7x, 7a; = 49 If 7a; = 49, a; = -f of 49, or aj = 7 4. There are three times as many girls as boys in a p.irty of 60 children. How many boys are there? How many girls ? 5. In a barrel of sugar weighing 200 lbs., there are three, varieties. A, B, and 0, mixed. There is 7 times as much of B as of A, and twice as much of C as of A. How much of A is there ? How much of each of the other kinds ? Ans., Of A, 20 lbs. ; of B, 140 lbs. ; of C, 40 lbs. 6. There were 4 kinds of liquor put into a cask, 2 times as much of the second as of the first, 2 times as much of the third as of the second, and 2 times as much of the fourth as of the third. The cask sprang a leak, and three times as much leaked out as was put in of the first kind, when it was found that there were 36 gallons remaining. How much was there put in of each kind ? Suggestion.— The equation is a! + 2a; + 4a; + 8a;— 3aj = 36. 7. In a pasture there are a certain number of cows and 23 sheep, in all 34 animals. How many cows are there ? Solution. — As it is the number of cows we seek, let x represent the numVjer of cows. Then ar + 23 is the number of animals in the pasture, and the equation is aj+23 = 34. Now the ar + 23 means just the same thing as the 34, that is x + 2d = 34. So, if we subtract 23 from each, there will be just as much left of one as of the other. Subtracting 23 from a* + 23, there remains a-, and subtracting? 23 from 34, there remains 11. Hence a; = 11 ; that is, there are 11 cows. 30 IKTEODITOTION". 39, The part of an equation on the left of the sign = is called the First Member, and that on the right, the Second Member. Note. — The pupil must not think because these examples are so simple that he can " do them in his head," that therefore algebra is a very clumsy method of solving problems. He will find by and by, that though the equation does not really help any in the solu- tion of such simple questions, it will solve a great many very diffi- cult ones about which he might puzzle his brains a great while to no purpose, if algebra did not come to his aid. Stick to it, then, and learn how to use this new instrument, the Equation, and you will by and by find it wonderfully useful. It is a grand patent for solving problems. 8. In a certain pasture there are three times as many horses as cows, and 20 sheep. In all there are 100 animals. How many cows are there ? How many horses ? Suggestions. The equation is a'4-3a; + 20 = 100 Subtracting 20 from each member, sj + Sj; = 80 Uniting the terms of the first member, 4a? = 80 Dividing each member by 4, x = 20 Hence there are 20 cows ; and, as there are three times as many horses as cows, there are 60 horses. 9. In a .basket of 60 apples there are 4 times as many red apples as yellow, and 10 green apples. How many yellow apples are there ? How many red ? 10. John and James together have 75 cents. James has 25 cents less than John. How many cents has John ? Suggestions. — Let x represent the number of cents which John has. Then, as James has 25 cents less, ir— 25 will represent what he has. But both together have 75 cents. Hence the equation is x-\-x—25 = 75. Now, if we drop the —25 from the first member, we make this member 25 greater than it now is, /. e., x + x is 25 greater than HOW PROBLEMS ARE SOLVED IN ALGEBRA. 31 x + x—25. Therefore, if we add 25 to the second member, making it 100, the members will still be equal.* This gives x-\-x— 100 or, %x = 100 Hence a? = 50 That is, John has 50^. 11. A merchant has 90 yards of cloth in two pieces. The longer piece lacks ten yards of containing 3 times as much as the shorter. How much in each piece ? Suggestion. — The equation is a; + 3a!— 10 = 90. 12. Divide the number 50 into two parts so that one part shall lack 10 of being 5 times the other. Suggestions. — The parts are represented by ar, and 5a5— 10. They are 10 and 40. 13. Divide the number 50 into 3 parts, such that the second shall be 5 more, and the third 15 less than the first. Tlie parts are 20, 25, and 5. Suggestion. — The equatiou is a^ + ir-f-5 + iC— 15 = 50. 14. There are 52 animals in a field. Twice the number of cows + 11 is the number of sheep, and 3 times the num- ber of cows — 13 is the number of horses. How many of each kind ? Ans., 9 cows, 29 sheep, and 14 horses. 16. A man said of his age, '* If to my age there added be One-half, t one-third, and three times three, Six score and ten the sum will be. What is my age ? Pray show it me.'* X X Suggestion. — The equation is x-k---\- --\-% = 130. * The word " Transposition " is purposely omitted from this introdnction. Nor la the idea designed to be presented. It will ])e better for the pupil to " think oat " the process, as above. t Meaning " one-half my rtg-g," " one-thinl my age."" 32 INTRODUCTION. Subtracting 9 from each member, Now, we can get rid of the fractions in the first member by mul- tiplying it by 6, the product of both the denominators. Thus, 6 times the first member is 6a? + 3aj + 2aj. Then, if we also multiply tl'c seconil member by 6, the products will be equal. For if two quantities are equal, 6 times one of them is equal to 6 times the other. Hence we have ^x-^Zx-\-^x = 726 Uniting terms, \\x = 726 Dividing by 11, a; = 66 16. Mary gave half her books to Jane, and one-third of them to Helen, when she had but 2 left. How many had she at first ? Suggestions. — Let x represent the number of books Mary had at first. Then she gave Jane - , and Helen - books. And what she gave the other girls, added to what she had left, makes all she had in the first place. Hence the equation is x X ^ Multiplying each member by 6, 3a;-f2aJ4-12 = 6aJ Subtracting 5a; from each member, 12 = « That is, Mary had 12 books at first. 17. A boy lost 25 cents of some money which his uncle gave him, and gave half he had left to his brother. He then earned 50 cents, when he had just as much as his uncle gave him. How much did his uncle give him ? Suggestions. — Let x = the number of cents his uncle gave him. Then he had ic— 25 cents after losing 25 cents. After giving away 2» 25 half of this, he had the other half, or — -— cents, left. He then 2 earned 50 cents, and the amount he had was equal to what his uncle gave him. HOW PROBLEMS ARE SOLVED IN ALGEBRA. 33 Hence the equation is -— — + 50 = a; Multiplying each member by 3, «— 25 + 100 = 2x Uniting, —25 + 100 makes 75, and a;+75 = 2x Subtracting x from each member, 75 = « 18. A boy being asked how many marbles he had, said, " If I had five more than I have, half the number sub- tracted from 30 would leave twice as many as I now have." How many marbles had he ? Suggestions. — Letting x represent the number of marbles the boy had, the equation is Now there is a little peculiarity about this equation, which the pupil must be careful to notice whenever it occurs, or he will make a great many mistakes. It is this : When we multiply each mem- ber by 2, to get rid of the fraction, we must write 60— «— 5 = 4a;. The mistake would be to write 60— d; + 5 = 4a;. The explanation aj + 5 is, that the — sign before the fraction — -- does not belong to the a;, but to the fraction as a whoU. The sign of x in the fraction — — — is + , since when no sign is expressed + is understood. 2 What then becomes of the — sign before the fraction, if it is not the same as the sign of a; in the equation 60— a;— 5 = 4a;? It has been dropped, since the thing signified by it has been performed, and the — sign before the x is the sign of that term in the original equation, changed. In like manner we subtract +5, by changing its sign. The boy had 11 marbles. 19. What is the value of x in the equation '6x— Suggestions.— Multiplying each member by 3, We have 9a;— 2 + 2a; = 64 Hence, a; = 6 34 IlsqTRODUCTIOlS'. 20. Find the value of x in the equation x-\ ^r — = 12 ~ Ans.,, X = 6. 21. What is the value of x in the equation x—1 X-^4: ^„ ^+3^ . .^. — — — = 15 -r- ? Ans.f X = 40^. 22. Show that in ic— 1 ^ 23—0; 4+a; 23. Two boys were to divide 32 marbles between them so that 1^ of what one had should be 5 less than what the other had. How many was each to have ? Suggestions. — Letting x — what one had, then 32— a; = what the other had. The equation is o + ^ = 32— a;, 2 32— a: + 5 =x. ^uory. — Why will either equation answer the purpose? 24. What number is that to which if 7 be added, half the sum will be 8 more than \ of the remainder of the number after 3 has been subtracted ? Equation — — — - = 8. Ans., x = Id. 25. The sum of two numbers is sixteen, and the less number divided by three is equal to the greater divided by five. What are the numbers ? Suggestion. — Let x and 16— a; represent the numbers. 26. Divide twenty-two dollars between A and B, so that if one dollar be taken from three-fourths of B's share, and three dollars be added to one-half of A's money, the sums shall be equal. How many dollars will each have ? HOW PROBLEMS ARE SOLVED IN ALGEBRA. 35 27. The sum of two numbers is thirty-three. If one- sixth of the greater be subtracted from two-thirds of the less number, the remainder will be seven. What are the numbers ? 28. The sum of A's and B's money is thirty-six dollars. If five-eighths of B's, less two dollars, be taken from three- fourths of A's, the difference will be seven dollars. How many dollars has each ? 29. The difference between two numbers is twenty-five ; and if twice the less be taken from three times the greater, the remainder will be eighty. What are the numbers ? 30. A and B gain money in trade, but A receives ten dollars less than B. If A's share be subtracted from twice B's, the remainder wiU be fifty-seven dollars. How much money did each receive ? 31. One number is four less than another, and if twice the less be subtracted from five times the greater, the remainder will be thirty-eight. What are the numbers ? 32. Two farms belong to A and B. A has twenty acres less than B. If twice A's number of acres be taken from three times B's, the remainder will be one hundred. How many acres has each ? 33. One number is seven less than another, and if three, times the less be taken from four times the greater, the remainder will be six times the difference between the two numbers. What are the numbers? 34. Anna is four years younger than Mary. If twice Anna's age be taken from five times Mary's, the remainder will be thirty-five years. What is the age of each ? 35. One number is ten less than another. If three times the less be taken from five times the greater, the remainder will be seven times the difference of the two numbers. What are the numbers ? A BRIEF SURVEY OF THE OBJECT OF PURE MATHE- MATICS AND OF ITS SEVERAL BRANCHES. /. Pure Mathematics is a general term applied to several branches of science, which have for their object the investigation of the properties and relations of quantity — comprehending number, and magnitude as the result of extension — and of form. 2. The Several Branches of Pure Mathematics are Arithmetic, Algebra, Calculus, and Geometry. 5. Arithmetic, Algebra, and Calculus treat of number, and Geometry treats of magnitude as the result of extension. 4. Quantity is the amount or extent of that which may be measured ; it comprehends number and magnitude. The term quantity is also conventionally Jipplied to sym- bols used to represent quantity. Thus 25, m, xi, etc., are called quantities, although, strictly speaking, they are only representatives of (juan titles. It is not easy to give a philosophical account of the idea or idea.^, represented by the word Quautity as used in Mathematics; antl doubtless, different persons use the word in somewhat different senses. It is obviously incorrect to say that " Quantity is anything which can be measured." Quantity may be affirmed of any such 2 INTKODUCTION. concept ; nevertheless, it is not the thing itself, but rather the amount or extent of it. Thus, a load of wood, or a piece of ground, can be measured ; but no one would think of the wood or piece of ground as being the quantity. The quantity (of wood or ground) is rather the am,ount. oi extent of it. The word is very convenient as a general term for mathematical concepts, when we wish to speak of them without indicating whether it is number or magnitude that is meant. Thus we say, " m represents a certain quantity," and do not care to be more specific. As applied to number, perhaps the term conveys the idea of the whole, rather than of that whole as made up of parts. It is, there- fore, scarcely proper to speak of multiplying by a quantity ; we should say, by a number. On the other hand, when we apply the term quantity to magnitude, it is with the idea that magnitude may be measured, and thus expressed in number. The distinction between quantity and number is marked by the questions, '' How much ? " and " How many ? " 5. Number is quantity conceived as made up of parts, and answers to the question, *^How many?" Thus, a distance is a quantity ; but, if we call that distance 5, we convert the notion into number, by indicating that the distance under consideration is made up of parts. Again, m may mean a value, as of a farm. We may or may not conceive it as a number (as of dollars). If we think of it simply in the aggregate, as the worth of a farm, m represents quantity; if we think of it as made up of parts (as of dollars) it is a number. 6* Number is of two kinds. Discontinuous and Con- tinuous. 7. Discontinuous Number is number conceived as made up of finite parts ; or it is number which passes from one state of value to another by the successive additions or subtractions of finite units ; i. e., units of appreciable magnitude. THE OBJECT OF PC7RE MATHEMATICS. 3 8, Continuous Number is number which is conceived as composed of intinitesimal parts ; or it is number which passes from one state of value to another by passing through all intermediate values, or states. Number, as considered in Arithmetic, and in this volume, is Discontinuous Number. Thus 5 grows till it becomes 9, by taking on additions of units of some conceivable value ; as when we consider it as passing from 5 to 9, thus, 5, 5 + 1 or 6, 6 + 1 or 7, 7 + 1 or 8, 8 4- 1 or 9. If the increment were any fraction, however small, t/ie fonn of the cojiceptit/n would he the same. Time affords a good illustration of Continuous Number. We usually conceive time as a discontinuous numher^ as when we think of it as made up of hours, days, weeks, etc. But it is easy to see that such is not the way in which time actually grows. A period of one day does not grow to be a period of one week by taking on a whole day at a time, or a whole hour, or even a whole second. It grows by imperceptible increments (additions). These incon- ceivably small parts, by which time is actually made up, we call infinitesimals; and number, when conceived as made up of such infinitesimals, we call Continuous Number. 9, Arithmetic treats of Discontinuous Number,— of its nature and properties, of the various methods of combining and resolving it, and of its application to prac tical affairs. The leading topics of Arithmetic are : 1. Notation; i. e., methods of representing number, as by the characters, 1, 2, 3, 4, etc., or by letters, as, a, 5, m, n, x, y, etc. 2. Properties of Numbers or deductions from the methods of Notation. 3. Reduction, as from one scale to another, from one denomina- tion to another, from one fractional form to another, or, in short, from any one form of expression to another equivalent form. 4. The various methods of combining number, as by addition, multiplication, and involution. 5. Resolving Number, as by subtraction, division, and evo- lution. 4 INTRODUCTION^. And all these processes as affected by the use of any notation, and upon integral or fractional discontinuous numbers of any kind. Arithmetic, therefore, philosophically considered, embraces much that is usually classed as Algebra. Thus all that usually precedes Simple Equations, and all that is embraced in this Part I. is simply a repetition and extension of the processes of Arithmetic with a new notation — the literal. Again, logarithms are nothing but a new scheme of notation, by means of which certain com- binations are more readily effected ; and the making of logarithms is but a reduction from one form of expression to an equivalent one in another notation. In the ordinary notation, a certain number is represented thus, 256 ; in the logarithmic notation it is 2.40834. 10, Algebra treats of the Equation, and is chiefly occupied in explaining its nature and the methods of transforming and reducing it, and in exhibiting the manner of using it as an instrument for mathematical investigation. The whole province of the relations of quantity, continuous or discontinuous number, is covered by Algebra, so far as the equation can be made the instrument of investigation. Much, therefore, of what is found in our Arithmetics can be more expeditiously treated by Algebra. Such are the subjects of Ratio, Proportion, the Progressions, Percentage, Alligation, etc. In fact^ the equation is the grand indrimient of mathematical investigation^ and demonstrates its efficiency in every depa.rtment of the science. To hope to get on in mathematics without Algebra, is to expect to walk without feet. 11, Calculus treats of Continuous Number, and is chiefly occupied in deducing the relations of the infini- tesimal elements of such number from given relations between finite values, and the converse process, and also in pointing out the nature of such infinitesimals and the methods of using them in mathematical investigation. 12, G-eometry treats of magnitude and form as the result of extension and position. tX)OICO-MATHEMATlCAL TERMS. 5 The.principal divisions of the science of Geometry are : 1. The Ancient, Special, or direct Geometry (the common Geome- try of our schools), including Trigonometry, Conic Sections, and all other geometrical inquiries conducted upon these methods. 2. The Modern, Indirect, or General Geometry (usually called Analytical), and 3. Descriptive Geometry. SYNOPSIS. Subject of Section. Quantity : Uses of the term. Definition of Pure Mathematics: Number: Illustration of : Kinds Several Branches of. of: Illustration of kinds. Subject Matter of the Several Arithmetic : Topics of. Branches. Algebra— Calculus— Geometry. mon II LOGICO-MATHEMATICAL TERMS. 13. A Proposition is a statement of something to be considered or done. Illustration.— Thus, the common statement, "Life is short," is a proposition ; so, also, we make, or state a proposition, when we say, "Let us seek earnestly after truth." — "The product of the divisor and quotient, plus the remainder, equals the dividend," and the requirement, " To reduce a fraction to its lowest terms," are examples of Arithmetical propositions. 14. Propositions are distinguished as Axioms, Theorems, Lemmas, Corollaries, Postulates, and Problems. 15. An Axiom is a proposition which states a princi- ple that is so simple, elementary, and evident, as to require no proof. Illustration. — Thus, "A part of a thing is less than the whole ot it," " Equimultiples of equals are equal," are examples of axioms. 6 INTRODUCTION. If any one does not admit the truth of axioms, when he understands the terms used, we say that his mind is not sound, and that we cannot reason with him. 16, A Theorem is a proposition which states a real or supposed fact, whose truth or falsity we are to determine by reasoning. Illustration.— "If the same quantity be added to both numerator and denominator of a proper fraction, the value of the fraction will be increased," is a theorem. 17 » A. Demonstration is the course of reasoning by means of which the truth or falsity of a theorem is made to appear. The term is also applied to a logical statement of the reasons for the processes of a rule. A solution tells how a thing is done ; a demonstration tells why it is so done. A demonstration is often called proof. IS. A Lemma is a tlieorem demonstrated for the purpose of using it in the demonstration of another theorem. Illustration. — Thus, in order to demonstrate the rule for finding the greatest common divisor of two or more numbers, it may be best first to prove that " A divisor of two numbers is a divisor of their sum, and also of their difference," This theorem, when proved for such a purpose, is called a Lemma. The term Lemmxi is not much used, and is not very important, since most theorems, once proved, become in turn auxiliary to the proof of others, and hence might be called lemmas. 10, A Corollary is a subordinate theorem which is suggested, or the truth of which is made evident, in the course of the demonstration of a more general theorem, or which is a direct inference from a proposition. Illustration. — Thus, by the discussion of the ordinary process of performing subtraction in Arithmetic, the following Corollary might be suggested : " Subtraction may also be performed by addition, as we can readily observe whnt number must be added to the subtra- hend to produce the minuend." LOGICO-MATHEMATICAL TERMS. 7 20, A Postulate is ii proposition which states that somethiug can be done, and which is so evidently true as' to require no process of reasoning to show that it is possible to be done. We may or may not know how to perform the operation. Illustration. — Quantities of the same kind can be added together. 21, A Problem is a proposition to do some specified thing, and is stated with reference to developing the method of doing it. Illustration. — A problem is often stated as an incomplete sentence, as, " To reduce fractions to forms having a common denominator." 22, A Rule is a formal statement of the method of solving a general problem, and is designed for practical application in solving special examples of the same class. Of course a rule requires a demonstration. 23, A Solution is the process of performing a problem or an example. It should usually be accompanied by a demonstration of the process. 24, A Scholium is a remark made at the close of a discussion, and designed to call attention to some particular feature or features of it. Illustration. — Thus, after having discussed the subject of multi- plication and division in Arithmetic, the remark that "Division is the converse of multiplication," is a scholium. SYNOPSIS. Subject of Section. Lemma. III. — Why the term is Proposition. /^/.— Varieties of. unimportant. Axiom. lU. — One who will not Corollary. III. admit the truth of. Postulate. III. Theorem. lU. Problem. How stated. III. Demonstration. Difference be- Rule. tween a solution and a demon- Solution. stration. Scholium. lU. PART I jl^^Anjuui: -^^^^^'^^^^^ larLfmJX'^/^ ^ FUNDAMENTAL RULES. Section ». ^'— w- ^atio^^ 25, A System of Notation is a system of symbols by means of which quantities, the relations between them, and the operations to be performed upon them, can be more concisely represented than by the use of words. SYMBOLS OF QUANTITY. 26» In Arithmetic, as usually studied, numbers are rep- NOTE.—Part First treats of the familiar operations of Addition, Subtraction, Multiplication , Division, Involution a»d, Evolution, and the tlieor// of Fractions, The only difference hettveen the jn'occsses here developed and those tvith which the 2)Hpil is already familiar, grotvs out of the notatittn. Hence appears the appositeness of the term Literal Arithmetic. Hence, also, the teacher should he careful that the pupil see the unity of purpose, and the reason for any difference in tnethod of execution. NOTATION. 9 resented by the characters, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, called Arabic figures, or, simply, figures. 27, In other departments of mathematics than Arith- metic, numbers or quantities are more frequently repre- sented by the common letters of the alphabet, a, b, c, . , . m, n . , . X, y, z. These letters may, however, bo used in Arithmetic ; and the Ai*abic figures are used in all departments of mathematics. This method of representing quantities by letters is often called the Algebraic method, and the method by the Arabic characters, the Arithmetical. It would be better to call the former the Literal method, and the latter the Decimal. 28. The Literal Notation has some very great advan- tages over the decimal for purposes of mathematical reasoning : 1st. The symbols are more general in their significa- tion; and 2d. We are enabled to detect the same quantity any- where in the process, and even in the result. Thus it happens that the processes become general formulas, or rules, instead of special solutions. Illustration. — To illustrate the first statement, suppose we say a boy has 7 apples ; you know just hoic many are meant. But wb? \ we Bay a boy has b apples, nobody can tell how many he has. In fact, it is not designed to tell the exact number, but only to say that he has some number. Again, 7 represents the same number of units always; but a letter may be used to represent any number of units we please; or, it may be used, as we have just said, without our caring to specify any precise number of units. This may seem to be a very unsatisfactory kind of notation; but with patience its advantages will appear. The following examples will illustrate the general, or comprehensive character of the literal notation more fully. 10 FUNDAMENTAL RULES. EXAMPLES. Ex. 1. A boy has 8 marbles which he sells for three cents each, and takes his pay in pencils at 6 cents each. How many pencils does he receive ? Suppose we answer that he receives 3 times 8 divided 3x8 by 6, or — - — pencils, without giving the number more 6 explicitly. Now take a similar example, nsing the literal nota- tion ; thus, A boy sells a certain number of marbles which we will represent by c, for a number of cents each, which we will call m, and takes his pay in pen- cils at l cents each. How many pencils does he receive ? We will answer as before, and say he receives c times w, divided by 5, or — — pencils. The pupil will notice this diflference between the answers ; both, as they now stand, simply tell what operations to perform in order to get the answers; but, in the former case, we can perform the operations and get the explicit answer, 4, while in the latter case, we can only leave it as it is. Sucn an answer as — r^ — may seem to the pupil to be no answer at all ; and indeed it is not an answer in the same sense as he has been accustomed to think of answers ; nevertheless it is often more useful. Notice that the answer 4 is only true for the specific example, while the answer — = — is true in every like example. We also observe that the quantities 3, 8, and 6 do not appear distinctly in the numerical answer, 4 ; but the c, m^ and h do in the literal, and would, in general, however complicated the problem. The literal answer is equivalent to the rule. Multiply the price of one 7narUe lyy the number of marlles^ and divide the product hy the price of a pencil. KOTATION. 11 2. One boy sold 5 pears at 3 cents each ; another sold 6 apples at 2 cents each ; and a third sold 3 melons at 8 cents each. How much did they all receive ? Atis., 51 cents. 3. One boy sold b pears at c cents each ; another sold m apples at n cents each ; a third sold d melons at g cents each. How much did they all receive ? Ans., bxc-\-mx7i-{-dxg cents. Suggestions. — Notice that in the 3d example the several quanti- ties of the problem are distinctly seen in the answer, but not so in the answer to Mc. 2. Moreover, the answer to Ex. 3 is equally true for any and all values of &, c, m, n, d, and g. Consider in like manner the two following : 4. If I buy 5 cords of wood at 4 dollars per cord, and pay for it in cloth at 2 dollars per yard, how many yards are required ? Ans., 10 yai'ds. 6. If I buy a cords of wood at b dollars per cord, and pay for it in cloth at c dollars per yard, how many yards are required ? . axb ^ Ans., yards. 6. A man had a flock of m sheep. He lost 2n of them and raised 10a. After which he sold the flock at $c per head, taking his payment in cloth at $b per yard. What operations must be performed on these numbers in order to ascertain the number of yards of cloth received ? And how will this number be represented ? 7. A man bought 3 horses. He gave for the first twice as much as for the second, and for the third c times as much as for both the others. If x represents the price of the first, how much did he give for the third ? Ans., (•^-fo) ^- 12 FUNDAMENTAL RULES. 20, In using the decimal notation certain laws are established in accordance with which all numbers can be represented by the ten figures. Thus, it is agreed that when several figures stand together without any other mark, as 435, the right hand figure shall signify units, the second to the left, tens, the third, hundreds, etc.; also that the sum of the several values shall be taken. This number is, therefore, 4 hundreds + 3 tens -}- 5 (units). 30, In like manner, certain laws are observed in repre- senting numbers by letters. FIRST LAW. Known Quantities, that is such as are given in a problem, are represented by letters taken from the first part of the alphabet; while Unknown Quantities, or quantities whose values are to be found, are represented by letters taken from the latter part of the al]3habet. Illustration.— A grocer has two kinds of tea, one of which is worth a cents (any given number being meant by a) per pound, and the other h cents. How many pounds of each must he take to make a chest of c pounds, which will be worth d dollars ? In this problem, a, &, c, and d are the given or known quantities, and hence are represented by letters from the first part of the alphabet. The unknown or required quantities are the nuraber of pounds of each of the two kinds of tea. We therefore represent the number of pounds of the first kind by «, and of the second kind by y. Scholium. — This law is not very rigidly adhered to, except that letters after and including t), are generally used to represent un- known quantities, while the others are used for known quantities. But it is sometimes convenient to use a different notation. Thus, in problems in Interest, the principal may be represented by j^, whether it is known or unknown, the interest, in like manner, by i, the rate per cent, by r, the time by i, etc. NOTATION. 18 Accented letters, as «', «", a", «"", etc., (read "« prime," " a second," " a third," etc.), and letters with subscripts, as fli, fljj «3, cL^i etc., (read "a sub 1," "a sub two," etc.), are sometimes used. This form of notation is used when there are several like quantities in the same problem, but which have different numerical values. Thus, in a problem in which several walls of different heights, breadths, and lengths, are considered, we may represent the several heights by a', a", a", etc., or «, , a^, a^, etc.; the thick- nesses by b', i", V" , etc., or J,, h or - . x'" 18 — , etc. Also 8 is -— , or - . w * is — 2' 8 af' gf 4 - 44, The Radical Sign, ^, is also used to indicate the square root of a quantity. When any other than the square root is to be designated by this, a small figure speci- fying the root is placed in the sign. Thus V^ signifies the 3d, or cube root of 5, and is the same as 5^. \/Mal]^ indicates the 5th root of 34a^ and is the same as (34a^^)^. Scholium. — Read \/5ac ("the square root of 5ac," not "radical 5ac''). The latter expression is generic, and applies as well to \/5ac, or \/5nc. Besides it is inelegant. Let the pupil read the following examples and give the signification of each. 18 FUNDAMENTAL RULES. EXAMPLES. Ex. 1. 25a~^k Bead, *'25, a exponent —2, h exponent |." It means 25 multiplied by -g, multiplied by the square of the cube root of h. (See articles 109, 110.) 2. x^~\ Read, "ic, exponent 1." Since 1 is ^ n n — ; — , x^~ is the same as x~^y and hence means that x is to be raised to a power indicated by m—n, and the nth. root of this power extracted. 3. Read and explain 2a^b~^. V27f*^y « . SYMBOLS OF RELATION. 45, The Sign of Geometrical Ratio is two dots in the form of a colon, : . Thus a : h, is read "a is to h," or, "the ratio of a to &." It means the same n,^ a-—h. 46, The Sign of Arithmetical Ratio is two dots placed horizontally, •• , Thus a -• h is read, "the Arith- metical ratio of « to 5 and is equiyalent to a—b. 47, The Sign of Equality is two parallel horizontal lines, = . Thus, 2cx ■= xy, is read, " 2cx equals xy.^^ 6ac —2by = dx^ is read, " oac minus 2by equals 3a:l" Four dots in the form of a double colon, : : , is the sign of equality between ratios. Thus, a:b::c:d. read, "a is to 5 as c is to rZ," means that the ratio of a to 5 equals the ratio of c to cl, and may just as well be written a : b :=: c :d, a c OY ~=z ~ all of which expressions mean exactly the same thing. 48, The Sign of Inequality is a character somewhat like a capital V placed on its side, <, the opening NOTATION. 19 being towards the greater quantity. Thus ay b is read, "a greater than ^." m ." d a SYMBOLS OP AGGREGATION. 50. A Vinculum is a horizontal line placed over several terms, and indicates that they are to be taken together. The parenthesis, ( ), the brackets, [ ], and the brace, i I have the same signification. Illustration. a-\-l)y.cd—e means that (a + 5) is to be multiplied by {cd—e). (a + &) x {cd—e) also means the product of {a + l) and {cd—e). Brackets and braces are used when one parenthesis would fall within another. Thus, \z-it[a-k-{])-\-c)x\y\ w, signifies that the product of (& + 6") multiplied by x, is to be added to a, and this sum multiplied by y\ to this product z is to be added and the sum multiplied by u. ol, A vertical line after a column of quantities, each having its own sign, signifies that the aggregate of the col- umn is to be taken as one quantity. Thus -}-a x is the same as {a—b-^c)x. —h SYMBOLS OP CONTINUATION. 32. A series of dots, , or of short dashes, , written after a series of expressions, signifies ''&c." Thus a : ar : ar^ : ar^ ar^ means that the series is to be extended from ai^ to ar^, whatever may be the value of n 20 FUNDAMENTAL RULES. SYMBOLS OF DEDUCTION. 53, Three dots, two being placed horizontally and the third above and between, . • . , signify therefore, or some analogous expression. If the third dot is below the first two, •.-, the symbol is read ^* since," *^ because," or by some equivalent expression. POSITIVE AND NEGATIVE QUANTITIES. 54, Positive and Negative are terms primarily ap- plied to concrete quantities which are, by the conditions of a problem, opposed in character. Illustration. — In estimating the value of a person's estate, bis property may be called positive, and his debts negative. Distance up may be called positive, and distance down, negative. Time before a given period may be called positive, and after, negative. Degrees above on the thermometer scale are called positive, and below, negative. 55, The signs + and — are used to indicate the char- acter of quantities as positive or negative, as well as for the purpose of indicating addition and subtraction. (See article 57.) 56, In problems in which the distinction of positive and negative is made, each quantity is to be considered as having a sigti of character, expressed or understood, besides the plus or minus sign, which indicates whether it is to be. added or subtracted. The positive sign need not be written to indicate character, as it is customary to consider quantities whose character is not specified us positive. Illustration I. — In the expression aJ)-\-m—cx, let the problem out of which it arose be such, tliat a, m, and x, tend to a positive result, and b and c to an opposite, or a negative result. Giving these NOTATIOT^. 21 quantities their signs of character, we have, ( + a) x (~h) -f (+?//) — (-c) X i + x,) which may be read, ''positive a multiplied by negative b, plus positive m, minus negative c multiplied by posi- tive a;." Suppressing the positive sign, this may be written, a {-b) + 771 — (-c) x^ by also omitting the unnecessary sign of multiplication. Illustration 2. — As this subject is one of fundamental importance, let careful attention be given to some further illustrations. We are to distinguish between discussi(ms of the relations between mere abstract quantities, and problems in which the quantities have some concrete signification. Thus, if it is desired to ascertain the sum or difference of 468, or //^, and 327, or 7i, as mere num- bers, the question is one conceniiug the relation of abstract numbers, or quantities. No other idea is attached to the expres- sions than that each represents a certain number of units. But, if we ask how far a man is from his starting point, who has gone, first, 468, or tt?, miles directly east, and then 327, or n, miles directly west ; or if we ask what is the difference in time between 468, or m, years B. C, and 327, or ??, years A. D., the numbers 468, or 7//, and 327, or ??, take on, besides their primary signification as quantities, the additional thought of opposition in direction. They therefore become, in this sense, concrete. Again, a company of 5 boys are trying to move a wagon. Three of the boys can pull 75, 85, and 100 pounds each ; and they exert their strength to move the wagon east. The other two boys can pull 90 and 110 pounds each; and they exert their strength to move the wagon west. It is evident that the 75, 85, and 100 are quantities having an opposite tendency from 90 and 110. Again, suppose a party rowing a boat up a river. Their united strength would propel the boat 8 miles per hour if there were no current; but the force of the current is sufficient to carry the boat 2 miles per hour. Which way will the boat move, and how fast ? The 8 and 2 are quantities of opposite character in their relation to the problem. Once more, in examining into a man's business, it is found that he has a farm worth m dollars, personal property worth n dollars, and accounts due him worth c dollars. There is a mortgage on his farm of & dollars, and he owes on account a dollars. The m^ «, and c are quantities opposite in their nature to b and a. This opposition in character is indicated by calling those qiuintities which contribute to one result positive^ and those which con- tribute to the opposite result negative. 32 FUNDAMENTAL RULES. ^7. Purely abstract quantities have, properly, no dis- tinction as positive and negative ; but, since in such prob- lems the plus or additive, and the minus or sub tractive terms stand in tlie same relation to each other as positive and negative quantities, it is customary to call them such. Illustration. — In the expression, 5ac—3cfZ-|-8a'y—2a^, though the quantities, «, c, d, x and y be merely abstract, and have no proper signs of character of their own, the terms do stand in the same relation to each other and to the result, as do positive and negative quantities. Thus, 5ac and ^xy tend, as we may say, to increase the result; while —Zed, and — 2atf? tend to diminish it. Therefore the former may be called positive terms, and the latter negative. *5S, Scholium. — Less than zero. Negative quantities are fre- quently spoken of as "less than zero." Though this language is not philosophically porrect, it is in such common use, and the thing signified is so sharply defined and easily comprehended, that it may possibly be allowed as a conventionalism. To illustrate its mean- ing, suppose in speaking of a man's pecuniary affairs it is said that he is worth " less than nothing ;" it is simply meant that his debts exceed his assets. If this excess were $1000, it might be called negative $1000, or -$1000. So, again, if a man were attempting to row a boat up a stream, but with all his effort the current bore him down, his progress might be said to be less than nothing, or nega- tive. In short, in any case where quantities are reckoned both ways from zero, if we call those reckoned one way greater than zero, or positive, we may call those reckoned the other way " less than zero," or negative, SO, The value of a Negative Quantity is conceived to increase as its numerical value decreases. Illustration. — Thus -3 > -5, as a man who is in debt $3, is bet- ter off than one who is in debt $5, other things being equal. If a man is striving to row up stream, and at first is borne down 5 miles an hour, but by practice comes to row so well as only to be borne down 3 miles an hour, he is evidently gaining; i. e., -3 is an in- crease upon -5. Finally, consider the thermometer scale. If the mercury stands at 20° below 0, (marked -20°) at one hour, and at -10' the next hour, the temperature is increasing; and, if it in- crease sufllciently will become 0, ja^dng which it will reach +1°, NOTATION. 23 + 2°, etc. In this illustration, the quantity passes from negative to positive by passing through 0. This is assumed as a fundamental truth of the doctrine of positive and negative quantities, viz. : That A QUANTITY IN PASSING TUROUGH MAY CHANGE ITS SIGN. [It appears in geometry, that a quantity may also change its sign in passing through infinity. Thus the tangent of an arc less than 90° is positive; but if the arc continually increases, the tangent becomes infinity at 90°, passing which it becomes negative.] NAMES OF DIFFERENT FORMS OF EXPRESSION. 00. A Polynomial is an expression composed of two or more parts connected by the signs plus and minus, each of which parts is called a term. 61. A Monomial is an expression consisting of one term. A Binomial is a polynomial having two terms. A Trinomial is a polynomial having three terms. Illustration. 5a''h—cd»+x—4.{a + l) is a polynomial of 4 terms. The first three are monomial terms, and the last is a binomial term. 5a4i-~ef and x^ + y' are examples of binomials. 'Za'^x^y— 125(^h—m. ^aV^-l^. 3. Write three times a into h, plus the binomial a minus h, divided by the sum of a square and h cube. 4. Write the fraction, the product of the sum of a and I into the sum of x and y, divided by the square root of a diminished by the cube root of h. va—\h 6. Write the fraction, a fifth power diminished by 3 times a square h cube, divided by the square root of the binomial x square diminished by y square. 6. Write the square root of the sum of x and y equals c minus the square root of the sum of x and h. 7. Write the fraction, the binomial 3 times x plus 1, divided by 5 times x, minus the fraction 3 times the bino- NOTATION. 25 mial X minus 1, divided by the binomial 3x plus 2, is greater than 9 divided by 11a:. 8. Write the square root of the fraction, h divided by a plus Xy plus the square root of the fraction c divided by a minus x, equals the 4th root of the fraction, 4 times the product of b and c, divided by a square minus x square. 9. Write a exponent |, minus b exponent —m. Write the result in three different forms. 10. Write, a exponent — , is to the binomial h minus x exponent — |, as 5 times c square plus rf, is to the 5th root of X 4th power. * 1 _ Result, an : -jt==- : : 5^ + (7 : \^x*, W(b—xY What binomials are there in the last result ? EXERCISES IN READING AND EVALUATING EXPRESSIONS. Read the following expressions, and find the value of ejich, when a = 6, S = 5, r = 4, and ^ = 1. 1. a'^ + Ub—c-\-d. Result, 3G 4-60—44-1, or 93. 2. 2«3_3fl2^ + c3. Result, —44. 3. 3(a2-//J)-«(cf-h^). Result, 15. 4. Between the expression -^ — -. and wba^—{(^-{-^b), which of the signs, =, >, or <, is correct ? Read and evaluate the following, calling « = IC, & = 10, c =z Q, m r= A, X z=z 5, y = 1. 5. {b—x){Va + b)-\-y/{a—b)(x-\-y), Result, 76. 2 26 PtJNDAMEKTAL RULES. 6. VcJa + b) — ^/¥{a—b). T^esw//, 6.49, nearly. 7. 50«a;'^4-4a^-100[^y, the parenthesis and the vinculum having the same signification. 12. If a = 2, ^ = 3, 2; = 6, and y = 5, show that V\{a-^b)y\-}-\^\{a + x){y-2a)\-hv'\{y-bya\=9. 13. With the same values show that {ay)^ (^a; + «^ + 3)~« _ \b(x-y )-^-[(axY- U2] \ 20ay _ 14. Find the value of ^/W^^ — {lO-]-n)^ if w = 6. 15. Find the value of (5771^ + 5^/^)^+^/^+0;^, if 771 := 4, and a; = 9. Test Questions. — Wliat are the chief points of difference between the Arabic or Decimal notation, and the Literal or Algebraic ? What is meant by the terms positive and negative as applied to quantity ? What is the meaning of the negative sign when prefixed to an exponent ? Read oc , or =, or <, or > y wa?— y" : 4m^. NOTATION. 37 SYNOPSIS FOR REVIEW. DEFINITION. SYMBOLS OF QUANTITY. Arabic. Literal. See Arithmetic. Advantages. Examples, Ist Law. 2nd Law. SYMBOLS OF , „„„^ OPERATION AND \ Exponent. DEFINITIONS. *^ Infinity. Meaning of. + , — vT, X, - Definitions of Power, Root, j 1. More general. i 2. Can (race quantities. ( Known quantities. Unknown quantitie>i. Accents., t^ubccriptt:, and Greek Udtefrs. Difference beticeen Alge- l>raanct Arithmetic. Letters in connection. Two points of difference. Figures with letters. b)a{c, a\b V. SYMBOLS OF RELATION AND DEFINITIONS. SYMBOLS OF AGGREGATION. Integral Exponent, Fractional Exponent, Negative Exponent, Radical Sign. :, ••, =, ::, > <, a. Geometrical Ratio. Arithmetical Ratio. Equality. Inequality. Variation. r — . (). []. iK I. J Vinculum. l^ Vertical Line. How rt-fu\. Examples. SYMBOLS OF CONTINUATION. SYMBOLS OF DEDUCTION. POSITIVE AND NEGATIVE QUANTITIES. FORMS OF EXPRESSION. Definition. Two signs of every quantity. Plus and minus terms become positive and negative. ( Meaning. " Less than zero." -< How negatives increase. ' How a quantity changes sign Polynomial. Monomial. Binomial. Trinomial. CoeflRcient. Similar Terms. Term. Illustrations. EXERCISES IN. EXERCISES IN READING AND EVALUATING EXPRESSIONS. 2.8 FUNDAMENTAL RULES. 64. Addition is the process of combining several quan- tities, so that the result shall express the aggregate value in the fewest terms consistent with the notation. 60, The Sum or Amount is the aggregate value of several quantities, expressed in the fewest terms consistent with the notation. Illustration. — To add 346, 234, and 15, is to find an expression for their aggregate value in the fewest terms consistent with the decimal notation. The sum or amount is 595, because it is such simplest expression for the aggregate. In like manner the sum of 4ac + 51} + 2x, idac + 2h + Sx, and 121) + 9x, is 17«c+19a;+19&, be- cause it is the simplest expression for the aggregate value consistent with the literal notation. If the pupil is acquainted with other scales of notation he knows that with radix 100, 595 would be represented by 2 figures, since all numbers less than 100 would be represented by one figure. 00. Prop. 1. — By Addition similar terms are united into one. Demonstration. — Let it be required to add 4«c, 5ac, —2ac, and — 3ac. Now 4«c is 4 times ac, and 5«c is 5 times the same quantity {ac). But 4 times and 5 times the same quantity make 9 times that quantity. Hence, 4«c added to 5«c make ^ac. To add —2ac to 9ac we have to consider that the negative quantity, — 2ac. is so opiJosed in its character to the positive, 9«c, as to tend to destroy it when combined (added) with it. (As if 'doc were property, and —2ac debts.) Therefore, —2a^ destroys 2 of the 9 times axi, and gives, when added to it, llac. In like manner — 3«c added to lac. ADDITION. 29 gives 4^c. Thus the four similar terms, iac, 5ac, —2ac, and —3ac, have been combined (added) into one term, Aac ; and it is evident that any other group of similar terms can be treated in the same manner q. e. d. EXAMPLES. Ex. 1. Add 13m%, —lOmhi, —Cmhi, omhi, and —im^n. Model Solution.— Adding together 13m'^/i and —10m'/?, the — lOw'w destroys 10 of the 13 times m^n and gives 3m'w. Adding 3m'/i and —Qm^n, the 'dm^n destroys 3 of the —Q/u^n and gives — dm'?}. —Sm'^n added to 5m^n destroys 3 of the 5 times m'/i and gives 2?n'^n. 2m^n added to — 4m'n destroys 2 of the — 47?t'n and gives — 2m'w. Hence the sum of 13m"^r?, —lOm^n, — 6m'w, 57n'n, and —4'm?n is —2m'^n. 2. Add ISax^, —5ax^, —lOax^, 4:ax^, and —6ax^, ex- plaining as above. Result, ax^. 3. Add —bSx?, —)lc^x\ %ch^, Sch^, and —ic*x% ex- plaining as before. Result, 0. 4. Add 3fl.r, Qax, —ax, 2ax, —7 ax, and 6ax, 5. Add 2%^ —iibf, —hyS Sby\ 3b f, and —2by\ 6. Add 5ax-, —2ax^, dax^, —9aa^, and ax^. Sum, —2ax\ 7. Add 6 J, —6x^, — lOar^, Sx^, and lla;i 8. Add —6a^ 2a% —oa^, \a?, — 3«2, and ^2. 9. Add —2a^/x, a^/x, —SaVx, la^fx, and —^a^/x. Sum, —a^/x. 10. Add — 2am^, 4a\/wi, 3am^, and — aa/w. /S^wm, 4«'v/wi. 11. AddlOflW, -^a^^/x, -2aW, and 4a*^i. 12. Add \\am, %\am, —3am, and am. Sum, 2am, 30 FUNDAMENTAL RULES. 13. Add 27^^ —a\ — 28^2, and — 4^^. Sum. — 6««. 14. Add 3?n2, — fm^, m^, and — |m2. ^9i/w, 2-^m\ 16. Add 7a/5, — 5a/^, 1%Vx, and — 3v^. 6^2^77?, 11 V^. 16. Add 9^, }J, — |J, —85, and — 15. .%m, .— ^. G7» Cor. 1. — 7n adding similar terms, if the terms are all podtive, the sum is positive; if all negative, the sum is negative ; if some are positive and some negative, the sum takes the sign of that kind (positive or negative) which is the greater. Scholium. — The operation of adding positive and negative quan- tities may look to the pupil like Subtractiou. For example, we say +5 and -3 added make +2. This looks like Subtraction, and in one view, it is Subtraction. But why call it Addition? The reason is, because it is s,imp\j putting the quantities together — aggre- gating them— not finding their difference. Thus, if one boy pulls on his sleigh 5 pounds in one direction, while another boy pulls 8 pounds in the opposite direction, the combined (added) effect is 3 pounds in the direction in which the first pulls. If we call the direction in which the first pulls, positive, and the opposite direction negative, we have +5 and -3 to add. This gives, as illustrated, +5i. Hence we see, that the sum of +5 and -3 is +2. But the diff'erence between +5 and —3 is 8, as appears in the following illustration : Suppose one boy is drawing his sleigh for- ward while another is holding back 3 lbs. If it takes just 10 lbs. to move the sleigh itself, the first boy will have to pull 13 lbs. to get it on. But if instead of holding dacJc 3 lbs., the second boy pushes 5 lbs., the first boy will only have to pull 5 lbs. Thus it appears, that the difference between pushing 5 lbs. (or + 5) and holding back 3 lbs. (—3) is 8 lbs. In like manner the sum of $25 of property and $15 of debt, that is the aggregate value when they are combined, is $10. +25 and -15 are +10. But the difference between having $25 in pocket, and being $15 in debt, is $40. The difference between + 25 and - 15 is 40. 17. A thermometer indicated +28° (28° above 0), it then rose 10°, then fell 3°, then rose 2°, and again fell 7°. What was the sum of its movements ; or, how did it stdJid at last ? ADDITION. 31 Model Solution. — Calling upward movement + and downward -, the movements were +10, -3, +2, and -7, the sum of which is + 2. Hence it rose 2°. As it originally stood at 28° above 0, and, in the whole, rose 2\ it stands at last 30° above 0. 18. A party are rowing up a stream, and alternately row and rest. During 3 periods of rowing they advance Smn, 2mnj and 6mn rods. But during the corresponding periods of resting, they float down 6mn, mn, and ^mn rods. What was the result ; did they, on the whole, ascend or descend, and how much? In other words, what is the sum of -r'Smiij +2m/?, -f-6mn, — omn, — mn, and — Amn? Arts. They ascended mn rods. {-\-mn.) 19. A man has a farm worth llOOc^, on which there is a mortgage of tlbcd ; he has personal property worth $8a/, and accounts due him of %%cdy but owes on account ^ocd, and on note ^Icd. What is the sum of his effects ? Or what is the sum of +100cc?, —Ibcd, -\-Scd, -{-2cd, —bed, and —Kcd? Ans. He is worth $Sdcd. {-}-H'3cd.) OS, Cor. 2. — The sum of two quantities, the one positive and the other negative, is the numerical difference, with the sign of the greater prefixed. GO, Cor. 3. — It appears that addition in mathematics does not always imply increase. Whether a quantity is increased or diminished by adding another to it depends upon the relative nature of the two quantities. If they both tend to the same end, the result is an increase in that direction. If they tend to oppo- site ends, the result is a diminution of the greater by the less. 70, Prop. 2. — Dissimilar terms cannot he united into one by addition, but the operation of adding is represented by writing them in succession, the positive terms being preceded by the + sign and the negative by the — sign. 32 . FUNDAMENTAL RULES. Demonstration. — Let it be required to add +4cy', -\-Mb, —2xt/, and —mn. Acy"^ is 4 times cy^, and Sal is 3 times ab, a diflferent quantity from cy^ ; the sum will, therefore, not be 7 times, nor, so far as we can tell, any number of times, cy'^ or" ab, or any other quantity, and we can only represent the addition thus : 4:Ci/^ + Sab. In like manner, to add to this sum —2xy we can only represent the addition, as 4:cy^ + Sab + (— 2xy). But since 2xy is negative, it tends to destroy the positive quantities and will take out of them 2xy. Hence the result will be A:mf + Sdb—2xy. The effect of —mn will be the same in kind as that of —2xy^ and hence the total sum wdll be 4:cy^ + dab—2xy—mn. As a similar course of reasoning can be applied to any case, the truth of the proposition appears. Q. E. D. Scholium. — In such an expression as 4:cy^ + dab—2xy — mn, the — sign before the um does not signify that it is to be taken from the immediately preceding quantity ; nor is this the signification of any of the signs. But the quantities having the — sign are consid- ered as operating to take away so much from atiy which may have the + sign, and vice versa. EXAMPLES. Ex. 1. Add together bax, —lOcy, Sb, and —n. IVIodei Solution. 5ax and lOcy being dissimilar will not unite into one term, since one is 5 times ax, and the other is 10 times cy, a different quantity ; therefore I can only rej^resent the addition, as bax-\- {—\Qcy). But the lOcy being negative tends to offset positive quantities, and will take out of such its own value. Hence bax + (— lOcy) is ^ax — lOcy. To this adding 8&, which is positive and hence will go to increase the result, I have ^ax — lOcy + 8&. Finally, as n is negative it diminishes the result by its numerical value, and I have for the sum ^ax—lQcy + Sb—n. 2. Add together 4«m, — 2c%, —8x, and 5Jn, explaining as above. 3. Add — 2c2m*, 4:cm, — 6cW, and lOc^m, explaining as before. ADDITION. 88 4. Add la'^, — 32ir"^, Qmn, and —6a^, explaining as before, and find the numerical value of the result if a = 3, J = 18, a; = 8, m = 2, and n = 5. Result, 21. 7 J. Cor. — Adding a ner/aiive quantity is the same as sub- tracting a numerically equal positive quantity ; that is, m + (— n) is m — n. 7*^. Prob. — To add Polynomials. Rule. — /. Write the polijnoniials so that similar terins shall fall in the same coluimi. II. Combine each column into one term, and write the result underneath luith its own sign. TJie polynomial thus found is the sum sought. Demonstration. — As the object is to combine the quantities into the fewest terms, it is a matter of convenience to write similar terms in the same column, as such, and only such, can be united into one. (66, "70.) Now, since in polynomials the plus and minus terms Ktand in the same relation to each other as positive and negative quantities (57), they may be considered as such, and united by 67. The partial sums will then be dissimilar terms and will be added by connecting them with their own signs (70). Q. e. d. EXAMPLES. Ex.1. Add together 16ac—'2m-{-xy, 'dm — 5xy—d—2aCf '-Sxy—4:ac—0m, and 2mn—3ac-\-Sxy. Model Solution. — Writing the first polynomial as it stands, I arrange the others so that 16ac — 2m + xy similar terms shall stand — 2ac + 3m — 6xy — d in the same column, for — 4ac — 6m — 3xy convenience in uniting — 5^^; + ^xy + 2mn ti^^.,^ rpj^^j.^ ^^^^^ ^^^ lac,— 5m, + '^'y?— d,-^ 2mn term similar to ^%mn I lac — 5m 4- xy — d -\- 2mn bring it down, and in like 34 FUNDAMENTAL RULB8. manner — d. 8a^ and —3xi/ are 5xt/. 5x1/ and —5xy are 0. and ity, or simply -{-xy, is the sum of the similar terms in xy. Writing this result I pass to the next column. —Qm and +3m are —Zm. — dm and —2m are —5m which being the sum of the similar terras in m, is written down In like manner the sura of the terms in ac is lac. The partial sums are, therefore, 7ac, —5m, +xi/, —d, and + 2wn. But these being dissimilar terms are added by connecting them with their own signs (70) ; whence, the sum of the several polynomials is '7ac—5m-\-xy—d + 2mn. In like manner solve and explain the following : 2. Add Qx + 5at/, ^3x-\-2ay, x—6ay, 2x-{-ay. Sum, Qx-\-2(ty, 3. Add day— 7, —ay*-\-8, 2ay—9, —Say— 11, and lOay —13. Sum, Hay— 32. 4. Add —Sab + lx, dab—lOx, 3ab—6x, —ab-\-9x, and 2aJ)-\-^x, Sum, 4:ab-^4:X. 5. Add —6a^-^2I), —3d + 2a% -ba^-Sh, 4:a^—2b, and 95— 3a2. Sum, —8a^—2b. 6. Add SaW — 7a¥ + ^axy, — 7aW — 2a¥ — axy, a¥ —laxy + 8am, —l{)a¥-[.a^^+3axy, and —5a^b^-\-\8a¥. Sum, 0. 7. Add ldax^—Uy^-\-3a(^—mn^, 4.ax^-{-15y^—3a% ^y"^ — 17ax^-^2ac^-j-2m^n-{-3mn^, and lOy^—a^c. Sum, 15y^-^6ac^-]-2mn^—4:ah-^2m^n. 8. Find the sum of 2a^-\-4:l^x—c^x% 2(^x^-{-^a^—6b% SLnd21^x—4:C^x^-]-2aK Sum, Sa^-Sc^a^. 9. Find the sum of 8a^x^ — 3xy, 5ax—5xy, 9xy—5ax, 2a^x^ + xy, siiad 6ax—3xy. Sum, lOa^x^-^bax—xy. 10. Find the sum of 2bx—12, 3x^—2bx, bx^—3Vx, Wx + 12, x^-\-3, and ox^—lVx, Sum, Ux^-Wx + 3. 11. What is the sum of 20a^c^x-{-16ah—15a^(^x—23ah ? Ans., 5a^(^x—Sah. ADDITION. 35 12. What is the sum of 16x?/ — Um + 17xy -f- Wi7n? Ans.y 33xy + Sdhm. 13. What is the sum of 5c* — 4dan + 10c* f Uan ? Ans., 15c* — 36an. 14. Wliat is the sum of lOx^y^ — 176^- + l^x^y^ + bsk — 4a:V^? Ans., ^Ix^y^ — nsk. Scholium I. — In practice, the expert will not take the trouble to arrange the polynomials, but will simply select and combine the similar terms, writing each result in the total sum, at once. Thus, in solving Ex. 7, when the object is simply to find the sum, and not, as above, to explain the jjrocess, we proceed as follows : Noting the 13aa;^, we cast the eye along till we find the similar term, +4aa!^, and say "4-17aa;'';" again, casting the eye along till we find — 17aa;', we say "0." Therefore nothing is written in the sum for these terms, as they mutually destroy each other. Again, looking to — 14y^, and then on to I5y'\ we say "y''," and, passing on to + 4y', say "oy'V' ^^^ again passing on to lOy^ and say " 15y ." This being the sum of the terms in y^, it is then written in the answer. In the same manner the work is carried on to completion ; /. e., only naming results^ and writing them in the total sum. In this manner, write the answers in the following ex- amples : 15. Add 'dx — 6y -\- 4c, 2x — 2y — 3c, and —x-\-dy-i-c. Stim, ix — 4y + '2c. 16. Add lla-j- i3x — 7d, 4:a — 10x — 2d, and —9a — z -\- 3d. Sum, 6a -{- 2x — iSd. 17. Add dax — 4tby + 2?nn — 16, 2by — 57nn + 11, 'd7n7i — 2ax + 5, and — by -\- m7i — ax. Su77i, rriTi — Sdy. 18. Add 6a7nx — 3b + ^cxy — 2ax^, ib — 3cxy + bax^ llcxy —xy — SaTUx — 3ax^, and bxy — 6 — b. SuTTif 12cxy — 2amx -f ^xy — 0, 36 FUNDAMENTAL RULES. 19. Add '^xy — 2x^, 3x^ + xy, x^ + xy, and 4:X^ — dyx. 20. Add 2ax — 30, Sa^ — 2ax, 5x^ — 3x^, and SV^+IO. Sum, 8a;2 — 20. 21. Add Sd'x^ — 3«a;, 7 ax — 6xy, 9xy — bax, and 2d\x^ 22. Add ^ax^-\-bVx, —2axi^6xK 3aa^—10x^y —Hax^ 4- 'SVx, and «a;2 -f 11 Va;. 23. Add 6a;y — 12a;2, — 4ii;2 + Sxy, ^x^ — 2xy, and — 3xy + 4a;2. 24. Add 4:ax — 130 + dx^, bx"^ + dax + 9^2, 7:cy — Wx + 90, and V^ + 40 - 6a:2. 25. Add 3r?-2 + Uc — e^ + 10, - 5«-2 + 6bc-\-2e' - 15, and _ 4a-2 — 9^c — lOe^ + 21. 26. Add la—bif, ^^/x + 2a, by^—^/x, and ^^a-^-l^/x together. Sum, iWx. 27. Add 4:m7i-^3ab—4:C, 3x—4:ab -\- 2mn, and Sm^ — Ap together. Sum, 6mn — ad — 4:C i- 'Sx -{- 3m^ — 4^. 28. Find the sum of 3d^ + 2ab + 4:b% ba^ - Sab + ^, — ^2 _|- 5ab — b\ 18a2 _ 20a5 — 19^*2, and 14a2 _ Zab + 20^2. 29. Find the sum of ^x^ — ba^ — bax^ + (^a^^, O^^ + S^c^ + 4:ax^ + 2a^x, — llia^ + 19aa;2 — lba% 13ax^ — 27a^x + 18^3, Sa^x — 20a^ + 12a:3^ and SW^x — 2x^ — dlax- — 7x^ Sum, — W — a^ 30. Find the sum of 2ab + 12 — x% x^y + xy + 10, dxy^ + 2x^y — xy, bxy + 11 + xVy, and 17 — 2x^y — x^y. Sum, 2ab + 50 + bxy — x^y -\- 4a^\^y. ADDITION. 87 31. Add x^ + ax—ab, ab—Vx-h^^y, ax-\'Xy—^ab, x^ -{-Vx—x, and xy + xy-\-ax. Sum, 2x^ 4- Sax — ^ab + ^xy — x. 32. Add 7a;i?/— 2a:V^-|-7, Vxy + Sxy^-^2, dyVx—Vyx —6, 9?/V^— 4i/*a;— 3, and l-\-7xy^—2yxK Sum, l8xhj-{-3xy^ + l. Scholium 2. — The object and process of addition, as now ex- plained, will be seen to be identical with the same in Arithmetic, except what grows out of the notation, and the consideration of positive and negative quantities. For example, in the decimal notation let it be required to add 248, 10506, 5003, 81, and 106. The units in the several numbers are similar terms, and hence are combined into one : bo also of the tens, and of the hundreds. To make this still more evident, let u stand for units, t for tens, h for hundreds, th for thousands, and t.th for ten thousands. 248 is then 2h + 4t + Su, 10506 is lt.th-\-5h + Qu, 600S is 5^^ + 3m, 81 is 8^ + lt^, and 106 is lh-\-Qu. Writing these so that similar terras shall fall in the same column, we have the arrangement in the margin. Whence, adding, we get the sum. The process of carrying has no analogy in the literal notation, since the relative values of the terms are not supposed to be known. 2h -{- 4:t -\- 8u Again, there is nothing usually ILt/i -j- 5h -\- 6u found in the decimal addition 5^^ +3?^ like positive and negative quan- ^^ i~ i^ titles. With these two excep- tions the processes are csseu- U -f 6w It.ih -\- 5th -f 9/i -f- 4^ + 4w tially the same. The same may 25944 be said of addition of compound numbers. 73. Prop. 3. — Literal terms, which are similar only with respect to part of their factors, may be united into one term with a polynomial coefficient. Demonstration. — Let it be required to add 5ax, —2cx, and 2mx. These terms arc similar, only with respect to x, and we may say 5a 38 FUNDAMENTAL RULES. times X and —2c times x make (5a— 2c) times x, or (5a—2c)x. And then, 5a— 2c times x and 2m times x make (5a— 2c + 2m) times x, or (5a— 2c + 2m)a?. Q. E. D. EXAMPLES. Ex. 1. Add ax^, —hx^, —2cx^, and 4:mx^, with respect to x^. Sum, (a—h—2c-\-4:m)x^. 2. Add 4:xy, Saxy, —lOmxy, and cxy, with respect toa;^. >S'wm, (44-3a— 10m-f-c)a;?/. 3. Add amx-\-2di/, 2cx~'ddy, and 3dx-\-5y, with respect to a; and ^. /S'?c + |«-2a;-t. 8. Add 2x^y^ + 2x-^y^—2x^, — 5sx^y^ + 2r^x-^y^ — a + 6a;8, 3a;y— 2^»2a;-»»yt4.3«, and — 2a^ + c«/. 74. Prop. 4. — Terms which have a common com- pound, or polynoinial factor, may he regarded as similar and added with respect to that factor. Demonstration. 5{x^—y^), 2{x^—y^) and — 8(2;'-^'^) make, when added with respect to {x^—y"^), 4:{x^—y'^), for they are 5 + 2—3, or 4 times the same quantity (x'^—y^). In a similar manner we may reason on other cases, q. e. j>. ADDITION. 39 EXAMPLES. Ex. 1. Find the sum of Vx-^n—dx^, 5a^—2 Vx-i-n, SVx-i-n — Hx^, and 10x^+8 Vx^i. Sum, 10 Vx -h w + 6x^. 2. Find the sum of 6a— 6(a— J) + 7, 3a + 12(a—b)—8, and 2(a—b)— 3a— 20. Sum, Ga— 21-^8 [a— i). 3. Find the sum of 7(7n + 3)— 16 (m— 3), 8(w + 3) + 7(m— 3), and 3 (m— 3)— 4 (m + 3). Sum, ll(m + 3) — 6(m— 3). 4. Find the sum of i\/a^Sh—i\/a—3h + i\/a^^^ -^\^a-3b. Sum, ^^a-U. 5. Add 3(a-c)ix-^f)-i, — ^-~ — L==, and (x-{-y^P Vx-\-y^ 6(a + c) (.T+^2)-i. g^^^ 3(3a + g) 6. Find the sum of a(a-\-h)-\-3 va—x, — 4a(a-}-5) + 1a{a~x)^, —GaVa — x-\- lla {a + h), —2a{a + h) —2{a—x)^, and 5a (a + J) +14 a/^— a;. 7. Add a v^— y + ^a:^ + c(a + a;)2, _ Ja;i/ + (a + c)(a + ic)2 + (a:— y)i, 2ia;y + (a— 1) Vx—y—a {a+xf. Sum, 2a (x—y)^ + 2bxy + 2c (a + a;)^. 8. Show that Vx + hy -\- ax — z-\-amy-j-c Vx -^dz-\-y = (am-^b-\-l)y-\-(c + l)x^-\-{d—l)z-{-ax. Test Questions. — Does addition always imply an increase? When does it not? When does it? What is addition? How are similar terms added ? How are dissimilar added ? Give the Rule for adding polynomials and demonstrate it ? 40 FUNDAMENTAL RULEU. O Q a < DEFS. SYNOPSIS r Addition. [Sum, or Amount. r I. Similar Terms, dem. PROPS. Cor. l.—Siffn of sum. Cor. 2. —Sum of Positive and Negative. Cor. 3 —Addition not al ways incj'ease. { Sch.— Sign of term. I Cot.— Addition of Nega- ] tive - Subtraction of i Podtive. , Sch. 1.— Practical method. PROB. To add Polynomials, rule. Dem. -! Sch. 2.— same as in Led- ^ 2. Dissimilar Terms, dem. PROPS \ 3. Terms Partially Similar. i 4. Compound Similar Terms. I mal Notation. 75. Subtraction is, primarily, the process of taking a less quantity from a greater. In an enlarged sense, Subtraction comes to mean taking one quantity from another irrespective of their magnitudes. Subtraction also comprehends all processes of finding the difference between quantities. The terms Minuend, Subtrahend, and Remainder, are used as in Arithmetic. 40 BtJBTRACTION. 41 76. The Difiference between two quantities is, in its primary signification, the number of units which lie between them ; or, ii is what must he added to one in order to produce the other. When it is required to take one quantity from another, t!ie difference is what must be added to the Subtrahend in order to produce the Minuend. Scholium. — The most compreliensive and fundamental notion of difference is this : Having reached any Bpecified point in a scale of numbers, or in estimating magnitude, in what direction, and how far must we pass to reach another specified point in the scale of numbers, or in the value of the magnitude. Illustration. — When we ask, "What is the difference between 3 and 8 ? " we ordinarily mean, " Over how many units must we pass in reci• Minuend, 5ax—Qh-Sd— ^m ) Difference, 3aa;—8& -1-2(^—1 3m If the three quantities included in the brace are added together, the sum will evidently be the minuend. If, therefore, we add the second and third of them (that is the subtrahend, with its signs changed, and the minuend) together, the sum will be what is necessary to be added to the subtrahend to produce the minuend, and hence is the difference sought, q. e. d. SUBTRACTION. 43 EXAMPLES. Ex. 1. From 4:ab-\-3c^—xy subtract 3xy—2z-\-2ab—c^. Model Solution.— Writing the subtrahend under the minuend so at similar terms shall of combination, I have that similar terms shall fall under each other, for the convenience 4a5 + 3c' — xi/ 2ah— c" + 3a^ — 2e 2ab + 4c' — ixy + 2z The difference sought may be considered as consisting of two parts: 1st, the subtrahcml with its signs changed, and 2nd, the minuend itself. The sum of these two parts is the difference, since it is what is necessary to be added to the subtrahend to produce the minuend. Therefore conceiving the signs of the subtrahend to be changed, and adding, I have 2aJ>+4:c^—4:xy-\-2z, as the difference sought. A more detailed Explanation is as follows: The first question is, What must be added to —2z to produce the corresponding term in the minuend? This is evidently +2z, as there is no corresponding term in the minuend, and I have only to write a term in the differ- ence which will destroy —2z when added to it. Passing to the next term, I inquire. What must be added to + 3a^ to produce —xyl First, I must add —Sxi/ (the term with its sign changed) in order to destroy the + Sxy, and then I must add —xy in order to make the —xy of the minuend. So in all I must add —Sxy and —xy, or —ixy. —^xy is, therefore, the difference sought. Passing to the next term. What must be added to — c' to make +3c'? I must add +c' (to destroy — c') and +3c' (to make up the required term in the minuend), or in all +c' and +3c', or +4c' ; which is the difference. Finally, the term to be added to 2oib in order to make kcib is composed of the two parts —2ab and +4«5, which make 2a5. It thus appears that 2a5 + 4c'-4icy + 23 is the difference, since it is what m ust he added to tfie subtrahend to produce the minuend. 2. From ^— 311^-^ %xy -^11 take 2x-\-l^^-'2xy—b. Rem., 22'-4Z.2_G. 44 FUNDAMENTAL RULES. 3. From 15fla;-f2%—6a^a;2 take — 5«a;— 4%— 3aW Rem., 20«^-f G%— 3a2a;a. 4. From 8«^J-2_3^^_^15_2v'^ take 8 + 10a:z/— 3tt^J-2 i?em., llflV^— 13a;?/4-7— 4Vic^. 5. From ia^x^—'dc^l'^hhj^ take 8%i4-4a2.Ti_2d 6. From «2—2aZ> + ^>2 take ^2_j_2aJ + ^>2. ^em., — 4ai. 7. From a2_j2 take a^—'lab—W. Rem.y 2ab. 8. From 24a:/— 14m?/ + 18a;y— 14 + 27a;;22 take 17a;/ — 10my—4.xy-{-20xz^—S. 9. From 17pmx^ — ISn^ + lOm* — 24 take 7pmx^—An^ + 10m4— 17. 10. Yrom a^-^2abi-i^ take a^—2ab-^ hi 11. Froma3+3a2^,4-3«JH^nakefl3_3^2j_^3«j2_53. 1 2. From a:^ + 2a:%^ -\- yi take a:t — 2x^y^ + ^i 13. From eVxTy + ^a^^ take 3 (.t + «/)^— 46?2a;2. i>?J., SVx + y + SaHl Suggestion. — Regard \/x + y as one quantity, and observe that 'J (x + y)^ is the same as 3 '\/x + y. 14. From 6a;— 3 Vxy 4- 17 (a + b) take 2x + 1l Vxy + 4 («+*). />^y.^ 4a:-10A/.^ + 13(« + ^). 15. From 10— a-\-b—c-]-x take ft + i— c— a;— 90. 16. From 6m + i^n— frz — 1 take 2?7i— 4w + -i« — 2. Diff; "km-^n — a + 1. SUBTRACTION. 45 Scholium I. — When several polynoniials are to be combined, some by addition and some by subtraction, it will be found expe- dient to write them so that similar terms will fall. under each other, writing the several subtrahends with their signs changed, and then add the quantities as they stand. 17. From the sum of 2ay'^—2b-\'3ax, 2ay^ + b—aXy and 'Sb—ay^+d—y, subtract 3ay--{-ib--ax-\'i/. Resu It, — 2i + 'dax -{-'6—'Zy. 18. From the sum of 'da^-\-xy^—2by, 6a^-\-3xy^^—3byy and 3xy^-\-4:a^-{-by, subtract 2a^^xy^—5by + 5. Result, l0a^-{-8xf-{-by—b. 19. From the sum of a^b^—3y'^ + 6xy, 3a^l^-\-3y^—2xy, and 5^' + 3aW—xy -f- 6, subtract a^b'^ -f xy—y^. Result, QaW + ^y^+xy^^. 20. To 5aa;2_7c^»-f 8mi— 2c-^ add 3cb—4:C-^ + %ax'^, then subtract IQm^ —bxy ■\-3cb—l)lc-^, add Qm^—l\cb + 3n —iax^, subtract —16ax^—2m^ + 4iCb, add 5m^-|-6c-^ and subtract 4:Xy—2n. Result, ldax^—22cb + llmi + 12c-"' ^-xy-\- 5n. 21. To lcy-^-\-Sax—bb, add ^b—2cy"^+m, then sub- tract 5a2;— 4m + 3 and —3ax-\-bcy~^—Q, add 10ax—2h 4- 8m — 3, and subtract 3m—10cy~^—2m. Result, li)cy~^^-^\Qax—3b + l2m. Scholium 2. — The proficient solves such examples as the above, and, in fact, all kindred ones, without re-writing the quantities. Thus, to obtain the result in Ex. 20, he looks at Saa:", casts his eye along till he sees 2aar'', says lax^ ; then looks forward to — 4aa;', and says Zax^ ; then notices —IQax^ in a subtrahend, and hence tliinks of it as -f IGtfa?^, and says l^ax^. Again taking up — 7e5, he looks along noticing the similar terms and says, mentally, — 4cJ, — 7c&, — 18c&, — 22c&. Again, he takes up Ba/^I, and running through with the similar terms, says, — 2/ni, 4;»i, 6mi, Umi. And so for all the other terms. 46 FUNDAMENTAL RULES. This is the prcbctical way of solving such examples ; and the pupil should exercise himself till he can write out the result ai once. He can go over the preceding examples thus, and should also solve, with- out writing, those which follow. 22. From ldx^—2ax—W take hx^—'lax—}^, 23. From 20ax —^x^M take ^ax + 5a;^ — d. 24. From hab^W—c-\-hc-h take I^—^ab + bc. 25. From ax-^—ax-'^-\-cx—d take a^—ax-^^—ex—'M. Diff., a (x-^- x^) J^{c + e)x + d. Suggestion. — This would at first become ax-^—ax^-\-cx-\-ex-\-d But this may evidently be written as above. 26. From x'^y—^ ^/xy—^ay take 3:^:2^ + 3 (xy^—^ay. 27. From the sum of 4o5a;— 150-f-4a;^, hx^-\-Zax^V)^x, and 90— 2«a;— 12\/ic; take the sum of %ax—'^^-\-'^x^, 7a;i_8«a;— 70, and 30— 4a:^, — 2a:2 + 4ttV. Result, llfl!2; + 60— rri— 4a2ic8. 78. Cor. 1. — When a parenthesis, or any symbol of like signification (50), occurs in a polynomial, jrreceded by a — sign, and the parenthesis or equivalent symbol is removed, the signs of all the terms which were within must be changed, since the sign — indicates that the quantity within the parenthesis is a subtrahend. 28. Eemove the parenthesis from the polynomial 3a^x -\-2i^y^ — {oa^x — 2mh-\-8b'^y^) and represent the result in its simplest form. Do the work mentally, writing only the result. EesuU, 2m'^z—2ah-eby. 29. Remove the parenthesis from ba—^b-\-oc—{—3a '\-2b—c). EesuU, 8a— 6^4- 4c. SUBTRACTION. 47 30. Remove the parenthesis from 4:a — 5x — {a — 4:x) -i-(x—Sa). Result^ —5a, Queries. — In Ex. 28 is the sign of 5a^x changed ? What is its sign as it stands in the parenthesis ? Is the — sign which appears in the result before 5a^x, the same as the one before the parenthesis in the example ? No, What became of that before the parenthe- sis? Ans. The operation which it indicated having been per- formed, it was dropped. In Ex. 30, why are not the signs of the terms in the last parenthesis changed ? 79. Cor. 2. — A parenthesis preceded hy the — sign can be placed around any number of terms by changing the signs of all the terms. The reaso7i of this is evident, since by removing the parenthesis according to the preceding corollary, the expression would return to its original form. 31. Introduce within a parenthesis the 3d, 4th and 5th terms of the following expression: 6ax—2cd—8m-{-ox—2y -fa;— 4«. Result, (jax—2cd—(8m—5x-^2y)-\-x—4:a. 32. Introduce within a parenthesis the last three terms of ixy-{-2cb—8x—5-^2b. Result, 4:xy-{-2cb—{Sx-j-5—2b). 33. Include in brackets the 3d, 4th, and 5th terms of 5ax—2a^-{-3x—12ay-{-16. Also the 4th and 5th. Also the 2d and 3d. Form of the first and last, bax—23^-\-{dx—12ay-\-\b), bax—{2x'^—3x) — 12ay-\-\6. Query.— In the last is the sign of 2a;' changed ? 34. Prove that (3x—Qy) + (4y— 4a:) -^ 2 {x + 2y) = x + 2y. 35. Prove that i(« + 5—c)4-i(^ + c—«) = b. 36. Prove that Qa—ib—2 (a + b) = 2 (2a— 3b). 4» FUNDAMENTAL RULES. 37. Prove that x^+ex^y-\-d — {x^-i-4:X^t/-{.l)z=2(x^y-{-l). *38. Prove that i{a—5b-{-ic) i- i {5b — 2a -]- ^c) = ^c 39. Prove that ^ (a^—a-\- 1 ) + i (2a2_« + 2) = ^ (10«2 -7« + 10). 40. Prove that ^a + ib—{ia—}b) = b. 41. Prove that i{9—15x)—i(l2-20x)—i{4:bx-20) = 1— 4.T. 42. Prove that i{a-ib + ic)-hi {a-ib+}c) = ^(90a —365 + 25c). 80, Cor. 3. — When sevei-al parentheses occur, included the one within the other, begin the removal with the inside one. 43. Remove the parentlieses and other marks of aggre- gation from ^a—\ — Yc—d-\-{4:X^—\)—xy'] — '6y\. Result, 4^ + SYNOPSIS FOR REVIEW. Subtraction. PEnN.T.ONS.J,"-*' _.«. j SCHOLIUM. I. uinerence. \ illustration, diagram. O GENERAL 1 j Sch. 1. -Both add. and Bub. <^ PROBLEM. |R"'e. »EM. 1 sch. g.-Practical method. F f Cor. I. — To remove. S BRACKETS. ] Cor. 2.— To introduce. 3 [ Cor. 3.— Several. TERMS PARTIALLY SIMILAR. Sch. 3. Test Questions.— What two answers can you give to the ques- tion, " What is the difference between 10 and 6 ? " Why do you change the signs of the subtrahend in subtracting? Why do you add the subtrahend, with signs changed, to the minuend ? When do you change the signs in removing a parenthesis ? Wliy ? What becomes of the sign before the brackets ? In removing a parenthe- sis preceded by a — sign, is the sign of the first term changed as well as the others ? State in the briefest manner the tlieory of sub- traction ? Rejjly. Subtraction is finding the diflFerence between quantities, that is, finding what must be added to one quantity to produce the other. This diflference may always be considered as 3 60 FUNDAMENTAL RULES. consisting of two parts, one of which destroys the subtrahend, anc the other part is the minuend itself. Hence, to perform subtrac- tion, we change the signs of the subtrahend to get that part of the difference which destroys the subtrahend, and add this result to *he minuend, which is the other part of the difference.' JSL icatinq^ 81, Multiplication is the process of finding the sim- plest expression for a quantity which shall be as many times a given quantity, or such a part of that quantity, as is represented by a specified number. The quantity to be multiplied is called the Multiplicand. The number by which we multiply is called the Multiplier. Taken together, the multiplier and multiplicand are called Factors. The result is the Product, 82, Cor. 1. — The multiplier must ahvays he conceived as an abstract yiumber, since it shows how many times the mul- tiplicand is to be taken. Thus, to propose to multiply $13 by $5 is absurd. We can under- stand that 5 times $12 is $60 ; but what is meant by 5 dollars times? 83, Cor. 2. — TJie product is always of the same kind as the multiplicand. 84, Scholium. — It is frequently convenient in practice to speak of the multiplier as positive or negative, although, literally under- stood, this is a contradiction of Cor. 1, which requires the multi- plier to be conceived as mere number. In a strict analysis, the MULTIPLICATIOlf. 51 multiplier in such cases is to be cousidered, first, without reference to its sign, i. «., as abstract, and then the sign is to be interpreted as indicating what is to be done with the product, when it is taken in connection with other quantities. So. Prop. 1. — The product of several factors is the same in whatever order they are taJcen. Demonstration. — 1st. ax6, is a taken & times, or a-\-a + a-^a + a to b terms. Now, if we take 1 unit from each term (each a), we shall get h units ; and this process can be repeated a times, giving a times 6, or 6 x a. .*. axb = bx a. 2nd. When there are more than two factors, as abc. We have shown that ab = ba. Now call this product /», whence abc = mc. But by part 1st, mc = cm. .*. abc = bac = cab = cba. In like man- ner we may show that the product of any number of factors is the same in whatever order they are taken, q. e. d. Scholium- — If the multiplicand is concrete, the reasoning is still the same. Thus $a x & = $a + $a + $« + |a, - etc. to b terms. Now take $1 from each of the terms of %a each, and we have %b ; and this process can be repeated a times, giving %b x a. .-. |a X ft = %bxa. Notice that in each case the multiplier is abstract. SG, Prop. 2. — When two factors have the same sign their prod^uet is positive ; when they have d,if- ferent signs their product is negative. Demonstration.— 1st. Let the«factoi*s be +a and +&. Consider- ing a as the multiplier, we are to take +&, atimes, which gives +«ft, a being considered as abstract in the operation, and the product, + aft, being of the same kind as the multiplicand ; that is, positive. Now, when the product, + a&, is taken in connection with other quantities, the sign + of the multiplier, a, shows that it is to be added; that is, written with its sign unchanged. .". ( + Z»)x( + a) = -\-ab. 2nd. Let the factors be —a and —b. Considering a as the mul- tiplier, we are to take —5, a times, which gives —ab, a being con- sidered as abstract in the operation, and the product, —ab, being of the same kind as the multiplicand ; that is, negative. Now, when this product, —aft, is taken in connection with other quantities, the 52 FUNDAMENTAL RULES. sign — of the multiplier shows that it is to be subtracted; that is, written with its sign changed. .-. (—5) x (—a) = +a6. 3d. Let the factors he —a and +5. Considering a as the multi- plier, we are to take +5, a times, which gives -\-ab, a being consid- ered as abstract in the operation, and the product, +ab, being of the same kind as the multiplicand; that is, positive. Now, when this product, -\-ab, is taken in connection with other quantities, the sign — of the multiplier shows that it is to be subtracted; that is, written with its sign changed. .-. (-I-&) x {—a) = —ah. 4th. Let the factors be +a and — &. Considering a as the mul- tiplier, we are to take —5, a times, which gives —ab, a being considered as abstract in the operation, and the product, —aft, being of the same kind as the multiplicand ; that is, negative. Now, when this product, —db, is taken in connection M'ith other quantities, the sign + of the multiplier shows that it is to be added; that is, written with its own sign. .-. (—5) x( + a) = —ah. Q. E. D. 87. Cor. 1. — The product of any number of positive fac- tors is positive. Thus (+a) X { + h) X ( + c) x ( + d) — ahcd, since ( + a) x (-f &) = +ah, which, in turn multiplied by -\-c, gives i-dbc, etc. 88, Cor. 2. — The product of an even number of negative factors is positive ; since we can multiply them two and two, thus obtaining positive products, which positive products multi- plied together make the complete product positive. Thus (-a) X i-h) x {-(^ x{—d)x (-e) x (— /) = ( + ab) x( + cd) X ( -1- ef) = + ahcdef or ahcdef. 89* Cor. 3. — The product of an odd number of negative factors is negative ; since, by the last corollary, the product of all but one (an even number) of such factors is positive, and then this multiplied by the remaining negative factor gives ( -f-) X ( — ), and hence is negative. 90, Prop. 3. — The product of two or more factors consisting of the same quantity affected with expo- nents, is the com^mon quantity with an exponent MULTIPLICATIOK. 53 equal to the sum of the exponents of the factors. That is, rt*^* X a'* = «''*'■"; or ft'^^-a'^.a* == a^+"+*, etc., whether the exponents are integral or fractional, positive or negative. Demonstration. — 1st. When the exponents are positive integers. Let it be required to multiply a^ by a", a* = aaa, and a^ = aa. .'. a^xa'^ = cum • aa = a'. That is, there are three factors each a, in u^, and tico like factors in a- ; and, as the product consists of all the factors in both multiplier and multiplicand, it will contain Jke factors each a, and hence is a\ In general : To multiply a'" by a" and a*, a*" = a/m to m factors, ft" = aaaaa to n factors, and a* -— aaaaa to « factors. Hence the product, being composed of all the factors in the quantities to be multiplied together, contains m -\-n-\-s factors each a, which is represented a'"+"+». Since it is evident that this reasoning can be extended to any number of factors; the proposition is proved in ^e case of positive integral exponents. 2nd. When the exponents are positive fractions. Let it be required to multiply 64^ by 64^ Now 64^ = 4-4, /. «., 2 of the 3 equal factors which make 64. In like manner 64^ is 2 ■ 2 • 2 2 • 2. And since 4-4 is 2 ^^ 2- 2, 64^ x 64^ = 2-2- 2- 2 x 2- 2- 2 2- 2 = 2», or 9 of the 6 equal factors into which 64 is resolvable, and may be represented 64« .-. 64tx64t = 2" = 64^ = 64t + t. m em In general, let a» be multiplied by d^. a* means m of the n equal factors composing a. Now, if each of these n factors be resolved into h factors, a will be resolved into hn factors, and to make the quantity a» we shall have to take hm instead of m factors. >H hm Hence a"* = a^. In like manner a^ may be shown equal to c* m c hm en a^ ; whence a" x a^ = a^ x a^. This now signifies that a is to be resolved into hn factors, and hn + em of them taken to form the me bm en bm + en me product. . •. a*xa^ = a'^xc^ = a"~J»", or a » ^. Finally, as it is evident that this reasoning can be extended to any number of fac- tors, the proposition is proved in the case of positive fractional exponents. 54 FUKDAMENTAL RULES. 3d. When the exponents are negative. Let it be required to tntilti- ply 2- by 2- 2- is 1, and 2"^ is t .-. 2^x2- = \,>^\, 1 « . = - , or 2-^ 2^' In general, «-™ x «-" = a-^-^. For a-"* = — , and a-" = (43). Whence «-'" x a-» = — x — = = ^— by the pra- ceding parts of this demonstration; and by (43) = «-'»-». This reasoning may also be extended to any number of factors. Q. E. D. EXAMPLES. Ex. 1. Prove as above that 81^x81^ = 81^ and that 8lV- =: 8ll 2. Prove that m<^ x m* = m<^+*. ^ 3. Prove that 16"^ x 16"^ = 16"^. 4. Prove that 25"^ x 25* is 1. 6. Prove that a~^ x a^ is a. Scholium. — The student must be careful to notice the difference between the signification of a fraction used as an exponent^ and its common signification. Thus f used as an exponent signifies that a number is resolved into 3 equal /actors, and the product of 2 of them taken ; whereas f used as a common fraction signifies that a quantity is to be separated into 3 equal pa/rts^ and the sum of two of them taken. 91. Prob. — To multiply Monomials. Rule. — Multiply the numerical coefficients, and to this product affix the letters of all the factors, affecting each with an exponent equal to the sum of all the exponents of that letter in all the factors. MULTIPLIOATIOK. 56 The sign of the product will he + except when there is an odd nwinher of negative factors; in which case it will he —. Demonstration. — Tliis rule is but an application of the preced- ing principles. Since the product is comi^osed of all the factors oi' the given factors, and the order of arrangement of the factors in the? product does not affect its value, we can write the product, putting the continued product of the numerical factors first, and then grouping the literal factors so that like letters shall come together. Finally, performing the operations indicated, by multiplying the numerical factors as in the decimal notation, and the like literal factors by adding the exponents, the product is completed. EXAMPLES. Ex. 1. Multiply together Sa^bx, 'Zch^y, and 5ach!, Model Solution.— Since the product must contain all the factors of the given factors, and the order of arrangement is immaterial (85), 'da^hx X 2cb^y x 5ac^x = 3 • 2 • 5 x a^a xW xcc^ xxxxy, which, by performing the operations indicated, becomes 30a^&Wy. 2. Multiply together Sabxy, 2a^ba:^f lOcx^, 4?/^, and a. Prod., UQxi^hx^if. 3. Multiply together bax^, —'^bij, —3a^c, ax^, and —2y\ Prod., —eOa^bcx^f. 4. Multiply together mz'*, 2rtvh?, —3ax^, — 5m«, and 4ta^x^. Prod.y 120a%3+*:c5+r+n 6. What is the product of —2d^d^ —I0ac% —d% — -ia'wx", and — c^. Ans., —SOa^^h'>^^'^-^^d^xi^-^K 6. Multiply 3a;i by 2a;i Prod., 6a:i 7. Multiply 60at by Sa*. Prod,, 480«i 8. Multiply 3a^b^ by —laW' Prod., —2\aUh 66 J-tJNDAMENTAL RULES. 9. Multiply 6ah^ by '^a^-^c. Prod., 35ah. 10. Multiply 10a2 by 3a~^. Prod., 30. 1 . - _ , H-2" 11. Multiply Sy"" by 62/2. p^od., ISy 12. Multiply 13a;^ by 5ic^. Prod., 65a: ««. 13. Multiply — 50a^ by —4ta^. Prod., 200ak 14. Multiply 309a^/;^ by 9«'^^>^. Pro^., 2'7Sla^b^+'^. 15. Multiply —6x-^y-'^ by — Sa:^^^. Pro^., ISa:^/. 16. Multiply x^ by ic~^. Prorf., a;i 17. Multiply .^ by ar-^. Prod., 1. 18. Multiply «2j-c by a^bc. Prod., a\ 19. Multiply Va^ by «%. Prod, a^x^. 20. Multiply v^2 by ^a\ Prod., a'^\ 92. Prob. — To multiply two factors together when one or both are polynomials. Rule. — Multiply each term of the multiplicand hy each term of the multiplier, and add the products. Demonstration. — Thus, if any quantity is to be multiplied by a + &— c, if we take it a times (/. e. multiply by a), then & times, and add the results, we have taken it a + 5 times. But this is taking it c too many times, as the multiplier required it to be taken fl^ + J minus c times. Hence we must multiply by c, and subtract this product from the sum of the other two. Now to subtract this pro- duct is simply to add it with its signs changed (YV). But, giving the — sign of c its effect as we multiply, will change the signs of the product, and we can add the partial products as they stand. Q. E. D. MULTIPLICATION. 5? EXAMPLES. Ex. 1. Multiply 2a—Sb-\-4:C by 'Sa-\-2b—5c. Model Solution.— Writing the multiplier under the multiplicand, IS a matter of convenience, I have 2« — 35 + 4c da + 2b — 5c 6a« — Qab -\- 12ac + iab _ 6&' + 8&C — lOar, + 15bc - ZOc" 6a' -5ab + 2ac — 65' + 235c — 20c' Now taking the multiplicand 8a times, I have 6a'— 9a5+12ac. Taking it 26 times, I have 4«5— 65' + 85c. I have thus taken it too many times, by 5c times. Hence I am to take it 5c times, and then subtract this partial product from the others. Therefore I multiply by 5c, and change the signs as I proceed, and finally add the three partial products. I thus obtain 3a + 26— 5c times the multiplicand. 2. Multiply x-\-y by j'-^j/. Prod., 7^-\-2xy + i/. 3. Multiply 5a;-|-4y by Sx—2y. Prod., 15a^-\-2xy—Sy\ 4. Multiply x^-}-xy—y^ by x—y. Prod., a^—2xy^+y\ 5. Multiply 2a^-Uy by 2^3-3^1 Prod., ^ac^ — 6bc^y — Qac'^y'^ + %f. 6. Multiply a^-i- 52^ ^2_^^_^^_^^ l^y ^_|_^_|_^, 7. Multiply «H3«2a; + 3«:r2 4-2:3 by a^—Za^x-Jr^a3?—7?. Prod., a«—3a*x^-\-3ah.^—xf^. 8. Multiply «2_^ by a^-\-b^. I n m 9. Multiply together (a + b), (a^-\-ab-\-l^), (a—b), and (a^^ab + l^). 68 FUNDAMENTAL RULES. Suggestion. — Perform this first by multiplying together I and II, and then III and IV, and taking the product of these products. 2d, By multiplying the product of I and III into the product of II and IV. 3d, By multiplying the product of I and IV by the pro duct of II and III. Result in each case, a^—¥. 10. Multiply a^—Q^ by 7n^—n\ Prod. , a^m^ — rn^x^ — ahi^ -\- n '^oi?, 11. Multiply 2a2+ 2a + 5 by a^—a. Prod., 2«4 + 3a2— 5«. 12. Multiply 2s;3^4^2_^8a; + 16 by 3a;— 6. Prod., 6^^4—96 13. Multiply fl^ -I- &—C by m—n. Prod., am — an-\-hm — h7i — cm-\-cn. 14. Multiply x:^-\-x^y'^-^y^ by x^—y^. Prod., x^—y^. 15. Multiply a;2— 4a; + 16 by x + b. Prod., a:3_f_a;8— 4a;H-80. 16. Multiply a^—c^y-\-c^y'^—ay^-\-y^ by a-\-y. Prod., a^-\-y\ 17. Multiply a;2— 50a;— 100 by a; + 2. Prod., a;3_48a;2_200a;— 200. 18. Multiply 2aj8 + 3a;— 1 by 2a;2— 3a;4-l. Prod., 4a;*— 9a;2 + 6a;— 1. 19. Multiply x^—h^ by x^ + h. Prod., x^—b^x^-^-bx^—iK 20. Multiply a^—h~^ by a^—b. Prod., d—a^-^—a^b + bi. 21. Multiply 2a-i-db^ by 2^-^ + 35^. Prod., 4a-i— 9^1 MULTIPLICATIOK. 5d 22. Multiply x^ + x^y-^-^-xiy-^ + y""^ by x^—y'^. Prod., x^—y~\ 23. Multiply (^■\-b"' by a^' + b**. Prod., a'«+^ + «"^>»»4-a'"Z>«4-^>'w+n. 24. Multiply a^a-f-ys by x^—yK Prod., x^—y\ 25. Multiply ?/i5 + ?/iW + w^ by w^ — w^. Prod., m^?i, 26. Multiply Sa'^-i— 2^>'^ by %a—W. Prod., 6a"»— 4aJ^-2— 9a'^i*2_|_6^n, 27. Multiply ^aP^Plr^P + a-'^PbP by da'^l^P—a^Pb-^P. Prod., 6a^-P -f- Sa'^^p j4p _ 2a^ + s/^^^-^Si) _ aPb-P. OS, Definition. — When an indicated operation is performed the expression is said to be expanded. 28. Expand (a + b){a—b); also (x + y) i'^'^ + y) ; also (x—a){3>^-\-ax-{-a^); also {?n + n)(7n-\-n) — {m — n)(m — n). Last result, 4mw. THREE IMPORTANT THEOREMS. 94, Theorem. — The square of the sum of two quantities is equal to the square of the first, plus twice the product of the two, plus the square of the second. Demonstration. — Let x be any one quantity and y any other. The sum is a; + y ; and the square is, the square of the first, x^, plus twice the product of the two, 2xy, plus the square of the second, y\ That is {x-\-yy =z x" ^2xy^-y\ For {x + yf ^ {x+y){x-\-y) which expanded becomes a;' + ^xy + y"^. Q. e. d. 60 I-UNBAMEKTAL RULES. EXAMPLES.' Ex. 1. What is the square of 2a^i-3x^? Model Solution. — This is the sum of the two quantities 2a' and 3a;^, and hence by the theorem the square equals the square of the first, 4a\ plus twice the product of the two, 12aV, plus the square of the second, 9x\ . •. {2a'' + Zxy = 4a* + l^a'x" + 9^*. Scholium. — The pupil should give mentally the squares of the following expressions : 2. Square 4a^ + 2a:. Resnlt, 16a + 16a^x-\-4:X^ 3. Square x~^-^xy. Result, x~^-\-^x^y-[-x?y^. 4. Square hrr^^-UW. Result, ^ha-^ + ^OaV^-^^a^K 5. Square ^h-'^ + ^x-^b-^^ Result, \a%'^-\-^ab-^x-^ + ^x-'ib-\ OS, Theorem. — The square of the difference of two quantities is equal to the square of the first, minus twice the product of the first by the second, plus the square of the second. Demonstration. — Let x and y be any two quantities. The dif- ference is x—y. Now {x—yy — (x—y) {x—y) which expanded gives x^ — 2xy + y'^. q. e. d. 'examples. Ex. 1. Square 2x—Sy. Result, 4:X^—12xy-\-9yK 2. Square x~^y—2y^. - Result, x'^y^—4:X~^y^-i-4:y\ 3. Square 2«"— 3^ " Result, 4«"— 12«"6 "-I-9J ". 4. Square m-P—n-i. Result, m-^—2m-Pn-^-\-n-^. 5. Square 3a "'—2b ". Result, 9a '»— 12a '"b »H-4Z> «. MULTIPLICATIOK. 61 90. Theorem. — The product of the sum and dif- ference of two quantities is equal to the difference of their squares. Demonstration. — Let x and y be any two quantities. Their sum isa;+y, and their difference is x—y. Now {x + y) multiplied by (x—y) gives, by actual mulplication, ar^— y^ or the diflference of the squares of the two quantities, q. e. d. EXAMPLES. Ex. 1. Find the product of 2a^-\-db and 2a^—db. Model Solution. — Here I have the sum of the two quantities 2a'' and 3ft, to be multiplied by their diflference. Now (2a' + 36) x (2«' — 3J) is, by the theorem, the square of 2a^ or 4a*, minus the square of the second, or 96^ . •. (2a« + 35) (2a' - 36) = 4a* - 96' 2. Find the product of «-f-26 and a— 2b. Prod., a^—4:tfi. 3. Find the product of 2a-\-3b and 2a— 3b. Prod,, 4fl2-9*2. 4. Find the product of 1a-\-2b and 7a— 2b. Prod., 49«2_4J2. 6. Find the product of 6a^-{-6b^ by 5a^—6b^. Prod., 25a«— 36Z»*. 6. Find the product of m* + w^ and m^—n^. Prod., m—n. 7. Find the product of 2^ + 3^?/^ and 2W— SV- Prod., 2x—3y. 8. Find the product of 3a^lfi+2aH^ and 3aW—2aH^. Prod., 9a4J«-4ajt. 62 FUNDAMENTAL RULES. SYNOPSIS FOR REVIEW. < O UJ I- o a. z O =3 QC Li. 0. r Multiplication. Multiplier. Multiplicand. Product. Factors. COR. ' I. Order of Factors, dem. 1. Multiplier, abstract. 2. Product like multiplicand. ScH. — How the sign of a mul- tiplier is to be understood. 1. Two factors. 2. More than two. ScH.— Multiplicand concrete. + ^ +iGive+. 2. Law of Signs. DEM. I COR. X X + Give Positive Factors. Even No. Negative Factors. Odd No. Negative Factors. r 1. Positive Integer. 3. Laws of Exponents, dem. \ ^- Positive Fraction. j 3. Negative Integer. I Negative Fraction. TO PERFORM MULTIPLICATION. THREE IMPORTANT THEOREMS. Prob. I. WHAT? Rule. Demonstration. Prob. 2. WHAT? Rule. Demonstration. r I. Square of Sum. Demonstration. ^ 2. Square of Difference. Demonstration. 13. Product of Sum and Difference Dem. Test Questions. — State and demonstrate the law of the signs in multiplication. How are expressions consisting of the same quan- tity affected by exponents, multiplied ? How many cases arise 'i Demonstrate each. What is the difference between the square of the sum, the square of the difference, and the product of the sum and difference of two quantities? Demonstrate. What is the product of «^ and a* ? Prove it. What of a^ and a-"^ ? Prove it. What of «l and «t ? Prove it. What of a"' and a-"'' ? Prove it. DIVISION. 63 .97. Division is the process of finding how many times one quantity is contained in another. The Dividend is the quantity to be divided. The Divisor is the quantity by which we divide. The Quo- tient is the result, which shows how many times the divisor is contained in the dividend. The Remainder is what is left of the dividend after the integral part of the quotient is produced. 08. Scholium I. — Division is the converse of multiplication. Since a product consists of (contains) as many times the multipli- cand as tLere are units in the multiplier, the multiplier shows how many times the multiplicand is contained in the product. The product, therefore, corresponds to the dividend, the multiplicand to the divisor, and the multiplier to the quotient. But, as in mul- tiplication, multiplicand and multiplier may change places without affecting the product, either of them may be considered as divisor and the other as quotient, the product being the dividend. 00. Scholium 2.— In accordance with the last scholium, the problem of division may be stated: Given the product of two fae^ tors and one of the factctrs, to find the other ; and the sufficient reason for any quotient is, that multiplied h/ the divisor it given the dividend. 100. Cor. 1. — Dividend and divisor may both he multi- plied or both he divided hy the sam£ numher without affecting the quotient. 64 FUNDAMENTAL RULES. 10 !• Cor. 2. — If the dividend he multiplied or divided hy any number^ while the divisor remains the same, the quotient is multiplied or divided by the same. 102. Cor. 3. — If the divisor be multiplied by any number while the dividend remains the same, the quotient is divided hy that number ; but if the divisor be divided, the quotient is mul- tiplied. 103. Cor. 4. — The sum of the quotients of two or more quantities divided hy a common divisor, is the same as the quotient of the sum of the quantities divided by the same divisor. 104. Cor. 5.— The difference of the quotients of two quan- tities divided hy a common divisor, is the same as the quotient of the difference divided by the same divisor. These corollaries are all immediate consequences of the notion of division. They need no demmistration^ but it is well that they be fully illustrated. Thus Cor. 1, may be illustrated as follows: If a given number of apples are divided among any number of boys, each boy will receive just the same number as if twice or thrice as many were divided among twice or thrice as many boys, or as if 1^ or ^ as many were divided among § or ^ as many boys. Cor. 4 may be illustrated thus : As 3 is contained in 8 four times and in 6 three times, it is contained in 8 + Q four + three, or 7 times. 105. Cancellation is the striking out of a factor com- mon to both dividend and divisor, and does not affect the quotient, as appears from (100). 106, Lemma l.-^When the dividend is positive the quo- tient has the same sign as the divisor ; but when the dividend is negative, the quotient has an opposite sign from the divisor. Demonstration. — This proposition is a direct consequence of the law of the signs in multiplication, since the dividend corresponds to the product, and a positive product arises from like signs in the factors, and a negative product, from unlike signs. DIVISION. 65 Scholium. — In both multiplication and division, Like signs give 4-, and Unlike signs — resfulls. 107. Lemma 2. — When the dividend and divisor consist of the same quantity affected by exponents, the quotient is the common quantity ivith an exponent equal to the exponent in the dividend, minus that in the divisor. That is, a^^-r-a"' _ (jm-n^ whether la and n be integi*al or fractional, posi- tive or negative. Demonstration. — This is an immediate consequence of the law of exponents in multiplication, since, in the corresponding case the exponent of the product was found to be the sum of the exponents of the factors. Now, as the dividend is the product of the divisor and quotient, it follows that the exponent of the quotient is the exponent of the dividend minus that of the divisor, q. e. d. EXAMPLES. Ex. 1. Divide a^ by a\ Model Solution. «^~a^ = a*; since a* x a' = «^, and division is finding a factor which multiplied into the divisor produces the dividend (99). 2. Divide in^ by m^. Quot., m^. 3. Divide w" by n~K Quot., tv" , or w *» . 4. Divide (ad)*"* by (abf. Quot., (a J)'" ", or (ab) - . 6. Divide d^ by a*. Quot., a-\ or ^ 6. Divide a~^ by a*. Quot., «-«, or - 7. Divide x~^ by ar^. Quot., x^. 8. Divide ar-a by x"^. Quot., x"^, or — j x^ 108* Cor. 1. — Any quantity with an exponent is 1, smce it may he considered as arising from dividing a quantity by itself Thus x° may be considered as if'-^oT which is 1, because divi- dend and divisor are equal. But by the law of exponents af"H-af" 66 FUNDAMEKTAL RULES. = x'^. .-. x'^ = 1. In like manner the significance of any quantity with an exponent may be explained. 6° = 1, for ^2 = 1, and also 5 ~ = 5\ .'. 5° = 1. 109, Cor. 2. — Negative exponents arise from division vjhen there are more factors of any number in the divisor than in the dividend. Thus a«-^a^ = a'-^ (lOV) = a-^ Again, ««-=-a^ = -' = 1 . or a^ .'. a-^ = — , which accords with the definition of negative expo- nents (43). 110, Cor. 3. — A factor may he tramfenrdfrom dividend to divisor (or from numerator to denominator of a fraction, which is the same thing), and vice versa, by changing the sign of its exponent. Thus TT- = -5r. ; for —- = - = --^&^ = — - Thus also ■=-. = a-^b-^, since "tt = ^~^ ^ ^ ' ^^^ ^ — ^~^- EXAMPLES Ex. 1. Show that , , — „ ^ - and ar-^ = — , I have — r a2 x^ x-^ Model Solution.— Since «-' = - and ar-" = - , I have ^2 ^ _a-' _¥ jf^ _¥ x" _ W T~^ " f ~ a' ' x''~a^^t~ a^y""' X' X' « T Tl 1 XU i. ^^ ^^ ^V ^^^ 2. In like manner show that — - — Dcd~^x~^ „ r — 5-5— T from nega the process, 3. Free -„ - from negative exponents and explain DIVISION. 67 111. Prob. 1. — To divide one monomial by another. Rule — Divide the nuTnerical coefficient of the dividerul by that of the divisor, and to the quotient annex the literal factors, affecting each with an exponent equal to its exponent in the dividend minus that in the divisor, and suppressing all fac- tors whose exponents thus become 0. The sign of the quotient will be + when dividend and divisor have lihe signs, and — when they have unlike signs. Demonstration. — The dividend being the product of divisor and quotient, contains all the factors of both ; hence the quotient con- sists of all the factors which are found in the dividend and not in the divisor. Or, the correctness of the rule appears from the fact that it is the converse of the corresponding operation in multipli- cation, so that quotient and divisor multiplied together produce the dividend. The law for the sign of the quotient is demonstrated in (106). Q. E. D. EXAMPLES. Ex. 1. Divide ^ah^b by 'daW. Model Solution. Operation. 3aV ) 12aV6 Explanation. — In the divisor there is a factor 3, hence there must be a factor 4 in the quotient, to produce 12 in the dividend. So, also, since there are 3 factors of a in the dividend, and 2 in the divisor, there must be 1 in the quotient in order that tiie product of divisor and quotient may be the dividend. In like manner 4 factors of x in the dividend, and 2 in the divisor, require 2 iu the quotient. There being 1 factor & in the dividend, and none in the divisor, one factor of h must appear in the quotient. Hence 12a'aj*& -^3aV = 4aa;"^&, which quotient containing all the factors of the dividend not found in the divisor, will, when multiplied into the divisor, produce the dividend, 68 FUNDAMENTAL RULES. 2. Divide IWcY ^1 ^^^f- Quot., S¥c. 3. Divide Scdx by dx. Quof., Sc, 4. Divide lOa^cx^y by 5a^. Qtcot, 2acy. 6. Divide ISa^b^x by —3a^x. Quot, —6ab. 6. Divide —20a^y^ by —bx^. Quot., 4.xyK 7. Divide —^2a^mY by '7aW. Quot, —6af. 8. Divide -Sla^bx^y by 9«%. Quot., —9aa^. 9. Divide — 13a:y by —ldxy\ Quot, x, 10. Divide 8aV by — S^j^c^. gwo^f., — 1. 11. Divide 3a^m by 3a%. §wo2^., 1. 12. Divide —x^y^ by a^-^y-^, §mo^., — a:^^^ 13. Divide af^ by ic^ Quot, x^-^. 14. Divide 5a-^^ by 2a&. ^woif., ^. 15. Divide 6ah^ by 3«^Ji Quot, 'Zar^bk 16. Divide 17«-^H.t2 by lla-^-^x^ Quot, ^^i^x-K 17. Divide aPb^c-' by «^>V. C^o^., aP-^b^-^c-i'+S). 112. Prob. 2. — To divide a polynomial by a monomial. Rule. — Divide each term of the polynomial divi- dend hy the monomial divisor, and write the results in connection with their own signs. Demonstration. — This rule is simply an application of the corol- laries (103, 104) ; since to divide a polynomial is to find the quo- tient of the sum or difference of several quantities, which by these corollaries is shown to be the sum or difference of the quotients of the parts, q. e. d, l)iv^isioK. 69 EXAMPLES. Ex. 1. Diyide 16a^xi-\-24:a^a^b—12a'^xc by Sasfi. Model Solution. Operation. ^aa^ ) 15aV + 24aV& — 12a^xc 5a'ar-» + Sah — 4a*z-^c Explanation. — I write the divisor on the left of the dividend, and the quotient underneath as a convenient form. Considering the first two terms, the quotient of their sum is the sum of their quotients (103) ; hence I divide each separately and add the re- sults, obtaining 5a^x~^ + Sdb. Again, the quotient of the difference between the sum of these two terms and the third is the difference of the quotients (104) ; hence I subtract from the quotient of the first two the quotient arising from dividing 12a^xc, which is 4a*x-'% and have for the entire quotient 5a^x-^ + Sab—4:a*x-^c. 2. Divide da^&^—lSa^iti-^6a^ by Sah. Quot, a^—Qa% + 2a. 3. Divide Ux'iy^—%7^f—24^y^ by 8a:. Quot, Zxy^—Qfif—Sy\ 4. Divide 210^7^— la^x^-^Uax by lax. QuoL, 3a^x^—ax-\-2. 6. Divide 4:2a^-na^+28a by Ha. QuoL, 6fl2_-J^^.4. 6. Divide m*—24]i:^c-h^Sk^ by 3M QuoL, 3k^^^Sk-^c+16k, 7. Divide 72a^(^—^Sa^c*—S2a^c^ by I6a^(^. Quot., ia?c-^ — Sa^c — 2flt«c». 8. Divide 36m*— 48m* by 4mi. Quot., dm^-~12mA 9. Divide m^—mJn^ by m^. Quot., m^—mi^n^\ 10. Divide a^—a^i^-i-a^ by ai Quot., a^—t^-{-aK 70 FUNDAMENTAL RULES. 11. Divide lla^-Sda^ by ll«i Quot, ai-daS\ 12. Divide 72mi^— 60mW by 2imK 13. Divide 2aH4«3J + 2a2^ by 2a2. 14. Divide lla^a^—Sab-j-lSa^ by 3«a;. Quot, i^ax^—--{-6^' 15. Divide a^^+i— a»»+2_aw+3_^m+4 ^y ^a, m 1 16. Divide bx^—lOx-"" -\-15aFy by 6x\ 113. Definition. — A polynomial is said to be arranged with reference to a certain letter when the term containing the highest exponent of that letter is placed first at the left or right, the term containing the next highest exponent next, etc. Illustration. — The polynomial 6a;y + 4xy* + 4«^y4-2/* + «*, when arranged according to the descending powers of y, becomes y*-\-4:xy^ ■\-^x^y'^-\-^x^y + x*. In this form it also chances to be arranged with reference to the ascending powers of x. 114. Prob. 3. — To pepform division when both divi- dend and divisor are polynomials. Rule. — /. Arrange dividend and divisor with ref- erence to the same letter. II. Divide the first term of the dividend hy the first term of the divisor for the first term^ of the quo- tient. III. Subtract from, the dividend the product of the divisor into this term of the quotient, and bring down as many terms to the remainder as may be necessary to form a new dividend. Divide as before, and continue the process till the worh is complete. DlVlSION^. 71 The demonstration of this rule will be more readily com- prehended after the solution of an example. Ex.— Divide 6a^^ + x^'-^a:^-\-a*—4:a^ by a^-{-a^—2ax. Model Solution. DIVISOR. DIVIDEND. QUOTIENT. a* — ^a'x + gV — 2a'a; + 5aV — 4aa;' — ^a'x + 4aV — 2ax^ a'x' — 2ax^ + x* a'x' — 2ax' + a!' Explanation. — Having arranged the dividend and divisor with reference to the descending powers of a, and placed the divisor on the left of the dividend, I divide a\ the first term of the arranged divi- dend, by a", the first term in the arranged divisor, and get a- as the highest power of a in the quotient. Now, as I want to find how many times a''—2ax + x'^ is contained in the dividend, and have found it contained a' times (and more), I can take this a" times the divisor out of the dividend, and then proceed to find how many times the divisor is contained in what is left of the dividend. Hence I mul- tiply the divisor by a' and subtract it from the dividend, leaving —2a^x-\-^a^x'^—4ax^ + x*. The same course of reasoning can be ap- plied to this part. Thus I know that the next highest jDower of a in the quotient will result from dividing the first term of this re- mainder by the first term of the divisor, etc. When this pro- cess has terminated I have taken a', and —2ax, and a?" times the divisor out of the dividend, and finding nothing remaining, I know that the dividend contains the divisor just a''—2axi-x- times. We will now give the demonstration of the rule. Demonstration. — The arrangement of dividend and divisor according to the same letter enables us to find the term in the quo- tient containing the highest (or lowest if we put the lowest power of the letter first in our arrangement) power of the same letter, and 80 on for each succeeding term. 72 FUNDAMENTAL RULES. The other steps of the process are founded on the principle, that the product of the divisor into the several parts of the quotient is equal to the dividend. Now by the operation, the product of the divisor into the Jirst term of the quotient is subtracted from the dividend ; then the product of the divisor into the 8eco?id term of the quotient; and so on, till the product of the divisor into each term of the quotient, that is, the product of the divisor into the wlwle quotient, is taken from the dividend. If there is no remain- der, it is evident that this product is equal to the dividend. If there is a remainder, the product of the divisor and quotient is equal to the whole of the dividend excejyt the remainder. And this remainder is not included in the parts subtracted from the divi- dend, by operating according to the rule. EXAMPLES. 1. Divide x^—3ax^-\-Sa^x—a^ by x—a. 2. Divide 2y^—l^y'^-\-^%y—lQ by ?/— 8. Qiiot, 2f—3y-\-2. 3. Divide x^—a^x^-\-2a^x—a'^ by x^—ax-j-a\ QuoL, x^-j-ax—a^, 4. Divide a*+4:¥ by a^—2ab-\-2i^. Operation. a^ _ 2ah + 21' ) a* + 4h' {a^ + 2ab + 2¥ a* — 2a'b + 2aW 2a'b- 2a'b- - 2aW- + 45* + 4«5=' 2a'l)' 2a'¥ -4a5« -4a5« + 45* + 45* 6. Divide 8fl2_26aJ + 15^ by 4.a—U. Quot.f 2a— bh, 6. Divide a^—¥ by a-h. Quot., a^-^ai^l^. DIVISIOK. 73 7. Divide a^—ia^x-^6ah^—4^a^-^x* by a^—2ax-{-x^. QtioL, d^—'Zax + x\ 8. Divide «3^:c8 by a-\-x. QuoL, a^—ax + 3^. 9. Divide i83^—76ax^—6'kt^-\-106a^ by 2x—3a. Quot, 24x^-'2ax—36a\ 10. Divide 2a^-\-a—6 by 2fl— 3. C^o^v « + 2. 11. Divide 2:2_^7a;^io by a; +2. Qnot., a; + 5. 12. Divide a:3— Sa^^— 46a:— 40 by a; + 4. Quot, a:2— 9a;— 10. 13. Divide a:8_93;2_,. 27.^—27 by x—3. Qvot., a:2_6a; + 9. [Note.— In the following the pupil will need to observe whether the terms are properly arranged or not before commencing the division.] 14. Divide bx^y-\-y^-{-j(^-\-hxy'^ by 4:xy + y^-\-x\ QuoL, x-\-y. 16. Divide Qx^y'^—-^xy^—4:a^y-\-y^-\-x^ by x—y. Quot., x^ — 37?y-\-3xy'^—y^, 16. Divide 6a?*— 96 by 3a;— 6. Quot, 2a;3+4a;2+8a; + 16. 17. Divide 3a^lA-3a^y^-b^-\-a^ by a^-l^-^3ay^-3a^b. Quot, a^ + 3a^ + '6al^ + h\ 18. Divide Q^—y^—i>x^y-\-bxy^-ir\Ox?y'^—\OxY by x—y. Quot, (x—yy, 19. Divide aP—li^ by a—h. 20. Divide m*-\-n* by m-{-n. 21. Divide 32a;5 + 243 by 2a; + 3, 22. Divide ¥—3y* by b—y. QuoL, b^ + lfiy^by^-\-f-^^' 7*4: , FU2!fDAMEKTAL RULES. 23. Divide x^-\-px-^q by x-\-a. x-\-a 24. Divide aH^* by a^-\-db^j'%-Y'b\ Quot., a^—abV^+l^. 25. Divide |2;3+i2^ + |:c+f by \x-\-l. Quot, ic^^j. Operation. ia; + 1 ) |«« + «^ + fa; +| (^M; j ^x^ + x" l« + f 26. Divide a:*+^ by a; + ?/. §wo^., x^—x^y-^xy'^—y^+^-' 27. Divide a^wi^ga^Z^^+Z^^n by a^+h\ Quot, a^-\-1)\ Operation. 28. Divide 2a3'*— 6a2«^>/i4.6a»^2/i_2^« by «'»— J« Quot, 2a^—Wb''-^2L^. 29. Divide ^a^s^j^^iS-fx+J by ix + d. Quot., a^—^x-^-l. 30. Divide x^—y^ by x^—y^- Quot, x+x^y^ + y. 31. Divide «— ^ by a^—bi. Quot, a^ + ah^ + a^b^ + bl DIVISION. ib 32. Divide x^—x^^^x^ + 6x—2x^ by a;^— 4a;^ + 2. Qtiot., x—x^. 33. Divide a27n_3am^n^2c2« by a'^— c^ 5 1 1 34. Divide oi?-\-bx-\ 1--3 by x-\--' X Xr X Quot, 2:2 + 4 4-^2- 35. Divide x^—-. by x — ;• Quot., qi?-\-x-\ f--^* X^ X X QCr 36. Divide ^--2x^-^4- + — by 2a-^x. X 2 a -^ n» T^• -J 0^ 5X^ 39 , X 1 37. Dmde^4-j^^+j^by-,+ -. ^ , 3a:2 13^ 39^^ 38. Divide a2^(a_l)a:24-(a— 1) ic3^(^_l) a4_a4i by a—x. Operation. a— aj ) a" + (a— !)«' + (a— !)«' + (a-l)aj*— 35* ( a + a; + aj' + «M;^ oar +(a- ax — -IK a;' ffa:» + (a-l)af« ax"- 7? aa?»+( M LUZ7 O o • . CO. 2 V-1078«3Z>4c3_^1232a3JV. Factors, UaW(^ and 66«— 77J + 88c. 123* Prop. 3.— If two terms of a trinom^ial are POSITIVE and the thir^d term is twice the product of the square roots of these two, and positive, the trino- mial is the square of the sum of these square roots. If the third term is negative, the trinomial is the square of the DIFFERENCE of the two roots. Demonstration. — This is a direct consequence of the theorems thit, "The square of the sum of two quantities is the sum of their squares j!>^M.« twice their product;" and " The square of their dif- ference is the sum of their squares minm twice their product." (94, 95.) 80 FACTORING. EXAMPLES. Ex. 1. What are the factors of a^ + 2ab-\-i^? Model Solution. — I observe that the two terms a' and ¥ of this trinomial are both positive, and that the other term, 2a&, is twice the product of the square roots of a^, and &*, and positive. Hence a^ + 2ab + I)'' = {a + if = {a + l){a-\-I>). Therefore, the factors are a + 1) and a + 6. 2. What are the factors of a^—2ax-\-a^ ? 3. What are the factors of m*-^n^-\-2mH^ ? Suggestion. — Here m* and n* are both positive, and 2m';i' is the product of their square roots. Ans., im^-\-ji^) and (m^ + n^). 4. What are the factors of Ua^—Sa-f-1? 5. What are the factors of m -f- 2 Vnm -{-n? Ans., (Vm + Vn) and (Vm-i-Vny 6. What are the factors of a;^+y^— 2a;%t ? Ans., x^—y^ and x^—y^, 7. What are the factors of a%-\-ab^-^2aU^ ? 8. What are the factors of x^-\-2xy—y'^'^ Query. — Can the last be factored according to this Propo- dtion ? Why ? 9. If 4«2j 16^2^ and IQab are the terms of a trinomial, what must be their respective signs so that the trinomial can be factored ? 10. Factor 16a^hn^—SaWm-\-l. Factors, (Umm—l) and (4a3^m — 1). 11. Factor Qa^^^ab-^^i^. Factors, (3a -{-^b) and (3« + i*). 1«. Factor ^da^b^—y^ab^-}-i¥. Factors, {Hab-lb^) and (7«^>~|J3), FUNDAMENTAL PROPOSITIONS. 81 13. Factor J + 2 + J. Factors, g + ^) and g + ^). 14. Factor Tt^:i^ + i^y^—\^y^- Farfor.% (^x^—:^i/) and (ix^—iy*). 15. Factor ^+-^-2. 1 6. Factor a^ + 2a\/x + x. Factors, (a + Vrr) and (a-\-Vx), 17. Factor re— 2*^/5+^2. Factors, (Vx—h) and (Va;— d). 18. Factor x—'ZVx'-y-^-y. Factors, {Vx—yy) and (a/5— V^). 19. Factor a%—'ilax^/h + x^. Factors, (a ^/h—x) and {a Vh—x), 20. Factor a + 2 Vfl + 1. Factors, (V« + l) and (V«-f-l). -i^4. Prop. 4. — T7ie difference betweeti two quanti- ties is equal to the product of the sum and difference of their square roots. Demonstration. — This is an immediate consequence of the theorem, that '' The product of the sum and difference of twoqian- tities is the difference of their squares." Thus a'— 5' = (a + 5) (a—b). Also, if we have m—n, in is the square of w», and n, of /i^ ; .-. m ^n = (m» + 7i^ (in»— n^), etc. ' EXAMPLES. Ex. 1. Factor Ux^-^y^. Factors, (^x—^y) and (^x-\-^y). 83 FACTORING. 2. Factor a^f—l^x^. Factors, (ay—hx) and {ay-\-hx), 3. Factor 16«%2_25^^2. Factors, (Aax—^by) and (4:ax + 5hy). 4. Factor x^—y^. Factors, (x^—y^) and (a^^+y^), 5. Factor x^—y^; also a^—y^ ; also ic^— y* ; also a;~* — y-*; also 4«-«— 9J-4. Factors of last three, (x^—y^) and (x^ + y^) ; (x~'^—y~^) and (x-^+y-^); and (2a~^~db-^) and {2a-^-^3b-^). 6. Factor m—n. Factors, {"s/m—Vn) and (a/wz + V^O- 7. Factor 2«— 4J2. Factors, 2, (V«— V^J) and (V'«+ a/2J). 8. Factor 26a^—Sy^^. Factors, {ox^^ — VSy"^) smd (bx'i -{- VSy^). 125. Prop. 5. — When one of the factors of a quan- tity is given, to find the other, divide the given quan- tity by the given factor, and the quotient will he the other. Demonstration. — This is the ordinary problem of division, since the divisor and quotient are the factors of the dividend. Any problem in division affords an example. EXAMPLES. Ex. 1. Resolve 2a^—dQa% into two factors one of which is 2«2. 13 19 2. Remove the factor — „ ^ from — , ^^ x^ 1/3 a:* 'ip 3. Remove the factor i- 4- ^r- from a^h'"^ -, — • h 2m 4 9 12 4 4. Remove the factor 3a~2-f-2:2;~i from ~4+ ~3i;+ ^* FUNDAMENTAL PROPOSITIONS. 83 126. Prop. 6. — The difference between any two quantities is a divisor of the difference between the same powers of the quantities. The SUM of two quantities is a divisor of the differ- ence of the same even powers, and the sum of the same odd powers of the quantities. Demonstration. — Let x and y be any two quantities and n any positive integer. Flrst^ ^—y divides x^ — y". Second, if n is eve7i^ x + y divides af—y\ Third^ iin is odd, x-\-y divides af + y". riBST. x—y) af'—y^ ( a?"-' + a?«-V + «"~ V + a?"-*^'' + etc. a^—x^-^y X'^-^y^ - x^-*y* af^*y* — y» Taking the first cast we proceed in form with the division, till enough terms to determine the law are found. We find that each remainder consists of two terms, the second of which, — y", is the second term of the dividend constantly brought down unchanged ; and the first contains x with an exponent decreasing by unity in each successive remainder, and y with an exponent increasing at the same rate that the exponent of x decreases. At this rate the exponent of x in the wth remainder becomes 0, and that of y, n. Hence the nth remainder is y—y" or 0; and the division is exact. SECOND AND THIRD. x-\-y)X" ±y" ( a^-' —3^'-^y + x"-^y^—xn-*y^, etc. — a?*~*y ± y" — af»-V— ^"V x^-y ± yn flf-V+iC'-'y" —x"—y ± y* —3f*~y—x^-*y* nf-y ± tr 84 FACTORIN-G. Taking xi-y for a divisor, we observe that the exponent of a; in tlie successive remainders decreases, and that of y increases the same as before. But now we observe that the first term of the remainder is — in the odd remainders, as the 1st, 3rd, 5th, etc., and + in the even ones, as the 2nd, 4th, 6th, etc. Hence if n is even, and the second term of the dividend is —y", the ?ith. remainder is y'^—y" or 0, and the division is exact. Again, if n is odd, and the second term of the dividend is +y", the nth remainder is —y" + y'' or 0, and the division is exact, q. e. d. 127» Scholium. — The pupil should notice carefully the form of the quotient in each of the above cases, and be able to write it without dividing. He should also be able to tell at a glance, whether such a division is exact or not, and if not exact, what the remainder, or fractional term of the quotient is. EXAMPLES. Write the quotients in the following without dividing : 1. {a^ - «/5) ^ (x — y). 2. («3 ^_ 58) ^{aJr h). Quot, a^-ab + l^. 3. («3 _ j3) -^ (a- h). 4. {a^ — m^) -7- (a + tn). 6. (a^ — 771^) -^ (a — m). 6. {nf + n') -^ {m -\- n). 7. {m^ — ¥) -=- (?n -\- b), also by (m — h). 8. (x^ - ^12) ^ {x^ 4- 2f). Suggestion. — Notice that x^"^ is the 4th (an even) power ofx^, and y^'^ is the sa)ne power of y^. It may be best for the pupil to obtain this quotient thus. Put x^ = a, and y^ = h. Then x^^ = a* and f" = l\ But (a*— &')-h (a + 5) = a^—a''l) + db''-'b\ Whence restor- ing the values of a and &, we have {x^'^—y^'^)^{x^-\-y^) =:x^—xhf +x^y^—y^, since a^ = x\ —a^l) = —x^y^, +db'^ = -\-x^y^, and —¥ = -/. Quot, m^ — m^n^ -f m^n^—m^n^ -\- n^, 10. (a;i8_2/i8)-j-(;7;3_^3)^ also by (x^ + y^). 1 1 . (ahn^ — y^) -f- (atn^ —y)- FUNDAMENTAL PROPOSITIONS. 85 Suggestion. — Notice that d^m^ is the third power of am"^, as y^ is of y. The quotient may be obtained by first using a single letter for aw", as x, writing out {x'^—y^)-^(x—y) = x^ + xy-^-y^, and finally restoring the value of x. After a little practice the pupil will not need to go through such a process, but can write the quotient at once. This quotient is a-m* -\-aiiry-\-y-. 12. (i-if)^(l-y). 13. (/_l)-^(y + l), also by (y-1). 14. {a^7p-^¥Y^)-^(ax + Wy^). 15. (162:4-81y8)^(2a;-h3y2)^ also by (2a;— 3^2). [Note. — In the following examples if the division is not exact, the quotient should be written with the fractional term.] 16. Is a}-{-h' exactly divisible by « + J, or a-- J? 17. Is m^—y^ exactly divisible by m + y, or m—y ? 18. Write the following quotients (x^^—y^^)-^(x-\-y), also hj x—y. (x^'^ + ^ -f- y^w +\^^(^x-\-y), m being an in teger. Query. — Is %m even or odd? Is 2m + l even or odd? FOR REVIEW OR ADVANCED COURSE. [Note. — The following corollary and examples may be omitted till after the pupil has been through the chapter on Powers and Roots, if thought desirable.] 128. Cor. — Proposition 6 applies equally to cases involving fractional or negative exponents. Thus, a;^— y^ divides x^-y^, since the latter is the difference between the 4th powers of «' and y*. So in general x ^—y »• an lu divides x "*— y •■, a being any positive integer. This becomes evident by putting x "^ — % and y •" = wj ; whence a?"'* = «", and at _£? _£f y T = w"". But v^—W is divisible by v—w^ hence x ♦"— y "■ is divisible by a; *"— y '. FACTORING. EXAMPLES. 19. What is the quotient of (tr^— /;-5)-^(«-l— ^>-i) ? 20. What is the quotient of {x~^—y~^)-^{x~^-^y~^)? Ans., x'^—x'^y"^ + a;~%"» — «/~5. 21. What is the quotient of [^-^j-^ [|+^^]? 22. Is x~^—y~'^ divisible by x~^—y~^, ov x~^ -{- y~'^^ 23. Is x^—y^ divisible by Vx-j-Vy, or Vx—Vy? 24. Is x^-i-y^ divisible by \/x-\-\/y, or Vx—Vy ? 25. What is the quotient of {ax^ + i^y) -^ (a^ \/x 129, Prop. 7. — A trinoTviial can he j^esolved into two binomial factors, when one of its tenns is the product of the square root of one of the other two, into the sum of the factors of the remaining term. The two factors are severally the algebraic sum of this square root, and each of the factors of the third term. Illustration. — Thus, in a;^ + 7a; + 10, we notice that llx is the pro- duct of the square root of a?", and 2-1-5 (the sum of the factors of 10). Therefore, the factors of x^ -I- 7a; -f- 10 are a!-f-3 and a?-|-5. Again a;- —3a;— 10, has for its factors a!-l-2 and a;— 5, — Sa; being the product of the square root of a;^ (or a?), and the sum of —5 and 2 (or —3), which are factors of —10. Still again, a;''-f-3a!— 10 — {x—2) (a7-i-5), determined in the same manner. Demonstration. — The truth of this proposition appears from considering the product of x^a by a;-i-&, which is x^ ^- {a -\-'h)x -\- db. In this product, considered as a trinomial, we notice that the term (a + 5)a!, is the product of 'y/x'^ and a-h&, the sum of the factors of FUNDAMENTAL PROPOSITIONS. 87 ab. In like manner (x-\-a) (x—h) = ^'^ -\-{a—l)x—ah^ and (pt—a) {x—h) = a^—(a+b)x + afK both of which results correspond to the enunciation, q. e. d. [Note. — In application, Proposition 7 requires the solution of the problem : Given the sum and product of two numbers to find the num- bers, the complete solution of which cannot be given at this stage of the pupil's progress. It will be best for him to rely, at present, simply upon inspection.] EXAMPLES. Ex. 1. Show that, according to this proposition, the fac- tors of x^-\-Sx-\-2 are x-\-l and x-{-2. Verify it by actual multiplication. 2. In like manner show the following : x'i - 7x H- 12 = {x — S){x- 4). x^- x — U = ix + 3) (x — 4). x^ -\- x—12=:{x — S)(x-{- 4). 3. Factor xi—5x-{-6, x^+Sx—^S, ar^— lla;-f28, xi-\-2x —35, 2^2— 2a:-8, 2:24-20;— 8, T2—y—y\ 5-42—^2. 130, Prop. 8. — We can often detect a factor by separating a polynomial into parts. Demonstration. — The correctness of this process depends upon the principle that whatever divides the parts, divides the whole. EXAMPLES. Ex. 1. Factor a^+12x—2S. Model Solution —The form of this polynomial, suggests that there may be a binomial factor in it, or in a pwt of it. Now «' — ix-^i is the square of x—2, and (a-'— 4a; + 4j-|-(16«— 32) makes a;-^ 4. 12a;— 28. But (x''-4x + 4) + (iex—Z2) = {x-2){x—2) + (x-2) 16 = {x-2) (aj— 2-1-16) = (x—2) (x-\-U). Whence «— 2, and aj-f 14 are seen to be the factors of a;^-|-12a;— 28. 2. Resolve 12a'^b-\-dby^—loaby into its factors. Solution. 3& is seen to be a common factor, and removing it, we have 4u'^ + y'^—6ay. This latter polynomial may be separated into 88 PACTOEiNa. the parts a'-'%ay-\-y'^^ and 3«'— Say. Hence, 4a^ + y2— Say = {a" -2a2/ + y'^) + (3a^-3ay) = {a-y) {a-y) + {a-y)M = («-y) {a-y 4-3a) = {a—y){4ta~y). Hence the factors of 13a-fe + 3%''— 15<%, are 35, a—y, and i.a—y. 3. Factor 6««+15fl54^— 4^3^2—1 O^^dc^. Factors, o?, Sa^—2o^, and 2a + 5 J. 4. Factor 3a2_2«5— 1. Factors, a—1 and Sa + l. 6. Factor 6a^—Sax-\-3xl Factors, 5a— 3x and a—x. 6. Factor 20aa;—25«?^— 165a; + 20^72. 7. Factor Sd^-^22-ab-i-15bK 8. Factor 12:»3_8^^_9^y _|_6^3 9. Factor ^2x^—2xy—20y\ 4:2x^—6Sxy-\-20f, and ^2x^ 10. Factor 64:xy—10ax + 96by—16ab. 11. Factor Aa; — a^mn — a^xy-\-a^my. 12. Factor ^5a^3?-{-3Wbx^y—21a:kxhj — l^abcxy\ 13. Factor 12a;5_8a;3^2^21a;2y— 14^3, :^^2. The Highest Divisor of a literal number is the divisor of tlie highest degree with reference to the letter of arrangement, i. e., the divisor wliich involves the highest exponent of that letter. The Highest Common Divispr of two or more literal numbers is the divisor of highest degree which is common to all of them. HIGHEST COMMON DIVISOR. 89 Scholium. — While, with reference to decimal numbers, the term greatest divisor is appropriate, it is scarcely proper to apply it to literal quantities, for the values of the letters not being fixed, or specific, great or small cannot be affirmed of them. Thus, whether a^ is greater than «, depends upon whether a is greater or less than 1, to say nothing of its character as positive or negative. So, also, we cannot with propriety call a^—y^ greater than a—y. If a = ^, and y — ^, a^—y^ = t^, and a—y — i ; .-. in this case a^ —y^, Lemma 3. — A divisor of any number is a divisor of any multiple of that number. Illustration. — This is an axiom. If a is contained in h, q times, it is evident that it is contained in n times 6, or nb, n times j, or nq times. 130, Lemma 4. — A common divisor of two numbers is a divisor of their sum and also of their difference. Demonstration. — Let a be a C. D. of m and n, contained in m, p times, and in n, q times. Then (m -\- n) -i- a = 2^ + q. Q. E. D. Illustration. — If this appears a little abstract and unsatisfactory to the learner, let him illustrate it thus : 4 is contained in 30, 5 times, and in 8, 2 times; hence it is contained in 20 + 8, 5 + 2, or 7 times, and in 20—8, 5 — 2, or 3 times. 92 FACTORIN^G. 137, Prob.— To find the H. C. D. of two polynomials without the necessity of resolving them into their prime factors. Rule. — /. Arranging the polynomials with refer- ence to the same letter, and uniting into single terms the like powers of that letter, remove any common factor or factors which may appear in all the terms of both polynomials, reserving them as factors of the H. C. B. II. Reject from each polynomial all factors luhich appear in it and not in the other. III. Talcing the polynomials, thus reduced, divide the one with the greatest exponent of the letter of arrangCDient, by the other, continuijig the division till the exponent of the letter of arrangement is less in the remainder than in the divisor. IV. Reject any factor ivhich occurs in every term of this remainder, and divide the divisor by the remainder a^s thus reduced, treating the remainder and last divisor as the former polynomials • were. Continue this process of rejecting factors from each term of the remainder, and dividing the last divisor by the last remainder till nothing remains. If, at any time, a fraction ivould occur in the quo- tient, multiply the dividend by any number that will avoid the fraction. The last divisor multiplied by all the first reserved common factors of the given polynomials, will be the H. C. D. sought. Demonstration. — We will first give a demonstration of this rule by means of a particular example. Let it be required to find the H. C. D. with reference to «, of the following polynomials: 12a^^>' + 35V-15a%+13«% + 35y''-15«&y^ and QaW—^o'Vy HIGHEST COMMON DIVISOR. 93 • Operation. t2a'b' + Wy' - ^5ab'y + 12«% + 3&y» - I5ahy' (A) 6a'b'-6a'h -' y-2bY + ^^Y + 6a%- 6a%'-26y« + aofty " (B) 4a''b + Y- ^aby + 4a V + y' - 5ay'' (^) Sa^^ - 3a% - g^' + «&y' + 3a'y - 3aV - y* -j- ay» (2)) (46 + 4y) a' - (5&y + Sy'') a + (Jt/^ + y'') {E) (36 + 3y) a» - {Zby + 3y^) a'' + Qyy^ + y") « - (^y" + y*) (/?') (G') (ff) 4a'' — 5ya 4- y'^ ) 3rt^ — 3ya' + y'^ci — y ' 4 13a»-12ya''+ 4y''a- 4y'(3a 12a»-15ya^+ 3y^« 3ya'+ y''^- 4y'' 4 (Jf) 12ya'+ 4y'a-16yH3y 12ya''-15y'^a+ 3y=' {0) Reject 19y». 19y^«-19y' ((?) a— y)4«''— 5ya + y'(4a— y 4a»-4y« - ya+y' — ya+y"" .: The H. C. D. of (A) and (B) is (&)(6 + y) (a-y) = ah'^+atry -h'y-by\ Reasoning. 1st. Removing h from both {A.) and (B), and reserving it as a fector of the H. C. D. (Lem. 1), and rejecting 3 from (A), and % from (B), since they are not common factors, and hence cannot enter into the H. C. D., we get (C) and (D). 2nd. Arranging (C) and (D) with reference to a, the letter with respect to which the H. C. D. is sought, both for convenience in dividing, and to observe if any other common factor appears, we have (E) and (F). In these we readily discover and remove the common factor (h + y), reserving it as a factor of the H. C. D., and get (G) and (H). 3rd. Having now found two of the common factors of (A) and (B), and removed some which were not common, it remains to determine whether there are any more common factors, that is, whether (G) and (H) have a C. D. 94 FACTORING. (G) is its own H. D., Lem. 2 ; hence if it divides (H), it is the H. C. D. of (G) and (H). We, therefore, try it. But, to avoid fractions, we multiply (H) by 4, since, as 4 is not a factor of (G), the H. C. D. of (G) and 4 times (H) is the same as of (G) and (H). We thus obtain (I). .-. If (G) divides (I) it is the H. C. D. of (G) and (H). Trying it, we find the remainder (L). Now, any C. D. of (G) and (I) is the divisor of (K) [a multiple of (G)], Lem. 3, and of (L), Lem. 4. We now have to find the H. C. D. of (G) and (L), upon which we reason just as upon (G) and (H). Thus, as (G) is its own only divisor of the 3nd degree, if it divides (L), or (M) [since (M) = 4 (L) and 4 is not a factor in (G)], it is the H. C. D. of (G) and (H). Trying, we find a remainder (O). Now, any divisor of (G) and (M) is a divisor of (N), Lem. 3, and of (O), Lem. 4. The question is there- fore reduced to finding the H. C. D. of (G) and (O). Upon these we reason as before. Rejecting 19y^ since it is not a factor of (G), and hence, cannot enter into the H. C. D., we have (P). The H. C. D. of (G) and (P) cannot be higher than (P), and as (P) is its own only divisor of its own degree, if it divides (G,) it is the H. C. D. sought. It does. .-. a-y is the H. C. D. of (G) and (tl). Finally, &, (& + y), and (a—y) are all the common factors of (A) and (B), and hence, & (b^y){a—y)~db'^-\-alyy—¥y—ly^ is their H. C. D. EXAMPLE. Find the H. C. D. with respect to x, of x^—^a^ + ^lx^ — 20a: + 4, and ^7^—l^x^ + ^lx—lO. Model Solution. — Calling the 2nd, (A) and the 1st (B), I have the following : {A) Operation. {B) %x^ — l^x" + 21a; - 10 ) «* — ^x^ + 21a!- — 20a; + 4 2 ((7) 2a;* - 16a;^ + 42a!^ _ 40a; + 8 (a; - 2 2a!* — 12a;^ + 21a;'' - 10a! IT 4a,3 + 21a;'' _ 30a; + 8 — 4a!« + 24a!^ - 42a; + 20 {D) Reject -3 - 3a;^ + 12a; - 12 {E) x" - 4a; + 4 HIGHEST COMMON DIVISOR. 95 {E) (A) J53 _ 4a. + 4 ) 2aj' - 12a;' + 21a; - 10 ( 2« ,- 4 2x' - 8a;' + 8a; - 4a;' + 13a; - 10 - 4a;' + 16a; - 16 (F) Reject -3 — 3a; + 6 (E) (O) X— 2)x^ - 4x + 4:(x — 2 a;' — 2a; — 2a; 4- 4 — 2a; 4- 4 Reasoning. The two given polynomials being arranged with reference to a;, and no common, or other factor, appearing, I proceed at once to determine by successive divisions their H. C. D. The H. C. D. cannot be higher than (A), the lower of the two ; and, as it is its own only divisor of the 3rd degree, if it divides (B), it is the H. C. D. But to avoid fractions I multiply (B) by 2, and, since 2 is not a factor of (A), the H. C. D. of (A) and (B), is the H. C. D. of (A) and (C). Now if (A) divides (C) it is the H. C. D. Trying it, I find a remainder", (D). Again the H. C. D. of (A) and (C) is also a divisor of (D), for (D) is the difference between (C) and (a;— 2) times (A), both of which are divisible by the H. C. D. of (A) and (C). The question is now reduced to finding the H. C. D. of (A) and (D). Upon these I reason exactly as before upon (A) and (B). Thus, since —3 is a factor of (D), and not of (A), it can be rejected ; and the H. C. D. of (A) and (D) is the H. C. D. of (A) and (E). This cannot be higher than (E) ; and as (E) is its own only divisor of the 2nd degree, if it divides (A), it is the II. C. D. Trying it, I find a remainder, (F). Upon this remainder, and (E), I reason as before, upon (D) and (A). Thus, the II. C. D. of (E) and (A) is a divisor of (E) and (F), since (F) is the difference between (A) and a multiple (2a;— 4 times) of (E). The question is then reduced to finding the H. C. D. of (E) and (F). Upon these I reason as before, rejecting —3, which is not a common factor, and hence, forms no part of H. C. T). of (E) and (F), and finding by trial that (G) is a divisor of (E). Therefore, x—2 is the H. C. D. of (A) and (B). 96 FACTORIJ^G. GENERAL DEMONSTRATION OP THE RULE FOR FINDING THE H. C. D. Let A and B represent any two polynomials whose H. C. D. is sought. 1st. Arranging A and B with reference to the same letter, for convenience in dividing, and also to render common factors more readily discernif)le, if any common factors appear, they can be removed and reserved as factors of the H. C. D., since the H. C. D. consists of all the common factors of A and B. 2nd. Having removed these common factors, call the remaining factors C and D. We are how to ascertain what common factors there are in C and D, or to find their H. C. D. As this H. C. D. consists of only the conmion factors, we can reject from each of the polynomials, C and D, any factors which are not common. Having done this, call the remaining factors E and F. 3rd. Suppose polynomial E to be of lower degree with respect to the letter of arrangement than F. (If E and F are of the same degree, it is immaterial which is made the divisor in the subse- quent process.) Now, as E is its own only divisor of its own degree (Lem. 2), if it divides F, it is the H. C. D. of the two. If, in attempting to divide F by E to ascertain whether it is a divisor, fractions arise, F can be multiplied by any number not a factor in E (and E has no monomial factoi'), since the common factors of E and F would not be affected by the operation. Call such a multiple of F, if necessary, F'. Then the H. C. D. of E and F', is the H. C. D. of E and F. If, now, E divides F', it is the H. C. D. of E and F. Trying it, suppose it goes Q times, with a remainder, R. 4th. Any divisor of E and F' is a divisor of R, since F'— QE = R, and any divisor of a number divides any multiple of that number (Lem. 3), and a divisor of two numbers divides their differ- ence. The H. C. D. divides E, hence it divides QE, and, as it also divides F', it divides the difference between F' and QE, or R. Therefore, the H. C. D. of E and F', is also the H. C. D. of E. and R. 5th. We now repeat the reasoning of the 3rd and 4th paragraphs concerning E and F, with reference to E and R. Thus, R is by hypothesis of lower degree than E ; hence, dividing E by it, reject- ing any factor not common to both, or introducing any one into E, which may be necessary to avoid fractions, we ascertain whether R is a divisor of E, • HIGHEST COMMOX DIVISOR. 9? 6th. Proceeding thus, till two numbers are found, one of which divides the other, the last divisor is the H. C. D. of E and F, since at every step we show that the H. C. D. is a divisor of the two num- bers compared, and the last divisor is its own H. D. 7th. Finally, we have thus found all the common factors of A and B, the product of which is their H. C. D. q. e. d. EXAMPLES. Ex.1. Find tlie H. 0. D. of Uax—Sa-\-aa:^—7ax^ jind 16a^x^-\-(jah:^—2Sa^a:^, and give the reasoning as above. T/ie H. a D. is ax—^a. 2. Find the H. C. D. of a^-\-Sa^b + 2aIri + ^ and 6(fi + 5^, giving the demonstration. The H. C. D, is a-\-l). 3. Find the H. 0. D. o['^Qifi-\-W—^W—l^a^ and ^la^lf^ -Wl^-l^a^yi. The K C, D. is 9«4_9a8. 4. Find the H. V. D. of ^xy'^—^f + Qx^y and ^x^y-{-^a^ —^xy\ The K C. D, is 2x + 2y, 5. Find the H. 0. D. of 3cs^—10x^-^15x-^S and :^—2x^ — ^-^^x^+13x-{'(J. llie H. a D. is7^-\-3x^-\-dx-^\. 6. Find the H. C. D. of W'a^-la^hT^y + 2alf^xY—^i^f and ^a^l^y^—2ab^xY—2bh^y^' The H. (L D. is %ax^—2hxy. 1:^8. Prob.— To find the H. C. D. of three op more polynomials. Rule. — Find the H. C. T>. of any two of the given polynomials hy one of the foregoing methods, and, then find the H. C. D. of this H. C. D. and one of the remaining polynom^ials. Continue this process till all the polynomials have been used. Demonstration. — For brevity, call the several polynomials A, B, C, D, etc. Let the H. C. D. of A and B be represented by P, whence P contains aU the factors common to A and B. Finding the n. C. D. of P and C, let it be called P' P', therefore, contains 98 FACTORING. all the common factors of P and C ; and as P contains all that are common to A and B, P' contains all that are common to A, B and C. In like manner if P" is the H. C. D. of P and D, it contains all the common factors of A, B, C, and D, etc. q. e. d. EXAMPLES. Ex. 1. Find the H. C. D. of ''Za:^ + Q:^-\-^^, 3.t3+9:cH9^ + 6, and d:^-^Sx^ + Qx + ^. The H. C\ D. is x + 2. 2. What is the H. 0. D. of 10a^+10aW + 20a% 2a^-\-2b% and i¥ -j- 12^2^ + 4:a^ + 12ab^ ? Atis., 2{a + h). 139 • The Lowest Multiple of a given literal number is the number of lowest degree with reference to some specified letter^ which contains the given number as a factor. The Lowest Common Multiple of several numbers is the number of lowest degree which is a multiple of each. [For the impropriety of the term Least Common Multiple as applied to literal numbers, see Art. 131.J 140, Prob.— To find the L C. M. of two op mope numbeps. Rule. — /. Tahe the literal numher of the highest degree, or the largest decimal number, and multiply it by all the factors found in the next lower which are not in it. II. Again, iriultiply this product by all the factors found in the next lower number and not in it, and so LOWEST COMMO^r MULTIPLE. 99 continue, till all the niutibers are used. TJie product tlius obtained is the L. C. M. Demonstration. — Let A, B, C, D, etc, represent any numbers arranged in the order of their degrees, or values. Now, as A is its own L. M., tlie L. C. M. of all the numbers must contain it as a factor. But, in order to contain B, the L. C. M. must contain all the factors of B. Hence, if there are any factors in B which are not found in A, these must be introduced. So, also, if C contains factors not found in A and B, they must be introduced, in order that the product may contain C, etc. Now it is evident that the product so obtained, is the L. C. M. of the several numbers, since it contains all the factors of any one of them, and hence can be divided by any one of them, and ifany factor were removed it would cease to be a multiple of some one or more of the numbers, q. e. d. Ex. 1. Find the L. C. M. of {x'—l), (x^—l), and (a; + l). Model Solution. — The L. C. M. must contain x^—\, and as it is its own L. M., if it contains all the factors of the other two, it is the required L. C. M. The factors of x^—\ are {x—V) and («* + «+ 1). But the product of these does not contain the factors of {x^—1) which are (a^ + l) and («—!). Hence we must introduce the factor (« + !), giving {x^—V} («+l), as the L. C. M. of a?^ — 1 anda;'*— 1. Now as this product contains the third quantity, it is the L. C. M. of the three. 2. Find the L. C. M. of {a^lf, a^-V", (a-Vf, and «» Suggestions. — The last is (a + J)^ which contains the factors of the 1st, but neither of the factora of the 3rd. Both lactors of («—&)' must, therefore, be introduced, giving {a^-Vfia—'by as the L. C. M. of the 1st, 3rd and 4th. And as it contains both factors of «--&^ viz.: (a + &) (a— &), it is the L. C. M. of all. 3. Find the L. C. M, of {x^—^a% (a:-f2a)3, and (x—%d)\ Z. C. if., (a:2_4a2)3. 4. Find the L. C. M. of Qc^-i^%xy^y^ and o^—xy\ L. C. M., (x^y) {a^-xy% 5. Find the L. C. M. of l^^m:(^, S5a^m% and 7b*x^. L. a M., 70a^bWx^, 100 FACTORING. 6. Find the L. 0. M. of Uxy, 10a% and ^Oa^x\ L. G. if., 80a%y. Scholium. — In applying this rule, if the common factors of the two numbers are not readily disceraed, apply the method of find- ing the H. C. D., in order to discover them. 7. Find the L. 0. M. oi a^—'Zax^-^-^a^x—Wy x^-]-^aa'^ -\-4:a^x-\-%o^, and x^—^a\ Model Solution. — The L. C. M. of these numbers must contain x^—^ax^ + Wx—^a^-^ and as it is its own L. M., if it contains all the factors o^ x^-\-2ax'' + ^a'x + 9,a\ it is the L. C. M. of these two polynomials. But as the common factors of these numbers, if they have any, are not readily discerned, we apply the method of H. C. D. and find that a;^ + 4a^ is the H. C. D. of the two. Since, then, x' — 2aa;' + 4a'^a;— 8a^ contains the factor a;^ + 4a^ of the second number, it is only necessary to introduce the other factor in order to have the L. C. M. of the two. Now {x"" + %ax''-\-4:(i'x^Sa^)^{x'' -\-^a') = « + 2«. B.QY\CQ {x^'—'iax'' ■\-ia^x—^a^) {x-\-%a) or a;*— 16a* is the L. C. M. of the first two numbers, since it contains all the factors of each, and no more. Now, to find whether a;*— 16«* is a multiple of the remaining number, a;"— 4«*, or, if it is not, what factors must be introduced to make it so, we proceed in the same way as with the first two numbers. But our first step (or 124) shows us that a;* — 16a* is a multiple of x''—id\ .'. «* — 16a* is the L. C. M. of the three given numbers, 8. Find the L. C. M. of Qx^—x—1 and '^x^+^x—%. L. a M„ (2a;24-3a;-2){3a; + l). 9. Find the L. 0. M. of x?—^x^-\-^'^x—lb and x?—dx + 7. L. C. M., (a;3_9^2_|.232--15) {x—1) = x^-16a^+S6x^ ~176a; + 105. 10. Find the L. C. M. of 2a;-l, 4rr2-l, and 4a;2 + l. 11. Find the L. 0. M. of a^-x, a^-1, and a^-{-l. L. a M., x{a^-l) = x'^-x. 12. Find the L. C. M. of x^-exi-]-Ux—6, x^-9xi-h2(jx —24, and ar^— 82^^-f 19^;— 12. L. 0, M., {x-1) {x-2) (x-^) (a;-4) = 0^-10:^3^35^2 —50^ + 24. BYN0P8TS. 101. SYNOPSIS FOR REyiEW. Z cc o I- o < to oq. zo LL.a. r Factor. 1 common. Divisor. \ (|. C. »[ distinction. Measure. u. c. D. COMMON. Multiple. ] LEAST C. M. Idistdiction ( LOWEST C. M. I 1 Numbers. Scholium L PROPS. COMPOSITE. PRIME. PRIME TO EACH OTHER. HOW THESE TERMS ARE APPLIED. 1. A MONOMIAL. Dem. 2. A POLYNOMIAL WITH MON. FACTORS. Dem. 3. a»±2a6 + ^»^ Dem. 4. a»-6^ Dem. 5. ONE FACTOR GIVEN. Dejt. iSc/i. Form of quotient. Cor. Fractional and neg- ative exponents. 7. A TRINOMIAL. WHEN ? HOW '( Dem. 8. SEPARATING INTO PARTS. r Distinction between G. C. D. and H. C. D. Lemmas. 1. DEM. 2. DEM. 3. DEM. 4. DEM. General j PROB. l. Rule. Dem. Method. 1 PROB. 2. Rule. Dem. I Distinction between Least and Lowest C. M. Problem, ritle. dbm. Scholium, by means of h. c. d. Test Questions.— What are the factors of a^—b'"i Ofa' + 2a5 + ft-^? Of «'-2a5+6''? or l-2x + x"i Of 3aH Ba'^a; + 3a V ? Orx'' + y--\-2xy'i Of a;^— a!-12? Of 7n? Of a'&'y? Ofa;'-y»? State the general rule for testing the divisibility of the sum or diflfereuce of like powers. Prove one of the cases, as (a"*—l'")^(a —b). Distinction between H. C. D. and G. C. D. Beivfeen Lowest and Least C. M. Explain the process of finding each by factoring. Give the General Rule in each case, and its demonstration. FRACTIOlfS i -g fejAF TlII III. * ECTiON I DEFINITIONS AND FUNDAMENTAL PRINCIPLES. 1J,1, A Fraction, in the literal notation, is to be con- sidered as an indicated operation in Division. A fraction is written, as in common arithmetic, with one number above another and a line between them. The number above the line, i. e., the dividend, is called the Numeratoi' ; and the number below the line, i. e., the divisor, is called the Denominator. Thus 3c + 4a;3 — means nothing more than (3a— 5ma;^)-^(3c+4a;^). Taken together numerator and denominator are called the Terms of the Fraction. 142. Scholium. — In the literal notation it becomes imprac- ticable to consider the denominator as indicating the number of equal parts into which unity is divided, and the numerator as indicating the number of those pai'ts represented by the fraction, since the very genius of this notation requires that the letters be not restricted in their signification. Thus in -, it will not do to say, h represents the number of equal parts into which unity is divided, since the notation requires that whatever conception we take of these quantities should be suflSciently comprehensive to include all values. Hence & may be a mixed number. Now suppose & =4f. It is absurd to speak of unity as divided into 4| equal parts. DEFINITIONS. 103 14ii, The Value of a Fraction is the quotient of the numerator divided by the denominator. 144, Cor. 1. — Since numerator is dividend and denom- inator divisor, loe have (100, 101, 103) the following : 1. Dividing or multiplying both terms of a fraction by the same quantity does not alter its value. 2. Multiplying or dividing the numerator multiplies oi divides the value of the fraction ; and 3. Multiplying or dividing the denominator divides or mul- tiplies the fraction. 145, Cor. 2. — A fraction is multiplied by its denominator by simply removing the denominator. This is simply cancelling equal factors from numerator and 3 3x4 denominator (Art. 105). Thus - x 4 = — - — ~ 3. In like manner 4 4 a ax -Y.x — ^ = a. X X 14(u The terms Integer or Entire Number, Mixed Num- ber, Proper and Improper, are applied to literal numbers, but not with strict propriety. Thus, whether m-\-n is an integer, a mixed number, or a fraction, depends upon the values of m and n, which the genius of the literal notation requires to be understood as perfectly general, until some restriction is imposed. For convenience, we adopt the following definitions : 147. A number not having the fractional form is said to have the Integral Form; as m-f-n, 2o^dSa~^x + 3x5^4. 148. A polynomial having part of its terms in the fractional and part in the integral form, is called a Mixed ,WT 1 . ccP-2y^, Number; as a—x-\ /—' 104 FRACTIONS. 149, A Proper Fraction, in the literal notation, is an expression wholly in the fractional form, and which cannot be expressed in the integral form without negative exponents. By calling such an expression a proper fraction, we do not assert anything with reference to its value as compared with unity. Thus - is a proper fraction, though it may be greater or less than unity. It may also be written ab-\ 150, An Improper Fraction is an expression in the fractional form, but which can be expressed in the integral or mixed form without the use of negative exponents. Inus, -—— =E 2a — ^ ; the former of which is called an 71— dab n—dab improper fraction. 151. A Simple Fraction is a single fraction with both terms in the integral form. _^ 2x—m^ ^ 2cmx^ . , ^ ^. Inus -. -, and — -- are simple fractions. 15 2. A Compound Fraction is two or more fractions connected by the word of; but the expression is not gener- ally applicable in the literal notation. Thus we may write o ^f 7 ^'ith propriety, but not ^ of — , unless a and h are integral, so that the fraction y may be considered as 2 representing equal parts of unity, as - does. If the word of is con- 3 sidered as simply an equivalent for x , the notation is, of course, always admissible. But it is scarcely a simple equivalent. FUNDAMENTAL PRINCIPLES. 105 153. A Complex Fraction is a fraction having in one or both its terms an expression of the fractional form. c ma^ ^"^'d ~cd Thus , and kj^ ^J-re complex fractions. 5 — a y 154. A fraction is in its Lowest Terms when there is no common integral factor in both its terms. 155. The Lowest Common Denominator is the number of lowest degree, which can form the denomi- nator of several given fractions, giving equivalent fractions of the same values respectively, while the numerators retain the integral form. 156. Reduction, in mathematics, is changing the form of an expression without changing its value. SIGNS OF A FRACTION. 157. In considering the signs of a fraction, we have to notice three things, viz.: the sign of the numerator, the sign of the denominator, and the sign before the fraction as a whole. This latter sign does not belong to either the numerator or denominator separately, but to the whole ex- ^, . ,, . 4:a—6cd . ,. pression. Thus, m the expression — XT' ^^ "^® numerator the sign of 4fl is +, and of 5cd — . In the denominator, the sign of 2x is -f , and of 4^^ _|_ j^jgQ^ The sign of the fraction is, — . These are the signs of operation. (56, 33, 34.) 158. The essential character of a fraction, as positive or negative^ can be determined only when the essential character of all the numbers entering into it is known. It may then be determined by principles already given. (80, 106.) 106 FRACTIONS. EXAMPLES. Ex. 1. Is the fraction — o.^ , . t essentially positive, or negative, when a, m, x, and y are each negative ? Model Solution.— I. The square of —a is +, since it is the pro- duct of an even number of negative factors (S8). . •. 4a'^ is positive. For the same reason mx is +• .'. 3w2a? is in itself positive, but it is made negative by the — sign before it. 2. The cube of —x is — , since it is the product of an odd number of negative factors (§J)). .•. '^x^ is negative. The square of —y is + (88). .-. 4«/^ is positive. 3. Finally, the essential sign of the fraction depends upon the relative values of a, ?w, x, and y. If 4a''' > 3ma;, and 2x'^>4:y'^^ the numerator and denominator have unlike signs, whence the value of the fraction (143) would be — ; but it is made + by the — sign before the fraction. Similarly, if ia^ < Zmx, and 2x^>4:y'^, the signs of the numerator and denominator are alike, whence the sign of the fraction would be +, but the — sign before the fraction would change it to — ; and the essential character of the fraction would be negative. 2. What is the essential sign of ' ^ , when «= — 3, X = —2, p = —4, and b z= —o? Ans., —. 3. What is the essential siern of v^ — when ° 3a^x X = —2, a = —1, and y = 3? Ans., — . 4. What is the essential siffn of ^^-^7 — t. — , when ® —2y^—3cx X = 4:, a = —2, y — —1, and c = Q? Ans., +. IfjO. There are five principal reductions required in operating with fractions, viz. ; To Loivest Terms, — From Improper Fractions to Integral or Mixed Forms, — From In- tegral or Mixed Forms to Improper Fractions, — To Forms having a Common Denominator, — and from the Complex to the Simple Form. Prob. 1. — To reduce a fraction to its lowest terms. Rule. — Reject all cojnrnon factors from both terms ; or divide both terms by their H. C. D. Demonstration. — Since the numerator is the dividend and the denominator the divisor, rejecting the f^ame factors from each does not alter the value of the fraction (100). Having rejected all the common factors, or, what is the same thing, the H. C. D. (which contains all the common factors), the fraction is in its lowest terms (154). EXAMPLES. g^3 Qax^ Ex. 1. Reduce -^^-^^^^-^ to ite lowest termB. Model Solution. — Resolving the terms of the fraction into their pnrae factors, I have — -r — — ^ — .r~,-;j = ^ ^ ^ - tt^-^- Now, cancelling the common factors, 3, a, and a + x, which is dividing dividend and divisor by the same quantity and hence does not alter the value of the fraction, I have — , or -— . Since in a{a-\-x) a^ + ax this there is no factor common to numerator and denominator, it is in its lowest terms (154). op, 15b^y 110m.TV I2a^x^ irya^r^x^ their lowest terms. 108 .FKACTTONS. ^ ^ ^ 15¥xy llOinx^if Ua^x^ ^ 4:ba^c^x^ , 2. Keduce ^^ /„ , ^^^ . % , ^^ o . , and ^ ^ „ „ to their lowest terms. 1 iX!' 36^ iJ«.»?<. (not given in order), 3^, — ,, 3a^a:,and -^^. 3. Reduce -,— „ ^-j^^, and g^.^^^;^— 9. to their ic 2 1 lowest terms. Results (not in order), t-o, ?7, and ^ h^ 3 a + x 4. Reduce -^^^-, Ub + W ^ "^^ .tH^^"^"^^ *" . , . - . , T» 1. x^—hx n—\ , 3a2_3aJ their lowest terms. Results, — — r-, — -^, and rr 5. Reduce » »— to its lowest terms. 6. Reduce TnrrT — ^^rV^ ^^ ^^^ lowest terms. 3a;2 + ^xy + 3«/2 7. Reduce — ^„ — ^r— - to its lowest terms. ^ ., 5(^-1) Result, z~, — —A* ' 7 (w + 1) 8. Reduce » ^^ ; — ^ to its lowest terms. y^—%xy-\-a? Result, t±m±^. y—x 9. Reduce -s — : 7 to its lowest terms. n^—An-\-A: rS ^6 10. Reduce -j ^ to its lowest terms. Result, s— -^ Scholium I.— Whenever the common factors of the term are not readily discernible, use the process for finding their H. C. D. (13t). 11. Reduce ^—„ ^-^ , ^J, and ^-^^^ to their lowest terms. ?, a2 4.«J_|_^,2 ^JT^ __:r_^jr^ and -2T-:72* REDUCTION. 109 ,« T. J 6ar* — 7a; — 20 Sax^ — Wax -\- 3a , 12. Reduce -r-^ -r^r — -- r , ^-^-5 ^-9 577-, , and 242^4-220:2 + 5 ^ ^, . , to their lowest terms. 48a:*H-16a;2_15 Results, -: „ , » , iTT— r-T — ? i"ld 4a?'+3' 10« + 5«rc 2a?J-f5a;— 1 Scholium 2. — Tlie opposite process is sometimes serviceable, viz. : the introduction of a factor into both terms of a fraction which will give it a more convenient form. x-4-1 x^ 1 13. What factor will change „ - to -3—:; ? Ans.j a;— 1. Ans., a-i-b. ih'dt factor will change -^ ^— ^ — fl8 + 8rc3 14. What factor will change 7 — to -^ — r, ? 15. What factor will change ^-^^^^+^— to 16a:*' Ans.f a-\-2x. 16. What factor will change ^ — ^r-^- — jtk to ° a^—2ab-\-4:if^ -16^^P fl3 + 8^ • [Note. — It requires no special ingenuity to solve such problems, since, if the factor does not readily appear, it can be found by divid- ing a term of one fraction by the corresponding term of the other.] 160. Prob. 2. — To reduce a fpaction from an improper to an integral or mixed form. Rule.— Perform the division indicated. (141.) Demonstration- — The operation is explained in the same manner as the corresponding case in division. EXAMPLES. Ex. 1. Reduce 9"~~~ *^ ^^ integral or mixed form. 110 FRACTIONS. Model Solution. — This being an indicated operation in Division, I have but to perform the division. Now, since the sum of the quotients is equal to the quotient of the sum, I have but to divide each term of lOax + c—h by 2x, or divide such as I can and indicate the division of the others, and add the results (112). Thus I find that = 5a + — ~ . 2x 2x 2. Reduce — — ! , — ^—^ ! — , and to integral or mixed lorms. r, T, n 4:a — 6" ^ ah , „ ^ ^ Results, 3a — , 6y-\-l-\-— , and a— 3c— 2 +— • oa X oa 3. Reduce — ~ — , — ' -—, — — , ^, and a:4-3 ' a + 'Zh ' ^-\-y to intearal or mixed forms. a-\-x ° Results, x^—xy-\-y'^ ^, a; + 9 -^ or 6 + 2a; — X -p y X -\- o x^ c -, and a-\-2b-\ -r^ x-{-3' ^ ^a + 2b 4. Reduce , ^ , ^ ^ ^ — t?— , and ^ a—V x—y' a?-{-'2ah + l^ ' x-\-y to integral or mixed forms. Results, a-^b, x^-\-x^y-{-x^y^-i-xy^-\-y^, a^—x^y-{-xy^—y^, and a^-j-a-{-l. 161, Cor. — By means of negative indices (exponents) any fraction can be expressed in the integral form. Til 1 1 771 Thus — = mx-: and by (43) - = ar-\ .-. — = mx-\ X X ^ ^ ■' X X 5. Express -^-^ , and =- in the integral form. ' x^y^ mx^ ° 6. Show that -^4v-o = m-%^ + 3/>rVZ» + 3w-%^ m (a + b)~^ + m-^b% REDUCTIONS. Ill integral form. Results, (x-\-y)~^, a^a^y, mH^—m^nx^—mn^x^-{-nH^, and 7 X %-hr'y. 162, Prob. 3. — To reduce numbers from the integral OP mixed to the fractional form. Rule. — Multiply the integral part by the given denominator, and annexing the numerator of the fractional part, if any, write the sum over the given denominator. Demonstpation.— In the case of a number in the integral form, the process consists of multiplying the given number by the given denominator and indicating the division of the product by the same number, and hence is equivalent to multiplying and dividing by the same quantity, which does not change the value of the num- ber. The same is true as far as relates to the integral part of a mixed form, after which the two fractional parts are to be added together. As they have the same divisors, the dividends can be added upon the principle that the sum of the quotients equals the quotient of the sum (103). EXAMPLES. Ex. 1. Reduce ^a—x^A to a fractional form. a—x Model Solution. — Multiplying 2a— x^ by a— a;, I have 20^— ax^ —2ax-^x^. Now indicating the division of this result by 3a— x', I 2a'-aa''—2ax + x^ „ , dax-Aa"" 2a'-ax'-2ax-\-x' nave . .\2a—x^-\ = a—x a—x a—x 3^23! 4a' H . But, as the sum of these two quotients equals the a—x quotient of the sum, I have, after uniting similar terms, ar"— Gur' + aa:— 2a' a—x 2. Reduce a—h-\ j to the form of a fraction. Result, — -v- 112 FRACTIOITS. 2x 3. Eeduce 1 -\ to the form of a fraction. y—x Result y^^' y—x 4. Reduce a-\-h 7 — to the form of a fraction. a—o Result, 7, or -, a—V b—a 5. Reduce x-\-'Z-\ ^— to the form of a fraction. Result. ^' x—% O/p 5 6. Reduce bx — to the form of a fraction. o Result, — o 7. Reduce x-{-l-\ to the form of a fraction. X Result, ^-±^2. X 8. Reduce 2a— 3^' + 4c H ~J~b ' ^' J. T> 1 2abx—2b^x , . , . „ x(a~b) 9. Reduce .'c ^—ni — to a fraction. Res., -^ ^. a^—b^ a-\-b 7 10. Reduce x-{-y ^—^ — Result, 11. Reduce x^ + %xy + f-'^:^^^y^''-t=t. Result, ^l^±f). x+y d^x 0^7? 12. Reduce -^ -^—{a?x?-\-a}x). Result, -, ^• 13. Reduce 3«-9- ?^-|?. Result, ^ a+3 ' «+3 163, Prob. 4. — To reduce fractions having different denominators to equivalent fractions having a common denominator. REDUCTIONS. 113 Rule. — Multiply hot) i terms of each fraction by the defio7}Unators of all tlie other fractions. Demonstration. — This gives a common denominator, because each denominator will then be the product of all the denominators of the several fractions. The value of any one of the fractions is not changed, because both numerator and denominator are multi- plied by the same number (100). EXAMPLES. Ex. 1. Reduce the fractions -, -^^, and --^^ to eauiv- y' a + J a—h ^ alent fractions having a common denominator. Model Solution. — Multiplying both terms of the fraction - by a +6 and «— J, or by a"— IP, I have r^ which has the same value 2J as -, since the numerator and denominator have been multiplied by the same number. In like manner multiplying both terms of , by y and a-5, 1 have 2^^|^iV , the value of whic^is 2—5 the same as = , since, etc. Finally, multiplying both terms of by y and rt + 6, I have — ~'~-^- — j;— -, which has the same value as ^ ? since, etc. These fractions have the common a—b denominator a^y—Vy, as in each case the new denominator is the product of all the old ones. 2. Reduce ^r-> ^ > w-^ and -o to forms havinff a 0. D. Queries.— By what are both terms of-- to be multiplied? By what both terms of — ? By what both terms of --^? 2c Ix results, 28c/t2a:y' 2Scn^xy' 28^^y' '''''^ 28^^* 114 FRACTI01!?^S. />j iC -I— 1 1 X 3. Reduce ^ , — r— , and to forms having a C. D. n h 4. Reduce 7, and — , to forms having a C. D. „ ^^ a^—db T al-^W- Results, r, — 75 , and — ^. Scholium.— Practically, this method consists in multiplying all the denominators together for a new denominator, and each numer- ator into all the denominators except its own for a new numerator. But it is much better to repeat the rule as given above, and let that be the form of conception, as it keeps the principle constantly before the mind. 1G4:» Cor. — To reduce fractions to equivalent ones having the Lowest Common Denominator, find the L. C. M. of all the denominators for the new denominator. Then multiply both terms of each fraction by the quotient of thai L. G. M. divided by the denominator of that fraction. Demonstration. — The purpose in getting the L. C. M. is to get the lowest number which can be divided by each of the denominators. That the process does not change the value of the fractions is evident from (100), the same as under the general rule. EXAMPLES. a €? cti^ 5. Reduce :; , -. rr, and 7- r^ to equivalent frac- 1—a (l—ay (1—^) tions having the L. 0. D. Model Solution. — By inspection I observe that (1—ay is the L. C. M. of the denominators, since it is the lowest number which con- tains itself, and it also contains each of the other denominators. Now, to make the denominator of- , (1— a)^ I must multiply it by (1— «)^-f-(l — a) ; ^. e., by (1 — ay. But to preserve the value of the fraction, I must multiply the numerator by the same quantity. Thus a _a(l — <^y _ a — 2a'^ + a^ izia-'ii:^-'-^ . ..o— ; etc. REDUCTION'S. 116 6. Keduce ^ , to forms having the L. C. D. x—y x-\-y Results, -'-^y^-^y+y\ .^^ t±^±^p:t. 7. Reduce , ;.rz~^ "IT ^^ ^^^^^ having the L. C. D. Results -^ h(x^-xy^f) c{x^-xy + f) 8. Reduce mn, ~ , — ^^ to forms having the L. C. D. m-\-n m—n „ ,, mhi — mn^ (m — 7iY ^ (m + nY Results, — 7 — ^, --^ — ^, and ^ ^-' 9. Reduce , - — -^, - — -3 to forms having the L.C.D. Tlie L, a D. is l-\-x—3^—x^. 10. Reduce -, — ^- , and -. 777, to forms having the L. CD. a^—¥ {a—hy 11. Reduce , —z 5, to forms having the L. CD. m + n m^-\-n^ m-\-n ° 12. Reduce -, ^ :; , and -7-^ — - to forms having the L.CD. /6VT. Prob. 5. — To reduce Complex Fpactlons to the form of Simple Fractions. Rule, — Multiply numerator and denominator of the complex fraction by the product of all the denom- itiators of the partial fractions found' in them. Or, multiply by the L. C. M. of the denominators of the partial fractions. Demonstration. — This process removes the partial denominators, since each traction is multiplied by its own denominator, at least, and this is done by dropping the denominator. It does not alter the value of the fraction, since it is multiplying dividend and divisor by the same quantity. 116 FRACTIONS. EXAMPLES. Ex. 1. Reduce -— to a simple fractional form. Model Solution. Explanation.— In order to free the numerator of its denominator, 2x 3&', I multiply the numerator -^ by d¥ ; but in order that this OCT may not change the v^lue of the fraction, I also multiply the denominator by the same. In like manner to free the denominator 4v -~ of its denominator, I multiply numerator and denominator by 0(1 5a^ Indicating these operations I have j . To multiply %,x5a''xSl>' o« -- by 36" I drop its denominator and have 2a;, which multiplied oO by 5a' gives for the new numerator lOa'^x. So, also, I obtain the new denominator by dopping 5a^ and multiplying 4y by 3i^, getting lOa'^x thereby 12&'y. Therefore the simple fraction is j^r^- which reduced to lowest terms is -^r-. 2. Reduce — — to a simple form. Result, —-- — "^ cm 3. Reduce — -| to a simple fonn. Result, ac+l ^ ' c 4. Reduce — ■ to a simple form. Result, ^^ x^— cnx-^cm n ADDITION. 117 5. Reduce to a simple form. y 5 3^ 6. Reduce 777 — -~ to a simple form. lO + ^a; ^ a X 7. Reduce - — ^ to a simple form. m n h—a fl + 8. Reduce uh—a^ ^^ ^ simple form. 'HON IIL 166, Prob. — To add Fractions. Rule. — Reduce them to forms having a comm^on denominator, if they have not such forms, and then add the numerators, and write the sum over the com- mon denominator. Demonstration. — The reduction of the several fractions to forms having a common denominator, does not alter their values (163), and hence does not alter the sum. Then, when they have a common denominator (divisor), the sum of the several quotients is equal to the quotient arising from dividing the sum of the several dividends by the common divisor, or denominator (103). 118 FRACTIONS. EXAMPLES. Ex. 1. What is the sum of , :,, and -• Model Solution. Operation.~The L. C. M. of 1-x, 1—x'' and 1—x^ is (l—x^) x{l+x) = l+x—x^—a;\ (l+a;)x (1 + 2x + 2x'-\-x^) _ l + dx + ix'' + Bx^ + x* (1— a;)x(l + 2.x + 2«^+ ic^ "~ 1+x—x^—x* (\+x'')x(l+x + x^) _ l + x + 2x^ + x^ + x* {i—x'')x{l+x + x^ ~ 1+x—x^—x* (l+a;")x(l4-g^) _ 1 + x+x^ + x* (1— «") X {1+x) ~ iTx—x^—x*' 1+x l + x"" 1+x^ l + Sx + 4x'' + Sx^+x* • l—x ' 1—x' 1—x' t+x-x^—x* l-{-x-\-2x'^ + x' + x'- 1+x + x^ + x* (A.) 1+x—x^—x* 1-^x—x^—x* ~ 1+x—x^—x* Explanation. — Explain the reduction to a common denominator as under (163) unless that is already sufficiently familiar. Having reduced the fractions to the L. C. D, I find (read A). Now since the sum of these quotients is equal to the quotient arising from dividing the sum of the several dividends, or numerators, by the common divisor, or denominator (103), I add the numerators and write the sum over the common denominator, which gives d + 5x + 6x''-[-5x^ + dx* ^, ,. 1 + x 1+x' , 1 + a?' '-—-—- ^ r — foJ* the sum ot , -, and -„. 1 + x—x^—x* l—x 1—x^ 1—x^ « *m ^ c—a , c-\-a ^ 6x-\-'7c—ff 2. Add ^r-, — — , and— — Sum, — „ . .. a c b ,rt ^ a^b-rcb-j-il^+-a^ 3. Add Y, -I, ~, and r-.. Sum, ^ b ab a I^ aW 4. Add 3^ ^j 12' 18' V ^^^ 9' ^^'^' ^' 5. Add — r— and — ^- Sum, — - — ADDITION. 119 6. Add :r-— -, :; , lUld :— — • Sum, l+« 1— « l+fl I— a 7. Add . and -- — -„• aSmw, - — ^. 8. Add - — - and , Sum, \-\-x 1-^x ' 1— a^ 9. Add -5 — ^ and 7 — -775- /S'wm, -^ ^y iTTTS* 10. Add ::^-,-f- ^»d — — • a^—y^ x-\-y ^—y Sum, ?^f-te. :/67. Cor. — Expressions in the mixed form may either be reduced to the improper form and then added, or the integral parts may he added into one sum, and the fractional into another, and these results added. Ex. 1. Add 7a:+^ and Sa;4-^^- o ox 17B8T FOBM OF OFBBATION. «— 2 22a;— 2 7x -I- 8a; + 3 3 3a;+4 40x' + 3a;+4 5a; 5x (22x-2)x5a; _ llOx'- lOx 3 X 5.r "" \5x (40x'4 -3x+4)x3 _ 1^20^H9« + 1_2 5a? X 3 ~ 15x llOx'-l Ox 120^ jf ?^ 13 __ 230a;''-x4-12 15x 15x ~ 15a; SECOND FOBM OF OPERATION. 7.r + 8x = 15x (a;— 2) X 5a5 _ 5x'— lOx 3 X 5x 15x (3x + 4)x3 _ 9ic + 12 5x X 3 ~ 15aj 120 FRACTIONS. ... 7. + — + a. + ^- = 15X+ _-- + ,— = 15,,. "^ 15x Explanation. — Since the sum of several numbers is the same in what ever order their parts are added, I may add the integral parts first. Adding 7.r and Sx I have 1 5x. Reducing the fractions to forms havin;^- a common denominator, ' becomes — -- — -, and — - — becomes ;j 15;r 5x — — — . Adding these I have .' ~ which added to 15x, the sum of the integral parts, gives for the entire sum 15.r 5a;'— a;+12 ■*" 15^ • 2. Add 2x, 3.^■-f-— , and x-\--^- iSum, ^^-^tt' « Ajj Sa^ ^ . 2ax ., , , , 2abx—3cx^ 3. Add a 5- to Jh ^^ x—y x+y Model Solution. Operation, (x—y) (x-\-y) = x^—y"^ {x-\-y)x(x+y) _ x' + 2xy+y* (x-y)x{x + y) x^-y^ (x—y) X (x—y) _ a;' — 2.ry + y» (x-\-y)x{.r—y) x^—y^ (x' + ^.ry^ 2/-^._(.r»— 2 ry -f- f) = 4xy G 122 FRACTION'S. x + y x—y _ ^xy x—y x + y x'^—y' Explanation. — The L. C. M. of x—y and x+y is their product, since tbey have no common factor. Hence x^—y" is the L. C. D. To reduce X + 11 — ^to a form having this denominator,! multiply both its terms bv x—y x^ + 2a?y + 11^ x+y^ which gives y~ ^ t - ^^ like manner multiplying both X y x-^y ^ XI x'^ — 2xy + y'^ _, tenns of by x—y, I have — v- o . I have now to subtract x+y x'—y^ /J.2 ^xv + v'^ x'^ + 2^ + v^ ^ \ from ^ „ ^ . Since the difference of the quotients x^—y^ x^—y^ ^ of two numbers divided by the same number, is the same as the quotient arising from dividing the difference between those numbers by the common divisor, I take the difference of the numerators (the quantities to be divided) which is 4a^, and dividing it liy x'^ — y^ I have ixy x + y X- for the remainder of — - less — 2. From ^-^ take l-\-x 1 a-\-x x-\-3 3. From 4. From 1—x 1 a — x X x—3 5. From dx take take take X Sa-{-12x Remainder, Remainder, Remainder, Remainder, 4:X 1—x^ 2x q2 — ^2 9__ x^—3x 3x—Sa 6. From dy take ^-^• 1 , , 1 7. From 8. From take a^—2ab-hb^ Remainder, — ^^ — o a—b—1 a — i _1_ ^^^^ ^—x^—3x 0^ + 4 ^^^ x^^\fdx-\-%^ Rem., Rem., {a-hf . ^ 3x + 2 , , lax—AOa „ . , 4(3-.r) 9. From take Remainder, -^ <- SIBTRACTIOX. 133^ 34.22; 2— 3a; 10. Combine the following fractions 2—x 2-\-x H 5 — 7-' Result, -— --■ ^ aH— 4 x-\-2 1 1 11. Combine a {a — b) {a — c) b {b — c) (b — a) + —, w IT- Result, -J-' c(c — a){c — b) abc ,^ ^, ,. Sa—^b 2a—b—c 15a— 4c a—ib 12. Combine —^ + -^ ^^ • Sla-4:b Result, 84 13. Combine — -^ + a-\-b ' rt2__52 a^^b^ Result, —. — :,-' 14. Combme ^- — .--rr^ 7-5 — r* Result, 0. -2a; 14- 2a; ^7?—\ ' 15. Combine ^t-t — — -r — ^ttt-, r^ — ^-7;^ — --^ • 2(a; + l) 10 (a;— 1) 5(2a;4-,3) Result, - 16. Combine -^ 5 + ;.r2-l) (2a; + 3) 2f?.9. Cor. — Mixed numbers may be subtracted by annexing the sid)lraJiend with its sign^ changed, to the minuend, and then combining the terms. The reason, for the change of signs is the same as in ichole numbers (77). EXAMPLES. 17. From 3a; + ?i^ take x - ^^. m m 2x Remainder, 2x -\ • m iU FRACTIONS. 18. From x jr— subtract 7a; Remamder, — - -— 6a; — a. 19. From 'da take dx + l)la Remainder^ 3a— Sx 20. From 2a; 4 ^r— take 3a; -^ — Remainder, 16a; + 23 42 ultinlicatinn ECT10N ¥= 170. Prob. 1. — To multiply a Fraction by an Integer. Rule. — Multiply the numerator or divide the denominator. Demonstration. — Since numerator is dividend and denominator divisor, and the value of the fraction is the quotient, this rule ia a lirect Nionsequence of (101, 102). EXAMPLES. Ex. 1. Multiply -X by m-{-n. Model Solutlpit Operation. ^— x(m+«) = -^^ . Explanation. — Since the value of a fraction is the quotient of the numerator divided by the denoc.iinator, and multiplying the divi- MULTIPLICATION. 1^5 dend multiplies the quotient, I multiply the numerator of ^~ bym + w, and have — ;; , which is therefore m+»timeg m—n 3xy • 2. Multiply 1^ by 3«. Model Solution. Operation. 3^,jiX3«=^r- Explanation.— Since 27W« is to be divided by 3)2. 174. Prob. 2. — To divide by a Fraction. Rule. — Divide by the niuiierator and multiply the quotient by the denominator. Or, what is the same thing, invert the terms of the divisor and proceed as in multiplication. Demonstration. — The correctness of the first process appears from the fact that division is thereverseof multiplication, which requires that we multiply by the numerator and divide by the denominator. The process of inverting the divisor and then multiplying by it is seen to be the same as the other, since this also multiplies the dividend by the denominator of the divisor and divides b^ the numerator. 132 FRACTIOlSrS. Again, this process may be demonstrated thus : Inverting the divisor shows how many times it is contained in 1. Then if the given divisor is contained so many times in 1, it will be contained in n, n times as many times; or in any dividend as many times the number of times it is contained in 1, as is expressed by that divi- dend, whether it be integral, fractional or mixed. Scholium I. — Since to multiply one fraction by another we may multiply the numerators together for the numerator and the denom- inators for the denominator, and since division is the reverse, we may perform division by dividing the numerator of the dividend by the numerator of the divisor, and the denominator of the dividend by the denominator of the divisor. This method will coincide with the others when they are worked by performing the operations by division as far as practicable, and this is worked by performing the multiplications equivalent to the divisions when the latter are not practicable. EXAMPLES. Ex. 1. Divide ^^, by — f-. Model Solutions. OPBBATION BT THB FIRST METHOD. x^ + y' 2y X' + ': X («+y) = iJ Explanation.— I first divide ^-^ by y, by dividing the numera- x^+y tor (101). But the given divisor is y divided by x + y; and as dividing the divisor multiplies the quotient (102), I must multiply this quotient, -J-—, by x-\-y. Performing this by dividing the X •\-y denominator, I have for the true quotient -^ -„ ' ^ x^—xy + y^ OPERATION BT THB SEOOKB METHOD. y __ ^'' ^ x^y _ 2y x^+y^ ' x + y x^+y^ ' y x^—xy+y"" biVisiox. 133 Explanation. — By inverting the divisor and indicating the multi- plication of tiie dividend by it, I indicate that the dividend is to be multiplied by xA-y and divided by y, which are the opei-ations required, lu this instance I perform the multiplication by ^-f-y, by dividing the denominator, and the division by y, by dividing the numerator. The operation by the third method is of the same form as the last. BXFLAKATION BT THS THIRD KBTHOD. I am to find how many times -^ is contained in ~^-. Now x^y aj' + y" -^— is contained in 1, — - times: since y is contained in 1, - «+y y y 1 X '\-'U times, it is contained mx-\-y.x-{-y times -, or — - times. Hence if y y V . . . . . x+v . . . . , . 2^^^ 2v* -^— IS contained m 1, — - times, it is contained m , ^ „ -r^—^ «4-y y x'+y'' x'+y' y times —^ 2. Divide , . ,„ by -^ — .• Q'^^*-^ / , \' « ^. .J 4 (a^—ab) ^ Qab /, ^ 2 (a— by 4. Divide «-+3a^+3..^ + ^ , J^^ Quot, — i— . x—y 3 ^^"'^ 12a: 6. Dmde — ^ by -^- C^o^., fl T^- -^ ^^ u ^ ^ / 2a;(a; + l) 6. Divide ^-— ^ by _• q^ot, -^^. 8. Divide -^£7. I^v '^. C^o^., 1. a*— 2«d4-62 " a^b a 134 tRACTION^S. 9. Divide 3^---^^ by j--^. Qmt., ^-J^. ,- ^. ., 1x . 2x ^ , x—\ 10. Divide x-\ ~ by x Quot., -- x—d *' x—^ ^ x—5 11. Divide xf^ •. by x • x^ -^ X Suggestion. — This quotient should be written by inspection in the same manner as {x*—y*)-7-{x—y), and is x^ + x + - + —■ Or it may be performed as follows 1 X X X* — - x^ + X -\- - + - -1 XX* X* x-" - 1 L _ L X^ X* L _ L X^ X* 12. Divide j-~ + j^ by f^-j~^- Quot., I. Suggestion. 1 X _l+x'' _1 X __ 1+x* 1+x'^ \—x ~ 1—x^ 1—x 1+x ~ 1—x^' 13. Divide — ^ + — -^ by —^— -^• a;— ^^ x-\-y '' x—y x-\-y «-■■ ^- DirisioK. 135 16. Divide :— ^ by -7; ?< Quoty 3x^ — xy — 2y2. Scholium 2. — It is sometimes conveniedt to write the divisor under the dividend in the form of a Complex fraction, and then reduce the result to a simple fraction by (165). 16. Divide 1-f- by 1—^. L aj (T + d a Operation. [-yx"' " a=»_l a-i Operation. (ib* + a¥—ab* U + ^-llxaJ" _ &=> + ! _*+l -a5='(6»-6+l) ~ a5« ' 18. Divide Trhj—if by xy-^-^x'^y. 3 . r^,-2/^1xxy , 3 y—y _ La;- J _ y'— x'g Suggestions. 3 r- = :i ^ = —. 5 • 19. Free ,3 '^ _^ from negative exponents. Result, — ^— Ts-i* axy^-\-l^y^ d-s J-2 20. Free _^ , _3 from negative exponents. a^b^ + a^ 136 FRACTIOKS. 21. Free — Tjf-o ^^^"^ negative exponents. mn~*e I ^-2 yZ 22. Free r- ^ ' -^ from negative exponents. ^^•' ~^^-a^-ha^xy^ ' 23. What is the reciprocal of --^^J Solution. — The reciprocal of a quantity being the quotient of 1 divided by that quantity, the reciprocal of -^— ^ is 1-^ -.r"^2> ^' 17o, Scholium 3. — Thus it appears that the reciprocal of a fraction is the fraction inverted. The reciprocal of a quantity with a negative exponent is the same quantity with a positive exponent. -2./-S x^ifi 24. What is the reciprocal of — ^^-r- Ans., -f-- 25. What is the reciprocal of 7 ~^,? (a—b)-' [To be taken in review.] 26. Multiply ^~- by ^- • Prod., ^--. «» — lA a^—h* a»—o» SYNOPSIS. 137 SYNOPSIS FOR REVIEW. z O 2i < I fiC I it o "Si FRACTION. FORMS. ( Terms, NUMBRATOB. Value of fraction. J Cot. 1. — Multiplying or dividing Denomxmatob. I numerator or denominator. I Cor. 2.— Removing denominator. Fractionai.. i Proper. J Mixed. I Improper \ Simple. Compound. Complex. Lowest terms.— L. C. D.— Redaction. Of numerator, of denominator, effraction. Essential sign effraction. PROB. 1.— To lowe&t termt<. Rule. Bern. Sch. By H. C. D. PROB. 2.— Prom Improper to integral or mixed forms. Ruuk Dem. Cor. Negative exponents. PROB. 3.— Prom integral or mixed to fractional forms. Rule. Dem PROB. 4.— To common denominators. Rule. Dem. Cor. L. C. D. Dem. PROB. 5.— Complex to simple. Rule. Dem. ADDITION. Pkob. Ritle. Dem. Cor. Mixed Numbers. SUBTRACTION. Pbob. Rule. Dem. Cor. Mixed Numbers. (Pbob. 1.— Fraction by integer. Rule. Dem. Sc/i. MULTIPLI- 3 CATION. Pbob. 2.— Any number by fractions. Rule. Dem. { Sr-h. '\>r. Mixed Numbers. (Pbob. 1.— Fractions by integer. Rule. Dem. DIVISION. -nProb. 2.— An V number by fraction. Rttlb. Dem». ( 1,2,8. 5Wi*. 1,2,3, Test Questions. — Upon what five principles in Division are moat of the operations in fractions based? Why does the process of reducing to forms having a common denominator not change the value of a fraction ? Give the rules for Multiplication and Division of Fractions, and the reasons for them. c//; lY. A^^ Section i INVOLUTION. The operations in Radicals are all based upon the most elementary principles of Factoring.* If the student learns how to use this key, he can unlock all the mysteries of the subject. GENERAL DEFINITIONS. 170. A Power 'n a produd iirisiug from multiplying a number by itself. The Degree of the power is indicated by the number of factors taken. Thus 2, 4, 8, 16, and 33 are, respectively, the 1st, 3nd, 3d, 4th, and 5th powers of 3. Scholium. — It will be seen that a power is a species of comi3osite number in which the component factors are equal. 177, A Root is one of the equal factors into which a number is conceived to be resolved. The Degree of the root is indicated by the number of required factors. * The eubjects treated in this chapter are among the most difficult, if not actually the most difficult for the pupil in the whole science. In the examination of hundreds of students from all parts of the country, the author has found that the rule is that they are deficient in knowledge of Badicals. An attempt is here made to assist the teacher in remedying this defect, by constantly holding tha attention to this one central principle. INVOLUTION. lS(i Thus, 2 is the 1st root of 2, tlie 2nd root of 4, the 3rcl root of 8, the 4th root of 16, the 5th root of 32, etc. 178, Scholium I. — P/ are correlative terms. Thus 32 is the 5th power of 2, and 2 is the 5th root of 32. 170. Scholium 2. — The Second Power ia &\so c&Wed the Square ; the Third Power ^ the Cube ; and, sometimes, the Fourth Power ^ the Biquadrate. In like manner the 2nd root is called the square root ; the third root, the cube root ; the fourth root, the biquadrate root. These are Geometrical terms which have been transferred to other branches of mathematics. The second power is called the square, because, if a number represents the side of a square, its second power represents its area, or the square itself. Conversely, if a number represents the area of a square, the square root represents the side. Also the third power represents the volume of a cube, the edge of which is the first power or cube root. Biquadrate means twice squared, and hence the fourth power. 180. An Exponent or Index^ is a number written a little to the right and above another number, and indicates 1st. If a Positive Integer, a Poiver of the number ; 2nd. If a Positive Fraction, the numerator indicates a 'Poiver, and the denominator a Root of the number; 3d. If a Negative Integer or Fraction, it indicates the Reciprocal of what it would signify if positive. Illustration. 4' is the 3rd power of 4, or 64. a"' is the with power of a, if m is an integer. 4"' (read " 4, exponent -3 ") is j^, or — . . 1 2 a-"» IS --. 8* is the cube root of the square, or the square of the cube root of 8, or 4. a» is the mth power of the nXh root of a, or the nth root of the mth power, if m and n are both integers. a~^< is — . Scholium. — It is obviously incorrect to read 4t, " the ^ power of 4." There is no such thing as a 2-fifths power, as will be seen by considering the definition of a power. Read 4l, " 4 exponent | ; " 140 POWERS AKD ROOTS. m m also a», " a exponent ^ " ; a »', "a exponent — f." These are abbre- viated forms for, " a with an exponent — f ," etc. In this way any exponent, however complicated, is read without difficulty. 181. Cor. — A FACTOR can be transferred from numerator to denominator of a fraction, or vice versa, bij changing the sign of its exponent, without altering the value of the fraction. Thus ^'"^" = ?!^- for ^'"^~" — __?! _^_cr y » _ «V 182. A Radical Number is an indicated root of a number. If the root can be extracted exactly, the quantity is called Rational ; if the root can not be extracted exactly, the expression it called Irrational, or 8urd. Thus the radical ^25a" is rational, but ^^\Za is surd. 183. A Root is indicated either by the denominator of a fractional exponent, or by the Radical Sign, y'. This sign used alone signifies square root. Any other root is indicated by writing its index in the opening of the v part of the sign. Thus '^am, \^am, are the 3rd and 5th roots of «m, and the same as (aw)i, («w)i. 184. An Imaginary Quantity is an indicated even root of a negative quantity, and is so called because no number taken an even number of times as a factor, produces a negati\ e quantity. Thus \/—4: IS imaginary, because no number multiplied by itself once produces —4. Neither +2 nor —2 produces —4 when squared. For a like reason y^— Sa'*, \/—5x, or y^ —14:0xy^ are imaginaries. ISS, All quantities not imaginary are called Real. INVOLUTION. 141 1S6. Similar Radicals are like roots of like quantities. Tims 4 V^j '^^ V^"> ^^^ {a^—x^) \/5a are similar radicals. IS't. To Rationalize an expression is to free it from radicals. ISS, To afifect a number with an Exponent is to perform upon it the operations indicated by that exponent. Thus to affect 8 with the exponent | is to extract the cube root of the square of 8, or to square its cube root, and gives 4. 1S9, Involution is the process of raising numbers to required powers. li)0. Evolution is the process of extracting roots of numbers. 191, Calculus of Radicals treats of the processes of reducing, adding, subtracting, or performing any of the common arithmetical operations upon radical quantities. INVOLUTION. 192. Prob. 1. — To raise a number to any required power. Rule. — Multiply the tiitniber by itself as inany times, less one, as there are units in the degree of the power. Demonstration. — Since the number of factors taken to produce a power, is equal to the degree of the power (1T6), it follows that to obtain the 2nd power we take two factors, or multiply the number by itself once; to obtain the 3rd power we take three factors, or multiply the number by MoeXHwke ; and in like manner to obtain the nth power we take n factors, or multiply the number by itself n— 1 times. EXAMPLES. 1. What is the 3rd power of 'Za^ ? 142 POWERS AND ROOTS. Model Solution. — Since the 3rd power of 2a^ is the product arising from taking it 3 times as a factor, I have 2a' x 3a' x 3a' = 8a^ 2. What is the 4th power of —lOah (_10«i) X (-lOa^) X (- lOai) x (- 10«4) =3 lOOOOa^, Answer. 3 to 6. What is the square of —bmhi ? Of 6aH ? Of --a-n? Of-—? 5 5^>2 • a^ 9 9 Answers, 257nhi^, S6j- , — ^-- , ^r^a^b-^. 7 to 10. What is the cube of —2al^? Of 12^"^? Of _^^.. Of-^% 3 6n Answers, —Sa^¥, 172S7rr^- = ^^ , -^.a^ ^* ^yit 27 ' 216n3 11. What is the squareof 2a— 3a;? or, {2a— Sx) x (2«— 3^:)? 12. What is the 3rd power of 2d^^Zx ? Ans., 8rt«-f-36a^a; + 54a2a:2 + 27a:3. 13 to 17. Expand the following: {\-^2x-^%^f, {l^x j^x'^—^Jf, (a-^b—cf, (l+2a; + :^'")^ and {\~^x^'6x'^—7?f. Results, l-{-4tx + l^x^ + 12x^-\-^a^, l—2x-\-dx^—4.a^-\-Zx^ —2x^ + x^, l + Qx^lbx^ + 2^x^-^lbx^-itQ7^ + x^, and l — iSx 4-15a;2-20a;3^15^_6^_^^6. 1Q Q!U (27a4_18a2Z,2_J4)2 (9,,2_52)3(^^_^2) 18. Show that ^ ^-^^ 1 +^ 64^^^— =^' 19. What is the square of 9ar + - ? ^ns. , 81a^2 ^ is _(. _ . 20. Expand as above (4—^^)1 Result, 16-8^7^ + ^7, 21. Expand (a^-\-c^Y. Eestilt, a^-\-4:ah^-i-6ac-}-4:ah^-^c\ INVOLUTION. 143 193. Cor. — Since any number of positive factors gims a positive product, all powers cf positive monomials are positive. Again, since an even number of negative factors gives a posi- tive product, and an odd numbei- gives a negative product, il follows that even powers of negative numbers are positive, and odd powers negative. 104, Prob. 2. — To affect a monomial with any exponent. Rule. — Per form upon the coefficient the operations indicated hy the exponent, and multiply the exponents of the letters hy the given exponent. Demonstration. — 1st. Wheri the exponent in a positive integer. — Let it be required to affect 4a'"6"aj-* with the exponent p ; or in other words raise it to the pth power. It is obvious that, if ^nH'x-' is taken p times, each of its factors will be taken p times. Now 4 taken p times as a factor is represented thus, 4** Again, a"' taken p times as a factor becomes a'^P because a'" signifies a taken m times as a factor, and if this group of m factors is taken p times, a will have been taken mp times, giving a'^P. n 1 In like manner, &~ signifies &' taken as a factor n times, and if this grouj^ of n factors is taken p times, V will have taken np times, making pr Lastly, X— signifies — (180, 3d) ; and this taken as IP a factor p times give — . But as f = 1, this fraction is by (1§1) x-P* Collecting the factors we have ............... ^Pa"^Pb *" x-p^ Q. E, D, 144 POWEKS AND ROOTS. 2nd. When the exfponeni u a positive //'action. Let it be required to affect 4:a"'b'x-% with the txponeiit ' This means that 4a'"5~ar-» is to be resolved into q equal factors and j9 of them taken. Or that we are to find the qth root of the quantity and then raise it to the joth power (180, 2d). If we then separate each of the factors of n 4a»»&Ta!-* into q equal factors, and then take p of each of these, wi shall have done what is signified by the exponent -• By definition, one of the q equal factors of 4 is 4:5 Also by definition, one of the q equal factors of a«», or m the ^h root of a"* is represented ( 1 83) by «« To separate ft" into q equal factors, we notice that &~ is n of the r equal factors of h. Now, if we resolve each of these r factors into q equal factors, h is resolved into rq equal factors; doing the same with each of the n fac- . tors represented by 5~, and taking one from each set, we 2L have ft**« which is therefore one of the q equal factors of J>~. To resolve x-* = — into q equal factors, we consider that a fraction is resolved by resolving its numerator and denominator separately. But one of the q equal factors of 1 is 1 ; and o?ie of the q equal factors of x" h X9 as seen in the resolution of a"\ Hence otie of the q equal factors 11 -i of or*, or — , is — = x « a;' •- X9 Collecting these factors we find that one of the q equal « 1 pi n s factors of 4taH~ ar-* is 4:«ay h''*ix 7 And finally p of these being obtained according to Case 1st, gives 4:9a 'i'brqx ?, as thc expression for A:aH~x-^ affected with the expo- nent - ; which result agrees with the enunciation of the rule. INVOLUTION. 145 3rd. When the exponent is negative and either integral or fractional. u Let it be required to affect Aa'"h~'jr* with the exponent —t. This, by definition of negative exijoiients, signifies that we are to take the rcoii^rocal of what the ex|vcssi(>n. would be if^ were positive. But 4a"'&'ar-' aftected with the exponent t (positive) is 4:'a'"'b^x-*% whether t is integral or fractional, as shown in the preceding cases. The reciprocal of this is . But since these factors can be transferred to the numerator by changing the signs of their expo- nents, we have 4-'a-""J ~aj", as the result of aflFecting 4a'"5'ar-* with the exponent —t, which result agrees with tl»e enunciation of the rule. [Note. — The above demonstration contains the fundamental principles of the whole subject of the Theory of Exponents, and it is of the highest importance that it be made perfectly familiar. The application of the rule is so simple in practice as to afford no diflSculty ; but the reasoning should be given in full in a suflicient number of the following examples to jvx in the mind these principles of the theory. After this is done, the pupil needs only to perform the operations. The danger is that the liow being so simple, the why will be disregarded.] EXAMPLES. Ex. 1. Affect 2a2jfc-* with the exponent 5: that is, raise it to the 5th power. Model Solution. Operation. {M'lic-y = 32a'°&nrc-". Explanation. {2a^h^c-*y is 2a''l>h-* x 2a^l)^ 7. /I x„ , n(n—l) ^ 7i(n—l) (n—2) „ Result, (i-^xy'=: i-^nx + -^ — ^-x^-{ — ^^ f^ -V Scholium. — TLis expansion is in itself a very useful formula, and should be memorized. 7. Expand (3-?/2)i by the B. F. Result, (3-2/^)i = 3i-f^2^2_p^_3^2/« -y^—, etc. 128 8. Expand (l-\-x^Yhj the B. F. Result, 1 + 5a:2_^ 10^:4 _|_i0a;6^5;^^^ ^10. 9. Expand (l—a^)-i by the B. F. 13 5 35 Result, l4--a2+-a4+_a6_^__a8^^ etc. 10. Expand Va^—a^e^ by the B. F. Result, Va^—a^e^ = aVl^^ = a{l—e^)^ = «(1— h ^ 1 . 1.3 . i-3:i_^_ ^^ . Qp" — , etc.;. 2.4 2.4-6 2.4-6.8 11. Expand ^^^ by the B. F. (c-\-xy ^ ^ 1 ,, 2a: 3ic2 4a:s + etc. = -(!_-- + —__ +, etc.). IKVOLUTIOK. 151 12. Expand — ^ — r by the B. F. H , etc. ^625 ' I 13. Expand (x + y + c)* by the B. F. Suggestions.— Put (x -{■ y) = z. .-. (x + y + c)* = (« + c)* = «* + 42V + 62V + 4«c' + c* = (restoring the value of z) (x + y)* +4(x4-y)''c + 6(x4-y)V + 4(a5 + y)c» + c\ But (.r + y)* = a;*+4a;'y + 6xy + 4.ry^ + y*, (j- + y)« = r3 + 3x'y + 3xy' + y^ and (x + y)-^ = :c^ ^-2xy^-y^ Whence by substitution we have (x + y + c)* = x* + 4x''y + 6xy + 4xy' + y* + 4cx« + 12cxV + 12cxy' + 4cy» + 6c'V + 12c''ajy + 6cy + 4c'x + 4c='y + c*. 14. Expand (2a—b-{-(^Y by the B. F. Result, Sa^ — 12a^ -\- 6ai^ ^ I^ -\- 12aV — 12«^c2 ^ 3j3c3 196, Cor. 1. — Uie expansion of a binomial terminates only when the exponent is a positive integer, since only when m is a positive integer will a/actor of the form m(m— 1) (m— 2) (m— 3) etc. become 0. 197. Cor. 2. — When m is a positive integer, that is when a binomial is raised to any power, there is one more term in the development than units in the exponent. Since the first coeffi- cient is 1 ; the 2nd, m ; the 3rd, -^ ; the 4th, m{m-l){m-2)_ ^^^ ^^^^ m(,m-l) (m-^) (m-3) , ^^^ ^ we notice that the last factor is m — (the number of the terni —2) ; and the number of the term, therefore, which has m—m as a factor is the (7?i + 2)th term. But this is 0. Hence the (m + l)th term is the last. 152 POWERS AND ROOTS. 198, Cor. 3. — When mis a positive integer the coefficients equally distant from the extremes are equal ; since (a -f by* — (J _l_ a)^ ; the former of which gives a^ + ma"^^ ^i7i|m— ^^^_2^2_|.^ etc. Whence it appears that the first half of the terms and the last half are exactly symmetrical. 199, Cor. 4. — The stem of the exponents in each term is the same as the exponent of the power. Scholium. — The last two corollaries apply to the form (x-^-y)"^, and not to such forms as (2a^ — 35^)'", after the latter is fully expanded. 200, Cor. 5. — A contenient rule for writing out the POWERS of binomials may he thus stated : 1. The FIRST term contains only the first letter of the binomial, and the last term only the second, while all the other terms con- tain both the letters. 2. The exponent of the first letter of the binomial in the first term of the development is the same as the exponent of the required potver and diminishes by unity to the right, while the exponent of the second letter begins at unity in the second term of the expansion and increases by unity to the right, becoming, in the last term, the same as the exponent of the power. 3. The coefficient of the first term of the expansion is unity ; of the second, the eocponent of the required j^ower ; and that of any other term may be found by multiplying the coefficient of the preceding term by the exponent of the first letter in that term, and dividing the product by the exponent of the second letter +1. 4. There is one more term in the development than there are units in the exponent of the power. INVOLUTION. 163 This rule is a deduction from the formula (« + &)'" = a'" + ina"'-'b m (m-1) w(^-l)(^-2) _ wi(m-l)(7n-2)(m-3) a'"-*6* etc. The 1st point appears from Cob. 8. The law of the exponents is directly observable from the formula. The coefficients of the first and second terms are seen in the formula to be as stated. The coefficient of the third term may be written m x — - — , which is the coefficient of the second^ or preced- ing term (m), multiplied by the exponent {m—\) of the first letter in that term, and divided by 2 which is the exponent (1) of the second letter + 1. In like manner noticing any other term, as the 5th, its coefficient may be written — ^^ ^^- ^ x — -— . [3 4 But — ^^ —- ' is the coefficient of the 4th term, m—S is the exponent of the first letter in the 4th term, and 4, the divisor, is the exponent of the second letter (3) -1- 1. The 4th point appears from Cor. 2. 16 to 20. Write out by the above rule the expansions of the following : (m + ny, (x + y)^, (a -f- ^ )^ (x -f- m)% (iri^ + ^3)3^ Suggestions upon the last. — Regard a^ and rm as simple num- bers represented by letters without exponents. Thus (a^ + w^)* = (al)« + 4 {a^y {m^) + 6 (J)' (m^)' + 4 {a^) (/7ii)' + (mi)*. Now performing the operations indicated, we have (a» 4- mir)* = af + 4a»mi + 6a^m + iaJni + m'. 201. Cor. 6. — If the sign between the terms of the bino- mial is minus and the exponent is a pof-itive integer, as (a — b)''*, the odd terms of the expansion are + and the even ones — . This arises from the fact that the odd terms involve even powers of the second or negative term of the binomial, and the even terms involve the odd powei^s of the same. 154 POWERS AN^D ROOTS. Thus the second term involves (—5) which makes its sign -- ; the 4th term has (—5)', the 6th term (-&)', etc. But the first term does not involve (— &), and the 3rd has {—bf or h'\ the 5th has {-by or b\ etc. 21 to 24. Write out the expansions of {m—ny, (x^yy, Result of the last, {a^—h^y — a2_4alji ^ ^db^—^c^l + it EVO LUTI N. 202. Prob. 1. — To extract the ^th (any root) of a per feet power of that degree. Rule. — Resolve the numher into its prime factors, and separate these into m equal groups; one of these groups is the root sought. Demonstration. — Since the wvth root (i. e., any root) of a num- ber is one of the m equal factors of that number, if a number is resolved into m equal factors, as the rule directs, one of them is the mtla. root. EXAMPLES Ex. 1. Extract the cube root of 74088. Model Solution. — Resolving 74088 into its prime factors I find them to be 2-3-3-3-3-3 7 • 7 • 7. These arranged in 3 equal groups give 3-3-7x2-3-7x2-3-7. Hence 3 • 3 7 = 42 is the cube root of 74088, since it is one of the 3 equal factors. 2 to 5. Extract in this manner the following : V 492804, 'V^592704, ^248832, ^456533. Roots, 702, 84, aad 12. 6. Extract the square root of Sla^x~~^yiz~^. Model Solution. — The two equal factors of 81 are 9-9; of a* a^a^; of x-^, ar^-x-^-^ of ys, y^ yi ; of z~^^ z~ts • s~tV, Hence EVOLUTION. 166 8ia*x-^y^z~^ = Qd'x-^y'z'^o x Qa^ar-^y^z~TG^ and consequently, 9«'a^'y»«~iff is its square root. V: 7 to 11. Extract \/2la^f^, ^Uar^x^, V^9xy^^ ^/U^ahn^ f~- Roots, ±ba%^, ±%a-H^, ±7ic%i, ±\2ahn% 12 to 15. Extract \/l25m^x^, "^ll^Sx^y^, \^—32a^^y-^y Boots, 5?n^x^, VZx^^, —2a^y-\ and ±2n-^y\ Query.— Why the ambiguous sign to the last ? 203. Scholium. — The sign of an even root of a positive num- ber is ambiguous (that is, + or — ) since an even number of factors gives the same product whether they are positive or negative (87, §8). The sign of an odd root is the same as that of the number itself, since an odd number of positive factors gives a positive product and an odd number of negative factors gives a negative product (88, 89). 204, Cor. 1. — The roots of monomials can be extracted by extracting the required root of the coefficient and dividing the exponent of each letter by the index of the root, since to extract the square root is to affect a number with the exponent ^, the cube root ^, the nth root i, etc. (194.) EXAMPLES. 16 to 21. In this manner write V^25«^^, v^— 343aSr«, ^^^^- ^^^"^' /S' -^ \/iS^- 243%- 3a^x^ Roofs, ±6ah, -'7x^y-% ±3mhi'-^x^, 3(^m-^y^, ^7,, 2a^^ and — 3itw-2 156 POWERS AND ROOTS. 205. Cor. 2. — The root of the product of several numbers is the same as the product of the roots. Thus, \/ahcx = v^a • \^b • v^c . V^:r, since to extract the mth root of abcx we have but to divide the exponent of each letter by m, which gives ah^c^x^ or v^a • v" J • v^ • v^. 206, Cor. 3. — The root of the quotient of two numbers is the same as the quotient of the roots. Thus, A / — is the same as — — , since to extract the rth root of — we have but to ex- "' n n tract the rth root of numerator and denominator, which operation is performed by dividing their exponents by r. TT r/m m^ \/m Hence \ / - = — r = — V ^ 71' a7 EXAMPLES. 22. Show that ^8"^^ = v^8 x v^27. Model Solution. — We may show this in two ways. 1st. — Experi- mmtally. Thus ^8x27 = ^216 = 6. Again v^8 x v^27 = 3x3 =-.6. Hence ^^8^^ (or the cube root of the product) = ^8 x v^27 (or the product of the cube roots). 2nd. — Analytically. \^^ x 27 signifies that the product of 8 and 27 is to be resolved into 3 equal factors, which is accomplished by resolving 8 into its prime factors, and 27 into its prime factors, and then separating these factors into three equal groups (202). This will give the same result as resolv- ing the product of 8 and 27, or 216, into 3 equal factors, since the prime factors of 216 are the same as those of 8 and 27. 23. Show that '^'ar-'^ = ^^- x ^^. 1 _m 2. _2 J. Suggestions. — The 5 equal factors of «-""&» are a ^&6~ a~'^b^% m 1 _m l_ -Jl L oT^t^, a "^b^^ and a ^h^", since by the rules of multiplication these EVOLUTION. 167 1 multiplied together make «-*"&". But a *&"• is the product of one of tlie 5 equal factors of a-"" by one of the 5 equal factors of 5^ or ^f^ by j/ft*. 24. Extract the square root of a^^-\-^d^M-\-aW^(?, Solution. — The factors of this are readily seen to be a', c', and a'' + 2a^ + &', which separated into two equal groups givea<^a + &) and ac(a +&). Hence ac{a + &) or a^c + (the is the required root. 25. Extract the square root of m'^—'lm^x-\-m'^:i^. Root, tf? {^—x) or m^—mhi» Scholium. — The extraction of roots by resolving numbers into their factors according to this rule, is limited in its application for several reasons. In the case of decimal numbers we can always find the prime factors by trial, and hence if the number is an exact power, can get its root. But in case the number is not an exact power of the degree required, we have no method of approximating to its exact root by this rule, as we have by the common method already learned in arithmetic. In case of literal numbers the difficulty of detecting the polynomial factors of a ])olynomial is usually insuperable. Hence we seek general rules which will not be sub- ject to these objections. 207. Prob. 2. — To extract the square root of a poly- nomial. Rule. — /. Ai^ange the polynomial with reference to one of its letters, as for division. II. Extract the square root of the first left hand term. Hiis root is the first term^ of the required root. Subtract the square of this teim of the root from the polynomial. III. Double the root already found for a Trial Divisor. By this trial divisor divide the first term of the remainder of the polytwmial, and write the quotient as the second term of the root. 158 POWERS AKD ROOTS. IV. Complete the divisor by adding to the trial divisor the last term in the root. Multiply the Ti^ue Divisor thus formed by the last term in the root, and subtract the product from the last remainder, bring- ing down such terms as may be necessary. V. Repeat the process of dividing, completing the divisor, multiplying and subtracting, in the same may till the polynomial is exhausted, or until thepe is no term of it remaining which can be exactly divided by the first term of the trial divisor. Demonstration. — 1st. The polynomial is arranged as in division, since sucli is the order which the terms assume in squaring any polynomial root. 2nd, In squaring any polynomial, the first term of the square is found to be the square of the first term in the root ; hence, in extracting the square root, the square root of the first term in the given polynomial is the first term in the root. 3rd. To prove the process of finding the divisors and the subse- quent terms of the root, we observe the following operations : (1) (2) B. {a + l + cy = [_{a + l)) + cf = (a + &)^ + [2(a + 5) + c> (1) (2) (3) = a'^ 4- (2« + &)& + [3(fl^ + 5) + c]c. * C. (a + l + c + df = [{a + 'bJrc) + dY = (a + h + cY (1) (2) + [2{a + d + c) + d:]d = a^ + (2a + b)h (3) (4) + [2(a + h) + c]c + [2(a + b + c)+d]d. Hence it appears; 1st, That the square of a polynomial (as a + h, a + i + c, or a + h + c + d) is made up of as many parts as there are terms in the root ; 3nd, That the first is the square of the first term in the root ; 3rd, That the second part is Ticice the first term of the root (the part already found), + the second term of the root, multi- plied by the second term ; 4th, That any one of these parts, as the /7th, is Twice that portion of the root previously found, or Twice the n—1 preceding terms of the root, + the wth term of the root, multiplied by this last or «th term. * This expaneion is made by treating (a + b) + c as a binomial, giving as its square (a + 6) ' + 2 (a + 6) c + c' . Expanding' and factoring we have the form in B, EVOLUTIOlf. 159 Scholium I. — If the first term of the arranged polynomial is not a perfect square, the root can not be extracted. Scholium 2.— If at anytime no term of the remainder can be exactly divided by the first term of the trial divisor, the root can not be extracted. EXAMPLES. Ex. 1. Extract the square root of ^9a^a?i—30cu7^ + 2bx^ Model Solution. OPERATION. 16a* Sa'^—Sax ! — 24«»« + 49aV -24a'a!+ 9«V Sa^-6ax-\-5x^ \ 40aV-30aa;« + 25iB* 40aV-30qa;H25a^ Explanation. — If this polynomial is a perfect square, the term containing the highest power of a, or of «, is the square of the first term in the root. Hence I place 16a* first in the arrangement. (25x* would do as well.) And, as the terms arising from squaring a polynomial (e. g. the root of this given number), are arranged according to the leading letter of the root, I arrange the whole polynomial according to the powers of a, as this will be the leading letter of the root, when I put 16a* fii*st in the power. Having arranged the given number, I know that the square root of the first term, 16a*, or 4a'' is the first term of the root, since in squaring any polynomial (e. g., the root sought), the first term in the square is the square of the first term in the root. Now, having removed by subtraction the square of the first term of the root, I double the root already found, obtaining 8a' for a trial divisor. This is the trial divisor, since the second part of any square (the square of the first term of the root being called the first part) is twice the root already found + the next term of the root, multiplied by this next term. I now find that the trial divisor is contained in the first term of the remainder —Sax times, which is, therefore, the second term of the root. But the true divisor is twice the root previously found plus this last term; hence I add —'da.x 160 POWERS AND ROOTS. to the trial divisor and have Sa"—Sax as the trm divisor. Multi- plying this true divisor by the last term of the root I obtain — 24a^« + 9aV, which is the second part of the given power. Subtracting this second part of the power, the third part is twice the root already found + the next term of the root, multiplied by this next term. In this case, therefore, the new trial divisor is Sa'^ —Qax. [Proceed just as in the last paragraph, and complete the explanation.] Finally 4a^ — 3aa; + 5«^ is the required root, for, indeed, I have actually squared it and subtracted this square from the given number and found no remainder. [The pupil should notice that the sum of the several subtrahends is the square of the root.] 2. Extract the square root of Sx-{-^-j-x*-{-^a:^-\-Sx^ Boot, x^-\-2x + 2. 3. Extract the square root of 25«V— 12«x3 + 16«4 + 4a:* —24:a^x. Root, 2x^—dax-\-4:aK 4. Extract the square root of x^—6aa^-^16a^x^—20a^x^ -^16a^x^—6a^x + a^. X? 4 5. Extract the square root of x^—7?-\-j -f-4rr— 2-f -^ Root, x^-^+-' 2 X 6. Extract the square root of 9x—2^x^y^-{-12x^-\-16y'^ — 16yt_f-4. Root, Sx^—4:yi-{-2, 7. Extract the square root of 9a-^-{-12a-^i^—ea + 4:b' —^a^^-\-a\ . Root, 3a-^-\-2b^—a^. Query.— In arranging the last with respect to a, why should ih* come before — 6a ? EVOLUTION. 161 20S. Scholium.— Since the square root of a quantity is either -f or — , all the signs in the above roots may be changed, and they will still be the roots of the same polynomials. Thus, in the 3rd Example, if we call the square rootof 4aj*, minus 2x', {^2x*,) which it is, as well as +2x^, and then continue the work as before, we get for the root '-2x^ + 3aa-4a\ 209, Prob. 3.— To extract the Square Root of a Deci- mal Number either exactly op approximately. Rule.—/. Separate the numhers into periods by placing a marh over units and over each alternate figure therefrom, calling the marked figure with the one at its left, if any, a pei^iod. The number of figures in the root is equal to the number of periods thus formed, II, Take the square root of the greatest square in the left hand period, and write it as the highest order in the root. Subtract the square of this figure from the period used, and to the rem^ainder annex the next period for a new dividend. HI. Double the root already found for a Trial Divisor, by which divide the new dividend, rejecting in the trial the right hand figure of this dividend. The quotient is the next figure in the root or a greater one. To obtain the True Divisor annex to the Trial Divisor the last root figure. Multiply this True Divisor by the last root figure, subtract the product from the last new dividend, and to the remainder annex the next period of the given number for another new dividend. IV. Double the root already found for a new Trial Divisor, and repeat the processes given in the 3rd paragraph till all the periods are brought down. If the number is a perfect square, the last remainder is 162 POWERS AN^D ROOTS. zero. If not, annex periods of two O's each, and covj- tinue the process till the required degree of accuracy is attained. All the root figures arising from decimal fractional periods are decimal fractions. Scholium I. — In separating decimal fractions, or the fractional part of mixed numbers into periods, make full periods of two figures each, annexing a if necessary. Scholium 2. — If at any time the Trial Divisor is not contained in the dividend to be used, annex a to the root and also to the Trial Divisor, and then bring down the next period and divide. Scholium 3. — When the work does not terminate with the last period of significant figures it will not terminate at all, and the given number is a surd. This is evident, since the process makes the unit's figure in each subtrahend, the square of the last figure in the root, but no figure squared gives in unit's place. The process can, however, be carried to any given degree of accuracy. Demonstration. — 1st. That this method of pointing gives the number of places in the root, is made evident by squaring a few numbers. Thus the square of 1 is 1, and the square of 10 is 100, hence the squares of all numbers between 1 and 10 have 1 or 3 figures ; that is, Twice as unany figures as the root, or one less than twice as many. Again, the square of 100 is 10000 ; hence the square of all numbers between 10 and 100 have 3 or 4 figures; that is, Twice as many as there are in the root, or one less. In like manner it is readily seen that the square of' any number consists of Ticice as many figures as the root, or one less. Hence the method of pointing indicates the number of figures of which the root consists. In the case of decimal fractions, since the number of decimals in a product equals the number in both the factors, there are always twice as many decimals in the square as in the root. Hence if the number of decimal places in the given number is odd, they are to be made even by annexing 0. 2nd. That the greatest square in the left hand period is the square of the highest order in the root, appears from the facts that the square of any number oi units between 1 and 9 falls in the first right hand period, the square of any number of tens between 1 and 9 falls in the second period, the square of any number of hundredshetwe&ii 1 and 9 falls in the third period, etc. Moreover, though the left hand period usually contains more than the square of the highest EVOLUTION. 163 order in the root, it can not contain the square of a unit more of that order, since all the figures that can follow this highest order in the root can not make another unit of this order. Thus the square of 3999, can not be as great as the square of 4000 ; but the square of 4000 gives 16 in the highest period of the power, hence the square of 3999, must give less than 16 in that period, 3rd. To prove the method of finding and using the Trial Divisor, suppose, in any given case, the pointing shows that the root con- sists of 4 figures. Represent the thousands of the root by T, the hundreds by A, the tens by t, and the units by v. The number itself will be (T -^h-ht + uy =-. r + {2T+h)h + [2(T+?i) + t]i + [2{T-\-h-{-t) + u]u. (207, 3rd paragraph in the Dem.) Whence having found and removed the square of the thousands, 7^, the next part of the power is {2T+h)h. But as the lowest order of this is hundreds multiplied by hundreds, or 10,000'8 we need not bring down anything below 10,000's, or the next period, for none of this part is contained in the lower periods. Again, for trial, considering this part as 2Txh, the product contains nothing lower than hundred thousands ; hence the ten thousands figure is to be omitted in the trial division. But the true divisor is 2T+h\ hence the root figure is annexed to the trial divisor, or really added to it, regarding the k'cal values. 4th. It is evident that this process is merely repeated as we increase the number of figures in the root ; and as the law of nota- tion is the same when we pass the decimal point into a fraction, no special exemplification is needed in such a case. EXAMPLES. Ex. 1. Extract the square root of 7284601. Model Solution. Operation. 728460i ( 2699 46 ) 328 276 529 ) 5246 4761 5889 ) 48501 48501 164 POWERS AND ROOTS. Explanation. — As the highest order in this number is millions, the highest order in the root is thousands, for the square of thousands is millions or ten millions, and the square of anything above thousands is more than ten millions. The root, therefore, consists of thousands, hundreds, tens, and units, as indicated by the pointing. The square of the thousands figure in the root evidently must b3 sought in the 7 millions, or the left hand period. The greatest square in this being 4, the square root of which is 3, the thousands figure of the root is 2, which I therefore place in the root. Now the root may be represented by T+h + t + u, T standing for the thousands, h for the hundreds, t for the tens, and u for the units. Hence the number itself will be represented by {T+h-\-t Having removed the T'^ (2 thousands squared) from the number, the second part is {2T+h)K the lowest order in which is hundreds multiplied by hundreds, which gives ten thousands; wherefore I bring down no order lower than ten thousands, or simply the next period. As this new dividend contains {2T+?i) h, I yvi\],fora trials consider it as simply containing {2T)xh, and as hundreds into thousands produce only orders aibove ten thousands, I may omit the 8 which is ten thousands, in making this trial. Using 2jr or 4 (thousands) for the trial divisor, I find it contained in 32 (hundred thousands) 8 times. But, as the trial divisor ii too small by this new figure, it is evident that adding it, thus making the divisor 48, it will not be contained 8 times. Neither will it be contained 7 times. Thus I find 6 to be the next figure in the root, and the true divisor 2T+h, to be 46 {i. e., 4 thousands and 6 hundreds). Multiplying 46 by 6, and thus forming the part [2T+h] /i, I find it to be 276 (ten thousands), which subtracted leaves 52 (ten thousands). Again, these two parts, viz., T'- and {2T+h)h having been re- moved, the next part of the power is [2 (T+h) + t] t, or [52 hundreds + t]t. As the lowest order of this part is tens squared, I need bring down nothing below hundreds, or the next period. The pupil can now fill out the demonstration as in the preceding paragraph. 2. Extract the square root of 7225. Boot, 85. * 3 to 7. Show that V9801 = 99, V47089 = 217, \/553536 = 744, a/43046721 =: 6561, ^5764801 = 2401. 8 to 11. Show that V^ = .7071+, VS == 1.73 + , ^50 = 7.071 +, V5 = 2.236 + . EVOLUTlOK. 165 210. Cor. — In extracting thr roots of common fractions, if the numerator and denominator are perfect squares, extract the root of each separately ; if they are not, it is usually best to convert the fraction into a decimal and then extract its root. 12 to 15. ^/%^'^, y/l^-n^+, \/i--577+. V I = .8164+. 211. Prob. 4.— To extract the Cube Root of a Poly- nomial. Rule. — /. AiTange the polynomial with reference to one of its letters, as for division. II. Extract the cube root of the first left hand term. Tliis root is the first term of the required root. Subtract the cube of this term of the root from the polynomial. III. Tahe three times the square of the root already found for a Trial Divisor. By this trial divisor diiride the first term of the remainder of the polyno- mial, and write the quotient as the second term of the root. IV. Complete the divisor by adding to the trial divisor 3 times this last term multiplied by the part of the root previously found, and also the square of the last term found. Multiply the ti^ue divisor thus formed by the last term in the root, and subtract the product frowj the last remainder, bringing down such terms as may be necessary. V. Repeat the process of forming Trial Divisors, dividing, completing the divisor, multiplying and subtracting, till the polynomial is exhausted, or until there is no term of it remaining which can be exactly divided by the first term of the Trial Divisor, 166 POWERS AND HOOTS. Demonstration. — 1st. The polynomial is arranged as in division, since this is the order which the terms assume in cubing any poly- nomial, similarly arranged. 3nd. In cubing any polynomial, the first term of the cube is found to be the cube of the first term of the root; hence, in extracting the cube root, the cube root of this term is the first term of the root. 3d. To prove the process of finding the divisors and subsequent terms of the root, we observe the following operations : (1) (2) A. (a + dy = a^ + 3a«5 + 3a5' + &' = «' + [3a' + Sai + ¥] 6. B. {a + h + cy = [(a + l>) + cY (1) (2) (3) = a^ + [Ha'' + dab + l*'\J) + [d (a + hy + 3 (a + h) c + c^] c. C. (a+h+c+dy = [ia+d + c) + d]' = (a + I}^cy + [d(a + l> + cy + 3(a + l} + c)d + d'']d (1) (2) (3) = a» + [3a' + 3a& + 6'']& + [3(a + &f + 3(a + 5)c + c''](j (4) + [d(a + l + cy + d(a + l+c)d + d']d. Hence it appears ; 1st, That the cube of a polynomial is made up of as many parts as there are terms in the root ; 2nd, that the Jirst part is the cube of the first term of the root ; 3d, That the second part is three times the square of the Jirst term of the root + 3 times the first term into the second term + the square of the second term, multi- plied lyy the second term of the root ; 4th, That any one of the parts of the power, as the nth, is Three times the square of the n—l preceding terms of the root, + 3 times the product of these terms into the next, or jith term., + tJie square of this last a?' nth term, all these terms heing multiplied by the last, or nth term of the root. Finally, it is evident that, if the work does not terminate by this process when the letter of arrangement disappears from the remain- der, it can never tern#nate, since the divisor always contains this letter. Scholium I. — If the first term of the arranged polynomial is not a perfect cube the root cannot be (ixtracted. Schoh'um 2. — If at any time no term of the remainder is exactly divisible by the first term of the trial divisor, the root can not be extracted. EVOLUTIOK. 167 + + eg & ^ o ^ 1 J> 1, <7i c o + 3 ^^ o e <0 CO CO ^ J a> O ■3 ^ §1 II ^ + + .2 + I CO eo I + I I to I > > Ss rO + + 1 1 1 8 1 ■k Sa Si "e *b 06 OS + + ^ -§ "^ ^^ tH T-t + + 1 1 «o 1 1 1 i :§ ^ •»• 'e "S i> t* 5 1 «N + 1 1 ^ 1 1 1 + 2© + 1 <: u eg 1 t V) n « T— 1 5 + -1 1—1 :^ 1 eg C3 II II II n CO i2 1—1 c ^ ^ ^1 ©» 1 f ^ *« II II II ^i 1 I -^^ 'e n ..t S 1 c '^ .. -"^ 1 ..^ jl 1 1 1^ il 1 5 X 2 ^ 1 > 5 |„ S» ^ CO O CO 1 1 s 168 POWERS AND ROOTS. Explanation. — 1st. I arrange this polynomial with reference to«, and thus see at once the first two terms. But the terms 36aV and 27a*6c'^a! are of the same degree with respect to a, and hence to determine which is to have the precedence, I notice that the first term in the root will be Sa^c, and as the second term of the poly- nomial divided by 3 times the square of this gives the second term of the root, I observe that the terms containing a and c are all to have precedence over those containing & and x. Uence I write dQa*c—8a^ next. The remaining terms I arrange, giving a the precedence and noticing that as x will be in the last term of the root, its higher powers will stand last. 2nd. As 27a"c' is the cube of the fii*8t term of the root, that term is ^\ which I consequently place in the root, and subtract the term 27 a^c^ from the polynomial. 3rd. As the second part of a cube of a polynomial is 3 times the square of the first term of the root, plus other terms, into the second term of the root, I take 3 times the square of this first term of the root or 27a*c', for a trial divisor. Dividing, I find the second term of the root to be —2a. But the True Bwiaor, or leading factor in this second part of the power, is 3 times the square of the former part of the root, + three times that part into the last term found, + the square of this term. Hence I add 3 times Sa^c multi- plied by —2a, and —2a squared, to complete the divisor. Having completed it, I multiply it by the last term of the root found, —2a, and thus form the second part of the power of the root, which I subtract from the given polynomial. 4th. The explanations of the next and succeeding steps, when there are more, are identical with the last, and can be supplied by the student. 2. Extract the cube root of a^—Sb^ + 12aIyi—{ Root, a— 21). 3. Extract the cube root of boi?—l—doiP + x^^Zx. Root, x^—x—l. 4. Extract the cube root of 66a^ + 1 — Q3a^ — ^x + ^ofi —mx^+^ZxK Root, 2a;2_3a:+l. EVOLUTION. 169 % ^ to CO -^ u 1 1 + + 1 + CO 1 00 ^ + "^ + 1 <>? «^ c^ T s ^- s "o + (§ 4- ^ + 1^ \5 I rH 1 o i> 1 1 1 « ^ ?^ 1— 1 00 CO T-H CO 1 + + 1 •0)9J CO •^ CO + + 1 1 •? CO CO 1 00 tH T*< 1 "fe + 1 00 1^ 1 1 *t. ^ ^ l^ (~i 2> CO ^ o* «M «M •s o O o -M -fS 2 © 0) . 170 POWERS AJ^D ROOTS. 2 12, Prob. 5.— To extract the cube poot of a decimal number, either exactly op apppoximately. Rule. — /. Separate the nitmber into periods hy pla- cing a marh over units and over each third figure therefrom, calling the figure marked, together luitli the two at its left (if there are so many), a period. The number of places in the root is the same as the number of periods. II. Take the cube root of the greatest cube in the left hand period, and write it as the highest order in the root. Subtract the cube of this figure from the period used, a,nd to the remainder annex the next period for a new dividend. III. Take three times the square of the root already found, regarding it as tens, for a trial divisor, by which divide the new dividend. The quotient is the next figure in the root (or a greater one). To obtain the True Divisor add to the trial divisor 3 times the product of the last root figure by the preceding part of the root, regarded as tens, and also the square of the last figure in the root. Multiply this true divisor by the last root figure and subtract the product from the last new dividend, and bring down the next period. IV. Repeat this process till all the periods have been brought down. If the number is a perfect cube the remainder is zero. If not, annex periods of 3 zeros each and continue the operation till the required degree of accuracy is attained, marking the figures thus obtained as a decimal. Scholium I. — In pointing off decimal fractions, or the fractional part of mixed numbers, make full periods of three figures each, annexing O's if necessary. Scholium 2.— If at anytime the trial divisor is not contained in the dividend to be used according to the 3rd paragraph in the EVOLUTIOK. 171 rule, annex a to the root and also two zeros to the trial divisor, bring down the next period, and then divide. Scholium 3. — When the work does not terminate with the last period of significant figures it will not terminate at all, and the number is a surd. This is evident since the right hand figure in any subtrahend arises from cubing the corresponding digit in the root, and the cube of no digit produces in unit's place. Demonstration. — 1st. That this method of pointing gives the number of figures in the root is made evident by cubing a few num- bers. Thus the cube of 1 is 1, and of 10 is 1000 ; hence the cubes of all numbers between 1 and 10 have 1, 2, or 3 (cannot have 4) figures. The cube of 100 is 1,000,000; hence the cube of numbers between 10 and 100 have 4, 5, or 6 figures, but can not have 7. Again, the cube of 1000 is 1,000,000,000; hence the cube of any number between 100 and 1000, i. e., of any number represented by 3 figures, contains 9, or one or two less than 9 figures. In like manner it appears that the cube of any integral number contains either three times as many Jig ures as the root, or one or two leas. In the multiplication of decimal fractions the number of fractional places in the product is equal to the number in both or all the factors used, hence the frac- tional part of any cube must have three times as many figures as the root. 2nd. That the greatest cube in the left hand period is the cube of the highest order in the root, appears from the facts that the cube of any number of units between 1 and 9 falls in the 1st period ; the cube of any number of tens between 1 and 9, falls in the second ; of any number of hundreds, in the 3rd, etc. Moreover, though the left hand period usually contains more than the cube of the digit in the highest order in the root, it can not contain the cube of a unit more of that order, since all the figures that can follow this highest order in the root can not make another unit of that order. Thus the cube of 3999 can not be as great as the cube of 4000. But the cube of 4000 gives 64 in the highest period. Hence the cube of 3999 must give less than 64 in that period. 3rd. In any given case, suppose the pointing shows that the root consists of thousands, hundreds, tens, and units. Represent the thousands by T, the hundreds by A, the tens by t, and the units by u. Then the number is (T-{-h + t-{-uy = T^-\-[ST'' + STh-\-h^]h ^[diT+hy +d{T-{-h)t-\-t']t-]-[^{T+h-{-ty-\-S(T-\-h + t)u-\-2r]u. But having removed the cube of thotliousands, T'^ the next part of the power is [37" + 3rA + A'J/i. No part of this can fall in either 172 POWERS AND ROOTS. of the two lowest periods of the power, since its lowest order arises from h^ which is 1,000,000, at least. Hence we need only bring do^\Ti one period. For a tiial, we consider this part as 37"^ x h, and hence the Trial Divisor is 3T^, or '^timesthe square of the root already found. Again, regarding this thousands figure as tens, n\akes the T, which squared and multiplied by the next figure of the root which is also hundreds, give millions, the same order as the new dividend. But the True Divisor is ^T'^ -\-^Th + h^\ hence we add to ZT\ '6Th + ¥ ; i. e., 3 times the root previously found multiplied ly the last figure^ and the square of this last figure. In making this correc- tion we are to remember to call the thousands so many tens^ which reduces it to hundreds, the order of the root which we are seeking ; whence the correction becomes the square of hundreds, or of the same order as the trial divisor, and can be added to it. 4th. It is evident that this process is merely repeated, as we pro- ceed to obtain other figures in the root ; and, as the law of notation is the same as we pass the decimal point, no special exemplification is needed in that case. EXAMPLES. Ex. 1. Extract the cube root of 99252847. Model Solution.— Operation. h t u 99252847 I 4 6 3 64 ' IHa? Divisor 3 (40)^ = 4800 Corrections I ^(,^?)«= ^3^^ True Divisor . 5556 35252 33336 Trial Divisor 3 (460)^ =634800 4140 9 True Divisor 638949 Corrections j^(^3^^^>^^ 1916847 1916847 Explanation. — As the highest order in this number is ten 7nillion8, the highest order in the root is hundreds, since the cube of a hun- dreds figure falls in millions period, while the cube of thousands falls in billions. Moreover, the cube of the hundreds figure is the greatest cube contained in 99, /. e. 64, the root of which is 4, which is, therefore, the hundreds of the root. That the hundreds figure is not greater than 4, is evident, since the cube of 5 hundreds is greater than the given number. EVOLUTION. 173 Therefore the cube root of the given number is h-\-t + u, and the number itself is (h + t + uy = h'-\- [3 h' + dht -^.t']t + [B{h + ty-\- (3A + t)u+u^]u. But having removed the h^ by subtracting the 64 (millions), the next part of the power is [3A* + 3A< + «"]«. Now the lowest order of this is t\ or the cube of tens, whicl) cannot fall below thousands, so that I need only bring down thousands period, /. €. the next lower. For a Trial I now consider this part of the power (35352) as 3^" x <, or 3[40J'^ xt. I reduce the 4 hundreds to the same order as the root figure which I am seeking, so that the product of its square by this root figure shall be of the same order as the new dividend. Therefore, reducing the 4 hundreds to tens it becomes 40, whence 3(40)^ = 4800, which being hundreds, goes into the new dividend, which is thousands, tens times. This trial divisor is really contained in the dividend 7 {tens) times, but as the corrections to be made upon it for the true divisor are so great, the true divisor is contained but 6 times, as I find by trying 7 for the tens of the root. Having thus found 6 to be the tens of the root, I correct my trial divisor, which by the formula is 3^" + 3A x ^ + ^'^, by adiling 3A x t or 3(40) x 6, and t" or Q^, and find the true divisor to be 5556 (hundreds). This multiplied by the 6 (tens) gives the second part of the power, ^. e. (3A* + 3/*< + 0^ = 33336 (thousands), which I therefore subtract from the given number. [The next step is exactly like the last, and the pupil can supply the demonstration.] 2. What is the cube root of 74088? Ans., 42. 3. What is the cube root of 12326391 ? Ans., 231. 4. AVhat is the cube root of 122097755681 ? Ans., 4961. 6. What is the cube root of 2936.493568? Ans., 14.32. 6. What is the cube root of 12.5 ? Ans., 2.321 nearly. 7. What is the cuhe root of .64 ? Ans., .8617 + . 8. What is the cube root of .08 ? Ans,, .43084-. 9. What is the cube root of .008 ? Ans., .2. 10 to 12. Show that \^ = 1.2599 +; \/5 = 1.7099 + ; V^ = 2.08008 + . 13 to 15. Show that ^| = . 87+; \^Wh = iiy <^^m 174 POWERS AND ROOTS. [Note. — With regard to the various shorter methods for extract- ing roots, the various methods of approximation and the like, no mathematician thinks of using them, or even those here given, but resorts at once to the table of logarithms. It is better that the student should spend his time in becoming perfectly familiar with the demonstration of a single method, than to cumber the memory with a multiplicity of processes which he will not remember, and which if we were to remember he would never use.] FOR REVIEW OR ADVANCED COURSE. 213, Prob. 6. — To extract roots whose indices are composed of factors 2 and 3. Solution.— To extract the 4th root, extract the square root of the square root. Since the 4th root is one of the 4 equal factors into which a number is conceived to be resolved, if we first resolve a number into 2 equal factors (that is, extract the square root) and then resolve one of these factors into 3 equal factors (that is, extract its square root) one of the last factors is one of the 4 equal factors which compose the original number, and hence the 4th root. In like manner the 6th root is the cube root of the square root, etc. EXAMPLES. Ex. 1. What is the 4th root of l^a^-maH-^^Ua^o^ —216«rr3^ 81:^4? Ans., ±(2a— 3^;). 2. What is the 6th root of lbx^—%Qio^ + ofi—Qo^-\-l — Qx + 15^:4? Ans.y ±{x—l), 3. What is the 6th root of 2985984? 4. What is the 8th root of 1679616 ? 214. Prob. 7.— To extract the mX\\ (any) root of a decimal number. Solution. — Any root can be extracted by a process altogether similar to those given for the square and cube roots, or by a simple inspecticm of the corresponding power of a binomial. Thus to extract the fifth root, point off by placing a point over units and every EVOLUTION. 175 fifth figure therefrom, for the 7th root over every 7th figure, etc. Extract the required root of the largest power of the 7//th degree in the let\; hand period for the first figure in the root. Subtract, and bring down the next period. To form the trial and true divisors, and hence to find the other figures of the root, consider the corres- ponding power of a binomial. Thus for the 5th root, we have (^-f &)'* = a^ + 5a^&-f lOa^ft' -H lOa'ft^ + 5aJ* + J" = a"^ + [5a* + lOa^J + load's' + 5a5'']&. The trial divisor is 5a*, i. e. five times the 4th power of the root already found regarded as tens. The corrections are 10«'&4-10a'&^ + 5a&'', regarding a as the root already found and as tens, and & as the next figure, i. e. the one sought by the trial. In the 7th root the trial divisor is 7a'', and the corrections are 21a»& + 35a*6» + 35a='&^ + 21a^&* + laV + h\ But in these cases, and much more in the case of higher roots, the trial divisor differs so much from the true divisor that the process is little better than guess-work. 216. Prob. 8.— To extract the iwth root of a polynomial. Rule. — /. Having arranged the polynomial as for division, take the root of the first teimv, for the first term of the required root. II. Subtract the power from the given quantitij, and divide the first term, of the remainder hij the first term of the root involved to the next inferior power, and multiplied by the index of the given power ; the quo- tient will be the next term of the root. III. Subtract the power of the terms already found from the given quantity, and using the same divisor proceed as before. Demonstration.— This rule demonstrates itself, as the final opera- tion consists in involving the root to the required degree. Scholium.— This rule may also be used for decimal numbers. EXAMPLES. Ex. 1. Find the fourth root of 16a*-ma^x-]-216a^xi 2. Find the fifth root of a:« + 5a;* -f- 10a:» + lOar^ + 5a; + 1. 176 POWERS AND ROOTS. + Si CO + Is o CO (?^ (?^ ?S O CO CO to CO + «4H o o o t- t- tH CO O O O S ^ » O ^ «v II S w ^ Si^ „ II lo o o e , .. rH -r-l »0 rO ''t^ ^* ^2 'P ^ iO r-( C^ CO '^ I> «> iO JO « CO Ol (N t> T— 1 1-1 i> I> CO CO CO 1 ^ o o ^ ^ TH ^ o o o lO o JO o o o CO -rt* o o o 05 '^ (M CO »o t> ■<^ i> -^ O iO ^ 00 05 o »o t- CO ?0 00 C* «M tH CO 00 Oi i> CO Tj< CO li li CO X X II « ,^-v £» o o X tH iO -.^ ,• lO o ^ (^ T— 1 o 1—1 lO 1^ II »o II 1 w rO rC> «"« « « ?b 1 II s o s •I— 1 JO rC S ^' d ^6 X ; to m c t-l * T-t (M CO ^ • u u g 2 g c • r-) '> > • rH Q 1 Q •§ fcl ^ o C*" i o ^ 2> I rA I CO <«*< £ o tv. M ^ 1 CO 05 fi O 2 ^ 6 CO o 1— ( t, 00 i ■^ p. »o V rH ^ i>- OB r-\. -M y—^ § CO 1 CV. c t- lO ^ 00 ^ s i §1 00 '-" Oi 00 00 ' (M OJ c o r-{ '" -cJ c c ^ c c c 2 1 c c c o c o « 00 oc ,_ T— CO «c Ol CO ^ -^ c oc oc «> S oc ri« «c tH t- JO o oc '«* lO «c "^ tr- Oi t- yz ■^ y-f tH ■rf in ee OJ o t> rH ta e> t» OJ CO + + IK «: «c T-t 05 cc CO l> 1 1 1 11 1 1 ^ ^rC S: Ic HC ^ ■^ rC rC Ss 5: e e t> 'e "J t 1 1 *« '^ t 1 II II 1 OJ CC cc c (l 4- > ■*• + t- 1 OJ cc ^ ir. ^ ^ u . v_ * OJ CO ^f ir. «c e^ "TT ■>^ 1 i 1 5 .2 1 i OD 1 p "t J S y 1 e ^ ^ 6 'i 1 5 + + s •2 e §11 -•J (h o o (h 73 + P a C CO -S 4- -f^ a 'M TS w c O ;© a o + Tli :t3 IC REDUCTIONS. 217, Prob. 1.— To simplify a radical by removing a factor. Rule. — Resolve the number under the radical sign into two factors, one of which shall he a perfect power of the degree of the radical. Extract the required root of this factor and place it before the radical sign as a coefficient to the other factor under the sign. Demonstration. — This process is simply an application of Cor. Art. 205, which proves that the product of the roots is equal to the root of the product. Thus V^Sa'^ = A/l6a*&' x Zdb = Vl6«¥ EXAMPLES. Ex. 1. — Reduce to its simplest form v^TSG^^c^. Model Solution. Operation ExpFanation. \/lS9a*W indicates that 189a^&V into 3 equal factors. 27aW and 7ac\ one of which is a perfect cube. Now I resolve each of these two factors into 3 equal factors, and taking one of each three I multiply them together. This product will constitute one of the three equal factors of the given number. 37a'&' = da^b >' Sah ''is to be resolved Therefore I first separate it into two factors, I REDUCTIONS. 179 X Sab, and Inc^ = v^7^ x v^W x ^7a^. Hence 'iab\/7^^ is one of the 3 equal factors of 189a*JV; or \/ma*b'7- = 'Sab\/Ta^. 2. Reduce V^1!a^^ to its simplest form. Result i Sax^V^cuT' 3. Reduce VSOa^afi to its simplest form. 4. Reduce A/3?5r*^^ to its simplest form. Suggestion. — If the factor of the decimal number under the radical is not aj^parent, it can readily be found by a few trials. Thus in the 3rd, we could try the square of 2, or 4 ; and then the square of 3, or 9 ; and then of 4, or 16 ; of 5, or 25 ; of 6, or 36 ; of 7, or 49 ; of which we should find 16 to be the greatest square factor. We need not try farther than 49, since this is more than ^ of 80, and no larger number can be a factor. 5. Simplify VllSS^^^ Suggestion. — Try the square numbers from 4 upward till you find the required factor. But a little judgment will save labor. Thus, we need not try 4, for no numl)er multiplied by 4 gives a 3 in units' place. For a like reason we would try 9, but not 16, 25 or 36. Then again we would try 49, but not 64 ; 81, but not 100; 121, but not 144. Finally 169 meets the case, and we have ^/TlS3x^ = Vl69urVx"7^ = ^/WxY x \/7^ = Ida^y^^Txy. 6. Simplify V^H^aW. 7. Simplify v^96a^9. Result, Ux^dcM. 8. Simplify ^'fmj^y^^^. Result, Qa^y'^^ \/Tof, 9. Simplify (11ll5a^-^I^^)i. Result, 7a3m^» v^5a^, or Ha^P^l^'f- 10. Simplify (352«tjio)i Result, 2a*^^^. Scholium. — Of course, by the use of fractional exponents, all the factors of such monomials may be written separately, as in (202). Thus, the result in Ex. 7, may be written (QQ)lai!^^, or 2(3)Wla^^, or 2ax(3)^af.r^, or, by taking the 5th root of the product instead of the product of the 5th roots of the last 3 factors, 2ax(Sa^x*)^, as above. The method of this rule is usually applied for removing factors which can be expressed without fractional exponents. 180 CALCULUS OF RADICALS. 11. Simplify "s/a^—ahc. Result, a y/a—x. 12. Simplify ^a^^^H^. Result, aby/a"^. 13. Simplify V{a^—IP){a-\-b). Result, (a + h) Va—b. 14. Simplify 3^50^- Result, Ux^V^. Suggestion. — When the radical has a coefficient, the factor removed from under the radical sign is to be multiplied into this coefficient. 16. Simplify {x-\-y) ^^a^—^x^y-^-xy^. Result, (x^—y^) Vx. 16. Simplify xyVx^y^—x^y^ Result, a^y^Vx—y. 17. Simplify ^x^^y-^^+K Result, x^y^^^z, 18. Reduce A/f to its simplest form. Result, ^a/6. Scholium. — A surd fraction is conceived to be in its simplest form when the smallest possible whole nuinber is left under the radical sign. The reason for this is, not only that the radical factor is thus made simpler, but, if a fraction were to be left under the radical sign the question would arise, What fraction ? Certainly not the least possible, for such a fraction can be diminished at pleasure. Thus, /I = /4 X ^ = 2/gU 2/16 X g^ = 8|/i. etc., with- out limit. Perhaps, if a fraction is to be left under a radical sign it will be proper to consider the expression as simplest when the fraction is nearest unity ; whence y - is to be considered as simpler than 84/1. 218* Cor. — The denominator of a surd fraction can always be removed from under a radical sign by multiplying both terms of the fraction by some factor which will make the denominator a perfect power of the degree required. REDUCTIONS. 181 sim- 19. Reduce W-, y ^, W^-, and W-^- to their plest forms. Scholium. — The root of a fraction having 1 for its numerator is equal to the same fraction into the root ot the next lower power of the denominator. Thus, |/i = 1 ^7, f"^ = ^ V^49, \/\ ^ \ V^343, etc. 20. Reduce \/ jk to its simplest form. 21. Reduce r to its simplest form. Result, — — - Va^—a^. a-\-x 22. Reduce ttx/ ^, SVIS' ^^^ V W ^^ ^^^^ ^^^' plest forms. n Remits, (not in order), ^ V\bx, t a/6«^, and — a/7. 23. Reduce -^ — Th\/ io — ■ to its simplest form. ^2— J2y 62 Result, -r-, — TTv v/«*. 24. Reduce V/o » ^"^ V 4 ^^ *^^^^ simplest forms. l..,...._,..........„ ^t fractions, but upon integral radicals only when the integer has a ^^yactor which is a perfect power of the degree of the radical. P . . 219. Prob. 2.— To simplify a radical, op reduce it to its lowest terms, when the index is a composite number, and the number under the radical sign is a perfect power of the degree indicated by one of the factors of the index. 18^ CALCULUS OF RADICALS. Rule. — Extract that root of the number which corresponds to one of the factors of the index, and write this root as a surd of the degree of the other factor of the given index. Demonstration.— The wwth root is one of the mn equal factore of a number. If, now, the number is resolved first into m equal factors, and tlien one of these m factors is again resolved into n other equal factors, one of the latter is the mni\\ root of the number. Illustration. — The 4th root of a number is one of the four equal factors of that number ; if we resolve the number into 2 equal fac- tors, and then one of these factors into 2 other equal factors, one of the latter is one of the 4 equal factors which compose the given number. EXAMPLES. Ex. 1. Reduce \/mi^. Model Solution. Operation. 'V^257?J^ = Vy^SS^? = V^^' = abV^. Explanation. — The 4th root of 25a*6" is one of the 4 equal factors of it. Hence I first resolve it into two equal factors, one of which is 5a^5^ Then I resolve 5«^5^ into two equal factors, or rather indicate it, as the operation cannot be fully performed, and have \/5a^¥. /^ba^h^ is, therefore, one of the 4 equal factors of 25a^&". But, by the last problem, \/^aFb^ = a'b\/'5b. Hence V^S^^F = db\/^. 2. Reduce ^27a^K Result, hVSab. Suggestions. ^27a'b' = V\^27a'b' = \^3ab' = h^/S^. 3. Reduce ^y—64:a^ Result, 2V^^. 4. Reduce \/266a^^. 5. Reduce VSln^mK 6. Reduce V^—^^y-^-y^. 220, Prob. 3. — To reduce any number to the form of a radical of a given degree. REDUCTIONS. 183 Rule. — Raise the nimibcr to a power of the same degree as the radical, and place this power under the radical sign with the required index, or indicate the same thing by a fractional exponent. Demonstration. — That this process does not change the value o£ the expression is evident, since the number is first involved to a given power, and then the corresponding rpot of tliis power is indicated, the latter, or indicdt^d operation, being just the reverse of the former. Thus, x — y^x"'. That is, raising x to the with power, and then indicating the wth root, leaves the value represented unchanged. EXAMPLES. Ex. 1. Reduce '^dh^ to a form of a radical of the 3rd degree. Operation. 7a V = ^{la'x^y = ^343^. Explanation. — If 1 cube 7aV and then extract the cube root of this cube, the result will evidently be the same as at first. Now ila^x^y = 343aV. But instead of performing the operation of extracting the cube root of 343aV, which would evidently return it to Ta^a;^, I simply indicate the operation, and have \^MZa^x*. 2. Reduce 2ay—S to the form of a radical of the second degree. Result, V^aV— 12fl«/ + 9. 3. Reduce a—x to the form of the cube root. ^/\ 4. Reduce A / 5 ^^ the form of the 4th root. 3 Reduce - to the form of the 3rd root. Result, 184 CALCULUS OF RADICALS. 6. Eteduce ^ to the form of the 4th root. 221. Cor. — To introduce the coefficient of a radical under the radical sign, it is necessary to raise it to a power of the same degree as the radical ; for the coefficient being reduced to the same form as the^ radical by the last rule, we have the pro- duct of two like roots, which is equal to the root of the pro- duct (194, and 205). EXAMPLES. 1. Introduce the coefficient in SxV^x^ under the radical sign. Model Solution. Operation. dx\/2x^ = ^/^fx" x \/2^= ^^Ix^ x ^x" = v^54^^ Explanation. — Cubing ^x I have 27a5^, the indicated cube root of which is 'v/27x\ This is evidently the same in value as 3a;. Hence Zx^^x^ = /v^27^^ X \/2^ = \/27^>^^ = ^54^ (209). 2. Introduce the coefficient in -a/2 under the radical sign ; in g-v/3; in ^Vi; in gv'O. Results, a/^, \/|, \/j. ^/l■ 3. Introduce the coefficients in the following expressions under the radical signs ; So? V2ax, (a—x) Va-i-x, ^ V^a, . c\^ (x—y) Vx—y, and ^v27a^. Ttvo of the results are V{(i^—o^^) (a—x), and V{x—yY. liKDUcTlOXS. 185 4. Introduce under the radical signs the coefficient in the following: i^, ^^, ox^/2b^, and ^^^. The last two are v'5"»+*^-*, and \/ — -r- 222, Prob. 4.--To peduco radicals of different degrees to equivalent ones having a common index. Rule. — Represent the numhers hy means of frac- tional indices. B educe the indices to forms having a common denominator. Perform upon the numbers the operations represented hij the numerators, and indicate the operation signified hy the detioniinator. Demonstration.— The only point in this rule needing further demonstration is, that multiplying numerator and denominator of a fractional index by the same number does not change the value a ma a 111 of the expression, /. t'., that jf = jP'. Now, a^ = .r^ x a? x ic^ to a equal factoi*8. If now we resolve each of these factors into m equal factors, x will be resolved into nib equal factors, and one of I a them will be represented xP>. But as in .r^ there are a factors, each x^, and as each of these is resolved into m factors, there will be in all ma factors, each .r^. Hence 3^ — .r^. [When f is used as a common traction, wc show that f = J thus: I signifies 3 of the 4 equal partx of some quantity. If now we separate each of these 3 parts into 2, the entire quantity will be separated into 8 parts, and in the three parts there will be 6. /. f = f. In an analogous manner when | is used as an exponent, we show that f is equivalent to J, thus : f as an exponent signifies 3 of the 4 equal / and —— to forms havinsr rational denominators. Owe of the results is _/y/i6. Z \/% (3)ix(6)i Vis SV^ 1 /5 Suggestion. -^ = — ^ , = ^ = -^ = sV^. V^ (6)ix(6)i 6 6 2^ Scholium. — This process is equally applicable to any form of radical factor in the denominator, whether monomial or polynomial. 7. Rationalize the denominator of ' - wa + X QMnnn.#!«« V^"^ _, {(l-X)\ ^ {a-^X)\ _ ^J o" -X^ Suggestion. , = \ 7 = ^ , ^ - \/a^-x 0? + a;)ix(« + 5cp ^"^* REDUCTIONS. 189 8. Reduce — r; .- to a form having a rational denom- inator. Result. — z~ — -> ., » 6c— 2x^ 224. Prob. 6. —To rationalize the denominator of a fraction when it consists of a binomial, one or both of whose terms are radicals of the second degree. Rule. — Multiply both terms of the fr actio ft by the denominator with one of its signs changed. Demonstration. — This rationalizes the denominator, since in any case it gives the product of the sum and diflFerence of the two terms of the denominator, which being equal to the difference of their squares, frees either or both from radicals, as the square of a square root is rational. EXAMPLES of Ex. 1. Rationalize the denominator of — ^ Operation. Model Solution. a{a + \^h) _ a' -I- a ^/b a— ^ft (a- y^)(a + y ft) «' -^ Explanation. — I observe that a—'s/hvnW be rationalized by mul- tiplying it by o+ \/^ since the product of the sum and difference of two quantities is the difference of their squares. Hence multi- plying both terms of the fraction, so as not to alter its value, I have tt'-f-av/ft 2. Rationalize the denominator of _ — - va—Vh a—b 190 CALCULUS OF RADICAL^?. 3. Reduce —to a form having a rational denom- 2a/2— aVa inator. ,, 3(2\/2 + 3\/3) _ 6a/2 + 9\/3 6a/2 + 9a/3 ^^^^^^' 8^2T~ ~ -19 ^^ 19 4. Reduce — to a form havius a rational denorai- 3-2V2 nator. Result ^ 4+^/3. 1-_V5 I + V2 5. Rationalize the denominators of — ;=, and 7=^' 2V6— Vl8 Va + ^+V«— aJ 3 + ^5' 2 + V2' i?^5i^/jfs, 2 -a/5, iV2^ 9 + l-A/iO, and ^ ^^^ — ' 6, Rationalize the denominator of Result^ x-^1 FOR REVIEW OR ADVANCED COURSE. 225, Prob. 7.— A factor may be found which will rationalize any binomial radical. Demonstration. — If the binomial radical is of the form \/{a ■^h)'!^^, m n — m or (a +5)*, the factor is (a + h) « , according to (223). If the binomial is of the form ■y/as-\- yJ' or «"» + &«, let a'" — x, REDUCTIONS. 191 and ft» = y ; whence w^ = x^, and ft* = y\ Also let y be the least common multiple of m and n, whence -xf and y'^ are rational. But ifr — a'", and y^ = J*. If now we can find a factor which will ren- der a^ + y, a?'^±y^ this will be a factor which will render am + ?»", rr'"±7>'» which is rational. To find the factor which multiplied by ,*^y' gives a^P ± y'/*, we have only to divide the latter by the former. Now ^= a-»(i>-l>— a^(J>-2)y' +aj»(l>-3)y2)-_a-»(p-4)y3r^ ±y'(p-l) (4), the + sign of the last term to be taken when y is odd, and the — sign when it is even (126). Therefore a!»^p-''— afy2r _2^p— i)y8r^ ±y'^-i>, is a factor which will render ^a'-\- ^Jf « r rational, of being understood to be a^^ and y = 6», and p the L. C. M. of m and n. If the binomial is J^a* — ^^b% the factor is found in a similar manner, and is aj»(i'-i)+«^^2^y''+a5»(^3)y2r+ +y(i>-i). EXAMPLES. Ex. 1. Rationalize Va-\- \/b or a^-^h^. Solution.— In this example «=1, /•=!, i>=6, a;=a», and y—h ■ Hence formula (A) becomes ah—a^lik + a^ht—ab + azh^—hs. This factor multiplied into a^ + ft^ gives a"— &', as the rational- ized product. 2. Find a factor which will rationalize V^ — \/t^ or Result, eV-4.A^f + 6^t;t-feV-t;l4.eVt;¥+eVt;V + ^Vt;¥ _j.eft;V^ + efy¥ + elv¥-|-efv^-fv^, is the required factor, and the expression rationalized is e®— v®. 192 CALCULUS OF RADICALS. 226. Prob. 8. — A Trinomial of the form Va -f ^/h -\- Vc 7nay he transformed into an expression with hut one radical term by multiplying it by itself with one of the signs changed, as Va+A/b — Vc. ITie product thus arising may then he treated as a hinomial radical hy conddering the sum of the rational terms as one term, and the radical term as the other. Thus, {Va-{-Vb-^Vc){Va+Vb—Vc)=a-^b—c + 2Vab. Again, [{a-\-b—c)-j-2\/ab]x[(a-\-b—c)—2Vab] = a^ + U^ -\-(^—2ab — 'Zbc~2ac, a rational result. Ex. 1. Rationalize VS—Vl — V^- Result, 4. COMBINATIONS OF RADICALS. ADDITION AND SUBTRACTION. 227. Prob. 1. — To add or subtract radicals. Rule. — // the radicals are similar the rules already given (72, 77) are siijficient. If they are not similar make them so hy (217-233), and eomhine as hefore. If they cannot he made similar, the comhinations can only he ind^icated hy connecting with the proper signs. Demonstration. — When the radicals are similar the radical factor is a common quantity and the coefficients show how many times it is taken. Hence the sum, or difference, of the coefficients, as the case may be, indicates how many times the common quantity is to be taken to produce the required result. If the radicals are not similar, the reductions do not alter their values; hence the sum or difference of the reduced radicals, when they can be made similar, is the sum or difference of the radicals. COMBINATION — ADiHTiON AND SUBTRACTION. 193 EXAMPLES. Ex. 1. Add Vl^ and \/242. Model Solution. Operation. ^18 = 2^/2, and ^^42 = 11^2. .-. V18+ V242 = ^^/2 + ny^2 = 14^2. Explanation. \/l^ = V^ ^ 2. But the square root of the pro- duct equals the product of the square roots ; hence V^ ^ 2=3/y/2. In liiie manner v'^242 = ^^121 x 2 = 11^2. Therefore ^18 + 'Y/242 = 3^^2+11^/2. But three times any quantity, as \/2, and 11 times the same are 14 times that quantity. .'. 3Y^2 + liy'2 = 14^2. 2. Add V2Wx^ and VT92xf. Sum, i7yVSx, 3. Add \/500 and v^lOS. 4. Add Va^y and Vc^y- Sum, (a + c)Vy» 5. From \/605 take the V^OS. Diff., 2\/5. 6. Add V605 and — \/405. 7. Add 3^/^ and 2y^l. 8. From 3a/^ take -3y ^- />ijf., ^\/lO. 9. What is the sum of \/ j and \ /q-'' ^^., ^V3' 10. What is the difference of V2a3^ — iax + 2a and V2aa?^H-4aa;-t-2a? viw-s 2V2a. 194 CALCULUS OF RADICALS. Why should no sign be given to the last answer ? If the problem read, From -\/2cta;^— 4a«+2a take \/2a^T^ax + 2aj why would the answer be —2^ 2a ? 11. What is the sum of (a—xy Vxy and {a-{-xY\/xy? Ans., 2(fl2+ic2)V^. 12. What is the difference between (a—xf^/xy and (a-^xYVxy"^ Ans., Aax^yi 3 5' Sum, WS. 13. Find the sum of 8\/|, V60, _2iVl5, and a/ 14. From W~ take W ., (3«.-|)v/; W Mh Rem., [Zax—~\\y-^- 16. Add «Y^l+[^]^ and ^\/l + [ff • Suggestions. a/^J = a>^l+| = ay^^ «f Vat + 6i In like manner &|/l +[^1^ = &t Vof + fti The sum is (at + 5f)Val+5f = (at+&t){al+&t)i = («l + &t)i 16. From (a-o;) V^F^^ take (a-a:) a /^±^. Rem., (a—x—1) Va^—x^, 17. Add and ; ^wm, 2a;. ay+Va:^— 1 a:— Va;^— 1 18. Add^^^ + ^^^ and ^^S-^^S- COMBINATIONS— Mr LTIPLICATION. 196 MULTIPLICATION. 228, Prop. 1. — The product of the same roots of two or more quantities^ equals the like root of their product. Demonstration. — That is \/x x \^y = ^xy. This is evident from the fact that \/xy signifies that xy is to be resolved into m equal factors. If now x and y be separately resolved into m equal factors and then one factor from each be taken to make a group, there will be m such equal groups in xy. Thus ^x is one of the m equal factors of x, and y'y is one of the m equal factors of y. Hence \\/x x y'y] • \/\/x x -y^y] • [/{/« x ^y'\ etc., to m fac- tors of ^x X ^y, makes up xy. Therefore ^x x \/y = ^vy. (See Abts. 205 and 202.) 22^, Prop. 2. — Similar Eadicds are multiplied by multi- plying the quantities under the radical sign and writing the product under the common sign ; Or by indicating the root by fractional indices, and, for the product, taking the common number with an index equal to the sum of the indices of the factors. Demonstration. 1st.— Since similar radicals are the same root of the same quantities, as \^x x y'a;, this is only a pailicular case under Prop. 1. 2nd, «"• X a?" signifies that one of the m equal factors of x is to be multiplied by another of the m equal factors, or by itself. This gives 2 of the m equal factors of x, which is what is indicated by a?». 230. Prob. 2.— To multiply radicals. Rule. — // the factors have not the same index, re- duce them to a common index, and then multiply 196 CALCULUS OF RADICALS. the mvmbers under' the radical sign and write the product under the common sign. Demonstration. — (This is the same as Prop. 1.) EXAMPLES. Ex. 1. What is the product of V2 and ^/^ ? Model Solution. Operation. \/2 = \/8, and v^3 = ^9. .-. '\/2 x \^S = ^ x^9 = ^72. Explanation. '\/2 = ^y^S, since the former is one of the two equal factors of 2, and the latter is three of the six equal factors of 3. In like manner \^d ~ v^9. Consequently, ^/2x ^y/S — ^/S X -y/O. Now since the product of the same root of two numbers is equal to the like root of the product, /y/8 x ^^ = ^^72. 2. Multiply \/dac by \/2ac. Prod., V433«V. 3. Multiply ^/a—x by "s/ a—x. Prod., ^a^—hahc ^AS)a^x^— VWx^ + hax^—7^. 4. Multiply y ^ by y ^. Prod., \ 6. Multiply ^/l by \/l. Prod., 3^+^ = 3^ = ''v/6561. 6. Multiply "V^Zax by '\/2ax. Prod., (2aa;)H = ""^Z A^Q^Qa^x^. 7. Multiply ^Jl by ^^\ Prod, A/|=:i^486. 5. Multiply 3a/2S by 2^xy, COMBINATIONS — M ULTlPLlCATIOK. W Suggestion. — Here we have the continuous product of 3, \/2aT^ 2, and yiri/. But, as the order of the factors is immaterial, we may write 3 X 2 X V^o* x ^xy=Q ^/Sa^xY- 9. Multiply 3a^| by ^y\' Prod., 6'\/236196. 10. Multiply 6a^ by 3ai 11. Multiply 2\/ab by d\^ab.. 12. Multiply ia^b^ by ba^bl 13. Multiply Sx^y^ by 2x^y^ and represent the product without fractional exponents. 14. Multiply A /- by \V^ and represent the product without the use of the radical sign and in its simplest form. Prod., ^(9000)1 15. Multiply cr by b". Prod., a^b* or, ^/aP'b'^, or («^^)'»». 16. Multiply |V5 by ^\^iO. Prod., ^v^250. 17. Multiply av^a;, b\/y, and c^z together. Prod., ahr^xf'Py^Pz'^^ 18. Show that 2\/9 x 16 = 16^/13. 19. Show that ^^24 x 6v^3 = 6\/l2. 20. Multiply 2\/a— 3v^ by SVa + Wb. Operation. 2 \/a — 3 yT 3 \/a + 2 v^ I 6a — 9 \/c^ + 4 \/qft — 6ft 6a — 5 \/ab — 6&, or 6(a— J)— S'y/aJ. 198 CALCt^LltS OP RADICALS. 21. Multiply 3 + a/5 by 2-\/5. Prod., l-^V?. 22. Multiply \/2-fl by V^-1. Prod,, 1. 23. Multiply iW^-Wl^ by Vs + Vs. . Pro^., 2^^3 — ^/10. 24. Multiply V^12 + Vl9 by '^12-a/T9. Prod., 5. By (22§) ^12+ Vl9x ^13-^19 = ^ {\^+ \/\^){l^-^/\^). 25. Multiply ^2_^V2 + 1 by a^-\-a\/^ + l. Prod., aHl- 26. Expand (x^ + l){x^—xV^^\)(x^^xV'^-\-l). Prod., cc6+l. 27. Multiply 3a/45-7V5 by Vii+2\/9i. Pro6?., 34. 28. Multiply ^/a-\-c^/l by Va~c\/h. Prod., a-^c^^. DIVISION. ;^5i. Prop. — The quotient of the same roots of two quanti- ties equals the like root of their quotient. Demonstration. — Let m be any integer and x and y any numbers ; we are to prove that v^^-^v^y, or 5^ = 4/^. Now, that ^' = y - is evident, since ^^ raised to the Twth power, that is y ^y Ux X yx X ya; X ytc to w factors x -iTF- — ^- — ^;:7- — ^i7= ==-; whence it appears y 2/ X yy x y 2/ x y y to m factors y that ^^ is the rnth root of- or equals i/ -. (Akts. 206, 202.) \/y y ^ y COMBINATIONS— DIVISION. 199 232. Prob. 3.— To divide Radicals. Rule. — // the radicals are of the same degree, divide the number under the sign in the dividend by that under the sign in the divisor, and affect the quotient witlo the common radical sign. If the radicals are of different degrees, reduce them to the same degree before dividing. Demonstration. — [Same as above ; or, it may be considered as the converse of the corresponding case in multiplication.] EXAMPLES. Ex. 1. Divide V^cfif by ^iay. Model Solution. Operation. a/3«V = \/27^y, and v^2^y = >v/4^'. Explanation.— Since V^^V = V^27aY, and ^2ny = ^4aY . _ = l-^-—-' And since the quotient of the 6th root of two numbers is the 6th root of their quotient, - ^ ^ = V zlJl-^ , which, by performing the operation indicated, becomes y — -— ; and by reducing so that the number under the radical sign shall have the integral form, this becomes ^ ^'kd2a*i/\ 2. Divide VV25ah^y by ^hahcy. Qvot., 5\/ax. 3. Divide a/3 by ^. Quot., ^^^ = ^\/^^. 4. Divide ^J\ by >y/|. Qmt., \^mA. 5. Divide V^^^ by \/2a^. QuoL, '\^I^. 6. Divide ^/72 by a/2. Quot., \/l. ^00 Calculus of radicals. 7. Divide x^y^ by x^y^, and represent the quotient with- out fractional or negative exponents. Quot., \ / -. V y 8. Divide 24:Vay by 6v^ay. Suggestion. '-^^^ = 4^ ^ 4-^5 = 4^^^^ = 4i/I 9. Divide 126V^y^ by 26\/^^, representing the quo- tient without the radical sign. Quot, 5x^y~K 10. Divide a/6 by Vi. Quot, V3V2. 11. Divide 20v^200 by 4^2. Quot, b'^l. 12. Divide ^^ by ^f. 13. Divide iVi by V^ + SVf Suggestions. ^/2 + 8 \/^ = 2 ^/\ + 3 Vi = 5 Vf. Wbence VS + SVi 5V^ 5 10 14. Divide v^a^H^ by v^^II*. C^^ojJ., v^^+^. 15. Divide {aWcf^ by («J)i G«^Oif., a^/lc. 16. Divide 200 by \/40. 0«^oJ^v lOVlO^ or (10)1 17. Divide aVx—\/hx-\-aVy—Vhy by V^+Vy. Suggestion. — Observe that a \fi— '\/hx + a \/y— \/by = a {y/x + \/y)-VKV^+Vy) = {a-V^{V^+Vy)' 18. Divide a + h-^c + 2Vab hj Va-\-V^—Vc, COMBINATIONS— INVOLUTION. 201 Operation. a + 'it^/ab + b — c \ \/a + \/ b — \/c a + \/ab — \/ac ^a + \/ft + yc ^ab + b — \/bc \/ac + \/bc — e 19. Divide bV^^^ by aVia-^-by. 20. Divide ^/c^—a? by rt— ar. ^wo/., v/-^-:- 21. Dmde my/— ^ by ^ — • G^.o^., -J^. INVOLUTION. 233. Prob. 4.— To raise a radical to any power. Rule. — Involve the coefficient to the required power, and also the quantity under the radical sign, writing the latter under the given sign. Demonstration.— This results directly from the principles of multiplication of radicals (230). Thus, to raise a/^b to the 7»th power, is to take m factors of ay'b^ which gives « y 6 x a-y^J x ay^ft, L'tc, to m factors. But as the order of the factors is immaterial (§5) this may be written aaa to wi factors x -{/& x ^yb x \/b to m factors. But ana to m factors is by notation a"*, and V^6x v^x v^6 - - - to m factors is by (230) v^. .*. The wth power of a\^& is ^"•'y^. q. e. d. EXAMPLES. Ex. 1. Raise Ja/| to the 3rd power. 202 CALCULUS OF RADICALS. Model Solution. Operation. J^VW = iVf • Wh Wi = i • i " i x VI Vl Vl - sV VtIt or Tfy Vl = -rfirVlO. Explanation. — The cube of ^■\/l- is a product three factors each = aVIi ^u^ as the order of the factors is immaterial this may be considered as 3 factors each = ^, or -^j, and 3 factors each = yf , or f Vf • Hence a Vf)« := ^ x | ^f = ih VI Which by removing the denominator from under the radical sign becomes 2. Raise 2a/3^ to the second power. 1 3 3. Raise -y4:X^y to the 6th power. 4. Raise — 3a/^ to the 3rd power. 5. Square 3 — ^2. /S^ware, 11— 6a/2. 6. Cube a/2- a/3. * Cube, ll\/2— 9\/3. 7. Cube 2Vx—y. Cube, S{x—y)Vx—y, or 8(2^—1/)^. ^54. Cor. — To raise a radical to a power ivJiose index is the index of the root, is simply to drop the radical sig7i. Thus, the square of Vab is ab, the cube of V^x^y is "^x^y, the square of V— 2a^^ is —^a^b, the 5th power of "s/a^—l^ is d^—b^. Scholium. — This process of involution is a special case under EVOLUTION. 23r^» Prob. 5. — To extract any required root of a mono- mial radical. Rule. — Extract the root of the coeffident, and of the quantity under the radical sign separately, affecting C0MBINATI0K8 — EVOLtJTiOX. 20t'i the latter with the giveii radical sign. Reduce the result to its simplest form. Demonstration — The nt\\ root of a-y^j, signifies one of the n equal factors oi a^^h. Hence if we resolve a into n equal factors, and y 6, into n equal factors, a group consisting of one from each is tae wth root o{ a^h. Thus one of the n equal factors of a is V^a, and of y ^ V^y ^ **or \^>^lx V^^ft x \^^b- --ton factors, is \^^h b'l-bton factors = ^& (233, 231). If now we take a group consisting of one from each set of factors, that is y^ V y b, we have the 7ith root of a y'ft, since we have one of its 7i equal factors. Q. B. D. EXAMPLES. Ex. 1. What is the 3rd root of WSc^? Model Solution. Operation. V^4 \/daFx = \/4'i^^/^a^x = y^4y^3a^ Explanation. — The cube root of 4:\/Sa^x is one of the 3 equav factors which compose it. In order to find this, I resolve each of the two factors 4 and \/da^x into 3 equal factors, and take one of each. 4 resolved into 3 equal factors becomes y^4 x \^i x -^4. (A process which in reality is only indicated.) In like manner \/3a*«, resolved into 3 equal factors becomes V \/da^xx \/da^xx \/da*x or V^Sa'xx Vy'^^xx V ^ Sa^x since the root of the product equals the product of the roots. Now taking one factor out of each of these groups I have y^fv -y/aia'i, and as three such factors could be formed from the number, \/4V'^'^*x is the cube root of 4 \/Sa*x. But this expression can be reduced to a more simple fonn, by observ. ing first that the square root of the cube root is the 6th root, and then reducing ^l to the same radical form. Thus I have \^4 V ^da*x = 'V^^S^ = ^16 ^da^ = ^4S^ by (233). 204 CALCULUS OP KADICALS. 2. Extract the cube root of a/^*^^ Root, x^c^. 3. Extract the square root of 32v 192a8:c2. Suggestion.— The square root of 32 v^IOSo^ = 4 ^/a x /v/l92aV 4. Extract the cube root of -a^^/h. 8 5. Extract the 4th root of l^c^^/x. Root, 2\/a^. 6. Extract the cube root of (a + x)\/a+x. Root, va-\-x. 7. Extract the cube root of :^\/ -o' Root, ^VSa. 8. Extract the square root of ^\/^' Root, «v^2. 236* Scholium. — This operation is but a special case of affecting a quantity with any given exponent (194), and the examples may be performed according to the rule there given. Thus, to extract the cube root of Q^dax'^ ; putting it in the form 8 x (Sax^)i and divid- mg exponents by 3 (multiplying by ^) we have 8^{Saa!'^)l = 2^Saa'. The pupil may solve the following in this manner : 9. Extract the 5th root of V32^. Suggestion. -^/SSa;'" = (32)ia^. Multiplying exponents by ^ we have (32)^'^a;. But (32)^^ = \/(32)i = >^2. .-. \/ ^^/^2^ = x^2. But the most simple way to solve this particular case is V y'SS*" = V^32^ = ^2x^ =- x^y2. Or by the rule given (235). 10. Extract the square root of \^4:9aK 11. Extract the cube root of 64^^80*. COMBINATIONS— EVOLUTION. 205 12. Extract the 5th root of 486fl\^4rt2. Root, 3v^. FOR REVIEW OR ADVANCED COURSE. 237. Prob. 6.— To extract the square poot of a binomial, one or both of whose terms are radicals of the second degree. Solution. —Such binomials have either Jhe fonn a±;i\/5or m^^± n^h. Now observing that {x ± yf — x^ ± 2xi/ + y^, we see that if we can separate the first term of any such binomial surd into two parts the square root of the product of which shall be ^ the other term, these two parts may be made the first and third terms of a trinomial (corresponding to x'^-\-2xy + y^), and the middle term being the second term of the given binomial, the square root will be the sum or difference of the square roots of the parts into which the first term is separated. EXAMPLES. Ex. 1. Extract the square root of 87— 12\/42. Solution. — Let x = one of the parts into which 87 is to be sepa- rated, and 87— a; the other. Then we have \/x{S7—x) = — 6\/42, or squaring, Slx—x"- = 1512, or «»— 87a; = —1512. .*. «= 63 or 24. and we have V87-12V'42 = V63-12\/42 + 24. Now \/Qd — 3\/7 and '\/24 = 2\/6, and as the middle term of the trinomial is negative and twice the product of these roots, its square root is 3 V7-2\/^- (123.) 2. Extract the square root of 3\/6-f-2\/i2. Suggestion. 3^/6 is to be separated into two parts. Let them be X and 3\/6— «• Then ar(3\/6— ar) = 12. Whence x = 2\/6 or V^6,and V3V6 + 2Vl2=V2V6 + 2yi2+y6=V2y6 + V'V/Q = rod., -60V4^V^, or -0Q>/^^, 208 CALCULUS OF RADICALS. 6. Multiply 24H-\/-49 by 24-a/— 49. Suggestion. — We have here the product of the sum and difference of two quantities which equals the difference of their squares. Hence (34 + ^1149) X (34-/v/^^^) = (24)'-(V^^^)'=576-(-49)=576 + 49 = 625. 7. Divide V— 16 by V— 4. Operation. ^/—lQ = 4 \/—lj and V— 4 = 2 -v/— 1- Hence .^_ 4 2V-1 Scholium. — A superficial view of the case might lead one to think that the quotient w^as ±2. Thus, noticing that the radicals are similar, he might conclude that "^ — = y ~—- = -y/I = ±2, an y^— 4 '^ — 4 incorrect result. 8. Divide 6\/^ by 2V^. 4>t^o^., |a/3. 2 9. Divide — V^ by — eV^. QuoL, ^Vs. 18 10. Divide l + V^ by 1 — V^. Operation. — Writing the example thus ^ • and multiply- 1 — V— 1 ing both terms of the fraction by 1 + ^/^—i there results, (see Ex. 6) l + 2\/^-l 2^/^ l-(-l) 2 = V-i. The example may also be performed thus : 1 + V^ I - A//^"! + 1 1 + y_i ^31 Since 1 divided by — ^^ gives y^, as — y^xy'^ SYN'OPSIS. 209 H O fic 5 I -J a I o I I CO u o 0. SYNOPSIS FOR REVIEW. Power.— Degree of. J g^H l.-Powers and Roots correlative. r. ) Root.— Degree of. A r 1. + Integer. | ,, -| 2. + Fraction. >■ . ■ I .'J. - Int. obFr. ) ScH. 2.— Square, Cube,etc. Exponen or Index Radical. 1. + Integer. + Fraction. - Int. or Fr. (' j Ratioual. Real. i , ^, ' Irrational. Imaoinaby. I How INDICATED. I Cor. Transferring a factor in a frac- ion from one term to another. \f 1. +m. 2. n 8. -men _m n* Sim. Rads. rationalization. To afiect witli Exponents. Involution, evolution. Calculns of radicals. Prob. I. To any power. Rule. Dem. Cor.— Signs. Prob. 2. To affect with Exponents. Rule. Dem.^ Cor. 1. — Terminates. Cor. 2.— Number of Terms. Cor. 3.— Coeflacients. Q)r. 4.— Exponents in ea. term. Cor. 5,— RuLK. l^ Cot: Q.-(a-b)'". Sch.— Signs. Cor. 1.— Same as (193). Cor. 2.— R. of Prod. = Prod. of Roots. Cor. 3 — R. of Quot. = Quot of Roots. Prob. 3. Bmojcuii Fosxula. Prob. I. Roots of Perfect Powers RuLB. Dem. I Sch. 1. Prob. 2. Square Root of Poly. Rulb. Dem. < 8071.2. ( Sch. 3. Prob. 3 Prob. 4 Prob. 5 Square Root of Dec. No. Rule. Dem. Cube Root of Poly. Rule. Dem Sch. 1. Sch. 2. Sch. 1. Sch. 2, Cube Root of Dec. No. Rule. Dem •\ Sch.l. Sch. 2, Sch. 8. Prob. 6. Prob. 7. Any Root whose index composed of fiictors, 2 or 8. Any Root. Solution, Gen^r^l Scholiqm, 210 SYNOPSIS. CO < o o < < z €0 O < z o o i o SYNOPSIS FOR WEVIEW,— Continued. Prob. I. Remove factor. Rxjle. j Sch. Simplest form. Dem. I Cor. Denomination of surd. Prob. 2. Index composite, etc Rule. Dem. Prob. 3. To given index. Rule. Dem. Prob. 4. To common index. Rule. Dem. j Prob. 5. To rationalize Monomial Denominator. Rule, Dem. Prob. 6. To rationalize Binomial Denominator. Bxrus. Dem. Prop. I. To rationalize any binomial. Dem. [^ Prop. 2. To rationalize any trinomial. Dem. f Prob. I. To add and subtract. Rule. Dem. 3 C Prop. I. Product of roots = equal root of product. Dem. ^ •< Prop. 2. Similar Radicals, how multiplied. Dem. S ' Prob. 2. To multiply Radicals. Rule. Dem. CO UJ £ < z < 11 Prop. Quotient of roots = root of quotients. Prob. 3. To divide Radicals. Rule. D&n. Involution of Radicals. Rule, Dem. Sch. Evolution of Radicals. Rule. Dem. Prob. 4. Prob. 5. Prob. 6. Va+ Vft and V Va+ Vh, etc. . j Sch. 1.— Called Impossible. Definitions, -j ^^^ 2._Reason for name. Prob. I. To add or subtract. Rule. Dem. Prob. 2. To multiply. Rule. Dem. Prob. 3. To divide. Rule. Dem. Test Questions. — By what must numerator and denominator of — ^ be multiplied to reduce it to a simple fraction ? Give the various significations of an exponent. Perform the operation VS X v^3, and explain the process. Repeat the Binomial Formula, and by means of it expand {l—x")^. Demonstrate the rules for square and cube root. Show that ; • What sign is given to a square root? Why? To a cube root? Wliy ? What is the value of a?" ? [Note.— Here ends the subject of Literal Arithmetic. The student is now prepared for the study of Algebra, properly so called ; i. e., The Science of the Equation.^ PART II. LGEBRA. iSiliEciiiif] ^ Sectjon I. EQUATIONS WITH ONE UNKNOWN QUANTITY. DEPINITIONS. 1. An liquation* is an expression in mathematical symbols, of equality between two numbers or sets of numbers. Illustration, ^x — 2a''y = - —- is an equation because it is an expression of equality between Sx—2a^y and ^-— • 2, Algebra is that branch of Pure Mathematics which treats of the nature and properties of the Equation and of * Do not pronounce this word " Equazion." For this common error there Is no authority. '' Equashun " is the correct pronunciation. 212 SIMPLE EQUATIONS. its use as an instrument for conducting mathematical investigations. 3, The First Member of an equation is the part on the left hand of the sign of equality. The Second Member is the part on the right. 4. A Numerical Equation is one in which the hnown quantities are represented hy decimal numbers ; as l%x^—^x = 48. 5, A Literal Equation is one in which some or all of the known quantities are represented by letters; as , ^, 4:00^ — 2 ax—c-^Uy = —^ 6. The Degree of an Equation is determined by the highest number of unknown factors occurring in any term. Illustration, ax—lx^ = c + ai^, is of the 3rd degree ; a^x—4:x = 12 is of the 1st degree ; x^y^ = 18 is of the 4th degree, etc. 7. A Simple Equation is an equation of the first degree. Illustration. y = ax-\-h is a simple equation, as also is — -— 8. A Quadratic Equation is an equation of the second degree. 9, A Cubic Equation is an equation of the third degree. A Biquadratic is one of the fourth degree. 10, Equations above the second degree are called Higher Equations. WITH ONE UNKNOWN QUANTITY. 218 TRANSFORMATION OP EQUATIONS. 11, To Transform an Equation is to change its form without destroying the equality of the members. 12. There are four principal transformations of simple Equations containing one unknown quantity, viz.: Clearing of Fractions, Traiisposition, Collecting Terms, and Dividi7ig by the coefficient of the unknown quantity. IS, These transformations are based upon the following AXIOMS. Axiom 1. — Any operation which does not affect the value of a term or member , mag be performed upon that term or member without destroying the Equaiion. Axiom 2. — If both members of an Equation are increased (yr diminished alike, the equality is not destroyed. 14, Prob. — To clear an Equation of Fpactions. Rule. — Multiply each member hy the least or lowest common multiple of all the denominators. Demonstration.— This process clears the Equation of fractions, since, in the process of multiplying any particular fractional term, its denominator is one of the factors of the L. C. M. by which we are multiplying; hence dropping the denominator multiplies by this factor, and then this product (the numerator) is multiplied by the other factor of the L. C. M. This process does not destroy the Equation, since both members are increased or diminished alike. EXAMPLES. Ex. 1. Clear the equation |4- 1+ | = -^^ of fractions. Model Solution. 6 is the L. C. M. of 2, 3, 6, and 3, the denomina- tors. Now, it is evident that, if the first member of this equation is equal to the second, 6 times the first member is equal to 6 times the 214 SIMPLE EQUATIOIn'S. second; hence I can multiply each member by 6 and not destroy the equality. Therefore taking the equation X X «_2« + 3 2 "^ 3 "^ 6 ~ "~3~ ' and multiplying each member by 6, I have ^x + 2x + x = 4a; + 6. The operation is performed thus : 2 times ^ isa;; and 3 times a; is 2 X X 3a?. 3 times - is x ; and 2 times x is %x. 6 times ~ is x. Hence 6 o 6 times the first member is 3.x -\-2x + x. 3 times — ^— is 2a; + 3; and 2 times 2a; + 3 is 4a; + 6. Hence 6 o times the second member is 4a; + 6. Thus the denominators have all been caused to disappear without destroying the equality of the members of the equation, as both have been increased alike. Illustration. — An equa- tion is aptly compared to a pair of scales with equal arms, balanced by weights in the two pans. Now, if the weights in the scale pans balance each other, that is, are equal, and we multiply the weights in each pan by 6 (or any other number), the balance (equality) will still be preserved. Or, if we increase or decrease the weights in both pans equally, the balance (equality) will not be destroyed. 2. Clear — — ~^i = ^oi fractions. Result, 8^— 15a; + 12 = QQ. 3. Clear 10 + ?^ = ^-^^ ^^ fractions. Result, 100 + 4.x— 1^ = b + bx—^Ox. . ^, x—l x—2 dx—1 4:—x „„ 4. Clear -^ -{-x = —-- + -~ of fractions. 4 6 bo WITH ONE UNKNOWN QUANTITY. 215 Suggestion. — The multiplier is 6. In multiplying the second term, «, 2 , it should be borne in mind that the — sign preceding this 3 compound term, shows that the term as a whole is to be subtracted. Hence when this term is written without any mark of aggregation, its signs are to be changed, as in removing terms from a bracket preceded by a minus sign. The equation cleared of fractions is Sx—d—2x + 4:-\-Qx=z 3a:— 1 + 8— 2a!. Why are not the signs changed in the last term, -— - ? Are all the signs changed in the term — — - ? o o Yes. What becomes of the — sign before this fraction in the given example ? It is dropped after the operations signified by it have been expressed in detail. We might write the equation cleared of fractions thus : 3j:— 3— (2a?— 4) + 6a; = 3a;— 1 + 8— 2a;, the term 2a;— 4 being still taken in the aggregate. Now removing the parenthesis (give the reason) we have 3a;— 3— 2a;4-4 + 6a; = 3a;— 1 + 8— 2a;, as above. Neglect to make this change of signs is one op the most common mistakes of beginners. 5. Clear 1 s = 4:C r — of fractions. m am tw a^m Result, a^mx + amx --dhi z=z ^a^cm^—^mx + mn. 6. Clear a~ = « ^^ fractions. Result, 3a;— 6a; + 18 = 2a. 7. Clear ^ ^r-^ -|- -, = 1 — a; of fractions. da %ah a^ Result, 4a5a;— 3«a; + 3a + 18§ = Wb—Qa^hz. 8. Clear ^ — x A ^ = ^^^ — 1 of fractions. a—o a-{-o a-{-o Suggestion. — The multiplier is a'— ft". Result, ax-irhx—ahi-^tl^x-^i^a—U = (a—by—a^-^l/^. 216 SIMPLE EQUATIOITS. 9. Clear = ^ of fractions. x—c x-\-2c ax-\-2aC'-bx—2bc = ax-\-hx—ac—hc, X = 3 -\- ' of fractions. Za—b Result, 4:ax^2bx = W—lbab + Qb'^-\-ax—2bx. 2iX X 10. Clear -^ = 3 + of fractions a— 2b 2a— b TRANSPOSITION. 15. Transposing a term is changing it from one member of the equation to the other without destroying the equality of the members. 16. Prob. — To transpose a term. Rule. — Dj^op it from the in ember in which it stands and insert it in the other member ivith its sign changed. Demonstration. — If the term to be transposed is +, dropping it from one member diminishes that member by the amount of the term, and writing it with the — sign in the other member, takes its amount from that member ; hence both members are diminished alike, and the equality is not destroyed. (Repeat Axiom 2.) 2nd. — If the term to be transposed is — , dropping it increases the member from which it is dropped, and writing it in the other member with the + signiwcreasesthatmember by the same amount; and hence the equality is preserved. (Repeat Axiom 2.) EXAMPLES. Ex. 1. Given the equation 3 + 2a;— 5 = 12— 4rr to trans- pose so that all the terms containing the unknown quantity, X, shall stand in the first member and the known terms in the second member. Model Solution. Operation. 3 + 2a:— 5 = 12— 4aj 2x + 4:x = 12—3 + 5 Explanation. — Dropping 3 from the first member diminishes that member by 3 ; hence to preserve the equality, I subtract 3 from the WITH ONE UNKNOWN QUANTITY. . 217 second member, or indicate the subtraction by writing it in the second member with the — sign. Thus the term 3 is transposed. Dropping —5 from the fii*st member incrmses that member by 5 ; and hence to preserve the equality I add 5 to the second member. Thus the term —5 is transposed. Dropping —4a; from the second member increases that member by 4x ; hence I increase the first member by adding Ax to it, and thus preserve the equality. I have thus arranged the terms so that all those containing the unknown quantity stand in the first member, and all known terms in the second member ; and yet I have preserved the equality. Illustration. — This operation can be illustrated by the scales on page 214. Suppose the positive terms to represent weights and the negative terms some forces lifting on the scale-pans. [Since the + and — terms represent quantities opposed in effects.] Taking the 8 from the first member corresponds to taking off so much weight. This can be compensated, so as to keep the scales in equilibrium, by applying a lifting power of to the other side, which is symbolized by —3 on that side. Again taking —5 from the first member is like taking away a lifting power of 5, which can be compensated by putting a «?e/^A^ of 5 on the other side ( + 5). In like manner the transposition of any term can be illustrated. In fact all operations ujion equations can he illustrated in a similar way. In the following examples transpose the unknown terms to the first member and the known to the second. 2. Given 5x—12a-^3c—2a;=:4:X—2x-^4:a to transpose as above. Result, 6x—2x—Ax-^2x = 4:a-\-12aSc. 3. Given 100-f 4a;— 6 = 5a; + 5— 30ir to transpose the terms as above. Result, Ax—bx-\- 30a; = 5 — 100 + 6. .1 m , 15 3a; ,^ 7 1 11 4. Transpose as above -^ — -^ +10— - = - — —x-^x. T, ,, 11 3a; 1 7 ,^ 15 Result,-x-x-- = -^^-^-X^^-. 10 218 SIMPLE EQUATIONS. 5. Transpose as above Sab 4:ax-{-lSbx^ = -~^—xK Eesult, ISbx^ j^ x^ — ^ax — — Sab. 5 c SOLUTION OP SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY. 17, To Solve an Equation is to find the value of the unknown quantity: that is, to find what value it must have in order that the equation be true. Illustration. — In the equation 4a;— 2 = 2aj + 4, if we call a*, 3, the first member is 10, and as the second is also 10 for this value of «, the equation is true. But if we try any other number than 3 for X, we shall find that the equation will not be true. Thus trying 4 for a?, we find the first member 14 and the second 12 ; and the equation is not true. Again, try 5. The first member becomes 18 and the second 14, and the equation is not true. Let the student see if he can ascertain by inspection what are the values of x in the following : JB + 3 = 3a7 + 1. 2a; = 30 — X. Though these equations are very simple, it is probable that it will take the student some time to guess out the values of a; which make them true. But, if it is so difficult to hit upon just the value of x which is required to make so simple an equation true, the task would be quite hopeless in such an one as 3x-l _ 13-^ _ Z^ _ ll(a; + 3) 5 2 ~~ 3 6 " Yet we have a very simple method of solving any such equation so as to tell certainly and easily what the value of a; is. This pro- cess is now to be explained, and is called Solving the Equation^ or sometimes the Eesolution of the Equation. WITH ONE UNKNOWN QUANTITY. 219 18, An equation is said to be Satisfied when a value is given to tlio unknown quantity which makes it a true equation : i, e., which makes its members equal. It is said to be Destroyed when its members by any process become unequal. W, To Verify an equation is to substitute the sup- posed value of the unknown quantity and thus see if it sat- isfies the equation. 20. Prob. 1.— To solve a simple equation containing one unknown quantity. Rule. — /. // the equation contain fractions, clear it of them by Art 14. //. Transpose all the terms involving the unknown quantity to the first memher, and the known terms to the second member by Art. 15. ///. Unite all the terms containing the unknown quantity into one by addition, and put the second member into its simplest form. IV. Divide each member by the coefficient of the unknown quantity. Demonstration. — The first step, clearing of fractions, does not destroy the equation, since each member is multiplied by the same quantity (Axiom 2). The second step does not destroy the equation, since it is adding the same quantity to each member, or subtracting the same quantity from each member (Axiom 2). The third step does not destroy the equation, since it does not change the value of the membei*s (Axiom 1). The fourth step does not destroy the equation, since it is dividing each member by the same quantity, and thus changes the members alike (Axiom 2). Hence, after these several processes, we still have a true equation. But now the first member is simply the unknown quantity, and the second member is all known. Thus we have what the unknown quantity is equal to ; i. e., its value. 220 SIMPLE EQUATION'S. 21* Scholium I. — It must he fixed in the pwpiVs mind that he can make hut two classes of changes upon an equation without destroying it : viz,, Such as do not affect the value of the members, or SUCH AS AFFECT BOTH MEMBERS EQUALLY. Every Operation must I>e seen to conform to these conditions. EXAMPLES. Ex. 1. Solve — ^ -\ h o = — o^ y a^d verify the /ii O o o result. Model Solution. x + 1 3a;-4 1 603 + 7 Operation. (1) _ + -^- + ^ = -- , (3) 20x + 20 + 24a;-33 + 5 = 30a?+35, (3) 20o? + 24«-30aj = 35-20 + 32-5, (4) 14«=42, (5) x=3. Explanation.— I first clear equation (1) of fractions by multiply- ing each member by the L. C. M. of its denominators, which is 40. aj + 1 This does not destroy the equation (Axiom 2). I multiply — ^r— 2 by 40 by dropping its denominator 2, thus getting a?+l, as the result of multiplying by 2, and then multiply x + 1 by 20, getting 20a; + 20. [In like manner explain the entire process of clearing of fractions.] Having cleared the equation of fractions I have (2). I now transpose the terms containing x to the first member and the known terms to the second member. Thus, dropping SOo? from the second member diminishes it by that amount, whence to preserve the equality of the members I subtract 30x from the first member, i. e., write it in that member with its sign changed. [In like manner explain the transpo- sition of each term.] I know equation (3) to be true, since I have changed both mem- bers alike, that is, have added to and subtracted from each member the same quantities, I now add together the terms of the first member, which does not affect the value of the member, and have 14«. In the same manner uniting the terms of the second member does not alter its value ; hence 14a? = 42, Finally, I divide each mem- ber by 14 and have a; = 3. This operation does not destroy the equation, since each member of the equality 14a; — 42 is divided by the same number (Axiom 2). Hence 3 is the value of «, WITH ONE UNKNOWN QUANTITY. 321 Verification. I will now see by actual trial if 3 does satisfy the given equa- tion. Substituting 3 for a:, I have 3 + 1 3x3-4 1 6x34-7 4 5 1 25 2^5 + 8 = ¥'"^ 2 + l+^ = 3|,or the members of which are identical ; and the value of a; is verified. 2. Solve and verify — - — =: — - — Result, x =z 1, ,, .« .. 3 + 1 7 + 5 3 12 ,,^ ., ,. ^ Venflcation. — ^r- = — --, or - = — • Whence it appears that 1 satisfies the equation. 3. Solve and verify 2x—- = — ^r 1 5— • Result, a; = 2. 4. Given -^ — - — -= = — = ^ to find x. 6 7 Result, X = 4:, 6. Given 7a; + 6— 3a; = 564-2a; to find x, and verify. 6. Given ^- + 6?/ = ^^^ to find y. ff n- o. . 3a;-ll bx-b , m-lx^ . , 7. Given 21 H -r — = —5 1 ^^ — to find x. Result, a: = 9. o -,. 5(a:-5) , . 284 - a; , ^ , 8. Given ^ ^^ 4- 6a; = — - — to find x. 20 5 222 SIMPLE EQUATIONS. 9. Given 25(lla:— 8) = 18(12a; + 2) to find x, X. Result, X = 10. Q/v. ^ 10. Given — + 2^+11 = 7 + 17 to find x 5 4 11. Given y-\ ^ = — g-- to find y. Result, y = 34. Verification. 3f + ^^^ = i^?^, or which is a true equation, since 20f-^3 is 6f. In verifying it is not well to go through the processes of clearing of fractions, transposition, etc. , but rather keep the terms as distinct as possible, and reduce each member separately to a form so simple that they can be seen to be equal. y o g I „ 12. Given — ^ \- 3z = — \- 1, U) resolve and verify as above. Result, z = 1-^. Verification. ^^ + 17 ~ ~~2 "^ ' ^^^^^^ reduces ^0 — — 75_55 17 72_72 ■^ 17 "" 17 "*" 17 ' °^ 17 ~ 17 ■ 13. Given % + ^-i^ — 3 =: ^~^^^ - 2^ to resolve and verify as above. j? u _ ^-'■ Kesmt, y — Tng* 14 Given ?(^zi) + ?(^^ - M.^_^). 14. Lriven 5 + 15-3 Result, x = —' WITH ONE UNKNOWN QUANTITY. 223 15. Find the value of a; m 3a; H z — = 5 H ^r 2 Suggestion. — Having cleared this equation of fractions, transposed the unknown terms to the first member, and the known to the second, there results — 31a; = —147. Hence to obtain a positive result divide each member by —31, and x = 7. 22* Cor. 1. — All the si^s of the terms of both members of an equation can be changed from -f to — , or vice versa, with- out destroying the equality, since this is equivalent to multiply- ing or dividing each member by — 1. 16. Find the value of ^ in ic H ^-^ = -^ 2. Result, X = L 97-7y , 6{y-l) ^, , 3«/-ll , . . 17. Given — ^ + -^-^ = 21 -f- ^ to find y. /i o lb Result, y = 9. 18. Find the value of ;z in 1 + ^±^ = Sz -^ ^-^. (See '^ Result, z = ~ 19. Find the value of x in —J- + -^ = 16 ^. Suggestion. — In clearing of fractions be careful to notice the term — • As this term is to be subtracted, when the sign of aggre- gation (the line between the terms of the fraction) is dropped, the signs of the separate terms must be changed. The equation when cleared is 6aJ4-6-|-4a?-|-8 = 193— 3aj— 9; and « = 13. It is well aJ4- 3 for the pupil to explain such cases thus : Having multiplied — — - by the L. C. D. 13, 1 have 3a; + 9 ; but this is to be subtracted, hence I annex it with its signs changed, as subtraction is performed by.--.. 2M SIMPLE EQUATIONS. 20. What value for z satisfies z -\ ^r— =12 ^ ? 21. Find the value of z which satisfies the equation ^"^ 4 ^-3 12 22. Given — r^ — — = —. 2f , to find x. 14 21 4 * Separating the fractions into parts, and reducing, j3^a;+|^—/xa; +i = i«-l-2f. Whence (^8^-^\-i)a; = _^- i-3|, or — ^o; = — If, or |a! = 5, and x = 35. 23. Given ^(7x + 9)-{-^{2^9-9x) = i(9x-ld)-}-\{Sx + 1), to find X. 24. Solve i(a;+l)+i{^ + 2) = lQ-i{x-\-S). 25. Solve j(3.2;-3)-i(3a;-4) == 5^-^(27+ 4a;). 26. Sol\e i{S—x)-\-x—li = i(x-^6)—^x, 27. Given — — 6x = 6, to find x, a Operation. — Multiplying each member by awe have 2a + x—5aa = 6a. Transposing, x—5ax = 3a, or (1— 5a)a: = 3a. [Note.— It is the common experience of pupils that they continue to find difficulty with Literal Equations even after they are quite familiar with Numerical ones, such as the 26 above given. This difficulty must be overcome. No one has caught the true spirit of mathematical reasoning till the literal notation is seen and felt to be more simple than the decimal.] 28. What value for x satisfies ax-\-h = cx-\-d? Verify it. d-b Ans., X = WITH ONE UNKNOWN QUANTITY. 225 Verification.— Substituting the value of x for x we have a-^^ a—c + 6 = <5 ^J. Performing the multiplications indicated, and ci — c reducing each member to an improper fraction, this becomes, ad—ab+ab—bc cd—hc + aJ—cd ... , , ^ , . = — Now —ah and +ao destroy each a—c a—c other in the first member, and +cd and — cd in the second. Hence we have = — , which is evidently true. a—o a—c '' 29. Given ax -\- i^ = bx-\-a^ to find the value of x. Result, X = a-\-b, 30. Given 057-1-771= 5a; -hn to find a;. Result, x= ^» a — 31. Given to-}- 2a;— a = 3a;— 2c to find x. Result, X =z -J — - . — 1 X 1 32. What value for x satisfies ax-\-b = — |-t? a 33. Solve — — = -^^^^ and verify the result. 771 c *' 34. Find X from ax aV^ •=ihx A a 2a -^±i?. Result x-'^^^-^. 4 4a— 35 35. Solve ?=f + ?Z:* + ?=£ = x-{a^+c)^ oca abc Result ^_ ^^c-\-al^-\-h(?-a-l)^c ac-\-ab-\-oc—\ 36. Given - -| — = c to find x. Result, x = a x ac — 1 226 SIMPLE EQUATIONS. «« ^. x—a 2x—db a—x ^. . .., , „ , 37. Given —5 h ^7- = lOo' + 11 J to find a;. O O /« Result, X = 26a + 24:b. 38. Given be = m -^ - to find a;. Result, X = 7 — : — • bc-\-m 39. Given -- 1 + Sab = — to find a;. a c „ _, ac(l— 3a5) Result, X = — ^ r-^' c—ad 40. What IS the value of x m — \- - — --5- = — ^ — 9 bo; 4-0 o 23» Scholium 3. — It is not always expedient to perform the several transformations in the same order as given in the rule. The process may often be much shortened by the exercise of a little ingenuity. The ultimate object is to so transform the equation that the unknown quantity will stand alone in the first member. An expeditious method of solving the last example is as follows: Multiplying by 9, we have 6a! + 7 + — — = 62: + 12. (The term tj~, -J o is multiplied by 3 by dividing the denominator, and by the other factor of 9, by multiplying the numerator.) Now dropping 2\x 39 Qx from each member and subtracting 7, we get — — = 5. Whence a; = 4. ., ^. 4a; + 3 , 7a:-29 80:4- 19, n -, 41. Given —- h = -—- — to find x. 9 52^—12 18 Suggestion. — Multiply numerator and denominator of the term ■^~- by 2, transpose and unite it to the second member, and there results 3— = 3-?: • Whence cc = 6. 5aj— 13 18 WITH ONE UNKNOWJf QUANTITY. 327 .„ _. Qx+5 , 8a;-7 36x+15 lOJ , „ , 42. Gmn — ^- + ^-^^ -^^ + ^ to find x. 43. Given «4tl _ |^2 ^ 3^ti to find ^. 15 7ic— 6 5 44. Given —irz = tt; = — l — to find x. and verify 15 7a;— 16 5 ' ^ the result. Result, x = —2. 9 —12 + 1 —4—4 —4—1 11 4 Verification. -^ ___ = ___, or- j^ -- = -1, 15 , 45. Given -^ — i-J H — = 8 to find x, and verify the value. 46. Solve for a; the equation - + a h—a h-\-a An elegant solution of this is obtained by first uniting the two terms of the first member, which makes the equation = — - a(b-a) aha --7- a^G)—a) = i , or -— — - X — -= Whence x = ^^ — { • 6+a a{b—a) O+a b(b-\-a) Suggestion. — Performing the multiplications and transposing, we I X X X a a a 6 2„^ ,. .,. ^^'^' 2 - 3 + r ^ 6 - 12 "■ 20 ' ^' 12'' = IS"^- ^'°'" ^'^'^'°S u 5 - 8 by — we have x = —a. 48. Solve "{-'.= '- ^-y- RuuU, y = ^. a a c ^ ad 228 SIMPLE EQUATIONS. 49. Solve 2x a^2b = 3 + X 2a-b • Result, X - = 2a- -5i + ^. a Suggestion. — Transpose - — ^ to the first member and unite the terms, giving -^-rz^^^-^.^ = 3. .'. x = 2«-5&+-. 50. Solve 3x — a = X — • o ei o 1 x(a—b) , ab x 51 Solve -^— ^ ~ ^ "^ T ~ 3' ax—hx X ab 3a— 3Z> + 2 ia+ab Suggestion. — g— + ^ = «+ -;|, or x=—^—. .-. _4a + a5 6 _ 3a(4 + 5) 3a-3& + 2 6(a-5)+4 52. Solve ^-« + ?^±^ = ^-^- „ ,, 8^— 4ac + «5c Result, y= ^^^^^_^^ » 53. Solve |(a;-a)— j(2a;-35)-^(«-a;) = ^•-fa. Result, x=.-rb. 4 54. If (^-% _ (P±pl + 3(«5 - %by) = Sab, show that 2/ = 0. 55. Solve ^^ = «. + f . iJ..««, X = t bx ^ Suggestion.— First divide each member by a, and then write the first member in two parts, reducing each fraction to its lowest terms. Drop zr from each member and - = c. WITH ONE UNKNOWX QUANTITY. 229 56. Solve ^^ + - = .. + 1. Suggestion. — An elegant solution of this is obtained by no- ticinff that ^4; 7- = -; — -- = x + 1—- — - ; whence the equa- tion becomes a?+l— .r 1+5- = «+!• Dropping a: + l from each member, transposing and changing signs, ^ = — , or Sx = 2x+l. .: x=l. 67. Solve - H -: = -^—' x+2 11 3 1 , _ Suggestion. ^ = 3 + -. .'. ^^^ = -, and ^ = 9. _^ ^ , 6a;+13 3a;+5 2a; , .. 58. Solve -^^ - --^ = --, and verify. Suggestion. — First destroy the term — • e#^ n 1 x—1 x — 2 x — 5 x—6 69. Solve ^ ^ = 2. ji,' x—2 x—3 x^6 x—7 Suggestion. — Reducing each term to a mixed number we have 1 + — --1 x = l + — „-l ^. Whence — x—2 x—d x—Q x—1 x—2 x—S = — . Reducing the terms in each member separately to 1 -I common denominators and adding, we get {x-2)(x-d) (a;-6)(ir-7)' Whence (x-2)(«-3) = (x-Q) (x-l), or a;'-13a! + 43 = x^-5x + Q, and x = 4^. 24, Scholium 4. — It often happens that an equation which involves the second or even higher powers of the unknown quantity is still, virtually, a simple equation, since these terms destroy each other in the reduction. 230 SIM1>LE :EQtATlOl!fS. 60. What value for x satisfies S—x—2{x^l){x-^2) = (x-3) (6-2x)? Am., X = 14. 61. Solve the equation (x—6){x—2) — (x—6)(2x-5) Ans., x = 2^. 62. Solve ^ - ? (3.-4) + (I^^U^^^zH =. .. — -- , and verify. 63. Solve (rc+ 1) (a;- 1) - (^+5) (x^3) + | = 0, and verify the result. 64. Solve (a+x) (h+x) = (c-\-x) {d-{-x). Result, X = a-\-h—c—d 65. Solve —^— = h x—c x—a x—b Result x - (^^J^^rh-^c) ■ Result, X - _-j-^^____. 66. Solve [a + x) {h + x)-a(h-^c) = y + x\ Suggestion. — Perform the multiplication indicated in the first member, and write the terms without clearing of fractions; this , XX ^^^ ixn ■ , Tx 0^0 + dbc (a + I})ac gives (a + d)x — ac=z —' Whence {a+l>)x = — j^ — = ^—-^-. Dividing by (a+b) we have ^ = ^- SIMPLE EQUATIONS CONTAINING KADICALS. 2S, Many equations containing radicals which involve the unknown quantity, though not primarily appearing as simple equations, become so after being freed of such radicals. IKVOLVING RADICALS. 331 26. Prob. 2.— To free an equation of Radicals. Rule. — /. Transpose the terms so that the radical, if there is bat one, or the more complex radical, if there are several, shall constitute one member. II. Involve each member of the equalion to a power of the same degree as the radical. III. If a radical still remains repeat the process, being careful to Tceep the members in the most con- densed form and lowest terms. Demonstration.— That this process frees the equation of the radical which constitutes one of its members is evident from the fact that a radical quantity is involved to a power of the same degree as its indicated root by dropping the root sign. That the process does not destroy the equality of the members is evident from the fact that the like powers of equal quantities are equal. Both members are increased or decreased alike. EXAMPLES. Ex. 1. Find the value of x in V'^x-^-W = 12. Model Solution. Operation. V^x + 16 = 13. 4a? + 16 = 144. 4x = 128. a; = 33. Explanation. — I first square each member of the equation. The first member, y^^c+lQ, is squared by dropping its radical sign, since the square of a square root is the number itself The square of the second member, 12, is 144. This process is equivalent to multiplying the first member by \/ix + lQ and the second by 13, hence as \/4:x+16 is equal to 12, both members have been increased alike. [The equation being freed from radicals the explanation becomes the same as before.] 2. Solve V5x-\-6 = 4, and verify. 232 SIMPLE EQUATIONS. 3. Solye VlO^ + 3 = 7. Result, xz=i~ o 4. Solve V'2a;-f 3 + 4 = 7, and veiify. Suggestion. — First transpose the 4 and unite it with the 7; otherwise, squaring will not free the equation of radicals. 6. Solve 8+ v3a;+6 = 14, and verify. 6. Solve and verify 3a/2^T6 + 3 = 15. Result, X = 6. 7. Solve Vax + 2ah—a = h. Result, x = a 8. Solve Vx + x^—x^:^ = 0. Result, x = -> 9. Solve «ic-|-a\/2aa;+a?* = a5. Result, x 2(a + 5) Suggestion. — Before squaring put the equation in the form /\/'^ax + x = ft— 05. 10. Given ^/l'^-\-y—^/y = 2 to find the value of y. Result, y = 4:. Queries. — If each member is squared as the equation stands, will the equation be freed from radicals ? Is the first member a binomial or a trinomial ? What is its square ? Which will give the most simple result, to square it as it stands or to transpose one of the radicals ? Which one is it best to transpose? Will once squar- ing free it from radicals ? 11. Given Vx—16 = S—\/x to solve and verify. Suggestion. — Solve this and the five following like the preceding by squaring twice. INVOLVING RADICALS. ^33 12. Solve Va + a;+V«^ = V«^. Result, X = -5--—: • 13. Solve Vx—a = Vx—^Va, Result, X = — r-« lb 14. Solve V^xVx-\-2 = V5X + 2, Result, ^ = KK- 16. Solve a + x = '^ d'^x^/W-^:i^, Result, x = —J — • 16. Solve 2VF+x—V^+x = \/x. Result, x {h-al 2a— h Query. — Why is it best to transpose one of the terms of the first member to the second before squaring ? 17. Given x-\-^/a—x = --p=- to find x. Va—x Scholium I. — It is frequently best to clear the equation of frac- tions first, even when it involves radicals, especially if the denomi- nator is of such a form as will not make the equation complicated. In this case, clearing of fractions we have x\/a—x + a—x = a. Whence x\/a—x = x, or ^Ja—x — 1. Finally x = a— 1. 18. Solve — -- = yx ^ Suggestion. — The pupil should exercise his ingenuity in seeing in what difierent ways he can solve such examples, and notice the most elegant methods. For example, compare the following : 1st Method. — Clearing of fractions x^—ax^ — x. Dividing by a;, , 1 1— a 234 SIMPLE EQtJATlO:frS. %nd Method. — Multiplying each member by y'ccwe have x—ax=\ as before. ^rd Method. — Dividing numerator and denominator of the second member by ^/x^ we have zz = — ^ • Whence multiplying each ^yx ^x member by y^«, x—ax = 1, etc. Mh Method. — Divide the numerator of the first member by its denominator and ^Jx—a^Jx = ^^. Dividing each member by 'y/a;, 1— a = - , or « = 19. Solve — -= — = —^-z= Result, — Vax-^h Svax-^6b ^ Suggestion. — The more elegant solution of this is to reduce each member to a mixed number, obtaining 1 := — = 1 = . \/ax + l) d^ax+5b 2 7 Dropping the 1, and dividing by —5, — z= — = = • Clear- y^ax + b S^ax + 5b ing of fractions Q\/ax + l()b = l^ax + lb. .'. ^Jax = Zb, etc. In a similar manner solve the two following : 20. Solve -—-^- = -^^—-L — Result, x = 4:, Vx-{-4: Vx-\-6 21. Solve -^^-—^ — = —=^ — Result, x = | -, )• Vx + b Vx + Sb ^^-^^ 22. Solve -^^- r= 1 4- ^^f~^ . Result, a; = 3. \/dx-\-l ^ Suggestion. — =iz — = -v/Saj— 1. ^%x+l 23. Solve -^^P^ = Wx-^+ ^• Va: + 2 ^ INVOLVING RADICALS. 2^5 Suggestion. — In this and the following use the same expedient as in the 22nd. Thus ^x—2=5^x—S + §\/x^ or 5^^x=Q, ox \/x = 13 . ^1^ 11' '•'^"121* 24. Solve -— — = 5— + Wa. VX H- V « ^ Suggestion. — We have v^— V<* = i V*+lf V^» or \^/x — 2f /y/a. .-. X = 16a. 25. Solve — = = c + -^^ 1/ c2 \2 Result, X = -(b-\ ) • a\ c—lj 26. Solve —=r^r =Vm. Result, x = - — ya-{-x—Va—x -• + m Suggestion. — Rationalize the denominator (224), and after reducing ^/a^—x* = x^/m—a. 27. Solve ^« ±^ = ,. Result, x = "-^l 28. Solve — ^ = 9. Result, x = -- ^4x-\-l-2Vx 9 29. Solve Vx+V^- Vx^Vx = l\ /— Va Suggestion. — Multiply each member by Vx-\-\/x, and after reduction y^x'—x = x—^\/x. Squaring, transposing, and com- 25 16' /-I 5 25 bmrng, ya?" = ^^- .'. x = 30. Solve V^/x-^^ - V^/is = V^l Result, a; = 9. 236 SIMPLE EQtJATTOKS. 31. Solve a/4^ + J'^ = \'/^- V a+x V a—x V « — ^ Result X = -\ - h—c 32. Solve - + - = V - 4- \/-^^ + -4- ^ = ^«. 33. Solve .15a: + 1.575— .875a; = .0625ic. x = 2, 1 Q/y. AK 34. Solve 1.2a:- , — = Ax+%.^. x =: 20. .5 lyo^ AK 35. Solve 4.8a;- ' \ — = 1.6a;+8.9. x = 6, .0 36. Solve 2.3a;- ^=^? = 3.2- |. 37. Solvea;-.24= ^^ • -« a; = 4.259 + . ./ii. SUMMARY OP PRACTICAL SUGGESTIONS. 27. — In attempting to solve a simple equation which does not contain radicals, consider, 1. Whether it is best to clear of fractions first. 2. Look out for compound negative terms. 3. If the numerators are polynomials and the denomina- tors monomials, it is often better to separate the fractions into parts. {Bx. 22, p. 224 is an illustration.) 4. It is often expedient, when some of the denominators are monomial or simple, and others polynomial or more complex, to clear of the most simple first, and after each step see that by transposition, uniting terms, etc., the equation is kept in as simple a form as possible. (See Fxs. 40 to 44.) 5. It is sometimes best to transpose and unite some of the terms before clearing of fractions. (See Exs. 46 to 54.) APPLICATIONS. 237 6. Be constantly on the lookout for a factor which can be canceled from each member of the equation {Ex, 55), or for terms which destroy each other (Exs. 60 to %Q), 7. It sometimes happens that by reducing fractions to mixed numbers the terms will unite or destroy each other, especially when there are several polynomial denominators. (See Ex. 59.) 28. When the Equation contains Radicals, con- sider, 1. If there is but o??e radical, by causing it to constitute one member and the rational terms the other, the equation can be freed by involving each member to the power denoted by the index of the radical. (See Exs. 1 to 9.) 2. If there are two radicals and other terms, make the more complex radical constitute one member, alone, before squaring. Such cases usually require two involutions. (See Exs. 10 to 16.) 3. If there is a radical denominator and radicals of a similar form in the numerators or constituting other terms, it may be best to clear of fractions first, either in whole or part. {^qqExs. 17, 18, 29.) 4. Frequently a fraction may be reduced to a whole or mixed number with advantage. (See Exs. 19 to 25.) 6. It is sometimes best to rationalize a radical denomina- tor. (See Exs. 26 to 28.) 6. Always watch for an opportunity to cancel a factor, or drop terms which destroy each other. APPLICATIONS. 29. According to the definition (3), Algebra treats of, Ist, The nature and properties of the Equation ; and 2nd, the method of using it as an instrument for mathematical investigation. The Simple Equation has been so thoroughly discussed 238 SIMPLE EQUATIONS. , * in the preceding part of the section, that the pupil will now be able to use it in solving problems. 30. The Algebraic Solution of a problem consists of two parts : 1st. The Statement, which consists in expressing by one or more equations the conditions of the problem. 2nd. The Solution of these equations so as to find the values of the unknown quantities in known ones. This process has been explained, in the case of Simple Equa- tions, in the preceding articles. 31. The Statement of a problem requires some knowl- edge of the subject about which the question is asked. Often it requires a great deal of this kind of knowledge in order to '^ state a problem." This is not Algebra; but it is knowledge which it is more or less important to have according to the nature of the subject. 32. Directions to guide the student in the Statement of ProUems : 1st. Study the meaning of the problem, so that, if you had the answer given, you could iwote it, noticing carefully just what opera- tions you would have to perform upon the answer in proving. This is called, Discovering the relations l)etween the quantities involved. 2nd. Represent the unknown (required) quantities (the answer) by some one or more of the final letters of the alphabet, as x, y, 0, or wj, and the known quantities by the other letters, or numbers, given in the problem. 3rd. Lastly, by combining the quantities involved, toth known and imJcnown, according to the conditions given in the problem (as you would to prove it, if the answer were known) express these relations in the form of an equation. PROBLEMS. Ex. 1. What number is that to the double of which if 18 be added, the sum is 82 ? APPLICATIONS. Ji39 Model Solution. Statement. — Let x represent the unknown number sought. Then the problem says that double this number, that is 2x, with 18 added, that is 2ic + 18, is 82. Hence 2a!+ 18 = 82 is the statement. Resolution of the Equation.— [With this the pupil is familiar.] Solving 2a;+18 = 82, we find x = 32. 32 is, therefore, the number sought. Verlflcatlon. 2 x 32 + 18 = 82. [Note. — In this example there are three members involved, the 18, 82, and the one to be found, which we call x. The relations between these numbers are explicitly stated in the problem, and the statement is easily made. This is not always so. These relations are often the most difficult thing to discover, and their discovery requires a knowledge of other subjects than Algebra. Suppose the problem to be : Given three masses of metal of equal weight, one of pure gold, one of pure silver, and one part gold and part silver. When they are immersed in water, the gold displaces 5 ounces, the silver 9 ounces, and the compound 6 ounces. What part of the last was gold and what part silver? Now in this problem the relations between the quantities are not explicitly stated. Yet by a knowl- edge of Natural Philosophy and a little of Memuration, they can be found out, and the statement of the problem made in an equation.] 2. What number is that, to the double of which if 44 be added, the sum is 4 times the required number ? Suggestion. — Suppose I guess that the number sought is 30, how will you tell whether I am right or not? You say: "Double 30 and add 44 to it, and, if you are right, the sum will equal 4 times 30." But 2x30 + 44 does not =4x30; so I am wrong. Now, call the number sought x, and use it as 30 was used in proving that it is not the answer, and the statement, 2x-\-A4 = 4x is obtained. Whence x = 22. Verify it. 3. What number is that, the double of which exceeds its half by 6 ? Ans., 4. 240 SIMPLE EQL'ATIOKS. 4. The property of two persons is $16000, and one owns three times as much as the other. How much has each ? Statement. Let x = the less amount. Then 3a; = the greater amount. And 3a? + a; = 16000. .'. x = 4000 and 3a; = 13000. 5. A man being asked his age replied : " If to my age you add its half, and third, and then deduct 10, the result is 100." What was his age ? Ans., 60. 6. After paying \ of a bill and \ of it, $92 still remained due. What was the bill at first ? Statement. — Let x = the amount of the bill ; then \x and \x had been paid. Taking these amounts from the bill we have x—\x—\x. But this, by the problem, was $92. Hence x—\x—\x = 93. .'. X = 140. 7. The sum of 2 numbers is 180 and the difference 10. What are the numbers ? Statement. — Let x = the less number : then a; + 10 is the greater, and a; + a; + 10 — 180. .*. x = 85, and a; + 10 = 95, and the two num- bers are 86 and 95. 8. The sum of two numbers is 5760, and their difference is ^ the greater. What are they ? Statement. — Let x = the greater; then 5760— a; is the less, and jT— (5760— a;) = ~. .'. x = 3456, and the less is 2304. 3 9. A man divided 80 cents among four beggars ; to the first two he gave equal amounts, to the third twice as much as to one of these, and to the fourth twice as much as to the third. How much did he give to each ? The equation is a;+a: + 2a; + 4a; = 80. Whence he gave the first and second 10c. each, the third 30c., and the fourth 40c. APPLICATIONS. 241 10. A, B, C, and D, invest $4755 in u speculation. B furnishes 3 timps as much as A, C as much us A and B together, and D as much as C and B. How much does each invest ? Ani<., A, *317 ; B, $951 ; C, $1268 ; and D, $2219. THE SAME PROBLEMS WITH THE LITERAL NOTATION. 11. What number is that, to n times which if m be added, the sum is s ? The equation is nx-^m = s, Ans., Scholium. — To make this conform to Prob. 1, we call « = 83, HO 1 .^ '*—'" 82—18 _„ r» .^x. 8—^ m = 18, and w = 2. .'. = — - — = 32. But the answer ' 71 2 n applies just as well to any other problem of the kind, no matter what numbers are involved. Thus, let the problem be, — What number is that to 5 times which if 20 be added, the sum is 100? -^o. ««. 1 ► «r, s—m 100—20 Now s = 100, m = 20, and n = 5. Whence = — = 16. n 5 We. therefore^ nee that is a general answer to all such problems. n 12. What number is that, to a times which if b be added, the sum is c times the number ? b Ans., Equation, ax-{-b = ex. .-. The number is c—a Queries. — How is this adapted to Prob. 2 ? What other problems can you state to which this value of x affords an answer ? 13. What number is that, a times which exceeds t times itbyw? Equation, ax — t = m. .-. x = -7 — - is the number. Queries. —How is this adapted to Prob. 3 ? What other problems 242 SIMPLE EQUATIONS. can you state to which -r — r affords an answer ? Repeat the same inquiries after each of the seven following exanfples. 14. The property of two persons is $m, and one owns n times as much as the other. How much has each? . m , 7?in A71S., --— and t—-- • l-f-/i 1-fn 15. What number is that to which if —th and -th of m n itself be added, and then a deducted, the result is J ? {a^h)mn Ans.f mn-^in-\-n 16. After paying — th and -th of a bill, a remained du^. What was the bill ? cnnn Ans., mn—n—m 17. The sum of two numbers is s, and their difference d. What are the numbers ? . s + d J s^d Ans., -ir- and — ;r — 33 • Cor. — Observe that the solution of this problem proves the very useful theorem : Half the sum plus half the difference of two quantities is the greater of the quantities, and half the sum minus half the dif- ference is the less. Thus, the sum of two numbers is 20 and their difference 13. What are the numbers ? h + o = 10 4-0=16, the greater; and ^^x = 10—6 = 4, the less, APPLICATIONS. 243 18. The sum of two numbers is s and their difference is — th of the greater. What are the numbers. m ° _ .. / \ ^ 4 "^* s{m—l) 19. A man divided ^i^^^ among 4 beggars; to the first two he gave equal amounts, to the third m times as much as to each of these, and to the fourth in times as much as to the third. How much did he give to each ? A ns., To each of the first two $- ^ , to the third am J i ii J. ii A «wi2 ,, and to the fourth 1^ 20. A, B, C, and D invest $s in a speculation. B fur- nishes m times as much as A; C as much as A and B together, and D as much as C and B together. How much does each invest ? A . . . , s „ fns ^ s{l-\-m) Ans^ Afurmshes 3^-^; B, 3^^-^; C, -^-^^■, 3 + 4m 21. At a certain election 943 men voted, and the candi- date chosen had a majority of 65 votes. How many voted for each? Ans., 439 and 504. 22. A farmer has two flocks of sheep, each containing the same number. From one he sells 39, and from the other 93, and then finds just twice as many in one flock as in the other. How many did each flock originally contain? Ans., 147. 23. A person spends J of his income in board and lodg- ing, -J in clothing, and ^ in charity, and has $318 left. What is his income ? Ans.j $720. 344 SIMPLE EQUATIONS. 24. From a bin of wheat J was taken, and 20' bushels were added. After this ^ of what was then in the bin was sold, and J as much as then remained + 30 bushels was put in, when it was found that the bin contained just ^ as much as at first. How much did it contain at first ? Suggestion. — Calling x the amount in the bin at first, after taking out I, there remains ^x. To this add 20 bushels and there is in the bin|x + 30. After selling i of this f remains, or |(|^a; + 20). J of this is Kf^ + 20). .-. |(^a;+20)+Kf'»+30) + 30 = ^aj. Whence X = 560, the answer. 25. A person has a hours at his disposal; how far may he ride iu a coach which travels b miles per hour, and yet have time to return on foot walking at c miles per hour ? abc Ans., J—-' 26. After paying -th of my money^ and then -th of what remained, I had %a left. How much had I at first ? Equation, x — x—{x a:)- = a, amn Ans.f 27. A boy, being asked his age, replied, 11 years are 7 years more than f of my age. How old was he ? Statement.— Let x = his age. Then |a;+7 = 11. .-. « = 10. 28. A boy, being asked how many sheep his father had, replied, that 40 were 5 less than | of his father's flock. How many sheep had his father ? Arts., 60. 29. If A can perform a piece of work in 10 days, and B in 8 days, in what time will they perform it together ? APPLICATIONS. Hr) Statement. —Let x = the number ot ICATI0NS. 261 mau beiiior engaged to work a certain number of days, continued in the service ^ of the whole time, but during this time was sick, and lost a number of days equal to | of the whole time, when finally he- had to break the engagement, having actually worked but G days. What was the whole period engaged for ?" 2. Form problems of which a; + 10a; =66 is the state- ment. 3. Give similar translations of each of the following eqr.:> X X ^ X X ^ ^ X ^ ^ ,, tions: --[-- + 2 = 2:; - -f - = 1 ; 2x---— = 7; x + ^x -760-r 600 = 2000; x—^^4^ = 8. b X X 4. Translate the equations 3a; -f- :c = 16000 ; ^r-f ^ + q -10 = 100. 5. Enunciate problems which will give rise to the follow- X X ing equations : x—-^ — ^=92; x-\-x-{-2x-[-4.x=SO. o 7 [Note. — The following examples give rise to equations found under \vi, 26.] 6. What number is that which gives the same quotient when 28 is added to its square root, and the sum divided by its square root +4, as when 38 is added to the same root, and this sum divided by the square root of the num- ber + 6? 7. A man has a certain number of square rods of land lying in a square ; if 12 rods be added, the whole being kept in the form of a square, his plat is increased by 2 rods on a side. How much land has he ? 8. What number is that to which if 12 be added its square root is increased by 2 ? 252 SIMPLE EQUATIONS. 9. If 4 times a certain uuinber be increased by 1, the square root of this sum + twice the square root of the num- ber itself, divided by the difference of the same quantities is 9. What is the number ? 10. There is a certain number to which if its square root be added, and the square root of the remainder be taken, and again the square root be subtracted from the number, and the square root of this remainder be taken, the differ- ence between the results is 1^ times the square root of the quotient of the number divided by the number 4- its square root. What is the number ? 11. What number is that from which if 32 be subtracted, the square root of the difference is equal to the square root of the number —^ the square root of 32 ? 12. What number is that from which if a be subtracted, the square root of the difference is equal to the square root of the number —^ the square root of a ? 25a Ans,, — . 13. What number is that whose square root H-2ff, divided by its square root +&, equals its square root -t-4«, divided by its square root -f-3J ? WITH TWO UNKNOWN QUANTITIES. ^fel€TS0M II. DEFINITIONS. SO, The preceding problems have all been solved by a single equation containing only one unknown quantity. In some of them several quantities have been sought, it is true, but we have managed to represent these quantities by the use of a single unknown quantity, x. There are, however, many problems in which this is not practicable. In such problems there are ttvo or more quantities sought, and the conditions are such as to give rise to two or more equations. Illustration. — To make this latter statement clear, consider the following problem: A says to B, "If | of my age were added to | of yours, the sum would be 19J years." But, says B to A, " It" f ot mine were subtracted from ^ of yours, the remainder would be 18f years." Required their ages. Here are two distinct quantities sought;, viz., A's age and B's age. Suppose we represent A's age by X, and B's by y. Now notice that there are also two sets of con- ditions. 1st, the statement which A makes to B ; and, 2i^d, the statement which B makes to A. According to the 1st, w^e have the equation ^x + fy — 19| ; and according to the 2nd, |«— |y = 18f. tV7. Independent Equations are such as express different conditions, and neither can be reduced to the other. 38, Simultaneous Equations are those which express different conditions of the same problem, and consequently the letters representing the unknown quantities signify the 254 SIMPLE EQUATIONS. same things in each. Each equation of a set of simultane- ous equations is, therefore, satisfied by tlie same values of the unknown quantities. Illustration. — Thus in the example above the two equations ix -f|?/ = 19^ and Ix—^y = 18] are independent equations, since they express different conditions^ and neither can be produced from the other. But, since these conditions are of the same problem, so that X in the first equation means the same as x in the second, and y in the first, the same as y in the second, they are simultaneous equations. It is evident that the true values of x and y must satisfy^ or ve7'ify, both equations. If, however, we were to write one equation from one problem, and one from anotlier, while they wotild be independent, they would not be simultaneous ; x and y would not mean the same things in the first equation as in the second. In fact, the equations would be so independent, that they would have nothing to do with each other. 39* Elimination is the process of producing from a given set of simultaneous equations containing two or more unknown quantities, a new set of equations in which one, at least, of the unknown quantities shall not appear. The quantity thus disappearing is said to be eliminated. (The word literally means putting out of doors. We use it as meaning causing to disappear.) 40, There are Three Methods of Elimination in most common use : viz., by Comparison, by Svbstltution, and 1^ Addition or SuUraction. There is also a very elegant method by Undetermined Multipliers, which is worthy of more attention than it generally receives, but which will be reserved for the advanced course. Scholium. — Any one of these methods will solve all problems; but some problems are more readily worked by one method than by another, while it is often convenient to use several of the methods in the same problem, especially when there are more than two un- known quantities. WITH TWO UNKNOWN QUANTITIES. 255 ELIMINATION BY COMPARISON. 41, Prob. 1. — Having given two independent, simul- taneous, simple equations between two unknown quantities, to deduce therefrom by comparison a new equation contain- ing only one of the unicnown quantities. Rule. — /. Find expressions for the value of the same utiknown quantity from each equation, in terms of the other unknown quantity and known quantities. II. Place these two values equal to each other, and the result will he the equation sought. Demonstration. — The first operations being performed according to the rules for simple equations with one unknown quantity, need no further demonstration. 2nd. Having formed expressions for the value of the same unknown quantity in both equations, since the equations are simultaneous this unknown quantity means the same thing in the two equations, and hence the two expressions for its value are equal, q. e. d. Scholium. — The resulting equation can be solved by the rules already given. EXAMPLES OF INDEPENDENT SLMULTANEOUS EQUATIONS. Ex. 1. Given ^x-\-y = 34 and 4y + a; = 16; to find x and y and verify the values. Model Solution. Operation. (1) 4.»4-y = 34 = 16 = 16-4y .-. y = 4 (2) iy+x = 16 .-. X = 16-4^ 84-y 4 ix + 2 = 34 .. x = S Explanation. — Transposing y and dividing by 4 I have from the 1st equation, x = — ~ . Transposing 4y in the 2nd equation I 4 have X = 16— 4y. Now, since these equations are assumed to be simultaneous, x means the same thing in both ; and since things that 266 SIMPLE EQUATIONS. 34— v are equal to the same thing are equal to each other, — -^ = 16 4 — 4y. From this equation I find y = 2. Finally, since y is found to be 2, putting 2 lor y in the first equation, I have 4a;+2 = 34 : whence x = 8. Verification.— Substituting for x, 8, and for y, 2, in the 1st equa- tion I have 32 + 2 = 34 ; and in the 2nd, 8 + 8 = 16, we see that both equations are satisfied. 2. Given 5x-\-4:y = 58, and 'dx-\-7y — G7, to lind x and y, eliminating by comparison. Verify the results. Results, \ ~ J \y = 7. 3. Given l\x-{-^y = 100 and 4:x—7y = 4, to find x and y, eliminating by comparison. Verify the results. Results, < ~~ / ly = 4:. A c^ 1 • o -^ + 3 ^ 3x—2y , 4. Same as above, given 8 — = 7 u— ^ and Suggestion. — Observe the effect of the — sign before the com- pound quantities. The value of y from the 1st hy = — - — , and from the 2nd, y = 160-30a; „ 5 + 7cc 160-30x , . Hence -g— = , and a;=6, and y = 5. 5. Given X + ay — h dx + y z=z c 6. Given ax + | = m ex - y = n 7. Given ax — by := m ex -\- dy •= n 8. Given ax-^-hy — m, and cx-\-dy = n, to find x and y, eliminating by comparison. Suggestion. — After having found the value of a; or y by compari- son, find the value of the other in the same way. WITH TWO UNJLNOWK QUANTITIES. 267 ELIMINATION BY SUBSTITUTION. 4'i. Prob. 2.— Having given two independent, simulta- neous, simple equations, between two unknown quantities, to deduce therefrom by substitution a single equation with but one of the unknown quantities. Rule. — /. Find frovi one of the equations the value of the unlcnowrh quantity to he eliminated, in terms of the other unknown quantity and knoivn quantities. II. Substitute this value for the same unknown quan- tity in the other equation. Demonstration.- -The first process consists in the solution of a simple equation, and is demonstrated in Art. 20. The second process is self-evident, since, the equations being simultaneous, the lettei*s mean the same thing in both, and it does not destroy the equality of the members to replace any quantity by its equal, q. e d. EXAMPLES. Ex. 1. Given the independent, simultaneous equations * ■^L3' _ ^iry = 8, and ^ + ?^i^ = 11, to find x and y, 2 6 o 4 eliminating by substitution. Verify the results. Model Solution. Operation. (1) ^ - fzJ^ = 8 3x4-3y— 2x + 2y = 48 x-\-5y = 48 X = 48— 6y ^^^ "a" "^ "4 ~ ^^ (8) ^8-5y+y ^ 48-5y-y ^ ^^ 3 4 48-4y U-3y _ 8 2 ~ 96-8y + 72-9;y = 66 -I7y = -103 y = 6 258 SIMPLE EQUATIONS. (4) ^«-^-i^ = 8 .-. a;=18 Explanation. — Taking equation (1) I clear it of fractions and solve it for cc, finding that x = 48— 5y. Now if I take the 2d equation and substitute this value for «, I shall have a simple equation with only the unknown quantity y in it. This substitu- tion does not destroy the equality, since the equations are simul- taneous (3§). Making the substitution I have — ^-7^ — - o H j^ — - = 11. Reducing this equation by the method for simple equations with one unknown quantity, I find y = G. Finally, resuming (1) I substitute this value for y, which evi- x-uQ X 6 dently does not destroy the equation, and have — — - = 8. Solving (4), which is now a simple equation with one unknown quantity, I find x = 18. (Instead of taking (1) in its first form it would be better, because so much shorter, to take its reduced form, X = 48— 5y. .-. X = 18.) [Note. — The student should keep a sharp lookout for oppor- tunities to efl^ect such reductions of terms as are made above in the equations following (3) and (4). In the latter the process consists in observing that — — - is - + 3, and — is — - -f- 2, hence the 2 2 o o first member becomes ^ + ^ " 5 + ^> ^^^ transposing the 3 and 2, 2 o X X we have - — - = 3, all of which can readily be effected mentally.] 2 o 2. Given 3x—2y = 1 and 3y—4:X = 1, to solve as above. Result, X — 6 and y = 7. 1 4- Ax Suggestion. — From the second, y = — - — , hence the first becomes 3i» — = 1. .-, a; = 5. Taking the reduced form WITH TWO UNKNOWN QUANTITIES. 259 of the second, y = - ^— , and putting 5 tor x,y = 7. Verify both equations. 3. Given ^-^+3 = ^±l,ancl S-^-^^1 = | + |. Eesul/j a; = 12 and y = 6. 4. Given | + | = l,aud | + | = 1. Verify. Result, X = —6 and y = 12. K /-,. a b , c <^ 5. Given - -\ — = m.and — I- - = n. X y ' X y Suggestion. — If we clear these equations of Tractions they will become quite complex. But multiplying the first by c and the , , . CLC J)c ^ ac ad -r^ ^-^ second by a, we have — H — = cm, and — | = an. From the X y X y former — = C7n , which substituted in the latter for — pdves X y X ^ he ad ad— be ctd—lc ,_^. em 4- — = an, or = an — cm. .-. y = . This y y y an-cm value of y may now be substituted in the first. To get the value of a? from such an equation as - = — -^ — r--, simply divide 1 by each member, i. c, take their reciprocals. 6. Given — h - = 1, and - H — = 1. X y X y Result, X = m-{-n, and y = m-{-n. 7. 6mn- + |=l, and 3^^ + 1 = 3. Result, X = 3a, and y = —25. 8. Given i - - = 5, and - + i = 7. X y X y 9. Given -^ = —^ , and x -\- iy = 3. 260 SIMPLE EQUATIONS. ELIMINATION BY ADDITION OR SUBTRACTION.- 43. Prob. 3.— Having given two independent, simulta- neous, simple equations between two unknown quantities, to deduce therefrom by Addition or Subtraction a single equa- tion with but one unknown quantity. Rul9. — /. Reduce the equations to the forms ax + by = m, and cx-fdy = n. II. If the coefficients of the quantity to he eliminated are not alike in both equations, make them so by finding their L. C. M. and then multiplying the mem- bers of each equation by this L. C. M. exclusive of the factor ivhich the term to be eliminated already con- tains. III. If the sigjis of the terms containing the quantity to be elUninated are alike in both equations, subtract one equation from the other, member by member. If these signs are unlike, add the equations. Demonstration. — The first operations are performed according to the rules already given for clearing of fractions, transposition, and uniting terras, and hence do not vitiate the equation?. The object of this reduction is to make the two subsequent steps practicable. The second step does not vitiate the equations, since in the case of either equation, both its members are multiplied by the same number. The 3rd step eliminates the unknown quantity, since,' as the terms containing the quantity to be eliminated have the same nurr^erical value, if they have the sarae sign, by suUrading the equations one will destroy the other, and if they have different signs, by adding the equations they will destroy each other. The result is a true equation, since, If equals (the two members of one equation) are added to equals (the two members of the other equation), the sums are equal. Thus we have a new equation with but one unknown quantity. (^. e. d. WITH TWO UNKNOWN QUANTITIEa 261 EXAMPLES. Ex. 1. Given 5x-\-6y = 28, and .r + 4^ = 14, to eliminate by addition or subtraction and find the values of x and y» Verify the results. Model Solution. Operation. (1) 5x + Qy = 2S (3) X + 4y = 14 (8) lOa; + 12y = 56 (4) dx + 12y = 42 (5) 7x = 14, .-. x = 2. (6) 2 + 4y = 14 4y = 12, .-. y = 3. Explanation. — The equations being in the required form need no reduction. To eHuiinate y I make its coefficients alike in both equations by multiplying the members of (1) by 2, and of (2) by 3, thus obtain- ing (3) and (4). This does not destroy the equations by (Ax. 2). Then subtracting the members of (4) from the corresponding members of (3) I have (5), which is a true equation since the mem- bers of (3) have been increased equally (Ax. 2). From (5) I have a- = 2. Finally, substituting 2 for a; in (2) I have (6), which is a true equation, since the value of the members of (2) is not altered by this substitution (Ax. 1). From (6) I find by previous methods y = 3. Veriflcation.— Substituting 2 for x, and 3 for y in both of the original equations, I have (1) 10 + 18 = 28 (2) 2 + 12 = 14 Whence I see that both equations are satisfied for x = 2, and J) =3. 2. Given 77a;— 12y = 289, and 55a; + 27?^ = 491, to find jc and y. Model Solution. Operation. (1) 77a; - 12y = 289 (2) 55a; + 27y = 491 693a; - lOSy = 2601 220a; + 108y = 1964 220a; + 108y = 1964 (3) 913a! = 4565 Ji62 simplp: equations, (3) 913a; = 4565 « = 5 (4) 275 4- 27y = 491 27y = 218 Explanation.— Since these equations are of tbe required form, I have only to make the numerical values of the terms to be elimi- nated alike. I will eliminate y, since its coefficients are smaller than those of a; and the process will not involve as large numbers. The L. C. M. of 13 and 27 is 108, hence I multiply (1) by 9, obtain- ing G9Sx-108y = 2601, and (2) by 4, obtaining 320a;+1082/:=3l964. This process does not vitiate the equations, since. Equals multiplied by the same number give equal products. I now observe that the signs of lOSy in the two equations are different, and consequently that by adding the corresponding members of the equations these terms will destroy each other and give an equation in x only. Add- ing gives a true equation, since. Equals added to equals give equal sums. I therefore have 913a; = 4565, or a; = 5. Substituting this result in (2), I have 275 + 27y = 491, whence y = 8. 44:. Scholium. — It is usually expedient, in examples involving two unknown quantities, to find the value of the second by substi- tution; but this is by no means always so. The pupil should perform the examples in several ways, if he can discern no choice of ways at first, and then compare the methods with reference to practicability. o ^. . 5a;-|-2y 3y— 12-f Sr?; , 15-f 2a;— 4y 3. Given y+ ^ ^ - ^ — ^-I— - = 4 -I-^ ^ , and ^ -}- a; = 2^ —^ , to find x and y. Suggestion.— Cleared of fractions and reduced to their proper form these equations become 6y -I- a; = 34 45a' — 31y = 25 Whence a? = 4, and y = 5. %' WITH TWO UNKNOWN QUANTITIES. 263 4. Given 1 H ^^ ^ = 10 j^ ^ and = 5a; ^-^ , to find the values of x and y. Results, ic = 3 and y = 7. 5. Given ^^i^ = ^^ and 8a;~5«/ = 1, to find the values of a; and y. Results, x = 7 and y = l\, 4«7. Scholium. — In practice it often requires considerable dis- crimination to determine which of the methods of elimination to employ. But, as any one method will solve all cases, the pupil need not hesitate too long in attempting to select the best one. If he sees any reason why one method will be better than another in the given case, he will of course use it; but, if no such ground for choice is apparent, it will often be well to try more than one method, and see if one is any more expeditious than another. EXAMPLES FOR GENERAL PRACTICE. Ex. 1. Given dx—by = 13 and 2x-\-7y = 81, to find x and y. Results y a* = 16 and y = l, X 11 X 1J 2. Given - -f- j = 9 and j -f- ^ = 7, to find x and y. Results, a" = 12 and y = 20. „ ^. 4:X—3y—7 3a; 2y 5 , ?/— 1 x 3v 3. Given Jf— = _ _ ^^ _ and-^+ - -^ - 1 = ^ + I + 10' *^ ^^^ ^ ^^^ ^' Results, a; = 3 and y = 2. *• «'- 1 - -tf = 1 - ^' + -i^ -^^-^%' to find X and y. Suggestion. — The first equation reduces at once to 7x = 7 + lly. In this case the pupil will see that the three methods of eliminating X are almost identical. Comparing the values of 7.r, we have \2y = 7 + lly: or we may call it suistituting the value of 72" found in 264 SIMPLE EQUATIONS the second equation ; i. e., 12y, for 7x in the first. Subtracting the second from the first we should have = 7—y. 6. Given h — = 10 and = 2, to find z and y, X y X y ^ 10 3 Suggestion. — The second becomes by dividing by 2, — = 1, 70 21 and by multiplying by 7, = 7. Now adding this to the X y 85 first we have — = 17, or a; = 5 and hence y = Z. 6. Given - -\ — = m and ■=. n. to find x and y. X y X y ^ ^ ^, ad Id , , he M , axl + bc , Suggestion. 1 = dm and = hi, •.-. = dm X y ^y X ■\-hn and x = -_ — . Instead of substitutinaj this value of x, it dm + on ^ will be less work to eliminate x from the two given equations as we T T rm- ^ ac le ^ ac ad i . did y. Thus we have 1 — — cm and = an, and sub- X y X y ic+ad bc+ad tractmg, = cm— an, .-. y = y cm— an 7. Given ix-^-^y = 14 and ix-{-^y = 11, to find x and y, and verify. Suggestion. — Do not clear of fractions. 8. Given x — -^ — n~^ ~ W~^ — 6 w— 5 lla; + 152 Sw + l ^ « ^ i a -^ — ^— — = -r ^ — , to find X and y, and veniy. 4 12 2 ' ^' -^ - ^. - 16 + 60a; lexy-W^ ^ ^ . « . o 9. Given 8a; — i— -- = --^— , and 2 + 62^ + 9a; Sy — 1 o-\-2y 27a;2-12^2 + 38 ,„,,,, . , = — ; TT^—^ — , to find the values of x and y. Sx—2y-\-l ^ Suggestion. — Multiply the 1st equation by 5 + 2y, and reduce before multiplying by 3y— 1. Clear the 2nd of fractions. Whence f—UOx = 187, and 15a; + 2y = 86. .-. x = 2,y = 3. WITH TWO UNKNOWN QUANTITIES. 266 10. Given 3r -^ Gv + 1 = — ? — r it-^ , and ^ 2a:— 4^-h3 ' 151— IGa: 9a;y— 110 , ^ , ^, , 3x z , — = —^ r— > to find the values of x and y. 4y— 1 3y— 4 ^ Result, X =z9, and ^ = 2. 11. Given 16a; 4- 6 v — 1 = ^ — — ~ , and ^ 8a; — 3y 4- 2 -^j ~ — ^r- = 5 — — -, to find the values of x 2a; + 2^ + 3 dx-}-2y—l and y. Result, x = 6. and y = 5. 12. Given (x-\-5) {y + H) = (^+1) (y— 9) + 112, and 2a;-f 10 = 3y + l, to find the values of x and y. Result, X = S, and y = 5. 13. Given bcx = cy-2b, and l^y + -i^^ = ?^ + c^a;, to find the values of x and y. Result, a; = 7-, y = • V 2 Suggestion.— From the 1st equation « = | . Substitutinpr c this in the 2nd, 6'y-)--^— — -'' = — + -JL ^ 2e\ Whence, trans- OC CO posing and uniting, y = -^ ^ + , — ^ or — ^ — y = X — :r-~ , and y — . Substituting this value of y in Co c the Ist equation and reducing, we find ^= j-- These equations can be solved by a variety of methods, but the pupil should con- stantly exercise his inventive genius to discover the most expedi- tious and elegant solutions. 14. Given 2a;-f-0.4y = 1.2, and 3.4^-0.021/ = 0.01, to find the values of x and y. Result, x = .02, y = 2.9. 12 •M6 SIMPLE EQUATIONS. Mc /.• \x-\-y = 18.73 ^^•^^^^^ I 0.56^+13.421^ = 763.4 APPLICATIONS. Ex. 1. There are two numbers, such, that three times the greater added to one-third the less is 36 ; and if twice the greater be subtracted from 6 times the less, and the remain- der divided by 8, the quotient will be 4. What are the numbers? Model Solution. Operation. Let x = the greater number, and y = the less number. Then (1) 3x + ^y = 36 (2) «X.-J? = 4 (3) Qy + 54:X = 648 (4) 6y - 2x z= 32 56a; = 616 x = n (5) 6y - 22 = 32 y = 9 Explanation. — As there are two unlcnown qunntities involved in this example, /. e., the two numbers sought, I let x represent the greater and y the less. There are also two sets of conditi(ms stated in the problem : 1st, 3 times the greater added to ^ the less is 36. This, according to the notation, is Sx + ^y = 36, which is the first equation. The 2d set of conditions is, that twice the greater is to he subtracted from 6 times the less, which is Qy—2x, and this differ- Qy 2X ence divided by 8, i. e., — - — . This quotient is equal to 4, Hence the second equation, -~ — = 4. WITH TWO rNKNOW>r QUANTITIES. 26f 2. Find two numbers, such, that if the first be increased by a, it will be m times the second ; and if the second be increased by &, it will be n times the first ? J, -. a-\-bm , b-\-an Result y T, and mn — 1 mn — 1 3. What two numbers are those, to ^ of the sum of which if I add 13, the result will be 17 ; and if from \ their differ- ence I subtract 1, the remainder will be 2 ? Verify. Ans.y 9 and 8. [Note. — In verifying the results in such examples as these, no attention s^liould be paid to the equations ; but the results should be tested directly by the statement. Thus, in this example, \ of the sum of 9 and 3 is 4. Adding 13 the result is, as the example requires, 17. Again ^ the difference of 9 and 3 is 3. Subtracting 1, the remainder is 2, as required.] 4. What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes f ; but the denominator being doubled, and the numerator increased by 2, the value becomes \ ? Ans,, |. Queries and Suggestions. — 'How many sets of conditions in this problem ? What are they ? How many unknown (required) quantities ? What are they ? Tliere must always be as many of one as of the other. The unknown (required) quantities here are the numerator and the denominator of the fraction. If these are called respectively x and y the fraction is -. 2x Now, by the first set of conditions, — ;= — |, and, by the second set, x+2 _ 2y *' 5. What fraction is that which becomes f when its nu- merator is increased by 6, and J when its denominator is diminished by 2 ? Ans., ^. 'MjH simple equations. 6. If 1 be added to the numerator of a certain fraction, its value is ^ ; but if 1 be added to its denominator, its value is }. What is the fraction ? Verify. Ans,, ^. 7. There is a certain number, to the sum of whose digits if you add 7, the result will be three times the left hand digit ; and if from the number itself you subtract 18, the digits will have changed places. What is the number? Verify. A?is., 53. Suggestion.— The two numbers sought are the two digits. Hence let y = the units digit, and x = the tens digit. The number then is 10a;+y. Just as when 6 is the units digit of a number and 5 the tens, the number is 10 x 5 + 6. Of course the number would not be represented by xy, for this would indicate the product of the digits. (See Part I., 30, Second Law, Scholium 1st.) The first conditions give 2x—y = 7, and the second lOcc + y— 18 = lOy + a;, i. e., the units becomes the tens figure and the tens becomes the units. 8. A certain number of two digits contains the sum of its digits four times and their product twice. What is the number? Ans., 36. 9. There is a number consisting of two digits ; the num- ber is equal to 3 times the sum of its digits, but if the num- ber be multiplied by 3, the product equals the square of the sum of its digits. What is the number ? Verify. 10. A number consisting of 2 digits, when divided by 4 gives a certain quotient and a remainder of 3 ; when divided by 9, gives another quotient and a remainder of 8. Now, the digit on the left hand is equal to the quotient which was obtained when the number was divided by 9 ; and the other digit is equal to ^ of the quotient obtained when the number was divided by 4. What is the number? Verify. WITH TWO L'NKNOWN QUANTITIES. 269 11. A farmer parting with his stock, sells to one person 9 horses and 7 cows for 300 dollars ; and to another, at the same prices, 6 horses and 13 cows for the same sum. What was the price of each ? A?is., *2-t and $12. 12. A son asked his father how old he was. His father answered him thus : If you take away 5 from my years, and divide the remainder by 8, the quotient will ])e J of your age ; but if you add 2 to your age, and multiply the whole by 3, and then subtract 7 from the product, you will have the number of the years of my age. What was the age of the father and son ? A71S,, 53 and 18. 13. A farmer purchased 100 acres of land for $2450; for a part of the land he paid $20 an acre, and for the other part $30 an acre. How many acres were there in each part ? Verify. Scholium. — Very many such problems can be solved equally well by means of one or of two unknown quantities. 14. At a certain election 946 men voted for two candidates and the successful one had a majority of 558. How many votes were given for each candidate ? Verify. 15. A jockey has two horses and two saddles. The sad- dles are worth 15 and 10 dollars, respectively. Now if the better saddle be put on the better horse, the value of the better horse and saddle will be worth ^ of the other horse and saddle. But if the better saddle be put on the poorer horse, and the poorer saddle on the better horse, the value of the better horse and saddle will be worth once and -^ the value of the other. Required the worth of each horse ? Result, 65 and 50 dollars. 16. A sum of money was divided equally among a certain number of persons ; had there been four more persons, each 270 SIMPLE EQUATIONS. would have received one dollar less, and had there been four fewer, each would have received two dollars more than he did : required the number of persons, and what each received? Verify. Suggestion. - = — T + l,and- = — -—2. Rence xu+4:x.- xy «« y y + 4: y y-4: ^ ^ +2/- + 4y, and xy—4:X = xy—2y^-\-Sy, or 4a; = y"^+4y, and —2x = —y^+Ay. Adding, 2x = Sy. 17. A farmer hired a laborer for ten days, and agreed to pay him $12 for every day he labored, and he was to forfeit $8 for every day he was absent. He received at the end of his time $40. How many days did he labor, and how many days was he absent ? Verify. 18. A boatman can row down stream a distance of 20 miles, and back again, in 10 hours, the current being uni- form all the time ; and he finds that he can row 2 miles against the current in the same time that he rows 3 miles with it. Required the time in going and returning. Result, 4 and 6 hours. Suggestion. — If aj and y are tbe times of rowing down and up, respectively, at what rate does he row down ? At what rate up ? Twice one of these rates equals 3 times the other. 19. A and B together could have completed a piece of work in 15 days, but after laboring together 6 days, A was left to finish it alone, which he did in 30 days. In how many days could each have performed the work alone ? Ans.f 50, and 2 If days. Suggestion. — If « represent the number of days A would require to do it alone, and y the number B would require, how much would each do in a day? How much both? How much would they do in 6 days ? How much would remain to be done by A alone ? How much would A do in 30 days ? In resolving these equations do not clear of fractions. 20. Two pipes, the water flowing in each uniformly, filled WITH TWO UNKNOWN QUANTITIES. 271 a cistern containing 330 gallons, the one running during 5 hours, and the other during 4 ; the same two pipes, the first running during two hours, and the second three, filled another cistern containing 195 gallons. The discharge of each pipe is required. Verify. 21. If I were to enlarge my field by making it 5 rods longer and 4 rods wider, its area would be increased 240 square rods ; but if I were to make its length 4 rods less, and its width 5 rods less, its area would be diminished 210 square rods. Eequired the present length, width, and area. Verify. 22. A farmer sells a horses, and b cows for ^m ; and at the same prices «i horses, and bi cows for ^//i,; what is the price of each? Apply the results to Ex. 11. See (30) Part 1. Ans., Of a horse $~^ r ; of a cow $ — j -r-- aOi—aiO axb—abx [Note. — Observe the symmetry of such results. Thus, in these numeriitors the a and 5 change places and in the denominators the subscripts change letters.] 23. A man bought s acres of land for %m. For a part he paid %a per acre, and for the rest $a, per acre. How many acres in each part ? Deduce from the getural answer obtained in this case the particular answers to Ex. 13. . m—ayS J m—as Am., and acres. 24. A waterman rows a given distance a and back again in b hours, and finds that he can row c miles with the current fore? miles against it: required the times of rowing down and up the stream, also the rate of the current and the rate of rowing? Ans., Time down, ^; time up, — -^ ; rate . , a(c3-^) . . . flr(c + fiO^ of current, -^^-^ ; rate of rowing, -^^-^^ Peduce from these answers those of Ex. 18, Smple'Eiiefims "M"^'-^'"^' """" "" ^^. WITH MORE THAN TWO UNKNOWN QUANTITIES. SCTI0N IIL 4(>, Prob. — Having given several simple, simultaneous, independent equations, involving as many unknown quanti- ties as there are equations, to find the values of the unknown quantities. Rule. — /. Combine the equations two imd two by either of the methods of elimination, eliminating by each combination the same uuJcnown quantity, thus producing a new set of equations, one less in number, and containing at least one less unknown quantity. II. Combine this new set two ajid, tiuo in like manner, eliminating another of the unknown quantities. III. Repeat the process until a single equation is found with but one unknown quantity. IV. Solve this equation and> then substitute the value of this unknown quantity in one of the next preceding set of equations, and there will result an equation containing ajvother single unknown quantity, the value of which can therefore be found. V. Substitute the two values now found in one of the next preceding set, and find the value of the remain - ing unknown quantity in this equation. Continue this process till all the unknown quantities are deter- mined. Scholium I. — If any equation of any set does not contain the quantity you are seeking to eliminate from the following set, this WITH MORE THAN TWO UN^KNOWN QUANTITIES. 273 equation can be written at once in that set and the remaining equa- tions combined. Scholium 2. — In eliminating any unknown quantity from a par- ticular set of equations, any one of the equations may be combined with each of the others, and the new set thus formed. But some other order may be preferable as giving simpler results. Scholium 3. — It is sometimes better to find the values of all the unknown quantities in the same way as the first is found, rather than by substitution. Demonstration I. — The combinations of the equations give true ^nations l)ecause they are all made upon the methods of elimina- tion already demonstrated. 2. That the number of equations can always be reduced to one by this process, is evident, since, if we have n equations and com- bine any one of them with each of the others, there will be n—\ new equations. Combining one of these ?i— 1 new equations with all the rest there will result «— 2. Hence n—\ such combinations will produce a single equation ; and as one unknown quantity, at least, has disappeared from each set there will be but one left. q. e. d. EXAMPLES. Sx. 1. Given (1,) '7x-2z-^^u = 17, (20 (3i) (50 find the values of t + 4.y-2z = 11, 5y—dx—2u = 8, -3u-^2t + 4:yz= 9, 3z-\-Su = 33, Xy y, z, t, and w. Operation. (Ig) (2«) (4«) Model Solution. t^iy-2z= 11 2< + 4y-3i^ = 9 3« + 8w= 33 35y_62_5w = 107 2nd set, from which X is absent. (1.) (23) (3a) 3e + 8?/= 33] ^ , ^ - 42,-4. + 3.= 13 [^ and « are absent. 1 (34) 32 + 8^= 33 1162-125W = -%! 4th set, from which «, t, and y^ are absent. 274 SIMPLE EQUATIONS. (Ig) 1303^ = 3909 .: u=S (IJ 3s + 34=33 .-. 2 = 3 (33) 42/-12 + 9 = 13 .'. y = 4 (Ig) ^ + 16-6 = 11 .-. t = l (3i) 20-3a!-6= 8 :: x = 2 Explanation. — I notice that I have 5 equations with 5 unknown quantities. From these I wish to produce a new set of 4 equations from which one at least of the unknown quantities shall be eliminated. I observe that x does not appear in (2^), (4^), and (5 J, hence I write these as three of the 2d set of equations. Then eliminating x between (Ij) and (SJ I have (4g), and thus obtain the 2nd set of 4 equations containing only 4 unknown quantities. Again, as t is contained in a less number of this set of equations than any one of the other unknown quantities, I eliminate it next ; i. e., I produce a 3rd set which does not contain it. As (^2) and (42) do not contain t, I transfer them at once to the 3rd set ; and then eliminating t between (I3) and (22) this set is complete, having 3 equations with 3 unknown quantities. Now eliminating y from this set by combining (23) and (33), and transferring (I3), I have the 4th set of two equations with only 2 unknown quantities. Combining these two so as to eliminate z I find u = S. Finally, substituting 3 for u in (I4), I find z = d. Substituting 3 for u and 3 for z in (33), I find y = 4. Substituting 4 for y and 3 for z in (Ig), I find t = i. Substituting the values of g^ and ^^ in (3i),I find x = 2. 2. Given j x-\-z = 26 ?/+ 2; = 15 3. Given- 4. Given Sx^4:y = 24— 2, 6x-\- y = 2; + 84, a; + 80 = 3«/ + 4;2. 3w+ a;4-2y— = 22 ^x— y + Sz = 35 4:U-^dx—2y = 19 2^ + 42^4-22; = 46 Values, - Values, ■ Values, x=:20, y = 10, z = 6. \x = n, y=z20, z = S. ' U = 4:, X = 6, z =7. WITH MORE THAN TWO UNKNOWN QUANTITIES. 275 5. Given 2^3^4 124, Results, • x= 48, y = 120, z = 240. Suggestion. — This may be solved in the ordinary way by clearing of fractions, etc., but the following is far more elegant: Dividing the Ist by 3 and the 2nd by 2 and subtracting, we have Subtracting ^ the Ist from the 3rd y z 17] + 72 60 '6 y + 2 = 14 80 24 2nd set. Or, j_ g _17 360 "^ 5 • 60 ~ 15 y ^ _ "^ 360'^2n2~6 Subtracting — — 2 6 1 6. Given n 1 y z = 1, and z = 240. 2 Values, ■ y a + h—c* 2 a—h-J^c' 2 2; = ^H-c— « Suggestion.— Do not clear of fractions. Having found the value of one unknown quantity, do not get the others by substitution, but return to the original equations and get each in the same manner. 7. Given r2 3__4_ J_ x"^ y z~12' 3_4 5_19 X y^ z" 24' x^ y^ z 2 Values, X = 6, % = 8. 276 SIMPLE EQUATIONS. x-\-a= y + z, 8. Given \ y-\-a = 2xi-2z, z-{-a =: 3x-\-Sy, r 9x-^6y-{-4:Z-\-3v-^2u' 16a;+4v-|-l = 0, 25x—5y-\-z -\- 6v — u + 1 = 0, x-}-2y-\-4:Z—v — 2u + 1 = 0, ^x—2y-\-z — 2v -\- u + 1 = 0. Values, z = a il' Tl' 7a 11* 9. Given ■ Values, ■ X = u = 10. Giv'^n 11. Given- u-{-v-{-x-\-y = 10, u-^v-{-x-\-z = 11, w + v + ?/ + 2; = 12, w + ^ + ?/ + ^ = 13, y ■^x-\-y-\-z = 14. 3j--l_65 X ■ 4 - 5 2"^ ^' Fa^i^e^?, - 6X 4:Z T"^ 3" , 5 = 2^ + 6' 3:^ + 1 7 14^6 2z '21 + ^ ^3 Values, 12. Given 11a;— lOi/ = 12y—llz x + z—2y _ 2;— ?/— 1 3 "" 2 ' 3^- =?/+;2; + 7. Values, ■ 169 "924' 220 924' 89 924' 445 924' 113 924* = 3, = 4, = 5, = 2, — 1, = 2, = 3, = 1. r?; = 10, L2! 11, 12. WITH MORE THAN TWO UNKNOWN QUANTITIES. 277 13. Given 10 15 9x-{-5y—2z 2x-{-y—dz 12 4 _7y±z±S - 11 " "^^^ 5y-\-3z 2x-{-3y— z 4 12 -\-2z :y-l + 3a; + 2y + 7 Values, ■ a; = 9, 2/ = 7, 2=3. 14. Given 4(a; + 2;) = 9-2/, i{x-z) = 2y-7. Values, x = 7, z=3. APPLICATIONS. Ex. 1. The sum of three numbers is 9. The sum of the first, twice the second, and three times the third is 22. The sum of the first, four times the second, and nine times the third is 58. What are the numbers? Aiis., 1, 3, and 5. Suggestion. — How many unknown quantities? How many sets of conditions ? What are they? Express the first in an equation, — the second, — the third, 2. Five persons, A, B, C, D, and E played at cards; after A had won half of B's money, B one-third of C's, C one- fourth of D's, and D one-sixth of E's, they each had $7.50. How much had each to begin with ? Ans., A, $2.75 ; B, $9.50 ; 0, $8.25 ; D, $8 ; and E, $9. 3. There are 4 men. A, B, C, and D, the value of whose estates is $14,000 ; twice A's, three times B's, half of C's, 278 SIMPLE EQUATIONS. 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, is $4,000. Kequired the property of each. Ans., A's, $2,000; B's, $3,000; C's, $4,000; D's, $5,000. 4. A number is represented by three figures ; the sum of these is 11 ; the figure in the place of units is double that in the place of hundreds, and when 297 is added to this num- ber, the sum obtained is represented by the figures of this number reversed. What is the number ? Ans., 326. Suggestion. — Letting x represent the hundreds figure, y the tens, and s the units, the number is represented by lOOx+lOy-^z. The number with the digits reversed is lOOz+tOy-^x. 5. A man worked for a person ten days, having his wife with him 8 days, and his son 6 days, and he received $10.30 as compensation for all three ; at another time he wrought 12 days, his wife 10 days, and son 4 days, and he received $13.20 ; at another time he wi'ought 15 days, his wife 10 days, and his son 12 days, at the same rates as before, and he received $13.84. What were the daily wages of each ? Ans.y The husband 75 cts. ; wife, 50 cts. The son, 20 cts. expense per day. Suggestion.— The value of the quantity representing the son's wages is found to be negative. Therefore the son produced the opposite effect from wages; i. e.,he was an expense. 6. Three masons. A, B, C, are to build a wall. A and B, jointly, can build the wall in 12 days ; B and C can accom- plish it in 20 days, and A and C in 15 days. How many days would each require to build the wall, and in what time will they finish it, if all three work together ? A71S., A requires 20 days; B, 30 ; and C, 60 ; and all three require 10 days. WITH MORE THAK TWO UNKNOWN QUANTITIES. 279 7. Three laborers are employed on a certain work. A and B, jointly, can complete the work in a days ; A and require h days, B and C require c days. What time does each one, working alone, require to accomplish the work, on the condition that each one, under all circumstances, does the same quantity of work ? And in what time would they finish it, if they all three worked together ? Am., A reqmres j^^^^z^ days, B ^^^„^_^ days, days. Deduce from these results those of the preceding example. 8. If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C together in 10 days, in how many days can each alone perform the same work? Ans.^ A in 14|f days, B in 17|f days, and C in 23-^ days. 9. A gentleman left a sum of money to be divided among his four sons, so that the share of the oldest was ^ of the sum of the shares of the other three, the share of the second \ of the sum of the other three, and the share of the third J of the sum of the other three ; and it was found that the share of the oldest exceeded that of the youngest by $14. What was the whole sum, and what was the share of each person ? Ans., Whole sum, $120 ; oldest son's share, $40 ; second son's, $30 ; third son's, $24 ; youngest son's, 126. 280 SIMPLE EQUATIONS. CO o Q. CO DEFINITIONS. SYNOPSIS. f Algebra. Equation. Members. Independent Equations. Simultaneous. Transposition. Elimination. Statement. Solution. With one unknown quMitity. With more than one unknown quantity. KINDS OF ^^""f\. EQUATIONS. ) Q"adratrc ^ Higher. AXIOiVIS. I. Clearing of fractions. Rm-is. Dem. m. TRANSFOR- J 2. Transposition, rule. Detn. ill. MATIONS. 1 3. Uniting terms. 4. Dividing by coefficient of unknown quan. Prob. I. To solve equations RuLB. Dem. Prob. 2. To free an equation of radicals RuLB. Dem. [ Practical suggestions. Rule. Dem. Practical suggestions. LJ f Number of methods. Reason for several. Prob. I. By comparison, rule. Dem. Prob. 2. By substitution, rule. Dem. Prob. 3. By addition and subtraction, rule. Dem. Prob. 4. With several unknown quantities. I „ ^ , RULE. Dem. [sch.l,2,a Test Questions. — Upon what principle is an equation cleared of fractions ? How is it done ? Upon what principle is elimination by addition and subtraction performed ? What comparison ? Sub- stitution? Give the seven Practical Suggestions upon solving Simple Equations. The six upon freeing of Radicals. Give the reason for changing the signs of the terms of a fraction having a polynomial numerator, preceded l)y a minus sign, when clearing of fractions. What is the general method of precedure in stating a problem ? Does the statement involve a knowledge of anything but algebra ? Illustrate. Upon what principle may all the signs of an equation be changed ? (This may be explained in at least four dif- ferent ways.) Having given the sum and difference of two quanti- ties, how are the quantities found ? Prove it. ^OPOF^TION F\OGF^ESSION RATI 0. 47. Ratio is the relative magnitude of one quantity as compared with another of the same kind, and is expressed by the quotient arising from dividing the first by the second. The first quantity named is called the Antecedent, and the second the Consequent, Taken together they are called the Terms of the ratio, or a Couplet, 48, The Sign of ratio is the colon, : , the common sign of division, -^, or the fractional form of indicating division. Illustration. — The ratio of 8 to 4 is expressed 8 : 4, 8^4, or j » a°y one of which may be read " 8 is to 4," or " ratio of 8 to 4." The antecedent is 8, and the consequent 4. The sign : is an exact equivalent for -j-, and by many writers, especially the Germans, the former is used exclusively. The sign : is, probably, a mere modi- fication of-!-, made by dropping the horizontal line, as unnecessary. Possibly the sign -i- finds its analogy in the fractional form for expressing division, by considering the upper dot as symbolizing a dividend, and the lower a divisor. 40. Cor. — A ratio being merely a fraction, or an unexe- cuted problem in Division, of which the antecedent is the 282 RATIO, PROPORTIOi?^, AND PROGRESSION. numerator, or dividend, and the consequent the denominator, or divisor, any changes made upon the terms of a ratio produce the same effect upon its value, as the like changes do upon the value of a fraction, when made upon its corresponding terms. The principal of these are, 1st. If both terms are multiplied, or both divided by the same number, the value of the ratio is ISOT changed. Illustration. 16 : 8 =2 4x16:4x8 = 2 2nd. A ratio is multiplied by multiplying the antecedent or by dividing the consequent. 8 = 4 8 = 8 Illustration. 32 2x32 32 = 8 3rd. A ratio is divided hy dividing the antecedent or by multiplying the consequent. Illustration. 24 :6 = 4 ¥ :6 = 2 24 2x6 = 2 50. A Direct Ratio is the quotient of the antecedent divided by the consequent, as explained above. (47.) An Indirect or Reciprocal Ratio is the quotient of the consequent divided by the antecedent, i. e., the recipro- cal of the direct ratio. Thus, the direct ratio of 6 to 3 is 2, but the inverse ratio is | or ^. When the word ratio is used without qualification it means direct ratio. The inverse or reciprocal^ it will be seen, is the ratio of the reciprocals. Thus the inverse ratio of 8 to 4 is the ratio of |^ to J, or f 51. A ratio is always writtek as a direct ratio. Thus, the inverse ratio of a^ to & is & : a, or - : ^ , the latter being expressed as the direct ratio of the reciprocals. RATIO. 283 52. A ratio of Greater Inequality is a ratio which is greater tlian unity, as 4 : 3. A ratio of Less Inequality is a ratio which is less than unity, as 3 : 4. 53» A Compound Ratio is the produpt of the cor- responding terms of several simple ratios. Thus, the compound ratio a : 6, c : ^, w : w, is acm : Mn. This term corresponds to compound fraction. A compound ratio is the same in eflfect as a compound fraction. 54, A Duplicate Ratio is the ratio of the squares, a triplicate, of the cuhes, a subduplicate, of the square roots, and a subtriplicate, of the ciche roots of two num- bers. Thus, «2 : ]^^ a^:i)S^ ^a : ^b, and \^a : ^/b. EXAMPLES. Ex. 1. What is the ratio of 3am^ to 6am^ ? Model Solution. — Since the ratio of two quantities is the quotient of the antecedent divided by the consequent, the ratio 3am^ : Qam^ is 3am» 1 - — - , or -m. 2. What is the inverse ratio of a— J to a^—l^ ? A71S., a + b, 3. What is the ratio of I to ^? Of|:|? Of ^^^=^ : a 3 2 6 3 a^-^a^ _^v Of — to?^? Of ""'-y' . ^-^ 3% 3by T^+f x^-xy+y^ Answers to threes 1, IJ, and 1-J-. 4. What is the triplicate ratio of 6 to 2, Arts., 27. 5. What is the subduplicate ratio of 64 to 16 ? Ans., 2. 6. What is the compound ratio of 3 to 4, 8 to 9, 2 to 6, and 4 to 2 ? Ans., 4 : 9, or ^. 7. Reduce 360 : 315 to its lowest terms. Suggestion. — This is the same as reducing a fraction to its lowest teniis. The result is 8 : 7. 284 RATIO, PROPORTION, AND PROGRESSION. 8. Reduce a^ -\- 2a^x : a^ to its lowest terms. Result, a-|-2a; : 1. 9. Which is the greater, 16 : 15, or 17 to 14? Suggestion. — To compare two fractions, reduce them to a common denominator. On the same principle these ratios become 224 : 210, and 255 : 310. Otherwise perform the division and compare the quotients. By this method we have 1.06 + , and 1.21 + . 10. Which is greater, a^—W\a—l, or a^-i-2ab + i^'. 11. Which is least of the ratios 20:17, 22:18, and 25 : 23 ? 12. Which is greater, « + 2 : J^-f 4, or « + 4 : Ja + S ? Ans.f a + 4: : ^a-\-6ya-{-2 : ia-\-4:. 13. What is the compound ratio of 15 : 12, 6 : 7, and 9:4? Ans., 135 : 56. 14. Compound the ratios a^—x^ia^, a + x:i, and h \ a—x. Result, The duplicate ratio of a-\-xto a. 15. Show that the compound ratio of x-\-y \ a, x—y:h, and h : — is 1. 16. Is the compound ratio of 3a + 2 : 6«4-l, and 2rt + 3 : a +2, a ratio of greater or of less inequality, if a is — f ? If a is 2 ? If « is — 2 ? ^^s.,— Zero, Greater, Infinity. 17. Compound the following : 7 : 5, the duplicate of 4 : 9, and the triplicate of 3 : 2. Result, 14 : 15. 7 ^.| $'$'$ ,. -,. Suggestion. 5 X ^^ X ^^^ = 14 . 15. 3 18. Compound the sub-duplicate of x^ : y^, and the tripli- cate of ^/x : \^y. Result, x^ : y^. PROPORTION. 285 19. Compound the inverse ratio of ^x-\-*Jy to ar— -y, and the direct ratio x-\-'l^xy-^y : Vx-{-Vy- Result, x—y, 20. Which is the greater, the inverse subtriplicate ratio of 8 to 64, or the direct duplicate ratio of 2 to 3 ? Pf^opof^tion. ^^csf^^^TMM IL 55, Proportion is an equality of ratios, the terms of the ratios being expressed. The equality is indicated by the ordinary sign of equality, =, or by the double colon : : . Thus, 8 : 4 = 6 : 3, or 8 : 4 : : 6 : 3, or 8-f-4 = 6h-3, or - = - all mean precisely the same thing. A proportion is usually read thus : " as 8 is to 4 so is 6 to 3." Scholium.— The pupil should practice writing a proportion in the form 5- = 3 , still reading it " a is to 5 as c is to dy One form d should be as familiar as the other. He must accustom himself to a c the thought that a:b :: c:d means t = -^ and nothing more. It will be seen that the language " 8 is to 4 as 6 is to 3," means simply that ft (\ - r= - , fbr it is an abbreviated form for saying that " the relation 4 8 which 8 bears to 4 is the same as (is equal to) that which 6 bears to 3 ; " that is, 8 is as many times 4 as 6 is times 3, or - = - • 50. The Extremes (outside terms) of a proportion are the first and fourth terms. Tlie Means (middle terms) are the second and third terms. Thus, in a : b = c : d, a and d are the extremes, and b and c are the means. 286 RATIO, PROPORTION, AND PROGRESSION. 57. A Mean Proportional between two quantities is a quantity to which either of the two bears the same ratio that the mean does to the other. Thus, if 7W is a mean proportional between a and 5, a bears the same ratio to m that m does to 5 ; i. e.^ a-.m :: m: K 58. A Third Proportional to two quantities is such a quantity that the first is to the second as the second is to this third (proportional). Thus, in the last proportion, 5 is a third proportional to a and m. So, also, a is a third proportional to 5 and m. Scholium. — Notice carefully the language used in the last two definitions. We do not say "a mean proportional ^," but "a mean proportional between,''^ two others. So, again, we say " a third proportional to two others." Moreover, it is necessary that the two others be taken in the order named in the statement. Thus, if y is a third proportional to m and n, m:n \:n:y. But, if y is a third proportional to n and m^ n-.m :: m:y. 59. A proportion is taken by Inversion when the terms of each ratio are written in inverse order. Thus, if a : b : : c : d, hy inversion we have h : a : : d : e. It is to be observed that in inversion the means are made extremes, and the extremes means. 60. A proportion is taken by Alternation when the means are made to change places, or the extremes. Thus, a : b :: c : d becomes by alternation either a : c :: h : d, or d : b :: c : a. The appositeness of the term alternation (taking every other one) is seen from the fact that the new order is obtained by taking the terms alternately ; that is, 1st and 3rd, 2d and 4th ; or 4th and 2nd, 3rd and 1st. 61. A proportion is taken by Composition when the sum of the terms of each ratio is compared with either term of that ratio, the same order being observed in both ratios ; or when the sum of the antecedents and the sum of PROPORTION. 287 the consequents are compared with either antecedent and its consequent. Thus, if a : h : : c : dy by composition we have a-\-b : a : : c-\-d : c, or a-\-b : b : : c-\-d : d, or a-^c : b-^d :: a : b, or a + c : b-\-d :: c : d. 02. If the difference instead of the su7ri be taken in the last definition, the proportion is taken by Division. 03. Four quantities are Inversely or Reciprocally Propor- tional when the 1st is to the 2nd as the 4th is to the 3rd, or as the reciprocal of the 3rd is to the reciprocal of the 4:th. Thus, if a and b are to each other inversely, or recipro- cally, as m and n, a \b \:n:m, or what is the same thing, « : : : — :-• m n 04. A Continued Proportion is a succession of equal ratios, in which each consequent is the antecedent of the next ratio. Thus, if a : b : : b : c : : c : d :: d : e. we have a continued proportion. OS. Prop. 1. — In a -proportion the product of the extremes equals the product of the means. Demonstration.— If a:h :: c:d then ad = ic. For a:h :: c:d is the same as £ = ^ , which cleared of fractions becomes ad = bc. a Q. E. D. 00, Cor. 1. — The square of a mean proportional equals the product of its extremei^y and hence a mean proportional itself equals the square root of the prodtict of its extremes. For, if a '. m '.: m : d, by the proposition m^ = ad. Whence ex- tracting the square root of each member, m = Vad. 07. Cor. 2. — Eithi^r extreme of a proportion equals the product if th' means divided by the olh^r extreme ; and, in like manner, either mean equals the product of the ewtremes 288 RATIO, PROPORTION, AND PROGRESSION. divided by the other mean. For, if a : b :: c : d, ad = he. ^ he he ^ ad ^ ad .'. « = — , « = — , =z — , and c = ^- • a d e h 68, Prop. 2. — // the product of two quantities equals the product of two others, the tivo former may he made the extrem^es, or the means of a proportion, and the two latter the other term^s. Demonstration. — Suppose my = nx. Dividing each member by ojy, we have — = - ; i. e.^ m-.x-.'. n\ y. In like manner dividing by mn we have - — —, i. e., y:n::x\m. Let the pupil determine how each of the following forms may be deduced from the relation my = nx. Given above. By what do you divide ? Griven above. Dividing mx by each mem- X TTl ber we have - = — y n By what do you divide ? How obtained ? 1. m : X : : n : y. 2. m : n :: X : y. 3. y ' n :: X : m 4. X : y :: m : n. 5. y ' X :: n : m. 6. X : m : : y ' n. 7. 71 : m '.: y : X. 8. n : y :•• m : X. TRANSFORMATIONS OF A PROPORTION. 6*.9. Prop. 3. — Proposition 1, together with the two principles that such changes in the terms of a propor- tion may he made, as, 1. Do not change the values of the ratios, 2. Change both ratios alike, are sufficient to determine in all cases what transfor- mations are possible without destroying the proportion. That these two principles are correct is evident from the nature of a proportion, as an equality of ratios. PROPOETION. 289 EXAMPLES. MULTIPLES. Ex. 1. li (I'.h :\x:y, prove that ma :mb I'.xiy. Solution. — This change does not alter the value of the first ratio, and hence the equality of ratios remains. 2. If a:b :: X :y is 7na :mh :: nx : ny ? Why ? Is the value of either ratio changed ? Why ? 3. If a\h'.\x\y is ma\h wnix-.yt Is the value of either ratio changed ? How ? Ans., This change does not destroy the proportion, because it multiplies both ratios by the same quantity. CL X 4. It a :b :: x:y. 18 a: mb :: x : my ? or — : b : : — : v ? or —:b:: x:my? m 6. If the first term of a proportion is multiplied by any number, in what four ways may it be compensated so as not to destroy the proportion ? Does multiplying the third term by the same number compensate ? Why ? Does dividing the 4th term ? Why ? Does dividing the second term? Why? 6. If the third term of a proportion is divided by any number, in what ways can the change be compensated so as not to destroy the proportion ? Give the reason in each case. CHANGES IN THE ORDER OP TERMS. 7. If a :b :: x : y is a i x -.: b : y? How are the ratios changed ? Solution. a:b :: x:y is the same as r- = - (55). Now multi- h y ^ ' plying each member of this equality by - , we have 13 290 RATIO, PROPORTION, AlsD PROGRESSION. a h X d =- X - = - X - X y X « ^ 1-. 1. • or - = - , which 18 X y' a:x ::i:y. Thus we see that a:i :: x:y is transformed into a:x ::h:y by multiplying both ratios by - . This does not destroy their equality by (69, 2). 00 Cf'. Hence we see that we can change the order of the means with- out destroying a j^roportion. 8. It m :x ::n:y, is n:7n:: y:x? How are the ratios changed ? 7Tl 0C71 Ans., Yes. The first, — , is multiplied bv -^, and the second, -, is multiplied by — , and — ^ = ^, since nx=my. y Xfh 7fl 71/X 9. If four quantities are in prcrportion are tbey in propor- tion by inversion ? How are the ratios changed ? Solution. a:l) '.'. c:d is the same as t = -j, by the definition of o di a proportion. Now dividing 1 by each member of this equality I have I d .... - = - , which 18 a c h:a:: d:c. The substance of this is that if two quantities are equal, their reciprocals are equal. 00b* Hence we see that we can take a proportion hy inversion with- out destroying it. 10. If 3a^ : W : : 6mx : lOm^x^, is 2 : a^ : : 6mx : &»? Solution. — Taking the proportion, and cancelling like factors from both terms of the same ratio, which does not change its value, and like factors from both antecedents, which divides both ratios and hence does not destroy the proportion, and like factors from both consequents, which divides both ratios, and hence does not destroy the proportion, we have PROPORTION. 291 or fl2 . ^ . . 2 : 5mx. Whence changing the order of the ratios 2 : 5mx : : a' : &', and finally changing the order of the means (69, 2), we have 2 : a' : : 5mx : ¥. 11. If ^03^ : i% : : a^x : %, show that J : i : : a^ : fa:. COMPOSITION AND DIVISION. 12. m a : b :: m : n, show that a-\-b : a : : m -{- n : m. Solution. a:b :: m:n is the same as ^ = — . Now add - to on the first member and - to the second, and we have - + - = — + -, n b n n or —V- = ; that is a + o :b :: m + n :n, h n 13. If a : J : : a; : y, show that a—h :b :: x—y : y. Suggestion. — Subtract from the first ratio -=^, and from the sec- ond^. y 14. If m \n :\x \ y, show that m-\-n \m—n :'.x-\-y IX— y. Suggestion. — By the method of Ex. 12, I have = — -^ and by that of Ex. 13, ^11^ = ^=^. n y Dividing the former by the latter, I have = — -\ m-n x—y^ that is, m-^n'.m—n :: x-\-y:x—y. SOc, Hence we see that a proportion may le taken by composition^ or by division^ or by both at once, without destroying it. 16. If ia—x : Ja + ic :: b—y : b + y, show that 2x : y :: a :b. 16. li a :b :: X : y, does it follow that a—y : b—x ::a:x? Ans., No. RATIO, PROPORTION^, AN^D PROGRESSIOl?". 17. From a\l\\x\y, prove that ma+nb : ma—nh ::mx-\-ny : mx—ny. Also that if a\l \\c\d, and m : n y J T a b c d \ix : y, am-.bnw ex \ ay, and —:-::-:-• m n X y IS. li a : b :: X : y is a^ : b^ :: x^ : y^? Is a^ : bn :: x^ : yn? [s Va : Vb : : V^^ : Vy ? Is a^ :b^ :: x^ : y^? Is a^ : b"^ :: x»^ : y^ whether m is integral or fractional, positive or negative ? Why is it that the ratios remain equal in each case ? How are they changed ? MISCELLANEOUS. 19. lia \b :: c : d, show that ?na + tz^ '.pa-\-qb :: mc-\-nd '.pc-\-qd, ' Suggestion. — The ratios to be compared when reduced to a C. D. acmjj + bcnp + admq + bdnq acmp + bcmqi-adnp + Mnq ' {ap-\-dq){cp + dq) {ap-^l)q){cp + dq) Now from the given proportion we learn that ad = he. Therefore, exchanging them in the two middle terms of the first ratio, the ratios become identical. This may also be shown as follows : Multiplying antecedents by m and consequents by n, ma :7il) :: mc: nd. By composition ma-{-nb:ma :: mc + nd:mc, or multiplying both ratios by wi, Tna + nb-.a-.imc + ndic. By changing the places of the means ma-\-7il):mc-\-nd :: a:c. In like manner it may be shown that pa + qh:pc + qd :: a:c. .'. ma + nhimc + nd :: pa + qb :pc + qd, or ma-{-ni:2>a-\-qb :: n),c + nd: pc-i-qd. The student should give the reason why each step does not vitiate the proportion, according to (69). 20. lt{a + b-}-c-i-d){a — b — c-j-d) = (a—b + c—d) {a-\-b—c—d) prove that a : b \: c \ d. Suggestion. — Performing the operations and reducing, 2ad—%bc == —%ad-\-%l)c, or ad = 5c. Whence ^ = -^ , or a:b:: c:d, ' a PROPORTION. 293 This may also be proved by writing according to Prop. 2, a+J + c-i-d : a—b-\-c—d : : a-{-b—e — d : a—h-c + d. Comparing the sum of each antecedent and its consequent with their difference, 2a -\-2c:2b + '-Zd::2a — 2c:2b—2d. Whence a + c: a-c :: b-\-d:h—d. Repeating the same processes we have a:b:: c:d. 21. If — - — = h, show that a—x :2a :: 2b : a + x. Pro- duce other forms of proportion from the given relation. How many can be produced ? 22. If r = s\/|, show that r : s : : 1 : ^2. 23. If (a-\-xY : {a—xY \'.x-\-y :x—y, show that a:x :: \/2a—y : \/y. Solution. a' + 2aa;4-aj' \a^—2m-^x^ :: x+y:x—y^ 2a' + 2x'» :^ax.:2x: 2y, a'-k-x^ : ckV : 2« : y, a? \^ \ : 2a— y : y, .'. a\x\: ^2a—y : ^y. Let the student give the reasons. 24. li a : h :'. c : d :: e : f '.: g '. h w i '. k, etc., show that - {a-\'C-\-e-\-g-\-i+,etc.) :{h-\-d-^f-\-h-\-k-\-,Qic.) ::a:b,or c : d, or e :f, etc. That is, in a series of equal ratios, the sum of all the antecedents is to the sum of all the conse- quents, as any antecedent is to its consequent. Solution. zr = rorab = ba, - = -^ or ad = be. b b b d a e ,, a g ., a i i i- . ,- = - or «/ = &e, =^ = Y or ah = bg, - = - or ak = bi, etc. b J b h OK Adding, a(&4-<^+/+^ + * + , etc.) = 6(a + c + e + ^ + i + , etc.); whence {a-\-c-\-e + g-\-i + ^ etc.) : (& + (i+/+A + ^ + , etc.) : : a : & or (since a:b = c:d^ etc.), as c:d:: e:f, etc. 25. Four given numbers are represented by a, h, c, dy what quantity added to each will give sums which are pro- portional. . he— ad a—o—c+d 26. If four numbers are proportionals, show that there is no number which, being added to each, will leave the resulting four numbers proportionals. Pf\OGf\ESSiON 70, A Progression is a series of terms which increase or decrease by a common difference, or by a common mul- tiplier.* The former is called an Arithmetical, and the latter a Geometrical Progression. A Progression is Increasing or Decreasing according as the terms increase or decrease in passing to the right. The terms Ascending and Descending are used in the same sense as increasing and decreasing, respectively. In an increasing Arithmetical Progression the common difference is added to any one term to produce the next term to the right ; and in i decreasing progression it is subtracted. In an increasing Geometrical Progression the constant multiplier by which each succeeding term to the right is produced from the preceding is more than unity ; and in a decreasing progression it is less than unity. This constant multiplier in a Geometrical Progression is called the Batio of the series. f 71. The character, . ., is used to separate the terms of an Arithmetical Progression, and the colon, :, for a like purpose in a Geometrical Progression. * This is the common use of the term. It Is aleio used to include what is called a Harmonical Progression. t This is an unfortunate use of the term Eatio, inasmuch as it is at variance with its use in proportion. To harmonize the uee of the term in proportion, with this use, may have led some writers to define ratio, as used in proportion, as the quo- tient of the consequent divided by the antecedent. But the definition has neither logic nor the common usage of authors, English or Continental, to support it. The French use rapport in proportion, and raisan in progression. ARITHMETICAL PROGRESSION. 296 Illustrations. 1 . .3. .5. .7, etc., is an Increasing Arithmetical Progression with a common difference 2. or +2. 15. .10. .5. .0. .—5, etc., is a Decreasing Arithmetical Progression with a common difference— 5. a. .a ±d. .a±2d. .a±Sd, etc., is the general form of an Arith- metical Progression, d being the com- mon difference. 2 : 4 : 8 : 16, etc., is an Increasing Geometrical Progression with ratio 2. 12 : 4 : 1 : 1 : TjSiy, etc., is a Decreasing Geometrical Progression with ratio J. a:ar:ar^ : ar* : ar*, etc., is the general fonn of a Geometrical Pro- gression, r being the ratio, and greater or less than unity, according as the series is increasing or decreasing. 72. There are Five Tilings to be considered in any pro- gression; viz., the tirst term, the last term, the common difference or the ratio, the number of terms, and the sum of the series, either three of which being given the other two can be found, as will appear from the subsequent discussion. 7 75. Prop. 1. — The formula for finding the nth, or last term of an Arithm,etlcal Progression; or, more properly, the formula expressing the relation between the first term, the nth term, the com^mon difference, and the nurriber of terms of such a series is I =^ a -\- (n — 1) d, in which a is the first term, f? the common difference, n the number of terms, and li\\Qni\\ orla.stterm, d being positive or negative according as the series is increasing or decreasing. 296 RATIO, PROPORTION, AND PROGRESSION. Demonstration. — According to the notation, the series is a. .a + d. .a-\-2d. .a + dd. .a + 4:d. .a + 5d^ etc. Hence we observe that as each succeeding term is produced by add- ing the common diflerence to the preceding, when we have reached the 7it\i term, we shall have added the common difference to the first term n—1 times ; that is, the nth term, or 1= a + {n—l)d. q. e. d. Scholium. — As this formula is a simple equation in terms of a, l, n, and d, any one of them may be found in terms of the other three. 74. Prop. 2. — The formula for the sum of an Arithmetical Progression, i. e., expressing the rela- tion between the sum of the series, the first term, last term, and nurrhber of terms is ra + li ' = b^h s representing the sum of the series, a the first term, I the last term, and n the number of terms. Demonstration. — If I is the last term of the progression, the term before it is l—d, and the one before that I— 2d, etc. Hence, as a. .a + d. .a + 2d. M + Sd ^, represents the series, I. .l—d. .l—2d. .1 — 3 LU CC O O CC Q. ^^ a. o QC a. O h- o o fiC CL < CC Definitions. Definitions. Fundamen- tal Propo- sitions. Transfor- mations ^ Definitions. A. P. G. P. f Ratio.— Terms of.— Antecedent.— Consequent.— t Conplet. Direct.— Inverse.— Greater Inequality, less. L Compound ratio.— Duplicate, sub-duplicate, etc. Sign of. Cor.— Changes in terms of. ' Proportion.— Extremes.— Means. Mean proportional.— Third proportional. Inversion. — Alternation. — Composition.— Division. . Inverse or reciprocal proportion.— Continued. ■ Prop. 1.— Cor. l.—Cor. 2. Prop. 2. Prop. 3.— Principles on which transformations are made. Equi-mnltiples.— Why proportion not destroyed. Chg. in order of terms.— " " " Composition, or division.— " " " Involution, or evolution.— " " *' Progression.— Arithmetical.— Geometrical. Increasing, or ascending.— Decreasing, or descend- ing. Common difference, positive, negative. Ratio, greater than 1, less than 1. Sign of Arithmetical Progression. — Of Geometri- cal. Five things.— Given, required. Two ftindamental formulas.— Produce them.— To insert means. Two fundamental formu- j To insert means. las.— Produce them. I Sum of infinite series. Test Questions.- -Give the various changes which can be made upon the terms of a ratio and tell how the ratio is affected, and Why f State the various transformations which can be made upon a proportion without destroying it, and Gim the reason in each case. Produce the two fundamental formulas of Arithmetical Progression. Also of Greometrical. APPLICATIONS. 305 APPLICATIONS. [Note.— Teacher and pupil should bear in mind that the object of this section is to teach the properties of Ratio and Proportion ; hence all the operations should be performed upon the proportion. The proportion should be kept in the form of a proportion, and not reduced to an equation. ] Ex. 1. Divide 60 into two parts which are to each other as 2 : 3. Suggestion. — Letting x and 60— x- be the parts, x : 60— a; : : 2 : 3. Hence x : 60 : : 2 : 5, or aj : 24 : : 2 : 2 ; and x=24. The pupil should give the reason for each transformation. What is the first trans- formation ? Composition. Why does it not destroy the propor- tion ? What the second transformation ? Why does it not destroy the proportion ? 2. A boy being asked his age said : John, who is 18, is older than I; but, if you add to my age \ of it, and from this sum subtract \ of my age, the result will be to John's a^e as 10 : 9. How old was the boy ? Verify. Operation. a;+^a;— i«: 18 :: 10 : 9, fx : 18 : : 10 : 9, a;:18:: 8:9, a? : 18 : : 16 : 18, .-. x = 16. Let the pupil tell, in each instance, just what the transformation is, and why, according to (69), the proportion is not destroyed. Verifloatlon. 16 + J/— V = ^^^ which is to 18 as 10 is to 9, the ratio being J^. 3. Two brothers being asked their ages, the younger re- plied, my age is to my brother's as 2 to 3 ; and if you add 18 to mine and 2 to his, the sums will be as 3 to 2. What were their ages ? Suggestion. — To solve with om unknown quantity we may repre- sent the younger brother's age by 2a; and the older's by 3a;. Then 2j; + 18 : 3x + 2 :: 3 : 2; Whence 2a;+18 : 27a;+18 :: 3: 18, 2x + 18:25a;:: 3:15:: 1:5. 2ic+18:«:: 5:1, 306 PROPORTION. 2a5 + 18:2iK:: 5:3, 18 : 2a; : : 3 : 3, 9 : X : : 9 : 6. .*. X = Q, 2x = 12 and 3a; = 18. [^o\e. — True, it gives a somewhat shorter solution of this example to put the first proportion immediately into the equation 4a; + 36 = 9a; + 6, whence ox = 30 and x = Q. But the object is to become familiar with the properties of a proportion.] 4. A man's ao:e when he was married Avas to his wife's as 3 to 2 ; but after 4 years, his age was to hers as 7 to 5. What were their ages when they were married ? Suggestion. — This may be solved with one unknown quantity, like the preceding, and that is the more elegant way. We may also use two. Thus, a;: y : : 3 : 3, and a;+4 : y + 4 : : 7 : 5. From the former a; : fy : : 3 : 3 . •. x = §y. Substituting in the latter |y + 4:y + 4 :: 7:5. Whence Sy + S :2y + S :: 7:5, y:2y + 8:: 3:5, 2y:2y + S ::4:5, 3?/ : 8 : : 4 : 1, and i/ : 1 : : 16 : 1. .'. y = 16, and a; = |y = 34. 5. A man is now 25 years old and his brother is 15. How many years before their ages will be as 5 to 4 ? Verify. 6. A man has two flocks of sheep, each containing the same number ; from one he sold 80 and from the other 20, when the numbers in the flocks were as 2 to 3. - How many were there in each flock in the first place ? Verify. 7. It is known to every one that a small body near the eye hides a large one farther off ; and it is a principle in optics so nearly axiomatic that we will take it for granted, that, in order to have the smaller body just cover the larger their distances from the eye must be in proportion to their breadths, or lengths. From this some very pleasing calculations can be made. The pupil may make the following : 1st. The breadth of a man's thumb is about 1 inch, and he can readily hold it at 2 feet from his eye ; how far off is the man who is 5 feet 8 inches high, when the breadth of the man's thumb at 2 feet from his eye just covers the man ? Ans., 136 feet. APPLICATIONS. 307 2nd. Wishing to know approximately the height of the top of a steeple from the ground, I found that my hand, which is 4 inches wide near the thumb, when held 2 feet from my eye, just covered the height of the steeple at a distance of 240 paces of 3 feet each. What was the height of the top of the steeple from the ground ? Ans.y 120 feet. 8. What number is that to which if 1, 5, and 13 be sev- erally added, the second sum will be a mean proportional between the other two ? Verify. 9. What ii umber is that whose | increased by 2 is to its | diminished by 1, as 6 is to 2J? A7is., 30. 10. The number of acres a farmer planted with corn is to the number he planted with potatoes, as | to 1 ; but if he had planted 6 acres less of corn, and J as many potatoes 4-15-J acres, the ratio would have been as f to f . How many acres of each did he plant ? [Note. — The five following examples are designed to be solved by using two or more unknown quantities.] 11. Find two numbers, the greater of which shall be to the less, as their sum to 42 ; and as their difference is to 6. Suggestion- Let x = one and y the other. Then (1) x:y::x+y:^2 (2) x:y::x-y:6, By equality of ratios x + y:4:2:: «— y:6, and x-\-y:x-y:: 7:1, 2x : 2y : : 8 : 6, x:2y :: i:Q, a^:|y::4:4, .-. ^ = ty. Substituting in (2) iyy-' |y-y:6 Or, f:l::iy:6, 4:1 :: y:6 1:1:: y : 24 .-. y = 24, a? = |y = 32. 308 PBOPOBTION. 12. Two numbers have such a relation to each other, that if 4 be added to each, they will be in proportion as 3 to 4 ; and if 4 be subtracted from each, they will be to each other as 1 to 4. What are the numbers ? Ans.y 5 and 8. Suggestion, x+4 : y+4 : : 3 : 4 and «— 4 : 2^—4 : : 1:4. Whence 4;c-|-4 4ic + 4 a? f 1 : 3 : : y : 4, or -— — :l::y:l. .-. y = — ^- . Substituting, aJ-4: ^ 4:: 1 : 4, a;-4 : 2x-4 : : 1:6, cc-4:a;:: 1 : 5, 4: x;: 4: 5. o . •. x = 5. Substituting, 6 : 3 : : y : 4, or 3 : 3 : : y : 8. .-. y = S. 13. Find two numbers in the ratio of 2|- to 2, such that when each is diminished by 5, they shall be in the ratio of 1^ to 1. Numbers, 25 and 20. 14. There are two numbers which are to each other as 16 to 9, and 24 is a mean proportional between them. What are the numbers ? Ans., 32 and 18. [Note. — The following examples may be solved by converting the proportion into an equation, at whatever stage of the solution it is found expedient.] 15. Find two numbers in the ratio of 5 to 7, to which two other required 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 those sums =16. Numlers, 30 and 42, and 6 and 10. 16. A farmer hires a farm for $245 per annum ; the arable land being valued at 12 an acre, and the pasture at 11.40 ; now the number of acres of arable is to half the excess of the arable above the pasture as 28 : 9. How many acres are there of each ? Ans.y 98 acres of arable, and 35 of pasture. 17. The quantity of water which flows from an orifice is proportioned to the area of the orifice, and the velocity APPLICATIONS. 309 of the water. Now there are two orifices in a reservoir, the areas being as 5 to 13, and the velocities as 8 to 7, and from one there issued in a certain time 561 cubic feet more than from the other. How much water did each orifice discharge in this time ? A7is., 440 and 1001 cubic feet. 18. At an election for two members of parliament, three men offer themselves as candidates, and all the electors give single votes. The number of voters for the two successful ones are in the ratio of 9 to 8 ; and could the first have had seven more without changing the numbers which the others had, his majority over the second would have been to the majority of the second over the third as 12 : 7. Now if the first and third had formed a coalition, and had one more voter, they would each have succeeded by a majority of 7. How many voted for each ? Ans.y 369, 328, and 300, respectively. 19. A man, driving a flock of geese and turkeys to mar- ket, in order to distinguish his own from any he might meet on the road, pulled 5 feathers out of the tail of each turkey, and 2 out of the tail of each goose, and upon count- ing them, found that the number of turkeys' feathers lacked 15 of being twice those of the geese. Having bought 20 geese and sold 15 turkeys by the way, he found that the number of geese was to the number of turkeys as 8 to 3. What was the number of each at first ? Ans., 45 turkeys, and 60 geese. PROBLEMS OP PURSUIT. 20. A fox starts up 120 feet ahead of a hound at exactly J^ past 2 o'clock P. M. ; the hound gives chase and gains 5 feet every 2 minutes. At what time will he overtake the iox? 310 t>ROPORTIOJSr. statement.— Letting x be the time wliich will elapse before the hound overtakes the fox, the problem becomes : If a hound gain 5 feet in two minutes, how long will it take him to gain 120 feet ? That is 5 : 120 : : 3 : a;. .-. a; = 48, and the hound overtakes the fox at 3 o'clock and 18 minutes. 21. A privateer espies a merchantman 10 miles to lee- ward at 11.45 A. M., and, there being a good breeze, bears down upon her at 11 miles per hour, while the merchant- man can only make 8 miles per hour in her attempt to escape. After 2 hours chase the topsail of the privateer being carried away, she can only make 17 miles while the merchantman makes 15. At what time will the privateer overtake the merchantman ? Ans., At 5.30 P. M. 22. A hare, 50 of her leaps before a greyhound, takes 4 leaps to the greyhound's 3 ; but 2 of the greyhound's leaps are as much as 3 of the hare's. How many leaps must the greyhound take to overtake the hare ? Suggestion. Let Sx = the number of the hound's leaps, whence 4a; = " " " hare's " in the same time. Then 2 :^ :: Sx: 4x + 50. .'. x = 100; and the hound takes 300 leaps. 23. The hour and minute hands of a clock are exactly together at 12 M. When are they next together ? Suggestion. — Measuring the distance around the dial by the hour spaces, the whole distance around is 12 spaces. Now, when the hour hand gets to 1, the minute hand has gone clear around, or , over 12 spaces. But as the hour hand ' has gone one space, the minute hand has gained only 11 spaces. Now as the minute hand must gain an entire round, or 12 spaces, to overtake the hour hand, we have the question: If the minute hand gains 11 spaces in 1 hour, how long will it take to gain 12 sj^aces? .-. 11 : 12 : : 1 hour : x hours ; and x = 1^\ hours, or 1 hour Oy\ minutes. APPLICATIONS. Sll Scholium. — 1. It is evident that the hands are together every l^^ hours; hence to find at what time thoy are together between any two hours on the dial, we have only to multiply 1^*^ by the number of the whole hours past 12 o'clock. Thus between 7 and 8 they pass each other at 7^^, or 7 o'clock and 3Sr^ minutes. Between 10 and 1 1 they pass each other at 10 o'clock 54^ minutes. 24. At what time between 6 and 7 o'clock is the minute hand jiirit J of the circle in advance of the hour hand ? Suggestion. — The question is : If the minute hand gains 11 spaces in one hour, how long will it take it to gain 6| rounds, or 75 spaces? Or, if it gains 1 round in -fij- hours, how long will it take it to gain 6| rounds? Ans., At 49y'T^ minutes past 6. 25. At what time between 4 and 5 is the hour hand of a watch just 20 minutes in advance of the minute hand ? Ans., At no time hetiueen these hours. The minute hand is within 20 minutes of the hour hand at 4 o'clock, and at 5^ minutes past 5. 26. Before noon, a clock which is too fast, and points to afternoon time, is put back 5 hours and 40 minutes ; and it is observed that the time before shown is to the true time as 29 to 105. Required the true time. Suggestion. — Letting «= the time pointed to, ar: : a! + 6| : : 29 : 105. Observe that to tuni the hands back 5h. 40m, is tlie same as to turn them forward 6h. 20m. x = 2h. 25m., and the true time was 2h. 25m. +6h. 20m., or 15m. before nine o'clock in the morning. 27. Two bodies move uniformly around the circumfer- ence of the same circle, which measures s feet. When they start, one is a feet before the other ; but the first moves m and the second M feet in a second. When will these bodies pass each other the 1st time, when the 2nd, when the 3rd, etc., supposing that they do not disturb each other's motion, and go around the same way ? Suggestion.— 1st. If3/">w, the second gains M—m feet a sec- ond, and having a feet to gain, overtakes the first, and does it in 312 PROPORTION. — seconds. The problem is then like the preceding, as the second gains a whole round every — seconds. Hence the second ° M—m o -^ s n -^ 2ji passing is at — from the starting, the third at -— — , the fourth ° M—m ° M—m . a + ds . at -- — , etc. M—m 2nd. If Jf < 7n, the second is overtaken by the first after the first has gained s—a feet, or in — seconds ; and in every seconds thereafter ; that is, from the m—M .. c ^ .- ' 2«-« 'da— a time or startmg, in , , etc. m—M m—M 28. The earth makes a revolution around the sun in about 365 days and Mars in about 687 days. How long is it from the time at which these planets are together on the same side of the sun till they are next together? That is, how long does it take the earth to gain an entire revolution? Ans., 687—365 : 365 :: 687 : x, ,\ x = 780 days, nearly. [Note. — The time required by a planet to go around the sun is called its Periodic Time^ or its year. When two planets are on the same side of the sun at the same time, they are said to be in Con- junction. The time from one conjunction to the next is the Synodic Period. The way in which this problem usually presents itself is this: We can obserT^e when a planet is in conjunction with the earth, and thus knowing the time of the earth's revolution (a year), we can find the Periodic Time of the planet, or how long it takes to go around the sun.] 29. Calling the earth's periodic time 365^ days, and observing Jupiter's synodic period to be about 399 days, how long is Jupiter in completing a revolution around the sun ; that is, what is its periodic time, or length of its year? Ans. ll^-J^-i^ of our 3'ears. 30. Saturn's synodic period is about 378 days ; what is its periodic time 9 A71S. 29^ of our years. APPLICATIONS. 313 PARTITIVE PROPORTION. S2a, Partitive Proportion is a term applied to the division of a number into parts which shall be in the ratio of given numbers. 31. Divide 350 into two parts which shall be to each other as 3 to 4. As 2 : 5. A^, 11 : 24. Suggestion. — Let 3a; represent one part, and ix the other ; whence dx-\-4x= 350, X = 50 ; and the parts are 150, and 200. 32. Divide a into two parts which shall be to each other as m : n. The parts are — -, — , and '- — - • ^ m-{-n m-\-n 33. Divide 560 into three parts which shall be to each other as 17, 23, and 16. As 43, 2, and 11. 34. Divide a into three parts which shall be to each other as m, n, and r. 35. Divide 23 into two parts which shall be to each other as f and f . The parts are 12^^ and 10|J. 36. Divide a into two parts which shall be to each other in the ratio — to -• As — : -• m n m n The parts in the last case are — , and m8-\-m ms-\-rn HAPTIII II ^Sectjon f. PERCENTAGE. S3. According to our definition, the Equation, of which it is the special province of Algebra to treat, is the grand instrument for investigating the relations of quantities. Now, in simple Percentage, there are four quantities to be compared ; viz., the Base, the Bate Ber Cent., the Bercent- age, and the Amount, and the problem is. To discover and express in equations the relations between these four quan- tities so that if a sufficient number of them are given the others may be found. [Note.— For Definitions see Practical Arithmetic, p. 225 et seq.] 84. Prob. 1.— To express the relation between base, pate pep cent., and percentage. Solution. — Let 5 represent the base, r the %, and p the percentage. If Now rfo means r lOOths of the base. Hence r % of 5 is -^ times 5, ?•& rh or — . .'. 7) = — • 100 ^ 100 85. Prob. 2.— To exppess the pelation between pate pep cent , amount or difference, and base. Solution. — Let s represent the sum or difference of the base and percentage. Then s = & ± rh 100 ¥l^^±ll , the + sign to be used 103 ' ^ PEKCKNTAGE. 316 when the base is increased by the percentage, and the — sign when the base is diminished by the percentage. Scholium. — The two formuloi expressing the relation between the four quantities J, r, /), and s, two of which must always be given to find the others, are in then.- selves suflBcient for the solution of all problems in Simple Percentage. EXAMPLES. Ex. 1. Bought a horse for $840, and sold it for 1560. How much did I lose per cent. ? Ans., 33^%'. Suggestion. — Here h and « are given to find r. Hence formula (2) is to be used. And as there is loss involved, the — sign is to be taken. Substituting in this formula, 560 = . Solving „ , 56000 ,,^^ 200 ,^^ for r we have -^vt^ = 100— r, or — - = 100— r; whence r = 100 840 3 _??? = "» = 33J. 3 3^ 2. A number being increased by 2 equals 14. Required the increase per cent. ? Atis., 16f^. rb Suggestion.— Formula (1) gives 2 = — - ; and (2) gives 14 = -— TTjA -^ • From which we are to find ;-. Multiplying (1) by 7, -iA '^rb ^^ 7;-& &(100 + r) „ ,^^ 100 14 = - . Whence — = ^-^ or 7. = 100 + r, or r = — = 16|. 3. A piece of cloth sold for $7?9, cash, which was b% off. Required the price of the cloth. Price, $820. 4. Sold 40^ of my wheat, and had remaining 981 bushels. How much had I at first ? Arts., 1635 bushels. 5. A man sold two horses at $420 each ; for one he re- ceived 25^ more, and for the other 25;^ less than its value. Required his loss. Loss, $56. 316 BUSINESS RULES. Suggestion. — Letting b and h^ be the values of the horses, we have ._. 6(100-25) , ... &j(100 + 35) ^^, ,„^- 420 = — - — ^ , and 420 = ^^ . .-. 756 = 125&i, or b K ft = -&j and 54-^1 =0^1, the value of the horses. Now h^, foupd O o Q from the 2nd, is 336, and as ~b^ — 840 = the loss, we have 896—840 o = 56 = the loss. Those who are not familiar with algebraic reasoning will prefer to find the values of J and h^ from the two formulas, and add them together. 6. A man sold 72 turkeys, which was 32% of the number he had remaining. How many had he at first ? Ans.y 297. 7. A farmer saved annually $145^, which was 33^% of his annual income. Required his income ? Ans.y $436^. 8. A merchant having 400 barrels of cider, sold at one time 4:5% of it ; at another time 20^ of the remainder. How many barrels did he sell in all ? Jns., 224 hbL rb 45x400 ... Operation. p ==TK?i= ^^^ = 180. 20x220 _ ~ioo~- **• •'• p+Pt=224:. 9. A housekeeper gave to her neighbor -J of a pound of tea, and had } of a pound remaining. What per cent, of her tea had she remaining? Ans., 85-Jf-^. « .. 3 rX ^ 7r 600 „^, Operation. 4 = 1^0' ' = lOO* '• " = ^ = ''^- 10. John has ^ of a dollar, and Henry has | of a dollar. What per cent, of John's money equals Henry's ? What of Henry's equals John's ? Ans., 120^, 83-J^. Operation. l = f^^« = w '■'■ = ''°- 1 ri ^ Sr 250 .„, [Note. — For other examples in percentage, see Olney's Practical Arithmetic] p 100 Pi 100 1I€TM>M IL SIMPLE INTEREST AND COMMON DISCOUNT. [Note. — For definitions see Practical Arithmetic] 86. Prob. 1. -To express the relation between princi- pal, rate per cent., time, and interest. Solution.— Let />, /■, t, and i represent respectively the principal, rate per cent, time and interest ; t being in the denomination for which the rate per cent, is estimated. Thus, if the rate per cent, is rate per cent, per year, t is to be underetood as years ; if the rate per cent, is per month, t is months, etc. Then as p is the base, the percentage for a unit of time is -^ (§4) ; and for t units of time it is —^ . .-. i = -^ . 87. Prob. 2.— To express the relation between amount, principal, rate per cent., and time. Solution. — Since the amount is the sum of principal and interest, representing t^e amount by o, we have a=p-\-i. But i= j^; 100 trp 100 + tr 100a hence « = />+-—, or a =p-——. r. p = 100' ^ 100 * " ^ 100 + tr 88, Scholium- — This problem embraces the common rule for Discount (see Arithmetic, 383). The pupil should be careful to understand the reasonableness of discount. For example, if I hold a note against Mr. B., which note is payable at any future time, if the note is drawing interest for all money is worth, tlie Present Worth of the note at the time it was given was its face. If the rate at which it is drawing interest is less than money is worth (or if it draws no interest) the Present Worth at its date is less than the face of the note. If it draws 7io interest, its Present Worth at any time is less than its face. Finally, if the rate which the note draws is greater than the market rate, the Present Worth of the note at its date is more than its face. Scholium.— The two formtdft (1) i = r^ and 100 + tr i2)a = p + i=p-^-, 318 BUSINESS RULES. are sufficient to solve all jjroblems in Simple Interest and Common Dis- count. EXAMPLES. Ex. 1. What is the interest on $250 for 1 yr. 10 mo. 15 da., at Q% per annum. Ans., $28.12|. Solution. — In this example I am to consider principal, time, rate per cent, and interest, the latter of which is the unknown quantity. Formula (1) expresses the relation between these quanti- ties. The rate per cent, being pet- anjium, the time must be in years. 10 5 15 days = |^ = .5 of a month. 10.5 months = — ^ =.875 of a year. .'. 1 yr. 10 mo. 15 da. = 1.875 years. Now (1) gives i =zrJL a 6 10" _ 1.875 x 0x,.._ - 100 - ^^•^^*- 2. What is the interest on $47.25 for 1 yr. and 6 mo., at %% per annum? Ans,, $4.2525. .3 3 S^.4^ Operation, i = j^ = 4.2525* 4 3. What is the interest on $145.50 for 1 yr. 9 mo. 24 da., at 6% per annum? Ans., $15.86 nearly. . 1.816x6x145.50 _ ^^ Operation, t = .^ = 15.86 nearly. 4. What is the interest on $123.75 for 2 yr. 8 mo. 12 da., at 6% per annum ? Ans., $20.0475. 6. W^hat is the interest on $475 for 2 yr. 7 mo. and 20 da., at Q% per annum ? Ans., $75.208 J. 6. What is the interest on $340.60 for 4 yr. and 5 mo., at 6^ per annum ? - Ans., $90,259. 7. What is the interest on $50.40 for 1 yr. and 10 mo., at "1% per annum? Am., $6,468, SIMPLE INTEREST AND Ct)MMON DISCOUNT. 319 Suggestion. — In solving this example the operation upon paper consists simply in multiplying 12.833 by .504. The pupil should always reduce the written details of his arithmetical work to the minimum. Thus, in this case, he sees mentally that 10 mo. are .833 of a year. Hence i = ^^J^^^^^""-! , But 1.833 X 7 he produces mentally, and writes 12.833. And 12.833 cance'lling the 100 from 50.40 makes it .504. Hence .504 the only written work should be as in the margin. 51332 In practice, nothing but this multiplication should be 64165 written down. 6^467832 Solve the following by thus reducing the written work to a minimum. The answers are given as in practice in business. 8. What is the interest on $49.80 for 2 yr. and 11 mo., at 1% per annum f Ans.^ $10.17. operation, i = 1?:!^^£H = 2.9166 x 7 x .498 = 10.17. 9. What is the interest on 195.40 for 3 yr. 9 mo., at S% per annum ? . 95.40x8x3.75 ^_ .„ _, ^ v, ,^ Operation, i = Ym\ = 28 62. The operation should be performed mentally. Thus 8 x 3.75 = 30. Dropping the 0, and one from the denominator, and for the 10 remaining in the denomi- nator removing the decimal point in 95.40 so as to make it 9.54, we have simply to multiply 9.54 by 3. All of this should be done at a glance, without writing more than is given above. 10. What is the interest on $196 for 5 yr. 7 mo., at 9^ per annum 9 Arts., $98.49. 11. What is the interest on $471.11 for 4 yr. 8 mo., at 1^% per annum f Ans., $164.89. 12. What is the interest on $18.60 for 3 mo. 12 da., at 3^ per mo.? Ans., $1.90. 3.4 X 3 X 18.60 , ^ „ „^^ «..« ^ , . «« Operation, i = ^ =1.7 x 3 x .372=.872 x 6.1 = 1.90. 13. What is the interest on $400 for $150 days, at %i% per month f Ans., $50. 320 BUSINESS EULES. Operation. i = j^^ = 50 14. What is the interest on $1,000 for 2 tyw. 12 da., at l\% per month? Ans., $36. 15. What is the amount of $432.10 for 5 yr. 4 mo. 24 da., at 1% per annum 9 Ans., $595.43. Operation, a = p^-^ = 432.10 '-^^ = 4.321 x 137.8 = 595.4338. 16. What is the amount of $325.25 for 2 yr. 9 rno. 12 da., at 6i% per annum ? Ans., $384.09. „«~ .► 100 + 2.784x6.5 „^, ^^ Operation, a = 325.25 — -^-^ = 384.09. 17. In what time will $13, at 6% per annum, give $0,975 interest ? Ans., 1 yr. 3 mo. Solution. — Since principal, rate per cent., time and interest are compared i = -r^ gives the relation. As time is required, I solve this equation for ^, and have t = . Substituting the given values, .025 6 .BkT«. 18. In what time will $45.25, at 6% per annum, give $1.81 interest ? Ans., 8 mo. 19. In what time will $70.50, at 9^ per annum, give $31,725 interest? Ans., 5 yr. 20. In what time will $140, at 7^ per annum,, give $10.861| interest? Ans., 1 yr. 1 mo. 9 da. SIMPLE INTEREST AND COMMON DISCOUNT. 321 21. In what time will $48.50, at ^% per annnmy amount to $56.187i ? Ans.y 2 yr, 7 mo. 21 da. Suggestion. — Subtract the principal from the amount to find the interest. 22. In what time will $248, at Q% per anuuni, amount to $282,224 ? Ans., 2 yr, 3 mo. IS da. 23. In what time will $700, at d%per ammm, amount to $712.35? Am., 2 mo. 10.5+ da. 24. At what per cent, will $325 produce $3.25 interest in 2 months? Arts., 6%. iff) Solution. — Same as above, finding r from the equation i — — J-, _ -lOOi _ 100 X 3.25 _ ~ <;> ~" |x325 ~ 25. At what per cent, will $40 produce $13.36 interest in 2 yr. 9 mo. 12 da. ? Ans., \%%. 26. At what per cent, will $125 produce $32,375 interest in 3 yr. 6 mo. ? Arts., 7|^. 27. At what per cent, will $124 produce $29. 17^ interest in 4 yr. 3 mo. 10 da. ? Ans., h\%. 28. At what per cent, will $2,360.25 amount to $2,470,395 iu 7 months ? Ans.^ %%. Suggestion.— Find the interest and then proceed as before. 29. At what per cent, will $230 amount to $249.83f in 11 mo. 15 da. ? Ans., 9%. 30. What princi}>al will in 3 yr. 8 mo. 15 da., at 6% per annitm, give $76,095 interest ? A ns., $342. BUSINESS RULES. Solution.— Solving i = -^ for^, I have _ im _ 100 X 76.095 _ 100 x 76.095 _ ^ ~ IT ~~ 6 X 3.7tV ~ 23.25 ~ 31. What principal will in 4 yr. 9 mo. 18 da., at 9^ jpgr annum, give 165.016 interest? ^ws., $150.50. 32. What principal will in 8 yr. 8 mo. 12 da., at 5% /?er amium, give $147.9435 interest.^ A7is., $340.10. 33. How long will it take for $200, at simple interest, at Q% per annum, to amount to $500 ? Ann., 25 years. Solution. — I have under consideration^, r, a, and t, the latter of which is the unknown quantity. The relation between these is a= p — T?m~ • Solving this for t, I find ^ _ 100( 120. .-. w > 120, which would make z negative. 12. I have two kinds of molasses which cost me 20 and 30 cents per gallon ; I wish to fill a hogshead, that will hold 80 gallons, with these two kinds. How much of each kind must be taken, that I may sell a gallon of the mixture at 25 cents per gallon and make 10 per cent, on my purchase? Ans., 58^ of 20 cents, and 21^^^ of 30 cents. 13. A lumber merchant has several qualities of boards ; and it is required to ascertain how many, at $10 and $15 per thousand feet, each, shall be sold on an order for 60 thousand feet, that the price for both qualities shall be $1 2 per thousand feet. Ans., 36 thousand at $10, and 24 thousand at $15. 14. How many ounces of gold 23 carats fine, and how many 20 carats fine, must be compounded with 8 ounces 18 carats fine, that the alloy of the three different qualities may be 22 carats fine ? Ans., 48 oz. of the 1st, and 8 oz. of the 2d. [Note. — These applications might be extended to much greater length, did space permit. The equation renders important aid in many problems in Compound Interest, but their discussion usually requires a knowledge of Quadratics, and some of them of Loga^ rithms. They must, therefore, be reserved for the future.] 335 UA D R ATI C ^ :^ aUATIONS p i^ LM>TEI\, IV< fSECJ lO¥i I. PURE QUADRATICS. 94. A Quadratic Equation is an Equation of the sec- ond degree (6, 8). Oo, Quadratic Equations are distinguished as Pure (called also Incomplete), and Affected (called also Complete.) 90. A Pure Quadratic Equation is an equation which contains no power of the unknown quantity but the second ; as ax^-{-b = cd, x^— 3b = 102. 97. An Affected Quadratic Equation is an equation which contains terms of the second degree and also of the first, with respect to the unknown quantity or quantities; as a:^—4^= 12, ^xy—x—y"^ = 16«, mxy + y = b. 98. A Root of an equation is a quantity which sub- stituted for the unknown quantity satisfies the equation. Note the difference between this use of the word root, and its former use as defined in Art. 39, Here we speak of *' the root of an equation," meaning the value of the unknown quantity ; in the former sense we speak of *'the root of a number," meaning one of its equal factors. 99. Prob.— To solve a Pure Quadratic Equation. Rule. — /. Transpose all the terms containing the unknown quantity into the first meniber, and unite fheTJi into one, clearing of fractions if vccrsanry, 336 QUADRATIC EQUATIONS. II. Transpose the known terms into the second inemher. III. Divide by the coefficient of the unknown quan- tity, and extract the square root of each member. Demonstration. — According to the definition of a Pure Quadratic, all the terms containing the unknown quantity contain its square. Hence they can be transposed and united into one by atlding with reference to the square of the unknown quantity (16). Extracting the square root of the first member gives the first power of the unknown quantity, i. e., the quantity itself. And taking the square root of each member does not destroy the equation, since like roots of equal quantities are equal. 100, Cor. 1. — Every Pure Quadratic Equation has two roots 7imnerically equal hut with opposite signs. This is apparent since every such equation, as the process of solution shows, can be reduced to the form x^ = a {a representing any quantity whatever). Whence, extracting the root, we have X = ± ^a \ as the square root of a quantity is both + and — (203). Scholium. — The question naturally arises, Why not put the ambigu- ous sign (the ±) before the a-, as well as before the second member? It is proper to ; but there is no advantage gained by it. Thus, if we write ±x = i^/^, we have -^-x — ±'\/a, or — .f = ± y^a. But the former is a? = ± -v/a, and the latter becomes so by changing the signs of both members. So that all we learn in either case, is that ar = + "V^a, and x = — ya. 101. Cor. 2. — The roots of a Pure Quadratic Equation may loth he imaginary, and both ivill he if one is. For if after having transposed and reduced to the form x^=za, the second member is negative, as x^=^ —«, extracting the square root gives x=-\-^/ —a, and x=-~^/ —a, both imaginary. EXAMPLES. Ex. 1. Given 3r>«- -10— .^2=12 + 4^:2-54 to find the value of a:. Model Solution. Operation. zx'-n-x' ^^n+^x'-u, (2) _2.r^ = -32, (3) «'' = 16, 0? = ±4, PURE QUADRATICS. 337 Explanation. — Transposing and uniting terms, I liave —2x*.— —32. Dividing each member by —2, I have x' = 16. Extracting the square root of each member, I find a; = ±4 (read, ''x equals plus and minus 4 "). Veriflcation. — Substituting +4 for x, the equation becomes 48 — 10-16 = 12 + 64—54, or 22 = 22. Substituting —4 gives just the same since —4 squared, (—4)'^, is the same as +4 squared. 2. Given a^-\-l = ~-\-4:,to find x. Roots, x= ±2. 4 3. Given -^^-^ — - = — ^ — , to find x. 10 o Roots, x= ±3. 5 5 8 4. Given 1- -; = - , to find the values of x. 4 + a: 4:—x 3 Roots, X = ±1, 6. Given a^—ah = d, to find the values of x. Roots, x= ^^d-\-db. 6. Given |^_3 + ^ = ^-;c2+^, to find the values of a;. Roots, x— ±3, 7. Given 13— v3ar5+16 = 5, to find the values of x. Roots, a; = ±4. 8. Given x -f- Vx^ + « = —, , to find the values of x. Roots, X = iiVSa. 15 9. Given --=^ =x-\-w^-\-^. Roots, x z= ±2, 10. Given V^+ri^-^i^^. Roots. x^^/2^{^h)^^. 338 QUADBATIC EQUATIONS. 11. GiTen = ax. Roots, x= + — — 12. Given ^-^=?^^-». X XX Boots, x= ±(Vm—Vn), 13. Giyen^,--+1 = --1 + -. Roots, X = ±\/a^—^, 14. Given 12— rc^ i^x^:: 100 : 25. Boots, x=z ±2. Operation. 12— x^ : ^a;' : 100 : 25 12-x^ : «" : : 50 : 25 : : 2 : 1 13:rB»::3:l 4:35" :: 1: 1 .: x = ±2. [Note.— Use the principles of proportion in solving these.] 16. Given j^x^-^-^x^S : ^x^-.:^x^^s : : 9 : 3. Boots, X = :t 4'v/2. 16. Given i{x^—5f : x^—6 : : 2 : 1. Boots, x = ±S. Operation. i(x^-5y : a;'-5 : : 2 : 1 «'— 5 : 1 : : 4 : 1 «'~5:5 ::4:5 «» : 5 : : 9 : 5 .-. aj = ±3. 17. Given |(11 +a;2) : \{4x^—2) : : 5 : 2. Boots, a; = ±2. 18. Given a;2+ 4 : a;2— 11 : : 100 : 40. Boots, X = ± \/2T. 19. Given ^±^ = 2 - ^+^. Roots, x = ±2. 20. Given ^—^ + -^-^^ = _. Boots, x = ±8. 21. Given (a;4-2)2 = 4:X-{-6. Boots, x = ±1, 22. Given i{x'i-12) = ix^-1. Boots, x = ±6, 23. Given ^ + ^ ^"^ ^ x^—7x x^ + Hx x^—7d Boots, X = ±9, PURE QUADRATICS. 339 24. Given a; + V^a/2— 4a; = 1. Roots, x = ±f 25. Given Va + x = Vx+Vo^-^-^. Roots, x= ±Va^—i^. 26. Given Vl + x^-h^l+a^+Vl — 3^= Vl—x^. Verify. Roots, x = ±|\/^^. 27. Given ^¥T^ = Vo^T^. Roots, x= ±^VHl^-a% 28. Given*— VaHa^^ Roots, X = ±-^^3-^^. 29. Given \ = *. Roots, x z=z ±- — -. 30. Given X^^±^±l = 3. Roots, x = ±2. 31. Given vW l Vg ^ ^ ^ ^^^^ ^ ^ ±iV5. 32. Given 7=^ H f=^ = x- x^\/2—xi x-V2-x^ Roots, X = ± Vs. 1 la 33. Given a— Va^—3^ a + Va^ -a:^ ^ Roots, X = ±iaV3. «>. r^- Va^+x^-\-x b „ ^ , a(b—c) 34. Given \ __ ---- = -• Roots, x = ± -— 7^- 340 QUADRATIC EQUATIONS. 35. Giyen — =L + —== — = ^ • Boots, X = ± jVs. 36. Given = a — l + x-i-Vl-i-x^ l—x-\-Vl+x^ Roots, x= ±V{a—2y—l, 37. Given ; H ; = ax. x-\-V2—a^ x—V2-^x^ Roots ■ A + 1 38. Given - — + -!— = a. Vl— a;+l vl + ^— 1 Roots. .-±V'4- APPLICATIONS. Ex. 1. What two numbers are those whose sum is to the greater as 10 : 7, and whose sum multiplied by the less pro- duces 270? Ans., 21 and 9. Suggestion. — Let 10a? = the sum of the numbers, and 7a? the greater. Scholium. — It is customary to omit the negative roots in giving answers to examples, the nature of which renders such answers impossible. In this case the question is about pure number, and hence the answers should be given without signs. 2. There are two numbers whose ratio is that of 4 to 5, and the difference of whose squares is 81. What are the numbers? Ans., 12 and 15. 3. What two numbers are those whose difference is to the greater as 2 to 9, and the difference of whose squares is 128 ? Verify. PURE QUADRATICS. 341 4. Find three numbers which bear the same ratio to each other as J, |, and J do to each other, and the sum of whose squares is 724. Numbers, 12, 16, 18. 6. Find three numbers in the ratio of m, u, and p, the sum of whose squares is equal to a. Numbers, ±\/ » , ^ , » , ±\/ . . ^t . . , and V rn^ + n pi 6. Divide 14 into two parts so that the greater part divid- ed by the less shall be to the less divided by the greater as 16 to 9. Suggestion. — Having — — : : : 16 : 9, it follows that 14 — X X x" : {U~xy : : 16 : 9, and x : 14-x : : 4 : 3, and a; : 14 : : 4 : 7. .'. x = 8, and 14— x = 6. 7. Divide a into two parts so that the greater part divided by the less shall be to the less divided by the greater as w to n. Parts, -—= ;=, and Scholium. — Example 7 is example 6 generalized. Tl^ pupil should deduce the results in the former from these. Thus, sub- stftuting a = 14, m = 16, and n = 9, _/ — —^= 8, etc. 8. What two numbers are they, whose product is 126, and the quotient of the greater divided by the less, S^ ? Gener- alize this. Ans., 6 and 21. 9. The sum of the squares of two numbers is 370, and the difference of their squares 208. Required the numbers. Numbers, 9 and 17. 342 QUADEATIC EQUATIONS. 10. Generalize the 9th, and show that Jv2(5 f^) and ^V'^(s—d) are general results. 11. For comparatively small distances above the earth's surface the distances through which bodies fall under the influence of gravity are as the squares of the times. Thus, if one body is falling 2 seconds and another 3, the distances fallen through are as 4:9. A body falls 4 times as far in 2 seconds as in 1, and 9 times as far in 3 seconds. These facts are learned both by observation and theoretically. It is also observed that a body falls IGy^^^ feet in one second. How long is a body in f alhng 500 feet ? One mile (5280 ft.) ? Five miles ? Ans., To fall 500 ft. requires 5.58 seconds. To fall 5 miles requires 40.51 seconds. 12. A and B lay out some money in a speculation. A disposes of his bargain for 111, and gains as much per cent, as B lays out. B succeeds in gaining $36 ; and it appears that A gains four times as much per cent, as B. Eequired the capital of each. Results^ $5 = A's capital, and $120 = B's. 13. A money safe contains a certain number of drawers. In each drawer there are as many divisions as there are drawers, and in each division there are four times as many dollars as there are drawers. The whole sum in the safe is $5,324; what is the number of drawers? Afis., 11. 14. Two travelers, A and B, set out to meet each other; A leaving the town C at the same time that B left D. They travelled the direct road from C to D, and on meeting it appeared that A had travelled 18 miles more than B; and that A could have gone B's journey in 15| days, but B would have been 28 days in performing A's journey. What is the distance between C and D ? Ans., 126 miles. C M D '- 1 . '• ^ 1 If x = CM = the distance A travelled, then «— 18 = MD = the dis- PURE QUADRATICS. 343 tance B travelled. = distance A travelled a day ; and — 15^ 28 = distance B travelled a day. Notice that the times are equal. 15. From two places at an unknown distance, two bodies, A and B, move toward each other till they meet, A going a miles more than B. A would have described B's dis- tance in n hours, and B would have described A's distance in m hours. What was the distance of the two places from each other ? ^/m-\- ^/n Ans., a X -pr yr- 16. A and B engaged to work for a certain number of days. A lost 4 days of the time and received $18.75. B lost 7 days and received $12. Now had A lost 7 and B 4 days, the amounts received would have been equal. How long did they engage to work and at what rates ? Ans., Whole time, 19 days. Suggestion. — ^Ifa;= the whole time, what represents A's daily wages ? What B's ? After the equation is formed, see if you can- not strike out a numerical factor from both members, and extract the root without expanding. 17. A vintner drew a certain quantity of wine out of a full vessel that held 256 gallons ; and then filled the vessel with water, and drew off the same number of gallons as before, and so on for four draughts, when there were only 81 gallons of pure wine left. How much wine did he draw each time ? Am., 64, 48, 36, and 27 gallons. Suggestion. — If he drew out - part of the contents of the cask x—l each time, there remained after the first drawing th of the x—l X 1 (x—iy wine ; after the second x or —, and after the fourth X X Q? ' {x-\y (x-i)* ,, x-\ « „„ 11 S^- ••-V-=^3^^'«'-^=f Whence - = ^. S41 QUADRATIC EQUATIONS. 18. A number a is diminished by the nth. part of itself, this remainder is diminished by the nth part of itself, and so on to the fourth remainder, which is equal to h. Required the value of n. ^^ A71S., - 19. There is a number such that, if the square root of three times its square + 4, be taken, the quotient of this root increased by 2, divided by the root diminished by 2, is 3. What is the number ? Query. — Which of the equations in the preceding j^art of this section does this give rise to ? 20. If the square root of the difference between the square of a certain number and 2, be both added to and subtracted from the number itself, the sum of the recip- rocals of the result is ^ of the number itself. What is the number ? Query. — Which of the equations in the preceding part of this section does this give rise to ? With what modifications ? AFFE f CTED, J^^TM>N II. 102. An AfiFected Quadratic equation is an equation which contains terms of the second degree and also of the first with respect to the unknown quantity. AFFECTED QUADRATICS. 345 x^-dx = 12, ^ + Sax^ = ??£+^^ ^^^^ cM __ ^^^ ^ g^ 5 0^ — 0, are affected quadratic equations. 103. Prob.— To solve an Affected Quadratic Equation. Rule. — /. Reduce the equation to the form x^+ax //. Add the square of half the coefficients of the second term to each mernher of the equation. III. Extract the square root of each member, thus producing a simple equation from, which the value of the unhnown quantity is found by simple transposition. Demonstration. — By definition an Affected Quadratic Equation contains l>ut tbree kinds of terms, viz.: terms containing the square of the unknown quantity, terms containing the first power of the unknown quantity, and known terms. Each of the three kinds of terms may, by clearing of fractions, transposition, and uniting, as the particular example may require, be united into one, and the results arranged in the order given. If, then, the first term, i. e. the one containing the square of the unknown quantity, has a coeffi- cient other than unity, or is negative, its coefficient can be rendered unity and positive without destroying the equation by dividing both the members by whatever coefficient this term may have after the first reductions. The equation will then take the form T}±ax= ±h. Now adding (-) to the first member makes it a perfect square (the squareof a;± -), since a trinomial is a perfect square when one of its terms (the middle one, a«, in this case) is ± twice the product of the square roots of the other two, these two being both positive (123, Part I). But if we add the square of half the coefficient of tlie second term to the first member to make * The characteriBticB of this form are, that the first member consists of two terms, the first of which is positive and contains simply the square of the unknown quantity, its coefficient being unity, while the second has the first power of the unknown quantity, with any coefficient (a) positive or negative, integral or frac- tional ; and the second member consists of known terms (b). 346 QUADRATIC EQUATION'S. it a complete square, we must add it to the second member to pre- serve the equality of the members. Having extracted the square root of each member, these roots are equal, since like roots of equals are equal. Now, since the first term of the trinomial square is x\ and the last i — J does not contain x, its square root is a binomial consisting of x ± the square root of its third term, or half the co- eflBcient of the middle term, and hence a known quantity. The square root of the second member can be taken exactly, approxi- mately, or indicated, as the case may be. Finally, as the first term of this resulting equation is simply the unknown quantity, its value is found by transposing the second term. EXAMPLES. X^ x^ Ex. 1. Given 6^x—^ = x—2l-\--, to find the value of o o X, and verify. Model Solution— Operation. 48_8a;-a!' = 8aj-184-«' -2x''-16x = -66 x'' + Sx = dS a;'' + 8a;+16 = 33 + 16 = 49 X + 4: = ±7, a? = ± 7—4 = 3, and —11 Explanation.— Clearing the equation of fractions, transposing and uniting, and dividing each member by —2, I have x^ + Sx = 33. Now since Sx contains the square root of a?" as one of its factors, the other factor, 8, is twice the square root of the other term of a trino- mial square (123, Part I). Hence ^ of 8 squared (16) is the third term. Adding this term I have x^ + Sx + 16, which is a perfect square. But as I have added 16 to the first member to make it a perfect square, I must add it to the second member to preserve the equality. This gives a;- + 8a; + 16 = 49. Extracting the square root of both members, I have a: + 4 = ±7. Finally, transposing the 4, 1 have x = —4 ±7, i. 6. a? = 3 and —11. Both are correct. Hence there are two roots (values ofx) of this equation, 3 and —11. AFFECTED QUADRATICS. 347 Verlflcatlon. — To verify the value a; = 3, 1 substitute 3 for a; in the given equation, and have 6-3— | = 3— 2J+|, or 3— | = |4-|, or ys. ^ y. To verify the value j^ — ll,l8ubBtitutefor«, — 11, in the given equation, and have 6 + 11— J^f^ = -ll-2| + if-i-, or 17— »-fL = -131 + ifi,orV=¥- 2. Given x^—Sx-\-5 = 14, to find the values of x, and verify. Result, re = 9, and —1. 3. Find the roots of the equation a:2_i2a:-f 30 = 3, and verify. Boots, 9 and 3. 4^; 9 4. Find the roots of a:— 2 = • X Suggestion.— This reduces to x'—Qx = - 9. Whence, completing the square, a:'-6a; + 9 = 9-9 = 0, and a:-3 = 0, or x = 3. In this case it appears that the equation has but one root. 6. Giyen i = ^^^—-^ , to find the values of x. 5 X 6. Find the roots of 3(a:— 4) = — i^^J. ^ X Suggestion.- This reduces to x'—^x = — 20. Whence, completing the square, a;'— 8./ + 16 = 16—20 = —4, and extracting the root, x— 4 = ± 2^/— li or « = 4 ± 2\/~^, two imaginary roots. 7. Find the values of a; in - = — -z* 5 X — Results, X = 5 + 2\/— 5, and 5— 2a/^. 8. Find the roots of x^^^ + n^ = 3— J5?. X X Roots, —30, and —40. 9. Find the roots of i(?|±l) = ?_a:. "Zx X Only one root, —2. 348 QUADRATIC EQUATIOXS. 10. Find the roots of 7(a:+7) + ^^^?^— ^ = 0. Roots, 5(— 1 + a/— 1), and 5(— 1 — V— 1), or as the same may be written, —5(1 — a/— 1), and — 5(1 -f a/— 1). 11. Find the roots of ^x^—^x + %^ = 42|, and verify. Boots, 7, and —6 J. Scholium. — This process of adding the square of half the co- efficient of the first power of the unkno^vn quantit}^ to the first member, in order to make it a perfect square, is called Completlng THE Square. There are a variety of other ways of completing the square of an afifected quadratic, some of which will be given as we l^roceed ; but this is the most important. This method will solve all cases : others are mere matters of convenience, in special cases. 12. Given x^—x-\-3 = 45, to find its roots, and verify. 13. Given 2ic2-|-8a;— 20 = 70, to find its roots, and verify. 14. Given 5 — = Id^x — - , to find its roots, and verify. 25 ^/^ 15. Given 6x-\ = 44, to find the values of x. X Suggestion. — Clearing of fractions, Qx^ + ^S—'Sx = 44a;, Transposing and uniting, Qx'^—4:7x = —35, Dividing by 6, x^—^x = — ^, Completing the square, x^— Ya; + (||)' = {iW—^ = tW^» Extracting the root, x—^ = ± ||, Transposing, a? = ff ± f|^ = 7, and |. 16. Find the roots of bx = 2x-\ — Suggestion. — Notice the compound negative term. Cleared of fractions and reduced, the equation becomes x'^—Sx = 4. .*. « = 4, and ~1. 17. Find the roots of ^-^ — = 3. X 4:X'^ Suggestion. — Multiply by 4x^ to clear effractions. AFFE(TKD QUADRATICS. 349 18. Find the roots of Sxi—20x—6'Z = 7:i'— 2.^2 + 100. Boots, 9, and — 3|. 19. Find the roots of 2a;— 2 = 2 + -. X Roots, 3, and —1. 20. Find the roots of ix^^^x + l = 8-|. Roots, I, and — |. 21. Find the values of x in the equation — —^ +5 ^^^ Result, X = 5|, and 5. a; + 10 22. Find the values of x in the equation — -^ — — = 6. Result, X = 10, and — f. 23. Giveni(a;-f4)-^ =4(4a;H-7)-l, to find the X-~~o values of x. Restdt, x = 21, and 5. 24. Given '"^^ + — f^ = 5^, to find the roots, and X X-f- L4 verify. 25. Find the roots of ^(8— a;)— ^"" = i(^ — 2), and X'^o verify. , ,2a?+9 4a;— 3 ^ . 3a;— 16 , 26. Find the roots of — ^ + j— 3 = 3 + --^—, and verify. 27. Find the values of x in the equation 3a;2_2aa; = b. J, ;, a±Va'-^Sb Result, X = -^ • 350 QUADRATIC EQUATIONS. Operation. Sx^—2ax = h; Dividing by 3, x'-^x = l, o 3 Completing the square, «— ^« + (q) =^ + - = , 3 \3/ 9 3 9 Extracting the square root, x—- = ± -^a^ + 3&, o 3 m . a a/o^ + 36 a± a/«' + 3J Transposing, x = - ± -^- — -- — , or ^ — ——. Scholium.— The form — ^^ signifies that there are two o values of «; i. e., that each of the signs + and — may be used. Thus the values in this case are X = ^ , and X = ^ 3 ' 3 28. Find the roots of Sx^-\-6ax = m. Roots, == — o 29. Find the roots of ^a^¥x^—Qa%^x = ¥. Roots, ZaW X (I 2 30. Find the yalues of x in the equation - + - = -. a X a Result, X = 1±a/1— ^. 31. Find the roots of ^:Z^ = t Sa—2x 4 Roots, ia, and ^a. 104, Cor. 1. — An affected quadratic equation has tivo roots. These roots may hoth he positive, loth he negative, or one positive and the other negative. They are hoth real, or hoth imaginary. Demonstration. — Let x'^-\-px — qhe any affected quadratic equa- tion reduced to the form for completing the square. lu this form p and q may be either positive or negative, integral or fractional, AFFECTED QUADRATICS. 351 Solving this equation we have a? = — | ± a/^ -\-q. We will now observe what diflFerent fonns this expression can take, depending up(jn the signs and relative values of y> and q. Ist. Wh^i p and q are both positive. The tfigns will then stand as given ; i. e., x = —^± a/ ^^ +q. Now, it is evident that i/ i" + 2 / a >^, for 4/^+2 is the square root of something more than ^. Hence, as ^ < y— +J, — | + y- + 5' is positive; but -^ — i/ J- +q ia negative, for both parts are negative. Moreover the negative root is numerically greater than the positive, since the former is the numerical sum of the two parts, and the latter the numerical difference. .*. When p and q are both + in the given Ibrm, one root is positive and the other negative, and the negative root is numerically greater than the positive one. See Example 1, above. 3nd. When p is negative and q positive. We then have x= ^ radical, x is positive ; but if we take the — sign, a; is negative, since /? + q>~. Moreover, the positive root is numerically the greater. . *. When p is negative and q positive, one root is positi ve and the other negative ; but the positive root is numerically greater than the negative. See Example 2, above. 8rd. When p avd q are hoth negative. We then have a? = ^ - \/^=f^ (-«) = f - /:--?• ^ «>» iff' > Vr - * '« real, and as it is less than ^, both values are positive. See Ex. 8. If ^ = J, 4/^ — q — 0, and there is but one value of «. and this is posi- tive. (It is customary to caU this two equal positive roots, for the sake of analogy, and for other reasons which cannot now be appre- 352 QUADRATIC EQUATIO]SrS. dated by the pupil.) See Examples 4 and 5. If ?• < ?, i/ j ~ 2 becomes the square root of a negative quantity and hence imaginary. See Examples 6 and 7. 4th. When p is positive aiid q negative. We then have x = — ~ ± |/ ^ — q. As before, this gives two real roots when ? < j- • When this is the case both roots are negative. [Let the pupil show how this is seen.] When 2 = V , the roots are equal and negative ; **. e. there is but one. When y < q both roots are imaginary. See Examples 8, 9, and 10. [Note. — It is not important that the pupil remember all these forms; but it is an excellent exercise to give the discussion. The ingenious student can put the results in a very neat analytical table.] Scholium. — It may be asked why, when we extract the square p^ p^ root of each member of the equation x'^+px + j- = q+^ , we write the ambiguous sign only before the root of the second member. The reason is the same as given under the Pure Quadratics, Art. 100, Scholium. Thus, in strict propriety, the square root of each mem- ber of this equation being taken or indicated gives ± l^+f ) = "^ \/ j-^q- But take these signs in any order we can, it amounts to taking the roots as having like signs (both + , or both — ) or unlike signs (one + and the other — ). Hence it is sufficient to give the ambiguous sign to one member only ; and it is most convenient to give it to the second. 32. Given V^ + 5 x ^/x^\^l'^=l'^, to find the values of x. Result y a: = 4, and —21. Suggestion. — First clear the equation of radicals. 33. Given A/4a;+5x\/7a; + l = 30, to find the values of X. Vahies, x = 6, and — 6^. 34. Given x-\-6 = \/x-\-6-\-6, to find the values of a: VahteSf x = 4, and —1. AFFECTED QUADRATICS. 363 35. Given x-\-16—7Vx-\-W=10—Wx-\-16, to find the values of x. Vaiucs, ar = 9, and —12. Suggestion.— Put the equation in the form x+6 = ^^x+lQ, and then square. 36. Given 3Vx+6-^2 = x-^Vx-^(), to find the vahic-- of X. Result, X = 10, and —2. 1 4 37. Given y-^- = -— :, to find the roots. ResuUy x= VS, iV3. 4 Suggestion. — Multiply by y and transpose, and y' :ry = — 1. 4 4 4 1 Completing the square, y' ~y+~ = -_l = Extracting the and /\/| = , Cor. 2. — An affected quadratic being reduced to the form x2 -f px = 4, i\\e. value of x is half the coefficient of the second term taken ivith the opposite sign, ± the square root of the sum of the square of this half coefficient, and the knoum term of the equation. This is observed directly from the form X = — ^ iA / j- + q, dnd more in detail in the demonstration of the preceding corollaiy. [Note. — The pupil should use this method in practice, but be careful that the complete method and its full demonstration is not lost sight of.] 38. Write out the roots of the following without going through the operations of completing the square, etc.; z^-{-4:Z = eO; y2_4y — 60; x^-f 16a- = — 60; .^2 — 16a: = -60. 39. Reduce the following to the form x^-{-px = q, and then write out the roots as above : 354 QUADEATIC EQUATIONS. Sx^-i-2x-\-6 = 11, gives x^ + ^x = J. Whence a; = — J ± V4+1 = -i±i = h and -|. —- + -—= —4, gives x^-\-~x =z — — . Whence x = 10 /lOO 64 -10±6 ,, ^ ^, -■3" ± V -9- - y = —3— =" -^*' ^^^ -"*• 4a; = 14, givesa;2— -2; = 7. Whence x = - y^g + 7=:i-^ = 4,and-lf. 9 ±23 8 40. Given 9x-^—12x-^ = —3, to find the values of x. ?, X = 3, and 1. Suggestion. 9ar-'' = — X 1 . 1 y + - 1 + - 41. Given -| + 1 = ^, to find the roots. ^ y y Suggestion. — Reduce the complex fractions to simple ones by multiplying numerator and denominator by y. Whence ^ — - -{-- — y -1 2/-1 = '■^. Multiply by 4(y'-l), and 4yV4 + 4y^ + 8y + 4 = ISy'-ld, .'. y = 3, and -\. 12a 42. Given \/a-\-x-j-V(t—x = — , to find x. 5'\/a-j-x Result, x = —, and — • o 43. GivenV^4-«— V^+^ = '\/2x, to find the roots. Result, x= - ^±- _}_ jV2a2 4-2Z^. Suggestion, \/x + a—\/2x=:^/x+b. x+a—2^2x^^2ax-\-2x - x+i: -2^JW^'2a^ = J)—a-2x. ^x^ + ^a^ = y^-^^db—^bx + a' + ^ax + 4:X^. AFFECTED QUADRATICS. 356 44. Given -: -—(a^—b^)x = -. r, to find a^ + b^ (a^)-*+(fl2Z>)-i the values of x. Result, x = a, and —b. Suggestion. -. t=—, r. Whence the given equa- tion becomes x'' — (a—b)z = ah. 45. Given — -= H — — —= = — -= , to find VX-\-ya — X \x — ya — x \x the roots. Roots, x = -^^ — Suggestion. — Add the fractions in the first member, and = — -. Whence 2a;' =? 2fea;-aft. 2x—a 106, Cor. 3. — Upon the principle that the middle term of a trinomial square is turice the product of the square roots of the other tux) (94, 95, Part I. ), u)e can often complete the square mxyre advantageously than by the regular rule. 46. Solve ^7?-\-Ux = m. Solution. — Dividing 16a; by twice the square root of 4a;', i. c, by 4a!, and adding the square of the quotient, (4)% to each member, 4a;' + 16x4-16 = 49. Extracting the root, 2x+4 = ± 7. .. x = f, and -V. 47. Solve 8a«-12a; = 36. Suggestion- — Divide the equation by 2, and proceed as above. 4^- — 6^ + (f)' = V-. 2.r— f = ±1, and x = 3, and -If In thia example the regular method is better. 48. Solve dj^-\-2xz=: 5. Suggestion.— Jlultiply by 3 and ^x' + Qx = 15. Whence 9a;' + 6a; + 1 = 16. a;=l, and— f. 356 QUADRATIC EQUATION'S. 49. Solve 110ic2— 21a: _ _i. Result, x = ^^ and ^. Suggestion.— Multiply by 110, and (IIO)V— 21 x 110a; = -110. Whence llOV-31 x 110a; + (-V-)^ = \, and llOx— ^ = ±f 50. Solve Z7?-\-6x — 2, Suggestion. (3)V + 3 • Sx + d)'' = 6 + ^ = \K Whence 3a; + |. = ± |. .-. x = ^, and —2. Scholium. — It appears that by this method the term to be added to complete the square is the square of | the coefficient of the first power of X. Therefore when this coefficient is odd, fractions arise. These can always be avoided by doubling the equation when this coefficient is odd, before completing the square. 61. Solve Zofi-1x = ^0. Solution.— Multiplying by 2 to avoid fractions, Qx^—lix = 80. (6) V— 6 • 14CC + (7)* = 49 + 480 = 529. 6aj-7 = ± 23. « = 5, and 52. Solve ^x{x-\-b) = 96 + 4a;(l— a;), and verify. Suggestion. — Perform as few multiplications as possible. When the square is completed, this stands (14)V + 14 • 22a; + (ll)'=14 • 192 + 121 = 2809. 53. Solve ^x^^ix-{-^ = 0. 54. Solve J- _ _?_ = |. a;— 2 xi-2 5 55. Solve (ax—b) (bx—a) = c^. Suggestion. abx''—{a'' + y)x = e—ab. 2a&a:'-(m) = 2c'-2aft, letting (m) stand for the term which becomes the middle term of the trinomial square and which disappears in the subsequent process. Then (2aZ>)-x- - (m) + {a" + &'0' = («' + 5')' + ^oM - 4:aW. .'. x - ^J^Ll \/(<^''-&')' + 4g^' ~'' "~" 2a5 56. Solve «- V^^^^ :z:g ^ _^ . Roots, x = a, and -|a. a-j-V^ax—x^ a— a; Suggestion- — Clear of fractions and condense. EQUATIONS SOLVED AS QUADRATICS. '6oa KCTI0M IIL EQUATIONS OF OTHER DEGREES WHICH MAY BE SOLVED AS QUADRATICS. 107* Prop. 1. — vdny Pure Equation (i. e., one con- taining the unknown quantity affected with hut one exponent) can he solved in a manner similar to a Pure Quadratic. Demonstration. — In any such equation we can find the vahie of the unknown quantity affected by its exponent, as if it were a simple equation. If then tlie unknown quantity is affected witli a positive integral exponent it can be freed of it by evolution; if its exponent be a positive fraction it can be freed of it by extracting the root indi- cated by the numerator of the exponent, and involving this root to the power indicated by the denominator. If the exponent of the unknown quantity is negative it can be rendered positive by multi- plying the equation by it with a numerically equal positive exponent. q. E. D. EXAMPLES. 1. Solve y-.-=-y. Suggestion, y' = 8. .-. y = 2. Why not put the ± sign before the 2 ? 2. Solve ^8 = - ~^-s' Result, V = -^• 3. Solve 3a;i— 5 = %x^. Result^ x = 125. 4. Solve |- 4- 1 = 5 (^ - l). Result, x = 32. 6. Soke 12arl 4. | = 1 + ^. Result, x = 27. o ^3 y S5S EQUATIONS OF OTHER DEGREES. 6. Given { x + {iax + 1 7«2)i \h — ^h^ ^i to find the value of X. Result, xz=z 16a. 7. Solve SaizF = 2ari;" + bb. Result, -^a/O"- 8. Solve a— 1 = Ja;" —a;" . -=(Mr 9. Solve a;t := 27. Also a;t = 4. Also 2/t = 32. i?oo/5, 81, ±32, and 8. Query. — Why the ± sign in one case and not in the others ? 108. Prop. 2. — Any equation containing one un- known quantity affected with only two different exponents, one of ivhich is twice the other, can he solved as an Affected Quadratic. Demonstration. — Let m represent any number, positive or nega- tive, integral or fractional ; then the two exponents will be repre- sented by m and 2m ; and the equation can be reduced to the form aj^'^-fl^af" = q. Now let y—x'^^ whence y'^=x^''\ whatever m may be. Substituting, we have y'^+py = q^ whence y = — ^ "^ i/ T +?• ^^^ y zizx"^] hence a? = | — | ± \/j + qT . Q. e. d. EXAMPLES. Ex. 1. Solve 3:z;3_^42:ct = 3321. Solution. — Let y = x^, whence y^ = x^; and 3y'' + 42y = 3321. From this y — 27, and —41. Taking the first value, x^ = 27. .*. x = 9. Taking the second x^ = —41. .\x= /v^ieSl. We there- fore find that x = 9, and \/lQ&i. SOLVED AS QUADRATICS. 359 2. Solve x^-h7x^ = 44. x = ±8, and ±(-ll)i Query.— How many values ? Which are imaginary ? 3. Solve ix^-^x^ = 39. x = 729, and (^V- 4. Solve 3a:« + 42:^-8 = 3321. a; = 3, and -\/U. 8 17 6. Solve ^+2 = -^. a; :^ 4^ and i\^2. Scholium. — It is not necessary to substitute am.ther letter for the unknown quantity as given in such examples. Thus, in Ex. 3, doubling, to avoid fractions, Sx^ + 2a;«^ = 78. Completing the square 8'a5* + (m) + l = 8 • 78+1 = 625. Extracting root, 8xi + 1 = ±25, X* = 3, and -^. .: x = 729, and /^V- [Note. — Solve the next six without substituting.] 6. Solve a;io + 31.r^ =32. x z= 1, and —2. 7. Solve a;" + 13/" = 14. x=h and (-14)»". 8. Solve 32:^ -4a;' = 7. x = Q^ and 1. 9. Solve 3a; + 2v^ = l. 10. Solve V2x—7x = —52. a; = 8, and ^. 11. Solve x-^2Vax-\-c=0. X— \^Va±Va—c}\ 12. Solve x\/ — X = — -_— a;= iVTiiA/I. 360 EQUATIONS OF OTHER DEGREES. Suggestion. — The first member may be written - — — — . Hence dropping the denominator, \/«» ^^^ squaring, Qx^—x*=:1-\-2x*+sbi*. 109* Prop. 3. — Equations may frequently he put in the form of a quadratic by a judicious grouping of terms containing the unknown quantity, so that one group shall he the square root of the other. Demonstration. — This proposition will be established by a few examples, as it is not a general truth, but only points out a special method. EXAMPLES. Ex. 1. Solve 2ai2 + 3a;— 5V2a;2_^3a;4.94_3 = o. Solution. — Add 6 to each member and arrange thus, (2a;^ + 3a: + 9) — 5(2x'' + 3x + 9)^ = 6. Put (2x'^ + 3a;4-9)i = y. and the equation becomes y'^-5p = G. Whence y = 6, and —1. Taking y = 6, 2a;''4-3aj+9 = 36. Whence x = S, and —4^. Taking y = — 1, 235'^ + 3aj+ 9 = 1. Whence x = ~^^Y~^ J . 4 2. Given (2a; + 6)^(20; + 6)^ = 6, to find the values of X, Suggestion. — Put y = (2x-f-6)^ ; whence y^ + y = Q, y = 2, and -3. .'. 2x + Q = 16, and also 2.r + 6 = 81. x = 5, and 37f Query.— Will the value x = 37^ verify ? Why ? Jns. — Since (2jr + 6)^ and (2«4-6)4 Hre'eveji roots, their signs are strictly ambiguous, though not so expressed in the example. Sub- stituting for .r, 37j^, the equation becomes (81)^"4-(81)i = 6. If now we regard (81)* — —3, as it is, as really as it is -f 3, the value verifies. Such cases arc frequent. 3. Given (.r + 12)^ =-- C> — {x + 12)^, to find the values of X, and verify both values. SOLVED AS QCADRATICS. J^fil 112 4. Given j^. -^^ = 5+71^ ru ' to find the values otx. (2a;— 4)* 8 (2a:— 4)^ Values, X = 3, and 1. Suggestion.— Put (2a;— 4)' = y. 5. Solve a; +5-^/3^+ 5 = 6. a; = 4, and— 1. 6. Solve 2V^-3a; + ll = a:2— 3a;4-8. Suggestion. z^—Sx+Q—2 \/x''—Sx+li = 0. Add 3 to each member and x^-Zx + ll—^^x^—dx+li = 3. Put \^x'—dx + li =-. y, and y'— 2y = 3. .-. .i- = 2, 1, and i(3 ± ^-31.) 7. Solve (a:3-9)2=:3 + ll(a:2_2). X = ±5, and ±2. 8. Solve (x-^^\x = ^2-^' a; = 4, 2, andi(-7±A/T7). 9. Solve x^f^l + ^^-idx^-^x) = 70. Suggestion. x*{l-\-—y = ^(Zx''+xy. Hence the equation may be written (3a;' + x)'— 9(3«'+a;) = 630. Results, X — 3, —31, and |(_1± V-^1). 10. Solve ^ - = "7 • i?oo^5, a: = 4, and 1. x—vx ^ Suggestion.— Divide by «+ /y/ir, and ^ = — j^ . Hence 4 = (x— ■^^xy, or ar— ' + 4c =■ 45; and t> = 5, und —9. a* = 8, and 4 ; y = 4, and 8. 8. Given ^ + 2- = GJI, and a^-^y^z= 65. Suggestion. — In the first, put — = r, whence v^ + 2vz= 9f|. Real and rational roots, x =z 4:, y =7. 0. Given x^+2xy-^y^-^2x = 120— 2^^, and xy—y* = 8. Roots, y = \y 4, —3 — a/5, and — 3-|-a/6; a; = 9, 6, —9 + ^5, and — 9— \/5. ^K4: SiMULTANEOLS EQUATIONS. 10. Given xY = 180— 8a;^, and x + dy = 11. Roots, X = 5, and 6 ; y = 2, and -J. 11. Given a;^ + 3.t + ^ = 73 — 2xy, and y^ ^- 3y + ic = 44. Suggestion. — Add the two equations together, and proceed as before. « = 4, 16, and -13^^58; and y = 5, -7, and — 1 ± V58. 12. Given xy-\-xy^ = 12, and x-{-xy^ = 18. 12 Suggestion. — From the first, x = ~- ; and from the second* 18 12 18 _. . . if = r- — ^. .'. -,i ^ = z i. Dividing denominators by l + i/, 1+y' 2/(1 + 2/) 1 + 2/ 2 3 and nunierators by 6, we have - = : whence 2 — 2^/ + 2i/* y 1— y + 2/^' = 3y. Hence x = 2, and 16 ; and y = 2, and ^. 13. Given x^-\-xy-^y'^ — 26, and a^-ficy+J/^ = 364. Suggestion. — From the first, x'^ + y"- = 2Q—xy; and from the sec- ond, by adding x^y^ to both members and extracting the square root, x^ + y'^ = '\/dQ4^ + x^y'^. Equating these values ofx'^ + y^, and squar- ing, we have GlQ—52xy + x'^y'^ = Siii + xy'^, whence xy = 6. Squar- ing this and adding it to the second, and extracting the square root, jc^-l-y' = 20. Also subtracting dx'^y^ = 108 from the second, and extracting the square root, x'^—y^ = 16. Whence x^ = 18, and 2/' = 3. ^ Another Solution. — Dividing the second by the first, we have x' —xy + y" — 14. Subtra(jting this result from the first, we have 2xy = 12. Whence the solution proceeds as above. [Note. — Though this field is illimitable, it is not thought necessary for the learner to pursue special methods farther, inasmuch as what is given will enable him to catch the spirit of such solutions, and no writer in discussing a problem involving processes even as complex as some given above, would fail to give hints at his methods of solution.] OF THE SECOXD DKGREE. 875 APPLICATIONS. [Note. — One of the most important things to be learned from the following examples is several devices frequently found serviceable in stating a problem, which make the equations arising more simple and easy of solution. These devices are of special necessity in examples involving progressions.] Ex. 1. What number is that which being divided by the product of its two digits, the quotient is 2, and if 27 is added to it the digits are reversed ? 2. There are three numbers, the difference of whose dif- ferences is 8 ; their sum is 41 ; and the sum of their squares is 699. What are the numbers ? Notation. — Let x be the second number, and y the diflference between the second and first, so that x—y represents the first. The first equation is 3.C+8 = 41, and the second (11— 2/)' + 131 + (19+y)' = 699. 3. There are three numbers, the difference of whose dif- ferences is 5 ; their sum is 44 ; and their product is 1950. What are the numbers ? 4. A grocer sold 80 lbs. of mace and 100 lbs. of cloves for 165 ; but he sold GO lbs. more of cloves for $20 than he did of mace for $10. What was the price of a pound of each ? 5. A and B have each a small field, in the shape of an exact square, and it requires 200 rods of fence to enclose both. The contents of these fields are 1300 square rods. What is the value of each, at $2. 25 per square rod ? Ans., One, $900; other, $2,025. 6. Find two numbers, such that the sum of their squares being subtracted from three times their product, 11 remain ; and the difference of their squares being subtracted from twice their product, the remainder is 14. (See 114.) 376 SIMULTANEOUS EQUATIONS 7. "What two numbers are those whose difference multi- plied by the difference of their squares is 32, and whose sum multiplied by the sum of their squares is 272 ? 8. The difference of two numbers is 2, and the square of their quotient added to four times their quotient is 9^. What are the numbers ? Ans., o and 3. 9. There are two numbers, whose sum multiplied by the less, is equal to four times the greater, but whose sum mul- tiplied by the greater is equal to 9 times the less. What are the numbers ? 10. Find two numbers, such that their product added to their sum shall be 47, and their sum taken from the sum of their squares shall leave 62. A7is., 5 and 7. 11. Find two numbers, such that their sum, their pro- duct and the difference of their squares shall be all equal to each other. Ans., f ± jVS, and i±iVS. 12. Find two numbers whose product is equal to the dif- ference of their squares, and the sum of their squares equal to the difference of their cubes. Ans., i\/5, and ^{5 + \/5). 13. A person has 11,300, which he divides into two ])or- tions, and loans at different rates of interest, so that the two portions produce equal returns. If the first portion had been loaned at the second rate of interest, it would have produced 136, and if the second portion had been loaned at the first rate of interest, it would have produced $49. Eequired the rates of interest. Ans., 7 and 6 per cent. 14. The fore wheel of a wagon makes 6 revolutions more than the hind wheel in going 120 yards; but if the periphery of each wheel be increased 1 yard, the fore wheel will make only 4 revolutions more than the hind wheel in going the OF THK ."SECOND DEGREE. 377 same distance. What i« the circiiniference of each wheel? Ans., 4 and 5. 15. The sum of two numbers is 8 and the sum of their nubes is 152 ; what are the numbers ? Suggestion.— Let x + y be one of the numbers, and a;—y the other. Then a; = 4, and 2y^-^Qxi/' ^ 152 (111). 16. The sum of two numbers is 7, and the sum of ihGir 4th powers is 641. What are the numbers ? 17. The sum of two numbers is G, and the sum of their 5th power is 1050. What are the numbers ? 18. The product of two numbers is 24, and their sum multi- plied by their difference is 20 ; find them. Aifs„ 4 and 6. 19. What two numbers are those whose sum multiplied by the greater is 120, and whose difference multiplied by the less is 16 ? A7is., 2 and 10. 20. What two numbers are those whose sum added to the sum of their squares is 42, and whose product is 15 ? Ans., 3 and 5. 21. A's and B's shares in a speculation altogether amount to $500 ; they sell out at par, A at the end of 2 years, B of 8, and each receives in capital and profits $297. How much did each embark ? A?is., A, $275 ; B. $225. Suggestion. — Letting j: )}e A's capital, and y B's, A gained 297 —X, and B, 297— y. And as the gains are proportioned to the pro- ducts of the respective tiraes into the capitals. 2x:8y :: 297— a; : 297-y. 22. What three numbers are those in A. P., whose sum is 120, and the sum of whose squares is 5600 ? Ann., 20, 40, 60. Suggestion. — In solving examples involving several quantities in arithmetical progression, it is usually expedient to represent the middle one of the series, when the number of terms \Hitdd, by .r, and let y be the c^mon difference. If the number of terms is 13& = Vl 15-4782 + = 10.74607 + |0t.o«« . = 11.54782- |Q9.0»37b = 124.09368 + ID'-"""" = Vi'-^-ooaes + = 11.13973 + ]^QI.U«13b = 10.74607 + lO^.o^.m^ 119.70845 + 101.0890636 _ 10.94113 + Whence 1.0390625 is log 10.94113; and proceeding with the com- putation, the logarithm of 11 maybe found with sufficient accuracy. Scholium. — The pupil will not fail to be impressed with an idea of the inmien.se labor involved in computing a table of logarithms. The common tables give the logarithms of numbers from 1 to 10,000, with provision, as will l)e seen hereafter, for using {hem to find the logarithms of much larger numbers, with sufficient accuracy for practical purposes. One page of such a table is given. (Page 386.) /?7. Prob.— To And the logarithm of a number from the table. 17 386 (a page of) a TABLE OF LOGARITHMS. N. 1 2 3 4 5 6 7 1 1 8 9 D. 280 447158 7313 7468 7628 7778 7933 8088 82421 8397 8552 155 281 8706 8861 9015 9170 9324 9478 9633 9787' 9941 .95 154 283 450249 1 0403 0557 0711108651 1018 1172 1326 1479 1633 154 283 1786! 1940 2093 2247 2400 2553 2706 2859 3012 3165 153 284 3318 3471 3624 3777 S930 4082 4235 4387 4540 4692 153 285 4845 4997 5150 5302:5454 5606 i 5758 5910 6062 6214 152 286 6366 6518 6670 6821 ! 6973 7125 7276 7428 7579 7731 152 287 7882 8033 8184 8336 8487 8638 \ 8789 8940 9091 9242 151 288 9392 9543 9694 9845 9995 • 146 1 .296 .447 •597 .748 151 289 460398 1048 1198 1348 1499 1649li;99 1948 2098 2248 150 290 291 2398 3893 2548 4042 2697 2847 4340 2997 4490 3146 3296 46S9 : 4788 3445 3594 4916 r>085 3744 5284 150 149 4191 292 5883 5532 56-0 5829 5977 6126 ! 6274 6423 6571 6719 149 293 6868 7016 7164 7312 7460 7608 7756 7904 8052 8210 148 294 8347 8495 8643 8790 8938 9085 9233 9880 9527 £675 148 295 9822 9969 .116 •263 .410 .557 .704 .851 -998 1145 147 29G 471292 1438 1585 1732 1878 2025 12171 2318 2464 2610 146 297 2756 2903 3049 3195 8341 3487 ' 3638 3779 8925 4071 146 298 4216 4362 4508 4653 4799 4944 5090 52:i5 5381 5526 146 299 5671 5816 5962 6107 6252 6397 6542 6687 6832 6976 145 300 7121 7266 7411 7555 7700 7844 7989 8138 8278 8422 145 301 8566 8711 8855 8999 9143 9287 9431 9575 9719 9863 144 302 480007 0151 0294 0438 0582 0725 0869 1012 1156 1299 144 303 1443 1586 1729 1872 2016 2159 : 2802 2445 2588 2731 143 304 2874 3016 3159 3302 3445 3587 : 3730 3872 4015 4157 143 305 4300 4442 4585 4727 4869 5011 5158 5295 5487 5579 142 306 5721 5863 6005 6147 6289 6430 1 6572 6714 6855 6997 142 307 7138 7280 7421 7563 7704 7845 ; 7986 8127 8269 8410 141 308 8551 8692 8833 8974 9114 9255 9396 9537 9677 9818 141 309 9958 ..99 .239 •380 •520 .661 .801 .941 1C81 1222 140 310 491362 1502 1642 1782 1922 2062 2201 2341 2481 2621 140 311 2760 2900 3010 3179 3319 3458 ' 3597 3737 8876 4015 lg9 312 4155 4294 4433 457214711 4850 4989 5128 5267 5406 139 313 5544 5683 5822 5960 i 6099 6288 6376 6515 6653 6791 139 314 6930 7068 7206 7344 7488 7621 7759 7897 8035 8173 138 315 8311 8448 8586 8724 ' 8882 8999 9137 9275 9412 9550 138 316 9687 9824 9962 ..99 i -286 .374 .511 •648 •785 •922 137 317 501059 1196 1833 1470 1 1607 1744 1880 2017 2154 2291 137 318 2427 2564 2700 2887 2978 3109 ! 8246 8382 8518 3655 1£6 319 3791 3927 4033 4199 i 4385 4471 4607 4743 4878 5014 186 320 5150 5283 5421 5557 5698 5828 5964 6099 6234 6870 136 321 6505 6640 6776 6911 7046 7181 7316 7451 7586 7721 135 322 7856 7991 8126 8260 8395 8530 i 8664 8799 8934 9068 135 323 9208 9387 9471 9606 9740 9874! ...9 .143 •277 .411 134 324 510545 0679 0818 094711081 1215 1349 1482 1616 1750 134 325 1883 20l7 2151 2284 ' 2418 2551 2684 2818 2951 3084 133 326 3218 3351 3484 3617 1 3750 3883 4016 4149 4282 4414 133 327 4548 4681 4813 4946 5079 5211 5344 5478 5609 5741 IS'oi 828 5874 6006 6139 6271 6403 6535 6668 6800 6932 7064 U'2 329 7196 7328 7460 7592 1 7724 7855 7987 8119 8251 8882 m 330 8514 8646 8777 8909 9040 9171 9303 9434 9566 9697 9 131 'sr 1 2 3 ^ 5 6 7 .^_ 3j LOGARITHMS. 387 Solution. — Pa2:e 386 is one page of a table of logarithms giving the logarithms of numbers from 1 to 10,000 directly^ and from which the logarithms of other numbers can also be found with little trouble. Thus, let it be required to find the logarithm of 325. Now, as the logarithm of 100 is 2. and of 1000 is 3, the logarithm of 325 must be between 2 and 3, i e. 3 and a fraction. The fractional part is all that is given in the table, as the integral can be known bj' simple inspection. Looking in the table down the column marked N (numbers), we find 325, and opposite it in the column headed 0, we find 1883, but just above this we observe 51, which belongs to this logarithm and which is simply omitted to save space in the table, since it really belongs as a prefix to all the logarithms clear down to the number 332 where it is replaced by 52. Prefixing the 51 to the 1883, we have .511883 as the decimal part of the logarithm. Hence log. 325 is 2.511883. In like manner the logarithm of any number represented hy three Jigures is found from the table. To find the logarithm of a number re])resented by four figures. Let it be required to find the logarithm of 2936. Looking for 293 (the first three figures) in the column of numbers, and then passing to the right until reaching the column headed 6, the fourth figure, we find 7756, to which prefixing the figures 46, which belong to all the logarithms following them till some others arc indicated, we have for the decimal part of the logarithm of 2936, .467756. But, as 3 is the logarithm of 1000, and 4 of 10,000, log. 29:56 is 3 and this decimal, or log. 2936 :^ 3.467756. Tofimi the logarithm of a number represented by moi-e than A^ figures. Let it be required to find the logarithm of 2845672. Finding the decimal part of logarithm of the first 4 figures 2845, as before, we find it to be .454082. Now^ the logarithm of 2846 is 153 (millionths, really) more than that of 2845. Hence, assuming that if an increase of the number by 1000 makes an increase in its logarithm of 153, an increase of 672 in the number, will make an increase in the logarithm of -jVy2^ or .672 of 153, or 103, omitting lower orders, and adding this to .454082, we have .454185 as the decimal part of log. 2845672. The integral part is 6, since 2845672 lies between ttie 6th and 7tb powers of 10. Hence log. 2845672 = 6.454185. Q. e. d. Scholium I. — If in seeking the logarithm of any number any of the dots noticed in the table are passed, their places are to be filled with O's, and the first two figures of the decimal of the loga- rithm taken from the column in the line b^low. Thus log. 3166 is 388 LOGARITHMS. 3.500511. This arrangement of the table is a mere matter of con- venience to save space. Scholium 2.— The column marked D is called the column of Tabular Difierences ; and any number in it is the difference between the logarithms found in columns 4 and 5, which is usually the same as between any two consecutive logarithms in the same horizontal line. The assumption made in using this difference ; viz., that the logarithms increase in the same ratio as the numbers, is only approxi- mately true, but still is accurate enough for ordinary use. 128, The Integral Part of a logarithm is called the Characteristic, and the decimal part the Mantissa. 12i). Prop. — Tlie Mantissa of a decimal fraction, or of a mixed number, is the same as the mantissa of the number considered as integral. Demonstration.— Above it was found that log. 2845673=6.454185. Now this means that l08-4 5 4i 8 5 ^ 3845673. Dividing by 10 suc- cessively we have 105-454185 — 284567.3, or log. 284567.2 = 5.454185, 10*-454185 — 38456.73 or log. 38456.73 = 4.454185, j^03-454185 — 3845.673 or log. 3845.673 =- 3.454185. 102-454t85 _ 284.5673 or log. 284.5673 --= 3.454185, ;|0l. 454185 — 28.45673 or log. 38.45673 = 1.454185, 100-484185 _ 3.845673 or log. 3.845673 = 0.454185. Now if we continue the operation of division, only writing 0.454185 — 1, T.454185, meaning by this that the characteristic is negative and the mantissa positive, and the subtraction not performed, we have 10T.454185 = .3845673, or log. .3845672 = T.454185, 107.454185 = .03845673, or log. .03845672 =^.454185, lOir.4 5 418 6 — .003845673, or log. .003845673 = 3.454185, etc. Q. E. D. Scholium. — The characteristic of an integral number, or of a mixed integral number and decimal, is one less tliau the number of integral places, as will appear by comparing such number3 with the LOGARITHMS. 889 powers of 10, as is done in cleinonHtrating (126). The character- istic of a number entirely decinial fractional, is negative, and one greater than the number of O's immediately following the decimal point, as appears from the last demonstration, or as appears from the fact that 10->=T'ff = -l; 10-« = t^ = .01; 10~^ = j^^^ = .001 ; etc. EXAMPLES. Find the logarithms of the following numhers : 285 ; 3145; 29056; 30942; 298.026; 32.56; 2.864; .3205; .00317 ; .00000328. Results, log. 298.026 = 2.474254 ; log. .00317 =3.501059. 130. Prob. — To find a number corresponding to a given logarithm. Solution. — Let it be required to find the number corresponding to the logarithm 5.515264. Looking in the table for the next less mantissa, we find .515211, the number corresponding to which is 8275 (no account now being taken as to whether it is integral, frac- tional or mixed ; as in any case the figures will be the same). Now, from the tabular difference, in column D, we find that an increase of 133 (millionths, really) upon this logarithm (.515211), would make an increase of 1 in the number, making it 3276. But the given loga- rithm is only 53 greater than this, hence it is assumed (though only approximately correct) that the increase of the number is -^Yy of 1, or 53-^133 = .3984 -h. This added (the figures annexed) to 3275, gives 32753984 + . The characteristic, being 5, indicates that the number lies between the 5th and 6th powers of 10, and hence has 6 integral places. .'. 5.515264 = log. 327539.84 + . EXAMPLES. Find tho numbers corresponding to the following loga- rithms : 3.467521 ; ^.467521 : 0.467521 ; 4.520281 ; 1.520281; 4.520281; 0. .520281 ; 1.520281; 2.490160; 2.490160 ; and 0.490160. Resnlfs, 2.490160 = log. 309. 1435 -f- ; ^.490160 = log. ,030914+; 0.490160 = log. 3.091435-i-, 390 LogaHithMS. 131. As logarithms are largely used to facilitate uiiiner- ical computations, it is important that the student be able to take any formula representing such operations and write at once the equivalent logarithmic operations. EXAMPLES. Ex. 1. If 28.035 : 3.2781 : : 3114.27 : x, what logarithmic operations will find x ? Suggestion. — The logarithm of the product of the means is the sum of their logarithms; and the logarithm of the quotient of this product divided by the first extreme, is the logarithm of said pro- duct minus the logarithm of the other extreme. .". log. .r = log. 3.2781 +log. 3114.27-log. 28.035 = 0.515622 + 3.493356-1.447700 = 2.561278. Having a table sufficiently extended, the number corresponding to this logarithm could be found, and would be the value of a?. 2. Find the product of 23 14, by 5.062, knowing that log. 23.14 is 1.364363, log. 5.062 is 0.704322, and log. 117.1347 is 2.068685. 3. How is 287 raised to the 5th power by means of loga- rithms ? How is the 5tli root extracted ? 4. Extract the 5th root of 31152784.1 by means of loga- rithms, knowing that log. 31152784.1 = 7.493497. Suggestion.— Log. ^311527843 = | log. 31152784.1 = 1.498699. The number, therefore, is 31.52 + . 6. What is the cube root of 30? Ans., 3. 107 4-. 6. What is the cube root of .03? Suggestion.— Log. .03 ="§.477121. Now to divide this by 8, we have to bear in mind that the characteristic alone is negative, i. e., 2:477121 = -2 + .477121, or -1.522879. This divided by 3 gives -.507626, or 0— .507626 = f.492374. But a more convenient method of effecting this division is to write for the —2, —3 + 1, whence we have for 2.477121, -3 + 1.477121, which divided by 3 givesT.492374, nearly. 7. Divide 3.261453 by 2, by 4, by 5. Last quotient, 1.4622906. E'C^^mi L DIFFERENTIATION. [This subject is inserted as the best method of reaching the demon- stration of the Binomial Formula and the production of the Logarithmic Series. While it is equally simple, to say the least, with the old method, it is more direct, and gives the student nothing but what is of fundamental importance in subsequent mathematical work.] 132. In certain classes of problems and discussions the quantities involved are distinguished as Constant and Variable. 133. A Constant quantity is one which maintains the same value throughout the same discussion, and is represented in the notation by one of the leading letters of the alphabet. 134. Variable quantities are such as may assume in the same discussion any value within certain limits deter- mined by the nature of the problem, and are represented by the final letters of the alphabet. III. — If X is the radius of a circle and y is its area, y — r.x^, as w.^ learn from Geometry, t being about 3.1416. Now if «, the radium, Note.— The preceding; part of this volume ftirnlshes a coun*e in Algebra quite as full as will be found practicable or desirable in mo!edient to give as much time to this subject as is required to master the University Algebra. There is nothing in the ordinary college course which requires more Al^bra than is found in this vohune. 392 DIFFERENTIATION". varies, y, the area, will vary ; but tt remains the same for all valued of X and y. In this case x and y are the variables, and tt is a constant. Again, if y is the distance a body falls in time a;, it is evident that the greater x is, the greater is y, i. e., that when x varies y varies. We learn from Physics that y = 16j*g!r', for comparatively small 25a^ (as, for example, 1, |, etc.), 2/ will be imaginary. This is the kind of limitation referred to in our definition of variables. 13ij» ScH. — The pupil needs to guard against the notion that the terms constant and variable are synonyms for knovm and unknown, and the more so as the notation might lead him into this error. The quantities he has been accustomed to consider in Arithmetic and Elementary Algebra have all been constant. The distinction here made is a new one to him, and pertains to a new class of problems and discussions. 130. A Function is a quantity, or a mathematical expression, conceived as depending for its value upon some other quantity or quantities. iLTi. — A man's wages for a given time is a function of the amount received per day, or, in general, his wages is a function of the time he works and the amount he receives per day. In the expression / = IGjJjic' (i.V4), second illustration, y is a function of x, i.e., the ipace fallen through in a function of the time. The expression 2ax^—Sx-\- 5b, or any expression containing x, may be spoken of as a function of x. 137. When wo wish to indicate that one variable, as y, is a function of another, as x. and do not care to be more specific, we write y = f{x), and read "^Z equals (or is) a DIFFRRKNTIATION. 393 function of a:."' This means nothing more than that y is equal to some expression containing the varia))le r, and which besides may contain any constants. If we wish to indicate several different expressions each of which contains x, wc write/(.r), q) (2:), or/ (;r), etc, and read '* the/ function of x" " the q> function of re," or '* the /' function of ic." III. — The expression /(;r) may stand for 0^—2^ + 5, or for 3(/z'— rr^V or for any expression containing r wmbined in any way with itself 01 with constants. But in the same discussion f{x) will mean the same thing tlirougliout. So again, if in a particular discussion we have a certain expression containing X{e. g., %x^—ax-\-Zab). it may be repre- sented by /(ir), while some other function of x, c. g.,5{n*—'j*) + 2x'^ . might be represented by/ (x), or (x). 138. In equations expressing the relation betweeen two variables, as in if = daafi—x^, it is customary to speak of one of the variables, as y, as a function of the other, x. Moreover, it is convenient to think of x as varying and thus producing change in //. When so considered, x is called the Independent and y the Dependent varial)le. Or we may speak of y as a function of the variable x. 130. An Inftnitesininl is a quantity conceived under such a form, or law, as to be necessarily less than any assignable quantity. Infinitesimals are the increments by which continuous number, or quantity (H), may be conceived to change value, or grow. III.— TYwe affords a good illustration of continuous quantity, or number. Thus a period of time, as 5 hours, increases, or grows, to another period, as 7 hours, by intinitesimal increments, i. t., not by hours, minutes, or even seconds, but by elements which are less than any assignable quantity. 140. Consecutive Values of a variable are values which differ from each other by less than any assignable quantity, i. e., by an infinitesimal. Consecutive values of a 394 DIPFERENTIATIOK. function are values which correspond to consecutiye values of its variable. 141, A Differential of a function, or variable, is the difference between two consecutive states of the function, or variable. It is the same as an infinitesimal. Ii^L- — Resuming the illustration y = IQ^x^ (/'Vt^), let x be thought of as some particular period of time (as 5 seconds), and y as the dis- tance through which the body falls in that time. Also, let x' represent a period of time only infinitesimally greater than x, and y' the distance through which the body falls in time x' . Then x and x' are consecu- tive values of x, and y and y' are consecutive values of y. Again, the difference between x and x', as cc'— .r, is a differential of the variable X, and y'—y is a differential of the function y. 142, Notation. — A differential of x is represented by writing the letter d before x, thus dx. Also, dy means, and is read "differential y.^' Caution. — Do not read dx by naming the letters as you do ax ; but read it " differential x." The d is not a factor, but an abbreviation for the word differential. 143, To Differentiate a function is to find an expression for the increment of the function due to an infinitesimal increment of the variable; or it is the process of finding the relation between the infinitesimal increment of the variable and the corresponding increment of the function. RULES FOR DIFFERENTIATING. 144, Rule I. — To dijferentiate a single variable, sim'ply write the letter d before it. This is merely doing what the notation requires. Thus, if x and x' are consecutive states of the variable x, ^. e., if x is what x becomes when it has taken an infinitesimal increment, x'—x is the differential of X, and is to be written dx. In like manner, y' —y is to be written dy, y' and y being consecutive values. DIFFERENTIATION. 395 145. Rule II. — Constant factors or divisors ap- pear in the differential tJie same as in the function. Dem. — Let ua take the function y = ax, in which a is any constant, integral or fractional. Let x take an infinitesimal increment dx, becoming x + dx \ and let dy be the corresjxjnding " increment of y, so that when x becomes x-k-dx, y becomes y + dy. We then have 1st state of the function . . y — ax\ 2d, or consecutive state . . y + dy := a(x+dx) = ctx + adx. Subtracting the Ist from the 2d dy = adx, which result being the difference b3tween two consecutive states of the function, is its differential {141). Now a appears in the difTei- ential just as it was in the function. This would evidently be the 1 same if a were a fraction, as - . We should then have, in like man- m ner, dy = — dx, for the differential of y = — a;. Q. E. D. 146, Rule III. — Constant terms disappear in dif- ferentiating ; or the differential of a constant is 0. Dem. — Let us take the function y = ax-\-b,m which a and & are constant. Let x take an infinitesimal increment and become x + dx ; and let dy be the increment which y takes in consequence of this change in x, so that when x becomes x + dx, y becomes y + dy. We then have 1st state of the function . . y = ax + b; 2d, or consecutive state . . y+dy = a(x + dx) + b = ax+ada+b. Subtracting the 1st from th • 2d dy = adx, which being the difference between two consecutive states of the function, is its differential {141). Now from this differential the constant b has disappeared. We may also say that as a constant retains the same value, there is no difference between its consecutive states (properly it has no con- secutive states). Hence the differential of a constant may be spoken of as 0. Q. B. D. * The word " contemporaneoaB " is often used in Ibis connection. o9v DIFPERENTIATIGK. 147, Rule IV. -— To differentiate the algebraic sum of several variables, differentiate each term sep^ arately and connect the differentials with the same signs as the terms. Dem. — Let u = x+y—z, u representing the algebraic sum of the variables x, y, and —z. Then is du — dx-^dy—dz. For let dx, dy, ind dz be infinitesimal increments of x, y, and z ; and let du be the increment which u takes in consequence of the infinitesimal changes in X, y, and z. We then have 1st state of the function .... u = x-^y—z ; 2d, or consecutive state . . . . u + dii z= x-^dx-\-y^dy—{z-\-dz). ^f u + du = x + dx + y + dy—z—dz. Subtracting the 1st state from the 3d du = dx+dy—dz. q. e. d. 148. Rule v. — The differential of the product of two variables is the differential of the first into the second, pias the differential of the second into the first. Dem. — Let i* = jcy be the first state of the function. The consecu- tive state is w + dii = {x + dx) {y + dy) — xy+ ydx + xdy + dx • dy. Sub- trading the 1st state from the consecutive state we have the differ- ential, i. e.,du = ydx + xdy + dx • dy. But, as dx • dy is the product of two infinitesimals, it is infinitely less than the other terms (ydx and xdy), and hence, having no value as compared with them, is to be dropped.* Therefore, du = ydx + xdy. Q. e D. * It will doubtless appear to the pupil, at first, as if this {?;ave a result only approximately correct. Such is not the fact. The result is absolutely correct. J\^o error is introduced by dropping dx • dy. In fact this term mnst be dropped accord- ins to the nature of infinitesimals. Notice that by definition a quantity which is Infinitesimal with respect to another is one which has no assignable magnitude with reference to that other. Hence we must so treat it in our reasoning. Now dx-dy is an infinitesimal of an infinitesimal (i.e., two infinitesimals multiplied together), and hence is infinitesimal with reference to ydx and xdy, and must b*» treated as having no assignable value with respect to them; that is, it must b« dropped. DIFFERENTIATION. 397 149. Rule VI. — The differential of the product of several variables is the sum of the products of the differential of each into the product of all the others. Dem.— Let u — xyz\ then du = pzdx + xzdy + xydz. For the Ist state of the function is u — xyz, and the 2d, or conaecutive state. H^da = (a? + dx) {y + dy) [z + dz), or n + du — xyz + yzdx + xzOy + xydz + xdydz + ydxdz+zdxdy + dxdydz. Subtracting, and dropping all infini- tesimals of infinitesimals (see preceding rule and foot-note), we have du = yzdx -f xzdy + xydz. In a similar manner the rule can be demonstrated for any number of variables. Q. E. D. 150. Rule VU.—TJie differential of a fraction having a variable numerator and denominator is tlie differential of the numerator multiplied by the denominator, minus the differential of the denom- inator multiplied by the numerator, divided' by the square of the denominator. Dem. — Let M = - : then is du = ~ . For, clearing of f rao- y y' tions, yu = x. Differentiating this by Rule 5th, we have udy + ydu — X xdv dx. Substituting for u its value -, this becomes — ^ + ydu = dx. y y ydx — xdv Finding the value of du, we have du = f-^. a?* + .Ete*+i7!r»-i- etc. (1) in which A, B, C, etc., are indeterminate coeflBcients independent of X (t. e. constants), and are to be determined. To determine these co- efficients we proceed as follows : Differentiating (1), we have m{a + x)'^-^dx = Bdx + 2Gxdx + SDa^dx + iJSufidx + Fxi^dx + etc. Dividing by dx, we have m(a + cc)"^» = B + 2Cx + SDx'- + 4.5^ + QFx!* + etc. (3) Differentiating (3) and dividing by dx, we have m(m—l)(a + x)"-^ = 2(7+ 3 • SJDx 4- 3 • 4^a;2 + 4 • 5Fx!' + etc. (3) Differentiating (3) and dividing 1 y dx, we have m{m-l)im-2){a + x)'^^ = 2 ' SI) + 2 ' d 4:Ex + d i - 5Fx^+ etc. (4) Differentiating (4) and dividing by dx, we have m{m-l)(m—2){m-'d)(a + ai)^'*-* = 3 • 8 • 4^+3 • 3 ' 4 ' 5Fx+ etc. (i5) DEMONSTRATION OF THE BINOMIAL THEOREM. 403 Differentiating (5) and dividing by dXy we have w(7n-l)(m-2)(m-3Xm-4Xa + a;)— » = 2 • 8 • 4 • aF-^ etc. (6) We have now gone far enough to enable us to determine the co- eflBcients A, B, C, D, E, and F, and doubtless to determine the law of the series. As all the above equations are to be true for all values of x, and as the coeflBcients A, B, C, etc., are constants, i. e., have the same values for one value of a; as for another, if we can determine their values for one value of x, these will be their values in all cases. Now, making x = 0, we have from (1) A =1 , from (3), B = m; from (3), C = m(m—l) - ,.^ _. m{m-l)(m-2) ^ ,^, „ r^ — ; from (4), D = -^ .^ ^ ; from (5), E = m(m-l)(m-2Xm-3) , ^ ,^^ w(m - IXw - 2)(77> - 3)(m - 4) j^ , from (0), Jf = — These values substituted in (1) give + fn(m-l){m-2)im-S)(m-i) ^.^^ ^ Which 18 the Binomial Formula. ISO, CoR. 1. — The nth, or general term of the series is m{m^l){ni-2) {m-n-^2) , , For we observe that the last factor in the numerator of the oo- eflBcient of any particular term is m— the number of tbe term less 2, i. e., for the nth term, m—(n—2\, or m—n + 2; and the last 404 THE LOGARITHMIC SERIES. factor in the denominator is the number of the term —1, i, e., for the nth term, ji—1. The exponent of x in any particular term is m— the number of the term less 1, i. e,, for the nth term, m—in—X), or m— 7i-h 1 ; and the exponent of y in any term is one less than the number of the terra, i. e., for the 7ith term, n—l. 100, Def. — In a series the Scale of Itelation is the relation which exists between any term or set of terms and the next term or set of terms. 101, Cor. 2. — The scale of relation in the binomial series is ( — — l)-, since the nth term multiplied hy this produces the (n -f- \)th term. This is readily seen by inspecting the series, or by writing: the (n4-l)th term and dividing it by the wth. Thus, substituting in the general term as given above, 71 + 1 for n, we have tn(m-l)(7n-2) ( ra-7i-\- \)^^^_^^^ \n asthe(7i + l)th term. This divided by the nth, or preceding term* m — n + lt/ /m + 1 \y gives -, or I 1 - • M3TIOM IIL THE LOGARITHMIC SERIES. 10^, The Modulus of a system of logarithms is a constant factor which depends upon the base of the system and characterizes the system. 103. JProp.—The differential of the logarithm of a THE LOOAHITHMIO '^KRIK!^. 40?) number is the differential of the number multiplied bif the modulus of the system^ divided by the number ; Ory in, the Napierian system, the modulus being 1, the differential of the logarithm of a number is the differetitial of the number divided by the number. Dem.— Let X represent any number, t. e. be a variable, and n be a :;onstant such that y = aJ». Then log y = » log x (1^3). Difl'eren tiating p - x^, we have dy = nx^-^dx ; whence dy n = -^-~ = -^- =1^ = JL a;»-'rf.r j?» , V ■, dx' (*) — dx -dx — X X X Again, whatever the differentials of log y and log x are, n being a constant factor we shall have the differential of log y equal to n times the differential of log x, which may be written d{\og y) = n'd (log x), whence n = j^~^ (2.) Now equating the values of ti as represented in (1) and (2), we have dy d{\og y) y dy TTT- -^ = rr ■ Whence d(logy) bears the same ratio to — , as d(log;r) dx \ &&' y X rf(log x) does to — . Let m be this ratio. Then rf(log y) = , and - ,, . mdx d{logx) = -^. We are now to show that m is constant and depends on the base of the system. To do this, take y — v^' , from which we can find as above n' = d^ (ogj^ = ^ . Now as m is the ratio of d (log y) to -^ , it is also the d Jog z) di ^ ^ " y ' 9 ratio of d([og z) to - ; and (f (log z) = — - . Thus we see that in any z case in the same system the same ratio exists between the differen- tial of the logarithm of a number and the differential of the number divided by the number. Therefore w is a constant factor. 406 THE LOGARITHMIC SERIES. Affain -^ — m- indicates the relative rate of chau<^e of a lo^a- ^ ax X rithm and ita number. Now it is evident that the larger the base the slower the logarithm will change with reference to the number. (See examples under Art 117.) But the factor - varies inversely in the number ; hence m must vary with, or be a function of the, base.* 164, Prob.— To produce the logarithmic series. Solution. — The logarithmic series, which is the foundation of the usual method of computing logarithms, and of much of the theory of logarithms, is the development of log (1 +ic). To develop log (1 +2!), assume log {l + x) = A + Bx+Cx^ + D^-¥E^-¥F3^+ etc., (1) in which a; is a variable, and A, B, C, etc., are constants. Differentiating (1), we have mdx\ 1 + x Dividing by dx. Bdx + 2Gxdx+dI)x^dx + ^m^dx+5me*dx+ etc ^ B + 2Cx+SDx' + 4^af^ + 5If\cU etc. (8) 1 +x Differentiating (3), and dividing by dx, we have -m^P-i— =2(7+2. 32)a;+ 3. 4^2 + 4. 5i?'a.'3+ etc (8) Differentiating (3), and dividing by 2 and by dx, we have W-— i-- = 3i) + 3.4^a;+2-3-5i^2^. etc (4) (1 +xf ^ ' Differentiating (4), and dividing by 3 and dx, we have -7» — L_ = 4^+4 5Fx+ etc (5) (1 + xf ^ ^ * What the relation of the modulus to the base is, we are not now concemed to know ; it will be determined hereafter. t The number is 1+aj; hence the differential is m times the differential of 1-f » divided by the number 1+x. THE LOGARITnMTC SERIES. 407 Differentiating (5), and dividing by 4 and dx, we have m TT-i-i = 5F+ etc.* (6) (I + «)» We have now gone far enough to enable us to determine the coef- ficients A, B, G, I), E, and F, and these will probably reveal the law of the series. As all the above e_ ^ 1 1 1 \ il.3>'"*'l3.3'»"^15.8»^"^ ^^^7 Th« numerical operations are conveniently performed as follows: 8 200000000 .66666(567 1 .66666667* 9 .07407 '07 8 .02469136 .0082804-. 5 .00164603 9 .00001449 7 .00013064 9 .000 10 1*^1 9 .00001129 9 .00001129 11 .00000103 9 .00000125 18 .00000009 .00000014 15 .00000001 .-. l0| ?2 = : .69314718 • ♦ Though the declmnl part of n lojrarithra Is penerally not exact, It 1* not cus- tomary to annex the + sipn. 410 THK LOGARITHMIC SKRIES. Second. To find log 'S, make 2 — 2, whence log 3 = log 2+2 Q + gi^, + _^ + Jg, + -Ig, ^ etc.). COMPUTATION. 5 2.00000000 25 .40000000 1 .40000000 25 .01600000 3 .00533333 25 .00064000 5 .00012800 25 .00002560 7 .00000366 .00000102 9 .00000011 .40546510 log 2 = .69314718 /. log 3 = 1.09861228 Thikd. To find log 4. Log 4 = 2 log 2 = 2 X .69314718 = 1. FouETH. To find log 5. Let x = 4, whence log 5 = log 4 + 2 Q + ^g-3 + Jj; + ^, + etc.). coMPm 'ATIO N. ^ 2.00000000 81 .22222222 1 .22222223 81 .00274348 3 .00091449 81 .00003387 5 .00000677 .00000042 7 .00000006 .22314354 log 4 = 1.38629430 •*• log 5 = 1.60943790 In like manner we may proceed to compute the logarithms of the prime numbers from the formula, and obtain those of the composite numbers on the principle that the logarithm of the product equals the sum of the logarithms of the factors. Thus, the Napierian logarithm of the base of the common system, 10, = log 5 + log 2 = 2.30258508. HIGHER EQUATIONS. 411 lOO, Prop, — The modulus of the common system is .43429448 + . Dem. — Since the logarithm of a number, in any system, divided hv tlie Napierian logarithm of the same number is equal to the modulus of that system {100), we have Com. log 10 A ^ , =^ r 10 ~ nioaulus of common system. But com. log 10 = 1, and Nap. log 10 = 2 30258508, as found above. Hence, Modulus of common system = K oi^KOKtvo = -43429448 *.oO*OoOUo 170. Prop.— The Napierian base is 2.718281828. Dem.— Let e represent the base of the Napierian system. Then by (165) com. log e : Nap. log e : : .43429448 : 1. But the logarithm of the base of a system, taken in that system is 1, since a' = a. Hence, Nap. log e = 1, and com. log e = .43429448. Now finding from a table of common logarithms the number corres- ponding to the logarithm .43429448, we have e = 2.718281828. IMCHOM I¥. HIGHER EQUATIONS. 171. Since every equation with one unknown quantity, and real and rational coefladents, can be transformed into one of the form :t^ + A3^-^ + Bx^-^-\-Cx''-^ L = 0, (1) this will be taken as the typical numerical equation whose solution we shall seek in this and the succeeding sections; 412 HIGHER EQUATIONS. and we shall frequently represent it by / (x) = 0, read "function x equals 0." The notation f(x) signifies in general, as has been before explained (157), simply any expression involving x. Here we use it for this particular form of expression. We shall also use/' {x) as the symbol for the first differential coefiicient of this function. 172, J^roiy. — Wlien an equation is reduced to the form x"-f Ax"-'4-Bx"-*4-Cx"-^ L = 0, the roots, with their signs changedy are factors of the absolute {known) terrUy L. [For demonstration see p. 363.] 173. Cor. — If a is a root of f (x) = 0, f (x) is divisible by X — a ; a7id, conversely, ifi(x) is divisible by x--a, a is a rootofi{x)=zO. Dem. — The first statement is demonstrated in the proposition, and the second is evident, since as f(x) is divisible by x—a, let the quotient be (l>{x); whence {x—a)(t>{x) — 0. Now x = a will satisfy this equation, since it renders x—a — Q, and does not render (p{x) Infinity, since by hypothesis x does not occur in the denominator.* 174* l?rop,—If the coefficients and absolute term in x° + Ax"-' + Bx"-'4-Cx"-* L = 0, are all integers, the equation can have no fractional root. Dem. — Suppose in this equation x := - \ - being a simple fraction V t in its lowest terms. Substituting this value of x, we have jtN a>(— 1 iM—i on — 3 * Could there be a term of the form — in

» ScH. — This proposition does not preclude the possibility of turd roots in this form of equation. These are possible. 116. ^Frop. — An equation f (x) = (171) of the nth degree, has n roots {if it has any), and no more. Dem. — Let a be a root of / {x) = 0, which is of the nth degree. Dividing /(x) by a; — a {173), we have f (x) = 0, an equation of the (71 — l)th degree. Let 6 be a root of

v/2, — V2, — 1, 3. 3. Roots 1, 2, 2, — 3, 4. 4. Roots - 3, 2 + V^^h 2 - V^^ 5. Roots 3, — 2, — 2, — 2, 1. 6. Roots I, i, - {. 7. Roots 1 ± V^, 2 ± \/^=r§". a Roots li, 2, >v/3, - V3. 9. Roots \/^^, - V^^, \/6, - \/5. 10. Roots 10, - 13, i, 1. 11. Roots 3 - 2 VS, 3 + 2 a/3, 2 - 3 sT^, 2 -f 8 \/=^, 1, - 1. 1S2, !Prop. — If the equation f (x) = has equal roots, the highest common divisor o/f{x) and its differential coeffi- cient* f ' (x), being put equal to 0, cotistitutes an equation which has for its roots these equal roots, and no other roots.\ * The differential coefficient of a function is sometimes called its first derived polynomial, The student must not suppope that the roots ot/(x) = 0, and it« flr»t differen- tial coefficient/' (sr) = 0, are necessarily alike, /'(z) = a series of l^rms some of which may be + and some -, and which may destroy each other, so as to render fix) = 0, for other values of x than puch as render/ (z) = 0, and not neoeM»rUy for any which do render/ («) = 0, except the equal roots of the latter. 416 HIGHEE EQUATIONS. Dbm. — Let a be one of the m equal roots of f{x) = 0, and let the other roots be b, e . . . . I; then f{oc) = {x — a}" {x ^b) {x — c) . . . , {x — l) {177)- Differentiating {149) and dividing by dx, we have /' {x) = m{x- ar-^ {x~b){x-e) {X'-l) + {x-a)>»{x-e) {x — I) + .... + {x — ay"{x — b) {x — c) . . . . -f etc. Now {x — a)'""* is evidently the highest common divisor of f{x) and f {x), and {x — a)^~^ = is an equation having a for its root, and having no other. In a similar manner, if f{x) = has two sets of equal roots, so that f{x) — {x — ay {x — by {x — c) (« — d) . . . .{x — I), differentiating and dividing by dx, we have f{x) - m{x — a)"-' {x — by{x-'C){x — d) (a? — ?) + (a; — «)"• r{x — 6)'^^ {x -^ c){x — d) . . . .{x — l) +{x — a)m(x — by{x — d) . . . .{x — n) + {x — a)m (x — by {x — c) . . . . {x — 1) + . . . . + {:x — a)"»(.'Z! — by{x — c) (.c — d) . . . . + etc. Now the highest common divisor of f{x) and f\x) is evidently {x — rt)*"-^ (x — by-K Putting this equal to 0, we have (.r — a)"*-^ {x — by~'^ = 0, an equation which is satisfied hy x = a and x — b, and by no other values. Thus we may proceed in the case of any number of sets of equal roots. 183, Sen. — In searching for the equal roots of equations of high degree, it may be convenient to apply the process of the proposition several times. Thus, suppose that /(ic) = has m roots each equal to a, and r roots each equal to b. Then the highest common divisor off{x) and /'(a;) is of the form (x — a)'^-^ {x — by-'^ ; whence {x—a)"^^ (x — 5/-1 = is an equation having the equal roots sought. Tliere- fore we can find the highest common divisor of {x — ay^-^x — b)'-^, and its differential coefficient which will be of the form (,/• — ay^—^v—by "^ and write {x — a)"*"^ {x — by-^ — 0, as an equation containing the roots sought. This process continued will cause one of the factors {x — a) or {x — 6) to disappear and leave {x — (Cy^' — 0, when m > r; (x — by~^ = 0, when r>m; or {x — a) {x — b) z= 0, when w = n From any one of these forms we can readily determine a root HIGHER EQUATIONS. 417 1S4:, Proj), — 1)1 an equation f (x) = 0, f(x) will chauyt sign when x jjasses through any real rooty if there ti< but am such root, or if there is an odd 7iumber of such roots ; biit if there is an even number of such roots, f (x) will not change sign. Let a, b,c . . . .e be the roots of /(^x) = 0, so that /(j) = {x—a) u—b) {x—e) .... {x-e) = {177)- Conceive x to start with some^ value less than the least root, and continuously increase till it becomes greater than the greatest root. As long as x is less than the least root, all the factors a'— a, x—b, etc., are negative; but when a- passes the value of the least root, the sign of the factor containing that root will become +,*and if there is no other equal root, this factor will be the only one which will change sign. Hence the pro- duct of the factors will change sign. But if there is an even number of roots, each equal to this, an even number of factors will change sign ; whence there will be no change in the sign of the function. If, however, there is an odd number of equal roots, the passage of x through the value of this root will cause a change of sign in an odd number of factors, and hence will change the sign of the function. Finally, as it is eiident that the signs of the factors, and hence of the function, will remain the same while .r passes from one root to another, and in all cases changes or does not change as above when x passes through a root, the proposition is established. Th« following example will be found very instructive : .z^ + 4ar*— 14a!='— 17x— 6 = 0. The least root of this equation is — 3. When x< — 3,/(aj) is — ; when x = —S,f{x) = 0; when x passes —3, increas- ing, /(j) changes from — to +, and remains + till x = —I, when it becomes 0, and changes sign as x passes —1, noUcithsta tiding they are equal roots. But there i.s an odd number of such roots, viz., three. But in a:*— 14a;' + 64a'— 96 = 0, two equal roots of which '.re 4, if we substitute 2 we get fix) = —16, and substituting 5, f{x) = — 1, the function not changing sign, although a root has been passed. * Suppose e be the lea«t root, and that e' iit the next state of x greater than c then ff—c is +. 418 HIGHER EQUATIONS. 185* Prop, — Changing the signs of the terms of an equation containing the odd powers of tlie unknown quan- tity changes the signs of the roots, DEJvr,— If x= a satisfies the equation af—Ax*+Bx^—Cx + D = 0, we have a^— Ao^ + Ba^— Ca + 1) = 0. Now changing the signs of the terms containing the odd powers of x, we have x^—Ax'^—Bq^+Gx + D = 0. This is satisfied by a; = —a, if the former is by a; = a. For, substituting —a for x, we have a^—Aa'^+Ba^—Ca + D = 0, the same as in the first instance. 180, Cor. — Changing the signs of the terms containing the even poivers will answer equally welly since it amounts to the same thing ; and if we are carefid to put the equation in the complete form, cha7iging the signs of the alternate terms will accomplish the purpose, III. — The negative roots of x^~-7x + Q = 0, are the positive roots of —3^ + 7x + Q = 0, or of x^—7x—6 = (0 being considered an even exponent) ; or, writing the equation ^'^±0.^2— 7a; + 6 = 0, changing the signs of alternate terms, and then dropping the term with its coeffi- cient 0, we obtain the same result. Again, the negative roots of a;*^ — 7a^ — 5-r* + 8-i'^— 133a;- + 508a5— 240 = 0, are the positive roots of a'« + 7.2'5-5.?"^-8.r8-182a;»— 508aj— 240 = 0,or of —x'-lx' + 53^ + Sx^ + lS2.T^ + 506x + 24Q = 0. 187, JProb. — To evaluate* f (x) for any particular value of Xj as X z=z a, more expeditiously than hy direct sub- stitution. Solution.— As /(a-) is of the form a;** + ^a;**-' + ^a;*-^ + Ca"*-^ . . X, let it be required to evaluate 'X* + Ax^ + Bx^ + Cx + I) tor x = a. Write the detached coefficients as below, with a at the right in the form of a divisor ; thus ♦ TWs means to find the valoe of. Thus, suppose we want to find the value of a*— 5a;* +2a;*— 5a5» +6a?'— «— 12, for x = 5. We might substitute 5 for a, of course, and accomplish the end. But there is a more expeditious way, as the solution pf this problem will show. HIGHER EQUATIONS. 419 ^A +B +C +D \a__ a a^ + Aa a^ -\- Aa^ -{- Ba a^ + Aa' + Ba^ + Ca a+A a^ + Aa + B a^ + Aa- -\- Ba -\- C a*-^Aa^ + Ba;^ + Ca + D Having written the detached coetficients, and the quantity a for which /(i") is to be evaluated as directed, multiply the first coetficieut 1 by a, write the result under the second, and add, giving a-hA. Multiply this sum by a, write the product under the third coefficient B, and add, giving a' + ^a + 5. In like manner continue till all the coeffi- cients (including the absolute term, which is the coefficient of sfi) have been used, and we obtain a^ + Aa^ + Ba^ + Ca + D, which is the value off(x) for x = a. Illustration. — To evaluate x^ —5a;' + Sar* — 3a^ + 6' —a;— 12, for r = 5: 1 _5 4-2 -3 +6 -1 -13 |_5 _6 _0 JO ^ 205 1020 2 7 41 204 1008 Now 1008 is the value of a-*— 6a^ + 2a:*— 3a:« + 6a-'-x— 12, for a; = 6; and it is easy to see that much labor is saved by this process. We are now prepared for the solution of the following important practical problem : 1S8» Prob, — To find the commensurable roots of numer- ical higher equations. The solution of this problem we will illustrate by practical examples. EXAMPLES. 1. Find the commensurable roots of a^— 27^^102? -\-^a?-^ C8a:+4:8 = 0, if it has any. Solution.— By {174), if this equation has any commensurable roots they are integral ; — it can have no fractional roots. Again, by {172), the roots of this equation with their signs changed are factors of 48. Now, the integral factors of 48 are 1, 2, 8, 4, 6, 8, 12, 16, 24, 48. Hence, if the equation has commensurable 420 HIGHER EQUATIOIS^S. roots, they are some of these numbers, with either the -f or — sign. We will, therefore, proceed to evaluate f{x) (i. c in this case jfi — 3a^-15a^ + 8a;-^ + 68;r + 48), for *■ = +1, a; = -1, x = +3, a; = -3, etc., \iy{187), as follows- 1 -2 -15 +8 +68 +48 [ +1 _J ^ -16 - 8 60 -1 -16 - 8 60 108 Hence we see that f or a; = + 1, f{x) = 108, and + 1 is not a root o^ f{x) = 0. Trying x= —1, we have 1 -3 -15 + 8 + 68 + 48 -1 3 13 -30 -48 -3 -12 30 48 Thus we see that for x = — l,/(a;) = 0, and hence that —1 is a root of our equation. We might now divide /(;i') by x + 1 (i 75) and reduce the degree of the equation by unity. But it will be more expeditious to proceed with our trial. Let us therefore evaluate /(a;) for a; = +3. Thus : 1 -3 -15 +8 +68 -30 -44 -15 -33 34 + 48 1 +3 +_48 96 ac© for X — have + 3, ,f{x) = 96, and +3 is not a root Trying « 1 -3 -3 -4 -15 +8 +68 8 14 -44 - 7 33 34 + 48 1 -2 -48 Hence for x = —2, f{x) = 0, and —3 is a root. Trying x =» +8^ we have 1 -2 -15 + 8 + 68 + 48 1 +8 8 3 ^36 -84 -48' 1 -12 -28 -16 * Of course it Is not necessary to retain the -i: sign, as we have don« In the pre- eeding operations : it has been done simply for emphasis. moHEB EQUATION.^. 421 Hence for r = ^3, /(t) — 0, and +3 is a root. Trying a; =r — 3, wa have 1 -a •-^15 +8 +68 + 48 I -8 -3 _15 _^ -24 -132 -5 8 44-84 Hence for x = ^S,f{x) = —84, and —3 is not a root. Trying a; = 4, we have 1 -2 -15 -f 8 +68 +48 |_4 _4 _8 -28 -80 -48 2 - 7 -20 -12 Hence for x = ^,f{x) = 0, and 4 is a root. We have now found four of the roots, viz., —1, —2, 3, and 4. Their product with their signs changed is 24. Hence, by (17^) 48+ 24 = 2 is the other root with its sign changed, i. €., there are two roots —2. That our equation had equal roots could have been ascertained by the principle in {1&2) ; but as the process of finding the H. C. D. is tedious, it is generally best to avoid it in practice. to 11. Find the roots of the following : - 2. ir*— a:S— 39ic24.24a; + 180 = 0; 3. 3^+bx^—dx—^5 = 0; 4. ic3+2.r2-23a;— G0 = 0; 6. a:*— 3x8_i4a;2-|-48a:— 32 = 0; 6. ^— 8arJ+13:r— 6 = 0; 7. x*—Uxi-\-lSx—S = 0;* 9. afi—ldx*-\-eW-'17lTi + 2l63'-10S = 0; 10. a:*— 452^— 40.rH-84 = 0; 11. a:5-3a:*— 9a*+21a«— 10a;+24 = 0. * In order to apply the process of evalaation, the coefficients of the missing powers must be supplied. Thus we have 1 + 0—11 + 18—8. 4:'22 HIGHER EQUATIONS, 12 to 17. Apply the process for finding equal roots (192, 188) to the following: 12. 2;3-f82:2 4.20a:-hl6 = 0; 13. x^—x^—^x^VZ = 0; 14. 2)3—50:2—8.^ + 48 = 0; 15. ir4_ii^2_|_i8^_8=3:0; 16. a:4 4.i3^4.33^2_^31^^10 = 0; 17. 3?—ldx^-\-%1x^—inx^+^\(jx—10% = {). 18 to 24. Having found all but two of the roots of each of the following by (187), reduce the equation to a quad- ratic by {17S), and from this quadratic 6nd the remaining roots : 18. x^—6x^-\-10x—8 == 0; 19. x^—4:ci^—8x + l]2 = 0; 20. a^—3x^-\-x-\-2 = 0; 21. 2.4—6.^3 + 240:— 16 = 0; 22. o:*—12o:8 + 50.2-2— 84.T + 49 = ;* 23. o:4_9a:8^i7^.2+.27o:— 60 = 0; 24. a:«— 40:4— 16.t3+112;?;2—208o: + 128 = 0; 25. 2o:8_3^2_|_ 2.^-3 = 0;t 26. 30,-8— 22-2— 60: + 4 = 0; 27. 8o.^-26o;2+llo: + 10 = 0; 28. or*— Jo; + f\ = 0; (Look out for equal roots.) 29. x^—Gs^+9ix^—dx-{-i^ = 0, Apply the method for finding equal roots. t We have x'—'^x''+x — ^ = 0. Pnt a; = |, whence J^ — ^r^y' + vy—^ =0, or y» — — y" + A'y — — = ^- ^^ ^^^ k = 2, we have y'—3y''+4y—12 = 0, which can he solved as before, for one value of y, and the equation then reduced to a quad- ratic and solved for the other values. Finally, rememhering that x = ^y, we have the valaes of x reqnirocL INTERPRETATION OF EQUATIONS. 4^5 J^ECTOON V. DISCUSSION, OR INTERPRETATION, OF EQUATIONS. ISO, To DiscusSf or Interpret^ an Eqvafiou or an Aff/ebraic Expression , is to determine its significance for the various values, absolute or relative, wiiich may be attributed to the quantities entering into it, with special reference to noting any changes of value which give changes in the general significance. Such discussions may be divided into two classes: 1st. The discussion of equations or expressions with reference to their constants ; and 2d. The discussion of equations or expressions with reference to their variables. The following principles are of constant use in such dis- cussions : 100. I*roj), — A fraction, when compared with a finite quantity, becomes: 1. Equal to 0, when its 7iumerator is a7id its denomina- tor finite, and tvhen its numerator is finite and its denomi- nator 00 . 2. Equal to oo , when its numerator is finite and its de- nominator 0, and lohen its numerator is oo and its dejiomi- nat or finite, 3. It assumes an indeterminate form when numerator and denominator are both 0, and when they are both oo .* Dem. — These facts appear when we consider that the value of a fraction depends upon the relative magnitudes of numerator and de- nominator. * By ttiis it is meant that - and — may have a variety of valnes, not that they neoeBsarily do have. 424 tKTE!tf»Rl5TATrois' OP JCQrATIONS. 1 . Let a be any constant and x a variable, then the fraction - a diminishes as x diminishes, and becomes when x is 0. Again, the a fraction - diminishes as x increases, and when x becomes oo, i.e., X I » a greater than any assignable magnitude, - becomes less than any X assignable magnitude or infinitesimal, and is to be regarded as in comparison with finite quantities. (See 130 and 148, Dem., and foot-note.) 2. As X increases, the fraction - increases, and hence when x be- a comes infinite, the value of the fraction is infinite. Also, as x dimin- a ishes, the value of - increases; hence, when x becomes infinitely small, or 0, the value of the fraction exceeds any assignable limits, and is therefore oo . X 3. Finally, if x and y are variables, - diminishes as x diminishes, y and increases as y diminishes. What then does it become when a?=0, and y = 01 i. e., what is the value of x? Simple arithmetic would lead us to suppose that ^ was absolutely indeterminate, i. e., that it might have any value whatever assigned to it, for a = 5, since = 5x0 = 0; 7^~'^i since = 7x0 = 0, etc. But a closer inspection will enable us to see that the symboi x is not necessarily indetermi- nate, or rather that the expression which takes this form for particu lar values of its components, has not necessarily an indefinite number of values for these values of its components. Thus, what the value X of - will be when x and y each diminish to will evidently depend upon the relative values of x and y at first, and which diminishes the x faster. Suppose, for example, that y — ^x; then = ■=- , Now, y ox suppose X to diminish ; the denominator will diminish 5 times as fast as the numerator, and whatever the value of x, the value of the frao INTERPRETATION OF EQIATIONS. 425 X X tlon will be ^. So if y = Ta;, - = ^, which is \ for any value of sc Hence, when a: = 0, and y = 0, we have - = ;; = ^- = r, or y sj ox o' ar a? 1 a? 2/~0~7i~7' ^'iy~0~ "^^ other value depending upon the rel- X x \ it y = Qx, we liave ^ = — = ^^ = ^ . And so if y =^ IOj, we have ;p* 00 aj 1 = ~ = Tq- = Tq . Thus we see that the mere fact that numerator . and denominator become 0, or become oo , does not determine the value of the fraction, i. e., gives it an indeterminate form. 191. A Real dumber or Quantity/ is one which may be conceived as lying somewhere in the series of num- bers or quantities between — oo and -f oo inclusive. hjj, — Thus, if we conceive a series of numbers varying both ways from 0, i. «., positively and negatively to «, we have - 00 .... -4, -3, -2, -1, 0, +1, +2, +8, +4, . . . . + oo. Now a real number is one which may be conceived as situated somewhere within these limits; it may be + , — , integral, fractional, commensurable, or incommensurable. Thus, +15624 and —15624 will evidently be found in this series. +V- may be conceived aa somewhere between +5 and +6, though its exact locality could not be fixed by the arithmetical conception of discontinuous number. So, also, —y^ is somewhere between —Sand —6. Again, + y'5 is some- where between +2 and +3, though, as above, we cannot locate it exactly by the arithmetical conception. The follo^ving Geometrical lUvMratum is more complete than the arithmetical. Thus, let two indefinite lines, as CD and AB, intersect (^cross) each other, as at 0. Now let parallel, ee —2, at c' —3, etc. Now conceive one of tliese lines to start from an infinite distance at the left and move toward the right. When at an infinite i%6 INTERPKETATION OF EQUATIONS. distance to the left of its value would be — oo , and in passing to it would pass tbrough all possible negdtim values. In passing it becomes at 0, changes sign to + as it passes, and moving on to .B infinity to the right, passes through all possible positive values. Hence we see how all real values are embraced between — O) and + oo, in- clusive.* 192, Ail Imaginary Number or Quantity is one which cannot be conceived as lying anywhere between the limits of — oo and + oo , as explained above. The algebraic form of such a quantity is an expression involving an even root of a negative quantity. f EXAM P LES 1. "What are the values of x and y in the exprestdons aV ^ a'h , , y = - J - , when V and « and a are a —a a — a * For example, the student who is acquainted with the elements of geometry knowp how to construct a line which is exactly equal to 4^5 (Geom., Part 1, 110). rhis line ho can locate between +2 and +3, and also between —2 and —3, since 4/5 is both + and-. t Transcendental functions afford other forms of imaginary expressions ; for example, sin-' 2, sec ^ X, log (—120), log (— w), etc. But our limits forbid the con- sideration of the interpretation of imaginaries, except in the most restricted sense, RB indicating incompatibility with the arithmetical sense of the problem. INTERPRETATION OF EQUATIONS. 427 unequal ? When b =z b' and a = a' ? Wlien a = a' and b and b' are unequal ? What are the signs of 2; and y when i > ^ and a > a', the essential signs of a, a', J, and h' being + ? When b y b' and a < a' ? If a' and ^ are essentially negative, and a = a', and b = b', what are the values of x and u? If a' and Z*' are each ? 2. What general relation between a and a' renders ^ ~-?; = ? What renders it oo ? 1 + a« Solution. — To render :; , = 0, we must have a'— a = 0, and 1 + aa' i + aa' finite or infinite; or else we must have \-\-aa' = , while a'— a is finite or {li}0). Now a— a = gives a' =a; whence, r = :; ,, which is for any value of a, finite or infinite. 1 + aa' 1 + a"^ '' Hence the relation a' = a fulfills the first requirement. Let us now see if 1+aa' = CO will also fulfill this requirement. This gives aa' = 00 , since subtracting 1 from oo would not mnke it other than cc . CO Thus we have a' = — . Hence, for all finite values of a (including 0) a' is 00 , and j- : = — ; = - : which can only be when o = oo . 1 + aa' oaf a "^ Therefore the particular values a' = = a = cc, render :j — — = ; but no general values do, unless a = a'. Again, in order that :; j = oo , we most have 1 + aa' = 0, and ° 1 + aa' a'— a finite or infinite; or else we must have a' — a = oo, and l + arf finite or 0. Now 1 +aa' = gives , 1^ ___!_ a'— a _^ "^a' _a"' + l _ a'^+l _ *^" a'' l + aa'~ a/^" a'—n' " ~^' a' • This reduct\on Is made by dropping a and 1, since the subtraction of a finite from an inflnile, or the addition of a finite to an infinite, does not change the char- acter of the infinite, Thuf>, in thic cape, to apsume that dropjilng a and 1 afl'ccted the relation between numerator and denominator, would be to assign to a and 1 some vaiuea with respect to the infinite a\ Dut tl.is is contrary tu the dcfinitioD of an Inftnits. 428 INTERPRLTATION OF EQUATIONS. for any value of a', finite or infinite. Therefore tlie general relation a =5 , between a and a' renders :; : = 00.* Let us now see if a 1 + aa' the relation a' — a = 00 will do the same. Now if a' — a = 00 , one or the other {a! or a) must be 00 . Let a' = 00 . We then have ~^ = — ; = - , which can only be co when a = 0. Hence the par- 1 + aa' aa' a iicular values a' = cx) and a = render :; , = 00 , but no general 1 + aa' ^ values meet the requirement unless a = ^ . 3. What general relation between a and a' renders I -^ aa' a' -\- a = ? What renders it 00 ? 4. In the expression y = — 2a: -f 4 ± v^-— 4x — 5, how many values has ^, in general, for any particular value of aj ? For what value or values of x has y but one value ? For what values of a; is y real ? For what imaginary ? For what values of a? is ^ positive ? For what negative ? Solution. — Writing the expression thus, we see that the value of y is made up of two parts, viz., a rational part — k%x — 4), and a radical part y'a;"^ — 4r — 5. But the radical part may be taken with either the -f or the — sign. Hence, in general, for any particular value of x there are two values ofy. 2d. But if such a value is given to x as to render the radical part 0, for this value of x, y will have but one value, viz., the rational part. But the condition ^x^—\x—^= gives a; = 5 and — L Thus, for a; = 5, y = —6, but (me value; and for x — — 1, y= +6, also but one value. 8d. To ascertain for what values of x, y is real, we observe that y is real when 3^ — 4a; — 5 is positive, and imaginary when x^—Ax — 5 is negative. Now for X positive, a;'— (4a;+5) is + when a;' >4r + 5 ; and for x nega- * It is to be observed that the relation a = , requires that a and a' have dif- ferent eseential signs ; while the relation a^ = a requires that ihey have the 8am« essential sij^nB' INTERPRETATION OF EQUATIONS. 4JW live, we have j^-f4r— 5, which is positive when x^ + 4i>5. The former inefiuality gives ar' — 4a;4-4>9, or x>o; aud the latter gives «* + 4r + 4 > 9, or x>\. Uenco lor positive values of x greater than 5, y is real, and for negative values of x numerically greater than 1, y is real. The 4th inquiry is answered by this : f, is imaginary for all valuesof a; between — 1 and +3. 5th. To ascertain what + values of X render y + , and what — , we observe that — (2a) — 4) ± /y/a* — ix — 5 can only be + when the -♦- sign of the radical part is taken and when y^ — kj— 5>2a* — 4. This gives x <2 ± \/—'3, i.e., an imaginary quantity. Hence y is never -f- for a? + . Taking? the negative sign of the radical, we see that both parts of the value of y are — , and conse- quently y is real and negative for all + values of x which render y real, t. e., for values greater than 5. Finally, for x — we have y = 2x4-4 ± ^x^ + ix — 5. Now when we take the + sign of the radi- cal, both parts are + ; hence this value of y is always plus. \Vlien we take the — sign of the radical, y is negative if 23" + 4 < y'a-* -f 4x — 5. But this gives a;< — 2 ± y^^. Hence y is never negative for any negative value of x. Therefore both values of y are positive and real for all negative values of x numerically greater than 1. 6 to 22. Discuss as above the values of y in the follow- ing; i. e., 1st. Show how many values y has in general, and whether they are equal or unequal; 2d. For what particular value or values of .r, y has but one value; 3d. For what values of .r, y is real, and for what imaginary ; 4th. For what values of a;, y is 4-, and for what — ; 5th. Also determine what values of x render y infinite: 5. t/2_j_2a:^— 2x^— 4y— a;4-10 = 0;* 6. y2—2a;y-f 2^— %4-2a; = 0; 7. y2_|_2xy-fr^— Gy + 9 = 0; 8. f-\-2xy-^Zx^—^x=iO\ 9. y'— 2a:y-|-3j2H-2y— 4r— 3ss0; 10. y3^2.Ty — 3:r2— 4:r = 0; 11. y'^—2xy-\-x^ + x = 0; ♦ In al) cases rolve the equation for y in the flrst place. In th1» ezampto 430 INTERPRETATION OF EQUATIONS. 13. y^—2xy-{-x^-\-2y-{-l = 0; 14 i/2_2a;2_2?/ + 6a;— 3 = ; 15. y^—2xy—3x^-2i/-{-'7x^l = 0; 16. y^—2xy—2 = 0; 17. y^—2xy + 2y + 4x^% = 0\ 18. 4?/2+4^ + 2^-3a;4-12 = 0; 19. 3/— 8:c2 = 12 ; 20. 12«/2-f4a;2 = 20; 21. x^ + y'^= 16; 22. ar^— z^2 _ 20. 193. Arithmetical Interpretations of Nega- tive and Imaginary Solutions. 1. A is 20 years old, and B 16. When will A be twice as old as B ? SuG's.— We have 20 + ;c = 3 (16 + x) ; whence a; = — 13. The arith- metical interpretation of this result is that A will never be twice as old as B, but that he was twice' as old 12 years ago, ^. e., when he was 8 andB4. 2. A is « years old, and B b. When will A be ^ times as old as B? ror7i>l what are the possible relative values of a and h consistently with the arithmetical sense of the problem ? Interpret for aynb, a =: nb, alc Also for 7^ = 1, aynb, ab, (3) when ab, x positive, which shows that the point at which they are together is at the right of B, i, e., in the direction which they X / <5 + X\ are traveling. The time-, t (or J, is positive, yrliich shows that they are together after passing A and B. For a<&, X is negative, and c + x, which equals -^^» *» also nega- tive. This shows that they were together at a point at the left of A. that is, before they reached the stations A and B. With this the expreeeions for the time also agree. Thus . beoomefl — r , and -— is also negative, since in this case «><>. 432 INTERPRETATION OF EQUATIONS. ---, . be be , ac ae When a = b, x — — -— -=c3o, and c + a; = r = - = oo a—b a—b which indicates that they are never together. When c = 0. In this case x = — \ = 0, and c + x = — - = 0, for a and 6 unequaL a—b a—b ^ indicating that they are together when they are at A and B. This is evidently correct, since A and B coincide in this case. When a = b, X — — V = ;; , and c 4- a; = TT , which shows that they are always to- a—b gether, - being a symbol of indetermination which in this instance may have any value whatever, as we see from the nature of the problem 194:* SCH.— The student should not understand that the symbol - always indicates that the quantity which takes this form has an indefinite number of values. It is frequently so, but not necessarily. The indetermination may be only apparent, and what the value of the expression is must be determined from other considerations. The Cal'mlus affords the most elegant general methods of evaluating such expressions. But the simple processes of Algebra will often suflBce. l_a^ 1— ar* Thus for a? = 1, z = x . But :; = 1 +x-¥x^, which, for a; = 1, 1— a? 1— a; 1— aj* is 3. Hence :; = 3, for .i; = 1. Here the apparent indetermina- \—x tion arises from the fact that the particular assumption (that x = 1) causes the two quantities between which we wish the ratio, \vl., the numerator and denominator, to disappear. Let the student find that , L~^ , = 2i for x = \. (See also 190, 8d part of demon- 1 — aj+a^ — ^ stration.) 4. Two couriers starting at 'he same time from the two points A and B, c miles apart, travel toward each other at the rates a and h respectively. Discuss the problem with reference to the place and time of meeting. (Consider when ayhf a<,bj and a = h.) IJSTEKPKhlATlu.N Of EQUATIONS. 433 5. Two (-ouriers, A jiiiil li, are traveling tlie same road in the same direction, tlie former at rate a, and the latter n times as fast They are at two places c miles apart at the same time. Discuss the problem with reference to place and time of meeting as in Ex. 3, adding the cunsiderations n > 1, ;/ < 1, /i = 1, 71 = 0. 6. Divide 10 into two parts whose product shall bo 40. Solution and Discussion.— Let x and y be the parts, then x + y - 10, xy - 40, and J? = 6 ± V— 15, y = 5 t y/^lS. These re- sults we find to he imaginary. This signifies that the prohlem in its arithmetical signification is impossible : this indeed is evident on the face of it. But, although impossible iu the arithmetical sense, the values thus found do satisfy the formal, or algebraic sense. Thus the sum of 5 + Y^— 15 and 5 — >y/ — 15 is 10, and the product 40. 7. The sum of two numbers is required to be a, and the product b: wliat U the maximum value of b which will ren- der the problem pur^sible in the arithmetical sense ? What are the parts forthi: value of b 'f 8. Divide a into two parts, such that the sum of their squares shall be a minimum. gUG's. — Let X and a — j be the parts, and m the minimum sum. Then .r* + (a — x\^ = 2j^ — 2cu + a*^ m; whence x — \a ± ^ y^/^ -^'. From this we see that if 2m>a'^ x is imaginary. Hence the least value which we can have is 2m = a^ or m = \a?. 9. Divide a into two parts, such that the sum of the square roots shall be a maximum. 10. Let d be the difference between two numbers: re- quired that the square of the greater divided by the lest shall be a minimum, 434 IN^TERPREIATION OF EQUATIONS. 11. Let a and h be two numbers of which a is the greater, to find a number such that ii a be added to this number, and b be subtracted from it, the product of this sura and this difference, divided by the square of the number, shall be a marximum. Sug's. — Let n be the number, and m the required maximum quo tient. Then by the conditions „- = m, whence we find _ a -h 's/a^ + ^ab -\-W — 4abm Prom this we see that the greatest value which m can have and ren- , , . (a + bf nru- • a — b 2ab der n real is m = ^ , , . This gives n = — ^r^ \ = -j . Aab *^ 2{l — m) a — b 12. To find the point on a line passing through two hghts at which the illumination will be the same from each light. SoiiTJTiON.— Let A and B be the two lights, and XY the line passing mmm >- — ^ © @ V D' A B D through them. Let a be the intensity of the light A at a unit's dis- tance from it, b the intensity of B at a unit's distance from it, c the distance between the two lights, as AB, and x the distance of the point of equal illumination from the light A, as AD (or AD'). Then, as we learn from Physics that the illuminating effect of a light varies inversely as the square of the distance from it, we have for the illumination of the point D by light A -^ , and for the illumination of the same point by light B, /-^;7-^2 • But by the conditions of the problem these effects are equal ; hence we have the equation to be discussed ; viz., a b sfi ^ (fl — »? * IXTERPRKTATION OF KQIATIONS. 4^)0 This gives ^^^ = ^ or ^-Z^ = ^ ,/ 6 = ^Vt or "L-i^t^/I^oTt^VllVi; DF, finally, a; = c — ^ , and x = c ^ which are the values of a; to be discussed. Discussion.— I. Let c be finite an(f>i). t When a >b,x = c — ~ > ic, since - -^ _ ^ 1 fnp « > &. This is as it should be, since for a> b the point of equal illumination will evidently be nearer to B than to A. Again, the otlier value of X gives x = c — =r^^ ~ > c, since — rr^^ -iB-i- and > 1, ^a — \/b A^a — Ajb when a >b. Hence we learn that there is a point beyond B, as at D', where the illumination is the same from each light. If we assume \/a = 2\/6, AD = f c, and AD' = 2e. 2. It is evidently unnecessary to consider the caee when a, for which the illumination is for each light. [Let the student give the reason.] II. When c = 0. In this case the original equation —. = -. x* [C—xy jeeomes -r = -= , whence a — b. We then have x=zc — ^ = ; A^a+'y/b \/a c \/a and a; = er!mit«tious abc, acb, c